Mechanisms

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Textbook for University

MECHANISMS AND MACHINE THEORY
(机械原理)
Y e Z h o n gh e ( 叶 仲 和 ) L a n Z ha o hui ( 蓝兆 辉 ) M. R . S m i t h

HIGHER EDUCATION PRESS
高等教育出版社

Abstract
T his tex t book is wr itten according t o“ T he Basic T eachi ng R equir emen ts for th e Course of Mec han isms a nd Mac hine T h eory in Adva nced Indust rial Colleges ” drawn up by t he National Min ist ry of E ducation . It con tains t he au t hors ’ results on t eaching a nd scien tific r esearch duri ng r ecen t years . I t aims a t cultivating st uden ts’ ba- sic design ability a nd c rea tive ability in design . An aly tical met hods and syn t hesis of mech anisms ar e em ph asized . T his tex t book r eflects new ach ievemen ts and develop men ts curr ent in mecha nism st udy . Altoget her th ere a re twelve ch ap ters : In t roduction , St ruct ur al Analysis of Pla nar Mec han isms , Kinema tic Analysis of Mec hanisms , Plan ar Linkag e Mech anisms , Cam Mec han isms , Gea r Mech anisms , Gear T rains , O th er Mecha nisms in Com- mon Use , Com bined Mecha nisms , Balanci ng of Mach inery , Mo tion of Mecha nical Systems and Its R egulation , Creative Design of Mecha nism Systems . At th e end of most ch apt ers , ma ny th inking problems a nd e xercises a re enclosed . T o h elp Chin ese st uden ts to read t he t ex t , a n E nglish-Chinese vocabula ry is a ppe nde d . T his book ca n be use d as a t ext book for t he course of Mec han isms a nd Machin e T heory or t h at of specialit y E nglish for undergraduates specializing in mac hine ry in a dvan ced enginee ring un ive rsities . I t ca n also be used as a r eference book for related teac hers , studen ts and engineeri ng technicia ns .

  图书在版编目 ( CIP) 数据     机械原理 : Mechan isms and Mach ine T heory/ 叶仲和 , 蓝兆辉 , —北京 : 高等教育出版社 , 2001 .7   高等工科院校本科机械类专业   ISBN 7 - 04 - 009234 - 4     Ⅰ . 机…   Ⅱ.①叶…②蓝…③史…   Ⅲ. 机构学 高等学校 - 教材 - 英文   Ⅳ . T H 111     中国版本图书馆 CIP 数据核字 (2001) 第 26188 号

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Mec han isms and Mach ine T heory ( 机械原理 ) Ye Z hongh e   Lan Zhaohui   M .R .S mit h 出版发行   高等教育出版社 社     址   北京市东城区沙滩后街 55 号 电     话   010 - 64054588 网     址   h tt p :/ / w w w .he p .edu .cn h tt p :/ / w w w .he p .com cn 经     销   新华书店北京发行所 排     版   高等教育出版社照排中心 印   刷  开     本   787 × 1092 1/ 16 印     张   16 字     数   380 000 版     次       年   月第   版 印     次       年   月第   次印刷 定     价   13. 90 元   邮政编码   100009 传     真   010 - 64014048

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P r e f ace
“ Mechanism s and Machi ne T heory ” is one of t he impor tan t technical foundation courses for m echanical st uden ts in u niversities . I t cove rs t he basic knowledge needed in analysis and syn t hesis of commonly-used mechan isms and t he dynam ics of m echanism syste ms . T his course gives studen ts an elem en tary abilit y to design or improve m echanical devices . This course plays an importan t par t in cultivating t heir creative abilit y and gr eat at ten tion has been paid t o t his course in engineering universities . Alt hough m any Chinese tex tbooks have been publis hed , English tex t books w rit ten by Chi nese teachers have not yet appeared . Faced with t he globalisation of t he economy , tech nology and education , Chinese st uden ts should be able to st udy and work in E nglish . T hey have, in fact , lea rnt E nglish for six years in middle school . Af ter en tering universit y , t hey learn more English courses bu t , because al- most all ot he r courses use Chinese tex tbooks and are taugh t in C hinese , t hey have few oppor tu- nities to learn ot her courses in English directly . In orde r to change t his stat us , more and more at ten tion has been paid to English or bilingual teaching . O f course , English tex t books are in st ock in English-speaking count ries . H ow ever , at t he initial stage , C hinese t eachers an d stu- dents could accep t text books w ritten by Chi nese t eachers more easily . I t is for t his r eason t hat we have produced t his t ex tbook . Mainly Ye Zhonghe and Lan Zhaohui of F uzhou U nive rsit y wrote the con ten ts of t his book . T he con ten ts fulfil t he funda m en tal teaching r equire men ts in China . D r . of t he U niversity of Newcastle upon Tyne in G reat Britain , has had ove r t hir ty years’ experi- ence of teaching t his subject and has checked and em bellished t he w hole book to ensure t he flu- ency of t he language . T his tex tbook is t her efor e a result of in terna tional cooperation . We would like t o ex pr ess our t han ks here to P rofessor Zou Huijun of S hanghai Jiao Tong U nive rsit y and Professor Zhang Ce of T ianjin U n ive rsit y for t heir encourage m en t and suppor t for t his tex tbook . We a re also grateful to F uzhou U nive rsit y for t he suppor t for its publication . We hope t hat t he publication of t his t ex tbook will promote t he teach ing in English of tech nical courses in universities t hroughout Chi na . We hope also t hat people outside C hina will find t his t ex tbook helpful, not least i n com ing t o know som et hing about the teaching of “ M echanisms and Machine T heory” in China . T his book can be used as a text book for t he course of M echanisms and Mach ine Theory or t hat of specialit y English for undergr aduat es specializing i n m achinery in advanced en gineering universities . It can also be used as a reference book for r elated t eachers , st uden ts and engineer- ing technicians . Ye Zhonghe L an Zhaohui M . R . Smit h

前     言
机械原理课程是高等学校机械类专业学生重要的技术基础课之一。这门课主要讲授常用机 构的分析和综合以及机构系 统 动力 学的 基本 知识 , 使学 生 初步 具备 设计 和改 进 机械 装置 的能 力 , 对培养学生的创新设计能力具有重要作用。这门课程在工科大学很受重视 , 已出版了不少 中文版教材 , 但中国人自己编写的英文教材目前尚未见到。 面对经济、技术和教育的全球化 , 中国大学生应该能够用英语学习和工作。事实上 , 中国 学生在中学已学过六年英语 , 进入大学后又进一步学习了英语课程。但他们很少有机会直接用 英语学习其它课程 , 因为其它的课程几乎都是用中文教材以中文讲授的。为改变这一状况 , 用 英文教学或双语教学越来越受到重视。当然 , 英 文教 材在 英语国 家是 现成的。 但在 开始阶 段 , 中国教师编写的教材应该更易于被中国师生接受。这就是我们编写这本教材的目的。 这本教材主要由福州大学叶仲和、蓝兆辉编写 , 内容符合中国的教学基本要求。英国纽卡 斯尔大学的 M . R . Smit h 博 士 对 本课 程 有 30 多年 的 教 学经 验 , 他 对 全书 进 行 了修 改 和 润 色 , 保证了语言流畅。这本教材是国际合作的结果。 感谢上海交通大学邹慧君教授和天津大学张策教授对编写这本教材的鼓励和支持 , 感谢福 州大学对出版本教材的支持。我们希望 , 这本教材的出版对国内大学在技术课程上用英文教学 能有所促进 ; 我们也希望国外的有关人士也能觉得这是一本有用的书 , 至少可从中了解机械原 理在中国的教学情况。 本书可作为高等工科院校本科机械类专业学生学习机械原理课程的教材及作为专业英语教 材 , 也可供有关教师、学生及工程技术人员参考。 叶仲和     蓝兆辉     M . R . S mit h    

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CON T E N T S
C h a p t e r 1   I n t r o d u c t i o n ………………………………………………………………………………… 1
    1. 1   St udy Object ……………………………………………………………………………………………… 1     1. 2   St udy Con ten t …………………………………………………………………………………………… 2     1. 3   Purpose …………………………………………………………………………………………………… 3

C h a p t e r 2   S t r u c t u r a l A na l y s i s o f P l a n a r M e c h a n i s m s
    2. 1   Plan ar Ki nematic Pairs a nd Pla nar Mec ha nisms     2. 2   T he Kinematic Diagram of a Mech anism

………………………………… 4

……………………………………………………… 4

……………………………………………………………… 6 ………………………………………………… 15 ………………………………………………… 19

    2. 3   Degree of F reedom of a Mech anism …………………………………………………………………… 12     2. 4   Points for At ten tion during t he Calculation of D O F     2. 5   T he Composition P rinciple and St ruct ural An alysis     Problems and E xercises

……………………………………………………………………………………… 23

C h a p t e r 3   K i n e m a t i c A n a l y s i s o f M e c ha n i s m s ……………………………………………… 26
    3. 1   T asks and Met hods of Kinematic A nalysis …………………………………………………………… 26     3. 2   Velocit y Analysis by t h e Met hod of Inst an t Cen tr es ………………………………………………… 26     3. 3   Kinematic An alysis by An aly tical Met hods …………………………………………………………… 31     Problems and E xercises ……………………………………………………………………………………… 39 ……………………………………………………… 43

C h a p t e r 4   P l a n a r L i n k a g e M e c ha n i s m s ………………………………………………………… 43
    4. 1   Cha ract eristics of Plan ar Link age Mec hanisms     4. 2   T he T ypes of Four-bar Link ages ……………………………………………………………………… 44     4. 3   Cha ract eristics Analysis of Four-ba r Linkages ………………………………………………………… 48     4. 4   Dim ensional Syn t hesis of Four-bar Link ages     Problems and E xercises ………………………………………………………… 56 ……………………………………………………………………………………… 65 ………………………………………………………………………… 72 ……………………………………………… 72

C h a p t e r 5   C am M e c h a n i s m s

    5. 1   Cha ract eristics and Classification of Cam Mecha nisms

    5. 2   Followe r Motion Curves ………………………………………………………………………………… 75     5. 3   Plate Ca m with T ra nslat ing Roller (or Knife-edge ) Follower ……………………………………… 81     5. 4   Plate Ca m with Oscillating Roller Follower …………………………………………………………… 93     5. 5   Plate Ca m with T ra nslat ing Flat-faced Followe r ……………………………………………………… 97  


5 .6   Plate Cam wit h O scillati ng Flat-faced Followe r ……………………………………………………… 99 ……………………………………………………………………… 107 …………………………………………………… 109

    Problems and E xercises ……………………………………………………………………………………… 102

Ch a p t e r 6   G e a r M e c h a ni s ms

    6. 1   T ypes of Gear Mec ha nisms …………………………………………………………………………… 107     6. 2   Fundamen tals of E ngag eme nt of T oo th P rofiles     6. 3   T he Involu te and Its Prop er ties ……………………………………………………………………… 110

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    6. 4   Sta ndard Involu te Spur Gears ………………………………………………………………………… 113     6. 5   Gearing of Involut e Spur Gea rs ……………………………………………………………………… 116     6. 6   Con tact Ratio of an Involu te Spur Gear Set ………………………………………………………… 119     6. 7   Manufact uring Met hods of Involute P rofiles ………………………………………………………… 120     6. 8   Adde ndu m Modificat ion on Involu te Gears ………………………………………………………… 124     6. 9   Helical Gears for Parallel Shaf ts ……………………………………………………………………… 127     6. 10   Worm Gearing ………………………………………………………………………………………… 132     6. 11   Bevel Gea rs …………………………………………………………………………………………… 136     Problems and E xercises ……………………………………………………………………………………… 139

C h a p t e r 7   G e a r T r a i n s ………………………………………………………………………………… 141
    7. 1   Gear T rains a nd T heir Classifica tion ………………………………………………………………… 141     7. 2   T rain R atio of a Gea r T r ain with Fixe d Axes     7. 3   T rain R atio of Elemen ta ry E picyclic Gea r T r ain ……………………………………………………… 142 …………………………………………………… 143

    7. 4   T rain R atio of a Com bined Gear T rain ……………………………………………………………… 146     7. 5   Applica tions of Gea r T r ains …………………………………………………………………………… 149     7. 6   Mec hanical E fficiency of Planeta ry Gea r T rains …………………………………………………… 151 …………………………………………… 153     7. 7   T oo th N um bers of Gears and Nu mbe r of Plan et Gea rs

    7. 8   I nt roduction to Oth er Kinds of Plan etary Gea r T r ains ……………………………………………… 154     Problems and E xercises ……………………………………………………………………………………… 157

C h a p t e r 8   O t h e r M e c h a n i s m s i n C o mm o n U s e ……………………………………………… 160
    8. 1   R atch et Mecha nisms …………………………………………………………………………………… 160     8. 2   Gen eva Mec hanisms …………………………………………………………………………………… 165     8. 3   Cam-T ype Index Mech anisms ………………………………………………………………………… 169     8. 4   U niversal Join ts ………………………………………………………………………………………… 171     8. 5   Screw Mecha nisms …………………………………………………………………………………… 173     Problems and E xercises ……………………………………………………………………………………… 175

C h a p t e r 9   C om b i n e d M e c h a n i s m s
    9. 1   I nt roduction

……………………………………………………………… 177

…………………………………………………………………………………………… 177

    9. 2   Met hods for t h e Classification of Com bined Mech anisms …………………………………………… 177     9. 3   Series Co mbin ed Mecha nisms ………………………………………………………………………… 178     9. 4   Parallel Combin ed Mech anisms ……………………………………………………………………… 181     9. 5   Co mpound Com bined Mech anisms …………………………………………………………………… 183     9. 6   Multiple Co mbin ed Mecha nisms ……………………………………………………………………… 185     9. 7   Feedback Co mbin ed Mecha nisms …………………………………………………………………… 186     9. 8   Mixed Co mbin ed Mecha nisms ………………………………………………………………………… 187     9. 9   Mat ters Needi ng At ten tion …………………………………………………………………………… 189     Problems and E xercises ……………………………………………………………………………………… 190

C h a p t e r 10   B a l a nc i n g o f M a c h i n e r y …………………………………………………………… 195
    10 .1   Purposes a nd Met hods of Bala ncing ………………………………………………………………… 195     10 .2   Balan cing of Disk-lik e Rot ors ………………………………………………………………………… 197     10 .3   Balan cing of Non-disk Rigid Rot ors ………………………………………………………………… 199

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    10 .4   Unbala ncing Allowa nce of Rotor

…………………………………………………………………… 204

    Problems and E xercises ……………………………………………………………………………………… 205

C h a p t e r 11   M o t i o n o f M e c h a n i c a l S y s t e m s a n d I t s R e g u l a t i o n ………………… 208
    11 .1   In t roduction …………………………………………………………………………………………… 208     11 .2   M otion E quation of a Mecha nical System ………………………………………………………… 208     11 .3   Solu tion of t he Mo tion E quation of a Mec ha nical System ………………………………………… 212     11 .4   Periodic Sp eed Fluctuat ion a nd its Regulation ……………………………………………………… 215     11 .5   In t roduction t o Aper iodic Speed Fluctua tion a nd its R egulation ………………………………… 219 ………………………………… 222     Problems and E xercises ……………………………………………………………………………………… 220

C h a p t e r 12   C r e a t i v e D e s i g n o f M e c h a n i s m S y s t e m s
    12 .1   In t roduction t o Design of Mech anism Systems     12 .3   Kinematic Paramete rs of t h e System

…………………………………………………… 222

    12 .2   Creative T hink ing and Crea tive Met hods …………………………………………………………… 224 ……………………………………………………………… 227     12 .4   Selection of Mecha nisms ……………………………………………………………………………… 228     12 .5   Coope ration of W orki ng Links ……………………………………………………………………… 230     12 .6   Evaluation of t he Mecha nism System ……………………………………………………………… 233     12 .7   Design E xam ple of a Mecha nism System …………………………………………………………… 234

V o c a b u l a r y …………………………………………………………………………………………………… 239 R e f e r e n c e s …………………………………………………………………………………………………… 245 前言 ………………………………………………………………………………………………… 247

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C h ap t e r 1 In t r oduction
1.1   S t u d y O b j e c t
T he re a re vast nu mbers of differen t kinds of mach ines in t he world , all wit h a wide variet y of const ructions , characteristics and uses . The single-cylinder four-stroke in ternal combustion en-

gine sho wn in Fig . 1-1 is a t ypical and very familiar m achine and works as follows . T he powe r from t he combusting gases pushes on t he piston 2 causing it t o t ranslate dow n wa rds . T hrough connecti ng-rod ( or coupler) 3 , the reciprocation of pist on 2 is t rans- form ed in to t he rotation of cr ank 4 and flyw heel 4″. ( The inertia of t he fly w heel 4″drives t he piston 2 t o move up wards on t he return stroke .) T he c ran k 4 , t he fly w heel 4″ , and t he pinion 4′ ar e fixed t ogether by keys , so t heir mo tions are t he sa m e . Also , t he cams haft ( wit h its tw o ca ms 5′and 5″ ) is keyed t o gear 5 . T he refore , gear 5 , ca m 5′and cam 5″ro- tate t oget her as one body . T he number of t eet h on t he gear 5 is twice t hat of t he pinion 4′so t hat t he pinion 4′ rotates t wice for every r evolu tion of t he gear 5 . This is t o coordinate t he motions of piston 2 , i n- let valve 6 and ou tlet valve 7 . T hrough t he con tact bet ween t he cam 5′ and followe r 6 , t he ca m ro tation
Fig . 1-1

is tr ansform ed in to a regula r reciprocation of t he inlet valve 6 so t hat t he valve opens and closes at a precise tim e . Similarly , t he ca m 5″ con trols t he regular r eciprocation of outlet valve 7 . In t h is way , t he in ternal combustion engine t ransforms t he heat energy from t he fuel and air mix- ture in to mechanical energy of t he crank rotation .     T he connecti ng-rod 3 is formed by t he rigid asse mbly of t he coupler body , big-end cover , nu ts , bolts etc . They a re t he basic ele ments of manufact ure , w h ich ar e called m ach i ne ele- m en ts . Afte r assembly , t he body , t he big-end cove r , t he nu ts and t he bolts form a rigid ・1・

st ruct ure and move t oget her . T he rigid st ruct ure is a basic elem en t in kine m atics analysis , w hich is called a li nk . Similarly , t he crank 4 , t he pinion 4′ , t he fly w heel 4″ , and t he keys , form anot her link . The gear 5 and t he t wo cams form a t hird link . In t he st udy of ki nem atics , we a re concerned only wit h t he motion of links , not of in dividual machine elem en ts . As desc ribed above , t he function of t he cylinder 1 ( or t he engine body or fra me ) , t he pist on 2 , t he coupler 3 and t he cr an k 4 is to t ransform t he r eciprocation of t he piston 2 in to t he rota tion of t he crank 4 . T hey constit u te a sli der-cra nk m echa nism . T he fu nction of t he engi ne fr am e 1 , t he pinion 4′and t he gear 5 is t o change t he direction and speed of rotation . T hey constit u te a gear mechan ism . T he function of t he fra m e 1 , t he cam 5′ an d t he follow er 6 is to t ransform t he con ti nuous rotation of t he ca m i nt o a regular reciprocation of t he follower . T hey constit u te a ca m mechan ism . ( The sam e applies to t he fra me 1 , the ca m 5″ and t he fol- lower 7 .) We see , t herefore , that a mechan ism is a system of link s w hich can t ransform or t ransmit force and mo tion . T he i nte rnal combustion engine s ho wn in Fig . 1-1 is t hus a syste m consisting of t hree ki nds of m echanism . I ts struct ural block diagr am is show n in Fig . 1-2 .

Fig . 1-2

M ach i ne is a m echanical system t hat can t ransmit or t ransform energy , m aterials or infor- m ation . As we shall see , many pr actical m achines contain mor e than one m echanism w hile m any simple machines ar e composed of only one basic mechanism . T he term “ m achi ner y” is used to cove r bot h mechanisms and m achines . T he st udy object of t his book is m achinery .

1.2   S t u d y C o n t e n t
T he in te rnal combustion engine is only one exa mple of t he m any diffe ren t k inds of m a- chines in t he world . M any mach ines , at first sigh t , appear t o have very differ en t const ruc- tions, tr ansmission characteristics and applications . Fort una tely , af ter exa mination of more exa mples in detail, we fi nd t hat most m achines , even very complicated ones , are built up from only a few t ypes of commonly used m echanisms , such as li nkage m echanisms ( t he slider-c ran k m echanism is one of t hese ) , gear mechanism s , cam m echanisms etc . T he functions and pro- files of m achines may be quite diffe ren t , but t he m echanisms used in t hem a re often t he sam e . T his sit uation is si mila r t o t he combination s of t he seven-piece puzzle . Wit h only seven plates , m any differen t pa tterns can be construct ed . T he most con ten t of t his book is for “ analysis” ・2・

and“ syn t hesis” of some com monly used mechanism s . Alt hough t her e are many kinds of m achine, it is no t possible or necessa ry t o st udy all kinds in t his book . We will exami ne t he necessary basic topics in“ machine ry t heory ” and study t he analysis and syn t hesis of some commonly used mechanism s . Dim ensions affecting t he motion of t he m echanisms are called k ine m a tics d i mensions . For exam ple , t he r adii of t he holes and t he shape of t he cross-section of t he coupler 3 have no effect on its motion . T hey a re not kine m at- ics di mensions . In t he coupler 3 , t he distance between t he cen t res of t he t wo holes is t he only kinem atics dimension . Syn t hesis is t he process of determini ng only t he kinem atics di mensions of a m echanism r equir ed to produce a pa rticular mo tion . T he term “ syn t hesis” differs from “ desig n ” w hich is t he process of pr escribing t he sizes , shapes , m aterial com positions , ar- ran gem en ts of pa rts , calculations of str eng th , met hods of production , etc . For exa mple , t he radii of t he holes and t he shape of t he cross-section of t he coupler 3 a re not con sidered in t he syn t hesis of t he m echanism , bu t must be considered duri ng t he design of t he machine w hen forces and st resses a re to be calculated .

1.3   P u r p o s e
T his t ex tbook “ M echanisms and M achine T heory” is an aid to t he st udy of m achines and m echanisms . I t is hoped t hat , after st udying t his tex t book , t he studen ts will grasp t he basic t heory and obtain t he basic knowledge and skills needed in m echanisms syn t hesis and kine matic and dyna mic analysis of machine ry . Th is should lead t o t he ability t o choose m achinery mo tion pat terns , analyze and syn t hesize m echanisms and develop designs for practical working m achin- ery . The kno wledge to be obt ained from t his tex t book is t he refor e a funda ment al i n analyzing existing m achines and designing new ones .

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C h ap t e r 2 S t r uc t u r al A nal y s i s o f P la na r Mec h a ni s m s
2.1   P l a n a r K in e ma t i c P a i r s a n d P l a n a r M e c h a ni sm s
2.1.1   Kinematic Pairs As mentioned in Chapte r 1 , a m echanism is a combination of lin ks w hich can t ransform or t ransmit a determined motion . In order to t ransmit and tr ansform motion , every lin k must be kept pe rmanently in con tact wit h ot he r li nks by som e ki nd of connection and have motion rela- tive to t he m . Such a mobile connection is called a ki ne m at ic p air . F or example , t he connec- tions between t wo links in Fig . 2-1 t o Fig . 2-6 a re kine m atic pairs . A pair t hat per mits only relative rota tion is called a revol u te pa ir , as show n in Fig . 2-1 . A pair t hat allows only rela- tive rectilinea r t ranslation is called a sli d i ng pa ir or p rism a tic p air , as s ho wn in Fig . 2-2 . T he kinem atic pairs in Fig . 2-3 to Fig . 2-6 ar e called gear pair , ca m pa ir , screw p air , a nd spherica l pair , r espectively . O ne kine matic pair can connect only t wo link s . T he par t of t he link surface w hich makes con tact wit h anot he r li nk and forms a kine matic pair is called a p air ele men t . T he combination of t wo such ele men ts on t he con nected lin ks constitu tes a kinem atic pair . For exam ple , t he cylindrical surfaces of t he hole and t he shaft in a revolut e pair ( Fig . 2-1) are tw o pair-elem en ts . T hose connections t hat joi n tw o m achine el- e ments firmly and do not allow t he connected machine ele ments t o move relative t o each o t her , such as welds , rivets , or nu ts and bolts , a re not ki ne matic pairs . If tw o lin ks con nected by a k ine matic pair can move relative to each ot her only on a plane , t he kinem atic pair is called a p lan ar k ine m a tic pair , ot her wise , a sp ati al ki nem atic p air . For example , t he r evolute pair in Fig . 2-1 , t he sliding pair in Fig . 2-2 , t he gear pair in Fig . 2-3 , and t he ca m pair in Fig . 2-4 are planar pairs , w hile the scr ew pair in Fig . 2-5 and t he spherical pair in Fig . 2-6 a re spatial pairs . If t wo pair-ele men ts of a k ine matic pair have surface con tact or its equivalent , t he ki ne- m atic pair is k no w n as a lower pair . T he revolu te pair in Fig . 2-1 and t he sliding pair in Fig . 2-2 a re plana r lo wer pairs , w hile t he sc rew pair in Fig . 2-5 and t he spherical pair i n Fig . 2-6 a re spatial low er pairs . Since t he tw o links of a lowe r pair ar e connected by a surface, t he pr es- sur e bet ween t he t wo links is lo wer . T his is a simple way of re me mbe ri ng w hy it is called a ・4・

Fig . 2-1

Fig . 2-2

Fig . 2-3

Fig . 2-4

Fig . 2-5

Fig . 2-6

lo wer pair . If t he connection takes place only at a poin t or along a line ( assum ing t he m aterials to be rigid ) , it is k no w n as a hi gher p air . T he gear pair i n Fig . 2-3 and t he ca m pair i n Fig . 2-4 a re plana r higher pairs . T he connection bet ween a ball and a plane is a spa tial higher pair . Since t he t wo link s of a highe r pair ar e connected by a poin t or a line , t he pressure bet ween t he t wo links is h igher . That is w hy it is called a higher pair . 2.1.2   Kinematic Chain and Mechanism W hen a numbe r of li nks are connected by mean s of ki ne matic pairs , t he resulting mobile system is a ki nem a tic cha in . If eve ry lin k in a kine matic chain has at least two pair-elem en ts , all links form a closed chai n , as show n in Fig . 2-7a and b . If one or more links i n a kine matic chai n have only one pair-ele ment , t hen t he kinem atic chain will be an open chai n , as show n in Fig . 2-7c and d . Most m achines use closed chains , w hile open chai ns ar e often used in robots and manipulat ors . In order to tr ansmit motion , one of t he lin ks in t he ki nem atic chain must be fixed to t he base of t he machine . The fixed link is also called t he f r a me and t he re is on ly one fra me for each m echanism . The fr am es of most m achines a re fixed to t he ground , w hile t he fra mes of ・5・

Fig . 2-7

som e m achines , e. g.cars , aeroplanes , etc ., can move relative to t he ground . T he relative reference syste m is of ten fi xed to t he fr am e w hen t he mo tion of lin ks is st udied . Some movi ng li nks have t heir o wn independen t motion cha racteristics . T hey are called dri vi ng li nks . T he ot her movi ng li nks a re called d riven li nks . A mong all t he driven links , one or more are used to gene rate t he expected out pu t motion . Such driven links a re called ou t- p u t links . For exa mple , in t he in ternal combustion engine in Fig . 1-1 , t he piston 2 is t he driving lin k , t he coupler 3 and t he crank 4 are t he driven lin ks . T he c ran k 4 is also an ou tpu t link . If all links of a mechanism move in planes t ha t r em ain par allel to each o ther , we say t hat t he m echanism is a pl ana r m echa nism . O t her wise , it is a spa tia l mechan ism . For exam ple , in t he int ernal combustion engine show n in Fig . 1-1 , all planes on w hich links move a re pa ral- lel, so it is a planar m echanism . Gear m echanism in Fig . 6-6 to Fig . 6-8 a re spatial mecha- nisms . All kine matic pairs in a plana r m echanism must be plana r kinem atic pairs , w hile t he m echanism con taining only plana r kinem atic pairs m ay be a spatial m echanism . F or exam ple , t he un ive rsal joint sho w n in Fig . 8-20 , w h ich con tains only revolu tes , is a spatial m echanism . If all kine ma tic pairs in a mechanism a re lo wer pairs , t he m echanism is called a lo wer pa ir mechan ism , or more often , a l inkage mechan ism . All mechanism s in Chap ter 4 ar e planar linkage m echanisms . The universal join t in Fig . 8-20 is a spatial lower pair mechanism . If a m echanism has one or more highe r pairs , it is called a h igher pair mechan ism . T he in ternal combustion engine in Fig . 1-1 is a plana r higher pair mechan isms . T he gea r mechanism in Fig . 6-6 to Fig . 6-8 are spatial higher pair m echanisms .

2.2   T h e K i n em a t i c Di a g r am o f a M e c h a ni s m
2.2.1   Definition and Purpose In order t o analyze an ex isting mechanism or design a new mechanism , it is helpful to draw a sim ple diagr am to indicate the kinem atic relationship bet ween links . Since t his diagra m is used on ly t o express the relationship betw een t he m otions of lin ks , it should be simple bu t provide all necessary ( not r edundan t ) informa tion determi ning t he relative motion of all links . Such a diagra m is called t he ki nem a tic d iag ra m of t he m echanism . T hose detailed st ructur es ・6・

irrelevan t t o t he motion t ransmission should be omit ted or simplified . For example, t he profile and section shape of a lin k , t he r adius of a revolu te , t he s hape of t he cross section of a sliding pair , t he num ber of machine elem en ts in a link and t heir m anner of connection a re irrelevan t to t he motion t ransmission . They should not be crowed in to a kinem atic diagram . In t he ki ne- m atic diagra m , links and kinem atic pairs should be repr esented by si mple and specified sym- bols . In t his way , t he kinem atic diagra m can reflect t he ki ne matic cha ract eristics of t he mech- anism more clea rly and more si mply . The abilit y to draw t he k ine matic diagram s of mecha- nisms from exa mina tion of r eal m achines or t he assembly drawing of a m achine is a basic techni- cal skill for en gineers engaged in designing , m anufact uring or m ain taining m achines . Since a mechanism is built of lin ks and kine ma tic pairs , we study first t he represen tations of lin ks and kine ma tic pairs in t he kine m atic diagra m befor e we star t t o draw t he w hole ki ne- m atic diagram of a mechanism . 2.2.2   Representation of a Kinematic Pair T he kinem atic function of a kinem atic pair is only t o constr ain t he t wo links it connects in order to m ain tai n a special r elative mo tion . For exa mple , t he k ine matic function of a r evolute (refe rring to Fig . 2-1 ) is to con st rai n t he t wo li nks con nected t o rotate abou t t he cent re of t he revolu te relative to each ot he r . A revolu te is convenien tly repr esen ted by a s mall circle placed at t he cen t re of t he revolu te no m atter ho w large its radius is , as sho w n in Fig . 2-8 . If a lin k is connected to t he fr am e by a r evolute , t he revolu te is called a f i xed p ivot . T he fra me is often indicated by shading , as show n in Fig . 2-8d . A fixed pivot can also be repr esented by a small circle wit h t riangle suppor t and shading , as show n in Fig . 2-8e . T he kinem atic fu nction of a sliding pair ( Fig . 2-2) is to constrain t he t wo connect ed lin ks t o t ranslate in t he direction of t he pat h way relative to each o t her . Therefore , t he act ual shape of t he cross section of t he sliding pair has no influence on t he kinem atics of t he m echanism . Sho wn in Fig . 2-9 a re some t ypical kine ma tic repr esen tations of sliding pairs . T he shaded lin ks represen t t he fr am e .

Fig . 2-8

For t he tw o lin ks in a sliding pair , t he motion cha racteristics of t he slidi ng pair will not change no mat ter w hich link is draw n as t he sliding block or t he guide ba r . F ur t hermore , t he kinem atic r elation ship betw een t wo lin ks in a sliding pair does not change so long as t he cen ter- line of t he slidi ng pair in t he kinem atic diagr am is pa rallel to t he pat h way in t he m echanism . ・7・

Fig . 2-9

T he refore , t he t wo links of a hydraulic cylinder show n in Fig . 2-10a can be r epresen ted in t he kinem atic diagram as show n i n Fig . 2-10b t o e .

Fig . 2-10

T he gear pair is represen ted by t wo chai n dot ted circles tangent to each ot he r ( t wo t oot h profiles m ay be added ) , as show n in Fig . 2-11 , since tw o gears are equivalen t kine matically to t wo friction w heels rolling wit hout slipping . In a ca m m echanism , t he cam con tour and t he end profile of the follower have an effect on t he motion charac- teristics betw een t he ca m and t he follo wer . Hence it is necessary to dr aw t he act ual ca m con t our and t he end pro-
Fig . 2-11

file of t he follower in t he kine matic diagra m , as sho w n in Fig . 2-4 .

2.2.3   The representation of a Link in the Kinematic Diagram     T he kinem atic function of a link is t o hold t he rela tive position of all pair ele men ts on t he link unchanged durin g t he motion of t he m echanism . F or exa mple , for a t wo-revolut e li nk in Fig . 2-12a , t he kinem atic function of t he link is to keep t he distance l A B bet ween t he cen ters of t he revolut es A and B unchanged . l A B is t he only kinem atic dim ension in t his link . T he refore , a link wit h on ly tw o revolu tes is illust rated m ere- ly by a st raigh t li ne join ing tw o sm all circles cen ter ed at t he cen ters of the revolu tes , as show n in Fig . 2-12b . A lin k wit h more t han t wo pair ele m en ts can be represen ted by a ha tched or w elded polygon wit h pair ele ments at corners , as show n in Fig . 2-13 . Not e t he difference bet ween t wo diagra ms in Fig . 2-14 . ・8・
Fig . 2-12

Fig . 2-13

Fig . 2-14

T he special represen tative symbols used i n a kine m atic diagra m for some common mecha- nisms are listed in T able 2-1 . Table 2-1   Special Symbols for Some Commonly Used Mechanisms
E lectric M ot or Con e G ea r Driv e

Bel t D ri ve

Rac k a nd Pi ni on

C hai n drive

W or m a nd W or m G ea r Drive

・9・

Continue    
Ou t e r-mesh ed Cy li ndrical G ea r Drive C am D ri ve

In ne r- mes he d C yli ndrical G ea r Driv e

Ra tc h et M ec ha nis m

    T he kinem atic diagra m should be draw n to scale so t hat t he diagra m has the sam e kine m at- ic characteristics as t hose of t he original mechan ism . I t can be used in gr aphical kinem atic and dyna mic analysis . If we want to e mphasize only t he struct ural characteristics and illust rate t he motion t ransmission , t hen t he diagra m does no t have to be dra wn exactly to scale . S uch a dia- gram is called t he sche m a tic d i ag ra m of t he mechanism . 2.2.4   Procedures for Drawing the Kinematic Diagram of a Mechanism (1) Run t he mechanism slowly , study car efully the struct ure of t he m echanism . Deter- mine t he t ypes of all kine matic pairs . Analyze t he t ransmission rout e from t he driving link to t he ou t put link . T hen st op t he mechanism at a suitable position for dra wi ng . ( 2) F or a planar m echanism , all moving li nks move in parallel planes . T hus , a plane par- allel t o t hese planes is chosen as a drawing plane . Sometim es , a local view may be draw n to clarify t he st ructure . (3 ) D raw t he sche matic diagra m of t he m echanism . Firstly , dr aw all fixed pair ele ments or t he pair ele ments on t he fram e . Be careful abou t t he relative positions bet ween t hese fixed pair ele ments . T his is an i mport an t step . T hen begin t o draw t he moving li nks . Dr aw t he drive rs first and t hen draw t he driven links according to t he rou te of motion t ransm ission . Be- fore dra wing a lin k , determ ine t he t ypes of all kinem atic pairs on it and its kinem atic dim en- sions . (4) F or conven ien t reference , t he lin ks a re num bered w hile t he kine matic pairs ar e let- tered . T he in pu t link is m arked wit h an arrow in t he dir ection of motion . Each lin k , no m at- ter how m any mach ine elem en ts it has , can have only one serial number . A n apost rophe to t he right of t he se rial number may be used to distinguish t he diffe ren t m achine ele men ts of t he sa me link . All mach ine elem en ts belonging to t he sa m e link in t he kinem atic diagra m must be con- ・ 10 ・

nected firm ly by welding symbols . (5) Calculat e t he degree of fr eedom of t he mechanism ( see Sec.2 .3 ) according t o t he schem atic diagra m , and check if it is t he sa me as t ha t of t he act ual mechanism . (6 ) M easure all and only kine matic dim ensions . T hose dim ensions w h ich will change dur- ing t he mo tion a re not kinem atic dim ensions . (7 ) Select a suitable scale μ l , t he fact or of w hich is μ l = t hen dr aw t he kinem atic diagra m of t he mechanism . Example 2-1 D raw t he k ine matic diagra m of t he in ternal combustion engine show n in Fig . 1-1 . Solution: T he crank 4 an d t he pinion 4′ar e connected firm ly wit h a key . T hey constit ute a sin gle link . T he refore , in t he kine matic diagram , t he crank 4 and t he pinion 4′ must be connected by a welding symbol and have t he sa me serial n umber . Similarly , t he gear 5 , t he cams 5′and 5″constit ute a single lin k . Not e the welding symbol between t he gea r 5 and t he ca m 5′. There are alt oget her one fr am e and six moving links : pist on 2 , coupler 3, crank 4 ( toget he r wit h pinion 4′ ) , gear 5 ( t oget her wit h cams 5′and 5″ ) , ca m follow er 6 , and cam fol-
Fig . 2-15

act ual lengt h   leng t h in diagra m  

, and

lo wer 7 . T he pist on 2 slides relative to t he fra me 1

along t he axis of t he cylinder . T he refor e , t he connection bet ween t he piston 2 and t he fra me 1 is a slidi ng pair . Although t he cross-section of t he cylinder is circular , it is not a r evolute pair . The type of any ki nem atic pair is determ ined by t he kine m atic relationship betw een t he t wo lin ks it connects . The circular section is used for ease of m anufacturing . D raw all fixed pair elem en ts first . F ollo wing t he procedures m en tioned above, t he kinem atic diagra m of t he mechanism is dra w n as sho w n in Fig . 2-15 . A local view ( Fig . 2-15b ) is draw n because of t he overlapping of t he tw o cam follow - ers . Example 2-2     T her e are t hree moving links in Fig . 2-16a . Circula r disk 1 ro tates rela tive to t he fra me 4 about a fixed axis A . T he circula r disk 1 m atches wit h a hole on anot her circula r
Fig . 2-16

・ 11 ・

disk 2 . The circular disk 2 matches wit h a large hole on t he link 3 . The lin k 3 oscilla tes about a fixed axis D . Dra w t he kinem atic diagra m of t he mechanism . Solution: O bviously , A is a fixed pivot bet ween t he fra m e 4 and t he link 1 . D raw first t he t wo fixed pivo ts A and D . Lin ks 1 and 2 are connected by a revolu te w hose cent re is at B . O n t he link 1 , t her e ar e only t wo revolu tes : A an d B . T her efor e, t he link 1 is r epresen ted by a st raight line joining tw o sm all circles cent ered at A and B . Similarly , t here a re only t wo revo- lu te pair elem en ts on t he lin k 2 . One is t he ele ment of t he revolu te B and t he ot he r is the ele- m en t of t he revolu te between t he links 2 and 3 w hose cen t re is located at C . T he link 2 is t herefore represen ted by a st raigh t line joining tw o sm all cir cles cen te red at B and C , Sim ilar- ly , t he link 3 is repr esen ted by a st raigh t line joi ning two sm all circles cen ter ed a t C and D , as show n in Fig . 2-16b .

2.3   D e g r e e o f F r e e d o m o f a M e c ha n i s m
2.3.1   Definition In t he four-bar m echanism show n in Fig . 2-17 , there ar e t hr ee moving li nks . There are m any kinem atic par am eters of link s and poin ts in th is m echanism , e. g . , φ 1 , φ 2, φ 3 . Bu t t hese ki ne matic pa ra mete rs a re dependen t on each ot her . In tuitively , we can see t hat among angles φ 1 , φ 2 and φ 3 , only one angle is i ndependent . In the five -bar mechanism A BCD E show n in Fig . 2-18 , if t wo angles ( say , φ1 and φ 4 ) ar e given , t he locations of all moving links will be determined . So t he number of independen t ki ne matic pa ra meters for t his mecha- nism is tw o . In t uitively , t he re is only one independen t kinem atic para m et er , say angula r coor- dinate φ 1 of t he ca m 1 , in t he cam m echanism show n in Fig . 2-4 . In t he in te rnal combustion engi ne show n in Fig . 2-15 , t he positions of all moving links can be dete rmined , so long as t he position of t he slide r 2 is given . T he number of independen t para m eters needed to define uniquely t he loca tions of all moving links wit hin a mechan ism wit h respect to t he fr am e is called

Fig . 2-17

Fig . 2-18

・ 12 ・

t he degree o f f reedo m ( DO F ) of t he m echanism . So t he DOF of t he mechanis ms in Fig . 2- 4 , Fig . 2-15 , Fig . 2-17 and Fig . 2-18 a re 1 , 1 , 1 , and 2 , respectively . 2.3.2   Degree of Freedom of a Link and the Number of Constraints of a Kinematic Pair     T he DOF of a mechanism can be calculated according t o t he nu mber of link s and t he num- ber and t ypes of ki nem atic pairs . If a moving li nk in a plane is not con nected t o any ot her lin k ( Fig . 2-19) , t hree indepen den t pa ram eters are required to complet ely define its exact location , e. g . tw o li near coordinates ( x A , y A ) t o define t he position of any poi nt A on t he link , and one angular coordinate ( φ) to define t he angle of a line on t he link wit h respect to t he axes . T he refore, an unconst rained li nk on a plane w ould have 3 DO F . T w o unconnected indepen- dent lin ks on a plane w ould have 6 DO F .

Fig . 2-19

Fig . 2-20

    When t he tw o lin ks are connected by a ki nem atic pair , t heir indepen den t mo tion will be const rained because t hey must re main in con tact at all ti mes . In ot her w ords , t heir DO F will decrease . T he numbe rs and t he t ypes of const raints a re differen t for differ en t kine ma tic pairs . For example , if tw o li nks a re con nected by means of a revolu te pair , as s ho w n in Fig . 2-20 , we need only 4 indepen den t para m eters to defi ne t heir exact locations . T hese m ay be th ree pa- ra meters ( x A , yA , and φ 1 ) to defi ne t he location of li nk 1 first , and ano ther para m eter (φ 2 1 ) t o define t he position of li nk 2 relative to link 1 . T hus , t he t wo-link system show n in Fig . 2-20 has only four DO F . Hence , con necting t wo planar lin ks wit h a revolu te pair has t he effect of re moving two DO F from t he syste m . In o t her words , a revolu te pair sets tw o con- st rain ts . I t const rains x and y relative t ranslation and perm its only pure relative rot ation be- tween t he links it con nects . Sim ilarly , a sliding pair ( r efe rring t o Fig . 2-2 ) const rain s relative t ranslation along t he common normal and r elative ro tation , perm it ti ng only a rectilinea r relative mo tion along t he pat h of t he slide way . T hus , one planar low er pair sets t wo const rain ts , or eli mina tes t wo DO F . If t wo links are connected by a planar higher pair as sho w n in
Fig . 2-21

・ 13 ・

Fig . 2-21 , the tw o links can no t have relative motion along t heir common norm al n-n at t he poin t of con tact . Ot he rwise , t he t wo link s will separat e or in terfere . Ho wever . one lin k can slide along t he common tangen t t-t and rotate relative to t he ot her link . T his highe r pair is t herefore also called a roll & sli de pair . T hus , a planar higher pair has only one const rain t or eli mina tes only one DOF . 2.3.3   Structural Formula of a Mechanism Each unconstr ained link has 3 DO F . Befor e an y connections a re m ade, a system of N moving links will have a to tal of 3 N DO F . Since each plana r low er pair eliminates tw o DOF and each plana r higher pair elimi nates one DOF , t he DO F of a plana r m echanism wit h N mov- ing links , Pl plana r lo wer pairs and P h plana r highe r pairs can be calculated as : F = 3 N - 2 Pl - P h w her e F denotes DOF . T his is called t he str uct u ral for m u la o f the mechan ism . N ote t hat N does no t include t he fra me . Using t he st ructur al formula of t he m echanism to determine t he DOF of t he ca m mech- anism in Fig . 2-4 . w e have : N = 2 , Pl = 2 , P h = 1 , F = 3 N - 2 Pl - P h = 3 × 2 - 2 × 2 - 1 = 1 , w hich agrees wit h t he r esult by in tuition . As a fur t he r demonst ration of t he use of t he st ructur al formula of a mechanism , refer to t he in ternal com bustion engine in Fig . 2-15 . We have : N = 6 , Pl = 7 ( four revolu tes , 2×7 - 3 = 1 . 2.3.4   Conditions for a Mechanism to Have a Determined Motion As m en tioned before , t he DOF of a m echanism is t he number of independent par am eters required to specify t he locations of all moving lin ks in a m echanism wit h r espect t o t he fram e . Independen t para met ers ar e supplied only by t he driving lin ks . Since t he driving lin ks are al- ways connected to t he fra me by lowe r pairs , one driving link ( toget he r wit h t he corresponding lo wer pair ) will provide only one independen t para m eter . For example , t he rot or of a motor has one independent rotation . T he pist on in t he hydraulic cylinde r has one in dependent t ransla- tion . T he refore , we can conclude t ha t t he links wit hin a mechan ism can have dete rmined rela- tive motions if and only if t he DOF of t he mechan ism is greater t han zero and equal to t he num- ber of t he driving lin ks . T his is called the con d itions f or a mechan ism to have a determ i ned m otion . If DO F is zero , it is a truss . If t he numbe r of driving links is less t han t he DOF of t he m echanism , som e driven lin ks will not have determined motion . F or exa mple , in t he five-bar linkage sho w n in Fig . 2-18 , F = 3 N - 2 P l - P h = 3 × 4 - 2 × 5 = 2 . If t he re is only one driving link , say link 1 , the positions of t he ot her moving li nks ( 2 , 3 , and 4 ) cannot be deter- mined . T hey may be a t t he positions indicated by the solid li nes or t he dashed lines or at any ・ 14 ・ A, B , C and D , and t hree sliding pairs , C , E , and F ) , P h = 3 , F = 3 N - 2 P l - P h = 3 × 6 ( 2 -1 )

o t her position . If t he numbe r of driving lin ks is greater t han t he DOF of t he m echanism , t he kinem atic chain cannot move due to t he conflict bet ween t he inpu t forces . In ext re me cases , t he weakest li nk in t he chain may be broken .

2.4   P o i n t s f o r A t t e n t i o n d u r i n g t h e C a l c ul a t i o n o f D O F
In t he developmen t of t he struct ural formula of m echanism Eq . was given to t he lin k dim ensions . It is not surprising to find many excep tions t o t he formula in practical cases wit h special geom et ric fea tures . T he designer needs t o be awa re of t hese possible inconsistencies . 2.4.1   Compound Hinge     In Fig . 2-22a , link 1 con nects wit h links 2 and 3 by revolu tes A 1-2 and A 1-3 , respective- ly . T he axes of t he tw o r evolutes coi ncide . I n t he k ine matic diagra m as show n in Fig . 2-22b , only one circle is draw n . T he t wo revolut es consti- tu te a compound hinge . T he numbe r of revolu tes i n a compound hinge is equal t o one less t han t he number of lin ks joined at t hat hinge . In a m echanism having gea rs or cams , car eful a tten tion must be paid t o t he existence of compound hinges . F or exa mple , in t he gear ed lin kage show n in Fig . 2-23 , t he bea rings of li nks 4 and 5 match t he pins on gear 2 . T her efore , t here a re t wo r evolutes C2 -4 and C2-5 at t he point C . They constitu te a compound hinge . Sim ila rly , D is also a compound hinge since t here ar e two revolu tes , D6-3 and D6-5 , at t he hinge D . T herefore , for t h is mechanism , we have N = 5 , Pl = 6 , P h = 2 , F = 3 N - 2 Pl - P h = 3 × 5 - 2 × 6 - 2 = 1 . Similarly , t here ar e two revolu tes and one sliding pair i n Fig . 2-24 .
Fig . 2-22

Fig . 2-23

Fig . 2-24

2.4.2   Passive DOF In t he plate ca m mechanism wit h oscillati ng roller follower sho w n in Fig . 2-25a , t he re are t hr ee moving links : ca m 1 , roller , and oscillating follow er 2 . There is a revolu te bet ween t he rolle r and t he follo wer 2 . T her efor e, according to Eq . (2-1) , F = 3 N - 2 P l - P h = 3 × 3 - 2 ・ 15 ・

× 3 - 1 = 2 . T he t wo DO F a re the angular coordinates of t he ca m and t he rolle r . H oweve r , t he rotation of t he rolle r abou t its cen t re does not alt er t he ou t put motion of t he follow er 2 . Such a DOF w hich does not change t he ou t put motion of t he m echanism is called a passi ve DO F . F rom now on , any passive DOF s hould be deleted before t he calculation of t he DOF of t he mechanism . If we delete t he passive DOF by welding t he roller to t he follower 2 as sho wn in Fig . 2-25b , t hen F = 3 N - 2 Pl - P h = 3 × 2 2 × 2 - 1 = 1 . T his is t he real DOF of t he m echanism . 2.4.3   Redundant Constraints
Fig . 2-25

    In m any cases , more t han one constr ain t m ay have exactly t he sam e kinem atic function . In t hese cases , on ly one of t he const rain ts should be coun ted during t he calculation . O t hers are called red u nda n t constrai n ts , w hich must not be coun ted during t he calculation . Redundan t const rain ts occur in m any sit uations as desc ribed belo w . (1 ) W hen two links ar e connect ed by more t han one parallel sliding pair , such as t he con- nection between t he followe r 2 and t he fra me 3 in t he ca m m echanism show n i n Fig . 2-26 , only one sliding pair can be coun ted during t he calculation , ot hers are redun dan t const rain ts and must not be coun ted .

Fig . 2-26

Fig . 2-27

    (2

y more t han one revolu te pair w hose axes coincide ,

such as t he connection bet ween t he crank and t he fram e in a multi-cylinder in te rnal combustion engi ne show n in Fig . 2-27 , on ly one of t he r evolu tes must be coun ted during t he calculation . O t hers a re r edundan t constr ain ts an d must not be coun ted . (3 ) W hen two links ar e connect ed by more t han one higher pair w hose common normals passing t hrough t he poin ts of con tact coincide ( such as t he t wo connections betw een t he ca m 1 and t he follow er 2 in t he ca m m echanism show n in Fig . 2-26 ) , only one of t he highe r pairs can be count ed duri ng t he calculation . Ot hers a re r edundan t con st rai nts , w hich should not be cou nted . Ho wever , if tw o lin ks a re con nected by t wo highe r pairs w hose common norm als passing t hrough t he poin ts of con tact do not coincide , as show n in Fig . 2-28 , t he tw o highe r pairs will ・ 16 ・

have tw o const rain ts . Fig . 2-28a is equivalen t to a revolu te , w hile Fig . 2-28b is equivalen t to a slidin g pair .

Fig . 2-28

Fig . 2-29

    ( 4) When t he distance between tw o poin ts on t wo lin ks r em ains constan t during t he mo- tion of t he m echanism , adding one lin k and t wo revolu tes with t heir cen ters at t hese t wo poin ts will create a redu ndan t const rain t .     In t he linkage A BCD show n in Fig . 2-29 (ignor e link E F and t he revolu tes E an d F for t he mom en t ) , A B = CD = E F , BC = A D and B E = A F . T her efore , t he distance betw een t he poin t E on t he link BC and t he poin t F on t he fram e A D is iden tically equal t o t he leng t h of li nk A B during t he motion of m echanism . If link E F and t wo revolu tes E and F are added , t h is one li nk and its tw o r evolutes will c reate one constr ain t , w hich is to keep t he distance be- tween t he poin ts E and F equal to t he lengt h of link E F . According t o Eq . , F=3N 2 Pl - P h = 3 × 4 - 2 × 6 = 0 . Zero DOF m eans t hat t he kinem atic chain can not move . In fact , t he li nkage can still move because of t he double-par allelogra m st ructure . T he reason is t hat t he const rain t from t he lin k E F and t he revolu tes E and F is exactly t he sam e as t ha t from t he original four-bar lin kage A BCD . T he refore , t he const rain t from lin k E F and t he revolut es E and F is redundan t . Befor e calculation , any redundant const rain t m ust first be deleted . T he DO F of t he mechan ism i n Fig . 2-29 should be calculated as F = 3 N - 2 P l - P h = 3 × 3 - 2 × 4 = 1 . (5 ) When t he locus of a poin t is a st raigh t line , adding one lin k wit h one fixed guide w ay par allel to t he st raigh t line and one revolut e wit h its cen t re a t t hat poin t will c rea te a redundan t const rain t . For example , in t he m echanism show n in Fig . 2-30 ( ignore sliding block 5 , fixed guide way A F and revolu te D for t he mo- m en t ) , A B = BC = B D . I t can be show n t hat t he poin t D will t race t he str aigh t-line A F pe rpendicula r to A E . Adding t he slidi ng block 5 , t he fixed guide-way A F and t he revolut e D will provide a const rain t, w hich is t o force t he point D move along t he straigh t line A F . Since t his con st rai nt is exactly t he sam e as t hat from t he
Fig . 2-30

original m echanism , t he const rain t is redundan t and t he mechan ism can still move . T herefore , aft er deleting t he redundan t const rain t , t he DO F of t he mechanism is calculated as F = 3 N ・ 17 ・

2 Pl - P h = 3 × 3 - 2 × 4 = 1 . (6 ) Form-closed cam mechan isms In order to ensur e con tact bet ween the ca m and t he follower , many kinds of form-closed cam m echanisms are used , such as show n in Fig . 2-26 and Fig . 5-6 . One of the tw o higher pairs bet ween t he cam and t he follower is a redundan t const rain t because of t he special dim en- sions . (7 ) Sym met rical struct ure or duplicated st ructur e In m any cases , e. g. in order t o balance forces and improve t he force conditions , many kinds of symm et rical struct ures or du- plicated struct ures a re used . S how n i n Fig . 2-31 is a gear t rain . From t he ki ne matic poin t of view , t he motion can be t ransmitted from gear 1 to gear 3 by just one gear 2 . Ho wever , for pr actical purposes , t hree gears 2 , 2′and 2″ a re dist ributed symm et rically in space . Thus , t he force conditions of the gears can be i mproved ,
Fig . 2-31

and a la rger torque can be t ransmitted by t he mechanism . During t he calcula tion of t he DOF of t he m echanism , only one of t he t hree gears 2 , 2′ and 2″s hould be coun ted w hile ot he rs a re re- dundan t . The connection at O is a compound hinge con sisting of O 4-1 and O 4 -3 . The DOF of t he gear t rain should t herefore be calculat ed as F = 3 N - 2 Pl - P h = 3 × 3 - 2 × 3 - 2 = 1 . Redundan t con st rai nts can i mprove t he rigidity of a mechanism , improve the force condi- tion in lin ks , over com e any undetermined motion , etc . and are widely used in m any mecha- nisms . T he so-called m et hod of“ deleting t he redundan t const rain ts during t he calculation of DO F” does not m ean t hat t he r edundant const rain ts should be omitted from real m echanisms . I t should be noted t hat all r edundan t constr ain ts require some special dim ension s . T herefore , at ten tion should be paid t o manufact uring accuracy w hen an y redundan t constr ain t is used . If m anufact uring and asse mbly errors in lin k leng t hs or pivot locations a re too la rge , t he redun- dan t constr ain ts will become r eal const rain ts and t he mechanism will jam . Example 2-3 Calcula te t he DOF of t he m echanism show n in Fig . 2-32a .

Fig . 2-32

Solution: T he re is a spri ng between t he fra m e 8 and t he slider 6 . The function of t he spring is to ・ 18 ・

force t he roller to m ain tai n con tact wit h t he ca m 7 at all tim es . I t does not kinem atically con- st rain t he r elative mo tion betw een t he tw o lin ks . T he refor e, t he spring should not be count ed in t he n umber of links . T he re are t wo par allel sliding pairs E and E′between t he fr am e 8 and t he slider 6 . O ne of t he slidi ng pairs is redundan t . T he hinge C is a compound hi nge of links 2 , 3 and 4 . T he rolle r has a passive DOF . Afte r deleting t he spring , t he redundan t constr ain t and passive DO F , t he mechanism can be redr aw n as show n in Fig . 2-32b . T her efore , t he DOF of t he m echanism can be calculated as F = 3 N - 2 P l - P h = 3 × 7 - 2 × 9 - 1 = 2 . T o t ransm it a dete r- mined motion , t wo drivers must be provided . Example 2-4 Calcula te t he DOF of t he m echanism show n i n Fig . 2-33 . Solutions: T he roller has a passive DO F . If t he roller is welded to link 5 , t he re is still a revolu te between t he link s 5 and 6 . A t poin t A , t he re are two revolu tes , A 7-1 and A 7 -2 . Similarly , C is a compoun d hinge of lin ks 4 , 5 and 7 . Wit h att en tion to t hese poin ts , t he DOF of t he m echanism can be calculated cor- rectly as F = 3 N - 2 Pl - P h = 3 × 6 - 2 × 8 - 1 = 1 .
Fig . 2-33

2.5   T h e C om p o s i t i o n P r in c i p l e a n d S t r u c t u r a l A n a l y s i s
2.5.1   Composition Principle of Mechanisms T he links and t he kinem atic pairs of a mechanism can be divided int o tw o pa rts . T he first par t consists of t he fram e , t he driver t he driver and t he ki ne matic pair con necting t he fram e and pairs belong to t he second par t . T he first par t

we will call the basic mechan ism and t he second par t t he system o f dri ven li nks . T he mecha- nism in Fig . 2-34 can for exa mple be divided i nt o tw o such parts as show n in Fig . 2-35 . D ur- ing such division and classification , t he sum of links , t he sum and t ypes of kine matic pairs do

Fig . 2-34

Fig . 2-35

Fig . 2-36

・ 19 ・

not change . T he sum of t he DO F of t he t wo par ts should t her efore be equal to t he DO F of t he original mechan ism . In Sec . 2. 3. 4 , we learned t hat in any m echanism w hich has a determined motion , t he number of drive rs must be equal to t he DOF of t he m echanism . I n t he basic m echanism , t he drive r is always connected to t he fra m e by a lowe r pair . Every driver ( and its corresponding lo wer pair ) has one DOF . T hus t he DOF of t he basic m echanism is equal t o t he nu mber of drive rs , or equal t o t he DO F of t he original m echanism . T he DOF of t he syste m of driven links must t hus be zero . In som e cases , t he syste m of driven li nks can be divided in to sm aller groups . If t he DOF of each group is zero and no group can be divided fur t her in to tw o or more ze ro-DOF groups , t hen such groups are called Assu r grou ps in me mory of Assur s cont ribu tion t o t his subject . F or exam ple , t he syste m of driven lin ks in Fig . 2-35 can be fur t he r divided in- to t wo Assur groups as sho wn i n Fig . 2-36 . In each Assur group , one or more pairs a re used t o con nect t he links wit hin t he group . Such a pair is called an i n ner pa ir . For exa mple , t he pair C i n t he group DCB and t he pair F in t he group G F E a re t he inne r pairs for t he groups concerned . Som e pairs in an Assur group a re used t o connect t he group to kinem atically determined li nks , Such pairs are called ou ter pa irs . For exa mple , t he group DCB is connected to the k ine matically det ermined links ( t he fra me and t he driver ) by low er pairs B and D . T he pairs B an d D a re t herefore t he ou ter pairs of t he group DCB . W hen t he group DCB is connected to t he determ ined links by t he out - er pairs D and B , a four-bar m echanism A BCD is cr eat ed and all lin ks in t he group DCB be- come ki ne matically determined . T he group G F E is t hen con nected to t he determi ned li nk BCE and t he fra m e by lo wer pairs E and G . T he pairs E and G ar e t herefore t he ou ter pairs of t he group GF E . N ote : t he revolu te E is not an oute r pair of t he group DCB . F rom t he asse mbly order of t he Assur groups , we can see t hat t he group DCB is t he first group , w hile the group G FE is t he second group . Hence , as m en tioned above , we can see t hat any m echanism w h ich has a determined mo- tion , can be assembled from a basic mechanism by connecti ng Assur groups to t he dete rmined links using ou ter pairs , group by group . T his is the co m posit ion pri nci ple o f m echa nism . On- ly afte r t he form er Assur group is assem bled , can t he la ter one be asse mbled . 2.5.2   Classification of Assur Group and Mechanism In a lowe r-pair Assur group , F = 3 N - 2 Pl = 0 . T herefore , Pl = ( 3/ 2 ) N . Since N and P l ar e in tege rs , t he num ber N of lin ks must be even . T he groups in Fig . 2-36 are t he simplest lo wer-pair Assur groups in w h ich t he re are tw o lin ks an d t hree pairs . If N = 4 , t he lower-pair Assur group has t wo diffe ren t constructions as show n i n Fig . 2-37 . In Fig . 2-37a , lower pairs A , B and C ar e used t o connect lin ks wit hin t he group . T hey a re t he inner pairs of t he group . T he group will be connected t o determined lin ks by lo wer pairs D , E and F . T hus , t he low er pairs D , E an d F are t he ou ter pairs of t he group . In Fig . 2-37b , low er pairs A , B , C and ・ 20 ・

D ar e the in ner pairs , w hile t he lower pairs E and F a re t he ou ter pairs . Assur groups have diffe ren t gr ades according to differen t number of links and diffe ren t st ructur e . T he groups in Fig . 2-36 , Fig . 2-37a and Fig . 2-37b are classified as grade Ⅱ , Ⅲ an d Ⅳ Assur groups , re- spectively . For t he sa me kinem atic chain , t he composition can be changed if t he fra me and/ or t he driving link is changed . F or exa mple , t he ki ne matic chain in Fig . 2-38 is t he sa me as t hat in Fig . 2-34 bu t t he driver i n Fig . 2-38 is t he link G F . T he m echanism i n Fig . 2-38 is t hen composed of a basic mechan ism and a gr ade Ⅲ Assur group , as sho wn i n Fig . 2-39 .

Fig . 2-37

Fig . 2-38

Fig . 2-39

    The gr ade of a mechanism is defined as t he highest grade of t he Assur group in t he mecha- nism . H ence t he mechanis m in Fig . 2-34 is a gra de Ⅱ m echa nism , w hile t he m echanism in Fig . 2-38 is of grade Ⅲ . T he basic m echanism is som etim es called t he gr ade I mechan ism , e . g ., a ceiling fan ( consisting of on ly a single rot ati ng li nk ) is a gr ade Ⅰ mechan ism . If all pairs in Assur group ar e r evolute pairs , t he group is called t he basic for m o f Assur grou p . If one or more r evolute pairs is replaced by sliding pairs , som e derivative forms of As- sur groups will be c reated . The group nam e , sche matic diagra m , inner pair and oute r pairs of som e commonly used grade Ⅱ Assur groups a re show n in T able 2-2 . T he links in dashed lines a re t he kine ma tically determined ones . 2.5.3   Structural Analysis As m en tioned above , a m echanism is assembled st arting wit h t he basic m echanism and addi ng Assur groups t o t he determined lin ks using t he ou ter pairs , group by group . The pur- pose of st ruct ural analysis is t o disconnect t he Assur groups from t he mechanism and t o deter- mine their t ypes and assembly order . T he steps of st ruct ural analysis for grade Ⅱ linkage m echanisms ar e as follows . (1 ) Delete all redundan t const rain ts . (2) T he fr am e and t he a re undet ermined links . (3 ) From all undeterm ined li nks that a re connected t o dete rmined links , choose tw o con- nected lin ks . T hese tw o lin ks constitu te a grade Ⅱ Assur group g fails , it means ・ 21 ・ y deter mined links . O t her lin ks

Table 2-2   Some commonly used grade Ⅱ Assur groups
Group name RRR RR P R PR P RP

Schema tic diagr am

Inner pair

B

revolu te pair B A , sliding pair

sliding pair A , r evolute B

revolu te pair A sliding pair 1- 3 sliding pair 2- 4

O u ter pairs

C, A

t ha t t he mechanism is not a grade Ⅱ m echanism .) T he pair connecting t hese tw o link s is t he inner pair of t he group . T he t wo pairs by w hich t he group is con nected to t he determined lin ks a re t he t wo out er pairs of t he group . (4) W hen t he group is connect ed t o t he determined links by t he ou ter pairs , all lin ks in t he group becom e k ine matically det ermined . No w r epeat step (3 ) un til all links become ki ne- m atically determ ined . T his procedure is som etim es called g roup d iv id i ng . During group dividing , any lin k and kine matic pair can only belong t o one group and cannot appear twice in differen t groups . According t o t he steps m en tioned above , for t he m echanism sho w n i n Fig . 2-40 , t he asse mbly order of groups , t ype of group , lin k serial numbers, in ner pair and oute r pairs of each group a re listed in T able 2-3 . Si nce t he highest gr ade of group in t his m echanism is Ⅱ , t he mechanism is a gr ade Ⅱ m echanism . D uring kinem atic analysis of t he m echanism , t he first group must be analyzed first . O nly af ter t hat , can t he second group be analyzed . ・ 22 ・
Fig . 2-40

Table 2-3   Structural analysis for the mechanism shown in Fig . 2-40
T ype First group Second group R RR R PR Li nk ser ial nu mb ers 2, 3 4 ,5 Inner p air A sliding pair C4 -5 Ou ter p airs B, D F , revolu te C3 - 5

T he t heory of st ruct ural analysis reveals t he in ternal rule of t he m echanism composition . I t can help us to unde rstand the st ructure of a m echanism , to analyze t he t ransmission rou te in t he m echanism , and to improve our abilit y in mechanism design . Problems and Exercises 2-1   Fam iliarize yourself wit h t he following te rminologies: Kinem atic pair , Pair ele men t , Planar pair , Spatial pair , Lower pair , Higher pair , Kine matic chain , Closed chai n , O pen chai n , F ram e , Driven lin k , O u tpu t link , Plana r m echanism , Spatial mechanism , Linkage m echanism , Higher pair m echanism , Kine ma tic diagram , Schem atic diagra m , Kinem atic di- m ension , DOF , Const rain t of kinem atic pair , Com pound hinge, Passive DO F , Redundan t const rain t, Basic mechanis m , Syst em of driven lin ks , Assur group , Inner pair , O u ter pair , G roup dividing . 2-2   W hat is t he kinem atic function of a kinem atic pair ? W hat is t he kine ma tic function of a link ? What const rain ts will a r evolute set ? W hat const rain ts will a sliding pair set ? W hat con- st rain ts will a highe r pair set ? W hy is a pair havin g surface con tact na med a lowe r pair ? W hy is a pair having poin t or line con tact na med a highe r pair ? 2-3   W hat poin ts should be con sidered during t he calculation of DOF of a m echanism ? 2-4   W hat is t he condition for a m echanism t o have a determined motion ? W hat will happen if t he num ber of driving links is less t han or gr eate r t han t he DO F of t he m echanism ?W hy are re- dundan t const rain ts used widely in pr actical mechan isms ? What should be considered during de- sign an d manufact ure w hen redundan t const raints a re used ? 2-5   W hat is t he composition principle of mechanism ? How is t he grade of a m echanism deter- mined ? 2-6   D raw t he k ine matic diagra ms and then calculat e t he degree of freedom of t he m echanisms in Fig . 2-41 to Fig . 2-46 . 2-7   Calcula te the degree of freedom of t he m echanisms show n in Fig . 2-47 to Fig . 2-58 and Fig . 4-8 . Indicat e all poin ts for att en tion before t he calculation of t he DO F . 2-8   Ca rry ou t t he struct ural analysis for t he m echanisms in Fig . 2-51 , Fig . 3-16 , Fig . 3- 17 , and Fig . 4-8 . List t he asse mbly orde r of Assur groups , t he type of group , t he gr ade of group , the gr ade of t he m echanism , t he lin k serial numbe rs , t he inner pair and t he ou ter pairs of each group in each mechan ism . ・ 23 ・

Fig . 2-41

Fig . 2-42

Fig . 2-43

Fig . 2-44

Fig . 2-45

Fig . 2-46

・ 24 ・

Fig . 2-47

Fig . 2-48

Fig . 2-49

Fig . 2-50

Fig . 2-51

Fig . 2-52

Fig . 2-53

Fig . 2-54

Fig . 2-55

Fig . 2-56

Fig . 2-57

Fig . 2-58

・ 25 ・

C h ap t e r 3 K i n e ma t ic A n a l y s i s o f Mec han i s ms
3.1   T a s k s a n d M e t h o d s o f K in em a t i c A n a l y s i s
    The tasks of kine ma tic analysis a re to fi nd angular positions , angula r velocities and angular accelerations of driven links and/ or positions , linea r velocities and linea r accele rations of poin ts on driven lin ks , according t o input para m eters of drivin g link ( s ) and t he dim ensions of all links . In order to determ ine w het her or no t all lin ks will i nte rfer e wit h each ot her , or t o deter- mine t he st roke of a driven lin k , or t o find t he locus of a poin t , w e must analyse positions of t he lin ks and/ or t he poin ts of i nt erest . P osition analysis is also t he first step in velocit y and ac- celeration analysis . In orde r to calculate t he stored ki netic energy by for mula E = m v / 2 or E = Jω/ 2 , or to determ ine t he po wer P of a mot or by form ula P = F v , we carry out t he velocity analysis , w hich is also a step on t he w ay to t he dete rmination of t he accele ration . Designe rs must en sur e t hat t he st resses in t he mate rials of t he proposed m achine are kep t well below t heir allowable levels . To calculate t he st resses , w e need to know t he static and dy- na mic forces on t he links . To calculate t he dynam ic forces , we need t o know t he accelera tion because t he iner tia forces ar e propor tional t o acceler ation . Ki nem atic analysis of m echanisms can be carried out by gr aphical or analy tical or experi- m en tal m et hods . By geom etric drawing , t he positions of all lin ks in a gr ade Ⅱ lin kage can be determ ined easily accordi ng to t he asse mbly order of Assur groups , step by step . In t his chap- ter , w e will i nt roduce one of t he gr aphical met hods , nam ed t he inst an t cen tr es m et hod , for velocit y analysis . The analy tical met hod has many advan tages ove r graphical m et hods . In t his chap ter , we will pu t t he st ress on t he analytical met hod .
2 2

3.2   V e l o c i t y A n a l y s i s b y t he M e t h o d o f In s t a n t C e n t r e s
3.2.1   Definition of the Instant Centre Sho wn in Fig . 3-1 are t wo bodies 1 an d 2 having r ela tive plana r motion . At any instan t ・ 26 ・

t here ex ists a pair of coincident poin ts , e. g. P1 and P2 ( ex tend t he tw o bodies , if necessary , so t hat t he poin t P is“ in ” bot h bodies ) , t he absolu te velocities of w hich a re t he sam e , in bot h m agnit ude and dir ection , i. e. 纈 P 1 = 纈 P 2 . At t his instan t , t her e is no relative velocit y between t his pair of coinciden t poin ts , i .e. 纈P 1 P2 = 纈 P2 P 1 = 0 . T hus , at t his instant , ei- t her lin k will have pure rotation r elative to t he ot he r link abou t t he poin t . T his pair of coinciden t poin ts wit h t he sa me velocities is defined as t he instan taneous cen t re of relative rotation , or more briefly t he in- stan t cen t re , denoted as P1 2 or P 2 1 . If one of lin ks is t he fr am e , t he instan t cent re is called an absolute instan t cent re , ot her wise, a rela- tive i nstan t cent re . T he absolute instan t cent re is t he zero-velocit y poin t on a moving link , bu t its accelera tion may no t be zero . A t t he position show n in Fig . 3-1 , t he tw o links rotate relative to each ot he r about t he in- stan t cent re P 1 2 . So any ot her pair of coi nciden t points , e. g. A 1 and A 2 , will have relative velocities , i .e . 纈A 1 A 2 and 纈A 2 A 1 . The directions of 纈 A 1 A 2 and 纈A 2 A 1 are perpendicular to P A . T he refore, if t he dir ection of t he relative velocit y of a pair of coinciden t poi nts , e.g . A 1 and A 2 , is kno wn , t hen t he instant cent re must lie som ew her e on t he norm al t o t he relative veloci- ty passi ng t hrough t he coi nciden t poin t A . 3.2.2   Number of Instant Centres of a Mechanism Each pair of lin ks i and j has an instan t cen t re and P ij is iden tical to P j i . Thus t he number N of instan t cen tres of a m echanism wit h k lin ks is N= k( k - 1 ) 2
Fig . 3-1

Not e: T he fra me is included in t he nu mber k . 3.2.3   Location of the Instant Centre of Two Links Connected by a Kinematic Pair (1 ) Revolu te pair If tw o lin ks 1 and 2 a re connected by a revolu te pair , as sho wn in Fig . 3-2a , t he cen tre of t he revolut e pair is obviously t he instan t cen t re P1 2 or P 2 1 . (2 ) Pure-rolling pair T he pur e-rolling pair is a special case of a highe r pair , as sho w n in Fig . 3-2b . There is no slippi ng bet ween t he t wo con tacti ng poin ts A 1 and A 2 , i .e. , 纈A 1 A 2 = v A 2 A 1 = 0 . T hus t he poin t of cont act A is t he instan t cent re P 1 2 or P 2 1 . Kinem atically , t he tr ansmission bet ween a pair of gears is equivalent to rolling wit hou t slipping bet ween a pair of circles . So t he con tact poin t of t he tw o pitch circles of t he gears 1 and 2 is t he in stant cen t re P 1 2 for the gears 1 and 2 , as sho w n in Fig . 3-2c . (3 ) Sliding pair ・ 27 ・

Fig . 3-2

As can be seen in Fig . 3-2d , r ela tive tr anslation is equivalen t to r ela tive rotation abou t a poin t located at infinity in eit her dir ection perpendicular to t he guide-way . T her efor e, t he in- stan t cen t re of t he t wo links con nected by a sliding pair lies at infinit y in eit her direction pe rpen- dicular to t he guide-way . T he inst an t cent res m en tioned so far a re called observable inst an t cent res and should be lo- cated and labeled before any ot hers ar e found . (4 ) Higher pair (rolli ng & slidin g pair ) Sho wn in Fig . 3-2e are tw o links 1 and 2 connected by a hig her pair . Their con tact poin t is poin t A . The direction of relative velocities , 纈 A 1 A 2 and 纈 A 2 A 1 , between A 1 an d A 2 must be along t he common tangen t . O t her wise t he re will be a relative velocity componen t along t he common norm al n-n w hich will make t he t wo lin ks sepa rate or in terfere . So t he instan t cen tre P 1 2 or P 2 1 must lie so mew here on t he com mon norm al n-n t hrough t he poin t A of con tact . Its exact position can be located by the aid of t he t heorem of t hree cen tr es as described below . A t- ten tion : Their instan t cen t re is not located at t he poin t of con tact or infinity . 3.2.4   Theorem of Three Centres ( Aronhold-Kennedy Theorem) A ny t hr ee lin ks , e.g . links 1 , 2 , and 3 , have t hr ee instan t cen tr es: P 1 2 , P 1 3 , and P 2 3 . According to t he t heore m of t hree cen t res , t he t hree inst an t cen t res P 1 2 , P 1 3 and P 2 3 must lie on a str aigh t line . T his t heor em can be proved as follows . Suppose t hat t he positions of P1 2 and P1 3 ar e know n , as show n in Fig . 3-3 . L et us con- sider any poin t , e. g . poin t C , ou tside t he line P1 2 P 1 3 . Since P 1 2 ( A ) is t he in stant cen tr e of t he lin ks 1 and 2 , t he link 2 rotates relative to the link 1 about t he poin t A . So 纈C 2C 1 ⊥ AC . Sim ilarly , 纈C 3 C1 ⊥ BC .Since 纈C 2 = 纈C 1 + 纈 C2 C 1 , t hen 纈C 2C 1 = 纈C 2 - 纈C 1 .Sim ilarly , 纈 C3 - 纈C 1 . Obviously , for any poin t C ou tside t he line P 1 2 P 1 3 , t he dir ections of t he vectors 纈 C2 C 1 and 纈 C3 C 1 are not t he sa me, i .e. 纈 C2 C 1 ≠ 纈C 3 C 1 . T herefore ( 纈C 2 - 纈C 1 ) ≠ ( 纈C 3 - 纈C 1 ) from w hich one obtains 纈C 2 ≠ 纈C 3 . H ence , according to t he definition of i nstan t cen tr e , t he poin t C can not be t he instan t cen tr e P 2 3 betw een t he links 2 and 3 . In o t her words , an y poin t outside t he st raigh t li ne P 1 2 P 1 3 can- not be t he instant cen tr e P2 3 . T hus t he t heorem of t hr ee cen- ・ 28 ・
Fig . 3-3
C3C1

=

tr es is derived : t he t hree i nstan t cen t res of any t hree i ndependent links in general plane mo tion must lie on a com mon str aigh t line . 3.2.5   Applications of Instant Centres Example 3-1 For t he four-bar m echanism show n in Fig . 3-4 , t he angular velocit y ω 1 of crank 1 is given . For t he position sho w n , ( 1 ) locate all i nstan t cen t res for t he m echa- nism , (2 ) find t he r atio ω3/ ω1 of the angular velocity of lin k 3 to t hat of link 1 , (3 ) find t he velocit y 纈 F of poi nt F on link 2 . Solution: (1 ) There ar e six instan t cent res in t his four-bar linkage m echanism . In order t o locate all instan t cen tr es i n a m echanism , we should first t ry to locate all obse rvable instant cen t res . O t her instant cen t res can then be located by t he t heor em of t hr ee cen tr es . T he re are four observable instan t cen t res ( P1 4 , P 1 2 , P 2 3 , and P3 4 ) an d t wo unobservable instan t cent res ( P 1 3 and P 2 4 ) in t his m echanism . According t o t he t heor em of t hr ee cen t res , P 1 3 will lie no t only on t he line P 1 4 P 3 4 , but also on the li ne P 1 2 P 2 3 . Since P2 3 is at infin it y per- pendicula r t o t he guidi ng bar 2 , line P 1 2 P 2 3 passes t hrough t he poin t P 1 2 and is perpendicular t o B F . Hence t he i nte rsection E of t he lines P 1 4 P3 4 and P1 2 P 2 3 is t he instan t cen t re P 1 3 . Sim ilarly , line P 2 3 P 3 4 passes t hrough t he poin t P 3 4 and is perpendicula r t o B F , t he in ter- section G of t he lines P 1 2 P 1 4 and P 3 4 P 2 3 is t he instant cent re P 2 4 . T hus it can be seen t hat it is usual to apply t he t heore m of t hr ee cen t res twice t o determi ne t wo lines , t he in tersection of w hich will be t he unobse rvable i nstan t cent re . Instan t cen tr es P 1 4 , P3 4 , and P 2 4 are absolute instan t cen tr es , w hile t he o t hers a re r elative in stan t cent res . (2 ) In order to find ω 3 for t he given ω 1 , we should take advant age of t he fram e 4 . T heir t hr ee inst an t cen tr es ( P 3 4 , P 1 3 , P 1 4 ) lie on a common st raigh t line . T he moving links 1 and 3 ro tate relative to the fr am e 4 abou t t he absolu te instan t cen t res P 1 4 ( A ) and P 3 4 ( D ) re- spectively . In link 1 , v E 1 = ω1 l A E . In link 3 , v E 3 = ω3 l D E . (E xtend t he t wo lin ks . if neces- sary . so t hat t he poin t E is“ in ” bot h links .) Since t he poin t E is t he in stan t cen t re P 1 3 be- tween t he links 1 and 3 , 纈 E 1 = 纈 E 3 . Therefore , ω1 l A E = ω3 l D E from w h ich i3 1 = ω 3/ ω 1 = l A E/ lD E = P1 4 P 1 3/ P 3 4 P 1 3 . The leng t hs of l A E and l D E a re measured directly from the kine matic diagr am of t he m echanism . The direction of ω3 is cou nte r-clock wise at t his instan t . From above , it is show n t hat t he ratio ω i/ ω j of angular velocities bet ween any tw o moving ・ 29 ・
Fig . 3-4

links i and j is equal to t he inverse ratio of t he tw o distances bet ween t he relative instant cen tre P ij and t wo absolu te i nstan t cent res P f i and P f j , t hat is , ω i P f j P ij = ωj P f i P ij (3-1)

w her e t he subsc rip t f r epresen ts t he fr am e . If t he relative instan t cent re P ij lies between t he t wo absolut e instan t cen tr es P f i and P f j , t hen t he dir ections of ωi an d ωj a re diffe ren t . O t her- wise , t he dir ections of ω i and ω j a re t he sa me . (3 ) Since t he links 2 and 3 a re connected by a slidin g pair , they canno t rotate r elative to each o t her . Th us , ω2 = ω3 = ω1 lA E/ l E D . Since P 2 4 is t he absolute instant cen tr e , t he lin k 2 ro tates ( relative to t he fram e 4 ) about t he poi nt P 2 4 ( G ) at t his instan t . T herefore , v F = ω 2 ・ l G F . Its dir ection is pe rpendicular t o G F , as sho wn in Fig . 3-4 . N ote : Alt hough t he velocit y of t he poin t P2 4 ( G ) is zero , its accele ration is not ze ro . Example 3-2 In t he ca m mechanism wit h t ranslating roller follow er show n in Fig . 3-5 , t he ca m is a circula r disk . Supposing t ha t t he angu- la r velocity ω1 of t he cam is kno wn , t he velocit y 纈2 of t he fol- lower 2 is to be found for t he position show n . Solution: As men tioned in Sec . 2 .4 .2 , t he roller has a passive DOF . T he velocity of t he follower will no t change if t he roller is welded t o t he followe r 2 . In such a case , t he ca m 1 and li nk 2 a re con- nected by a higher pair . The instan t cent re P1 2 bet ween t he ca m
Fig . 3-5

1 and link 2 must lie som ew her e along t heir common nor mal n-n t hrough t he poin t of cont act C . According t o t he t heor em of t h ree cent res , P 1 2 must lie on t he st raigh t line connecting P 1 3 and P 2 3 . Since t he lin ks 2 and 3 are con nected by a sliding pair and t heir instan t cen t re P 2 3 is at i nfinity perpendicula r to t he guide w ay , t he line P 1 3 P2 3 passes t hrough P 1 3 and is perpendic- ula r t o t he guide w ay . T hus t he i nte rsection B of t he common norm al n-n and t he line P 1 3 P 2 3 is t he in stant cen t re P1 2 and 纈B 1 = 纈B 2 . N ote : N eit her t he cen t re O of t he circle nor t he con- tact poin t C is t he instan t cent re P 1 2 . O n t he ca m 1 , 纈B 1 = ω 1 lA B . Since t he follo wer 2 is t ranslating , all poin ts on t he follower 2 have t he sa me velocity 纈 2 . So 纈2 = 纈B 2 = 纈B 1 = ω 1 lA B . Example 3-3 In Fig . 3-6 , gea r 3 rolls on t he fixed r ack 4 wit hou t slipping . Assuming t he velocit y 纈 1 of slide r 1 is know n , t he velocit y 纈 D of t he cent re D of t he gear 3 is to be found . ・ 30 ・

Solution: In orde r t o find t he velocity of a poin t on t he gear 3 , angula r velocit y ω 3 of t he gear 3 should be found first . As m en tioned before , i n order to fi nd ω3 of t he gear 3 for t he given velocit y 纈1 of t he slider 1 , w e al- ways take advan tage of t he fra me 4 . Thus we should t ry t o locate t he t hree i nstan t cen t res , P 3 4 , P 1 4 and P 1 3 , bet ween t he th ree link s , 1 , 3 and t he fra me 4 . T he gear 3 rolls on t he fixed rack 4 wit hou t slipping . So t he cont act poin t C is t heir instan t cen t re P3 4 . P 1 4 lies at infi nit y pe rpendicular t o A B ( no t AC !) . P 1 3 must lie on bo t h lines P1 4 P 3 4 and P2 3 P 1 2 . So the int ersection E of t he lines P 1 4 P 3 4 and P 2 3 P 1 2 is t he instant cen tr e P1 3 bet ween t he lin ks 1 an d 3 . T her efor e 纈 E 3 = 纈 E 1 . Since t he slid- er 1 is tr ansla ting , 纈 1 = 纈 E 1 = 纈 E 3 = ω 3 l C E . T hus ω 3 = 纈1/ l C E from w hich 纈D = ω 3 lC D = 纈 1 lC D/ l C E . T he dir ection is as show n in Fig . 3-6 . 3.2.6   Advantages and Disadvantages of the Method of Instant Centres T he met hod of i nstan t cen t res offers an excellen t tool in t he velocit y analysis of simple m echanisms . H oweve r , i n a complex m echanism , some instan t cen t res may be difficult to find . In some cases t hey will lie off t he paper . Lastly , it should be poin ted out t hat an instan t cen t re , in gene ral , changes its location on bo t h lin ks during motion . T he acceleration of t he instan t cen t re is not zero ( excep t for fixed pivo ts ) . T he refor e , t he instan t cen tre m et hod can- not be used in acceleration analysis .
Fig . 3-6

3.3   K in e ma t i c A n a l y s i s b y A n a l y t i c a l M e t h o d s
In graphical m et hods , none of t he information ob tained for t he first position of t he mecha- nism will be applicable t o t he second position or to any o thers . T he kinem atic diagram of t he m echanism must be redraw n for each position of t he driver . T his is very tedious if a mechan ism is to be analyzed for a complete cycle . F ur t he rmore , t he accuracy of t he graphical solu tion is limited . In con t rast , once t he analytical solu tion is derived using an analytical met hod , it can be evaluated on a compu ter for diffe ren t dim ensions and/ or at differen t positions wit h very little ef- for t . T he accuracy of t he solu tion far surpasses t hat required for mechan ical design problems . T hus , i n t his chapte r we will pu t st ress on t he analytical met hod . G raphical m et hods can be used if necessa ry as a check on t he analytical solutions . T he re exist m an y kinds of analytical met hods for k ine matic analysis of lin kages . T he ki ne- m atic analysis of a m ulti-ba r li nkage m echanism seem s to be a hard task at first sigh t . H owev- ・ 31 ・

e r , it becom es easier if t he Assur-group m et hod is used . As m en tioned in Sec. 2. 5 , most link- age m echanisms a re built up by adding one or more commonly used Assur groups to t he basic m echanism . Since t he DO F of an Assur group is zero , Assur groups have kinem atic determina- tion . T ha t is , t he motions of all lin ks i n an Assur group can be determined so long as t he mo- tions of all oute r pairs are know n . T aking t his fact in to accoun t , one can set up subroutines in advance for some com monly used Assur groups . T hen t he kinem atic analysis of a multi-bar linkage mechan ism is reduced t o two sim ple steps : first , dividing t he mechanism in to Assur groups and secondly , calling t he corr esponding subrou tine for each Assur group according to t he t ype and t he asse mbly order of t he Assur group . T his m ethod is called t he Assu r-g rou p m et hod for kinem atic analysis . In t he nex t sections , we will set up som e com monly used k ine matic analysis subroutines before analyzing a six-bar lin kage m echanism . 3.3.1   The LINK Subroutine Suppose t hat t he x and y componen ts of position , velocity and accelera tion of a poin t A ( i . e . x A , yA , v A x , v A y , a A x , a A y ) , t he angular position , t he angular ve- locit y , an d the an gular acceler ation of lin k A B ( i .e.θ, ω, ε) , and t he lengt h ( lA B ) of t he lin k A B are know n , as show n in Fig . 3-7 . T he x and y componen ts of position , velocit y , and acceleration of poin t B ( i .e . x B , yB , v B x , v B y , aB x , aB y ) can be calculated as follows . In a Ca rtesian co-ordinate syst em , x B = x A + l A B cos θ  and   y B = yA + l A B sin θ velocit y analysis can be derived . vB x = v A rived . aB x = a A x - l A B ( sin θ)ε - lA B ( cos θ) ω   an d   a B y = aA y + lA B ( cos θ)ε - lA B ( si n θ) ω
2 2 x

Fig . 3-7

Diffe ren tiating t he above position analysis formulae wit h respect t o ti me , t he formulae for - lA B (sin θ) ω   and   vB y = v A y + l A B ( cos θ) ω

Diffe ren tiating agai n , t he formulae for analyzing t he acceleration of t he poin t B can be de-

T hese six formulae can be progra mm ed in a subroutine . In t he T RU E BASIC compu ter language , any subrou tine must begin wit h statem en t S UB SUBRO U T I N E- NA M E ( T ABL E O F PARA M E T E RS ) and end wit h state ment E ND S UB . Let us na me t he subroutine LI NK . T hen t he LI N K subroutine is as follo ws: S UB LIN K ( XA , YA , VAX , VAY , AAX , A AY , Q , W , E , L AB , XB , YB , VBX , VBY , ABX , ABY )     L E T   XB = XA + LAB * COS ( Q )     L E T   YB = YA + LAB * SI N ( Q )     L E T   VBX = V AX - L AB * SI N ( Q ) * W     L E T   VBY = VAY + L AB * COS ( Q ) * W ・ 32 ・

    L E T   ABX = AAX - LAB * SI N ( Q ) * E - LAB * COS ( Q ) * W^2     L E T   ABY = AAY + LAB * COS ( Q ) * E - L AB * SI N ( Q ) * W^2 E ND SUB Every evaluating sta tem en t must begin wit h L E T . In order to facilitate t he unde rstanding of t he progr am , para mete rs should have easily-recognized nam es . For example, VAX . T he table of t he pa ra mete rs corr esponds to x A , y A , ε, lA B , xB , yB , vB x , ra meters of t he poin t B will be know n . 3.3.2   The RRR Subroutine In t he R R R group show n in Fig . 3-8 , t he kinem atic para m eters of t he tw o ou ter poin ts A and C and t he lengt h s of t he t wo link s A B and CB a re know n . The angula r position s , angular velocities , and an gular accelera tions of t he lin ks A B and CB can be calculated as follo ws . W hen x A , yA , O n t he link CB , yC + lCB sin θ CB O n t he link A B , y A + l A B sin θ AB Combini ng t hese tw o sets of equa tions , one ob- tains : x C + lC B cos θ CB = x A + l A B cos θ AB yC + l CB sin θ CB = yA + l AB sin θ AB (3-2)
Fig . 3-8

v A x is na m ed

vA x ,

v A y , a A x , a A y , θ, ω,

vB y , a B x , aB y . Afte r t he subroutine is called , t he kinem atic pa-

x C , y C , lA B and l CB are determined , t here a re tw o assembly modes for x B = x C + l CB cos θ C B and yB = x B = x A + l A B cos θ A B and yB =

t h is group , as show n in Fig . 3-8 , one i n solid lines and t he o t her in dashed lines .

T he re are tw o unk no w ns , θ A B and θ C B , in t his

set of equations . Since t here are t wo assembly modes for t his group , t here will be t wo sets of solutions . Alt hough some ma them atical skill can be used to solve the above t rigonom etric non- linear equations to ob tai n t wo sets of formulae for θ A B and θ C B , t he calculation process is t edious and t he form ulae de rived w ould be very com plicated . I t is hard to judge w hich set of formulae corr esponds t o a specific asse mbly mode . T he following is a si mple met hod to overcome t his difficult y . ( a )   lAC = ( x C - x A ) 2 + ( y C - yA ) 2

(b )   cos θ A C = ( x C - x A ) / lA C   and   sin θ AC = ( y C - yA ) / l A C T he subrou tine may be used for any combinations of t he positions of poin ts A and C . Note t hat si n θ A C may not be equal to 12 ( cos θ A C ) since sin θ A C may be negative . T he m agni-

tude of θ A C can be calculated according to t he values of bot h cos θ A C and sin θ A C by t he AN G L E function in T R U E BASIC . N ote again t hat θ may not be equal to A T N ( sin θ / cos θ) since θ m ay be greater t han ・ 33 ・

is only from - π / 2 t o +π / 2 .No te :θ differen t from θ A C is 180° CA .
2 2 2 ( c ) cos θ B A C = ( l A B + l A C - lC B ) /

( 2 lA B lA C ) 12 ( cos θ B A C ) . If l A C > l A B + l CB , t hen cos θ BA C >

Since 180° >θ , sin θ B A C > 0° BAC =

1 . T his m eans t ha t t he distance lA C between t he tw o out er poin ts is gr eat er t han t he sum of l A B and lC B . If l A C < | lA B - l CB | , t hen cos θ B A C < - 1 and t he distance l A C is less t han t he differ- ence bet ween l A B and l CB . In t hese cases , t he R R R dyad can not be assem bled . T he calcula- tion of 12 ( cos θ B A C ) will fail and compu tation will be stopped .

(d ) As m en tioned before , t her e a re tw o assem bly modes for t he R R R group . For t he as- sem bly mode show n by solid li nes , θ AB = θ AC - θ B A C . During motion of t he mechan ism , t he assembly mode does not change as a r esult of change in position . ( e ) x B = x A + lA B cos θ A B   and   yB = y A + lA B sin θ AB (f) cos θ CB = ( x B - x C ) / lC B   and   si n θ CB = ( yB - yC ) / l C B T he m agnitude of θ C B can be calcula ted accordi ng to t he values of bot h cos θ C B and si n θ CB by t he A NG L E function in TR U E BASIC . Velocity analysis can be progr essed only after t he position analysis is finished . T he angular velocities , ω A B and ω C B , of t he li nks A B and CB can be found by differentiatin g Eqs . ( 3-2 ) wit h respect to tim e . vC x - lC B ( sin θ CB ) ω CB = v A x - lA B (sin θ AB ) ω AB vC y + l CB ( cos θCB ) ωC B = v A y + lA B (cos θ AB ) ω AB (3-3)

Bot h velocit y and accele ration equations of a gr ade Ⅱ Assur group ar e dualistic linea r equa- tions . The explicit expressions for ωA B and ωC B can be found easily by solving t he two equations simultaneously . By diffe ren tiating Eqs . ( 3-3 ) wit h respect to ti me , anot her set of dualistic li near equations wit h tw o un know ns , i. e . the angula r acceleration s ε A B and ε C B of t he links A B and CB , is derived . The explicit expressions for ε A B and ε C B can be found easily by solving t he two equa- tions si mult aneously . Using t he above explicit expr essions , not t he equations , t he subrou ti ne na med R R R for kinem atic analysis of t he R R R group can be writt en as follows:     SUB R RR ( XA , YA , V AX , V AY , AAX , A AY , XC , YC , VCX , VCY , ACX , ACY , L AB , LCB , QAB , WAB , E AB , QCB , W CB , E CB ) L E T   L AC = SQR ( ( XC - XA )^2 + ( YC - YA )^2) L E T   COSQAC = ( XC - XA )/ L AC L E T   S I NQ AC = ( YC - Y A )/ L AC L E T   Q AC = AN GL E ( COS QAC , S I NQ AC ) L E T   COSQBAC = ( L AB^2 + LAC^2 - LCB^2)/ ( 2 * LAB * L AC ) L E T   S I NQBAC = SQ R ( 1 - COSQBAC^2) L E T   QBAC = A NG L E ( COSQBAC , S I NQBAC ) ・ 34 ・

L E T   Q AB = QAC - QBAC L E T   XB = XA + LAB * COS ( QAB ) L E T   YB = YA + L AB * SI N ( Q AB ) L E T   COSQCB = ( XB - XC )/ LCB L E T   S I NQCB = ( YB - YC )/ LCB L E T   QCB = A NG L E (COSQCB , S IN QCB )     …… L E T   WAB = …… L E T   WCB = ……     …… L E T   E AB = …… L E T   ECB = …… E ND SUB A tten tion should be paid to t he sequence of t he revolut e′ s let ters in t he table of t he pa ram e- ters w hen t he subrou tine is called . F or t his subrou tine , t he t hree let ters A , B , and C a re ar- ran ged in CC W . 3.3.3   The RPR Subroutine Sho wn in Fig . 3-9a is an R P R Assur group . The revolu tes A an d C ar e ou ter r evolute pairs . The eccen t ric A B is perpendicular to guide-bar B D . The ki ne matic par am eters of t he cen ters , A and C , of t he tw o ou ter revolu te pairs and t he lengt h of eccen tric A B a re know n . T he re ar e two assembly modes for t his group . O ne is sho w n in solid lines , t he ot her i n dashed lines . T he angula r position , angular velocit y and angula r acceler ation of t he guide-ba r BD (θ B D , ω, ε ) can be calculated as follows:

Fig . 3-9

lAC =

( x C - x A ) 2 + ( y C - yA ) 2

cos θ ( x C - x A ) / lA C , AC = ・ 35 ・

sin θ ( y C - yA ) / l A C AC = l BC =
2 l2 AC - lAB

If lA C < l A B , t hen t he group can not be assembled . In t his case , t he calculation of lA C - lA B will fail and t he compu tation will be stopped . θ A CB = arctan lAB l BC
2 2

θ BD = θ AC + M θ ACB θ AB = θ BD - M π / 2 w her e M is t he coefficien t of t he asse mbly mode . F or t he solid mode , das hed mode , M = + 1 . For t he M = - 1 . During mo tion of t he mechan ism , t he asse mbly mode does not

change as a result of change in position . We can determine t he value of M accordin g to t he as- sem bly mode at any angle of t he drive r . From Fig . 3-9 , we have x C = x B + lB C cos θ B D = x A + l A B cos θ A B + l BC cos θ BD y C = yB + lB C sin θ B D = y A + lA B sin θ A B + l B C si n θ BD     Diffe ren tiating t he above equations wit h r espect to tim e r esults in ( yC - y A ) ω + cos θ B D × vl B C = v C x - vA x ( x C - x A ) ω + sin θ B D × v l BC = v C y - v A y w her e v lB C is t he de rivative of lB C wit h respect t o time . Solving t he dualistic linear equations Eqs . ( 3-4 ) simultaneously , t he explicit expr essions for t he t wo u nknow ns ω and v l B C can be found . Diffe ren tiating Eqs . ( 3-4 ) wit h respect t o time ( note t hat v l B C is a va riable ) , anot her set of dualistic linear equations wit h t wo unknow ns ( one of t he un know ns is ε) is derived . T he explicit expr ession for ε can be found easily by solving t he tw o equations simultaneously . T he subrou tine for t he kinem atic analysis of t he R P R group , w hich we will nam e R P R , can be written as follows:     S UB R P R ( M , XA , Y A , V AX , V AY , AAX , A AY , XC , YC , VCX , VCY , ACX , ACY , L AB , QBD , W , E )     L E T LAC = SQ R ( ( XC - XA ) ∧ 2 + ( YC - YA ) ∧ 2 )         ……     L E T   QBD = Q AC + M * Q ACB         ……     L E T   W = ……         ……     L E T   E = …… E ND SUB ・ 36 ・ (3-4)

If l A B = 0 , t hen t he R P R group in Fig . 3-9 a is simplified in t o anot he r R P R group sho w n in Fig . 3-9b . F or t he R P R group i n Fig . 3-9b , l A B = 0 and θ A C B = 0 . T he value of M can be set as any value . Ki nem atic analysis subrou tines for o t her grade Ⅱ Assur groups , e.g . R R P , P R P groups in T ab . 2-2 , can also be derived and established in a si mila r met hod . 3.3.4   Main Program To analyze any mechanism , a m ain progra m is required . In t he main program , suitable kinem atic analysis subrou ti nes are called accordi ng to t he t ypes and t he assembli ng sequences of Assur groups . Example 3-4 T he six-ba r linkage show n in Fig . 3-10 is t he sam e as t hat in Fig . 2-40 . T he driving c ran k 1 rotates at a constant angular velocity ω 1 of 10 rad/ s . The know n di mensions of t he m echanism are : x E = 0 , y E = 0 , x B = 41 mm , yB = 0 , x F = 0 , y F = - 34 mm , l E D = 14 m m , l D A = 39 m m , lB A = 28 m m , ∠ ADC = 35° , l D C = 15 m m , l F G = 55 m m . A m ai n progr am is required t o analyze t he ou tpu t motions of lin k FG and point G . T he m echanism will be analyzed for t he w hole cycle w hen t he driver ED rotates from 0°to 360° wit h a st ep size of 5°. Solution: ( a ) Group dividing T he com position of t his mechanism has been analyzed in Sec . 2. 5. 3 . T he t ypes and t he assembly orde rs of Assur groups ar e listed in T able 2-3 of Chap ter 2 . I n t his linkage , link ED is t he driver . Links 3 an d 2 form s an R R R dyad . Afte r t his dyad is connected to t he driver and t he fra me , t he motion of bot h links 3 and 2 ar e det ermined . Thus w e can det ermine t he motion of point C on t he link 3 . R ocke r 4 an d block 5 forms a R P R dyad . Th is dyad can be assembled only af ter t he mo tion of t he poin t C is determ ined . (b ) M ain progra m R E M The m ain progr am for t he linkage mechanism in Fig . 3-10 FOR Q 1 = 0 T O 360 S T EP 5 CA L L   LI NK (0 , 0 , 0 , 0 , 0 , 0 , Q1 * P I/ 180 , 10 , 0 , 14 , XD, Y D , VDX , V DY , ADX , ADY ) CA L L   R R R ( XD , YD , V DX , VDY , ADX , ADY , 41 , 0 , 0 , 0 , 0 , 0 , 39 , 28 , Q 3 , W3 , E3 , Q2 , W 2 , E2 ) ・ 37 ・
Fig . 3-10

L E T   QDC = Q 3 + 35 * P I/ 180 CA L L   L IN K ( XD YD , VDX , VD Y , ADX , ADY , QDC , W 3 , E3 , 15 , XC , YC , VCX , VCY , ACX , ACY ) CA L L   R P R ( 0 , 0 , CA L L   L IN K ( 0 , - 34 , 0 , 0 , 0 , 0 , XC , YC , VCX , VCY , ACX , ACY , - 34 , 0 , 0 , 0 , 0 , Q 4 , W 4 , E4 , 55 , XG , YG , V GX , 0 , Q 4 , W4 , E4) V GY , AGX , AGY ) P R I N T   Q1 , Q 4 * 180/ P I , W 4 , E4 , XG , YG , VG X , VG Y , AGX , AGY N EX T   Q1 E ND T he subrou tine for t he first Assur group R R R should be called before t hat for t he second Assur group R P R is called . In t he first Assur group R R R , t he point D should be a dete rmined poin t . So we use the first CA L L LI NK state men t to calculate t he kinem atic pa ra mete rs of t he poin t D before R R R is called . T he para m eter P I i n t he first CAL L line is set to π by T RU E BASIC aut om atically . Emp hasis should be pu t on t he asse mbly mode , t he sequence and t he corr espondence of para mete rs in t he table of pa ra meters w hen calling a subrou tine . Correspond- ing data a re t ransferred according to t he sequence , not according to t he na m e . As m en tioned before , an R R R group has two assembly modes , as show n in Fig . 3-8 . T he R R R subroutine is w rit ten for t he assembly mode show n in solid li nes . In orde r to use t he R R R subrou tine , t he revolu tes D , A and B of t he R R R dyad in Fig . 3-10 must corr espond to t he revolu tes A , B and C of t he R R R dyad in Fig . 3-8 , respectively . In t he second Assur group R P R , t he revolu te cen t re C should be a dete rmined poin t . So t he second CAL L LI N K sta tem en t must be used to calculate t he kinem atic pa ram eters of t he poin t C after RR R is called and before R P R is called . T he m ain progr am ends wit h t he state ment E ND after w hich all subroutines ar e list ed in any order . Pu tting t he related subrou tines (LI N K , R R R , R P R ) after t he E ND statem en t of t he m ain progra m and runni ng it on a com pute r , produces t he following ou tpu t values w hen φ : 1 = 65°
Q1 65° Q4 64. 2° W4 3. 165 E4 - 2 .475 XG 23 .9 YG 15 .5 V GX - 156 .7 VGY 75 .76 A GX - 117 .3 A GY - 555 .4

3.3.5   Check on the Output Data Faced wit h a vast a moun t of digits print ed on t he scr een , it is ha rd t o judge w het he r or not t he ou tpu t results ar e correct . T he out pu t values from t he analy tical m et hods can be checked as follows . F or som e non-special position of t he m echanism , t ry t o dra w t he k ine matic diagra m of t he lin kage m echanism by Au toCAD or t ry to dr aw it on paper as exactly as possible . Measure t he x and y coordinates of t he out pu t poin ts and/ or t he angular position φ of t he out pu t link . ・ 38 ・

T hen t he m easur ed da ta are compa red wit h t he ou tpu t position dat a ob tained by t he analy tical m et hod . Minor differences between t he m easured values and t he analyzed values ar e accep table and can be conside red as drawing errors and m easuring errors . If t he out pu t position at t he non-special position of t he m echanism is correct and t he out pu t position data change smoot hly duri ng t he w hole cycle , t hen t he position analysis can be considered to be correct . Af ter t he ou tpu t position data are checked , t he ou tpu t velocity data obt ained in t he analy t- ical m et hods can be checked by the instant cen te rs m et hod or by ot he r met hods . A simple r w ay to check ou t put velocit y is to exa mine t he qualitative natur e of t he data . For example, if t he displacem en t is increasing , the corresponding velocit y must be positive ; o t her wise, nega tive . When t he displace men t reaches its lim it , t he corr esponding velocit y must be ze ro . Sim ilarly , if the velocit y is inc reasing , t he corresponding acceler ation must be positive ; o t her wise , negative . When t he velocit y reaches its limit , t he corresponding acceler ation m ust be zero . Problems and Exercises 3-1   W hat is instan t cen tr e ? W hat is absolut e instant cen tr e ? What is r elative inst an t cen t re ? W hat is t he t heore m of t hree cent res ? 3-2   Locate all instant cen t res of all linkages in Fig . 3-11 for t he position show n .

Fig . 3-11

・ 39 ・

3-3   Locate all instan t cen tres of mechanisms for t he position show n in Fig . 4-5 , Fig . 4-7 , Fig . 4-9 , Fig . 4-11 , Fig . 5-34 , Fig . 5-36 , and Fig . 5-37 . 3-4   W hy do we always t ake advan tage of t he fra me w hen t he met hod of instan t cen t res is used t o find t he ratio between t he angular velocities of t wo moving links ? 3-5   For the position sho wn of a gea red linkage in Fig . 2-23 , det ermine t he ratio ω 3/ ω 1 of t he angular velocity of gea r 3 to t hat of gear 1 , using the m et hod of instan t cent res . 3-6   I n Fig . 3-12 , gear 4 rolls on fixed r ack 5 wit hou t slippi ng . Determine ratio ω4/ ω 1 for t he position show n , using t he m et hod of instant cen t res . 3-7   For a pivo t four-ba r lin kage si mila r to t he one in Fig . 4-23b , l A B = 65 mm , l A D = lB C = 125 mm , lD C = 90 mm , ∠ BAD = 165°and ω A B = - 10 rad/ s . Usin g t he met hod of instan t cen tr es , ( a ) find t he velocit y of poi nt C . (b ) locat e the poi nt E on t he line BC (or its ex tension ) w hich has t he minimu m velocit y of all poin ts on t he li ne BC and its ex tension , and t hen calculate its velocity . ( c ) locate t wo positions of t he c ran k A B corresponding t o v C = 0 . 3-8   Recoun t t he advan tages and t he disadvan tages of t he m et hod of instant cen tr es . Can t he m et hod of instan t cen t res be used in acceler ation analysis ? 3-9   Compa re t he advan tages and t he disadvan tages bet ween t he grap hical met hod an d the ana- lytical met hod for kinem atic analysis . 3-10   W ha t is t he kinem atic determi nation of any Assur group ? What a re t he advan tages of t he Assur group m et hod in t he kine ma tic analysis of linkage by compu ter ? 3-11   Can t he assembly mode of a grade Ⅱ Assur group change duri ng mo tion ? 3-12   Complete t he R R R subroutine in Sec . 3. 3. 2 and store it in a file na med R R R . 3-13   Complete t he R P R subrou ti ne in Sec . 3. 3. 3 and st ore it in a file na m ed R P R . 3-14   S how n in Fig . 3-13 is a simplified R R P dyad composed of lin ks 1 and 2 . T he guide w ay for the slider 2 is fixed and horizon tal . T he revo- lu te A is t he oute r revolu te . There a re tw o as- sem bly modes for t his dyad . For t he asse mbly mode show n in solid lines , let M = + 1 . For t hat in dashed lines , let M = - 1 . Suppose that t he coefficient M , t he y coordinate yB of t he
Fig . 3-13 Fig . 3-12

revolu te B , t he leng t h l A B of t he link A B , and t he kine matic par am eters ( x A ,

yA ,

vA x ,

v A y , a A x , a A y ) of t he ou ter poin t A ar e know n . Derive explicit expr essions to calcula te t he angular position θ A B , angula r velocit y ω A B , angula r accele ration ε A B of the lin k A B . Write a subrou tine for t his si mplified R R P dyad and store it in a file nam ed R R P . T he first line of t he ・ 40 ・

subrou tine is: S UB R R P ( M , YB , LAB , XA , YA , V AX , VAY , AAX , AAY , QAB , WAB , E AB) For exercises from 3-15 t o 3-24 , a m ain progra m is required for each m echanism to analyse t he ou tpu t motion w hen the driving crank rotates 360°wit h a step size of 5°. Combine the main progr am wit h the related subrou tines and t hen run it on a com pu ter . 3-15   S uppose t ha t in t he four-bar linkage A B3 C3 D show n in Fig . 4-34 , t he lin k DC3 is a drive r w hich runs a t a constan t speed of 10 rad/ s . T he dim en sions of t he linkage ar e : x A = 0 , yA = 0 , x D = 200 mm , y D = 0 , l A B = 300 mm , lB C = 365 mm , l D C = 360 mm . A nalyse t he ou tpu t motion of t he driven link A B 3 . 3-16   S ho w n in Fig . 4-42 is a six-ba r dw ell li nkage . T he dashed curve is produced by a cou- pler poi nt E on a c ran k-rocker m echanism A BCD . The E 1 E 2 E 3 portion of t he curve approxi- m ates to a circular arc . T he len gt h of lin k E F is equal to t he radius of t he a rc . T hus , t he ou t- put link G F will dw ell w hile t he couple r point E moves along t he circula r arc E 1 E 2 E 3 . T he m echanism has the following dim ensions : x A = 0 , y A = 0 , x D = 300 mm , yD = - 200 mm , x G = 60 mm , y G = - 20 mm , lA B = 100 mm , lB C = 400 mm , lD C = 300 m m , l B E = 200 mm , l E F = 150 mm , lG F = 120 m m . S ho w t hat , at t he position φ , t he angular velocit y A B = 74 .824° ωG F of t he ou tpu t link GF is alw ays zero for any angular velocit y of t he crank A B . 3-17   In t he oscillating guide-bar m echanism as show n in Fig . 4-9 , xB = 0 , yB = 100 mm , x A = 0 , y A = 0 , lB C = 50 mm , lA E = 200 mm . T he driving crank BC rotat es an ti-clockwise at a constant speed of 10 rad/ s . A nalyse the ou t put motions of link A E and poin t E . 3-18   In t he crank and oscillating block m echanism as show n in Fig . 3-14 , x A = 100 m m , yA = 0 , l A B = 30 m m , xC = 0 , yC = 0 , lB D = 50 mm , l D E = 40 mm , and ∠BD E = 90°. T he xA = 0 , yA = 0 ,

cr ank A B rotates at a constan t speed of 10 rad/ s . Analyse t he ou tpu t motion of poin t E . 3-19   T he m echanism show n in Fig . 3-15 has t he following di men sions: tates a t a constan t speed of 10 rad/ s . Analyse t he ou tpu t motion of point E . x D = 200 mm , yD = 0 , lA B = 80 m m , lC D = 60 mm , and lB E = 380 m m . T he cr ank A B ro-

Fig . 3-14

Fig . 3-15

3- 20   In t he offset slider-c ran k mechanism sho wn in Fig . 4-5 ,

xA = 0 ,

yA = 0 , l A B = 100 ・ 41 ・

mm , lB C = 300 m m , e = 70 m m . The c ran k A B ro tates at a constant speed of 10 rad/ s . Anal-

yse t he out pu t mo tion of poin t C . 3-21   In t he six-bar m echanism sho w n in Fig . 3-16 , x A = 0 , yA = 0 , x D = 450 mm , yD = 0 , l A B = 150 mm , lB C = 400 mm , lD C = 350 mm , ∠ CDE = 30° , l D E = 150 m m , l E F = 400 mm . T he crank A B rotates at a constan t speed of 10 rad/ s . Analyse t he ou tpu t motion of poin t F . 3-22   In t he six-bar mechan ism show n in Fig . 3-17 , x A = 0 , y A = 0 , l A B = 28 m m , lB C = 66 mm , ∠ DBC = 60° , lB D = 33 mm , x E = 94 mm , y E = 15 mm , l E F = 120 m m , T he c ran k A B ro tates at a const an t speed of 10 rad/ s . A nalyse t he ou tpu t motion of t he poin t F .

Fig . 3-16

Fig . 3-17

3- 23   The m echanism show n i n Fig . 4-8 has t he following dim ensions : x A = 0 , constan t speed of 10 rad/ s . A nalyse t he out pu t mo tion of poin t F .

yA = 0 ,

x B = 0 , y B = 50 mm , lBC = 100 m m , l A D = 50 mm , l D E = 150 mm . T he crank BC rotates at a 3-24   T he crank-shape r m echanism show n i n Fig . 4-10 has t he following dimensions: x A = 0 , yA = 100 m m , x C = 0 , yC = 0 , y F = y G = 186 .6 mm , lA B = 50 mm , l C D = 200 mm , l D E = 100 m m . T he driving c ran k A B rota tes clock wise at a constan t speed of - 10 rad/ s . A nalyse t he ou t put motion of t he poin t E .

・ 42 ・

C h ap t e r 4 P la n a r L i n k ag e Mec h a n i s m s

4.1   C h a r a c t e r i s t i c s o f P l a n a r L in k a g e M e c h a ni s ms
Linkage mechanism s are lowe r-pair mechan isms , i n w hich all k ine matic pairs are low er pairs , i .e. revolu tes or prism atic pairs . T he m ain practical advan tage of lower pairs ove r higher pairs is t heir bet ter ability t o trap lubricant betw een envelopi ng surfaces . T he load spreads on t he w hole surface of t he lower pair . T her efor e, the cont act pr essur e betw een tw o link s is low - er . As a r esult , t he linkage is preferred for low wear and heavy load sit uations . The kine matic pair elem en ts a re cylindrical or planar w hich are easy to manufact ure . The disadvan tage of link- age m echanisms is t hat it is difficult to pe rform precision mo tion as will be explained lat er . A plana r f our-bar m echan ism is t he simplest planar linkage m echanism wit h one degree of freedom . T he re a re four li nks and four lo wer pairs in a four-bar mechan ism , as show n in Fig . 4-1 and Fig . 4-2 . F our-ba r m echanisms a re ex t rem ely versatile and useful devices . M any quite complex motion con t rol proble ms can be solved wit h four-ba r m echanisms . For t he sake of sim- plicit y , designe rs should always first t ry to solve t heir proble ms wit h t his device . I n t his

Fig . 4-1

Fig . 4-2

・ 43 ・

chap ter , we pu t e mphasis on plana r four-bar m echanisms .

4.2   T h e T y p e s o f F o u r-b a r L in k a g e s
4.2.1   The Basic Form of Four-bar Mechanism If all lower pairs in a four-bar m echanism are revolu te pairs , as show n in Fig . 4-1 , t he m echanism is called a revol u te fou r-bar mechan ism , w hich is t he basic form of t he four-bar mechanism . In a r evolute four-ba r m echanism , t he links connect ed to t he fra me are called side li nks . Usually , one of t he side links is an inpu t link , an d t he ot he r side link is an ou tpu t lin k . T he floating link couples t he in- pu t t o t he out put . T he floating link is t her efore called t he cou p ler . If a side link can ro tate con ti nuously t hrough 360°relative to t he fram e , it is called a crank ; ot her wise , a rocker . According to t he types of t he tw o side links , t he types of t he r evolute four-bar m echanisms can be divided in t o : ( a ) Cr ank-rocker m echanism : In Fig . 4-1a , one side lin k A B can rotate con tin uously th rough 360°relative t o t he fram e w h ile t he ot her side lin k DC just
Fig . 4-3

rocks . T he refore , A B is a c ran k w hile DC is a rocker . T his mechanism is called a cra nk-rocker mechan ism . T he in pu t link m ay be the cr an k or t he rock-

er . In t he foo t-operat ed sewing m achine show n i n Fig . 4-3 , t he oscillation of the driving rock- er DC is t ransformed in to t he con tinuous rotation of t he driven crank A B . ( b ) Double-cr ank mechanism : In Fig . 4-1b , bot h t he side lin ks A D and BC can m ake complete revolu tions relative t o t he fra me A B . Thus , bot h A D and BC a re c ran ks . T his m echanism is called a double-crank mechan ism or a d rag l ink mechan ism . If one c ran k rota tes at a constan t speed , t he o t her c ran k will rotate in t he sam e direction a t a varying speed . (c ) Double-rocker mechanism : I n Fig . 4-1c , bot h t he side links D A and CB can only rock t hrough a limited angle relative t o t he fr am e . T herefore , bot h D A and CB are rock- ers . T his mechan ism is called a double-rocker mechan ism . T he cr ane sho w n in Fig . 4-4 is a fa mous use of t he double-rocker m echanism . In order to avoid raising or lowering t he load w hile movi ng it , t he cent re E of t he w heel on t he coupler should t race a horizon tal line . 4.2.2   Variation of Revolute Four-bar Mechanisms T he four-bar mechan ism has some sim ple and useful va riations . In t he follo wi ng we dis- cuss four such variation s . ・ 44 ・
Fig . 4-4

(1 ) Replacing a r evolute pair wit h a sliding pair If t he r evolute pair D in a cr ank-rocker m echanism ( Fig . 4-1a ) is replaced by a sliding pair , t he revolut e four-ba r mechanism t urns in to a sli der-crank m echan ism ( Fig . 4-2a ) . T he slider 3 t ranslates back and fort h w hile t he crank A B rota tes con tinuously . If t he ex tended pat h of t he cen t re of r evolute C goes t hrough t he cen tr e A of the crankshaft , t he mechanism is t hen called an in-li ne sl ider-cr ank mechan ism ( Fig . 4-2a ) , o t her wise , an eccen tric ( or o f f - set ) sli der-cr ank mechan ism ( Fig . 4-5) . The distance from t he c ran kshaf t A to t he pat h of t he cent re of t he revolu te C is called t he o f fset , denot ed as e . N ote : The tw o m echanisms in Fig . 4-5 are equivalen t to each ot her . If t he c ran k is a driver , t he mechanism can be used as a compressor . If t he slider is an input , t he m echanism can be fou nd in most in te rnal combustion engi nes . In Fig . 4-6 , t her e are tw o sliding pairs . T he ou tpu t displacem en t x of t he t ranslating guide-ba r 3 is t he si ne function of t he inpu t angle φ of t he crank A B , i. e. x = r sin φ . Thus , t h is crank and tr ansla ting guide-bar m echanism is often used as a si nusoi d gener ator .       In Fig . 4-7 , t he ou tpu t displace ment y of t he slider 3 is t he tangen t function of t he inpu t angle φ of t he oscillating guide-ba r A C , i. e. y = l tan φ . Thus , t his oscillating guide-bar m echanism can be used as a t angen t genera tor .

Fig . 4-5

Fig . 4-6

Fig . 4-7

   

g diffe ren t links as t he fr am e If two lin ks connect ed by a r evolu te can rotate 360°relative to each o t her , t he revolu te is

called a f u lly rota ti ng revol u te ; ot her wise , a p artia lly rota ti ng revol u te . T he revolu tes A and B in Fig . 4-1a an d Fig . 4-2a are fully rotating revolu tes , w hile t he revolu tes C and D in Fig . 4-1a and t he revolu te C in Fig . 4-2a are par tially rot atin g revolu tes . For t he sa me ki ne- m atic chain , different ki nds of li nkage m echanisms will be gene rated by holding differ en t lin ks fixed as t he fra me . Such kinds of va riations are called in versions . T hus t here ar e as many in- versions of a given linkage m echanism as t he re a re lin ks . Holding a lin k fixed as t he fr am e is simila r to fixing a viewer to t he link and obse rving t he ot he r links . Thus , it is of impor tance to note t ha t in version o f a mechan ism i n no w ay changes t he type o f revol u te an d the rel ati ve m o- ・ 45 ・

t ion between its li nks . This propert y will be applied oft en in t his book . Let’ s conside r inve rsions of t he cr ank-rocker m echanism A BCD in Fig . 4-1a by holding o t her lin ks fixed as t he fra me . If t he link A B is held as t he fra me as show n in Fig . 4-1b , t hen t he lin ks BC and A D become side links . Since revolu tes A an d B ar e fully rota ting revolu tes , bo t h side lin ks BC and A D can ro tate 360°r elative to t he fra me A B . Thus bot h side lin ks are cr anks , and t he mechanism becom es a double-crank m echanism . If t he link CD is held as t he fra me , t he m echanism becom es a double- rocke r m echanism , as show n in Fig . 4-1c . If t he link BC is held as t he fram e as sho w n i n Fig . 4-1d , t hen t he mechanism becomes anot her cr ank-rocker m echanism . Fig . 4-2 sho ws t he four inve rsions of t he slider-cr an k kinem atic chain . Fig . 4-2a is t he basic slider-crank m echanism . If t he link A B is held as t he fra me ( Fig . 4-2b ) , t hen bot h side links BC and A E can rotate 360°relative to t he fra me A B because bot h fixed pivots A and B a re fully rotating r evolutes . W hen t he cr ank BC rotates at a constan t speed , t he rotating guide-ba r A E will rotate i n t he sam e direction at a varying speed . T his mechanism is called a rot at in g gu ide-ba r mechan ism . The sm all plane r s ho w n in Fig . 4-8 is one of its applica tions . If t he crank BC is shor ter t han t he fra me B A ( Fig . 4-9) , t he guide-ba r A E can only oscillate . T he lin kage m echanism in Fig . 4-9 is called an oscill ati ng gu i de-bar m echa nism . T he quick- ret urn mechan ism in a shaper show n in Fig . 4-10 is one of its applications .

Fig . 4-8

Fig . 4-9

    Fig . 4-2c is obtained by holding t he lin k BC as t he fram e . Since t he revolu te C is a par- tially ro tating revolu te, t he slider 3 in Fig . 4-2a becomes an oscillating block 3 in Fig . 4-2c . T he m echanism in Fig . 4-2c is called a crank an d oscilla ti ng block mechan ism . The hydraulic cylinder show n in Fig . 4-11 is one of its applications . The hydraulic cylinder is used widely in practice . T he self-tippi ng vehicle show n in Fig . 4-12 is an exa mple . T he last inversion ( Fig . 4-2d) is obtained by holdi ng t he guide-block 3 as t he fr am e . I t is called a tra nsla ti ng sli d i ng-rod mechan ism and used in hand-opera ted w ell pump mecha- nisms ( Fig . 4-13 ) , in w hich t he handle is t he lin k A B ( ex tended ) . ・ 46 ・

Fig . 4-10

Fig . 4-11

Fig . 4-12

    W hen t he link 3 , w h ich con tains two sliding pairs , in Fig . 4-6 is fixed , t he linkage m echanism becomes an elli p tic tra m m el as show n in Fig . 4-14 . T his na me comes from t he fact t hat any poi nt on li nk A B tr aces out an ellipse . If t he link 1 in Fig . 4-14 is fixed as t he fra me, one obtains a m echanism know n as t he O ld ha m coup li ng or double rot at in g b lock m echan ism , as sho w n in Fig . 4-15 . T his mechanism is used to connect t wo rotating shafts wit h pa rallel bu t non-collinea r axes .

Fig . 4-13

Fig . 4-14

Fig . 4-15

・ 47 ・

   

ging a r evolute pair In a slider-cr ank m echanism ( Fig . 4-2a ) , t he lengt h l A B of

t he crank A B is determi ned accordi ng t o t he kine m atic r equir e- m en ts, w hile t he r adii of t he revolu tes are determined by the t ransmitted po wer . W hen t he power t ran smitted by t he li nkage m echanism is quite large , t he radii of revolu tes s hould be in- cr eased . The appea rance of t he cr ank A B can be affected if t he ra- dius of the revolu te B on t he cr ank is larger t han t he leng t h l A B of t he c ran k as show n in Fig . 4-16 . T his enlarged crank pin is called an eccen tric and can be used to replace the cr an k i n t he original mechanism . N ote : Enla rgi ng a revolut e pair in no w ay changes t he motion r elationship bet ween any links . (4 ) In terchanging guide-ba r and sliding block . T he re m ay be many differen t kinem atic diagr ams for a sliding pair . Any link in a sliding pair can be dra wn as a guide-ba r , and t he ot he r lin k as a sliding block . F urt hermor e , t he cen- tr e line of any sliding pair can be t ranslated wit hout changing any relative motion . T herefore , all t he kinem atic diagra ms in Fig . 4-17 a re equivalen t to each ot her kine matically .
Fig . 4-16

      a )             b )               c)             d )               e )

Fig . 4-17

4.3   C h a r a c t e r i s t i c s A n a l y s i s o f F o u r-b a r L in k a g e s
4.3.1   Grashof Criterion (Conditions for Having a Crank) In a r evolute four-bar m echanism , t he inpu t mo tion is usually ob tai ned t hrough a side lin k driven by an electric mo tor directly or indirectly t hrough belt mechanism or gears . T herefore a designer must ensure t hat one side link is a crank , w hich can be used as t he drivi ng link . T his can be checked by t he Grashof crit erion as follows . Suppose w e wish t o design a c ran k-rocker m echanism A BCD as show n in Fig .4-18 , in w hich the side lin k A B is an input c ran k , w hile t he side link DC is a follo wer rocker .T he ・ 48 ・

lengt hs of the four link s ar e a , b , c and d , as show n .According to t he theory of t he composition principle of m echanism discussed in Sec .2 .5 , t his revolu te four-bar m echanism is built up by as- sem bling a R R R Assur group ( links BC and DC ) on t he basic mechanism ( t he fra me and t he drive r A B ) .T he distance bet ween t he points B and D on t he basic m echanism is deno ted as f ( = lB D ) .If t he R R R Assur group can be asse mbled on to t he basic mechan ism by t he t wo ou ter revolu tes B and D , t he lengt hs of t he t hree sides in △ BCD must obey t he t riangle inequality , t hat is , t he leng t h of one side of a t rian gle must be less t han t he sum of t he lengt hs of t he ot her t wo sides .A pplyi ng t he triangle inequalit y to △ BCD in Fig .4-18 , we ob tain t he following rela- tionships: b+ c> f c+ f> b b+ f > c T hese can be r ewritten as b+ c> f b - c< f c- b< f T he distance f is a va riable value during the mo tion of t he m echanism . T he inequalities (4-1 ) should be satisfied for any value of f if t he R R R Assur group can always be assembled on to t he basic m echanism for any position of t he driving c ran k A B . O bviously , fm ax = a + d and f m i n = d - a w hen t he crank A B and t he fra me A D a re collinear as show n in Fig . 4-18 . T he refore , t he follo wing inequalities should be satisfied if t he inequalities ( 4-1 ) a re t o be satis- fied for any value of f . ( 4 -1 )

Fig . 4-18

b + c > f max = a + d b - c < f mi n = d - a c - b < f mi n = d - a ・ 49 ・

w hich can be rew rit ten as a+ d< b+ c a+ b< c+ d a+ c< b+ d A dding pairs of t hese i nequalities leads t o t he result a< c a< b a< d T hus from t he inequalities ( 4-3 ) , we can see t hat t he crank in a crank-rocke r mechan ism must be t he shortest link . Again , from t he inequalities ( 4-2) , we can conclude t hat the su m of t he shor test li nk a and any ot her link is less t han t he su m of t he r em ainin g t wo links . In o t her words , t he sum of t he s hort est and t he longest li nks must be less t han t he sum of t he re- m aining t wo lin ks . T his is called Grasho f criterion . Suppose t hat l m a x is t he lengt h of t he longest lin k , l mi n is t he leng t h of t he shor test link , and lb and lc a re t he lengt hs of in term ediate links . T he G rashof criterion can be expressed as: l m a x + l mi n < lb + l c . A linkage mechan ism w hich satisfies t he G rashof criterion is sometim es called a Grashof l inkage mechan ism . No w we consider inversions of t he cr ank-rocker m echanism by holding o t her lin ks fi xed as t he fr am e . T he lengt hs of all lin ks ar e not changed so t hat all inequalities m en tioned above are satisfied i n each case . As m en tioned in Sec.4 .2 .2 A B is t he fra me ( Fig . 4-1b ) , a double-c ran k mechan ism results . If t he link CD opposite to t he shor test lin k A B is t he fr am e ( Fig . 4-1c ) , a double-rocker mechanism results . If t he link A D or BC adja- cen t to t he shortest lin k A B is t he fra me ( Fig . 4-1a or d ) , then a cr ank-rocker m echanism re- sults . (4-3) (4-2)

Fig . 4-19

Fig . 4-20

    If l m a x + l m i n > lb + lc , t he linkage m echanism is a non-Gr ashof linkage m echanism , as show n in Fig . 4-19 , in w hich no link can rotate t hrough 360°relative to any o t her link and all inversions are double-rocker mechanism s . How ever , in a Gr ashof double-rocker mechan ism ( Fig . 4-1c ) , t he couple r can rota te 360°wit h respect to o t her links . A well-know n example of ・ 50 ・

t he Gr ashof double-rocker mechan ism is t he swi ng mechanism of a swing fan ( Fig . 4-20 ) . In Fig . 4-20 , t he worm is fixed to t he fan vane , w hile t he w orm gea r is fixed to t he couple r A B . W hen t he fan vanes rot ate , t he worm drives t he w orm gear so t hat it rotates continuously wit h t he couple r A B r elative t o t he link A D about t he cen tr e of revolu te A . T hus , t he link DC will rock relative t o t he link A D . In t his way , t he fan case fixed t o t he side link A D will oscillate relative t o t he fra me DC abou t D . If l m a x + l mi n = l b + lc , t he cen t re lines of t he four li nks can become colli near . At these po- sitions, t he out pu t behavior m ay become indeterm inat e . T hese positions ar e called change- poi n ts . Such linkage m echanisms are called chan ge-poi n t m echan isms . F or exa mple , t he driving lin k A B and t he driven link DC in Fig . 4-21 a re of equal lengt h , and t he coupler BC is equal in lengt h to the fr am e A D . At t he change-poin t A B1 C1 D , t he rotation direction of t he driven lin k DC becomes indeter minat e . W hen t he driving crank A B rotates from A B 1 t o A B 2 , t he driven link DC m ay continue rot atin g in t he original direction t hrough t he change-poin t from DC1 to DC2 , or rota te in t he reve rsed direction from DC1 to DC2′. The configura tion A B 2 C2 D is called a para llel-cr ank mechan ism w hile t he configuration A B 2 C2′D is called an a nt ipa ral lel -cr ank mechan ism . The change-poin ts can be handled by ine rtia of fly w heel fixed wit h t he driven crank or providin g t he duplicate lin kage 90°out of phase, as show n in Fig . 4-22 . As a consequence, each linkage carries t he ot her t hrough its change-poin ts so t hat t he ou tpu t rem ains determinate at all positions . T able 4-1 sum ma rizes t he type criteria for t he revo- lu te four-ba r mechanis ms .

Fig . 4-21

Fig . 4-22

Table 4-1   Type criteria for the revolute four-bar mechanisms
Link as t he frame l m ax + l m in < lb + lc Grashof T h e shor test link Opposite t o shortest link Adjace nt t o shortest link Double-cra nk Double-rock er Crank-rocke r Double-rocker l m ax + lm i n > l b + lc Non-Gr ashof lm ax + l m in = lb + lc Ch ange-poi nt

From t he above , w e know t hat t he Gr ashof criterion l m a x + l mi n < l b + lc is only a necessary ・ 51 ・

condition , no t sufficient condition for having a crank . T o det ermine the t ype of a r evolute four-bar m echanism , we must check not only w het her t he necessary condition is satisfied bu t also w h ich li nk is the fr am e . In an offset slider-cr an k mechanism sho wn in Fig . 4-5 and Fig . 4-26 , t he sum of t he lengt h a of t he crank A B and t he offset e must be less than t he leng th b of t he couple r BC , if t he c ran k A B is to ro tate 360°relative t o t he fra me . 4.3.2   Pressure Angle α and Transmission Angle γ In a crank-rocke r mechanis m A BCD as sho w n in Fig . 4-23 , t he crank A B is an input lin k and the rocker DC is an ou t pu t li nk . T he force F applied to t he out pu t link DC is tr ansmitt ed t hrough t he coupler lin k BC . If t he friction for ce and gr avit y force are neglected , t hen t he di- rection of t he force F in t he coupler is along t he line of its pi n join ts . The poin t receiving t he force on t he follower lin k DC is t he poin t C . T he acut e angle bet ween t he directions of t he force F and t he velocity of t he poin t receiving t he force on the follower is defi ned as t he p res- su re an gle α of t he mechanism at t hat position . T he com plem en t of t he pressure angle α is called transm ission an gle γ . If ∠ BCD < 90° , t hen γ= ∠ BCD , as sho w n in Fig . 4-23a . If ∠ BCD > 90° , t hen γ= 180°- ∠ BCD , as sho wn in Fig . 4-23b .

Fig . 4-23

T he force F applied t o t he followe r has t wo componen ts: the radial componen t Fr and t he t angen tial componen t F t , as show n in Fig . 4-23a . Only t he tangen tial componen t F t can cre- at e t he ou t put torque on t he driven link DC . The radial component Fr increases pivot friction and does not con t ribute to t he ou tpu t torque . F r = F sin α= F cos γ and F t = F cos α= F sin γ . For t his reason , it is desir able t hat α is not too grea t or γ is not t oo sm all . α and γ change duri ng mo tion . For smoo t h ope ration of any m echanism without jerky move men ts , t he m axi- mum value of α s hould be less t han t he allowable pressur e angle [α = 40° value of γ should be large r t han t he allowable tr ansmission angle [ γ = 50°. T hus w e s hould find t he ex t rem e values of α and γ . ∠ BCD reaches its ex t re me w hen t he driving cr ank and t he fra m e lin k are collinear , as ・ 52 ・

show n i n Fig . 4-24a and b . γ m i n ( and α m a x ) will occur in eit her of t he tw o positions . The ex- tr em e values of γ are illustra ted in Fig . 4-24a and b by γ m ′ i n and γ m ″ i n , r espectively , I t is com- mon practice t o calculate bot h values and t hen pick t he w orst case γ ″ mi n } . T he α an d γ m ust be dra wn on t he driven link . For t he sa me kinem atic chain , t he posi- tions and t he values of α and γ will change , if a differ en t li nk is chosen as t he driver . For ex- a mple, if t he rocker DC is a driver ( Fig . 4-25) , t he α and γ of t he mechanism at t hat position a re diffe ren t from t hose in Fig . 4-23a . In a slide r-crank m echanism , if t he cr ank is an inpu t li nk and t he slider is an ou tpu t , as show n in Fig . 4-26 , t hen t he acu te angle be- tween t he coupler BC and t he slider pat h is t he pr essur e angle α at t hat position . γ = 90°- α . T he ex t re me values of α and γ, α m a x and γ m i n , occur w hen t he c ran k A B is perpendicula r to t he slide r pat h , i .e. , α m a x = 90° - γ mi n = a rcsin a+ e b .
Fig . 4-24

γ ′ m i n = min { γ mi n

Fig . 4-25

Fig . 4-26

4.3.3   Toggle Positions and Dead-points In a crank-rocke r mechanism ( Fig . 4-27 ) , t he rocker DC r eaches its tw o li miting posi- tions DC1 and DC2 , w hen t he cr an k A B and t he coupler BC become overlappi ng collinear ( A B 1 C1 D ) and ex tended collinear ( A B2 C2 D ) . Note : compa ri ng Fig . 4-27 wit h Fig . 4-24 , we can see t hat t he limitin g position of t he rocker DC is different from that w here γ m i n occurs . Fig . 4-28 shows t he tw o lim iting positions C1 and C2 of t he slider in an offset slider-c ran k m echanism . In Fig . 4-27 , if t he link A B is a drive r , t hen near t he limiting positions of t he rocker DC , a small t orque applied t o t he link A B can gener ate a h uge t orque on t he follow er rocker DC . In t his sen se , t he li miting positions a re called toggle positions . If t he rocker DC is a driver , t hen at its limiting positions , t he for ce applied t o t he follower A B passes t hrough t he fixed pivot A of t he follo wer . In t hese sit uations , t he m echanism cannot move by applying ・ 53 ・

a t orque on the driving rocker DC . In t his sense , t he lim iting positions are called dead poi n ts .

Fig . 4-27

Fig . 4-28

    By definition , a c ran k can ro tate con tinuously relative to t he fra me . T her efore , in any four-bar mechanism ( except t he change-point mechanism s ) , t he dead poin t will not occur if t he cr an k is a driver . Change-poin ts ( position A B 1 C1 D in Fig . 4-21 ) of a change-poin t m echanism a re also dead poi nts . Since a rocke r or a slider can only move back and for t h be- tween its t wo limiting positions , t he dead poin ts will occur if t he rocker or t he slider is a driv- er . The dead points occur w hen t he driver reaches eit her of its tw o limiting positions . T here- fore , if a rocker or a slide r is t he driver , a fly w heel on t he driven cr an k will be required to car- ry t he mechan ism th roug h t he dead poin t , as show n in Fig . 4-3 . In Fig . 4-22 , the dead poin ts are overcome by providing t he duplicate li nkage 90°ou t of phase . In some circum stances , t he dead poin t is ve ry use- ful . It can provide a self-lockin g feat ure . An exa mple of t he applica tion of a dead poin t is t he cla mping device on mach ine tools , as show n in Fig . 4- 29 . Near t he toggle position , a sm all force F P applied on t he coupler
Fig . 4-29

BC will produce a very la rge torque on t he rocker DC and consequently t he w orkpiece is clam ped firmly .

T hen t he lin kage is moved slightly beyond t he toggle position and against a fixed stop . T he m echanism is at t he dead poin t under t he force from t he clam ped w ork- piece . Any at tem pt of re moving w orkpiece t hen causes t he clamp to ja m ha rder against t he stop . Sho wn in Fig . 4-30 is a landing mechan ism i n airplane . W hen t he w heel is at its lo west position , links BC and CD are collinear . If t he st rut A B suppor ting t he w heel would rotate coun ter-clock wise under t he weigh t of t he airplane, the force t ransmitted to t he link CD t hrough t he lin k BC passes t hrough t he pivot D . T he refore t he m echanism is at a dead poin t . A sm all t orque on t he link CD is enough to preven t t he li nk DC from ro- tating . T his landing mechanism can give t he airplane h igh landing reliabil- ity . ・ 54 ・
Fig . 4-30

4.3.4   Quick Return Characteristics W hen designing m achine tools such as s hapers or po wer-driven saws , it is desirable t o give t he working stroke of t he follo wer a slow er speed because of t he higher resistance , and t o give t he r eturn stroke a faster speed in orde r t o r educe idletim e . Such a mechan ism is called a qu ick- ret ur n mechan ism . In t he c ran k-rocker mechanism show n in Fig . 4-27 , ∠ C1 DC2 is called t he an gu la r stroke of t he rocker , denoted as φ m a x . ∠ C 1 A C 2 is called t he crank acu te angle between the t wo li m - it in g positions , denoted as θ . If t he input cr ank A B rot ates coun te r-clock wise , t he rocker will rock from its righ t limitin g position DC2 to its left limiting position DC1 , w hen t he c ran k t urns t hrough t he angle ( 180° + θ) from A B2 to A B 1 . T hen t he rocker will rock back from DC1 t o DC2 w hen t he crank t urns th rough t he angle ( 180°- θ) from A B 1 to A B 2 . If t he inpu t cr ank A B rotates coun ter-clock wise at const an t speed , t he average angular velocities of t he rocker DC in t he t wo strokes of t he follower are diffe ren t . T he ratio of t he faster aver age angu- la r velocit y ω f t o t he slow er one ω s is called t he coef f icien t o f tr avel speed varia tion , denot ed as K . So , K= ωf φm a x/ t f t s 180° +θ = = = ωs φ m a x/ t s tf 180°- θ

w her e t f and t s are t he time duration s for t he faste r stroke and t he slowe r stroke , r espectively . From t he above , we can see t hat K is also t he tim e ratio of t he slowe r stroke to t he faster st roke . If t he li nkage m echanism in Fig . 4-27 is used as a quick-ret urn mechanism , t he coun ter- clock wise st roke of the follo wer rocker should be t he wor king st roke, and t he clock wise st roke should be t he ret urn stroke . If t he clock wise stroke of t he follower rocker is needed to be a wor king st roke , t hen t he rotation direction of t he cr ank s hould be r eversed . A cr ank-rocker m echanism wit h special dim ensions may no t have quick ret urn char acte ris- tics . F or example , in Fig . 4-31 , t he driver A B ro tates constan tly . At t he t wo limiti ng posi- tions of t he rocker , points B 1 , A , B 2 , C1 , and C2 are locat ed on t he sa m e st raigh t line . T he cr ank acu te angle θ between t wo limitin g positions is zero . T herefore t he coefficien t K of t ravel speed variation is equal t o one . T her efor e t he cr ank-rocker m echanism in Fig . 4-31 has no quick-r eturn characteristics . In t he offset slider-c ran k m echanism show n in Fig . 4-28 , t he distance C1 C2 is t he st roke H of t he slider . ∠ C2 AC1 is t he cr an k acu te angle θ bet ween t wo limiting positions . If t he driving c ran k A B rotates coun ter-clockwise wit h constan t angula r velocit y , t he slider will take a longer time in its righ twa rd st roke t han in its leftw ard stroke . T he coefficient K of t he t ravel speed varia tion , or the time ratio , is θ)/ ( 180°- θ) . Since θ = 0° , an in-line slider- cr ank mechanism ( Fig . 4-2a ) has no quick-r eturn characteristics .     In an oscillating guide-ba r m echanism ( Fig . 4-32 ) , two limiti ng positions CD 1 and CD 2 ・ 55 ・

of t he follower guide-bar CD occur w hen t he driving crank A B is perpendicula r t o t he oscillat- ing guide-bar CD . ∠ D1 CD2 is t he angular st roke φm a x of t he follo wer . T he acu te angle be- t ween A B 1 an d A B 2 is θ . For t his lin kage m echanism , θ happens to be equal to φm a x . If t he cr ank A B runs coun ter-clock wise at a const an t angular velocit y , t hen t he left ward st roke of t he followe r CD is t he slower st roke w hile t he righ twa rd st roke is t he faste r st roke . The coefficien t K of t he travel speed variation , or t he tim e ratio , is θ)/ ( 180°- θ) .

Fig . 4-31

Fig . 4-32

4.4   D i me n s i on a l S y n t h e s i s o f F o u r-b a r L i n k a g e s
Dimensional synt hesis of a li nkage mechan ism is t he determination of t he k ine matic dim en- sions of t he lin ks necessary to achieve t he required motion . Usually , differ en t proble ms will need different m et hods . 4.4.1   Body Guidance     pivots have not been det ermined A revolu te four-ba r mechanism A BCD is to be de- signed t o guide a li ne segmen t E F on t he coupler passing t hrough t hree specified positions E 1 F 1 , E 2 F 2 and E 3 F 3 , as sho wn in Fig . 4-33 . Such a syn t hesis proble m is called body g ui da nce . The know n positions of t he li nk E F are draw n in bold lines and t he ot he r unkno wn lines are dra w n in fine or dashed lines . T w o moving revolu te cen t res B and C are chosen ar- bitra rily on t he couple r . F or example, poin ts B 1 and C1 a re chosen a rbit ra rily on t he first position of t he coupler . T he shape of t he quadrilater al BCF E should rem ai n t he ・ 56 ・
Fig . 4-33

sa me i n all position s . T he refor e , constructing quadrilate rals B 2 C2 F2 E 2 ≌ B 1 C1 F1 E 1 and B 3 C3 F3 E 3 ≌ B 1 C1 F 1 E 1 , we get poi nts B1 , B 2 , B 3 and C1 , C2 , C3 . Since t he locus of t he poin t B r elative to t he fr am e is a circle t he cen t re of w hich is t he fixed pivo t A , a cir cle is con- st ructed passing t hrough t he t hree poin ts B 1 , B 2 and B 3 . Bisect B 1 B2 and B 2 B 3 . T he in ter- section of t he tw o bisect ors is t he fixed pivo t A . Similarly , bisect C1 C2 and C2 C3 . T he in ter- section of t he tw o bisectors is t he fixed pivo t D . Som e li nes a re no t draw n for t he purpose of clarity . G raphical m et hods provide valuable insigh t in to t he syn t hesis process as t he solution is mor e easily unde rstood . The accuracy of t he graphical m et hods by hands is insufficien t . How - eve r , t he accuracy of t he graphical met hods is good enough if Au t o CAD is used . If the moving revolu te cent res B and C a re chosen arbitr arily on t he coupler , it m ay happen t hat t here is no crank i n t he synt hesized lin kage , or t hat t he α m a x and γ m i n m ay not be accep table , or t hat it is not possible to move t he coupler to all spec- ified positions wit hout disassembling a dyad and re- assemblin g it i n anot her asse mbly mode . That will obviously be unsatisfact ory . For exa mple, a r evo- lu te four-ba r linkage A BCD is t o be designed to guide a line segm en t E F on the coupler BC t hrough t hr ee positions E 1 F 1 , E 2 F2 and E 3 F 3 , as show n in Fig . 4-34 . If t he poin ts E and F ar e selected as t he moving revolu te cen t res B and C re- spectively , t he fixed pivots will be A and D . It can be seen t hat t he assembly modes of t he R R R dyad in positions 2 and 3 are differ en t no m atte r w hich side link is t he driver . T he syn- t hesized m echanism can no t t he refore move t he couple r t hrough all t hree specified positions in a continuous motion cycle . U nfor tunately , t he synt hesis techn iques have no cont rol over t he mode of a dyad . F or t his reason , t he mechan ism must be checked afte r syn t hesis t o see w het he r t he mode of t he dyad r em ains t he sa me in a con tinuous motion cycle . If not , one m ust t ry selecting differen t location s of t he movi ng revolu te cen t res and t he mechanism should be re- designed . T his check is called consistency o f the assem bly m ode o f a dyad . F ur t hermore , t he syn t hesized mechanism should be checked for α m a x an d γ mi n w hen required . (2 ) T w o fixed pivots have been dete rmined A revolut e four-bar lin kage A BCD is t o be designed t o guide a line segm en t E F on t he cou- pler BC to pass t hrough t hree specified positions E 1 F 1 , E 2 F2 and E 3 F3 , w hile t he positions of t he tw o fixed pivots A and D have been determ ined . Fig . 4-35 illust rates t he proble m . Ob- viously , in t his case, one can not choose arbitr arily t he locations of t he moving revolu te cen tr es B and C on t he coupler . O t her wise , t he cen t re of t he circle t hrough t he t hree positions of a moving revolu te cen t re will not happen to be t he corresponding fixed pivot . Such a moving ・ 57 ・
Fig . 4-34

revolu te cen t re , t he location of w hich is to be dete rmined , is called an un deter m i ned m ov i ng revol u te cen tre ( U M RC ) . T he U M RC B ( Fig . 4-35 ) connects t wo links A B and BCF E . T he coupler BCF E is said t o be de- termined because t he positions of a li ne segm en t E F on t he coupler a re know n . T he link A B is undet er- mined alt houg h t he poin t A is determined . T he de- termined link ( BCFE ) connected by U MR C ( B ) is selected as a reference li nk . A determined poi nt ( A ) on t he undetermined lin k ( A B ) connected by U M R C ( B ) is selected as a circu m ference poi n t . T he locus of t he cir cumference point ( A ) relative t o t he r efe rence link ( E F ) is a circle , t he cen t re of w hich is t he U M RC ( B ) and t he radius of w hich is t he leng th of t he undetermined lin k ( A B ) . In order to find t he locus of the circumference poin t A r elative to t he reference lin k E F , t he link E F should be stationary . Since t he link E F is a moving link , t he conver ting fra me m et hod should be used so t hat t he link E F is inver ted to be stationa ry , or an im aginary fra me . Triangles E 1 F 1 A , E 2 F 2 A and E 3 F3 A in Fig . 4-35 defi ne t he t hree relationships of t he poin t A relative t o t he lin k E F at t he t hr ee in stants . Sup- pose that t he link E F is fixed in position 1 and t hat E 2 F2 and E 3 F 3 are inver ted t o t he first po- sition E 1 F 1 . Const ructi ng △ E 1 F 1 A 2′ ≌ △ E 2 F2 A and △ E 1 F 1 A 3′ ≌ △ E 3 F3 A , we get t wo poin ts A 2′ and A 3′. Poi nts A , A 2′an d A 3′a re t he t hree position s of t he point A relative to t he lin k E F w hen the link E F is sta tiona ry at E 1 F 1 and t he li nks have sam e relative motions wit h respect to each ot he r . T he circle passin g t hrough t he poin ts A , A 2′and A 3′ is t he locus of t he poin t A relative t o t he first position E 1 F 1 of t he link E F . T he cen t re of t he circle is t he first position B 1 of t he U MR C B . A not he r U M R C C can be found by a simila r m et hod . Const ructing △ E 1 F 1 D2′ ≌ △ E 2 F 2 D and △ E 1 F 1 D3′ ≌△ E 3 F 3 D , we will get poin ts D2′ and D 3′. The cen tr e of t he cir- cle passing t hrough t he t hr ee poin ts D , D 2′and D 3′is t he first position C1 of t he U M RC C . 4.4.2   Function generation A revolu te four-bar linkage is t o be designed t he ou t put side link of w hich swings t hrough t wo specified angles φ 1 2 and φ 1 3 corresponding to t he t wo swing angles θ 1 2 and θ 1 3 of t he inpu t side link . Fig . 4-36 illust rates t he problem . The t wo fixed pivo ts A and D a re know n and t he moving revolu te cent res B and C a re t o be found . Such a synt hesis problem t hat involves coor- dina ting t he rota tional and/ or tr ansla tional orien tations of t he inpu t and ou t put is called f unc- t ion gener at ion . ・ 58 ・
Fig . 4-35

Fig . 4-36

Fig . 4-37

In t h is design proble m , t her e are tw o U MR Cs ( B and C ) . Since only t hr ee positions are required in t his design problem , B or C can be chosen arbitr arily . Suppose t hat t he first posi- tion B 1 , of t he U M RC B on t he inpu t side link is chosen arbitra rily . Constructing △ E 2 A B 2 ≌ △ E 1 A B 1 and △ E 3 A B 3 ≌ △ E 1 A B 1 , w e get poin ts B 2 an d B 3 . T he U MR C C con nects t he links BC and CDF . T he position of t he link BC is u nknow n w hile t he positions of t he li nk DF a re determ ined . So t he li nk DF is select ed as t he r efe rence lin k . The determ ined poin t B on t he un determi ned link BC is selected as t he circumfer ence poin t . Triangles DF 1 B 1 , DF 2 B 2 and DF 3 B3 ( Fig . 4-37 ) represen t t he t hree position relationships betw een t he cir cumference poin t B and t he reference lin k D F at t hr ee instan ts . U si ng inversion , let t he reference lin k DF be fixed at position 1 . D F2 and DF 3 ar e inver ted t o D F1 , w hile t he position relationships a re kep t unchanged at t he sa m e time , i .e.construct △ DF 1 B2′ ≌ △ DF 2 B2 and △ D F 1 B 3′ ≌ △ DF 3 B 3 . H ence we get tw o new poin ts B 2′and B 3′. T he poin ts B 1 , B 2′and B 3′a re t he t hr ee position s of t he poin t B relative to t he lin k DF 1 , if t he r eference li nk DF is stationary at t he first position and t he ot her links have r elative motion wit h respect to each ot her . A circle ・ 59 ・

passing t hrough t he t hr ee poin ts B 1 , B 2′ and B 3′ is t he locus of t he poin t B relative to t he first position DF 1 of t he link DF . T herefore , t he i nte rsection of t he bisectors of B 1 B 2′and B 2′ B 3′ is t he cen t re of t he circle , or t he first position C1 of t he U M RC C ( neit her C2 nor C3 ) . Constructing △ DC2 F 2 ≌ △ DC1 F 1 an d △ DC3 F 3 ≌ △ DC1 F 1 , we get poin ts C2 and C3 . Af ter syn t hesis , t he lengt h of t he undet ermined link should be checked to see if its leng t h is kept unchanged during motion , i .e ., B 1 C1 = B 2 C2 = B 3 C3 . Check also G rashof′ s criterion , α m a x or γ m i n , t he space limitation and t he consistence of t he assembly mode of a dyad . If not satisfied , t he location of t he moving revolut e cen tr e B should be selected again and t he mecha- nism should be redesigned . 4.4.3   Design of Quick Return Mechanisms (1 ) Cran k-rocker mechanism As m en tioned in Sec . 4. 3. 4 , t he c ran k-rocker mechanism show n in Fig . 4-27 can be used as a quick ret urn mechanism . W hen t he driving cr an k A B runs at a constan t speed , t he coefficien t of t ravel speed varia tion K = ( 180° + θ)/ ( 180°- θ) θ= K -1 180° K+1 So (4-4)

In t he limiting positions , AC2 = a + b and A C1 = b - a . T he leng t hs , a and b , of t he cr ank A B and t he couple r BC can be calculated if t he distances A C1 an d A C2 ar e know n , i .e. a= A C2 - A C1 2

AC2 + A C1 b= 2 4-38) , ∠ C1 A i C2 is constant . If ∠ PC1 C2 = 90° , t hen PC2 is t he diam eter of t he circle .

(4-5)

By a well-k no w n geom et rical t heorem , for any poin t A i on t he ar c C1 P of a circle ( Fig . Suppose that the leng t h c of t he followe r rocke r , t he angular st roke φm a x , and t he coeffi- cien t K of the tr avel speed variation have been specified . A crank-rocker mechan ism is t o be designed . T he follo wing design process of the cr an k-rocker mechan ism is illustr ated in Fig . 4-39 .

Fig . 4-38

Fig . 4-39

・ 60 ・

   

pivot D and dr aw t he two limiting positions , DC1 and DC2 , of t he (b ) Calculate θ wit h Eq . ( 4-4 ) according to t he specified value of K . (c ) Through C1 const ruct a line perpen dicula r t o C1 C2 . T hrough C2 construct a line so

followe r rocker wit h t he know n values of c and φ m ax .

t hat ∠ PC2 C1 = 90°- θ . The in tersection of t he two lines is labeled P . ( d ) D raw a cir cle wit h t he midpoin t of C2 P as t he cen t re and t he leng th of t he line C2 P as t he diam eter . ( e ) T he cir cle is t he locus of t he fixed pivot A . A ny poin t A on t he arc C1 P will satisfy ∠ C1 AC2 = θ . If t he lengt h d of t he fra m e is know n , t hen an arc is draw n wit h t he poi nt D as a cen tr e and d as a r adius . T he in tersection bet ween t he circle and t he arc is t he fixed pivo t A . If d is unknow n , t hen choose a suitable poin t on t he a rc C1 P as t he fixed pivo t A . Measure t he distances A C1 and AC2 . T he act ual leng t hs , a and b , of t he crank A B and t he coupler BC can be calculated from Eq . ( 4-5 ) . Since any poin t on t he a rc C1 P can be used as t he fixed pivo t A if t he leng t h of the fra me is unknow n , t he re is an infin it y of solu tions . Since ∠ AC2 D > γ mi n , t he position of t he poin t A can not be set too low . Check α m a x or γ mi n afte r syn t hesis . If K , φm a x and t wo of t he di mensions a , b and c a re kno wn , t he m echanism can be de- signed analytically as follows . In △ C1 A C2 ( Fig .4-27 and Fig. 4-39) , using t he cosine rule , one ob tai ns ( C1 C2 ) = ( A C1 ) + ( A C2 ) - 2 × A C1 × A C2 × cos θ T hus φm a x 2 c sin 2 from w hich , one ob tains a (1 + cos θ) + b (1 - cos θ) - 2 c sin
2 2 2 2 2 2 2 2

= ( b - a ) 2 + ( b + a ) 2 - 2( b - a ) ( b + a ) cos θ

φ m ax 2

=0

(4-6)

Af ter dete rmining any t wo of t he dim ensions a , b and c , t he t hird dim ension can be cal- culated easily according to t he values of θ and φm a x using Eq . ( 4-6 ) . Af ter t hat , d can be calcula ted as follows . In △ C1 AC2 , accordin g to t he sine rule , C1 C2 b- a = = sin ( ∠ AC2 C1 ) si n θ T he refore ∠ AC2 C1 = arc sin ( b - a ) sin θ φ m ax 2 csin 2 (4-7) 2 csin φ m ax 2 sin θ

∠ A C2 D = 90°In △ A C2 D , according t o t he cosine rule ,

φm a x - ∠ AC2 C1 2

(4-8)

・ 61 ・

d 2 = c2 + ( a + b ) 2 - 2 c ( a + b ) cos ( ∠ AC2 D ) (2 ) O ffset slide r-crank m echanism

(4-9)

A n offset slide r-crank mechanism ( Fig . 4-28) can be a quick-return mechan ism . Suppose t hat st roke H , tim e ra tio K , and offset e are know n , t he gr aphical design met hod for t his m echanism show n in Fig . 4-40 is similar to t ha t in t he last exa mple ( Fig . 4-40 is self-ex plana- tory ) . T his m echanism , however , can be easily designed analytically wit h som e equations de- rived as follo ws . In △ C1 AC2 ( Fig. 4-40 ) , according t o t he cosine rule , H 2 = ( b - a ) 2 + ( b + a ) 2 - 2 ( b - a ) ( b + a ) cos θ = 2 b ( 1 - cos θ) + 2 a (1 + cos θ) b- a H = . sin ( ∠ AC2 C1 ) sin θ In right t riangle A M C2 , si n ( ∠ AC2 C1 ) = e .Therefore , b+ a
Fig . 4-40
2 2

( 4-10)

In △ C1 AC2 , according t o t he sine rule ,

( b 2 - a 2 ) sin θ e = ( b + a ) sin ( ∠ A C2 C1 ) = H

( 4-11)

If K (from w hich θ can be determined ) and any t wo of t he four para m eters ( H , e, b and a ) ar e know n , t hen t he ot her two unk no wn s can be calculated by solving Eqs . ( 4-10 ) and (4-11 ) simultaneously . (3 ) Oscillating guide-ba r mechanism A n oscillating guide-bar m echanism ( Fig . 4-32 ) is a quick-ret urn m echanism . As m en- tioned in Sec . 4 .3 .4 , t he oscillatin g guide-ba r CD r eaches its ext re me position s w hen t he driv- ing crank A B is perpen dicula r to t he guide-bar CD . Suppose t hat t he tim e ratio K and t he lengt h of t he fr am e l A C are kno wn . The crank acu te angle θ bet ween tw o limiting positions happen s t o be equal t o t he angular stroke φm a x of t he guide-bar CD , i .e. φ m a x = θ= K - 1 180° K+1

In righ t t riangle A BC , l A B = lA C sin ( φm a x/ 2 ) . 4.4.4   Path Generation I t is of ten desired to syn thesize a li nkage mechan ism so t hat a poi nt on t he coupler will move along a specified pat h . This synt hesis problem is called pat h genera tion . Usually , t his t ask can be ca rried out by a four-ba r linkage . T he pa th gener ated by t he point on t he couple r is called a coup ler cu r ve and t he gener ati ng poin t is called t he cou p ler poi n t . Differen t poin ts on t he coupler , or on t he ex tension s of t he coupler , will t race in nume rable couple r curves as t he positions of the coupler poi nt on t he coupler and t he di men sions of t he li nks a re altered . ・ 62 ・

A very useful reference for t he designer in selecti ng a proper crank-rocke r m echanism to ob tain a required coupler curve is an atlas of four-ba r coupler curves . The book consists of a set of char ts con tain ing approxim ately 7300 couple r curves of cr ank-rocker mechanism s . A n exam- ple taken from t his a tlas is show n in Fig . 4-41 . Every couple r curve is produced by a coupler poin t on the coupler . T he sm all circle on t he coupler curve shows t he relative position of t he coupler poin t on t he couple r . Each dash on t he coupler curves repr esents 5°of t he inpu t c ran k ro tation to provide an indication of t he coupler poin t velocit y . T he longe r t he dash , t he faster is t he velocity . By scann ing t hrough t he atlas , t he designe r m ay be able t o find sever al coupler curves on differen t lin kage m echanisms t hat will approx imat ely accomplish t he sam e task . Af- ter considering t he practical sit uation , e.g . t ransmission angle , space limitation , etc ., an op ti mal linkage mechanism can be chosen from t hese li nkage m echanisms . Then w hat re mains is only to adjust t he scale for t he size of curve desired t o complete t he solu tion .

Fig . 4-41

Some coupler curves have sections t hat a re nearly st raigh t-line segmen ts; ot hers have cir- cular arc sections . O ne of t he most in te restin g uses of couple r curves having st raight -line or cir- cle-a rc segment is in t he syn t hesis of a si x-bar m echanism having a substan tial d well at t he ex- tr em e poin t of its out pu t mo tion . T he substan tial por tion E 1 E 2 E 3 of t he coupler curve show n in Fig . 4-42 approxim ates a circular arc . It is produced by a coupler poin t E on a c ran k-rocker m echanism A BCD . A con- necting link E F wit h a lengt h equal t o t he radius of t his ar c E 1 E 2 E 3 is now added . At t he posi- tion show n , poin t F coincides wit h t he cen tre of t he a rc w hile t he coupler poin t E moves t hrough poin ts E 1 , E 2 and E 3 . In Fig . 4-43 , t he fixed pivot G is located at t he in tersection of ex tension of the t wo st raight -line segm en ts of t he coupler curve . T herefore , t he ou t put link G F will have dw ells at bo t h ex t re mes . T he arrange men t show n in Fig . 4-44 uses a figur e-8 couple r curve having a st raigh t-line segm en t t o produce an in term ediat e d well lin kage mechanism . T he fixed pivo t G must be lo- cated on t he ex tension of the str aigh t-line segm en t . ・ 63 ・ pu t link G F will d well ,

Fig . 4-42

Fig . 4-43

Fig . 4-44

4.4.5   Limitations of Linkage Mechanisms Sho wn in Fig . 4-45 is a r evolu te four-bar lin kage a rranged to co-ordinate t he ro tations of t he inpu t link A B and t he ou t put lin k DC . S uppose t hat t he ou tpu t link DC is r equired t o rotate t hrough an angle φ 0 i from its i nitial position DC 0 w hen t he i n- put lin k A B rotates θ 0 i from its initial position A B0 . In ot he r words , t he revolu te four-ba r linkage is t o be syn t hesized to gener ate a given function φ 0i = f (θ 0i) . Since t he angular r elations hip does not change if all lin k leng t hs ar e multiplied by a common factor , t here ar e only t hree indepen den t len gt h dim ensions in t his syn thesis proble m . Let l A D = 1 . T hen t he leng t hs a , b and c are t he th ree independen t leng t h dim ensions . Also t o be consid- er ed are t he initial positions , θ 0 and φ 0 , of t he inpu t lin k A B and t he ou t put link DC . In all , t here ar e only five independen t design variables t hat must be calculated in designin g t he linkage t o gene rate t he function φ0 i = f θ 0 i ) , i .e. φ 0i = f a , b , c, θ 0 ,φ 0 ,θ 0i) ( 4 -12 ) Suppose t hat t he lin kage is used t o co-ordinate t he rotational angle of t he inpu t and ou tpu t for five positions , i .e.φ θ 0i = f 0 i ) , ( i = 1, 2, 3, 4, 5) φ 01 = f φ 05 = f P ut ti ng t hese five specified relationships bet ween φ 0 i and θ 0 i in to Eq. ( 4-12) , one ob tain s five equations as follows . a , b , c ,θ 0,φ 0 ,θ 01 ) a , b , c ,θ 0,φ 0 ,θ 05 )     …… Since t here are only five unknow ns ( a , b , c, θ 0 , φ 0 ) in t his set of equations , t he val- ues of t he five un know ns can be found by solving t hese equations simultaneously . ・ 64 ・
Fig . 4-45

Since t here are only up t o five independen t design variables in th is syn t hesis proble m , at most five equation s can be solved sim ultaneously . T her efore t his linkage can co-ordinate exactly only up to five relationships bet ween t he inpu t angle and t he ou tpu t angle . A t ot her positions , t here will be some e rror ( called str uct ur al er ror ) bet ween t he act ual function and t he r equir ed function . Suppose t hat a revolu te four-bar linkage A BCD is t o be designed so t ha t a coupler poi nt E will pass t hrough an ideal curve ( dashed curve in Fig . 4-46 ) . I t can be show n t ha t t he act ual couple r curve ( solid curve) can pass exactly t hrough up to nine poin ts on t he ideal curve , as show n in Fig . 4-46 . From t he last examples , we can see t hat a linkage m echanism can m atch t he function exactly a t only a limited num ber of positions . At o t her positions , t here will be st ruct ural errors . If t he number of t he r equir ed positions is larger t han 3 , t he algebraic syn t hesis met hod oft en leads t o a set of non-linear equations con taining t ranscenden tal functions of t he un know n angles . I t is ve ry t edious to solve t hese non-li near e- quations si mult aneously . Also , t he m et hod canno t r eally con t rol α s crite- m ax , γ mi n , G rashof’ rion , and t he st ructur al e rror betw een t he tw o pr ecision poin ts . In most indust rial synt hesis proble ms howeve r , designe rs of ten pr efe r to min imize t he m axim um struct ural error rat her t han t o satisfy a few poin ts exactly w hile losing con t rol over t he m axim um struct ural error . Op- ti mization m ethods , ho wever , can m inimize t he maximu m st ructur al error unde r some con- st rain ts by op ti mizing t he dim ensions of t he mechanism . optimization m et hods a re now widely used in t he syn t hesis of lin kages . Many optimization met hods need an initial linkage before op- ti mization . T hese can be syn t hesized by choosing a few t ypical r equir ed positions . S ynt hesis m et hods mentioned in t his chapte r ar e ext re mely useful in providing a reasonable initial linkage for fur ther op timization . Problems and Exercises 4- 1   According to t he kine ma tic diagr ams of the mechanism s w h ich you drew in Chapte r 2 , give t he nam es of t he m echanisms in figures from Fig . 2-43 t o Fig . 2-46 . 4-2   Listed in t he following table a re five sets of dim ension s of a revolu te four-bar linkage A BCD si mila r to t he one in Fig . 4-25 . Determi ne t he t ype of t he linkage and t he t ype of t he t wo side li nks A B and DC ( crank or rocker ) according to t he G rashof c rite rion . Can t he cou- pler BC rotate 360°wit h r espect to o t her lin ks ( Y es or No ) ?
Fig . 4-46

・ 65 ・

lA B 45 20 20 80 40

lB C 50 35 45 20 30

lD C 60 70 70 45 20

lA D 20 90 90 60 35

T yp e o f lin k a g e

Ty pe o f AB

T y p e o f DC

C a n BC r o t a t e 360° ?

4- 3   In a revolu te four-bar linkage A BCD sim ila r to t he one in Fig . 4-25 , lB C = 100 mm , l D C = 80 mm , l A D = 110 m m . ( 1) Find t he range of t he values for t he lengt h lA B of link A B if t he lin kage is eit her ( a ) a cr ank-rocker m echanism wit h crank A B , or ( b) a double-rocke r mechanism . (2 ) Can t he lin kage be a double-c ran k m echanism by choosing l AB suitably ? W hy ? 4-4   In t he r evolute four-ba r mechanism si mila r t o t he one in Fig . 4-23b , let lA B = 60 mm , lB C = 130 m m , l D C = 140 m m , l A D = 200 m m , and ∠ B A D = 135°. (1 ) Determ ine t he type of t he r evolute four-bar m echanism . (2 ) If t he side link A B is a drive r and rotat es at a con stan t speed , ( a ) find t he pressure angle α an d t he tr ansmission angle γ of t he mechanism at t hat posi- tion . (b ) find t he angular stroke φ m a x of t he link DC . ( c ) find t he crank acu te angle θ bet ween t he t wo limiting positions . (d ) calcula te t he tim e ratio K . ( e ) will any dead-poin t occur during t he w hole cycle of t he motion ? (f) find t he m axi mum pressure angle α m a x and t he minimum t ransmission angle γ mi n . (3 ) If t he side link DC is a driver , ( a ) find t he pr essur e angle α ′and t he tr ansm ission angle γ ′ of t he m echanism at t hat posi- tion . (b ) will any dead-poin t occur durin g t he w hole cycle of t he motion ? If so , w hen ? ( c ) find t he maximum pr essur e angle α m a x and t he mini mum tr ansmission an gle γ m in . 4-5   In t he offset slider-c ran k m echanism show n in Fig . 4-5 , t he driver crank A B rot ates at a constan t speed . L et lA B = 120 mm , lBC = 250 mm , offset e = 60 m m , and φ A B = 60° . Find (1 ) t he leng th of st roke H of t he slider , (2 ) t he c ran k acu te angle θ bet ween t he tw o limitin g positions , (3 ) t he time ratio K , (4 ) t he pressure angle α and t he t ran smission angle γ a t t ha t position , (5 ) t he m ax imum pressure angle α m a x and t he minimu m t ransmission angle γ mi n . 4-6   In an offset slide r-crank m echanism A BC sim ilar t o t hat in Fig . 4-5 , t he crank A B is a drive r . T he leng t h of t he crank l A B = 25 mm . T he offset e = 10 mm . T he m axi mum pressure angle α . Find t he st roke H of t he slide r and t he crank acute angle θ between t he two m a x = 30° ・ 66 ・

limiting positions . 4-7   In t he double-c ran k linkage A BCD show n in Fig . 4-1b , l A B = 80 m m , lBC = 180 mm , l D C = 230 m m , l A D = 200 mm . T he driver BC rotates at a con stant speed . Find t he minimu m t ransmission angle γ mi n of t he li nkage . 4-8   According to t he data in P roble m 4-3 , find t he range of values for t he driving c ran k A B , if t he linkage is to be a cr ank-rocker m echanism and t he mini mum t rans mission angle γ mi n is la rger t han 50°. Compa re t he result wit h t hat in Problem 4-3 (1 ) ( a ) . 4- 9   T he crank-rocker m echanism show n i n Fig . 4-31 has no quick-r eturn char acte ristics . Prove t ha t t he len gt hs of lin ks of t he linkage satisfy t he following equation : lA B + l A D = lBC + lDC 4-10   Show n in Fig . 4-47 are t he tw o positions , B 1 C1 and B 2 C2 , of t he coupler BC of a rev- olu te four-bar linkage A BCD . T he lin k A B is a drive r . T he pressur e angle α at t he first posi- tion is 0°. The second position of t he m echanism is a toggle position . Design t he linkage . 4-11   Design a cr ank-rocker mechan ism A BCD to move a line E F on t he couple r BC from po- sition E 1 F1 to position E 2 F 2 and then t o position E 3 F 3 ( Fig . 4-48 ) . x E 1 = 2. 18 mm , y E 1 = - 15. 1 mm , x E 2 = - 3 .19 mm , y E 2 = - 56. 8 mm , x E 3 = 67 .5 m m , y E 3 = - 32. 8 mm , θ E 1F 1 = 61. 7° , θ E 2 F 2 = 73 .4° , θ E 3F 3 = 108 .5° , l E F = 125 mm . (1) If t he locations of t he fixed pivo ts A and D have not been det ermined , design t he m echanism . Check w het he r t he side lin k A B is a c ran k . Can bo t h t he poin ts E and F be se- lected as t he tw o movi ng r evolute cen tr es ? W hy ? (2 ) Supposed t hat x A = 0 m m , yA = 0 mm , x D = 100 mm and yD = 0 m m . Find t he re- quired moving revolu te cen t res B 1 and C1 on t he coupler . 4-12   Design a revolut e four-ba r linkage A BCD so that a poin t E on the coupler B EC will pass t hrough t he t hree poin ts , E 1 , E 2 and E 3 , as show n i n Fig . 4-49 . T he first position B 1 of t he moving revolu te cent re B , and t he fixed pivots A and D have been specified . x A = 0 mm , y A = 0 mm , x D = 100 m m , yD = 0 mm , y E 1 = 128 .5 m m , x B 1 = 44. 2 mm , y B1 = 22 .2 m m . x E 1 = 50. 4 mm , x E 2 = - 44 .65 mm , y E 2 = 115. 4 mm . x E 3 = - 63 .4 mm . y E 3 = 29 .35 mm .
2 2 2 2

Fin d t he first position C1 of t he movi ng r evolute cen tr e C .

Fig . 4-47

Fig . 4-48

Fig . 4-49

・ 67 ・

4-13   In a revolu te four-ba r lin kage A BCD , as t he driver side link A B rotates from position 1 t o position 2 ( Fig . 4-50 ) , t he ot her side lin k DC ro tates 24. 6° coun terclock wise . As t he lin k A B goes from position 2 to position 3 , t he link DC rot ates 39 .3°coun te r-clock wise . l A B = 50 mm and lA D = 100 mm . S yn t hesize t he lin kage graph ically .

Fig . 4-50

Fig . 4-51

4-14   In an offset slider-crank m echanism similar to t he one in Fig . 4-5 , as t he crank A B ro- tates 100°coun ter-clock wise from position 1 t o position 2 , t he revolu te cen tr e C on t he slider will t ranslate from poin t C1 to poin t C2 ( Fig . 4-51 ) . As t he c ran k A B ro tates 100°coun ter- clock wise from position 2 to position 3 , t he revolu te cen tr e C t ranslates from poin t C2 to poin t C3 , Offset e = 10 m m , x A = 0 m m , y A = 0 mm , x C 1 = 28 .76 mm , x C 2 = 40 .88 m m , xC3 = 69 .08 m m . Locate t he t hird position B 3 of t he revolut e cen t re B . 4-15   In an offset slider-c ran k mechanism , a point F on t he slider will displace 23 .25 mm from poin t F 1 t o poin t F 2 w hen t he crank A B rotat es from position A B 1 t o position A B 2 ( Fig . 4-52 ) . As t he crank A B rotates from position A B2 to position A B 3 , t he poin t F dis- places 52. 13 m m from point F2 to poin t F 3 , lA B = 50 mm , Design the offset slider-cr an k mechan ism . 4- 16   In an offset slider-rocker mechanism A BC , t he position of t he fixed pivot A and t he tw o positions , C1 P 1 and C2 P 2 , of a li ne CP on t he coupler BC a re know n , as show n in Fig . 4-53 . x A = 0 m m , y A = 0 mm , x C 1 = 26. 05 mm , yC 1 = 45 m m , x C 2 = 66 .5 m m , yC 2 = 45 mm , θ , θ . T he moving revolu te 1 = 7° 2 = 11° cen t re B is on t he line CP Locate t he first position B 1 of t he poin t B .
Fig . 4-53 Fig . 4-52

4-17   In a revolu te four-bar lin kage A BCD , side link A B is a driver . T he positions of t he side link CD and a line segm en t CE on t he coupler CB E corr esponding t o t wo positions of t he link- age a re know n , as sho wn in Fig . 4-54 . T he first position of t he linkage is also a dead poin t . l A D = 100 m m , lC D = 64 .550 8 mm , ∠ A DC1 = 42 .39° , ∠ A DC2 = 133. 8° , θ , C 1 E 1 = 104. 7° θ , Find t he second position B 2 of t he revolu te B . C 2E 2 = 93. 7° ・ 68 ・

4-18   In a crank-slide r mechanism , t wo sets of corr esponding positions between t he slider and a line segm en t A E on t he c ran k A B E ar e know n , as show n in Fig . 4-55 . The position C1 of t he slide r is its lef t limiting positions . x A = 0 mm , y A = 0 mm , x C 1 = - 100 mm . yC 1 = - 28 .15 m m , x C 2 = - 65 .56 mm . Find t he first position B 1 of t he revolu te B .

Fig . 4-54

Fig . 4-55

4-19   In a crank-rocke r linkage A BCD , side link A B is a driver . T he positions of t he rocker CD corresponding to t wo positions of t he lin kage a re know n , as show n in Fig . 4-56 . l A D = 100 m m , lC D = 75 mm , ∠ A DC1 = 64 .3° , ∠ A DC2 = 134. 2°. At the first position of t he link- age , t he pr essure angle of t he lin kage is zero . Position DC2 is one of t he limit position s of t he rocker . Find t he first position B 1 of t he revolut e B . 4-20   In an offset slider-cr ank m echanism A BC , t wo sets of corr esponding positions betw een t he cr ank A B and a poi nt F on t he slider are know n , as show n in Fig . 4-57 . W hen t he c ran k A B is located at position A B1 , t he slider reaches its left limit position . l A B = 30 mm , ∠ B 2 A B 1 = 112 .8° , l F 1 F2 = 32. 96 m m . Find t he first position C1 of t he r evolute C on t he slid- er .

Fig . 4-56

Fig . 4-57

4- 21   W hen t he w heel of t he landing mechanism in an airplane is at its lowest position , as show n in Fig . 4-30 , lin ks BC and CD are collinea r . Suppose t he position of revolut e C is un- know n . Find t he position of t he revolu te C on line DCB so t hat t he link A B will rotat e 90° cou nte r-clock wise w hen t he lin k DC rotates 60° clock wise . T he know n dimensions are : x A = 0 mm , yA = 0 mm , x D = 103 m m , y D = 23 .55 mm , and l A B = 100 mm . 4-22   In a r evolute four-bar lin kage A BCD , cr ank A B is a driver . The positions of t he c ran k A B and a line segmen t DE on ano t her side li nk CDE corresponding to t wo positions of t he linkage are know n , as show n in Fig . 4-58 . T he first position of t he rocker is one of its t wo ・ 69 ・

limiting positions . l A D = 36 mm ,

l A B = 13. 5 m m , ∠ D A B 1 = 28. 6° , ∠ D A B 2 = 137. 9° ,

∠ A D E 1 = 153. 8° , ∠ A DE 2 = 100°. Find t he second position C2 of t he revolu te C .

Fig . 4-58

4- 23   In a crank-rocke r mechanism A BCD s ho wn in Fig . 4-59 , c ran k A B is a driver . T he first position of t he crank A B and t wo positions of t he rocke r CD ar e know n . A t t he second position, t he value of tr ansmission angle reaches its mi nimum . l A D = 42. 3 m m , revolu te B . lD C = 13 .5 mm . ∠ D A B 1 = 86. 9° , ∠ A DC1 = 46. 6° , ∠ A DC2 = 102°. Fi nd t he second position B2 of t he

Fig . 4-59

4-24   In a revolu te four-bar linkage A BCD , cr ank A B is a driver . T he coefficient K of t ravel speed va riation of t he rocke r CD is t o be 1 . W hen t he cr ank A B is located at position A B 2 , as show n in Fig . 4-60 , the rocke r CD r eaches its limit position . l A D = 100 mm , lA B = 25. 2 mm , ∠ D A B 2 = 26. 43°. Design t his rev- olu te four-ba r linkage . 4-25   Design a crank-rocker mechan ism A BCD similar to t he one in Fig . 4-23a t o give 45°os- cillation of t he rocker DC wit h equal time back and for t h , from a constan t inpu t speed c ran k A B . L et lA B = 30 m m and lB C = 80 mm . 4-26   Design a cr ank-rocker m echanism si mila r to the one in Fig . 4-27 t hat will give a K value ・ 70 ・
Fig . 4-60

of 1 .2 wit h an angular stroke of 45° for t he rocke r DC , from a constan t inpu t speed cr ank A B . T he lengt h of t he rocker DC is 100 mm . Check t hat t he minimum tr ansmission angle γ mi n be la rger t han 40°. 4-27   Design a cr ank-rocker m echanism si mila r to the one in Fig . 4-27 t hat will give a K value of 1 .25 wit h an angula r st roke of 32°for t he rocker DC , from a constan t inpu t speed c ran k A B . L et lA B = 75 m m and l D C = 290 m m . Fin d t he lengt hs of links BC and A D . 4-28   Design an offset slider-crank m echanism similar to t he one in Fig . 4-40 . T he offset e is 20 mm . T he coefficient K of travel speed variation is to be 1. 3 . T he working st roke H of t he slider is to be 50 m m . 4-29   Design an offset slider-c ran k m echanism similar to t he one in Fig . 4-40 . The r atio of t he lengt h of t he coupler BC t o t he leng t h of t he crank A B is t o be 4 . T he coefficien t K of t ravel speed variation is t o be 1 .2 . The working stroke H of t he slider is t o be 200 mm .

・ 71 ・

C h ap t e r 5 C am Me c h a n i s m s
5.1   C ha r a c t e r i s t i c s and Cl a s s i f i c a t i on o f C am Me c ha ni sms
As we learned in Chap ter 4 , linkage mechanisms ar e not suitable for t ransmitting compli- cated mo tion , especially w hen t he driven lin k should d well for a short tim e during a cycle . I n such cases , ca m mechanisms can be used . 5.1.1   Characteristics of Cam Mechanisms     Cam mechanisms a re t he sim plest m echanisms to t ran sfer a simple mo- tion in to any desired complicated motion . There a re only t hree lin ks in t his m echanism: a ca m ( drive r ) , a followe r ( t he driven ele ment ) and t he fra me . T he ca m has a curved or grooved surface w hich m ates directly wit h t he follower and im par ts motion to it . I t is very easy to design t he ca m cont our even if t he motion of the followe r is ve ry complicated and so they a re used ex tensively in modern m achinery . Fig . 5-1 and Fig . 5-2 illus- tr ate t wo exa mples of ca m applications . T he mo tion of t he valve 3 in an in ternal combustion engine ( Fig . 5-1 ) is determined by t he cont our of t he
Fig . 5-1

cam 1 . Show n in Fig . 5-2 is a t ool-feeding mechan ism in an aut om atic machi ne tool . W hen t he cylindrical cam 1 rotates at a const an t speed , t he follo wer 2 will oscillate and will force t he t ool fra me 4 t o t ranslate back and fort h . T he motion of t he oscillatin g followe r 2 is dete rmined by t he shape of t he curve on t he cylindrical ca m 1 . T he cost for cu tting ca m accurately is high . Ca m m echanism can not t ransmit heavy loads because of t he poin t or line con tact between t he cam and t he followe r . As a r esult , it is usually used for con t rol m echa- nisms or ligh t-load m echanisms . 5.1.2   Classifications of Cam Mechanisms Ca m m echanisms can be classified in seve ral ways . (1 ) By t he ca m shape ( a ) Plate ca m ( or disc ca m ) T his is t he most popular t ype of ca m ( Fig . 5-3a ) . T he ・ 72 ・
Fig . 5-2

cam usually ro tates con tinuously at a constan t speed . (b ) Translating ca m A t ranslating cam ( Fig . 5-3b ) is a plate ca m wit h an infinit e radius . It t ranslates back and for t h w hich produces a la rge iner tia for ce . So it is seldom used . ( c ) T hree-dim ensional ca m In a t hr ee-dimensional ca m m echanism , t he ca m and t he follow er move in t wo non-par allel planes . T he cylindrical cam s ho wn i n Fig . 5-3c is one kind of t hree-dim ensional ca m . T he cylindrical ca m can rot ate con ti nuously in one direction , and so can be used at high speed . (2 ) By t he motion type of t he follower ( a ) Translating follow er ( Fig . 5-4 )

Fig . 5-3

Fig . 5-4

    T his kind of follo wer t ranslates back and fort h along t he guide way of t he fra me . If t he cen t reline of t he follower passes t hrough t he cen tre of t he cam shaft , it is called the i n-li ne tr ansla ti ng follo wer , as show n in Fig . 5-4a ; ot her wise, t he o f fset tr ansla ti ng f ollower , as show n in Fig . 5-4b . T he perpendicular distance from t he cen- tr e of t he cam shaft t o an extension of t he cen t reli ne of t he fol- lower is called t he o f fset , denoted as e .     lating follo wer . (3 ) By t he shape of t he followe r end ( a ) K nife-edge followe r ( Fig . 5-5a , d) Alt hough it is sim ple , t he str esses at t he line of con tact ar e excessive and cause r apid wea r on t he follo wer and t he ca m sur- face . F or t h is reason , t his t ype of follo wer is seldom used in practice . H owever , it is t he t heoretical basis for all followers wit h ot he r end shapes . (b ) Roller followe r ( Fig . 5-5b , e )
Fig . 5-5

g follower ( Fig . 5-5d , e , f) T his type of follow er w or ks more smoot hly t han t he t rans-

T he roller followe r gr eatly reduces w ea r because t he contact is almost entirely rolling rat her ・ 73 ・

t han sliding . T his t ype of follower is probably used mor e t han any ot her type of follo wer . ( c ) Flat-faced follower ( Fig . 5-5c , f)     T his follo wer is simpler and less expensive t han t he roller follo wer . T he con tact st ress is generally sm aller since t he r adius of curvat ure of t he follower is infin ite . This followe r is used often in high-speed cam mechan isms si nce a dyna mic pr essure oil film can be form ed easily be- tween t he cam and t he follow er w hen the speed of t he cam is high . T he motion curve of t he followe r will be restrict ed since all sections of t he ca m con tour must be convex . (4 ) By t he manne r of keeping t he cam and t he follow er in con tact If a follower is to reproduce exactly t he motion t ransmitted by a cam , t hen it must re main in con tact wit h t he ca m at all speeds at all tim es . T his can be achieved by force or by t he form of t he follower or t he cam . ( a ) Force-closed ca m m echanism

Fig . 5-6

If t he con tact between t he ca m and its followe r is obtained by a preloaded spri ng , or by ・ 74 ・

gravit y , it is called a f orce-closed ca m m echa nism , as show n in Fig . 5-1 . (b ) Form-closed ca m m echanism If t he con tact is ob tained by letting t he rolle r follower sit in a ca m groove or by using a con- jugate condition , it is called a for m -closed ca m mechan ism , as show n in Fig . 5-6 . T he re is a large number of cam and followe r combina tions . T he designer can choose a suit- able combination according to t he following factors : type of inpu t ( ro tation or t ranslation ) , t ype of ou tpu t , relative position bet ween t he shafts of t he ca m and t he follow er , ro tati ng speed of t he ca m , loading on t he cam and t he follower , cost , etc . . Simplicit y is alw ays a gove rning factor in t he choice of a cam and follower . For t his reason , t his chapte r is devoted to the sim- plest t ype of cam , t he plat e cam .

5.2   F o l l o w e r M o t i o n C u r v e s
Befor e a ca m is designed , a motion curve of t he follower should be specified . A t ypical cam application w ould require t he motion of t he follow er somet hing like t hat show n in Fig . 5-7 , w her e one full revolu tion of t he ca m is expressed on t he abscissa w hile t he displacem en t of t he follo wer is show n on t he ordinate . T he followe r is required t o rise from its lo west position t o its highest position (from Fig . 5-8a to b ) , d well at the highest position (from Fig . 5-8b to c ) , ret urn to its low est position ( from Fig . 5-8c to d) , and d well at t he low est position (from Fig . 5-8d t o a ) before repeati ng t he cycle . The grea test dist ance t hrough w hich t he follow er moves is k no w n as t he tota l f ollower travel , or t he li f t , deno ted by h . T he rotation angle of t he ca m duri ng w hich t he follower rises from its lowest position to its highest position is called t he ca m ang le f or rise , denoted by δ 0 . T he angle during w h ich t he followe r d wells at t he highest position is called t he ca m an gle f or ou ter d well , deno ted by δ s . T he ro tation angle duri ng w hich t he followe r returns from its highest position to its lowest position is called t he ca m ang le for ret ur n , denoted by δ 0′ . T he angle during w hich t he follower d wells at the low - est position is called t he ca m ang le f or in ner d well , denoted by δ . The curve in Fig . 5-7 s′ shows t he relationship between t he displacem en t s of the follow er an d t he rota tion angle δ of t he ca m . T his is called t he d isp lacemen t cu rve of t he cam mechanism . Diffe ren tiating s wit h r espect to ti me t will r esult in velocit y v = d s d s dδ d s = = ω, or d t d δd t dδ

Fig . 5-7

・ 75 ・

a)

                            b)

c)                         d)

Fig . 5-8

ds v ds ds = . Since ω is a constan t , t he - δ curve has a shape similar t o t he v - t curve. So , dδ ω dδ dδ denot ed by s′ , is called t he qu asi-velocit y . Diffe ren tiating v wit h respect t o t r esults in acceler ation d ds ds ds ω d d dδ dδ dδ =ω =ω dt dt dδ
2 2

a=

dv = dt

dδ d s a 2 d s =ω 2 , or 2 = 2 dt dδ dδ ω ・ 76 ・

d s d s Since ω is a constan t , t he 2 - δ curve has a shape similar t o the α - t curve . So 2 , dδ dδ denot ed by s″ , is called t he quasi-acceler at ion . T he di mensions of bot h s′and s″ar e mm . Usually , in given design conditions , s is a function of δ ( no t of t ) . We shall t herefore follow t he usual practice and be concerned wit h s′and s″r at her t han v and a . For any given δ 0 and h , t he follow er m ay rise along diffe ren t mo tion curves and t he corre- sponding dyna mic char acte ristics a re differ en t . Som e kinds of motion curves ar e int roduced be- low . 5.2.1   Constant Velocity Motion Curve Constan t velocit y m eans t hat t he v - t (or s′- δ) diagram is a horizont al str aigh t line , so t he s - δ diagra m ( t he in tegr ation of t he s′- δ curve ) is an inclined st raigh t line and t he accel- er ation ( t he diffe ren tiation of t he velocit y) is zero , as show n in Fig . 5-9 . A t t he beginni ng of the rise , t he velocit y changes from zero to its m aximu m ( const an t ) value in zero tim e . The diffe ren tiation of veloci- ty ( i .e.acceleration ) at t his poi nt is t her efore positive infinite . At t he end of t he rise , t he velocity changes from its m axim um to zero in ze ro tim e and t he acceler ation is negative i nfinite . A ny r eal follo wer must , of course, have som e m ass , so at t hese tw o poin ts , infinite accele ra- tion will produce an infinite ine rtia force . Alt hough t he elasticity of
Fig . 5-9

2

2

links in ca m m echanisms prevents infin ite acceleration , t he values for t he accele ration are very high . The consequen t large i ner tia force will result in a very large a moun t of shock , w hich is called a rig i d i m p u lse . Th is will produce vibra tion , noise , high st ress levels , and serious wea r . Consequently , a constan t velocit y motion curve can not be used for t he full st roke . 5.2.2   Constant Acceleration and Deceleration Motion Curve In t his t ype of motion curve , t he follo wer is given a con stan t acceler ation during t he first half of t he rise and a constan t deceleration during t he second half of t he rise . The velocity curve and displace ment curve can be c reated by in te- gr ation of t he acceleration curve, as sho w n in Fig . 5-10 . A t t he beginning of t he rise , acceler ation changes from ze ro to a constan t positive value in ze ro tim e . There ar e tw o more abrup t changes in t he acceleration of t he follow er , at t he midpoin t and t he end of t he rise . An abrup t change in t he accelera tion causes an abrup t change in t he
Fig . 5-10

ine rtia force . Since every link in a cam m echanism has elasticit y , an

abrup t change in t he m agnit ude and/ or dir ection of i ner tia force w ould initiate undesired vibra- tion , w hich is called a so f t i m pu lse . Therefore t his motion curve is not satisfactory for high speed operation . Ho wever , because of t he lo wer m axi mum accelera tion compar ed wit h ot her ・ 77 ・

motion curves , t his motion curve can be used for lo w and in te rmediate speed ca ms . 5.2.3   Cosine Acceleration Motion Curve (or Simple Harmonic Motion Curve) Since t he accele ration is positive during t he first half of t he rise and negative during t he sec- ond half of t he rise , a cosine curve bet ween 0 to π can be used as t he accel- eration curve in t he rise, as show n in Fig . 5-11 . T he velocit y curve and t he displace men t curve can be created by in tegra tion of t he accelera tion curve . We can see t hat alt hough t he mo tion curve avoids t he sudden re- ve rse of accele ration a t t he middle-poin t of t he t ravel , t he acceleration will still change abruptly at t he beginning and end poi nts . T hat would produce soft impulses . So t his motion curve can be used only for low or in te rmedi-
Fig . 5-11

ate speed ca ms .

T hus it is very impor tan t w hen choosing a displace men t curve to ensure t hat t he velocit y and acceleration curves ar e con tin uous at all tim es . 5.2.4   Sine Acceleration Motion Curve ( or Cycloid Motion Curve) Since t he acceleration is positive during t he first half of t he rise and negative duri ng t he sec- ond half of t he rise , a si ne curve bet ween 0 t o 2 π can be used as t he acceler ation curve in t he rise , as show n in Fig . 5-12 . T he equa tion of t he quasi-acceleration in t he rise will be δ π s″ = c1 sin 2 δ 0 tegration of t he equation of s″wit h respect to δ, s′ = c2 - c1 δ 0 δ cos 2πδ 2π 0 ( 5-2 ) ( 5-1 )

T he equation of the quasi-velocit y s′can be created by t he in-

In tegrating again wit h respect t o δ results in t he displace men t equation ,
2 δ 0 δ s = c3 + c2 δ - c1 π 2 sin 2 δ 0 (2 π)

( 5-3 )

Fig . 5-12

T he constan ts c1 , c2 and c3 can be found by applying t he bounda ry conditions to t he equa- tions . In orde r to delete t he rigid i mpulse , s′should be zero at t he beginn ing and the end of t he rise , t hat is , s′ = 0 w hen δ= 0 . and s′ = 0 w hen δ= δ 0 , F urt hermore , s should be equal t o h w hen δ = δ 0 . Substit u ting the t h ree boundary conditions int o Eq . 5-1 t o Eq . 5-3 yields t hr ee equa tions and solving t he t hr ee equations simultaneously r esults in c1 = 2π h/ δ 0 , c2 = h/ δ 0 , and c3 = 0 . Substit uting t hese constan ts in to Eq . 5-1 t o Eq . 5-3 yields t he displace men t , quasi-veloci- ty and quasi-acceler ation equation s as follows . ・ 78 ・
2

s= h s′ = s″ =

δ 1 sin δ 0 2π

δ 2 π δ 0 δ 2π δ 0 (5-4)

h 1 - cos δ 0 2π h 2 sin δ 0

δ 2π δ 0

Since t he dim ensions of s′and s″ a re mm , all angles in all equations i n t his chap ter s hould be expr essed i n radian s , no t in degr ees . T he maximu m values of quasi-velocity and quasi-accel-
2 er ation , ′ sm a x and ″ sm  a x , for t he sine accele ration motion curve are 2 h/ δ 0 and 2 π h/ δ 0 , respec-

tively . T he acceleration is negative during t he first half of t he return and positive during t he sec- ond half of t he ret urn , as show n in Fig . 5-12 . T he equations for t he r eturn wit h a sine accel- er ation motion curve can be derived i n a si mila r way . T hey are as follo ws . s= h 1 s′ = s″ = (δ - δ 0 - δ s) 1 + sin δ ′ 2π 0 2 π 2π (δ - δ 0 - δ s) δ ′ 0 ( 5 -5 )

h 1 - cos δ ′ 0 2π h 2 sin δ ′ 0 2π

(δ - δ 0 - δ s) δ ′ 0

(δ - δ 0 - δ s) δ ′ 0

Not e t hat the displacem en t s i n t he ret urn is measured from t he lowest position of t he fol- lower . Angles δ in all equations in t his chap te r are always m easured from t he sta rting poin t of t he rise . In t his motion curve , t he velocity and t he acceleration always begin from zero , and change smoot hly . T here is no sudden change in velocit y an d accele ration at eit her end of t he st roke, even if t he re ar e dwell periods at t he beginning or t he end of t he t ravel . T her efore t here is neit he r rigid impulse nor soft impulse . T his results in ve ry good oper ati ng char acte ris- tics . Wea r , s hock , st ress and noise ar e quite low . T her efore t his mo tion curve gives a good dyna mic characteristic and is especially recom mended for high-speed ca ms . 5.2.5   3-4-5 Polynomial Motion Curve A not he r motion curve t hat is i ncreasingly being u tilized is t he polynomial motion curve . Since polynomials can be used t o approxim ate any function , it is not surprising to find t hat t he polynomial motion curve can fit almost any required motion . T he standa rd polynomial equa tion is
2 n s = c0 + c1δ+ c2δ + … + cnδ

w her e t he constan ts c0 , c1 , … , cn depend on t he bou ndary con ditions . I t is customa ry to specify certain geom etric proper ties at t he beginning an d t he end of t he t ravel . Suppose t ha t , in orde r t o delete any soft and rigid i mpulse , s′and s″of t he followe r are t o be zero at t he be- ginning and t he end of t he rise , as show n in Fig . 5-13 . T herefore , t he bounda ry conditions ・ 79 ・

for the rise ar e: s = 0 , s′ = 0 and s″ = 0 w hen δ= 0 ; and s = h , s′ = 0 and s″ = 0 w hen δ= δ 0 . Since t her e are six bounda ry conditions , a polynomial of six terms is required . Assum e t hat
2 3 4 5 s = c0 + c1 δ+ c2δ + c3δ + c4δ + c5δ

( 5 -6 )

Diffe ren tiating t he equation t wice wit h r espect t o δ will re- sult in equations for s′and s″. s′ = ds 2 3 4 = c1 + 2 c2 δ+ 3 c3 δ + 4 c4 δ + 5 c5 δ dδ
2

( 5 -7 ) ( 5 -8 )

d s 2 3 s″ = 2 = 2 c2 + 6 c3δ+ 12 c4δ + 20 c5δ dδ solvin g t he six equation s sim ultaneously r esults in c0 = 0 , c1 = 0 , c2 = 0 , c3 =

Fig . 5-13

Substit uting t he six bounda ry conditions in to Eq . 5-6 t o Eq . 5-8 yields six equations and 10 h 15 h 6h c4 = - 4   and   c5 = 5 . 3 , δ δ δ 0 0 0

Substit uting t hese constan ts in to Eq . 5-6 t o Eq . 5-8 yields t he displace men t , quasi-veloci- ty and quasi-acceler ation equation s as follows : δ s = h 10 δ 0
3

δ - 15 δ 0
2

4

δ +6 δ 0
3

5

h δ s′ = 30 δ δ 0 0

δ - 60 δ 0

δ + 30 δ 0

4

( 5 -9 )
3

δ δ h s″ = 2 60 - 180 δ 0 δ 0 δ 0

2

δ + 120 δ 0

T his motion curve is called a 3-4-5 polynomial mo tion curve based on t he po wers of t he re- m aining terms in t he displace men t equation . ′  s m a x and ″ sm  a x of t his mo tion curve are 1. 88 h/ δ 0
2 and 5 .77 h/ δ 0 , r espectively . For t he sam e δ 0 and h , ′ sm  a x and ″ sm  a x of t he 3-4-5 polynomial mo-

tion curve a re smaller t han t hose of t he sine accele ration motion curve . There is neit her rigid impulse nor sof t i mpulse . Therefore , t he 3-4-5 polynomial motion curve is t herefore used fre- quen tly for high speed . T he s - δ, s′- δ and s″- δ equations for ret urn wit h a 3-4-5 polynom ial motion curve can be derived similarly . T hey are : δ- δ 0 - δ s s = h 1 - 10 δ ′ 0 δ- δ 0 - δ s h s′ = 30 δ ′ 0 δ ′ 0 δ- δ 0 - δ s h s″ = - 2 60 δ ′ 0 δ ′ 0
3

δ- δ 0 - δ s + 15 δ ′ 0 δ- δ 0 - δ s - 60 δ ′ 0 δ- δ 0 - δ s - 180 δ ′ 0

4

δ- δ 0 - δ s - 6 δ ′ 0 δ- δ 0 - δ s + 30 δ ′ 0
2

5

2

3

4

( 5 -10 )
3

δ- δ 0 - δ s + 120 δ ′ 0

T he s - δ, s′- δ and s″- δ curves for t he rise and t he r eturn wit h 3-4-5 polynom ial mo- ・ 80 ・

tion curves a re show n i n Fig . 5-13 . 5.2.6   Combined Motion Curves As m en tioned above , t he si ne acceleration motion curve has neit her rigid nor soft impulse , bu t its m axim um acceler ation is quite large . T he maximum acceler- ation of t he constan t acceler ation and deceleration motion curve is lo w , but t he mo tion curve has a sof t im pulse . T he basic motion curves m en tioned in t he last sections can be combined in m any cases t o improve t he overall dynamic characteristics . T he motion curve show n in Fig . 5-14 is called the mod i f ied tr apezoi da l accelera tion m otion cu rve, w hich is a combination of t he sine acceleration curve and constan t accele ration and deceleration motion curve . This mod- ified motion curve is designed to min imize ex t re me accele ration and delete any impulse at t he sam e time . ′  sm a x and ″  sm a x of t his motion
2 curve are 2 h/ δ 0 and 4. 888 h/ δ 0 , r espectively .

Fig . 5-14

T he modified sine acceler ation motion curve ( Fig . 5-15 ) is anot her popular motion curve . The shape of t his motion curve is bet ween t hose of t he cosine and t he sine acceleration mo tion curves . Compared wit h t he cosine accele ra- tion motion curve , t his mo tion curve has no soft impulse . ′  s m a x and
2 ″ sm  a x a re 1. 76 h/ δ 0 and 5. 528 h/ δ 0 , r espectively . F or t he sam e δ 0

and h , ′  sm a x and ″  sm a x of t his motion curve are smaller t han t hose of t he sine acceleration motion curve and t he 3-4-5 polynomial mo tion curve . So t his t ype of motion curve is used in ca m m echanisms wit h high speeds and heavier loads . Choosing a suitable motion curve is one of t he key steps in t he
Fig . 5-15

design of a ca m m echanism . N o one motion curve is preferable for all applications . W hen choosing a motion curve , we should consider

t he speed of t he ca m , t he load on t he follower , cost accoun ting , etc . For higher speeds , mo- tion curves wit h lowe r maximu m accelerations should be chosen . W hen t he mass or load is la rg- er , t he motion curve w hich has a lower m axi mum velocity should be considered first . F or low speed and lightly loaded ca ms , dyna mic char acte ristics a re not critical and economy of produc- tion should be con sidered first , e . g . t he tangen t ca m show n in Fig . 2-4 and t he circular arc cam s ho wn in Fig . 5-20 can be chosen .

5.3   P l a t e C am w i t h T r a n s l a t i n g R o l l e r ( o r K ni f e-e d g e ) Fo llo wer
    T he cam profiles can be syn t hesized by gr aphical m et hods or analytical met hods . Gr aphical m et hods offe r clear physical concepts and can help us to understand t he analy tical met hods . ・ 81 ・

T he refore , a graph ical m et hod is in troduced first . 5.3.1   Graphical Synthesis of the Pitch Curve Sho wn in Fig . 5-16a is a plate cam wit h tr ansla ting offset knife-edge followe r ( ignore line A B B′ K for t he mom en t ) . I t is t o be designed to accomplish a motion as follows; rise t hrough a lift h = 80 mm wit h 3-4-5 polynomial mo tion curve during δ 0 of 140° , δ s = 40° , ret urn wit h sine acceler ation mo tion curve during δ ′of 100° 0 , and δ ′ s = 80° . T he cam rotates count er-clock- wise . T he min imum radius rp of t he cam is 100 mm . T he value of offset e is 40 mm . First we calculate t he displacem en t s of t he followe r for diffe ren t values of t he cam angle δ according t o t he given mo tion curve . For t he rise wit h 3-4-5 polynomial motion curve , Eq . 5- 9 is used . For t he r eturn wit h sine accele ration motion curve , Eq . 5-5 is used . T able 5-1 lists som e s vs . δ data . Table 5-1   Some s vs .δ data
No . δ / (° ) s/ m m 0 0 0 1 20 1 .86 2 40 11 .6 3 60 29. 4 4 80 5 0. 6 5 1 00 6 8.4 6 12 0 78 .1 7 ~9 1 40 ~1 80 80 10 2 00 7 6.1 11 22 0 55 .5 12 24 0 24. 5 13 260 3 .8 9 14 ~1 8 2 80 ~3 60 0

If t he cam rota tes an angle ∠ B 0 O B′ = δ count er-clock wise from t he beginning ( from Fig . 5-16a t o b) , t he follower will rise t hrough a corr esponding distance B′ B = s from t he lowest po- sition alon g its guide-way . Suppose t hat t he relative positions of all lin ks in t he m echanism in Fig . 5-16b are frozen and t he w hole mechan ism is rotated δ in t he direction opposite t o ω . T he cam will be restored to its original position and t he follo wer will move to position A B , as show n in Fig . 5-16a w he re t he angle bet ween A 0 B 0 and A B is δ and t he distance B′ B = s. T his m eans t ha t , if t he cam is held stationary and t he fra me and t he follower ar e rotat ed a round t he cen tr e O of t he ca mshaft in t he direction opposite to ω, t he r elationship betw een t he follo wer travelling distance along its guide-way and t he inverse angle is exactly t he sam e as t he desired motion curve . T his is called t he p ri nci p le o f i nversion . D urin g i nversion , t he perpen dicula r distance from t he cen t re O of t he ca mshaf t to t he ex- tension of cent reline of t he followe r , i .e. offset e , always stays the sam e . So during inversion , t he ex tension of cen t reline of t he follo wer always st ays tan gen t to a circle wit h radius equal to t he offset e and cen tr e a t t he cen tr e O of t he camshaft . This circle is called the o f fset circle . Since △ B 0 O K 0 ≌ △ B′ OK , ∠ B 0 O B′= ∠ K0 O K = δ . During inversion , t he knife-edge of t he follower always stays in con tact wit h t he ca m con tour . So t he locus t raced by t he knife- edge is t he ca m con tour , w hich is also called t he p itch cur ve in t h is exa mple . T he minimu m circle cen tr ed at O and t angent to t he pitch circle is called t he base circle of the ca m pitch curve , or p ri me circle . I ts r adius is denoted by r p . When t he knife-edge con tacts t he prim e ・ 82 ・

Fig . 5-16

circle , t he followe r is at its lowest position . T he procedure to design t he pitch curve using t he principle of inversion is as follows ( refer t o Fig . 5-16a ) . (1 ) Draw t he prime circle wit h radius rp and t he offset cir cle wit h radius e . D raw t he ini- tial position A 0 B 0 of t he follower according t o t he m agnitude and direction of the offset . T he t angen t poin t bet ween t he ex tension of t he line A 0 B 0 and t he offset circle is labeled K0 . (2 ) Divide t he offset circle in to a number of equal segm en ts in t he dir ection opposite t o ω and assign st ation numbers , K 0 , K1 , K 2 , K3 , etc ., t o t he boundaries of t hese segmen ts ac- cordi ng to t he position nu mbers in T able 5-1 , i. e.∠ K 0 O K = δ . (3 ) Lines K A a re dra wn from t hese poin ts , making t he m all t angen t to t he offset circle . The in tersections of t he lines K A and t he pri me circle a re labeled as B′. T he line K A can also be located by m aking ∠ B 0 OB′ = δ and t hen drawing line K B′ A tangen t t o t he offset circle . (4 ) T he corr esponding position B of t he k nife edge can be lo- cated by laying off lB′ B = s along t he guide-w ay of t he follower . (5 ) T he pitch curve is obt ained by dr awing a smoot h curve t hrough each poin t B , as show n in Fig . 5-16a . Not e: (1 ) T he pitch curve can not be dra w n in the direction of ω . (2 ) T he cen t reline of t he follo wer must always be tangen t t o t he offset circle . ・ 83 ・
Fig . 5-17

(3 ) T he poin t B can no t be located by laying off lBB′= s along OB′. (4 ) Alt hough t he curve B 0 B 7 is t he curve for a rise , ∠ B 0 OB 7 ≠δ 0 because ∠ B 0 O B 7 < ∠ B 0 OB 7 ′ = ∠ K 0 OK 7 = δ ′ O B1 4 = ∠ K 9 OK 1 4 = δ ′. 0 . Simila rly , ∠ B 9 O B 1 4 > ∠ B 9 0 (5 ) T he a rc B 7 B 9 is t he arc for out er dwell, but O B 7 < OB 7 ′ + B7 ′ B 7 = rp + h . If offset e = 0 , t he offset t ranslating followe r becomes an i n-line t ranslating followe r , as show n in Fig . 5-17 . The steps mentioned above can be simplified to syn t hesize t he pitch curve , or ca m cont our , of t he plate ca m wit h in-line t ranslating knife-edge followe r ( r efe r to Fig . 5-17 w h ich is self-explanat ory ) . 5.3.2   Graphical Synthesis of Plate Cam with Translating Roller Follower Suppose t hat a plate ca m wit h t ran slating roller follo wer ( Fig . 5-18 ) is t o be designed , using t he sam e mo tion curve as in t he previous section . O bvi- ously , t he given motion curve applies to t he rolle r cent re . A vir tual t ranslating k nife-edge followe r is welded to t he tr anslat- ing rolle r followe r and t he kn ife-edge coincides wit h t he roller cen tr e . Therefore , t he motion curve of t he knife-edge is t he sam e as t hat of t he roller cen t re , or that of t he t ranslating roller follower . The locus of t he knife-edge , or t he locus of t he roller cen tr e, relative to t he cam can be designed according to t he met hods men tioned in t he last section , as t he dashed curve sho w n in Fig . 5-18 . T he locus is called t he p itch cu r ve o f t he ca m . M any rolle r circles ar e draw n wit h roller radius r R and
Fig . 5-18

cen tr e at t he poin ts on t he pitch curve . T hese roller circles rep- resent t he positions of t he roller relative to t he ca m during inver-

sion . Since t he roller is tangen t to t he ca m con tour at all tim es , t he ca m con tour is t he enve- lope of t he fam ily of t he roller cir cles . So t he ca m con tour is found by drawing a smoo t h curve w hich is tangen t t o all roller circles , as t he solid curve show n in Fig . 5-18 . We now have tw o curves , t he pitch curve and t he ca m con tour , w hich a re par allel . T he minimu m radius of t he pitch curve , not t he cam cont our , is called t he radius rp of t he prim e circle . For t he knife- edge followe r , t he pitch curve and t he cam con t our a re t he sam e . T he tan gen t poin t T between t he roller an d t he cam con tour is a poin t on t he ca m con tour corresponding to t he rolle r cen ter B . Line B T is a common norm al n-n t o the pitch curve and t he cam con tour , as sho w n in Fig . 5-19 . N ote : T he poi nt T can not be loca ted by laying off lB T = r R along t he guide-way or OB . ・ 84 ・
Fig . 5-19

Example 5-1 Sho wn in Fig . 5-20 is a circular arc ca m wit h a t ranslating offset roller followe r . T he ca m cont our consists of four a rcs: T0 T1 , T1 T 2 T3 , F, D T 3 T 4 and T 4 T T 0 , as t he solid a rcs show n in Fig . 5-20 . T heir cen tr es a re poin ts E , and O , r espectively . The ca m h .       Solution: T he motion of t he roller cent re repr esents t he motion of the follower . T he required par am eters cannot be fou nd wit hout t he locus of t he roller cen t re relative t o t he ca m , or t he pitch curve . T he pitch curve and t he ca m con tour a re tw o pa r- allel curves for t he plate ca m wit h roller followe r . Since t he con tour of t his ca m is built up by four a rcs , t he pitch curves are also built up by four a rcs , as t he dot-an d-dash a rcs sho wn in Fig . 5- 20 . T w o corresponding arcs have t he sa m e cen- t re . rp is equal to l O B , no t lO T . Using t he principle of i nversion , t he fra m e
Fig . 5-20

rotates clock-

wise . Indicate rp , δ ′ , δ ′ 0 , δ s , δ 0 s , and t he lift

and t he follow er a re rotated coun terclock wise and t he roller cen t re will move coun ter-clock wise along t he pitch curve . D uring inversion , t he cen t reline of t he follower will alw ays be tangen t t o t he offset cir cle . W hen t he roller cen tr e B arrives at poin t B0 and B 2 respectively , t he fol- lower is a t t he sta rting position and t he h ig hest position of t he rise . T he cen tr elines of t he fol- lower are located on lines B 0 K 0 and B 2 K2 , respectively . So the curve B 0 B 1 B 2 is t he curve for t he rise . T hus t he angle t hrough w hich t he cen tr eline of t he follower is invert ed from position B 0 K 0 to position B 2 K 2 is δ 0 . This angle δ 0 is also equal t o ∠ K 0 O K 2 , or ∠ B 0 OB 2 ′. Note t hat ∠ B0 O B 2 ≠δ 0 . A t t he position B 2 , t he roller cen tr e has moved from its lowest position B2 ′to its highest position B2 along t he cen t reli ne A 2 B 2 of t he followe r . So B 2 ′ B 2 = h . Note t hat B 2 G ≠ h . W hen t he roller cen t re sta rts moving from poi nt B 2 , t he distance between the roller cen tre and t he cen t re O of t he ca mshaf t will dec rease i mm ediately , so t here is no ou ter d well for t his cam mechanis m , i .e.δ . Note t hat t he circular arc B 1 B 2 B 3 is not a curve for oute r d well . s = 0° W hen t he rolle r cen tr e a rrives at poin t B4 , t he follow er r et urns t o its lo west position . So t he curve B 2 B 3 B 4 is t he curve for t he ret urn . W hen t he roller cent re is at poin t B 4 , t he cen tr eline of t he follower is loca ted on li ne B 4 K 4 . T he angle t hrough w hich t he cent reline of t he follow er is inve rted from t he position B 2 K 2 to t he position B 4 K4 is δ ′. T his angle is also equal to 0 ・ 85 ・

∠ K 2 OK 4 , or ∠ B2 ′ OB 4 . N ote : ∠ B 2 OB 4 ≠ δ ′ 0 . O bviously , t he circula r a rc B 4 B B 0 is t he curve for t he inner dw ell and ∠ K 4 OK 0 = ∠ B 4 O B 0 = δ ′ s . 5.3.3   Analytical Synthesis of the Pitch Curve (1 ) Pitch curve for offset t ranslating followe r T he process of analytically generating t he pitch curve of a pla te ca m is closely parallel t o t he graphical m et hod . A Car tesian co-ordinat e syste m is fixed wit h t he ca m . I ts origin is located at t he cen tr e O of t he ca mshaf t and its y axis is pa rallel to the cent reline A 0 B 0 of t he follow er w hen δ= 0 , as sho wn in Fig . 5-21 . Compa ring Fig . 5-21 wit h Fig . 5-16a , we can see t hat in Fig . 5-21 , ∠ K 0 OK = δ, and B′ B = s . Since K B′= K 0 B 0 = s0 = pitch curve , can now be calculated as x B = O E = F E + O F = K G + OF = M [ ( s0 + s ) sin δ+ Ne cos δ] yB = B E = BG - EG = BG - F K = ( s0 + s) cos δ - Nesin δ w her e M an d N are iden tifiers of t he direction of t he cam rotation an d t he direction of t he offset , respective- ly . If t he ca m ro tates cou nte r-clock wise , M = + 1; o t her wise , M = - 1 . If t he sen se of t he mom en t of t he velocit y vect or vB of the roller cen tr e i n t he rise abou t t he cen t re O of t he ca mshaft is t he sam e as t hat of t he cam ro tation , as show n in Fig . 5-21 and Fig . 5-35 , t he m echanism is called a posit ive o f fset mechan ism and N = + 1 ; ot her wise , t he m echanism is called a negati ve of fset m echa nism and N = - 1 . In any case, t he value of e≥ 0 . Alt hough Eq . 5-11 is derived a t a par ticula r posi- tion in t he rise , it and t he following equations (Eq . 5- 12 t o Eq . 5-33) can be used in t he w hole cycle . (2 ) Pitch curve for in-li ne t ran slating follower If t he offset e becom es zero , t he offset t ranslating follo wer will becom e an in-line tr anslat- ing follow er ( Fig . 5-17) . So t he formulae of t he pitch curve of a plate cam wit h i n-line t rans- lating follo wer can be derived dir ectly from Eq . 5-11 by set tin g e equal t o zero , t hat is , x B = M ( r p + s ) sin δ y B = ( rp + s) cos δ 5.3.4   Analytical Synthesis of the Cam Contour For any pitch curve and specific radius r R of roller , t he re are two curves par allel to it .O ne is ・ 86 ・ ( 5 -12 )
Fig . 5-21
2 r2 K B = KB′ + p - e ,

B′ B = s0 + s . T he co-ordinates of the roller cen t re ( or k nife-edge ) B , or t he equation s of t he

( 5 -11 )

called t he ou ter envelope, and t he ot her is called t he i nner envelope, as s ho wn in Fig .5-22 and Fig .5-6a .If t he cam rotat es in a coun ter-clock wise dir ection , then t he pitch curve will be gener- at ed around t he cen t re O of t he ca mshaf t in t he opposite dir ection , clock wise in t his exam ple , because t he pitch curve is generated by t he principle of inversion .T he direction of t he tangen t line at t he poi nt B is from B t o E , as show n in Fig .5-22 .The slope of t he t angent line B E is tanθ= d yB d yB/ dδ yB ′ = = d xB d xB/ dδ x B ′

w her e xB ′and yB ′can be derived by differ en tiating x B and yB in Eq .5-11 wit h respect to δ, respectively . xB ′ = M [ s′ si n δ+ ( s0 + s) cos δ - Nesin δ] yB ′ = s′ cos δ - ( s0 + s) sin δ - Necos δ     If w e set K B =
2 2

( 5 -13 )

y′ B + x′ B , t hen sin θ = yB ′ / K B and
Fig . 5-22

cos θ= x B ′ / K B . Line T′ B T is a com mon normal t o t he t hr ee par allel curves . So ∠ B T D = θ . N ow , t he co-ordi- followe r , can be derived in righ t t riangle T B D . x T = x B ê M r R sin θ= x B ê M r R yB ′ / KB y T = yB ± M r R cos θ= yB ± M r R xB ′ / KB If t he ca m ro tates coun ter-clockwise , for t he in ner envelope . 5.3.5   Locus of Centre of Milling Cutter The norm al distance from the cen tr e C of milling cut ter t o t he ca m con t our is always equal t o t he radius r C of t he cu t ter ( Fig . 5-23 ) . So t he locus of t he cen tre of t he milli ng cu t ter , t he cam con tour and t he pitch curve of a plate cam wit h roller followe r a re t hr ee pa ral- lel curves . The locus of t he cen tr e of t he milling cu tter is t he out er envelope of t he cam con tour wit h distance r C . T he slope of t he tangen t line of the ca m con tour at t he poin t T is t he sa me as t hat of t he pitch curve at t he point B . W it h t he help of Eq . 5-14 , t he formulae of t he locus of t he cen t re C of t he milling cu tt er can be derived as follows x C = x T - M r C yB ′ / KB
Fig . 5-23

na tes of t he tangent poin t T , or t he equa tions of t he ca m con t our of t he plate cam wit h roller

( 5 -14 )

M = + 1 ; ot herwise , M = - 1 . N ote t he“ê ” and

“±” signs in Eq . 5-14 . The uppe r signs a re used for t he ou ter envelope , and t he lo wer signs

yC = y T + M r C x B ′ / KB

( 5-15) ・ 87 ・

5.3.6   Pressure Angle α     Sho wn in Fig . 5-24 is a plate ca m wit h t ranslating roller followe r . Ignoring friction , t he force F exer ted by t he ca m on t he follo wer is alon g t he com mon normal . According to t he definition of pressure angle in Sec . 4 .3 .2 , t he acute an- gle betw een t he line of t ravel of t he follower an d t he norm al to t he pitch curve at t he rolle r cen tre is t he pressure angle α of t he mechanism at t his position . T he force F does no t act along t he line of t ravel of t he followe r but will have a compo- nen t acting perpendicula r t o t he followe r cen treline . T his force is an undesirable side-t hrust on t he follo wer , w hich re- sults in deflection and ja mmi ng of t he follower stem . T he la rger t he value of α, t he large r t he side-t h rust will be . If t he side-t h rust is too large, t he followe r m ay ja m in t he guide-
Fig . 5-24

way . P ressure angle α changes during motion so it is necessary t o con t rol t he value of t he m axi- mum pr essure angle α m a x . Listed in T able 5-2 a re t he values of allo wable pr essur e angle [ α ] . For a force-closed cam mechan ism , it is t he spring force , no t t he force from t he ca m , t hat m akes t he followe r move in t he ret urn st roke . In t his case, t he followe r will never jam in t he guide-way during the ret urn . Hence t he allowable pressure angle [α] in t he ret urn for a force- closed cam mechan ism is quite large . The oscillating follo wer works mor e smoot hly t han t he t ranslating follow er . Table 5-2   The allowable pressure angle [ α] for cam mechanism
R eturn i n for m-closed cam mech anism or rise T ranslating followe r Oscillating follower 25° ~ 35° 35° ~ 45° R et urn in force-closed cam mech anism 70° ~80°

    In Fig . 5-24 , poin t P is the instan t cen t re bet ween t he ca m 1 and t he followe r 2 ( Refer t o Example 3-2 in Sec . 3 .2 .5 ) . Therefore , v P 1 = v P 2 , or ω1 × O P = v 2 . So , OP = v2 = ω1

d s/ d t d s = = s′. In rig ht t riangle BP K , t he gene ral form ula for t he pr essur e angle α can be dδ / d t dδ derived . tan α= | OP - e| = BK | s′- Ne | | s′- Ne | = 2 s0 + s rP - e2 + s ( 5-16)

w her e N = + 1 for t he positive offset m echanism and N = - 1 for t he negative offset mecha- nism . I n any case, t he value of e ≥ 0 . A suitable amoun t of positive offset will reduce t he ・ 88 ・

m aximu m pr essur e angle in t he rise . Bu t t he m axim um pressure angle in t he ret urn will be- come la rger . Ho wever , for force-closed ca m mechanism s , t his may very w ell be acceptable be- cause t he spring assists t he followe r to move in t he ret urn and [α] in t he ret urn for force-closed cam mechanism s is quit e la rge . T herefore a suitable a moun t of positive offset is high ly recom- m ended in the force-closed cam mechanism .     As can be seen from Eq . 5-16 , for t he t ranslating roller follower , an inc rease in rp will defi nit ely reduce α and hence α m a x . How ever , larger cams require mor e space and t his m eans mor e m ass and gr eate r iner tia forces w h ich may lead t o un wan ted vibr ations w hen speeds are high . Therefore we should t ry to use a sm aller r p . Bu t , t he smaller t he value of rp is , t he la rger t he value of α m a x will be . When r p is so sm all t hat α m a x is equal to [ α ] in t he rise or in t he ret urn , t he corresponding rp is called t he m i n i m u m r ad i us o f the pri me circle, deno ted as rp mi n . To calculate rp mi n by differen tia ting Eq . 5-16 is often very difficult because of t he result- ing complex equation . In practice , r p must satisfy st ruct ural requir em en ts . T he minim um val- ue of rp satisfying t hese require ments , deno ted by rp* , is given by one of t he following experi- m en tal st ructure formulae : r p* > 1 .1 r s + r R if t he ca m and t he shaf t ar e t he sa me elem en t , as show n in Fig . 5-25a , or r p* > ( 1. 6 ~ 2 .0 ) r s + r R ( 5-17b) if t he ca m and the shaft a re tw o different m achine elem en ts an d connected by a key , as sho wn in Fig . 5-25b . r s in t he i nequalities ( Eq . 5-17 a and b ) is t he radius of t he shaft .     D urin g t he design of a cam , rp* is first chosen as a t rial value for r p . D uri ng t he calculation of t he pitch curve and t he ca m cont our using a compu ter
Fig . 5-25

( 5-17a )

progr am , α is also calculated at t he sam e tim e usi ng Eq . 5-16 and t hen t he value of α is checked . If αis greater t han [α] at som e δ, rp should be en larged or a suit able a moun t of pos- itive offset adop ted . T hen t he pitch curve , t he ca m con t our , and α should be calculated again from t he zero value of δ . W hen rp is large enough , α m a x will be less than [ α] . Not e t hat t he lengt h of t he guide-way should be designed t o be as long as possible and t he amoun t of cantilever of t he follower should be designed t o be as s hort as possible . O t her wise , t he follo wer m ay still ja m in t he guide- way even if α max < [α] . 5.3.7   Radius of Curvature     Anot her impor tan t factor affecting cam size and perform ance is ca m curva ture . The for- mula for t he radius ρ of curvat ure of a ca m profile is: ・ 89 ・

2 2 3/ 2 ( x′ + y′ ) ρ= - M x′ y″- x″ y′

( 5-18)

w her e : x and y a re t he coordinat es of t he ca m profile ; x′and y′ a re t he first derivatives of x and y wit h r espect to δ, respectively ; x″ and y″ are t he second derivatives of x and y wit h re- spect to δ, respectively .     If t he calculated ρ for a cam angle δ is positive , it m eans t hat t he corresponding section of t he ca m profile is convex ; ot her wise , it is concave .     F or a plate ca m wit h t ranslati ng roller followe r , diffe ren tiating Eq . 5-11 with r espect to δ, respectively , results in xB ′ , yB ′ , xB ″and yB ″. Substit uting t he m in to Eq . 5-18 yields an easie r formula to calcula te t he radius of curvat ure ρ B of t he pitch curve of a plate cam wit h t ranslating roller followe r : [ ( s0 + s ) + ( s′- Ne ) ] ρ B = - ( s0 + s) ( s″- s0 - s) + ( s′- Ne ) ( 2 s′- Ne )
2 2 3/ 2

( 5-19)

    As mentioned above , t he pitch curve , t he ca m con tour and t he locus of t he cen t re of t he milling cut ter of a pla te cam wit h rolle r followe r a re t hr ee pa rallel curves . O n t he convex sec- tion , t he radius of curvat ure ρT of t he ca m con t our is equal to ρ B minus r R , i . e .ρ T =ρ B - rR . If ρ B > r R , t hen ρ T > 0 an d t he cam con tour is clearly defined as show n in Fig . 5-26a . If ρ B = r R , t hen ρ T = 0 and t he ca m con tour becom es poi nted forming a“ cusp ”as show n in Fig . 5- 26b . N at ur ally , t his condition produces very high st resses and is to be avoided . We t her efore st rive to make t he mi nimum value of t he radius ρ T on t he convex section la rger t han an allo w- able value , usually 3 ~ 5 mm . In ot her w ords , on the convex section ( w he re ρ B > 0) , the fol- lowing inequality must be satisfied . ρ B > r R + ( 3~5 ) m m ( 5-20)     If ρ B < r R , as show n in Fig . 5-26c , t he envelope of t he fa mily of t he roller circles is t he curve A E BCED , w hich int ersects itself at E .The m aterial BCE will be removed af ter m achining .T he result is a poin ted ca m con tour from w hich t he in tended motion w ould not be achieved .T his is called m otion d istortion or u n dercu tti ng .So inequalit y ( 5-20 ) must be satisfied for all convex sections of the pitch curve .If a cusp occurs on t he convex section , the condition can be rectified by eit he r increasing r p or r educing r R .H ow ever t he use of a sm aller roller has the disadvan tage of causing higher st ress .Also , if t he roller is t oo sm all , t he pin of t he roller may no t be st rong enough .Usually , r R is chosen bet ween 0. 1 rp and 0. 15 r p .     O n t he concave section of t he pitch curve , t he corresponding cam con tour is also concave and will no t becom e poin ted ( Fig . 5-27 ) . H oweve r , if t he mi nimum value of t he absolute val- ue of t he radius of curvatur e on t he concave section of t he cam con tour , | ρ T | m i n , is less t han t he radius r C of t he m illing cu tte r , t hen t he locus A E BCE D of t he cen t re of t he milling cu t ter will int ersect itself at E and undercu tting will occur in the bott om of t he concave section of t he cam con tour , as t he cross-hatched area show n i n Fig . 5-27 . So | ρT | mi n must be greate r t han r C . In ot her words , on t he concave section ( W he re ρ B < 0) , following inequality must be sat- ・ 90 ・

Fig . 5-26

isfied . |ρ ( r C - r R )   or   ρ ( rR - rC ) B | > B < ( 5-21)     If under cut ting occurs on t he concave section , t he condition can be rectified by eit her increasing r p or reducing r C . Example 5-2     F or t he plate ca m wit h t ranslating offset roller followe r sho wn in Fig . 5-18 , a progra m is required
Fig . 5-27

t o calcula te t he co-ordinates of t he pitch curve , t he ca m con tour and t he locus of t he cen t re of t he m illing cu tt er , α, an d ρ B . Use t he data i n Sec . 5 .3 .1 . The follo wer has a roller of r adius r R = 20 mm . T he ca m con tour is to be cu t by a milling cut ter wit h r adius r C = 30 m m . Solution:     Since t he motion curves in t he rise and t he ret urn are 3-4-5 polynomial and sine accelera tion motion curves , respectively , Eq . 5-9 an d Eq . 5-5 a re used to calculate s , s′ , and s″in t he rise and t he ret urn , respectively . T he co-ordinates ( x B and yB ) of t he pitch curve for t his ca m a re calcula ted according t o Eq . 5-11 . F or the cam mechanism show n in Fig . 5-18 , t he values of M and N i n Eq . 5-11 should be : M = + 1 and N = + 1 . T he co-ordinates ( x T and y T ) of t he ca m con tour are calculated according to Eq . 5-14 . T he co-ordinates ( x C and yC ) of t he cen t re of t he milling cut ter can be calculated according t o Eq . 5-15 . α is calculated according t o Eq . 5-16 . ρ B is calculat ed according t o Eq . 5-19 .     T he progr am w ritten in T R U E BAS IC compute r language is as follows . R E M : AL P HA is pr essur e angle . RH OB is radius of curvat ure of t he pitch curve .     LET M = + 1 ・ 91 ・

    LET N = + 1     L E T R P = 100     L E T E = 40     L E T R R = 20     L E T R C = 30     L E T H = 80     L E T D E L T A 0 = 140 * PI/ 180     L E T D E L T AS = 40 * P I/ 180     L E T D E L T A 01 = 100 * P I/ 180     L E T D E L T AS1 = 80 * PI/ 180     L E T S0 = SQ R ( R P ∧ 2 - E ∧ 2)     FOR I = 0 T O 360 S T E P 2       L E T D E L T A = I * P I/ 180       I F DE L T A < = DE L T A0 T H E N         L E T D 2 = DE L T A/ D E L T A 0         L E T S = H * (10 * D2 3 - 15 * D2 4 + 6 * D2 5 )         L E T S1 = H * ( 30 * D2 ∧ 2 - 60 * D2 ∧ 3 + 30 * D2 ∧ 4 )/ DE L T A0         L E T S11 = H * (60 * D2 - 180 * D2 2 + 120 * D 2 3 )/ D E L T A 0 2       E LSE IF DE L T A < = ( D E L T A 0 + DE L T AS ) T H E N         LET S = H         L E T S1 = 0         L E T S11 = 0       E LSE IF DE L T A < = ( D E L T A 0 + DE L T AS + D E L T A 01 ) T H E N         L E T D 4 = ( D E L T A - D E L T A 0 - DE L T AS )/ D E L T A 01         L E T S = H * (1 - D4 + 1/ ( 2 * P I ) * SI N (2 * PI * D4 ) )         L E T S1 = - H/ DE L T A01 * ( 1 - COS (2 * PI * D4 ) )         L E T S11 = - 2 * PI * H/ D E L T A 01 ∧ 2 * SI N (2 * PI * D4)       E LSE         LET S = 0         L E T S1 = 0         L E T S11 = 0       E ND IF       L E T XB = M * ( ( S0 + S ) * SI N ( DE L T A ) + N * E * COS ( DE L T A ) )       L E T YB = ( S0 + S ) * COS ( D E L T A ) - N * E * S I N ( DE L T A )       L E T XB1 = M * ( S1 * SI N ( D E L T A ) + ( S0 + S ) * COS ( DE L T A ) - N * E * SI N ( DE L T A ) )       L E T YB1 = S1 * COS ( D E L T A ) - ( S0 + S ) * SI N ( DE L T A ) - N * E * COS ( DE L T A )       L E T KB = SQ R ( XB1 2 + YB1 2 ) ・ 92 ・
∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧

      L E T X T = XB + M * R R * YB1/ KB       L E T Y T = YB - M * R R * XB1/ KB       L E T XC = X T - M * RC * YB1/ KB       L E T YC = Y T + M * RC * XB1/ KB       L E T A L PH A = A T N ( ABS ( S1 - N * E )/ ( S0 + S ) )       L E T N U M E = ( ( S0 + S ) 2 + ( S1 - N * E) 2 ) ( 3/ 2)       L E T D E NO = - ( S0 + S ) * ( S11 - S0 - S ) + ( S1 - N * E) * (2 * S1 - N * E )       L E T R HOB = N U M E/ D E NO       P RI N T I , XB , YB , X T , Y T , XC , YC , AL P HA * 180/ PI , R HOB     N EX T I     E ND     Some of the ou t put dat a a re listed in T able 5-3 for refer ence . Table 5-3   Some output data of Example 5-2
I (δ) 110 240 xB 142. 40 - 120. 60 yB - 94. 395 - 23. 443 xT 124. 15 - 101. 44 yT - 86.201 - 29.162 xC 151.52 - 130.19 yC - 98.492 - 20.583 α(° ) 4. 186 46. 618 ρ B 114 .77 175 .73
∧ ∧ ∧

5.4   P l a t e C a m w i t h O s c i ll a t i n g R o l l e r F o l l o w e r
    In t his mechanis m , as s ho wn in Fig . 5-28 , t he followe r oscillat es according to

t he given motion curve w hile t he cam rota tes con ti n- uously . T he angular displacem en t of t he follo wer is t he angle φ .The angula r lift of the follower is φm a x . T he motion equations derived in Sec .5 .2 can also ap- ply t o t he oscillating follo wer so long as s is replaced by φ an d h by φm a x . For exam ple , t he φ - δ formula for t he rise of t he oscillating followe r wit h sine accel- er ation motion curve can be tr ansla ted from Eq .5-4 as follo ws . φ = φm a x δ 1 si n δ 0 2 π δ 2π δ 0

5.4.1   Graphical Synthesis     Before designing t his cam , the following da ta should be provided : t he r adius of prime cir cle (or t he smallest r adius of pitch curve ) r p , t he distance lO A
Fig . 5-28

bet ween t he two pivots of t he ca m and t he follower , t he lengt h l A B of t he follo wer a rm , t he ・ 93 ・

initial stat us of t he follo wer ( left or righ t ) , t he direction of ω, t he mo tion curve φ - δ, and t he radius of roller r R . Similar to t he graphical m et hod used for t he tr ansla ting follo wer , t he cam is held stationary w hile t he fra me and t he follo wer ar e inver ted aroun d t he cent re O of t he cams haft in t he dir ection opposite to ω . During inve rsion , t he pivot A of t he follow er moves in the direction opposite to ω on a circle wit h cen t re at O , w hich is called t he p ivot circle , i .e. , ∠ A 0 O A = δ . At t he sam e tim e , t he follow er swi ngs ou twa rd t h roug h an angle φ corre- sponding t o t he δ rela tive to t he fra me according to t he know n mo tion curve , i. e. , ∠ B′ AB = φ an d l A B′ = lA B = l A 0 B 0 . No te : T he t hree poin ts O , B′ , and B , a re not located on sa me st raight line . T he locus of t he roller cen t re B relative to t he cam is called t he pitch curve , show n as t he dot -and-dash curve in Fig . 5-28 . T he ca m cont our is tangen t t o all roller circles . D raw m any rolle r circles wit h radius r R and cen t re at poin ts on t he pitch curve . T he cam con- tour is ob tained by drawing a smoot h curve tangen t to t he family of all rolle r cir cles , as t he sol- id curve show n i n Fig . 5-28 .     Since t he followe r arm must have som e ph ysical dim ensions , t he side edge of t he st raigh t a rm of t he follower m ay in terfere wit h t he ca m con tour . I t is necessa ry t o check t hat t his will not occur at all positions . To avoid in terference , t he arm may be designed as a ben t rod , as show n in Fig . 5-5e . 5.4.2   Analytical Synthesis of Pitch Curve     T he procedure t o syn t hesize the pitch curve analytically is closely parallel to t hat in t he graphical m et hod . A car tesian co-ordinate system is fixed wit h t he ca m . I ts origin is located at t he cen t re O of t he ca mshaft and its y axis coincides wit h t he cen t re line O A 0 w hen δ = 0 , as show n i n Fig . 5-28 . T he angle φ 0 bet ween t he lo west position A 0 B 0 of follower a rm and t he cen t re line O A 0 can be calcula ted as
2 2 l2 O A + l A B - rp φ 0 = arccos 2 lO A l A B 2 2 2 (2 lOA lA B ) 2 - ( l2 O A + l A B - rp ) 2 2 2 ( lO A + l A B - r p )

= arct an

( 5-22)

    As we learned during gr aphical syn t hesis , ∠ A 0 O A = δ, ∠ O A B′ =φ 0 , ∠ B′ A B = φ and l A B′= l A 0B 0 = lA B . T hus t he co-ordinates of t he rolle r cen t re B , or t he equations of t he pitch curve , can be derived in Fig . 5-28 . x B = O E - BG = M [ l O A si n δ+ Nl A B si n ( φ δ) ] 0 + φ- N y B = A E - A G = lO A cosδ - l A B cos ( φ δ) 0 +φ- N cien t of t he followe r . If t he cam rotates coun ter-clock wise, M = + 1 ; ot her wise, ( 5-23)

w her e M is t he coefficien t of t he direction of t he ca m ro tation and N is t he oscillating coeffi- M = - 1 . If t he swing direction of t he follo wer in t he rise is t he sa m e as t hat of t he ca m ( as show n i n Fig . 5-29 and Fig . 5-36 ) , N = + 1 ; ot her wise ( as show n in Fig . 5-28) , N = - 1 . 5.4.3   Analytical Synthesis of Cam Contour and Locus of the Centre of Milling Cutter     T he pitch curve, t he ca m con tour and t he locus of t he cent re of t he milling cu tter of a plate ・ 94 ・

cam wit h rolle r follo wer a re t hree parallel curves , no m atte r w het her t he follower tr anslates or oscillates . T herefore , for t he plate ca m wit h oscilla ting rolle r follow er , Eq . 5-14 can also be used to calcula te t he co-ordinates ( x T and y T ) of its cam cont our . For t he sa me reason , t he lo- cus of t he cent re ( x C and yC ) of t he milling cu t ter can also be calculated by Eq . 5-15 . Note t hat , for t he oscilla ting roller followe r , t he x B ′ and yB ′ar e not t hose in Eq . 5-13 . T hey m ust be derived by differ en tiating the x B and yB in Eq . 5-23 wit h r espect to δ, respectively . 5.4.4   Radius of Curvature of the Pitch Curve     Differentiatin g x B and yB in Eq . 5-23 wit h r espect to δ results i n xB ′ , yB ′ , xB ″and yB ″ , t hen substit uting t he m in to Eq . 5-18 yields an easier formula for t he radius of curvat ure ρ B of t he pitch curve of t he plate ca m wit h oscillatin g roller follow er :
2 3/ 2 [ l2 φ ′ ) 2 - 2 l O A l A B ( 1 - Nφ ′ ) cos ( φ O A + lA B ( 1 - Ν 0 + φ) ] ρ B = 2 2 3 l O A + lA B (1 - Nφ ′ ) - l O A lA B K

( 5-24)

w her e   K = ( 1 - Nφ) ( 2 - Nφ) cos ( φ 0 + φ) + φ ″sin ( φ 0 + φ) .     Similar to the plate ca m wit h tr anslati ng roller follow er , it is necessa ry t o check t he pres- ence of cusps and undercu tting of t he cam con tour usi ng t he inequalities 5-20 and 5-21 . 5.4.5   Pressure Angle α     In t he absence of friction , t he for ce F exer ted by t he ca m on t he follo wer acts along t he norm al n-n to t he pitch curve at t he roller cen tr e . By definition , the pressure angle α of a plate cam wit h oscillating roller follower is t he acute angle bet ween the norm al n-n to t he pitch curve a t t he roller cen tr e and t he line perpendicular to t he follo wer cen t reline A B , as show n in Fig . 5-28 . According to the t heore m of th ree-cen t res , t he in te rsection poin t P bet ween t he norm al n-n and t he fra m e O A is t he i nstan t cen tr e bet ween t he ca m 1 and t he follow er 2 , i .e . , v P 1 = v P 2 , or ω1 l O P = ω 2 lA P .     Since ω 1 N ( lA P - lOA ) =ω 2 lA P , lA P = φ wit h r espect to δ .     A line PD is draw n from t he poin t P pe rpendicular to t he follo wer a rm A B , as show n in Fig . 5-28 . In righ t t riangle PB D , ∠ B PD = α . tan α= lB D lA B - l A D l A B - lA P cos ( φ 0 + φ) = = lP D lPD l A P sin ( φ 0 + φ) | lA B (1 - Nφ) - l O A cos ( φ 0 + φ) | l O A sin ( φ 0 + φ) ω 1 lO A lOA = w here φ ′is t he derivative of ω1 - Nω 2 1 - Nφ ′

Substit uting l A P in t o t he last formula will derive t he general formula to calculate α . tan α= ( 5-25)

    During t he calculation of t he pitch curve and t he cam con tour using a compu ter progra m , α is also calculated at t he sa me time using Eq . 5-25 and t hen t he value of α is checked . T he value of [α] is listed in T able 5-2 . If α is grea ter t han [α] at some δ, rp and/ or lA B s hould be ・ 95 ・

changed . T hen t he pitch curve, t he ca m con tour , and α should be calculated agai n from t he ze ro value of δ . No te t hat , for t he oscilla ting rolle r follow er , α m a x m ay not be decr eased if on- ly r p is enla rged . T here is an optim al lA B correspondin g t o each rp so that α m a x is mi nimised . If rp is large enough and lA B is t he op timal lengt h correspondi ng t o t he r p , t hen α m a x will be less t han [α] . Example 5-3     Show n in Fig . 5-29 is a circula r cam wit h an oscillating rolle r followe r A B . T he ca m ro- tates abou t ca mshaft O clock wise . Indicate r p , δ 0, δ s, δ ′ 0 , δ ′ s , φ m a x , and α of t he mecha- nism at t he position show n . Solution:     Sim ilar t o Exa mple 5-1 , t he pitch curve must be dr aw n first since t he r equir ed par am eters can no t be found without t he pitch curve . T he pitch curve of t his cir cula r cam is also a circle , as t he dot-and-das h cir cle show n in Fig . 5- 29 . T he sm allest radius of t he pitch curve is t he ra- dius of t he prim e circle, i .e. , lO B 0 = r p .     F rom t he principle of inversion , the pivot A of t he oscillati ng follower moves on t he pivot circle in t he direction opposite to ω w hile t he roller cen- t re B moves on t he pitch curve in t he direction op- posite to ω . N ote t hat t he cen t re of t he pivot circle is t he cen tr e O of t he ca mshaft , not t he cir cle cen- t re D . When t he roller cent re B a rrives a t poin t B 0 and B1 , t he distance between t he rolle r cen tr e and t he ca mshaf t is t he sm allest and t he la rgest ,
Fig . 5-29

respectively , or t he follower is at t he lowest and highest position , respectively . W hen t he rolle r cent re arrives at point B 0 and B 1 , t he corresponding position of the followe r pivot is lo- cated at poin ts A 0 and A 1 , r espectively . W he re : l A 0B 0 = l A 1 B 1 = lA B . W hen t he roller cen tre moves from t he poin t B 0 to t he poi nt B 1 along t he pitch curve in t he direction opposite to ω, t he fra me rotates from O A 0 to O A 1 in t he sam e direction . So ∠ A 0 O A 1 = δ 0 . There is neit her ou ter nor inner d well , so t he r em ainin g angle is δ ′ 0 , as s ho wn in Fig . 5-29 . N ote : Alt hough t he half circle B 0 B B 1 of t he pitch circle is a curve for t he rise , δ 0 ≠∠ B 0 OB 1 = 180° .     Position A 1 B 1 is t he highest position of t he followe r . Let t he oscillating follo wer swing back around t he pivot A 1 . W hen t he rolle r cen t re con tacts wit h t he pri me circle at poin t B 1 ′ , t he follo wer is at its lowest position . So ∠ B 1 A 1 B1 ′ = φm a x .     T he normal t o t he circular pitch curve passes t hrough t he circle cen t re D . T he pressure ・ 96 ・

angle α of t he m echanism at t he position show n is indicated in Fig . 5-29 accordin g to t he defi- nition of t he pressure angle .

5.5   P l a t e C a m w i t h T r a n s l a t i ng F l a t - f a c e d F o ll o w e r
    For t he plate cam wit h t ranslating flat-faced follow er as show n in Fig . 5-30 , if t her e are t wo pa rallel sliding pairs between t he follo wer and the fra me , t hen one of t he guide-ways is a redundan t const rain t . The motion of t he follower re mains t he sam e no m atte r w hether the fol- lower is in-line or offset . It is convenient t o design t his kind of plate cam according to t he in- line configur ation . No te t hat t his is no t t rue for t he translating roller followe r . 5.5.1   Graphical Synthesis     Sim ilar to t he roller follow er , t he int ersection B of t he cen t reline of t he follo wer ste m and the face of t he follow er is tr eat ed as the knife-edge of a vir tual knife-edge follo wer w hich is fixed to t he flat- faced follo wer . In t his way , t he motion curve of t he flat- faced followe r applies also to t he vir tual k nife-edge follow - er . T he design m et hod for t he t ranslating in-line kn ife- edge followe r m en tioned in Sec . 5 .3 .1 can be used to de- sign t he locus of t he knife-edge of t he virt ual kn ife-edge followe r r elative to t he ca m ( refer t o Fig . 5-17) , show n as t he do t-and-dash curve in Fig . 5-30 . T he locus is also t he locus of t he in te rsection B relative to t he cam , w h ich is called t he pitch curve . I n Fig . 5-30 , ∠ B 0 O B = δ and B′ B = s . Since t he fla t face of t he follo wer is always per- pendicula r to t he cen tr eline of t he followe r , a line is draw n a t each of t hese poin ts B pe rpendicula r to t he cen- tr eline of t he follower to represen t t he position line of t he flat face of t he follo wer during inver- sion . D uri ng i nversion , t he ca m con tour is always tangen t to t he flat face of t he follo wer . So t he ca m con tour m ay be draw n as a smoot h curve tangent to t he fa mily of t he position lines of t he flat face, show n as t he solid curve in Fig . 5-30 . As can be seen , t he ca m con tour is t he lo- cus of tangen t poin t T relative t o t he cam . 5.5.2   Analytical Synthesis     Similar to the plate ca m wit h tr anslati ng roller follow er , a Ca rtesian co-ordinate system is fixed wit h t he ca m . I ts origin is loca ted at t he cen t re of t he cam shaft and its y axis coincides wit h t he cen tr eline A 0 B 0 of t he follo wer w hen δ= 0 , as show n in Fig . 5-30 . Since t he pitch curve of t he plate cam wit h t ranslating flat -faced follower is t he sam e as t hat of t he plate ca m ・ 97 ・
Fig . 5-30

wit h t ranslating in-line roller ( or knife-edge ) follo wer , Eq . 5-12 can be used to calculate t he co-ordi nates of t he pitch curve .     In Fig . 5-31 , t he in tersection D of t he common normal passing t hrough t he t angen t poin t T an d t he line P1 3 P 2 3 is t he i nstan t cent re P 1 2 of t he ca m 1 and t he follower 2 . So v D 1 = v D 2 , or ω1 l O D = v 2 . Therefore , lOD = v 2 d s/ d t d s = = = s′ ω 1 dδ / d t dδ l O D = s′ , and lB′ B = s .

    In Fig . 5-30 , ∠ B0 O B = δ,

So lD T = l O B = rp + s . The co-ordinates ( x T and y T ) of t he t angen t poin t T , or t he equations of t he cam con tour , can be derived by a met hod sim ilar to that used to derive Eq . 5-11 . x T = M [ ( rp + s) sin δ+ s′ cos δ] y T = ( rp + s ) cos δ - s′ sin δ If t he ca m rotates coun ter-clockwise, M = + 1 ; ot herwise M = - 1 . ( 5-26)
Fig . 5-31

    As distinct from t hose of plate cam s wit h rolle r follo wer , t he pitch curve and t he ca m con- tour of t he plate ca m wit h flat-faced followe r ar e not t wo pa rallel curves . In t he sections of out - er or in ner d well , l B T = l O D = s′ = 0 . T he poin t B coi ncides wit h t he tangen t poi nt T . T here- fore , t he pitch curve and t he ca m con tour are coincident in t hese two sections . In t he sections of the rise and t he ret urn , lB T = l O D = s′ ≠ 0 . T he poin t B does not coincide wit h t he tangen t poin t T . T her efor e, the pitch curve and t he ca m con t our are not coincident i n t hese t wo sec- tions . 5.5.3   Locus of the Centre of Milling Cutter     For a flat-faced follo wer , t he locus of t he cen t re of the milli ng cut ter is an ou ter envelop par allel to t he ca m cont our , not to t he pitch curve . T he formulae for t he locus of t he cen t re C of t he milling cu t ter can t he refore be derived direfctly by t ransforming Eq . 5-15 , x C = x T - M r C y T′ / KT yC = yT + Mr C xT ′ / KT w her e x T and y T a re t hose in Eq . 5-26 and K T = ( 5-27)

2 2 y′ ′ and y T ′ can T + x′ T . T he formulae of x T

be derived by differ en tiating Eq . 5-26 wit h respect to δ, r espectively . 5.5.4   Radius of Curvature of Cam Contour and Radius of Prime Circle     T he value of α for t his mechan ism is always zero , as show n in Fig . 5-31 . T her efor e, as distinct from t he plat e cam wit h t ranslating roller followe r , r p
mi n

is not restricted by [ α] .

Ho wever , if r p is t oo sm all , it will be impossible to design a cam cont our t o tr ansmit t he de- sired motion . F or example, in graphical drawing , w hen a sm all r p is used , the flat face of t he ・ 98 ・

followe r occurs at lines 1 , 2 , 3 , etc ., as show n in Fig . 5-32 . A practical cam con tour can- not be draw n tangen t to all t hese flat faces , since t he li ne 2 lies ou tside t he i nte rsection of t he lines 1 and 3 . T he cam con tour show n on t he compute r screen would be an in tersecti ng curve A E BCED . T he curve BC is concave . So t he act ual cam con tour A E D aft er machi ning cannot drive t he follo wer in to position 2 , i .e . , t he follow er cannot pe rform t he desired mo tion .     T he concavit y and int ersection of the ca m cont our can be checked by observing t he cafm con tour on t he compute r scr een or by calculating t he radius of curvat ur e ρ Τ of t he ca m con tour . Dif- fe ren tiating Eq . 5-26 wit h respect t o δ results in x T ′ , yT ′ , x T″ and y T″. Substit u tin g t he m in t o Eq . 5-18 yields an easier formu- la for ρ Τ: ρT = r p + s + s″ ( 5-28)
Fig . 5-32

w her e s″is t he quasi-acceleration w h ich is negative in t he second half of t he rise an d t he first half of t he ret urn . F rom Eq . 5-28 we can see t hat , if r p is too small, ρT may be nega tive at som e cam angles . In t hese cases , t he ca m cont our will be concave , just as t he curve BC in Fig . 5- 32 . T he refor e , in order to avoid concavit y and i nte rsection , rp must be large enough so t hat ρ T is always positive . In practical use, all ρ T should be larger t han an allow able value [ ρ T ] . y , [ρ T ] = (3 ~ 5) mm . rp is chosen by t he experi men tal struct ure formula Eq . 5- 17 . During t he calculation of t he co-ordinates of t he ca m con tour by compute r , ρ T is also calcu- lat ed at the sa m e ti me . If ρ T is less t han [ ρ T ] at some δ , rp should be enla rged and t hen t he cam con tour and ρT should be calculated agai n from t he zero value of δ . When r p is large enough , all ρ T will be la rger t han [ ρ T ] for all values of δ an d all sections of t he ca m con tour will be convex . 5.5.5   Width of the Flat Face     The wid t h of t he fla t face should be la rge enough to allow t he ca m con tour to be tangen t to t he flat face at any time ( refer to Fig . 5-30 ) . The flat face usually has a circular boundary w hose radius r F must be (3 ~ 4) m m large r t han t he maximum dist ance lB T ) m a x from any t an- gen t poin t T to t he poin t B (r efe r to Fig . 5-31) . T his distance can be easily picked off in t he cam dra wi ng . H owever , it can also be found easily by t he analy tical met hod . As show n in Fig . 5-31 , lB T = l O D = | s′ | . During t he calculation of t he coordina tes of t he cam con tour by comput er , all values of s′ correspondi ng t o all δin t he rise and t he r eturn have been calculated . T hen t he m axim um value of all | s′ | , | s′ | m a x , can be found .T he act ual value of r F should be 3 ~ 4 m m la rger t han | s′ | max .


5.6   P l a t e C a m w i t h O s c i ll a t in g F l a t - f a c e d F o l l o w e r

    In t his cam mechanism ( Fig . 5-33 ) , the follo wer oscillates according to a given ・ 99 ・

motion curve w hile t he ca m ro tates con ti nuously . If t he ex tension of t he flat-face of t he follow er pass- es t hrough t he pivot A of t he follower , t he follo w- er is called an i n-li ne oscill ati ng f l at f aced follo w- er ; o ther wise , an o f fset oscill at in g f l at - f aced f ollower . T he perpendicula r distance from t he pivot A t o t he ex tension of t he flat face of t he fol- lower is called t he o f fset , denoted by e . T he cir- cle cent red at t he pivo t A wit h t he offset e as a ra- dius is called o f fset circle . 5.6.1   Analytical Synthesis of the Cam Contour     Before syn t hesizing t he ca m con tour , the fol- lowing data should be provided : rp , t he leng t h l O A of t he fra me , e, t he initial status A 0 E 0 T 0 of t he follo wer , the direction of ω, and t he mo tion curve φ - δ of t he follo wer .
Fig . 5-33

    To help t he syn t hesis of t he ca m con tour , an auxilia ry line A H par allel to t he fla t face E T is fixed t o t he follo wer , as s ho w n i n Fig . 5-33 . A Car tesian co-ordinate system is fixed to t he cam . I ts origi n is locat ed at t he cent re O of t he ca mshaft and its y axis coincides wit h t he fra me O A 0 w hen δ = 0 , as show n in Fig . 5-33 . T he initial angle φ 0 betw een t he flat face of t he follo wer and t he fr am e O A can be calculated in righ t t riangle O A 0 H 0 . φ 0 = arcsin rp - Ze = ar ct an lO A rp - Ze
2 l2 O A - ( r p - Ze )

( 5-29)

w her e Z is t he coefficien t of in tersection . If t he line E T in te rsects t he line A O at a poi nt be- t ween A and O , as sho w n in Fig . 5-33 , Z = - 1 . O t her wise , as show n i n Fig . 5-34 , Z = +1 .     W it h t he principle of inve rsion , t he cam is held stationa ry w hile t he fr am e and t he flat- faced follower are invert ed a round t he ca m in t he dir ection opposite t o ω, i .e. , ∠ A 0 O A = δ and l O A = lO A 0 .     Since the line A H rem ains par allel to t he flat face E T at all tim es , t he mo tion curve of t he followe r applies not only to t he flat face E T but also t o t he line A H . During inver- sion , t he flat face E T and t he li ne A H swing an angle φ out ward relative to t he fra m e O A ac- cordi ng to t he given motion curve , i.e ., ∠ O A H = φ 0 + φ.     No w t he flat face E T of t he followe r is tangen t t o t he cam con t our at poin t T . A norm al t o t he flat face t hrough t he tangent poin t T in tersects wit h t he ex tension li ne of A O a t poin t P . Accordi ng to t he t heore m of t hree cen t res men tioned in Sec . 3 .2 .4 , t he poin t P is t he in- ・ 1 00 ・

stan t cent re of the cam 1 and t he follo wer 2 , i .e. , v P 1 = v P 2 , ω 1 l OP = ω 2 lA P . ω1 N ( l A P - l O A ) = ω2 l A P w here : N is t he oscillation coefficien t of t he follower . If t he swing direction of t he oscillating followe r in t he rise is t he sam e as t hat of t he ca m , as show n in Fig . 5-33 , N = + 1 . O t her wise , as show n in Fig . 5-34 , N = - 1 . Therefore we have lA P = lO A ω1 lOA = ω1 - Nω 2 1 - Nφ ′

w her e φ ′ is t he derivative of φ with respect t o t he ca m angle δ .     Since ∠ A 0 O A = ∠ O A G = δ and ∠ O A H = φ 0 + φ, ∠ E A D = ∠ JT E = ∠ H A G = ∠ O A G - ∠ O A H = δ - φ0 - φ . In right t riangle A P K , one can derive t he following l E T = l A K = l A P cos ( φ 0 + φ) = lO A cos ( φ0 + φ) 1 - Nφ ′ ( 5-30)
Fig . 5-34

    N ow t he x and y co-ordi nates ( x T and y T ) of t he t angen t poin t T on t he cam con tour cor- respondin g to t he ca m angle δ, or t he equations of t he ca m con t our , can be derived as follow . x T = - lA D + l A I + l E J = M { l O A sin δ+ N Ze cos [δ - N ( φ 0 + φ) ]                   - lO A cos ( φ 0 + φ) sin [ δ - N ( φ 0 + φ) ]/ ( 1 - N φ ′ )} y T = l O D + l E I - l J T = l O A cos δ - N Ze sin [ δ - N ( φ 0 + φ) ]                 - l O A cos ( φ0 + φ) cos [ δ - N ( φ0 + φ) ]/ ( 1 - Nφ ′ ) w her e M is t he coefficien t of t he direction of t he cam rota tion . If the ca m ro tates coun ter clock wise , M = + 1 , ot he rwise , M = - 1 . 5.6.2   Locus of the Centre of Milling Cutter     T he locus of t he cen tr e of t he milling cu tte r is t he ou ter envelope of t he ca m con tour . T he refore , Eq . 5-27 can also be used t o calculate t he x an d y co-ordinates ( x C an d y C ) of t he locus of t he cen tr e C of t he milling cu tte r for t his ca m . Not e: for t his cam , 5-27 should be t hose in Eq . 5-31 , not t hose in ot her formulae . 5.6.3   Pressure Angle α     By definition , α is t he acu te angle bet ween the direction of t he velocity v T of t he tangen t poin t T and t he normal K T to t he flat face , as show n in Fig . 5-33 . In righ t t riangle A T E , ∠ A T E is also equal to α . α= ∠ A T E = arct an lA E lE T = arctan e( 1 - Nφ ′ ) l O A cos ( φ 0 + φ) ( 5-32) x T and y T in Eq . ( 5-31)

5.6.4   Radius of Curvature of the Cam Contour     Diffe ren tiating x T an d y T in Eq . 5-31 wit h respect to δ, respectively , results in x′ T, x″ T, ・ 1 01 ・

y′ T and y″ T . Substitu ting t hem int o Eq . 5-18 r esults in an easier form ula for ρ T as follows: ρT = lO A [ ( 1 - Nφ ′ ) (1 - 2 Nφ ′ ) sin ( φ ″ cos ( φ 0 + φ) + φ 0 + φ) ] + Ze 3 (1 - Nφ ′ ) ( 5-33)

w her e φ, φ ′and φ ″a re t he angular displacem en t , quasi-angula r velocit y , and quasi-angular acceleration of the follo wer , respectively .     D urin g syn t hesis of t he ca m cont our of a plate ca m wit h flat-faced followe r , t he ca m con- tour must be convex in all section s , i .e. t he r adius of curvat ure ρ T of t he ca m cont our must be positive for any δ . In practical design , all values of ρ T should be la rger t han an allowable value [ρT ] . Usually , [ρ = 3 ~ 5 mm . T ] 5.6.5   Length of the Flat Face of the Follower     To ensure t he tangency bet ween t he flat face and t he cam con tour , t he flat face of t he fol- lower must be large enough . D uri ng t he calculation of t he co-ordinates of t he ca m con tour , t he distance l E T bet ween tw o tangen t poin ts E and T is also calculat ed at t he sam e ti me using Eq . 5-30 . The m ax imum value ( l E T ) m a x of all l E T values can be found easily . T he actual leng t h of t he flat face should be 3 ~ 5 mm longer t han ( l E T ) m a x . Problems and Exercises 5-1   Compare t he advan tages and t he disadvan tages of ca m mechanisms and lin kage mecha- nisms . 5-2   Recoun t t he advan tages and t he disadvan tages of t he following kinds of follo wer : knife- edge followe r , roller follow er , flat -faced follo wer . 5-3   W hat is rigid impulse ? What is sof t im pulse ? 5-4   De rive equations for s , s′ , and s″for a rise wit h polynomial motion t o satisfy t he follo w- ing bounda ry conditions: s = 0 , s′ = 0 , s″ = 0 , s = 0 w hen δ = 0 ; and s = h , s′ = 0 , s″ = 0, s = 0 w hen δ = δ 0 . 5-5   F or t he plate ca m wit h t ranslating offset rolle r followe r , is t he angle between two r adius vect ors at tw o ends of t he r eturn section of t he pitch curve equal t o δ ′ ? Is t he difference be- 0 tween t he maximum radius and t he minim um r adius of t he pitch curve equal t o t he lift h of t he followe r ? 5-6   W hy should we specify a value of [α] ? W hy is [α] in t he rise less t han t hat in t he ret urn for a force-closed ca m m echanism ? Why is [α] of t he t ranslating rolle r follo wer in t he rise less t han t hat of the oscillating roller followe r i n t he rise ? 5- 7   For t he plate ca m wit h tr ansla ting roller follower , w hat is positive offset configuration ? W hat m easure can be adopt ed t o reduce t he m axi mum pressure an gle α m a x in t he rise ? Is t he sta tem en t“ The la rger t he positive offset , t he bet ter ”correct ? 5-8   How is r p followe r ? ・ 1 02 ・
mi n

chosen initially ? H ow is rp checked in t he plat e cam wit h diffe ren t kinds of

5-9   For t he plate ca m wit h t ranslating offset roller follower as show n in Fig . 5-35 , arcs G H and IJ ar e t wo arcs wit h cen tr e at O . Indicate radius of pri me circle rp , offset e , ca m angle for rise δ 0 , cam angle for ou ter dwell δ s , cam angle for ret urn δ ′ , ca m angle for inner dw ell δ ′ 0 s , and lift h . F or t he position show n , indicate pressure angle α, displacem en t s and t he corr esponding cam angle δ . 5- 10   U nde r w ha t conditions will t he cam con tour becom e poin ted on t he convex section of a plat e ca m wit h roller follo w- er ? H ow can t his be rectified ? W hat should be consider ed w hen t he value of r R is t o be chosen ? 5- 11   U nder w hat conditions will undercu tting occur on t he concave section of t he cam con tour of a plate ca m wit h roller follo wer ? H ow can t his be recti- fied ? 5-12   A plate ca m wit h tr anslati ng offset rolle r follower similar to t hat i n Fig . 5-35 is t o have t he follo win g mo tion : a rise t hrough lif t h = 40 m m wit h a sine acceleration motion curve dur- ing δ 0 = 160° , δ s = 40° , a ret urn wit h a 3-4-5 polynomial motion curve during δ ′ 0 = 80° , and δ ′ s = 80° . T he dimensions a re to be : rp = 40 mm . r R = 12 m m , e = 12 m m and r C = 25 m m . ( 1) Const ruct t he pitch curve and t he ca m cont our gr aphically wit h a scale of 1 ∶1 . L abel in red in k the cent reline of the follo wer , s , t he roller and α corresponding t o δ = 70°an d δ= 0°. (2 ) Write a progra m to calculat e t he co-ordinat es of the pitch curve , t he ca m con tour and t he locus of t he cen tr e of t he milling cu t ter , α, and ρ Β . Show t he t hree curves on a com- put er screen . ( 3) Find t he maximum pressure angle ( α rise ) m a x in t he rise and t he maximum pr essure an- gle ( α r e t ur n ) m a x in t he r eturn , r espectively . ( 4) If t he offset e = 0 , find ( α ri se ) m a x and ( α r e t ur n ) m a x , respectively . (5 ) If t he follow er is offset at t he o t her side wit h t he sam e a moun t of offset , find (α r ise ) m a x and ( α r e t u rn ) m a x , r espectively . ( 6) Compa re ( α rise ) m a x and ( α r e t u rn ) m a x between problem s ( 3) , (4 ) , and ( 5) . ( 7) If δ ′ and δ ′ and r C = 50 mm , calculate ρ . 0 = 30° s = 130° Β corresponding to δ of 227° Sho w t he locus of t he cen t re of t he m illing cut ter on a compu ter screen . Does t he locus in ter- sect itself ? ( 8) If δ ′ 0 = 30°and δ ′ s = 130° and r R = 35 m m , calcula te ρ Β corr esponding to δ of 205° . Sho w t he cam con tour on t he compute r screen . Does t he ca m cont our in tersect itself ? 5-13   Will t he motion curve change i n a pla te cam wit h t ranslating rolle r follo wer if one of fol- lowing changes happens ? ( 1) t he direction of cam rota tion chan ges w hile t he pitch curve r em ains unchanged . ・ 1 03 ・
Fig . 5-35

( 2) t he radius of t he rolle r changes w hile t he ca m con tour rem ains unchanged . (3 ) bo th t he ca m con t our and t he radius of roller change w hile t he pitch curve re mains unchanged . ( 4) t he direction of offset changes w hile t he pitch curve and t he a moun t of offset re main unchanged . 5-14   Wh y should t he a rm of the oscillating roller followe r in som e ca m m echanisms be designed as a ben t rod , as sho w n in Fig . 5-5e ? 5-15   M ust α m a x be reduced if on ly r p is en larged in a plate cam wit h os- cillating roller follo wer ? Ho w can α m a x be reduced ? 5-16   F or t he plate ca m wit h oscillating roller follower as sho w n in Fig . 5-36 , a rcs G H and IJ ar e tw o arcs wit h cen tr e a t O . Indicate radius of prime circle r p , ca m angle for rise δ 0 , cam angle for oute r d well δ s, cam angle for ret urn δ ′ , ca m angle for inner d well δ ′ 0 s , and angula r lift φ m a x . For the position show n , indicat e pressure angle α , angular dis- placem en t of followe r φ and t he corresponding ca m angle δ . 5-17   A plate ca m wit h an oscillating rolle r follower similar to t hat in Fig . 5-36 is t o have t he following motion : an angular lift φ m a x = 20°with a sine accelera tion motion curve duri ng δ 0 = 130° , δ s = 30° , a ret urn wit h a 3-4-5 polynomial motion curve during δ ′ 0 = 140° , and δ ′ s = 60° . T he dim ension s a re to be: rp = 40 m m , l O A = 80 mm , lA B = 76 mm , r R = 12 mm , and r C = 16 mm . ( 1) Const ruct t he pitch curve and t he ca m cont our gr aphically wit h a scale of 1 ∶1 . L abel in red ink t he fra me O A , φ, cen tr eline A B of t he follower , t he roller and α corresponding to δ= 60°and δ= 0°. (2 ) Write a progra m to calculat e t he co-ordinat es of the pitch curve , t he ca m con tour and t he locus of t he cen t re of t he milling cu tter , t he pressure angle α, and t he radius of curva- t ure ρ B of t he pitch curve . Sho w t he t hr ee curves on a compu ter scr een . (3 ) If δ 0 ′ = 30° , δ ′ s = 170°and r R = 30 mm , calculate ρ B corresponding to δ = 165° . Sho w t he cam con tour on a compu ter screen . Does t he ca m con tour int ersect itself ? (4) If δ ′= 30°and δ ′ 0 s = 170°and r R = 12 mm and r C = 100 mm , calculate ρ B corre- sponding to δ= 184°. Show t he locus of t he cen t re of t he millin g cut ter on a comput er screen . Does t he locus in te rsect itself ? 5-18   F or a plate ca m wit h t ranslating fla t-faced follo wer , answer the follo wi ng questions :     (1 ) Suppose t hat t he ca m con tour is designed for t he in-line configuration accordi ng to a given motion curve bu t t he follo wer is asse mbled in a positive offset configuration . Does t he motion curve change ? Compare t he forced conditions in t he t wo configurations .       ( 2 ) A re t he pitch curve and t he ca m con tour t wo pa rallel curves ? Why does t he ca m   (3 ) Is rp ・ 1 04 ・ dete rmined by [α] ? By w hat is t he rp dete rmined ? cont our coincide wit h t he pitch curve in the sections of t he ou ter dw ell and t he in ner d well ?
m in m in

Fig . 5-36

 

 

( 4 ) S hould t he wid t h of t he flat face of t he follo wer be

enlarged , if the ca m is redesigned wit h a larger r p and t he sa me mo- tion curve ? Why ?       ( 5 ) U nder w hat conditions will t he ca m con tour i nte rsect it -   (6 ) W hat happens if t he wid t h of a flat face is too shor t ? self ? H ow m ay t his be rectified ? 5-19   For t he plate ca m wit h t ranslating flat-faced follower as sho w n in Fig . 5-37 , arcs GH and IJ ar e tw o arcs wit h cen t re at O . Indicate radius of prim e circle rp , ca m angle for rise δ 0 , ca m angle for out er dw ell δ s , cam angle for ret urn δ ′ 0 , cam angle for in ner d well δ ′ s , and lif t h . For the position show n , indicat e pr essure angle α, displace ment s and t he correspond- ing cam an gle δ . 5-20   A plat e cam wit h a t ranslating fla t-faced followe r similar t o that in Fig . 5-37 is to have t he follo win g mo tion : a rise t hrough lif t h = 40 m m wit h a sine acceleration motion curve dur- ing δ , δ , a ret urn wit h a 3-4-5 polynomial motion curve during δ ′ = 110° , and 0 = 140° s = 50° 0 δ ′ . T he dimensions a re to be : rp = 40 mm , r C = 20 mm . s = 60°     (1 ) Con st ruct t he pitch curve and t he cam con t our graphically wit h a scale of 1 ∶1 . L abel in red ink t he follower cen t reline , s , t he flat face, t he tangen t poin t T between t he ca m con- tour and t he flat face , and α corr esponding to δ= 60°and δ= 0°.       (2 ) Measur e t he m inimum leng t h of t he follower face on t he left side and t he righ t side   ( 3) Write a program to calculate t he co-ordinates of t he pitch curve , t he cam con tour of t he follower stem . and t he locus of t he cen tre of t he milling cu tte r , and ρ T . S ho w the t hr ee curves on a compu ter screen .       (4 ) Fin d t he m axim um quasi-velocity in t he rise and t he mini mum quasi-velocit y in t he   ( 5 ) If δ ′ 0 = 70°and δ ′ s = 100° , calculat e ρ T corresponding t o δ = 206° . S ho w t he ca m ret urn . Compare t he results wit h t hose from P roblem ( 2) . cont our on a compu ter sc reen . Does t he cam con t our in tersect itself ? 5-21   For t he pla te ca m wit h oscillating flat-faced followe r as sho wn in Fig . 5-34 , a rcs G H and IJ a re t wo arcs wit h cen tr e at O . Indicate radius of pri me circle r p , cam angle for rise δ 0 , cam angle for ou ter d well δ ′ , ca m angle for inne r dwell δ ′ s , cam angle for r eturn δ 0 s , and an- gular lift φm a x . F or t he position show n , indicate pressure angle α, angular displacem en t of fol- low er φ and t he correspon di ng ca m angle δ . 5-22   A plate ca m wit h an oscillating flat-faced follower similar to t hat in Fig . 5-34 is t o have t he following motion : an angular lift φm a x = 20°wit h a 3 - 4 - 5 polynomial motion curve during δ , δ , a ret urn wit h a si ne accele ration motion curve during δ ′ = 180° , and δ ′ 0 = 130° s = 20° 0 s = 30°. T he dimensions a re to be : rp = 30 mm , l O A = 80 mm , e = 12 .5 mm , r C = 15 mm .     ( 1 ) Const ruct t he cam cont our graphically . Label in r ed ink t he fr am e O A , φ, t he ・ 1 05 ・
Fig . 5-37

flat face , t he t angent poin t T , an d α correspon ding t o δ= 90°.     (2 ) Write a progra m to calculate t he co-ordina tes of t he ca m con tour and t he locus of t he cen t re of t he milling cu tter , α, t he distance l E T , and ρT . S how t he t wo curves on a compu ter screen .     (3 ) If δ and δ , calculate ρ . S ho w t he ca m con- 0 = 30° s = 120° T corresponding to δ= 26° tour on a com pu te r sc reen . Does t he ca m cont our in tersect itself ?

・ 1 06 ・

C h ap t e r 6 Gea r M ec h a n i s m s
6.1   T y p e s o f G e a r M e c h a n i s m s
Gear m echanisms are widely used in all kinds of machi nes to tr ansmit motion an d pow er bet ween rot atin g shafts . Circula r gears have constant t ransmission ratio w hereas , for non-cir- cular gears , t he ratio varies as t he gears rotate . In t his chap ter , only circular gea rs a re consid- er ed . Depending upon t he relative shafts positions , circular gear mechanism s can be divided in- to planar gear m echanisms and spatial gear m echanisms . 6.1.1   Planar Gear Mechanisms Planar gear m echanisms a re used to tr ansmit motion bet ween pa rallel shaf ts . The teet h of t h is kind of gea r m echanism a re dist ribu ted on cylindrical surfaces . Plana r gear m echanisms can be divided in to spur gear m echanisms ( Fig . 6-1 ) , helical gear mechanisms ( Fig . 6-2 ) and double-helical gear m echanisms ( Fig . 6-3) . The too t h t race of a spur gea r is par allel to its ax- is , and t herefore a straigh t li ne w her eas the t oot h t race of a helical gea r is a helix . A double- helical gea r , sometim es called he rri ngbone gea r , is a combination of t wo helical gea rs wit h equal and opposit e helix angles .

Fig . 6-1

Fig . 6-2

Fig . 6-3

    T here ar e t hree t ypes of cylindrical gear : t he ex ternal gear , t he in ternal gear , and t he rack . T wo m eshing ex ternal gears form an ex ternal gear pair wit h tw o gears rotating i n oppo- ・ 1 07 ・

site directions . as show n in Fig . 6-1 . The en gage men t bet ween an ex ternal gea r and an in ter- nal gear forms an in ternal gear pair , wit h t wo gears ro tating in t he sa me direction , as show n in Fig . 6-4 . T he smaller gear in any gea r pair is norm ally called t he pinion . A rack can be re- garded as a par t of a gear wit h infinite r adius . A r ack and pinion convert a rot ation in t o a recti- linea r t ranslation , or a rectilinear t ranslation in to a rota tion , as show n i n Fig . 6-5 .

Fig . 6-4

Fig . 6-5

6.1.2   Spatial Gear mechanisms Spatial gea r m echanisms ar e used t o transmit mo tion and powe r between nonpar allel shaf ts . They can be divided in to bevel gear m echanisms ( Fig . 6-6 ) , crossed helical gear m echanisms ( Fig . 6-7 ) and worm and worm w heel m echanisms ( Fig . 6-8) .

Fig . 6-6

T he axes of tw o m eshin g bevel gear mechan isms i nt ersect each ot her and t here ar e th ree diffe ren t t ypes: st raigh t bevel gea r m echanisms ( Fig . 6-6a ) , helical bevel gear m echanisms ( Fig . 6-6b) an d spiral bevel gear m echanisms ( Fig . 6-6c ) . A mong t hese , t he str aigh t bevel gear m echanisms are used most widely w hile spir al bevel gea r m echanisms a re suitable for t he case of high speed and heavy loads . T he axes of t wo m eshing crossed helical gears ar e neit he r pa rallel nor in te rsecting . as show n in Fig . 6-7 . T heir r elative positions can be a rbit ra ry . If t he values of t wo helix angles a re equal and t he directions of helixes a re opposit e for an ex ternal pair , or t he sa me for an in ter- ・ 1 08 ・

nal pair , t he axes becom e pa rallel and t he c rossed helical gear m echanism becomes a planar heli- cal gear mechanism . T he axes of t he componen ts of a w orm and worm w heel m echanism are perpendicular bu t nonin te rsecting , as show n in Fig . 6-8 . T he worm is normally t he driver and has t he sm aller dia mete r . T he w orm gea r or worm w heel is t he driven gea r . Usually , t he w orm and worm gear m echanism is used to provide la rge speed reduction .

Fig . 6-7

Fig . 6-8

    Among all t he kinds of gea r mechan isms , spur gea r mechanism s ar e t he funda men tal ones . T he refore , our discussion star ts wit h spur gea r mechanism s .

6.2   F un d a me n t a l s o f E n g a g eme n t o f T o o t h P r o f i l e s
6.2.1   Fundamental Law of Gearing T ran smission of mo tion and po wer between a pair of gears is accomplis hed by m eans of pushing t he teet h of t he driven gea r by t he t eet h of t he drivin g gea r . Fig . 6-9 shows a pair of m eshing teet h . The drivi ng pin ion rotates clock wise wit h angular velocit y ω 1 w hile t he driven gear ro tates counte rclock wise wit h angula r velocit y ω2 . T he profiles of t he teet h con tact at poin t K wit h t he com mon normal n-n . T he com mon norm al n-n i nte rsects t he cen t re line O 1 O 2 at poin t P . I t is know n from Chap ter 3 t hat t he poin t P is t he instan t cen tr e of velocity of t he gears , i .e . v P 1 = v P 2 or ω 1 O1 P = ω 2 O 2 P . T he t ransmission ratio i 1 2 of t he t wo gears can t he refor e be w ritten as i1 2 = ω 1 O2 P = ω O1 P 2 (6-1)

T his m eans t hat t he tr ansm ission ratio of tw o meshing gears is inversely proportional to t he ratio of t wo line segm en ts cut from t he cen t re li ne by t he common normal of t he toot h profiles t hrough t he con tact poin t . T his conclusion is called t he f u nda men t al l a w o f geari ng . T he poin t P given above is called t he pitch poin t . As t he cen t re dist ance O 1 O2 is con- ・ 1 09 ・

stan t , t he position of t he poin t P m ust be fi xed if a constant t ransmission ratio i1 2 is r equir ed . T his im plies t hat , w herever the t eet h cont act , t he common normal n-n of t he t oot h profiles t hrough t he con tact poi nt must in tersect t he cen t re line at a fixed poi nt P , if a constan t t ran smission ratio i1 2 is required . In t his case , t he locus of t he poin t P on t he motion plane of t he pinion is a circle wit h its cen t re O 1 and radius O 1 P . T his circle is called t he pitch circle of t he pinion . Similarly , t he pitch circle of t he gear is t he circle wit h its cen t re O 2 and r adius O2 P . T herefore , a pair of gears wit h const an t t ransmission ratio corresponds kine ma tically t o a pair of fric- tion w heels ( wit h diam eters equal to t he pitch circle diam eters of t he gea rs) w hich a re tangen t and roll wit hou t slippi ng . A pair of m eshing gears is t herefore dr aw n as t wo tangen t circles in t he k ine matic diagra m of a m echa- nism . If t he t ransmission ratio i1 2 is not constan t , t he pitch poin t P will move along t he cen tre line O 1 O2 accordingly . T he loci of P on t he motion planes of bot h gears are now called t he pitch curves w hich a re not circles . T his applies only for non-circula r gea r mechanism s . 6.2.2   Conjugate Profiles Meshing profiles of teet h t hat can yield a desired t rans mission ratio a re term ed conjugate profiles . F or cir cula r gears , t he con jugate profiles are t hose t hat provide t he desired constan t t ransmission ra tio . Gener ally speak ing , for any specific t oot h profile , we can fi nd its conju- gate profile . Theor etically . t here is an infinit y of pairs of conjugate profiles to produce any spe- cific t ransmission r atio . Never t heless , only a few curves have been used as conjugate profiles in practice . A mong t he m , involu tes a re used most widely since gears using involu tes as teet h pro- files , or involu te gears as t hey are called , can be manufact ured and asse mbled easily . In t his chap ter we discuss only involu te gea rs .
Fig . 6-9

6.3   T h e In v o lu t e a n d I t s P r o p e r t i e s
6.3.1   Generation of Involute A n involu te is the curve generated by any poin t on a st ri ng w hich is un wr apped from a fixed cylinder ( B K in Fig . 6-10 r epresen ts t he un w rapped lengt h ) . This is equivalen t ki ne- m atically t o a rigid rod B K rolling wit hou t slippi ng on t he cyli ndrical surface and any poi nt on t he rod generates an involu te . T he fixed circle is called t he base circle of t he involu te and t he st raight line B K is called t he generating line . r b is t he radius of t he base circle and t he angle θ K is called t he unfoldi ng angle of t he involute at poin t K . ・ 1 10 ・

6.3.2   Properties of the Involute A n involu te gener ated as above has t he follo wing properties . (1 ) T he leng t h of t he gene rating line segmen t unrolled from t he base circle is equal to t he arc leng th of t he base circle rolled , i .e . B K = AB (2 ) T he norm al of an involut e at any poin t is t angent t o its base circle .
Fig . 6-10

(3 ) I t can be show n t hat t he tangen t point B of t he genera ting line wit h t he base circle is t he curvat ure cen t re of t he involute at t he poin t K . In ot her w ords , t he lengt h of the segm en t B K is t he radius of curvat ure of t he involu te a t t he poin t K , i .e . ρK = B K = r K - rb
2 2

(4 ) According t o proper t y (2 ) , any tangent t o a base circle is t he common normal to t wo arbit rary i nvolu tes gener ated from t hat base circle . Mor eover , t he norm al distance between t hese t wo involutes r em ains t he sam e , no mat ter w het he r t hey unfold in t he sa m e direction or in t he opposite direction . T he norm al distance is always equal to t he a rc distance along t he base circle bet ween t he tw o star ting poin ts of t he t wo i nvolu tes . In Fig . 6- 11 , for exa mple , A 1 B1 = A 2 B 2 = A B an d B 1 E 1 = B2 E 2 = B E . (5) T he shape of an involute depends only on t he radius of
Fig . 6-11

its base circle . For t he sa me unfolding angle θK , t he smalle r t he radius rb of t he base circle , t he sm aller t he radius of curvat ure of t he involu te , as sho w n in Fig . 6-10 . As t he radius r b of t he base circle approaches infinit y , t he involu te becomes a str aigh t line . (6 ) No involu te exists inside its base circle . A n involu te begins from its base circle and goes ou t ward . T he above proper ties provide t he founda tion for t he t heory of involu te gearing . 6.3.3   Equation of the Involute Suppose t hat an involut e gear show n in Fig . 6-10 is t he driven gear in con tact wit h anot her gear at t he poin t K . T he force acts along t he norm al BK of t he involu te profile if friction force is no t considered . Therefore , accordi ng t o t he definition of pressure angle , t he angle betw een t he normal K B and t he velocity v K of t he poin t K is t he pr essur e angle α K , w hich is also equal t o ∠ B OK . In △ OB K , we have ・ 1 11 ・

tan αK =

BK A B r b ( αK + θK ) = = = αK + θ K rb rb rb rb = r K cos αK

T he pola r para m etric equation of the involu te wit h αK as a para m eter is t hus rK = rb cos αK (6-2)

θ K = tan α K - αK ・ w her e θK is defined as the involute function of t he pressure angle αK and can be w rit ten as θ K = inv αK .T he inverse function αK of θ K (or inv α K ) must be calculated n umerically . If t he roll angle β ( = αK + θ K ) is taken as a par am eter , we can derive t he corresponding par am et ric equa tion of t he i nvolu te i n a r ectangular coordinate syste m as x = rb ( cos β + βsin β) y = r b ( sin β - β cos β) 6.3.4   Gearing of Involute Profiles Fig . 6-12 sho ws a pair of m eshin g involut e profiles . According t o proper t y (2 ) of t he involu te in Sec . 6 .3 .2 , t he com mon normal to t he meshing invo- lu te profiles t hrough t heir cont act poin t K m ust be t he common tan- gen t t o t heir base circles . T he position of t his common tangen t re- m ains unchanged as bot h gears rotate , as does t he common norm al t o t he involu te profiles . T his r esults in a fixed pitch poin t P . T he refore, according t o t he funda mental law of gearing m en tioned in Sec . 6 .2 .1 , t he t ransmission ratio will re main const an t . T he locus of t he contact poin t of tw o m eshing profiles on t he fixed plane is called t he tra jectory of con t act . As the common nor- m al t o t he involu te profiles coincides wit h t he common tangen t to t he base circles , t he contact poin t of t he involut e profiles should lie on t he common tangen t to t he base circles . T he t rajectory of con tact of t he involute gea rs t herefore coincides wit h t he common tangen t to t heir base circles . T his t rajectory of con tact is also called t he line of action . I t can be seen t her efore t hat t he common norm al to the involu te profiles , t he common t angen t to t he base circles and t he line of action ar e indeed t he sa m e line for an involute gear pair . T w o involu tes cannot con tact ou tside t he common t angent to t he base circles . If one poin t on an involute is t o come in t o cont act wit h some point on anot he r involu te , bot h poin ts will not m eet un til t hey a rrive on t he common tangen t to t he base circles . T he reaction force bet ween t he m eshi ng involu te profiles is exer ted along t he line of action if t here is no friction . As t he position of the line of action st ays unchanged during motion for an involute gea r pair , t he direction and m agnit ude of t he r eaction force does not change during ・ 1 12 ・
Fig . 6-12

(6-3)

motion and t he refor e t he t ransm ission of po wer is smoot h . T his is one of t he advan tages of in- volu te gea rs . As show n in Fig . 6-12 , △ O1 N1 P is si mila r to △ O 2 N 2 P . The t ransmission r atio can t herefore be w rit ten as i12 = ω1 O2 P r b 2 = = ω2 O 1 P r b 1 ( 6 -4 )

T his m eans t hat t he t ransmission r atio depends on ly on t he ratio of t he r adii of t he tw o base circles . A change in cen t re distance does not t her efor e affect t he constan t t ransmission ratio of an involute gea r pair . Th is proper ty is called t he separabilit y of t he cen tr e distance in involute gearing and is very useful in t he asse mbly of involu te gear m echanisms .

6.4   S t a n d a r d In v o lu t e S p u r G e a r s
6.4.1   External Gears Fig . 6-13 shows a par t of an ex ternal spur gear . The toot h nu mber of a gear is denot ed by z , w hich must be an int eger . T he lengt h of t he too t h along t he axial direction is called t he f acew id t h , deno ted by B . O n t he t ransverse plane, the circle w her e t he tips of t he teet h lie is called t he ad den d u m circle (or ti p circle ) . The dia met er and radius of t he addendum circle a re deno ted by d a and r a , respectively . Sim ilarly , t he circle w he re t he roots of t he teet h lie is called t he deden d u m circle ( or root circle ) wit h diam eter d f and radius rf . T he space between t he profiles of tw o adjacen t teet h and wit hin t he addendum cir- cle is called the toot h space . T he arc leng t h of t he t oot h space along an arbit ary circle is called t he spacewi d t h on t his circle , denoted by e K . T he arc lengt h of t he too t h along an arbit ary circle is called t he toot h t hickness on t h is cir cle , denoted by s K . The ar c dis- tance between correspondi ng poi nts of adjacen t teet h along an a rbita ry circle is called t he p itch on th is circle , denoted by p K . Obviously , p K = eK + s K . If t he dia m et er of t his circle is d K , t hen π dK = z pK ( 6-5 ) Between t he addendum circle and t he dedendum cir- cle , t her e is an im portan t circle w hich is called t he ref er- ence circle . Par am eters on t he r eference circle a re st an- da rdized and denoted wit hou t subscrip ts , such as d , s , e and p . T he module m of a gear is int roduced on t he refer-
Fig . 6-13

・ 1 13 ・

ence circle as a basic pa ra meter , w hich is defined as m= T he refore p π (6-6) (6-7)

p = e + s =πm

T he units of m a re mm . The module m has been standardized as follo ws . Tab 6-1   Modules of Involute Cylindrical Gears ( GB / T 1357—1987)
First ser ies 1.5 , 2 , 2. 5 , 3 , 4 , 5 , 6 , 8 , 10 , 12 , 16 , 20 , 25 , 32 , 40 , 50 0.35 , 0.7 , 0. 9 , 1.75 , 2 .25 , 2 .75 , ( 3. 25 ) , 3.5 , ( 3.75 ) , 4.5 ,

mm

0.1 , 0 .12 , 0.15 , 0. 2 , 0 .25 , 0.3 , 0.4 , 0.5 , 0. 6 , 0. 8 , 1 , 1 .25 ,

Second ser ies

5.5 , (6. 5) , 7 , 9 , (11 ) , 14 , 18 , 22 , 28 , ( 30) , 36 , 45

Modules of t he first series are prefer able to t hose in the second se ries and t hose i n br ackets should be avoided if possible . Combini ng Eqs . (6-5) and (6-6) will give d = mz ( 6 -8 ) A too t h is no w divided i nt o tw o pa rts by t he refe rence circle . T he r adial distance betw een t he refer ence circle and t he addendum circle is called t he ad den d u m and deno ted by ha . T he radial distance bet ween t he refe rence circle and t he dedendum circle is called the deden du m and denot ed by hf . T he radial dist ance between t he dedendum circle an d addendum circle is called toot h dep th , denoted by h wit h h = h a + hf . Sizes of t he teet h and gea r ar e propor tional to t he module m . For st anda rd spur gea rs , t he following t hree relations should be satisfied sim ultaneously : ha = h a m hf = ( ha + c ) m e = s = 0. 5π m
* * * * * *

( 6 -9 ) ( 6 -10 ) ( 6 -11 )

w her e h a* is called t he coe f f icien t o f ad den du m an d c * is called t he coe f f icien t o f bot tom clea rance . Bot h ar e standa rdized : h a = 1 and c = 0 .25 for t he normal toot h ; h a = 0 .8 and c * = 0 .3 for t he shorte r t oot h . I t can be seen t hat module m reflects the size and t he bending st rengt h of t he too t h . T he following relationships t her efor e apply to st anda rd spur gears : h = ( 2 ha + c ) m d a = d + 2 h a = ( z + 2 ha ) m d f = d - 2 hf = ( z - 2 ha - 2 c ) m
* * * * *

( 6 -12 ) ( 6 -13 ) ( 6 -14 )

So far t he size of t he base circle and t herefore t he shape of t he involute profile of t he too t h have no t be consider ed . As is kno wn from equa tion ( 6-2 ) of the involu te , t he ratio rb/ r is equal to t he cosine of t he pr essur e angle α on t he refer ence circle . T her efore , pressure angle α is taken as a basic par am eter to determi ne t he base cir cle , i.e ., ・ 1 14 ・

d b = d cos α most commonly 20° (sometim es 15° ) .

( 6 -15 )

w her e d b is t he diam eter of t he base cir cle . The pr essur e angle α is also standa rdized . I t is T he pitch on t he base cir cle is called t he base p itch and denoted by p b ( Fig . 6-13 ) . T he distance between corr esponding sides of adjacent toot h profiles along t he com mon normal is called t he nor m al p itch and denoted by p n . According t o t he proper ties of t he involute , p n is equal t o p b an d pn = p b = A mong t he above par am eters , πd b πd = cos α= p cos α z z ( 6 -16 )

z,

m , α, h a* and c * are t he fu nda men tal par am eters

w hich dete rmine the size and shape of a standa rd involu te gear . A standard gea r he re implies t hat m , α, ha* and c * take t he st anda rd values and e = s . 6.4.2   The Rack Fig . 6-14 s ho ws a pa rt of a rack . A rack can be r e- garded as a special form of gear wit h an infinit e number of t eet h and its cen t re at i nfinity . The radii of all circles be- come infinite and all circles become st raight lines , such as t he reference li ne , tip line and root line . Since t he radius of t he base cir cle is infinite , t he involu te toot h profile be- comes a st raigh t line t oo and t he pressure angle re mains t he sam e at all poin ts on t he too t h pro- file , i .e . αK = α, w hich is also called t he no m i nal p ressu re ang le ( 20°or 15° ) . As t he teet h profiles are parallel st raigh t lines , t he pitch r em ains unchanged on t he refer- ence line , tip line or any ot her line , i .e . p K = p = π m . This implies t hat t he su m of eK and s K is constan t , alt hough neit her eK nor s K is constan t . T he addendum and dedendum a re t he sa me as for ex ternal gears . 6.4.3   Internal Gears Fig . 6-15 shows a par t of an in te rnal gear . T he teet h ar e dist ribu ted on t he in te rnal surface of a hollow cylinder . T he t oot h of an in ternal gear t akes t he shape of t he too t h space of t he corre- sponding ext ernal gear , w hile t he toot h space of an in ternal gear takes t he shape of t oot h of t he corr esponding ex ternal gea r . T he diam eters of t he addendum circle an d deden du m circle a re cal- culated as follows:
Fig . 6-15 Fig . 6-14

・ 1 15 ・

d a = d - 2 h a = ( z - 2 ha* ) m d f = d + 2 hf = ( z + 2 ha* + 2 c * ) m curve .

( 6 -17 ) ( 6 -18 )

w her e d a must not be less t han d b to ensur e t hat t he profile of t he toot h on t he top is an involute

6.5   G e a r i ng o f In v o l u t e S p u r G e a r s
6.5.1   Proper Meshing Conditions for Involute Gears Fig . 6-16 sho ws a pair of meshing gea rs . In t he position show n , t wo pairs of teet h con- tact simultaneously at poi nts M and L . To ensure a con tinuous t ran smission , t h is sit ua tion must occur w hen t he pair of t eet h on the left is coming ou t of m esh and t he pair of teet h on t he right com es in to mesh . According to t he prope rt y of t he involu te , t he common tangent to t he base circles coincides wit h t he common norm al t o t he teet h profiles . In ot her words , t he con- tact poin ts M and L bet ween t wo involu tes must lie on t he common tangent to t he base circles and M L is t he common norm al to t he teet h profiles . To m ain- tain t he proper m eshi ng of tw o pairs of profiles at t he sam e tim e, t he norm al distances of t he teet h M 1 L1 and M 2 L 2 on bo t h gears must be the sa me . By the proper ties of involu te curves , t hese tw o segments a re normal pitches (or base pitch- es ) of t he gea rs and we have p n 1 = p n 2 or p b 1 = p b 2 . According t o Eqs . (6-16 ) and ( 6-6 ) , t his results i n π m 1 cos α 1 =π m 2 cos α 2 ( 6-19) Since bot h modules and pressure angles ar e standa rd pa ra me- ters , t he above condition implies t he following t wo practical conditions : m 1 = m2 α 1 =α 2 ( 6-20)
Fig . 6-16

In ot he r words , t he modules and pressure angles of t wo m eshing gears should be t he sa me and th is proper ty is called t he proper m eshin g condition for involu te gears . 6.5.2   Centre Distance and Working Pressure Angle of a Gear Pair

T he re a re t wo requir em en ts i n design ing a gear pair . First of all, t he backlash should be ze ro t o pr even t shock bet ween t he gears . T he backlash is t he difference bet ween t he spacewid t h of one gear and t he too t h t hickness of ano t her , bot h measured alon g t he respective pitch cir- cles . I n ot he r words , to obtain zero backlash of a gear pair , t he toot h t hick ness of one gear on ・ 1 16 ・

its pitch circle should be equal to the spacewidt h of anot he r on its pitch circle , i .e . s′ 1 = e′ 2 and s2 ′ = e1 ′ . In practice , t here should be a sm all a moun t of backlash to keep an oil film betw een t he gear teet h and preven t ja mmi ng af ter t he gea rs becom e heated . T his is provided by toler- ances on toot h t hick nesses w hile t he nomi nal backlash should be ze ro . Secondly , t he bott om clea rance should take t he standard value c = c m t o prevent profile in terfe rence an d store some lubrican t between t he teet h . T he bot tom clea rance c is t he r adial distance from t he dedendum circle of one gear to t he addendum circle of anot her ( Fig . 6-17 ) . A gea r pair should be assembled wit h a corr ect cen t re distance . Bo t h t he backlash an d bo t- tom clea rance ar e rela ted t o t he cen t re distance . F or an ex te rnal gear pair , t hese increase wit h t he cen t re distance . If bot h are standard gea rs and t he bott om clear ance takes t he value c = c * m , t he cen t re distance becom es ( see Fig . 6-17) a = r a 1 + c + r f2 = rf1 + c + r a2 = r1 + h a m + c m + r 2 - ( h a + c ) m = r1 + r 2 =     m ( z1 + z2 ) 2 ( 6 -21 ) a is t he sum of t he reference r adii, as show n in Fig . As men tioned befor e, t he mes hing of tw o gears corresponds to t he pur e rolling of t heir pitch circles . We norm ally indicate t he para m eters on t he pitch circle by t he use of t he super- scrip t “′ ”. T he t ransmission r atio i1 2 can also be w rit ten as i1 2 = ω r2 ′ rb 2 r2 cos α r 2 z2 1 = = = = = ω r′ rb 1 r1 cos α r 1 z1 2 1
* * * * *

6-17a . This cent re distance a is t he refor e called t he r efe rence cen tr e distance .

T he refore , if t wo gea rs ar e moun ted wit h t he refer ence cen t re distance a , t heir reference circles a re tangen t to each ot her and roll wit hout slipping . In t his case , t he reference circles co- incide wit h t heir pitch cir cles . Moreover , if bo t h gears ar e st anda rd , w e have s1 ′ = s1 = e1 ′= e1 = s2 ′ = s2 = e2 ′ = e2 = π m/ 2 and t his results i n t he zero backlash . I t has been show n t hat t he t wo design r equir em en ts are fulfilled for a pair of standard gea rs if t hey a re mou nted wit h t he reference cen t re distance a . T he te rm w orki ng pressu re ang le is defined as t he angle bet ween t he velocity of pitch poin t P (on bot h gears ) and t he line of action and is denoted by α ′. F rom Fig . 6-17b , we can see t hat rb 1 = r 1 ′cos α ′and r b 2 = r2 1 ′cos α ′and α 2 ′= α 1 ′= α 2 ′. T herefore , t he pressure angles on t he tw o pitch circles are t he sa me and α ′can be t hough t of as t he working pressure angle or t he pressure angle on t he pitch circles . Ca re should be taken her e t o distinguish the pitch cir cle from t he reference circle and t he wor king pressure angle from t he pressure angle . T he r efe rence circle and pressure angle exist on a single gear and t hey ar e fixed af ter m anufact ure . The pitch circle and working pressure angle are defined only after a pair of gears come in to engage men t . T heir sizes depend no t only ・ 1 17 ・

Fig . 6-17

on t he gea rs bu t also on t he wor king cen t re distance a′ = r1 ′ + r2 ′ . If a′ = a , t hen t he pitch circles coincide wit h t he refer ence circles and α ′ = α . F or m any reasons however , t he working cen t re distance a′is som etim es not equal t o the reference cen tr e distance a . T hen r′ and α ′ are not equal t o r and α, respectively , as show n in Fig . 6-17b . Since r b = r cos α, we have t he following impor tant relationship : acos α= acos α = r b 1 + rb 2 ( 6 -22 ) For an in ternal gea r pair , t he refer ence cen tr e dis- m ( z 2 - z 1 ) , w hich fulfills t he 2 t wo design r equir em en ts in t he case of standa rd gears . tance is a = r 2 - r 1 = T he relation acos α= acos α rem ain s t rue . 6.5.3   Meshing of a rack and pinion Sho wn in Fig . 6-18 is t he meshing of a rack and pinion . Accordi ng t o t he proper ty of i nvolu te gearing , t he common norm al t o t he teet h profiles is t he tangen t t o t he base cir cle of t he pinion . T his line is also t he line of action . Since t he t oot h profile of t he rack is a str aigh t line , a line pe rpendicular to t he str aigh t profile of t he r ack an d tangen t to t he base circle of t he pinion is t he line of action . T he line of cen tr es passes t hrough t he cent re of t he pinion and is perpendicular t o t he reference line of t he rack . The i nte rsection bet ween t he line of action and t he line of cen t res is t he pitch poin t P and O P = r 1 ′ . Accordi ng t o t he defini- tion of working pressure angle , α ′ , t he angle bet ween t he line of action an d a line parallel to t he refer ence li ne of t he rack is t he w or king pressure angle α ′. F rom Fig . 6-18 , we can see ・ 1 18 ・
Fig . 6-18

t hat t he working pressure an gle α ′is equal t o t he nom inal pr essure angle α of t he rack . Since t he too t h profile of t he r ack is a st raigh t line, the positions of the line of action and t he pitch point P r em ain unchanged , no m atte r w here t he rack is placed (nea rer to or furt her away from t he pinion ) . T he refore , α ≡α ′. F urt hermor e, since r1 ′ cos α ′ = r1 cos α= rb 1 and α ≡α ′ , we have r1 ′≡ r1 . As m en tioned above , α≡α ′ , and r1 ′≡ r1 a re characteristics of r ack and pinion gearing and diffe r from t hose of t wo spur gea rs . If a r ack and pinion pair a re so mount ed t hat t he r efe rence line of t he rack is tangen t to t he reference circle of t he pinion , t he pitch li ne of t he r ack coincides wit h its refer ence line , bo t h a re tangen t t o t he reference cir cle of t he pinion . If t he rack is placed nea rer to or fur t her aw ay from t he pinion , t he pitch circle of the pinion and t he pitch line of t he rack rem ain unchanged . Ho wever , t he r efe rence line of t he r ack is now sepa rated from its pitch line .

6.6   C o n t a c t R a t i o o f a n In v o lu t e S p u r G e a r S e t
Sho wn in Fig . 6-19 is a pair of m eshing gea rs . The driving pinion ro tates wit h angula r ve- locit y ω 1 clock wise and t he driven gear wit h ω 2 coun te rclock wise . The li ne of action N 1 N 2 cuts t he addendu m circles of t he tw o gears at poin ts B 2 and B 1 , r espectively . T he meshing begins at poin t B 2 , moves along t he li ne of action and ends at poin t B 1 . The segm en t B 1 B 2 is t here- fore called t he act ual line of action . B 2 and B 1 approach N 1 and N 2 respectively as the r adii of bo t h addendum circles inc rease , bu t t hey cannot pass beyond N 1 an d N 2 as an involute does not ente r its base cir cle . N 1 N2 is t he refore t he limit of B1 B 2 and called t he t heoretical line of action . In Fig . 6-19 , tw o pairs of teeth con tact at poin ts B 1 and K , respectively . B1 K is t he norm al pitch p n ( or p b ) of t he gears . T he first pair of teet h is coming out of con tact . In orde r to get a continuous motion t ransmission , t he second pair of teet h must have m eshed along line B 2 K before t he first pair moves ou t of cont act . T he norm al pitch of t he teet h p n (or p b ) must t hen be not greate r t han t he lengt h of t he act ual line of action B1 B 2 . T he cont act ratio is defi ned as t he ra tio of t he leng t h of B 1 B2 t o p b an d deno ted as ε α . T he condition of con tinuous motion t ransmission is t hen expressed as ε α= Since B 1 B 2 = P B 1 + P B 2 = r b 1 ( tan αa1 - tan α ′ ) + r b 2 ( t an αa 2 - t an α ′ ) = m cos α[ z1 ( tan α ′ ) + z 2 ( tan αa2 - tan α ′ )] a 1 - tan α 2
Fig . 6-19

B1 B2 ≥1 pb

(6-23 )

( 6 -24 ) ・ 1 19 ・

and p b =π m cos α, ε α can be calculated as follows ε α= 1 [ z1 ( tan α a 1 - tan α ′ ) + z 2 ( tan αa2 - tan α ′ )] 2π ( 6-25)

w her e α ′ is t he working pr essur e angle , α a 1 and α a2 are t he pressure angles on the adden du m circles Sim ilarly , we have t he form ulae to calculate ε α for a rack and pinion pair : 2 ha 1 ε α= z ) + 1 ( tan α a 1 - tan α 2 π sin α cos α and for an in ternal gear pair : ε α= 1 [ z1 ( tan α a 1 - tan α ′ ) + z 2 ( tan α ′- tan α a2 ) ] 2π ( 6 -27 )


( 6 -26 )

T he value of t he cont act r atio in dica tes t he aver age number of toot h pairs in con tact during a cycle t o shar e t he load . T he higher the con tact ratio , t he gr eate r t he aver age numbe r of too t h pairs to share t he load and t he h ig her t he capacit y of t he gea r set to t ransmit t he pow er . In practice , an allowance is m ade so t hat t he con tact r atio ε α is la rger t han 1 .2 .

6.7   M an u f a c t u r i ng M e t h o d s o f In v o l u t e P r o f i l e s
6.7.1   Cutting of Tooth Profiles T he re are various m et hods for m anufact uring t oot h profiles , such as die casting , pr ecision forging, pow de r process , moulding , cu tting and so on . A mong t he m , cut ti ng is t he most widely used . There ar e generally t wo t ypes of cut ting m et hods , i .e . form cut ti ng and gener- ating cut ti ng . (1 ) Form Cut ting In form cu tting ( Fig . 6-20 and Fig . 6-21 ) , t he desired too t h profile is produced by pass- ing a cu t ter t hrough t he blank . F orm milli ng wit h a disk milli ng cut ter is show n in Fig . 6-20 . T he cut ter ro tates and t he blank is fed axially . End milling cu t ters ( Fig . 6-21 ) are used for la rge modules and double helical gears . T he section of t he cut ter has t he shape of t he too t h space of t he gea r to be cu t . Index ing is needed after a toot h space is cu t so t he accur acy of t he gear depends on t he accuracy of t he index plate as well as t he section shape of t he cu tte r . As t he shape of an involu te depen ds on its base radius , a differen t cu tte r is required for each num- ber of teet h even for t he sam e module and sa me pressure angle . Th is is divor ced from reality : in practice , only 8 to 26 cu tte rs are available for each module and pressure angle . Each cu t ter is used for several differ en t numbe rs of teet h and an e rror is t he refor e in t roduced . T he cu tting process is in te rrup ted by the indexing and t he production rate is lo w in form milli ng . I ts main advan tage is t hat it can be accomplis hed on common ly available millin g machines . ・ 1 20 ・

Fig . 6-20

Fig . 6-21

    ( 2) Generating Cu tting In a generating cu tting m et hod , t he edges of a cu t ter take t he form of a gear ( or rack ) wit h t he sa me module and pressure angle as t he gea r to be cu t . There is a gene rating mo tion bet ween the cut ter and t he blank as if t he cu tt er engages as a gear wit h t he gea r t o be cu t . By addi ng t he cu tting motion between t he cu t ter and t he blank , t he profile of t he t eet h is form ed gradually by a se ries of cu ts . Fig . 6-22 shows t he shapi ng process wit h a pinion-shaped shaper cut ter . If z 0 is t he num- ber of teet h of t he cu tte r and z t he num ber of teet h of t he gear t o be cu t , t heir transmission ra- tio should be i= ω 1 z = ω z0 2

w hich defines t he genera ting motion to be provided by t he kine m atic chain in t he cut ti ng m a- chine . T he cu tting motion is t he reciprocation of t he cu tte r w h ile t he feed is t he move ment of t he cu tte r towa rd t he blank . The blan k should ret reat a little as t he cut ter goes back t o pr even t scraping on t he finished flank by the cut ter .

Fig . 6-22

Ex ternal gears can be cu t wit h a rack-shaped shape r cu tte r as show n in Fig . 6-23 . T he ・ 1 21 ・

edges of the cut ter a re no w st raigh t lines w hich can be made accura tely . T hat is w hy involute gears ar e easy to manufact ure .

Fig . 6-23

T he cu tting met hod show n in Fig . 6-23 is not con tinuous in t he shaping of a gea r . In m ass production , hobbing show n i n Fig . 6-24 is used i nstead of shaping . T he shape of a hob is like a screw wit h some axial slots t o form t he cut ti ng edges . T he cut ting edge takes t he shape of a rack and t he hob is t her efore a rack-shaped cut ter . The rotation of t he hob provides t he cu tting motion and is also equivalen t to t he con tinuous t ranslation of t he rack-shaped cu tter . At t he sam e time , t he blank is driven at a ce rtain t ransmission r atio to t he cut ter by t he kine matic chai n in the cut ting mach ine and t his gives t he required generating motion . M ean w hile , t he hob is fed along the axial direction of t he blank to cu t t he teet h for t he w hole facewid t h . Cu t- ting is con tinuous and t he production ra te is t herefore improved .

Fig . 6-24

In the generating met hod , one cu tter is enough for each module and pr essur e angle, in spite of t he n umber of teet h of t he gea r to be cut . I t is t he refor e t he most widely used m et hod t o cu t gears . 6.7.2   Cutting a Standard Gear with Standard Rack-shaped Cutter T he profile of a st andard rack-shaped cu tte r show n in Fig . 6-25b is similar to t hat of a standa rd r ack show n in Fig . 6-25a bu t t he tip line is c * m higher t han t he addendum line for cu tting t he profile at t he root to provide t he bot tom clea rance . The cu tting edge betw een t he tip li ne and t he adden du m line is not a st raigh t line and t he corr esponding gea r profile cut is not ・ 1 22 ・

Fig . 6-25

involute . T he refor e, in discussing t he gene rating of an involu te profile , we simply regard t he standa rd r ack-shaped cu tter as having t he sa me profile as t he standard rack . To cu t a standard gear , t he t ransmission ratio of t he cu tter an d t he blan k in gene rating motion is t he sam e as t hat of t he rack and t he gear , i .e ., r1 ω1 = v2 . T his is accom plished by t he kinem atic chai n in t he cut ti ng m achine . F urt hermore , t he reference li ne of t he cu t ter should be t angen t t o t he refer ence circle of t he gear after t he feeding has fi nished , as in Fig . 6- 26 . Thus t he r eference line of t he cu tte r rolls wit h t he r eference circle of t he gea r wit hout slip- ping . T herefore , t he pitch , module , pressure angle , too t h t hick ness and spacewidt h of t he gear on t he reference circle are t he sa me as t hose on t he reference line . As t he t oot h t hickness and spacewidt h of t he cut ter on t he reference line a re equal to each ot her , t he gear is cut wit h s = e = p/ 2 =π m/ 2 . Since t he distance bet ween the reference line and t he tip line of t he cu t ter is ( h a* + c * ) m and t he dedendu m circle of t he gear is cut by t he tip line of t he cut ter , t he ra- dius of dedendum circle of t he gear rf1 = r1 be cut on a lat he wit h radius r a 1 = r1 + ha m .


( ha + c ) m . Note : t he addendum cir cle of





t he gear is not cut by t he dedendum line of t he cu tte r ! To get a standard gea r , t he blan k m ust

Fig . 6-26

・ 1 23 ・

6.7.3   Cutter Interference In a genera ting process , it is sometim es found t hat t he t op of t he cu tt er en te rs t he profile of the gear and som e par t of t he involute profile nea r t he root portion is re- moved , as show n in Fig . 6-27 . This is called cu tter i n ter f erence and will r educe t he con tact ratio as well as t he str eng th of the toot h . O bviously , cu tter in te rference should be avoided or min imized . F or a rack-shaped cu tt er , it can be proved t hat cu tte r i nte rfer ence will occur if t he addendum line of t he cut ter passes t he lim it poin t N1 of the line of action s ho wn in Fig . 6-28 . In cu tting a st anda rd gear , t he r eference line
Fig . 6-27

of t he cu tter is tangen t to t he r eference circle of t he gear an d t he adden du m

line of t he cut ter in tersects t he line of action at poin t B 2 T o preven t cu tte r in terfe rence , t he poin t B 2 should not pass point N 1 , as sho wn in Fig . 6-28 , i. e ., P N 1 ≥ P B2 from w hich we ob tain h a* m 1 m z sin α ≥ 2 sin α 2 h a* z≥ 2 sin α T his results in t he m inim al number z m i n of teeth of a standard spur gear cu t by a standard r ack-shaped cut - ter wit hou t cu tte r i nte rfer ence . z mi n 2 h a* = sin 2α ( 6 -28 )
Fig . 6-28

Values of z mi n for differ en t α and h a* values a re giv- en below .
α ha


20° 1 17 0 .8 14 1 30

15° 0. 8 24

z mi n

6.8   A d d e n d um Mo d i f i c a t i o n o n In v o l u t e G e a r s
6.8.1   Introduction of Addendum Modification Standard gears en joy in te rchangeabilit y and a re widely used in m any ki nd of mach ines . Ho wever, t hey also have som e disadvan tages . T he numbe r of teet h should not be less t han z m i n to preven t cu tt er in te rfer ence . More compact const ruction can not be achieved wit h la rger z values . T he w orki ng cent re distance a′s hould be equal t o t he reference cen t re distance a to ・ 1 24 ・

m eet t he tw o design requirem en ts . No ot her values can be used . T he curvat ure radius of t he t oot h profile an d t he toot h t hickness of t he pinion at the root are less t han those of t he gear . T he refore t he st reng t h of t he pinion is m uch low er t han t hat of t he gear . The st rengt h of t he standa rd gea r mechanism is t hus lower even if t he st rengt h of t he gear is adequat e . To improve t he pe rformance of gears , addendum modification is e mployed . Gears wit h addendum modification are also called cor rected gears . T he gear-cut ting m achines and t he cu t- ters ar e the sa m e as t hose for standa rd gea rs . In adition , t he tr ansm ission ratio bet ween t he cu tte r and t he blank r em ains u nchanged . T he difference from cut tin g standa rd gears is t hat t he reference line of t he cut ter m ay no t be tangen t to t he reference circle of t he gear . As m en tioned in Sec . 6. 5. 3 , r 1′ = r1 for r ack and pinion gea ri ng . The line tangen t to and rolling wit hou t slippi ng on t he refer ence circle of t he gear is t he pitch line of t he rack-shaped cu tt er . As m en- tioned in Sec . 6. 4. 2 , t he pressur e angle and t he pitch of a rack ar e t he sam e on any line . T he refore , pa ram eters z , m , r = m z/ 2) , α, and rb = ( r cos α) of t he corrected gea r re main t he sam e as t hose of standa rd gears . T his m eans t ha t differ en t por tions of t he sa m e involu te are e mployed for t he profiles of t he standard gear and t he corrected gear . 6.8.2   Geometric Dimensions of Corrected Gears In cut tin g t he corr ected gea r , t he r ackshaped cu tt er is located a distance x m from t he position used for cut ting t he standard gea r , i .e . t he distance bet ween t he pitch line and t he reference line is x m . H ere , m is t he module and x is t he modification coefficient . If t he cu tter is placed fur- t her away from t he position for cut ting a stan dard gear , x is positive an d t he modification is consid- x and t he modification er ed positive . Ot her wise , if t he cut ter is placed to- wards t he axis of t he blan k , a re negative . To preven t cu tte r in terference , the modifica- tion distance x m of t he cu t ter should be sufficien tly la rge i .e . x m ≥ ha m - N 1 G , as s ho wn in Fig . 6- 29 . T his results i n x≥ h
* a *

z sin α 2 ha* ( z m i n - z ) = z mi n

2

Fig . 6-29

from w hich t he minim al modification coefficien t x m i n can be derived using Eq . (6-28 ) : x mi n ( 6-29)

O bviously , x m i n > 0 if z < z m i n .W hen z > z m i n , t he gea r can be negatively modified , as x m i n < 0 . Since t he dedendum circle of t he gear is cu t by t he tip line of t he cu tter and t he cut ter has ・ 1 25 ・

ret reated x m , t he radius of t he dedendum circle rf for t he correct ed gear is now given by rf = r - ( ha* + c * ) m + x m reference cir cle ar e respectively πm π + 2 KJ = + 2 x tan α m 2 2 πm π e= - 2 KJ = - 2 x tan α m 2 2 s= T he bendi ng st reng t h of a positively modified gear is t herefore i mproved . 6.8.3   Gearing of a Corrected Gear Pair To keep zero backlash for a corrected gea r pair , t he following rela tions should hold , as in t he case of standa rd gea rs s1′ = e2′ s2′ = e1′ as sho wn in Fig . 6-30 . Then t he pitch on pitch circles of bot h gears is p′ = s1′ + e1′ = s2′ + e2′ = s1′ + s2′ The following formula can be derived from Eq . ( 6-33) inv α ′ =
Fig . 6-30

( 6-30)

As show n in Fig . 6-29 , t he t oot h t hickness and t he spacewid t h of a corrected gea r on t he ( 6-31) ( 6-32)

T he too t h t hickness of a positively modified gear is la rger t han t hat of a st anda rd gear .

( 6-33)

2 ( x 1 + x 2 ) tan α + inv α z1 + z2

( 6-34)

This is called t he gearing equation wit hout back lash . T his e- quation can be used t o calculate t he w orki ng pressure angle α ′ac- cordi ng t o t he modification coefficien ts x 1 and x 2 . T hen t he w or king cen t re distance a′ can be calculated from a′ cos α ′ = a cos α . For a given specific working cen tr e distance a′ , t he w ork- ing pressure angle α ′should be calculated first from a′ cosα ′ = a cos α . T hen t he sum of x 1 and x 2 can be calculated by Eq . (6-34 ) . If x 1 + x 2 ≠ 0 , t hen α ′ ≠α and a′ ≠ a . T o get t he stan- da rd bot tom clearance c = c * m as show n in Fig . 6-31 , t he following relationship should hold a′ = r a 1 + c * m + rf2 = r f1 + c * m + r a2 t hen t he radii of t he addendum cir cles can be derived as r a1 = a′- r f2 - c * m r a2 = a′- r f1 - c * m Attention : it can be proved t hat if x 1 + x 2 ≠ 0 , t hen                 a′ ≠ a + ( x1 + x 2 ) m                 r a 1 ≠ r1 + h a m + x 1 m ・ 1 26 ・


( 6-35)

Fig . 6-31

r a 2 ≠ r2 + h a* m + x 2 m Corrected gears a re not in terchangeable like standa rd ones . 6.8.4   Types of Corrected Gear Pairs A dden du m modification overcomes t he shortcomings of standa rd gears t o som e ext en t . By m eans of positive modification , cut ter in terference can be avoided for a gea r wit h a num ber of t eet h less t han z mi n . Correct ed gear pairs can take any working cen tr e distance wit hout back- lash . Positive modification can improve t he str eng th of t he pinion and t he refor e i ncrease t he st rengt h of t he w hole gear m echanism . T he re a re tw o types of corrected gear pairs , one wit h r efe rence cen t re dist ance and t he o t her wit h modified cen t re distance . A corrected gea r pair wit h refe rence cen t re distance is also called a height correction gear pair wit h x 1 + x 2 = 0 . N orm ally , t he pinion is positively modified and t he gea r nega tively . T hus , the t hickness of t he t eet h incr eases on t he pinion and decreases on t he gea r . T his m eans t hat t he str eng t h of t he teet h is incr eased on t he pinion and dec reased on the gear . A gear pair wit h modified cen tr e distance is also called an angular correction gea r pair wit h x 1 + x 2 ≠ 0 . This is a general case and toot h t hick nesses on bo t h gears can be i ncreased or de- cr eased . Usually , t he pinion should not be negatively modified . T he working pressure angle and t he working cen t re distance can m eet special require men ts . T he m ain task in designing a corrected gea r pair is t o determi ne t he modification coeffi- cien ts w hich should meet t he following require men ts . ( a ) The gearing equation wit hou t backlash m ust be satisfied . (b ) N o cut ter in terference, i .e . x 1 ≥ x 1 mi n and x 2 ≥ x 2 m i n . (c ) T he toot h t hick ness sa on t he addendum circle s hould be large enough , i .e . s a ≥ m . (d ) The con tact ra tio should be large enoug h , i. e . εα≥ 1. 2 .

6.9   H e l i c a l G e a r s f o r P a r a l l e l S h a f t s
6.9.1   Generation and Characteristics of Helical Teeth U n til now , gears have been discussed only in t he tr ansverse plane . In fact , gea rs always have a cer tai n facewidt h . The profile of a planar gea r is form ed by a st raigh t line K K on a gen- er ati ng plane as t hat plane rolls wit hou t slipping on t he base cylinde r . In t he case of a spur gear , t he str aigh t line K K is parallel to t he ax is of t he gear .T he t oot h surface of t he spur gear is t herefore an involu te cyli nder , as sho w n in Fig . 6-32 . T he toot h surfaces of t wo spur gears a re generat ed by t he sam e st raigh t line K K parallel to t he axes of the gears as t he gene rating plane rolls on two base cyli nders wit h pa rallel axes . The t oot h surfaces of spur gea rs cont act on ・ 1 27 ・

a st raigh t line par allel t o the axes of t he gears as sho w n in Fig . 6-33 . T his i mplies t hat too t h profiles go in to an d ou t of contact along t he w hole facewid t h at t he sa m e time . T his will t here- fore result in t he sudden loading and sudden u nloading on teet h as toot h profiles go int o and ou t of con tact . As a result , vibr ation and noise ar e produced . T hat is w hy t ransmission by spur gears is no t completely smoot h .     Helical gea rs a re produced to overcome t he shortcomings of spur gears . The st raight line K K on t he gener ating plane is no longer par allel to t he axis of t he gea r as sho w n in Fig . 6-34a . As t he gener ati ng plane rolls wit hou t slippin g on t he base cylinder , eve ry poin t on t he st raigh t line K K will produce an involu te . The curve con necti ng t he sta rting poin ts of t he involutes on t he base cylinde r is a helix . The surface profile of a helical gea r is t herefore called an i n vol ute helicoid .

Fig . 6-32

Fig . 6-33

Fig . 6-34

    The toot h surfaces of t wo helical gears for par allel shafts a re gene rated by t he sam e st raigh t line K K inclined to t he axes of t he gea rs as t he generating plane rolls on t wo base cylinde rs wit h ・ 1 28 ・

par allel axes , as s ho wn in Fig . 6-34b . The t oot h surfaces of two engaging helical gears con tact on a str aigh t li ne in- cli ned to t he axes of t he gears . T he lengt h of t he con tact line changes gr adually from ze ro t o m ax imum and t hen from m axi mum t o ze ro as sho wn in Fig . 6-35 . T he load- ing and unloading of t he teet h becom e gradual and smoot h . T ha t is w hy helical gears can oper ate at a h igher speed . 6.9.2   Parameters of Helical Gears T he re a re tw o sets of para mete rs for a helical gear .
Fig . 6-35

O ne set is on t he t ransverse plane and t he ot he r set on t he normal plane . Gea r teet h are cu t by moving or feeding t he standard t ool perpendicular t o t he norm al plane . T he cu tters a re t he sa me as t hose for spur gears wit h standard para met ers . T he refore , para m et ers on t he normal plane are t he standard values . On t he ot he r hand , it can be seen from t he genera ting process of t he profile show n in Fig . 6-34 t ha t t he t ran sve rse profile of a helical gea r is an involu te, t he sa me as in t he case of a spur gear . To make use of t he formulae for spur gears , t he par am eters in t he equations for spur gears should be r eplaced by t hose on t he tr ansverse plane of helical gears . T he refor e , it is necessa ry t o set up relationships bet ween bot h sets of pa ra meters . By un w rapping t he reference cylinder as s ho wn in Fig . 6-36 , it can be seen clea rly t ha t

Fig . 6-36

p n = pt cos β and m n = m t cos β

( 6 -36a ) ( 6-36b )

w her e subscript “ n ” i ndicates t he norm al plane and “ t ” indicates t he t ransverse plane . β is t he helix angle on t he refer ence cylinde r . It can be positive or negative , referring t o righ t hand- ・ 1 29 ・

ed or left handed . T he relationship between α n and α t can be seen clea rly i n a helical rack sho w n in Fig . 6-37 . △ abc is on t he t ransverse plane and △ a′ b′ c on t he norm al plane wit h ab = a′ b′ and a′ c = accos β . Then tan α n = ac accos β = = tan α t cos β ab ab ( 6-37)

T he addendum and dedendum of a standa rd helical gear can be calculat ed on t he normal plane as ha = h a*n m n hf = ( ha*n + cn* ) m n w her e ha n is t he normal coefficien t of addendum and cn is t he norm al coefficien t of bot tom clearance . Since t he cu t- ters for helical gears are t he sam e as for spur gea rs ,
* * * *

( 6-38) ( 6-39)

mn ,

h a n and cn take t he sam e standa rd values as for spur gears . T he refer ence dia mete r and cen tr e distance should be calculated on the t ransverse plane as           d = zmt = zmn cos β (6-40 )
Fig . 6-37

m n ( z1 + z2 ) 1           a= ( d 1 + d2 ) = (6-41 ) 2 2cos β I t can be seen from above t hat t he adjust m en t of t he cen tre distance can be done by not on- ly addendum modification bu t also changing t he heli x angle of t he gears . F ur t hermore , helical gears can also be modified as can spur gea rs if necessary . 6.9.3   Proper Meshing Conditions for Helical Gears As t ransverse profiles of helical gea rs a re involutes , a pair of helical gears moun ted on par- allel shafts should have t he sa m e module and pressure angle on t he t ransverse plane according to t he proper m eshi ng condition of spur gea rs , i .e . m t l = m t 2 , α t1 = α t 2 . In addition , teet h on bo t h gea rs should slan t at t he sam e angle . T his m eans β 1 = - β 2 for an ext ernal gear pair or β 1 =β 2 for an in ternal gear pair . Since cos β 1 = cos β 2 , m t 1 = m t 2 and α t1 = α t 2 i mply m n l = m n 2 and α n1 = α n 2 according t o Eq . (6-36 ) and Eq . ( 6-37 ) . T her efor e, t he modules and pressure angles of bot h gears on t he norm al plane ar e also equal t o each ot her , respectively . Bot h α n and m n should be standard values . 6.9.4   Contact Ratio for a Helical Gear Pair By t he definition of con tact ratio i n t he case of a spur gear set in Sec . 6. 6 , cont act r atio ε α = B 1 B 2/ p b , w here B 1 B 2 is t he lengt h of t he actual line of action , or t he distance moved by t he too t h pair along t he base circle during engagem en t ( Refer to Fig . 6-19 ) . p b is t he base pitch . S ho w n in Fig . 6-38 are unfoldi ng base cylinde rs for a spur gea r and helical gea r . Let us ・ 1 30 ・

compa re t he mesh ing processes of a helical gear pair and a spur gea r pair . Line B 2 B 2 indica tes t he sta rt po- sition of meshing and B 1 B 1 t he end position for t he spur gear pair . T he corresponding helical gears com e in to cont act also at B 2 B 2 , bu t gradually . O ne of t he end faces of t he helical gea r begins to come out of con- tact at B 1 B 1 , but t he teet h do not get ou t of con tact fully u ntil t he t oot h arrives at B 1 ′ B1 ′ . T his lasts for a distance Δ L as t he teet h are inclined at an angle β b on t he base cylinder . β b is t he helix angle on t he base cylinder wit h tan β b = π d b π d cos α t = = tan β cos α t l l
Fig . 6-38

( 6-42)

w her e l is t he lead of t he gear . I t can be seen in Fig . 6-38 t hat t he distance for t he helical gear t o move along base cyli nder during t he engagem en t is B 1 B 2 + Δ L . T he refore , t he con tact ra- tio for a helical gear pair is ε γ= B1 B2 + Δ L B1 B2 Δ L = + =ε α+ε β pb t p bt p bt ( 6-43)

w her e ε β is called t he face con tact r atio or overlap ratio . Hence ε β= B tan β cos α b t Δ L B tan β B sin β = = = p bt pb t pt cos α t πm n ( 6-44)

w her e B is t he facewid t h of t he helical gear . ε α is now called t he transverse con t act r atio and can be calculated wit h the para mete rs on t he t ransverse plane by ( 6-25) , i .e . ε α= 1 [ z 1 ( tan α ′ + z2 ( tan αa t 2 - tan α ′ a t 1 - tan α t ) t ) ] 2 π ( 6-45)

D ue to ε β , t he con tact ratio of a helical gear pair is m uch higher t han that of a spur gear pair . A large con tact ratio means t hat more gear teet h ar e engaged at t he sam e tim e w hich gives high st rengt h to t he gea rs . T his is one of t he advan tages of helical gea rs ove r spur gea rs . 6.9.5   Virtual Number of Teeth for Helical Gears Conside r t he normal plane n-n at poin t P on t he reference cylinder as sho wn in Fig . 6-39 . T he in tersection of t he reference cylinder wit h t he norm al plane is an ellipse wit h r as its shor t radius and r/ cos β as its long r adius . According to t he information in “ Advanced M at he m at- ics” , t he radius of curvat ure ρ of t he ellipse at poin t P is ( r/ cos β) 2 r ρ= = r cos2β A spur gear can be dr aw n using t he helical gea r values of ρ, h a n , cn ,
* * *

m n and α n t o re-

place r , h a , c * , m an d α, r espectively . Then t he number z v of teet h of t he spur gea r can be derived as ・ 1 31 ・

zv =

mt z 2ρ d z = = = 2 2 m n m n cos β m n cos β cos3β

( 6-46)

I t can be show n t hat t he t oot h profile of t he spur gear is equivalen t t o t hat of a helical gear on t he norm al plane . T he spur gea r is called t he v ir t ual gear o f t he hel ical gear . T he number z v of teet h of t he virt ual gear is called t he vir t ual numbe r of teeth for t he helical gear . zv is used t o select t he series number of milling cu tter if a helical gear is t o be cu t by a form cu tte r . It is also used to calculate t he st rengt h of t he too t h . F ur- t hermore , t he minimum num ber of teet h of t he stan- da rd helical gear wit hout cu tter in terfe rence can be de- rived from (6-46 ) as z mi n = z v m i n cos3β w her e zv
m in

( 6-47)

is t he min imal numbe r of t eet h of t he

standa rd vir tual spur gea r wit hou t cu tter in te rference . O bviously , st anda rd helical gea rs can have fewe r teet h t han standard spur gears . Compared to spur gears , helical gears oper ate mor e smoot h ly and have higher st rengt h . T he work- ing cen t re distance of a helical gear pair can be adjust- ed by t he helix angle β . H elical gea rs ar e t herefore widely used for h igh speed or heavy load tr ansm ission . T he main disadvan tage of helical gears is t hat t hey fore limited to wit hin 8°to 20°t o preven t excess axial t hrust .
Fig . 6-39

produce an axial t hrust w hich is ha rmful t o t he bearings of t he shaf ts . The helix angle is t here-

6.10   W o r m G e a r i n g
6.10.1   Worm Gearing and its Characteristics Worm gear drives show n in Fig . 6-8 ar e used t o tr ansm it motion and pow er between non- in tersectin g and non-parallel s hafts , usually c rossing at a righ t angle . Worms a re a kind of screw , usually righ t handed for convenience of cu t ting , or left handed if necessary . Worm w heels ar e the m ating gears of t he worms . Worms are usually drive rs to reduce t he speed . If not self-lockin g , a worm w heel can also be t he drive r in a so-called back-drivi ng mechanism to incr ease t he speed . T he following a re t he main cha racteristics of wor m gear drives: (1 ) Smoo t h silen t ope ration as scre w drives . (2 ) G reater speed reduction in a single step . T his mean s compact design s . (3 ) If t he lead an gle of a w orm is less t han t he friction angle, t he back-drivin g is self-lock- ing . In t h is case , only t he w orm can be t he drive r . T his function is sometim es required for ・ 1 32 ・

safet y . (4 ) Lo wer efficiency due t o t he greater relative sliding speed between t he teet h of t he wor m and worm w heel . The friction loss m ay result in overhea ting and serious w ea r . T here- fore , brass is usually used as t he mat erial for t he worm w heel to reduce friction and w ea r . 6.10.2   Types of Worms Cylindrical worms ar e most widely used in in dust ry . There a re str aigh t sided axial worms (ZA-worm , Fig . 6-40 ) , involute helicoid w orms ( ZI-worm , Fig . 6-41 ) , arc-con tact worms (ZC-w orm , Fig . 6-42) and so on . T he profile of a ZA- wor m on t he t ransverse plane is an A rch imedes spir al . The profile of a ZI- worm on t he tr ansverse plane is an involut e . T he profile of a ZC-worm on t he axial plane is a concave a rc . A mong t hese , t he ZA- worm is t he most sim- ple and called t he com mon worm toget her wit h t he ZI-w orm . T he ZC-worm has higher pow er capacit y and is more and more widely used in w orm reducers . In som e special cases , envelop- ing worms ( Fig . 6-43) an d spiroids ( Fig . 6-44 ) are used . T hese new kinds of w orm drives have higher power capacity and efficiency bu t a re complicated in m anufact ure .

Fig . 6-40

Fig . 6-41

・ 1 33 ・

Fig . 6-42

Fig . 6-43

Fig . 6-44

6.10.3   Proper Meshing Conditions for Worm Drives T he t ransverse plane of a w orm w heel passing t hrough t he axis of t he w orm is called t he mid-plane . T he engage ment bet ween a worm and a w orm w heel on t he mid-plane corr esponds t o t hat of a rack and pi nion as sho w n in Fig . 6-45 . T her efor e, t he modules and pressure an- gles of t he worm and worm w heel on t he m id-plane s hould be equal to each ot her . In ot her words , t he module m t 2 and the pressure angle α t 2 of t he w orm w heel on t he t ransverse plane should equal t he module m x 1 and t he profile angle α x 1 of t he w orm on t he axial plane r espective- ly , i .e . m t2 = m x1 α t2 = α x1 In addition , t he lead angle λ 1 of t he worm should be equal t o t he helix an gle β 2 of t he wor m w heel in t he case t hat t hey cross at a righ t angle , i .e . λ 1 =β 2 and t he directions of bot h helices should be t he sam e . T he above conditions are only those necessa ry conditions for a worm and a w orm w heel to m esh properly . In fact , t he worm w heel should be cu t by a cut ter wit h t he sa me par am eters as t hose of the m ati ng worm . F urt hermore , t he rela tive position betw een t he cu tter and t he wor m w heel a t t he end of cu tting s hould be t he sam e as t hat between t he worm and worm gear in engagem en t . 6.10.4   Main Parameters and Dimensions for Worm Drives (1 ) The profile angle of worm . The axial profile angle of w orm α x 1 is prescribed to be 20°. T his can be incr eased to 25°in powe r tr ansmission or decreased to 15°or 12°in indexing devices . (2 ) T he module . T he series of modules for worms is someho w diffe ren t from those for gears . Modules for wor ms ar e list in T able 6-2 . ・ 1 34 ・

Fig . 6-45

    ( 3) T he n umber of t hreads of t he worm z1 and t he number of teeth of a worm w heel z 2 . z 1 can be chosen from 1 to 10 and z 1 = 1 , 2 , 4 , 6 a re preferable . A sm all number is chosen if t he t ran smission ratio is h igh or self-locking in back-driving is required . A la rge number is tak- en t o increase t he efficiency . z 2 can be calculated from z2 = i 1 2 z1 and values bet ween 29 ~ 70 a re suggested for power t ransm ission . Table 6-2 Modules and Reference Diameters of Worms ( GB / T 10085—1988)
Module m Mid-diamete r d1 Module m Mid-diamete r d1 Module m Mid-diamete r d1 1 18 4 ( 31 .5) , 40 , (50) , 71 12 .5 (90) , 112 , (140 ) , 200 1. 25 20 , 22 .4 20 , 28 5 (40 ) , 50 , (63) , 90 16 (112 ) , 140 , ( 180 ) , 250 1.6 2 (18 ) , 22. 4 , (28) , 35 .5 6. 3 (50) , 63 , (80) , 112 20 ( 140 ) , 160 , ( 224) , 315 2 .5 (22. 4) , 28 , ( 35. 5) , 45 8 ( 63 ) , 80 , ( 100 ) , 140 3 .15 (28) , 35 .5 , (45 ) , 56 10 ( 71 ) , 90 , ( 112 ) , 160 25 (180 ) , 200 , ( 280 ) , 400

mm

    (4 ) The mid-diam eter d 1 of w orm and t he refer ence dia met er d 2 of w orm w heel .Since t he wor m w heel should be cut by a cu tt er wit h t he sa me shape as t he m atin g worm , t he mid-diam e- t er d 1 of t he worm is st anda rdized to reduce t he nu mber of cu tte rs . D1 should be chosen toget h- er wit h t he module from T able 6-3. T he rigidity and str eng t h of t he w orm shaft increase wit h d 1 .T he reference dia mete r of worm w heel can be calculat ed by d 2 = m z 2 . (5 ) The lead angle λ 1 of t he w orm , λ 1 can be calculated as follo ws : tan λ 1 = z 1π m m z 1 = πd1 d1 ( 6-48)

T he efficiency increases wit h inc rease of λ 1 .If λ 1 is less t han t he equivalent friction angle φ v , t he worm gear pair is self-locking in back-drivi ng . (6 ) The cent re distance a of t he worm gear pair .Th is can be calculated as follows . ・ 1 35 ・

a = r1 + r2 + x 2 m =

1 ( d1 + m z 2 + 2 x 2 ) 2

( 6-49)

Here x 2 is t he modification coefficien t of t he w orm w heel .Worm w heels a re equen tly mod- ified to get standa rd cen tre distances .Since t he shape of t he worm should be t he sa me as t he standa rd w orm w heel cu tte r , t he w orm should no t be modified .

6.11   B e v e l G e a r s
6.11.1   Types and Applications of Bevel Gears Bevel gears show n in Fig .6-6 are used to t ransm it mo tion and powe r betw een in tersecting shaf ts .T he teet h of a bevel gear a re distribu ted on the frustu m of a cone .T he corresponding cylinders in cylindrical gears become cones , such as t he r efe rence cone, addendum cone and de- dendum cone .The dim ensions of teet h on differ en t t ransve rse planes are diffe ren t .F or conve- nience , par am eters and dimensions at t he large end are taken t o be standard values .The pressure angle is nor mally 20°and modules should be chosen from T able 6-3 . Table 6-3   Modules of Bevel Gears( GB / T 12368—1990) mm

… , 1 , 1.125 , 1. 25 , 1. 375 , 1 .5 , 1. 75 , 2 , 2.25 , 2.5 , 2. 75 , 3 , 3. 25 , 3. 5 , 3 .75 , 4 , 4. 5 , 5 , 5 .5 , 6 , 6.5 , 7 , 8 , 9 , 10 , …

    The shaft angle of a bevel gear pair can be any r equir ed value .In most cases , t he t wo shafts in tersect at a righ t angle . As m en tioned above, t here are st raigh t ( Fig .6-6a ) , helical ( Fig .6-6b ) and spiral ( Fig .6-6c ) bevel gears .Str aigh t bevel gea rs a re most widely used as t hey are easy to design and manufac- ture .S piral bevel gears ope rate smoo t hly an d have higher load capacit y .F or h igh speed and heavy load , spiral bevel gears ar e preferable , such as in aeroplanes , au tomobiles and t ractors . Here , ho wever , only st raight bevel gears a re discussed . 6.11.2   Back Cone and Virtual Gear of a Bevel Gear T he bot tom of t he reference cone is called t he ref erence circle o f t he bevel gear wit h its ra- dius r = m z , as show n in Fig .6-46 .T he reference cone an d t he back cone of a bevel gear have t he sam e axis and t heir gene rat rices in te rsect perpendicularly at t he r efe rence circle .In Fig .6- 46 , O P A and O PB are t he refer ence cones of pinion 1 and gear 2 , respectively , w hile O1 P A and O2 P B a re t he back cones of pinion 1 and gear 2 , respectively .In Fig .6-47 , O P A and O 1 P A a re t he r efe rence cone and back cone of t he gea r 1 .The reference cone an gle of gea r 2 here is δ 2 = 90° and t he surface of t he reference cone becomes a plane .S uch a bevel gear is called a crow n gear .T he back cone of a crow n gea r is a cylin der . T he back cone can be u nfolded im agina rily in to a sector gear with z teet h as show n in Fig . 6-46 and furt her filled up to a full gear .T his i magi nary gear is called t he v irt ual gea r o f t he bevel gear .The too t h numbe r of t he virt ual gear is called the virt u al n u m ber of teet h z v of t he ・ 1 36 ・

bevel gear .T he t oot h profile of t he vir tual gea r is almost t he sa me as t hat of t he bevel gear at t he large end .The module and pressure an gle of t he virt ual gear ar e t hose of t he bevel gear .T he engage men t of t he virt ual gea rs is equivalent to t hat of t he bevel gears . T he reference radius r v 1 of t he vir t ual gear 1 is t he back cone distance O1 P as in Fig .6-46 for t he pin ion ,i .e . rv 1 = O1 P = r1 z1 m = cos δ 1 2cos δ 1

O n t he o ther hand , d v 1 = zv 1 m .T his results in

Fig . 6-46

Fig . 6-47

zv1 = Sim ilarly ,

2 rv1 z1 = m cos δ 1 z2 cos δ 2

( 6-50)

zv2 =

( 6-51)

T he en gage men t of bevel gears can now be u nderstood wit h t heir vir tual gea rs .For a pair of bevel gea rs to m esh prope rly , t heir respective modules and pressure angles at t he la rge ends should be equal .T he cont act r atio of t he bevel gea r set .The vir t ual num ber of t eet h z v s hould not be less t han t he minim um number of teet h of t he vir tual gear . 6.11.3   Parameters and Dimensions of Bevel Gears As men tioned above , most dim ensions of bevel gea rs a re m easured at t he la rge en d .First of all, t he reference dia mete r is d = m z = 2 R sin δ t ransmission r atio of a gear pair is ・ 1 37 ・ ( 6-52) w her e R is t he ou ter cone distance and δ t he refer ence cone angle as show n in Fig .6-48 .T he

i1 2 =

ω z2 d 2 si n δ 1 2 = = = ω 2 z1 d 1 si n δ 1

( 6-53)

W hen t he s hafts of tw o gears in tersect at a righ t angle , i .e . Σ = δ 1 +δ 2 = 90° , t he t ransm ission r atio becom es i1 2 = tan δ 2 = cot δ 1 and t he ou ter cone distance becom es R=
Fig . 6-48
2 r2 1 + r2 =

( 6-54)

m 2

2 z2 1 + z2

( 6-55)

T he addendum and dedendum ar e measur ed a- long t he genera trix of t he back cone and t her efore

t he addendum dia mete r an d dedendum dia m eter of a st anda rd bevel gea r becom e d a = d + 2 ha cos δ= d + 2 h a* m cos δ d f = d - 2 h f cos δ= d - 2 ( h a* + c * ) m cos δ For t he bevel gea r , h a* = 1 and c * = 0. 2 , not 0. 25 . T he apexes of t he dedendum cone and t he r efe rence cone coincide .T he dedendum angle θ f and dedendum cone angle δ f can be calculated as follows . θ f = arctan hf R ( 6-58) ( 6-59) ( 6-56) ( 6-57)

δ f = δ- θ f

If t he addendu m cone has t he sam e apex as t he reference cone , t hen t he bot tom clea rance changes along t he facewidt h .I n t his case , t he addendum angle θ a and addendum cone angle δ a can be calculated as follo ws . θ a = arctan ha R ( 6-60) ( 6-61)

δa = δ+ θ a     F or t he gea r pair to have constan t bot tom clea rance eve ry w he re along t he facewid t h , as show n in Fig . 6-49 , t he addendu m angle θ a should be t he sam e as t he dedendum angle θ f . Such a design can avoid t oo sm all bot tom clea rance at t he sm all end and ensure lubrication . The ad- dendum cone angle δ a in t his case becom es δ a = δ+ θ f ( 6-62) In such a design , the apex of t he addendum cone does no t coincide wit h t hose of t he dedendum cone and t he reference cone .
Fig . 6-49

・ 1 38 ・

Problems and Exercises 6- 1   U nder w hich conditions can a pair of gears ope rate uniformly , continuously and smoot hly ? 6-2   W hat cha racteristics do involute gears have ? 6-3   W hat is t he con tact r atio ? Ho w does it r elate to t he numbe rs of teet h z1 , z2 , module m , pressure angle α, coefficient of addendum h a and w or king cen ter dist ance a′ ? 6- 4   H ow can t he module , pressure angle , numbe r of teet h and modification coefficien t of a gear be obtained w hen it is cu t by a rack-s hape cut ter in t he gene rating m ethod ? 6- 5   W hat are t he differences betw een t he pitch circle and t he reference cir cle , bet ween t he wor king pressure angle and t he pressure angle ? Ho w ar e they related to each ot her ? 6-6   W hat is cut ter in terfe rence and how can it be avoided ? 6-7   W he re is adden du m modification useful and ho w do t he par am eters z , d , d a , df , d b , s , e change in addendu m modification ? 6-8   O n w hich plane ar e t he para mete rs of a helical gear defined as standard values and w hy ? 6-9   W hat is t he vir tual gear of a helical gea r and w hy is it in troduced ? 6-10   What ar e t he advantages and diadvan tages of helical gea rs ? 6-11   W hat is the mid-plane of a worm drive ? W hy is t he mid-diam eter of a w orm defined as a standa rd value ? 6-12   What ar e t he back cone an d vir tual gear of a bevel gear ? Why are t hey in troduced ? 6-13   An involu te has t he radius of base circle r b = 50 m m . Dete rmine t he follo wing .             g angle θK , pr essur e angle αK and radius of curvatur e ρ K of t he involute on gle α K and radius of circle r K if θ K = 20° . t he circle r K = 65 m m . 6-14   A st anda rd spur gea r has pressure angle α = 20° , number of teet h z = 40 and tip dia m eter d a = 84 mm . Determine t he module m . 6- 15   A n involut e standa rd spur gear has pa ram eters z = 26 , sur e angle on t he tip circle . 6-16   A pair of standard spur involu te gears has a module of 5 mm , pressure an gle α = 20° , cen t re distance a = 350 mm , t ransmission ratio i1 2 = 9/ 5 . Calculate t he numbers of teeth z 1 , z 2 , reference dia m eters d 1 , d 2 , addendum diam eters d a 1 , d a 2 , base diam eters d b 1 , db 2 , t oot h t hickness s and spacewid t h e . 6-17   H ow m any teet h w ould an ext ernal standa rd spur involu te gea r have w hen its deden du m circle and its base circle coincide ? W hich one is bigge r as t he numbe r of teet h increases ? 6-18   For a pair of ex ternal stan dard spur gea rs , α ′ = α = 20° , m = 5 m m , z 1 = 19 , z2 = 42 . Calcula te t he con tact r atio ε α . 6-19   A pair of exte rnal spur gea rs have t he numbe rs of teet h z 1 = 30 , z 2 = 40 , module m = 20 ・ 1 39 ・ m = 3 mm , ha* = 1 , α = 20°. Fin d t he radii of curvat ure of t oot h profile on t he refer ence circle and t he tip circle and t he pres- m , α, h a , hf ,


mm , pr essur e angle α= 20°. Determine t he working pr essure angle α ′ w hen t he w ork ing cen- tr e distance is a′ = 725 m m . If α ′ = 22° 30′ , find t he w orki ng cen tre distance a′. 6-20   Given a pair of ex ternal spur gea rs with z1 = z2 = 12 , m = 10 mm , α = 20° , ha* = 1 , z2 = a′ = 130 mm , x 1 = x 2 . Calculate x 1 and x 2 , check t he cu tte r i nte rfer ence . 6-21   There is a pair of ex ternal standard spur gea rs in a shaping machi ne wit h z1 = 17 , 118 , m = 5 m m , α= 20° , ha* = 1 , a′ = 337 .5 mm . T he pinion is worn ou t an d t he gear is worn to such an ex ten t t hat t he toot h t hickness is decr eased by 0 .75 m m . Ho w can t he gea r be repaired by addendum modification and a new pinion be design to m esh wit h t he repaired gear ? 6-22   A pair of standard ex ternal helical gears have t he following par am eters:


z1 = 20 , z 2 =

118 , m = 5 m m , α n = 20° , h a n = 1 , B = 30 m m , a = 350 mm . Find t he helix angle β, total cont act ratio ε γ and t he virt ual numbers of teet h z v 1 , zv 2 . 6-23   A standa rd w orm w heel has t he number of teet h z2 = 40 , refer ence diam eter d 2 = 200 mm . It m eshes wit h a single-t hreaded worm . Determ ine (1 ) Module of t he w orm gea r set on t he mid-plane m t 2 and m x 1 ; (2 ) A xial pitch p x 1 an d lead of t he w orm ; (3 ) Cent re distance a wit hout modification . 6-24   A pair of st raight bevel gea rs have para m eters z1 = 15 , z 2 = 30 ,


m = 5 mm , ha* = 1 ,

c = 0 .2 , Σ = 90°. Det ermine ot her di mensions d 1 , d 2 , d a1 , d a 2 , df 1 , df 2 , δ 1 , δ 2 , δ a1 , δ z v 1 and zv 2 with con stant bott om clea rance . a2 , δ f1 , δ f2 ,

・ 1 40 ・

C h ap t e r 7 Gea r T r ai n s
7.1   G e a r T r a in s a n d T h e i r C l a s s i f i c a t i o n
In t he previous chap ter , t ransmissions wit h only a pair of gea rs are conside red . In many practical cases , t his is not enough . In various kinds of m achines , it is of ten necessa ry to use series of gea rs to m eet different require men ts . Such a t ransm ission syste m by more than one pair of gears is called a gear t rai n . T he re a re differ en t kinds of gear trains as show n i n Fig . 7-1 . If t he position s of all gear axes in a gear t rain a re fixed , it is called an ord i nar y gear tr ai n or gear tr ai n w it h f ixed axes . T his can fur t he r be divided in to gea r t rains wit h fixed par allel axes and gear tr ains wit h fixed non - parallel axes . If at least one gea r axis i n a gea r t rai n rotates around t he ot he r axis , t he gear t rai n is called an ep icyclic gear trai n . F or example , t he axis of gea r 2 in Fig . 7-3 rota tes a round t he axis of gear 3 . T he refor e , t he gear t rain in Fig . 7-3 is an epicyclic gear tr ain . Epicyclic gear t rains can fur t he r be divided in to ele ment ary epicyclic gear t rains and combined gear tr ains . S how n in Fig . 7-3 is an elem en tary epicyclic gear t rai n . In t his gea r t rain , t he axis of gear 2 rot ates around t he fixed axis O - O . At t he sam e tim e, gear 2 also rotates rela- tive t o t he link H around its ax is A - A . T he move ment of gea r 2 is simila r to t ha t of a planet in the solar system . Such gea rs a re t he refor e called pl anet gears . T he link carrying planet gears is called t he p lanet ca rrier , denoted by H . T he gears engaging wit h planet gears and ro- tating around t he sa me fixed axis as t hat of t he planet ca rrier are called su n gears . T he re is one planet ca rrier , tw o sun gears , and one or more planet gears in an elem en ta ry epicyclic gear t rain . Su n gea rs and t he planet carrie r are called t he f u nda men t al m em bers i n an elemen t ary epicycl ic gea r trai n , as t hey rotate a round a fixed axis on t he fra me . In such an ele men tary epicyclic gear tr ain , if neit he r of the tw o sun gea rs is fixed as show n in Fig . 7-3 , t he degrees of freedom of t he m echanism a re given by F = 3 × 4 - 2 × 4 - 2 = 2 and it is called a di f f eren ti al gear tra i n . Ot he rwise , if one of t he sun gea rs is fixed wit h t he fr am e as sho wn in Fig . 7-4 , t he degrees of freedom of t he m echanism are given by F = 3 × 3 - 2 × 3 - 2 = 1 and t he gear t rain is called a p l anet ary gea r trai n . A combined gea r t rain is a com bi nation of sever al ele men tary epicyclic gear t rains or a com bi nation of at least one ele men ta ry epicyclic gea r tr ain wit h at least one ordinary gear train , as show n in Fig . 7-6 . In t his chap ter , the calculation of the train ra- tios , or t ransm ission ratios , of differ en t kinds of gear tr ains will be st udied first . Lat er , appli- ・ 1 41 ・

cations , efficiency and design proble ms will be discussed to som e ex ten t .
gea r t rai ns wit h fi xed a xes Gear t rains elemen tary e picyclic gear t rains epicyclic gear t rains combi ned gea r t rains Fig .7-1 gear t rains wit h fixed pa rallel axes gear t rains wit h fixed non-parallel ax es pla netary gear t rains differe ntial gea r t rai ns

7.2   T r a i n R a t i o o f a G e a r T r a in w i t h F i x e d A x e s
T he t rain ratio of a gea r train is t he r atio of t he angula r velocities of inpu t and ou tpu t m em- bers in t he gear tr ain . The t rai n ratio here includes tw o factors , t he m agnitude and t he relative ro tati ng dir ection of t he tw o m em bers . 7.2.1   Absolute Value of Train Ratio Conside r t he gear tr ain in Fig . 7-2 as an example . Gea rs 1 and 2 form a pair of ex ternal cylindrical gea rs and gea r 3 is an in ternal one . Gea rs 3′ and 4 form a pair of bevel gears . T he gears 4′ and 5 a re a w orm gea r pair . Suppose gea r 1 is t he driving w heel (i npu t m ember ) and gear 5 t he driven w heel ( ou tpu t me mbe r ) . T hen t he absolu te value of t he tr ain r atio of t he gear train is | i1 5 | = | ω1/ ω 5 | . T he absolute values of t he t ransm ission ratios of every pair of m eshing gears a re as follows | i1 2 | = | i2 3 | = | i3 4 | = | i4 5 | = ω3 ω4 ω4 ω5 ω1 z2 = ω2 z1 ω2 z3 = ω3 z2 = = ω3 z4 = ω4 z3 ω4 z5 = ω5 z4 ω1 ω 2 ω 3 ω 4 ・ ・ ・ ω2 ω 3 ω 4 ω 5 ω 1 ω 5 (a) (b ) (c) (d )
Fig .7-2

Multiplying t he above equations side by side, we get | i1 2 i2 3 i3′ 4 i4′ 5 | = = = | i15 | =

z2 z3 z 4 z 5 z1 z 2 z 3′z 4′

(7-1)

T his means t hat t he absolute value of t he tr ain ratio of a gear t rain wit h fixed axes is t he product of t he absolu te values of t ransmission ratios of every pair of m eshing gears composing t he gear t rain . Its absolute value is t he ratio of t he product of teet h numbers of all t he driven gears over t hat of all t he driving gears , i .e. | T rain r atio of gear t rain wit h fixed axes | = product of t oot h num bers of all t he driven gea rs product of too t h numbe rs of all t he driving gea rs (7-2) ・ 1 42 ・

7.2.2   Relative Rotating Directions of Gears In t he above gea r t rain , t he rotating direction of the gear 1 is given as show n in Fig . 7-2 w her e t he dir ection of t he arrow i ndicates t hat of t he pe rip her al velocit y of t he gea r on t he visi- ble side . Wit h t he help of arro ws , t he rotating directions of ot her gears can be easily deter- mined . For cylindrical and bevel gea rs , arrows of m eshing gears bot h poin t eit her t owa rds or away from t he pitch poin t . For t he w orm gea r pair , a m et hod is suggested to determi ne t he relative dir ection of ro tation as follows . T he lef t hand is used for righ t-handed worms and t he right hand for left-handed . T he hand grasps t he worm such t hat t he finge rs poin t in t he rot at- ing direction of t he w orm . T hen t he t humb indicates the direction of peripheral velocit y of t he wor m gear at t he pitch poi nt . In t his way , t he rotating direction of every gear in t he gear t rain can be dete rmined as show n in Fig . 7-2 . I t is easy t o see t hat tw o gea rs wit h pa rallel shafts rot ate in t he sa m e direction in an in ternal gear pair or in opposit e directions in an ex ternal gear pair . I n t he first case, t he t rain ratio is defi ned positive and in t he second case negative . I n a gear t rain wit h fixed parallel axes , t he sign of t rain ratio can be determined by ( - 1) m w here m is t he number of ex ternal gea r pairs in t he gear train . In ot he r words , if t he re ar e odd num bers of ex ternal gear pairs between t he in- put and ou t pu t gears , t he t rain r atio is negative , o t her wise positive . A tten tion should be taken he re t hat t he rotating directions of gea rs i n gear tr ains wit h fixed non-parallel axes can not be determ ined by ( - 1)
m

. T hey can only be determ ined by drawing

a rrows , as show n in Fig . 7-2 . When t he axes of bot h inpu t and out pu t gea rs ar e pa rallel to each ot he r , t he directions of t he t wo gea rs will be t he sam e or opposite . T he train r atio m ust be positive or negative . In ot he r words , t he “ + ” or “ - ” sign must be added befor e t he r atio of teet h numbe r t o indicate t heir relative ro tating directions . T he gear 2 in t he above gear t rain engages simultaneously wit h the gears 1 and 3 . It is t he driven gea r for gea r 1 and driving gear for gea r 3 . Its number of teet h can be cancelled in Eq . ( 7-1 ) . T he function of gea r 2 is to change t he rota ting dir ection of t he out pu t gea r , not t he absolu te value of t he t rain ratio . S uch gea rs are called i d le w heels .

7.3   T r a i n R a t i o o f E l eme n t a r y E p i c y c l i c G e a r T r a in
Sho wn in Fig . 7-3a is a typical elem en tary epicyclic gear t rain . It consists of t wo sun gears (gea r 1 and gear 3 ) , one planet gea r ( gear 2 ) , one planet ca rrier H , and t he fram e . Suppose t hat the planet carrier H rotates in an angular velocit y ωH . T he t rai n ratio of t his gear t rain can no t be calculated si mply as a gear t rain wit h fixed axes because of t he rotation of t he planet carrier . I magine t ha t we add an angular velocit y ( - ωH ) to t he w hole gear t rain . I t will keep t he relative motion betw een any t wo lin ks unchanged . N ow t he angular velocity of t he planet ca rrier H is ωH - ωH = 0 . T his m akes t he planet carrier H fixed , as show n in Fig . ・ 1 43 ・

7-3c . T he refore , t he position of all t he axes of t he gears a re fixed and t he gear tr ain is convert - ed in t o a gear tr ain wit h fixed axes , as show n in Fig . 7-3c . Such an imagina ry ordina ry gear t rain is called t he conver ted gear t rain of t he original ele men tary epicyclic gear train .

Fig . 7-3

Since t he convert ed gear t rain is a gea r t rain with fixed axes , its t rain ratio can be calculat - ed as for an ordina ry gear t rai n . In t he conve rted gear t rai n show n in Fig . 7-3c , t he angular velocities of t he gears 1 and 3 ar e ( ω1 - ωH ) and ( ω 3 - ω H ) , respectively , not ω 1 and ω 3 .
H H H H T hey a re denoted as ω1 and ω 3 , respectively , i . e . ω 1 = ω 1 - ω H and ω 3 = ω 3 - ω H . T he H t rain ratio between t he gears 1 and 3 in t he conver ted gear t rain is denot ed as i1 3 . It can be de-

rived in t he sam e way as for a gear train wit h fixed axes as follows . i
H 13 H ω 1 ω 1 - ω H z2 z3 z3 = H = = = ω z1 z2 z1 ω 3 - ω H 3

(7-3)

H T rain r atio i 1 3 is positive w hen gears 1 and 3 in t he conver ted gea r tr ain ro tate in t he sa me

direction and negative w hen opposit e . T hat is w hy t here is a “ - ” symbol befor e t he too t h number ra tio in t he above equation . This sign cannot be neglected and should be ca refully de- termined . Example 7-1 In t he gea r t rain in Fig . 7-3 , z1 = z 2 = 30 , z 3 = 90 , n 1 = 100 r/ min clock wise , n 3 = 100 r/ mi n coun te rclock wise . Find n H . Solution: Eq . ( 7-3 ) can be w rit ten as
H i1 3 =

n1 - n H z3 = n3 - n H z1

Not e again t ha t t he “ - ” sign befor e t he t oot h number ratio is because of t he opposite di-
H H rections of ω 1 and ω 3 in t he conver ted gear t rain , no t because of t he opposite directions of n 1

and n 3 . Usually , t he value of angular velocity is defined as positive w hen cou nte rclock wise . ・ 1 44 ・

T hen n 3 = + 100 r/ m in and n 1 = - 100 r/ m in . Inse rting t he values of z 1 , z 3 , n 1 and n 3 , we get - 100 r/ mi n - n H 90 = + 100 r/ mi n - n H 30 and t hen n H = + 50 r/ m in T his m eans that t he planet carrier H ro tates in 50 r/ min coun terclock wise . Be careful again her e t hat the values of n 1 and n 3 should be i nser ted in to the equation toget her wit h t heir sign s . In an ele men tary epicyclic gea r t rain , we s hould distinguish i1 3 from i1 3 , and ω 3 from
H H ω 3 , and so on . i 1 3 is dependent only on t he st ructur e of t he gear tr ain and has not hing t o do H

wit h n 1 and n 3 . How ever , i1 3 can be affected by t he inpu ts of t he gea r t rain s . T he directions
H H of ω3 and ω 3 can be opposite . T he sign of i1 3 , or t he sign before t he t oot h numbe r ratio , is

determi ned by t he r elation ship betw een ω1 and ω3 , not t he relation- ship bet ween ω1 and ω 3 . Example 7-2 In t he planeta ry gear tr ain sho w n in Fig . 7-4 , 101 , z2′= 100 , z3 = 99 . Find t he t rain ratio i H 1 . Solution: For t he planeta ry gear t rai n sho w n in Fig . 7-4 ,
H i1 3 =

H

H

z 1 = 100 , z 2 =

Fig . 7-4

ω1 - ωH ω z2 z3 1 - ω H = = ω3 - ωH - ωH z 1 z 2′

T he refore , ω1 z 2 z3 101 × 99 1 =1=1 = ωH z1 z2 ′ 100 × 100 10 000 iH1 = ω H = 10 000 ω 1

I t can be seen t hat t he t rai n r atio her e is ve ry la rge . Bu t only t he planet ca rrier H , not gear 1 , can be t he inpu t m embe r and the gea r t rain can be used only t o decr ease t he speed , not t o i ncrease it . Ot her wise , t he gea r t rain will be self-locking . If z 3 is 100 instead of 99 i n t his example , we will get ω 1 z2 z3 101 × 100 1 =1 =1 = ω z 1 z 2′ 100 × 100 100 H i H 1 = - 100 T his m eans t hat increasi ng z 3 by only 1 changes not only t he m agnitude bu t also t he dir ec- tion of t he angular speed of t he ou tpu t gea r 1 . ・ 1 45 ・

So far , on ly t he angular velocities of funda m en tal m embers have been consider ed . If t he axis of a planet gear is pa rallel t o t he axes of sun gea rs and t he planet carrier , t he kinem atic re- lation ships bet ween t he planet gea r and t he fundam en tal m embers can also be derived in t he sa me way . For exa mple, in t he gea r t rain sho wn in Fig . 7-3 , t he following relationships hold : i12 =
H

ω1 - ωH z2 = ω2 - ωH z1 ω 2 - ω H z3 = ω z2 3 - ω H

H i2 3 =

Bu t t hese t wo equations do not hold for t he ele men ta ry epicyclic gear t rai n show n in Fig . 7-5 , w hich consists of bevel gears . As t he axis of planet gear 2 is not par allel to t he axis of plan-
Fig . 7-5

et carrier H , t he algebraic value ( ω 2 - ω H ) is mean ingless . Neve rt heless , k ine matic relation- ships betw een t he fundam en tal me mbe rs can be determined by t he follo wing equation : i
H 13

ω1 ω1 - ωH z 2 z3 = H = = = -1 z 1 z2 ω3 ω3 - ωH

H

(7-4)

H T he value of i 1 3 is negative because t he rotation directions of t he gears 1 and 3 i n t he con-

vert ed gear train are opposite , as t he arrows show in Fig . 7-5 . No te t hat t he arrows indicate t he rota tion directions of t he gears 1 and 3 in t he conve rted gear tr ain , not t hose in t he original elem en ta ry epicyclic gear t rain . In t his differen tial gear t rain wit h t wo degrees of freedom , t he ro tation directions of tw o drivin g gea rs 1 and 3 may be t he sam e .

7.4   T r a i n R a t i o o f a C o mb i n e d G e a r T r a i n
Befor e calculatin g t he tr ain ratio of a com bined gear t rain , it is necessary to divide t he combined gear tr ain i nt o several elem en ta ry epicyclic gea r t rains and ordinary gear t rains and t hen to w rite t heir t rain ratio equations independen tly . First , find t he planet gears ( t he axes positions of w hich are not fixed ) and t heir corre- sponding planet carrie rs . Those gears engaged wit h t he planet gea r and havi ng t he sa me fixed axis as t hat of t he planet ca rrier ar e t he corr esponding sun gea rs . There may be more t han one planet gear in an elem en ta ry epicyclic gear t rain . If more t han one planet ca rriers ar e found , t hen t hey must belong t o diffe ren t ele ment ary epicyclic gear t rai ns . Some me mbers of a com- bined gea r t rain m ay belong to and play differen t roles in different subt rains and t hey become links among t he subt rains . To de rive t he t rain ratio of a combined gear t rain , the tr ain ratio equa tions of all subt rains should be w ritten first . I n every t rain r atio equa tion for an epicyclic gear t rain , t he angular ve- locit y of t he planet carrier should appea r and t hose of t he planet gea rs should be avoided . In t h is way , t he minimu m necessary tr ain r atio equations can be written easily . T hen t he t rain ・ 1 46 ・

ratio of t he com bined gear t rain can be found by solving t he equations sim ultaneously . T he fol- lowing exa mples illustr ate t he steps in analysing a combined gea r t rain . Example 7-3 T he numbers of teet h of t he gears in t he gea r t rain show n in Fig . 7-6 a re given as z1 , z2 , z 2′, z 3 and z 4 . Calculate t he t rain ratio i 1 H . Solution: (1 ) Dividing t he gea r t rain First , t he planet gear is searched for . The gea r 3 is found t o be a planet gea r . T he link H is its planet ca rrier . The gea rs 2′and 4 en- gage wit h t he planet gea r 3 and have t he sam e fixed axis as t hat of t he planet carrier H . T he refore , t he gears 2′ and 4 ar e t he corresponding sun gea rs . T he planet gear 3 , t he planet carrier H , and t he sun gears 2′ and 4 make up an elem en ta ry epicyclic gear t rain . T he r em aining gears 1 and 2 make up a gea r t rain wit h fixed axes . T he gea r 2′is fixed wit h the gear 2 . (2 ) Deriving t he tr ain r atios of subt rains independen tly T he t rain ratio equations of t he t wo subt rains ar e: i1 2 = i 2′ 4 =
H

Fig . 7-6

ω 1 z2 = ω z1 2

(a) (b )

ω 2′ - ω H ω 2 - ω H z3 z4 z4 = = = ω4 - ωH - ωH z2 ′z3 z 2′ ω 2 z4 =1+ ωH z 2′

From Equation ( b ) we have : (c)

(3 ) Fin d t he tr ain ra tio of the combined gear t rain Multiplying Eqs . ( a ) and ( c ) gives t he result : i1 H = ω 1 ω1 = ωH ω2 ω2 ωH = z2 z4 1+ z1 z 2′

Not e: t he t rain ratio of t he combined gea r t rain can not be calcula ted by taking t he w hole gear t rain as an elem en ta ry epicyclic gear train and t hen adding ( - ωH ) to t he w hole gear t rain . The following equation is wron g
H i1 3 =

ω1 - ωH z2 z3 z4 = ω3 - ωH z 1 z 2′z 3

because in such a conver ted mechan ism ( addi ng - ωH to t he w hole gea r t rain ) , t he gear 1 be- comes a planet gear an d t he conver ted m echanism is not a gear t rain wit h fi xed axes . It is t herefore necessa ry t o divide t he combined gear t rain in to several ele men tary epicyclic gear t rains and gear t rains wit h fixed axes befor e calculatin g t he t rain r atio of a combined gea r tr ain . ・ 1 47 ・

From Fig . 7-6 , we can see t hat t he combined gea r t rai n is formed by connecting t he gear t rain wit h fixed axes and t he ele men tary epicyclic gear tr ain in series . The total t rain ratio is t he product of t heir t rain ratios . Example 7-4 A r educer of a hoist is show n in Fig . 7-7a . N umbe rs of teet h of all t he gears a re given . Fin d t he tr ain ra tio i1 5 . Solution: (1 ) Dividing t he gea r t rain T he given combined gear train in Fig . 7-7a can be divided in to t wo sub trains as show n in Fig . 7-7c and d . Gears 2 and 2′ a re planet gea rs . T he gea rs 1 , 2 ( 2′ ) , 3 and 5 m ake up an elem en tary epicyclic gear t rain ( differen tial gear t rai n ) in w hich t he gear 5 serves as t he planet carrier on ly . T he gears 3′ , 4 and 5 constitu te an ordina ry gear tr ain . T he gea r 3′is fixed wit h t he gear 3 . T he st ructure of t he syst em can be illustr ated by t he block-sche me in Fig . 7-7b . (2 ) Deriving t he tr ain r atios of subt rains independen tly

Fig . 7-7

T he gears 1 , 3 , and 5 a re t he funda men tal m embers in t he differ en tial gea r t rai n sho w n in Fig . 7-8c . We t herefore derive the t rain r atio equation as i5 13 = ・ 1 48 ・ ω z2 z3 1 - ω 5 = ω z1 z2 ′ 3 - ω 5 (a)

In t he gea r t rain wit h fixed axes sho w n in Fig . 7-7d , we have i3′ 5 = ω 3′ ω 3 z5 = = ω 5 ω5 z3′ (b )

(3 ) Fin di ng t he tr ain ra tio of the combined gear t rain From Equation ( b ) we get ω 3 = z5 ω5 z 3′

Inse rting th is r esult in to Equation ( a ) , we have ω 1 - ω 5 ω1/ ω5 - 1 z2 z 3 = = z5 z5 z 1 z 2′ - ω5 - ω5 -1 z 3′ z 3′ from w hich i15 = ω1 z 2 z 3 z5 = +1 +1 ω5 z 1 z2′ z 3′ (c)

Substit uting t he numbe rs of teet h in to Equation ( c ) , we get i15 = 33 × 78 78 1+ + 1 = 28 .24 24 × 21 18

T he fundam en tal m embers 5 and 3 in t he differen tial gea r t rain sho wn in Fig . 7-7c are re- lat ed by t he ordina ry gea r t rain show n in Fig . 7-7d . T he degree of fr eedom of t he w hole gear t rain becomes one . Such a combi ned gear t rain is called a closed epicyclic gea r t rain .

7.5   A p p li c a t i o n s o f G e a r T r a in s
Gear t rains ar e widely used in m any ki nds of m achines t o pe rform t he following functions : (1 ) Br anching tr ansmission A gear tr ain m akes it possible for a mot or to drive mor e than one driven shaft at t he sa me tim e . In a machine tool , for example , t he spindle and t he feeding screw a re gene rally driven by t he sa me motor t hrough gea r t rains . (2 ) To get a large t rain ratio W hen a la rge t rai n ratio is needed , a gea r t rain is a bet ter choice t han only a pair of gears . Especially , an epicyclic gear t rain can provide a ve ry large t rain ra tio , as show n in Exam- ple 7-2 . (3 ) To change t he speed of rotation Fig . 7-8 shows a tr ansmission box . Besides t he differen- tial gear t rain , t here are a brake T to brake t he gea r 3 and a clu tch C t o connect t he gear 3 and t he planet ca rrier H . Gear 1 is t he input m ember and H t he ou t put . When t he brake T
Fig .7-8

・ 1 49 ・

is on and t he clu tch C is off , we get a planeta ry gear t rain an d t he tr ain ratio can be calculat ed as
H i1 3 =

ω ω z3 1 - ω H 1 - ω H = = ω 3 - ω H - ωH z1 i1H = ω1 z3 =1+ ωH z1

W hen t he br ake T is off and t he clu tch C is on , all t he moving m embers are fixed toget h- er . ω 1 =ω H and t he t rain ratio becom es i1 H = 1 . Such kinds of speed change a re easy t o handle and t herefore used widely in vehicles . (4 ) To combine or resolve t he motion T he re can be tw o inpu t funda men tal m embers in a differ en tial gea r t rai n as it has t wo degrees of freedom . T his means t hat a differen tial gear t rain can combine t he motions of t wo inpu t shafts t o get t he t hird one . For t he gear t rain sho wn in Fig . 7-5 , we get from Eq . ( 7- 4) t ha t ωH = 1 ( ω1 + ω3 ) 2 (7-5)

w her e ω 1 and ω 3 a re combined to get ω H . S uch gear tr ains a re used in som e machine tools like hobbing m achines . T he sa me differential gea r tr ain can also be used inversely t o resolve t he motion . Show n in Fig . 7-9 is a rear w heel differential gear t rain of an aut omobile . S un gears 1 and 3 drive t he tw o rea r w heels sepa- rately . T hey are bot h driven by bevel gea r 4 as t he planet carrie r H . If t he au tomobile runs st raig ht ahead , t he gea rs 1 and 3 turn at t he sam e angular velocit y , i .e ., ω1 = ω3 . Inser ting t his in to Eq . (7-5) , we get ωH = ω 1 =ω 3 T his means t hat t he differential gear t rai n turns as a w hole and t he friction loss in it is avoided . If t he ca r t urn s left wit h a r adius r as show n in Fig . 7-9 , t he r esistance of t he road w ould m ake t he w heel t urn wit hou t slippin g . T hat is ω1 r - l = ω3 r + l Solving Eqs . ( 7-4 ) and ( 7-6 ) simultaneously , we have ω1 = ω3 = ・ 1 50 ・ r - l ωH r r+ l ωH r
Fig . 7-9

(7-6)

Such an au t om atic resolution of motion preven ts t he ex t ra wear of t yres w hen t he veh icle t urns .

7.6   M e c h a n i c a l E f f i c i e n c y o f P l a n e t a r y G e a r T r a in s
T he m echanical efficiency of a gear t rain is impor tan t if t he gear t rain is used in pow er t ransmission . The m echanical efficiency of an ordinary gea r tr ain is t he product of efficiencies of every pair of gears , if t hey ar e connected i n series in t he gear tr ain . T he mechanical efficien- cy of an epicyclic gear train is not so easy t o calculate and w e discuss here only planetary gear t rains as t hey ar e often used in powe r t ransmission . Denoting t he input power , t he out pu t powe r and t he frictional power as N d , N r an d N f , respectively , t he m echanical efficiency η can be defined as follo ws : η= Nr = N r + Nf 1 Nf 1+ Nr (7-7)

or η= N d - Nf Nf =1 Nd Nd (7-8)

Since N d or N r is usually know n , t he key point here is t o det ermine N f . From Eq . ( 7-7 ) , we get Nf = Nr From Eq . ( 7-8 ) , we get N f = N d (1 - η) ( 7-10) T he frictional lost po wer in a m achine depends on t he reactions in k ine matic pairs , fric- tional coefficien ts and relative velocities between pair elem en ts . All of t hese do not change if an epicyclic gear tr ain is t urned in t o t he conver ted gear t rain by adding an ex t ra angular velocit y ( - ωH ) t o t he w hole gea r t rai n . In ot he r words , t he frictional lost powe r N f in an epicyclic gear train is t he sam e as t hat N fH in the corr esponding conve rted m echanism . T his is t he w ay we calculate t he efficiency of an epicyclic gear t rain . This m et hod is t he refor e called t he method o f the converted mechan ism . Conside r t he planeta ry gea r t rai ns s ho w n in Fig . 7-10 as examples . If t he mom en t acting on t he gea r 1 is M 1 , t hen t he po wer t hrough t he gear 1 is Nl = M l ω l In t he conve rted mechanism , t he power t hrough t he gea r 1 becom es NlH = M l ( ω l - ω H ) = N 1 (1 - i H 1 ) ( 7-11)
H H If N 1 > 0 , t hen t he gear 1 is a driving w heel i n t he conver ted mechan ism and N 1 is t he

1 - 1 η

(7-9)

inpu t powe r i n t he conver ted m echanism . According to Eq. ( 7-10) , t he frictional lost power ・ 1 51 ・

Fig . 7-10

N in t he conve rted mechanis m is t hen
H H H N fH = N 1 (1 - η 1 3 ) = N 1 ( 1 - iH 1 ) ( 1 - η 13 ) H w her e η 1 3 is t he m echanical efficiency of t he converted mechan ism , w hich can be calculated as

H f

in ordinary gea r t rains .
H H If N1 < 0 , t hen t he gear 1 becom es a driven w heel in t he conver ted m echanism an d N1 is

t he out pu t powe r in t he conver ted m echanism . According t o Eq . ( 7-9 ) , t he frictional lost po wer is Nf = | N 1 |
H H

1 1 = | N1 (1 - iH1 ) | H - 1 H - 1 η 13 η 13

H H 1 can be approxim ated by ( 1 - η 1 3 ) as η 1 3 is near to 1 . T he refore , for t he bo t h H - 1 η 13

cases , t he frictional lost pow er can be written as
H N f = N fH = | N 1 ( 1 - iH 1 ) | (1 - η 13 )

( 7-12)

If t he gear 1 is the driving w heel in t he epicyclic gear t rain , t hen N 1 is t he inpu t po wer of t he epicyclic gear t rain an d t he efficiency of t he epicyclic gea r t rain is η 1H = N 1 - Nf Nf H =1 = 1 - | 1 - iH 1 | ( 1 - η 13 ) N1 N1 ( 7-13)

If t he gea r 1 is a driven w heel in t he epicyclic gea r tr ain , t hen N 1 is t he ou tpu t power of t he epicyclic gear t rain an d t he efficiency of t he epicyclic gea r t rain becom es η H1 = | N1 | 1 = H | N 1 | + Nf 1 + | 1 - iH1 | ( 1 - η 13 ) ( 7-14)

I t can be seen no w t hat t he m echanical efficiency of a planeta ry gea r t rain is a function of ・ 1 52 ・

H its tr ain ra tio | i H 1 | , as well as a function of η 1 3 . If | i H 1 | is large enough , t hen η 1 H may be-

come negative and t he gea r t rain is self-locking . Therefore , it is necessa ry t o calculate t he m e- chanical efficiency according to Eqs. ( 7-13 ) and ( 7-14 ) w hen we design a planeta ry gear t rain . A planetary gear tr ain is called a positi ve mechan ism if t he train ratio of its convert ed
H m echanism i1 3 is positive . T he negative m echanism is similarly defined . I t can be show n from

Eqs . ( 7-13) and ( 7-14) t hat a negative m echanism has a highe r efficiency t han its convert ed m echanism . T ha t is w hy nega tive mechanisms a re more often used in power t ransmission . Usually , t he tr ain r atio of a negative m echanism can no t be very la rge due to t he li mits on t he practical dimensions . W hen a la rge t rain r atio is needed i n a po wer t ransmission , seve ral nega- tive mechanism s can be con nected in se ries to get a large t rai n ratio .

7.7   T o o t h Numb e r s o f G e a r s a n d Num b e r o f P l a n e t G e a r s
T he choice of t oot h num bers of gears and t he numbe r of planet gea rs in a planetary gear t rain s hould m eet t he follo win g four require ments . T he gear t rain show n in Fig . 7-10a is taken as an example to illust rate t he require ments . (1 ) T rain ratio condition Suppose t rain ratio i 1 H is given . From Eq. ( 7-3 ) , we have z3 = ( i 1 H - 1) z 1 (2 ) Concen t ric condition Gear 1 , gear 3 and planet carrier H should ro tate a round t he common axis . If all t he gears ar e standard spur gears , we have r3 = r 1 + 2 r 2 1 1 m z 3 = m z1 + m z2 2 2 z2 = 1 ( z3 - z 1 ) 2 1 z ( i - 2) 2 1 1H ( 7-15)

Inse rting Eq. ( 7-15) i nt o t he last equation , we get z2 = (3 ) Assem bly condition Alt hough one planet gear is enough kinem atically , more planet gears of t he sa me size are moun ted bet ween sun gea rs 1 an d 3 t o tr ansmit more power . To balance t he cen trifugal forces of t he planet gea rs , t hese planet gears should be spaced equally . If t he number of planet gears is k , t hey ar e separa ted from each o t her by an angle φH = 360° / k . Suppose t he first planet gea r has been moun ted at O 2 as s ho w n in Fig . 7-11 . T hen turn t he planet ca rrier H t hrough φH . T his makes t he gear 1 turn t hrough an angle φ 1 . ・ 1 53 ・ ( 7-16)

φ ω z3 1 1 = = i1 H = 1 + φH ωH z1 or φ1 = 1 + z3 z3 360° φ H = 1+ z1 z1 k

In orde r t o moun t t he nex t planet gea r at O2 , t he an gle φ 1 should con tain an exactly in tegr al number of teet h of gea r 1 , i .e . φ 1 = N 360° z1
Fig .7-11

Comparing wit h t he above equation , we get z1 + z3 = N k w her e N should be an in teger . (4 ) No-overlapping condition ( 7-17)

In Fig . 7-11 , O 2 and O 2′are the cen tr es of neighbouring planet gears . To avoid t he over- lapping of t he t wo planet gears , t he cen t re dist ance O2 O 2′ should be greater t han t he adden- dum dia m eter of t he planet gea rs . If all t he gears a re standard spur gea rs , we have 2( r1 + r 2 ) sin or ( z1 + z2 ) sin 180° > z 2 + 2 h a* k ( 7-18) 180° * > 2 ( r 2 + ha m ) k

Eqs . ( 7-15) to ( 7-18) should be sa tisfied in choosing t he toot h numbers of the gears and t he number of t he planet gea rs . F or ot he r t ypes of planet ary gear t rains , similar equations can be derived and should be fulfilled .

7.8   In t r o d u c t i o n t o O t h e r K i n d s o f P l a ne t a r y G e a r T r a in s
7.8.1   Planetary Reducer with Small Tooth Difference T he planetary gear t rain sho wn in Fig . 7-12 consists of fixed in ternal sun gea r 3 , planet gear 2 and planet ca rrier H . T he cen tr e distance bet ween gea r 2 and gear 3 is called offset e . T he t rain ratio can be calculated as follows: ω2 - ωH z 3 = ω3 - ωH z 2 Since ω 3 = 0 , we get iH2 = ・ 1 54 ・ ω H z2 = ω z3 - z2 2

T he t oot h difference ( z3 - z2 ) is chosen small and t he absolute value of t he t rain ratio i H 2 becom es gr eat . Such a gear t rain is t he so-called planeta ry gea r t rai n wit h small toot h differ- ence . Gear 2 pe rforms a planetary motion . A n out put m echanism is therefore needed t o ou tpu t t he rota tion of gea r 2 . The most widely used ou t pu t m echanism is show n in Fig . 7-12 . There is a plate on t he ou t put shaf t V and t here ar e seve ral pins wit h

Fig . 7-12

dia mete r d 2 on t he plate . T hese pins locate in t he holes made on t he planet gear wit h dia m eter d 1 . The relation between d 1 and d 2 is d 1 = 2 e + d 2 . Bo t h t he pins and holes a re dist ribut ed on t he circles wit h t he sa m e dia mete r D . As t he planet moves , t he kine ma tic relationship be- tween t he planet gear and t he ou t put pla te corresponds to t he par allel-cr ank mechan ism O 1 O 2 B A . T his m akes t he ou tpu t s haft ro tate wit h t he sa me angula r velocit y as t he planet gear . 7.8.2   Cycloidal-Pin Wheel Planetary Gearing T he cycloidal-pin w heel planetary gearing is a special planet ary gea r t rain wit h one too t h diffe rence w here t he in ternal gear is a pin w heel and t he ex ternal gea r is a cycloidal gea r as show n in Fig . 7-13 . T he t oot h profile of a cycloidal gear is for med as show n in Fig . 7-14 . T he pitch circle of gear 3 rolls wit hout slipping along t he pitch circle of gea r 2 w hich is fixed . T hen a poin t M on gear 3 ou tside its pitch circle draws a cur tate epicycloid M 1 ~ M 6 . If t he poin t M is t he too t h profile of gear 3 , t he curtat e epicycloid is t he too th profile of gear 2 . P ractically , a circle wit h radius rz and cen t red at t he poin t M is taken as t he t oot h profile of gea r 3 and t his circle gener- at es t he too t h profile of gea r 2 t hrough C1 ~ C6 . The gea r 3 is t herefore called a pin w heel and t he gear 2 a cycloidal gear . Compared wit h t he planetary gear t rain wit h one t oot h difference consisting of involute gears , t he cycloidal-pin w heel planetary gearing enjoys t he following advan tages . (1) There is no t oot h profile in terference bet ween t he in te rnal and t he ex te rnal gea rs , w hich may happen in t he case of involu te gears . ・ 1 55 ・

 

Fig .7-13

Fig . 7-14

(2 ) T he re are more pairs of teet h in con tact at t he sa me tim e t o tr ansmit t he po wer . T he cont act ratio and t her efor e t he t ran smit ting capacit y are high . (3 ) T he working pr essur e angle is s maller . This decreases t he load on t he bearing and in- cr eases t he mechanical efficiency . 7.8.3   Harmonic Drive Gearing A set of harmonic drive gearing consists of a rigid circular spline , a flexspli ne and a wave generat or , as show n in Fig . 7-15 . The rigid circular spline is a rigid in ternal gear . T he flexs- pline is a nonrigid hollow ex ternal gear . I t can also be seen as a special planet gear . There are t wo or t hree rolle rs moun ted on t he wave gener ator w hich is inser ted int o t he hole of t he flexs- pline . T he numbe r of rollers is called t he w ave number n . The toot h difference betw een t he rigid spli ne and t he flexspline ( z r - z s ) should be a multiple of n and is often taken as n . T he wave generat or is t he driving m ember . As it rotates , its rollers make the teet h on t he flexs- pline en gage wit h t he rigid spline . T he role of the wave gene rator is a planet carrier . The t rain ratio of t he drive can be calculated as for a planet ary gea r t rain . O ne of t he rigid splines and t he flexspline should not rotat e and t he ot he r becomes t he ou t- put me mber . If t he rigid spline is fixed and t he flexspline is t he ou t put m ember as s ho wn in Fig . 7-16a , t hen t he t rain ratio is iH s = ωH zs = ωs z r - zs

Alterna tively , if t he flexspline does no t rotate and t he rigid spli ne is t he out put me mbe r as show n in Fig . 7-16b , t hen t he tr ain ra tio becom es iH r = ・ 1 56 ・ zr z r - zs

Fig .7-15

High t rain ratio , high efficiency , smoot h t ransmission and simple const ruction ar e t he ad- van tages of t his gearing . I ts m ain problem is t he fatigue of flexspline .

Fig .7-16

Problems and Exercises 7-1   Ho w are gea r t rains classified and w hat ar e t heir m ain applications ? 7-2   Ho w can the t rain ratio of an ordinary gear t rain be calcula ted ? H ow can t he relative ro- tating dir ections among gea rs i n an ordina ry gea r t rai n be determined ? 7-3   W hat are t he st ruct ural characteristics of an elem en ta ry epicyclic gear t rain ? What is t he convert ed gea r t rain ? Ho w can t he t rai n ratio of an ele men tary epicyclic gea r t rain be calculat - ed ? H ow can t he sign in fron t of t he t oot h number ratio be determined ? Ho w can t he rotating direction of t he out put m em ber in an epicyclic gear t rain be det ermined ? 7-4   Ho w is a combined gear tr ain divided ? 7-5   H ow can t he m echanical efficiency of a planeta ry gear tr ain be calcula ted ? What are t he diffe rences between t he positive mechanism an d t he negative m echanism ? 7-6   W hich condition s should t oot h numbers of gears and number of planet gears i n a planetary gear train fulfill ? 7- 7   S how n in Fig . 7-17 is a hoist . T he teet h numbe rs of all the gea rs are given . Find t he ・ 1 57 ・

t rain ratio i1 5 and poin t out t he rota ting direction of t he handle t o r aise t he weigh t .

Fig .7-17

Fig .7-18

7-8   S ho w n in Fig . 7-18 is t he gear t rain in a clock . S ,

M and H deno te t he poin ters of sec- z 2 = 64 ,

ond , min ute an d hour , respectively . The given numbers of teet h are z 1 = z2 ′ = 8 , z3′= 12 , z4′= 15 . If t he modules of gear 4 and gear 5 are equal , find t he numbers of teet h z 3 , z 4 , and z5 . 7- 9   S ho w n i n Fig . 7-19 is a differ en tial gear t rain wit h z 1 = 15 , z 2 = 25 , z2′ = 20 , z 3 = 60 , n 1 = 200 r/ min , n 3 = 50 r/ min . Find n H w hen (1 ) n 1 and n 3 a re in t he sa me dir ection ; (2 ) n 1 and n 3 a re in t he opposite directions . 7-10   S how n in Fig . 7-20 is a gear t rain in w h ich gear 3 en gages wit h gea rs 2 and 4 sim ultaneously . W hat kind of gear train is it ? If z 1 = 34 , z 2 = z3 = 20 , and z4 = 74 , find t he t rain r atio i1 H . 7-11   Show n i n Fig . 7-21 is a gea r t rain wit h z1 = 6 , 25 , z 3 = 57 , and z 4 = 56 . Find t he tr ain r atio i1 4 . z2 = z 2′ =
Fig .7-19

Fig . 7-20

Fig . 7-21

Fig .7-22

7-12   In t he combined gear t rain sho w n in Fig . 7-22 , z 1 = 36 , z 2 = 60 , z 4′= 69 , z 5 = 31 , z 6 = 131 , ・ 1 58 ・ z7 = 94 , z8 = 36 , m agnitude and t he direction of n H .

z3 = 23 ,

z 4 = 49 ,

z9 = 167 , n 1 = 3 549 r/ min . Find bot h t he

7-13   Show n in Fig . 7-23 is a gea r t rai n wit h z 1 = z5 = z 6 = 17 , z 2 = 27 , z 2′ = 18 , z 3 = 34 , t rain ratio i1 6 . 7-14   S ho w n in Fig . 7-24 is a differ en tial gear tr ain in a t ex tile machine wit h z 1 = 30 , z 2 = 25 , z 3 = z4 = 24 , z5 = 18 , z6 = 121 , n H = 316 r/ min , n 1 = 48 ~ 200 r/ m in . n 1 and n H a re in t he sam e dir ection . Find n 6 . 7- 15   I n t he gear t rai n show n in Fig . 7-25 , gear 1 is z 2 = 20 , z 3 = 60 , z 4 = 90 , z 5 = 210 . Fi nd n 3 .    
Fig .7-23
3

z4 = 51 . Fin d t he

fixed wit h t he shaft of t he mot or and t he motor is mounted on gea r 3 . n 1 = 1 440 r/ mi n . z 1 =

Fig .7-24

Fig . 7-25

7-16   In t he gear t rain sho wn in Fig . 7-10a , z1 = 20 and i 1 H = 4. 5 . All t he gears are standard spur gears wit h h a* = 1 . Determine the numbers of teet h z 2 and z 3 and t he number of planet gears k . If t he mechanical efficiency of every pair of gears is η= 0 .9 , find t he mechan ical effi- ciency of t he gea r t rain η 1H .

・ 1 59 ・

C h ap t e r 8 O t h e r Mec han i s ms i n C o mm o n U s e
    The planar linkage m echanisms , cam m echanisms , and gea r m echanisms discussed in pre- vious chap ters are t he commonest basic m echanisms used in m achine design . In addition to t hese , m any m achines and in st ruments cont ain ot he r kinds of m echanism . In t his chap ter , som e of t he m in common use will be in troduced briefly t o broaden t he st uden t′ s knowledge .

8.1   R a t c h e t M e c h a ni sm s
8.1.1   Working Principle of Ratchet Mechanism A ratchet mechanis m consists of a drivin g rocke r 1 , a driving pawl 2 , a driven ratchet 3 , a holding pawl 4 , and t he fra m e 7 , as sho w n Fig . 8-1 . T he driving rocker 1 is on t he sa me shaft as t he ratchet 3 . T he holding pa wl 4 and t he drivi ng pawl 2 are forced in to con- tact wit h t he ratchet 3 by springs 5 and 6 , respectively . W hen t he drivi ng rocker 1 rotates coun te r-clock wise , t he driving pawl 2 will move int o t he toot h space of t he ratchet 3 and cause it to rotate . Mean w hile , t he holding pawl 4 will slide over t he too t h back of t he ra tchet 3 . W hen t he driving rocker 1 ro tates clock wise, t he holding pa wl 4 will move in to t he toot h space of t he ra tchet 3 to pr even t it from ro tating clock wise . T her efor e, t he ratchet 3 will
Fig .8-1

dw ell w hile at t he sa me tim e, t he driving pawl 2 will slide t he ra tchet 3 will ro tate in te rmitten tly coun te r-clock wise .

over t he t oot h back of t he ra tchet 3 . Thus , w hen t he driving rocker 1 oscillates con tin uously , T he motion t ype of t he inpu t link of t he ratchet mechanism is oscilla tion . T herefore , ratchet m echanisms ar e often used in series wit h mechan isms w hich can t ransform a con tinuous ro tation in to an oscillation , e .g . crank-rocker mechan isms , cam m echanisms , etc .

・ 1 60 ・

8.1.2   Types of Ratchet Mechanisms According t o t heir struct ure cha racteristics , ratchet mechanisms in common use can be di- vided in to t wo types: t oot h ratchet m echanisms and silent ratchet m echanisms . (1 ) Toot h r atchet mechanism s T he re are two sub- types for too t h ratchet m echanisms: ext ernally meshed ( Fig . 8-1 ) and in ternally meshed ( Fig . 8-2 ) . If t he dia m eter of t he ratchet becom es i nfinite , it becom es a ratchet rack ( Fig . 8-3) . T he in termit ten t rota tion of t he ratchet in one dir ection will become an in termitt en t t ranslation of t he ratchet rack i n one direction .

Fig . 8-2

Fig .8-3

    By using differ en t struct ures , t he in ter mit ten t rotating direction of t he ratchet may be fixed or changeable . ( a ) Fixed dir ection ( Fig . 8-1 ) : In t his mechanism , t he driven r atchet 3 can rot ate in ter- mitt en tly only in t he fixed dir ection ( coun ter clock wise in Fig . 8-1 ) w hen t he driving rocker 1 oscillates con tinuously . T he t oot h shape of t his kind of r atchet is asymm etrical . The com- monly used teet h for t his r atchet are saw toot h ( Fig . 8-4a ) , t riangular wit h st raigh t edges ( Fig . 8-4b) an d t riangular wit h curved edges ( Fig . 8-4c ) .

Fig .8-4

    ( b) Changeable dir ection ( Fig . 8-5 an d Fig . 8-6 ) : If t he direction of int ermittent rota- tion of t he driven ratchet is required t o be changeable , t he ra tchet 3 can have rectangular teet h ・ 1 61 ・

w hile t he pawl should be reve rsible, as show n in Fig . 8-5 . W hen t he pa wl is in position 2 , t he ratchet 3 will rota te in te rmitten tly coun ter-clock wise . Conversely , if t he pawl is turned over in to position 2′ , the ra tchet 3 will rotate in te rmitten tly clock wise . Sho wn in Fig . 8-6 is ano t her kin d of ratchet mechanism , t he ratchet rot ation direction of w hich can be changed . W hen t he pa wl 1 is loca ted at t he position as sho w n , t he r atchet 2 will ro tate i nte rmittently coun ter clock wise . If t he pawl is lifted , rotated 180°abou t its axis and dropped , t hen t he pa wl will i nser t agai n in to t he t oot h space of t he ratchet and t he ratchet 2 will rotate in term it ten tly clockwise . If t he pawl is lif ted , rota ted 90° abou t its axis and dropped , t hen t he sm all pin fixed wit h t he pa wl 1 will rest on t he pla tform on t he top of t he case . T he pawl and t oot h of the r atchet will com e ou t of mesh . In t his way , t he ratchet 2 will not rota te w hen t he drivin g pawl 1 oscillates .

Fig .8-5                                 Fig .8-6

    All t he ratchet mechanism s m en tioned above are single function , i.e . t he driving rocker can drive t he r atchet only in eit he r direction of t he oscilla tion . S ho w n in Fig . 8-7 is a double function r atchet mechanism . In bo t h direction s of t he oscillation , t he drivi ng link 1 can drive t he ratchet 3 t o ro tate in termitt en tly in t he sa me dir ection . T he shape of t he driving pawl 2 could be st raigh t ( Fig . 8-7a ) or hooked ( Fig . 8-7b ) .

         

a)

                        b)

Fig .8-7

Fig . 8-8

・ 1 62 ・

    The rotation angle of t he ratchet in one cycle can be adjusted by adjusti ng t he rocking angle φ of t he rocker . For exa mple , if t he ratchet m echanism is connected i n series wit h a c ran k- rocker mechanism , t hen t he angle φ can be adjusted by chan gin g the link len gt hs of t he c ran k- rocker m echanism . Alternatively , a cover pla te 4 over ratchet 3 may be used , as show n in Fig . 8-8 . By adjusti ng t he position of t he cover plate , t he rotation angle of t he ratchet in one cycle can be adjusted . (2 ) Silent ratchet m echanism In toot h ratchet m echanisms , rigid impulse and noise a re created w hen t he pawl st rikes t he too t h roo t . In addition , the str eng th of t he pawl is low and t he tip of pawl is subject to wea r . T o avoid t he noise , an alternative design is oft en used in w hich t he effect of friction pro- duces t he required mo tion . S ho wn in Fig . 8-9 is such a m echanism . The driving sector block 2 is held in con tact wit h driven w heel 3 by mean s of a retain ing spring at tached t o t he driving rocker 1 . W hen t he driving rocke r 1 rota tes an ti-clock wise , t he friction force bet ween t he sector block 2 and t he driven w heel 3 causes t he sector block 2 t o rotate r elative to t he drivi ng rocker 1 an ti- clock wise . The norm al force and t he frictional force in- cr ease . When t he frictional force is large enough , t he driving sector block 2 will drive t he driven w heel 3 so t hat t hey rotate toget her . T he driven w heel 3 is driven anti-clockwise t hrough an an gle . W hen t he rocker 1 ro- tates clock wise , t he w heel 3 would be caused to rotate clock wise bu t is pr even ted from doing so by locki ng sec- tor block 4 . The sector block 2 ro tates clock wise r elative t o the rocke r 1 against t he retaining spri ng and allo ws t he rocker 1 to move fr eely w hile t he w heel 3 is held locked by block 4 . In t his way t he w heel 3 is driven step- wise an ti-clock wise w hile t he inpu t rocker 1 oscillates back and fort h . Compared wit h t oot h ra tchet m echanisms , t he silen t ratchet m echanism can t ransmit mo- tion smoot hly and silen tly . T ha t is w hy t his kind of r atchet m echanism is called“ silen t ”. In one cycle, the t oot h r atchet must rota te at least one toot h space and its rotation angle m ust be a multiple of t he angle between tw o neighbouring teet h . How ever , in the silen t ra tchet mecha- nism , it is t he frictional force between t he driving sector block and t he friction w heel w hich t ransmits motion . Therefore , i n t he silen t ratchet m echanism , t he rota tion angle of t he driven w heel in one cycle can be changed fr eely by adjusting t he rota ting angle of t he drivi ng rocker . Ho wever , some sliding between t he drivin g sect or block and t he friction w heel is inevitable . T he refore , t he accur acy of motion is poor and t he t ransmitted torque is limited . In order t o increase t he t ransmitted t orque , more w orking planes are usually used in t he silen t ratchet m echanism , as show n in Fig . 8-10 . I t consists of an ex ternal ring 1 , a st ar ・ 1 63 ・
Fig . 8-9

w heel 2 , and som e rolle rs . N otches a re cut on t he star w heel 2 . T he roller is pushed to t he na rrow side of t he no tch by a small spring so t ha t normal forces and frictional forces will exist bet ween t he rolle r and t he ex ternal ring and between t he rolle r and t he sta r w heel . W hen t he ex te rnal ring 1 ro tates clock wise , t he frictional force causes the roller to move t owards t he nar- ro w side of t he notch . T he normal for ce and t he friction- al force inc rease . W hen t he frictional force is large e- nough , t he ex te rnal ring 1 will drive t he sta r w heel 2 so t hat t hey ro tate toget he r . When t he ext ernal ring 1 ro- tates coun ter-clockwise , t he roller will roll to wards t he wider side of t he notch . The norm al forces and t he fric- tional forces bet ween t he roller and t he ext ernal ring and bet ween t he roller and t he star w heel will dec rease and approach ze ro . T her efore , t he star w heel 2 will dw ell w hile t he ex te rnal ring 1 rota tes cou nte r-clock wise . If t he ex ternal ring 1 oscillat es con tinuously , t he star w heel 2 will rotate in termitt en tly clock wise . 8.1.3   The Applications of Ratchet Mechanisms A pa rt from its use for in te rmitten t motion , t he ra tchet m echanism can also be used as a safet y device in lif ting m achines . In t he hoist show n in Fig . 8-11 , t he holdin g pa wl 2 slides over t he t oot h back of t he r atchet 1 w hen weigh t F Q is lifted . If t he po wer is cu t suddenly , t he weight F Q t rends to drop freely and m ake t he r eel 1′and ratchet 1 rota te clock wise . At t his time , t he holding pawl 2 moun ted on t he fra m e will drop in to t he t oot h space of t he ratchet by spring for ce and its ow n weigh t . T hus , t he r atchet 1 is preven ted from rotating and t he weigh t F Q can not drop . In t h is circumstance , t he ratchet mechanism acts as a ratchet . Ratchet m echanisms a re often used in mach ines to allow ove rrun . O ne of t he most fa mous exa mples is t he so called“ free w heel ”in bicycles , as show n in Fig . 8-2 . If our feet st op ped- aling (or even back-pedal ) w hen t he bicycle is going for wa rd , t he chain and t he w heel 1 will also stop rotating . H oweve r , t he rea r axle 3 will con tin ue to rotate clock wise by iner tia force . T he pawl 2 on t he rear axle 3 will slide over t he t eet h back of t he in ternal r atchet ( a clear clat- ter can be heard w hen t he pawl 2 st ri kes t he too t h root of t he in ternal ratchet ) . T his is t he so- called over ru n . ( I t should be poin ted out her e t ha t overrun is t he unique characteristic of t he ・ 1 64 ・
Fig .8-11 Fig . 8-10

safet y device . To lo wer t he weig ht F Q , t he pawl 2 m ust be wit hdraw n from the teet h of t he

ratchet m echanism .) In modern m achine tools , t he t ool fra me is often requir ed to approach t he workpiece at dif- fe ren t speeds : fast for a quick approach an d slowly for an accurate feed . M achine t ools often use t he silent r atchet mechanism show n in Fig . 8-10 t o accomplish t his . Suppose that t he pow - er is inpu t to t he ex ternal ri ng 1 t h roug h a w orm and w orm gear m echanism . W hen t he ex ter- nal ring 1 ro tates clockwise, it forces t he rolle rs to be pressed tigh tly in the narro w side of t he notch of t he sta r w heel 2 . T hus , t he ex ternal ring 1 drives t he sta r w heel and t he ou tpu t shaft at a lo w speed clock wise . W hen t he ou tpu t shaf t is required t o be run at high speed , t he pow er drives t he ou tpu t shaft and t he sta r w heel dir ectly clockwise t hrough anot her tr ansmission chain (not show n in t he figur e ) . H oweve r , t he exte rnal ring 1 is still driven t hrough t he worm and wor m w heel m echanism and runs at low er speed . Since t he speed of t he sta r w heel 2 exceeds t hat of t he ex ternal ring 1 , t he m echanism acts as an overrun clu tch and it is not necessary to cu t t he low speed t ransmission chai n w hen a high speed is r equir ed . T he speed in te rchange t herefore becom es quite easy .

8.2   G e ne v a M e c h a n i s m s
8.2.1   The Working principle of the Geneva mechanism A Geneva m echanism consists of a drivi ng pla te 1 wit h a roller G , a driven Geneva w heel 2 wit h r adial slots , and a fra me , as sho w n in Fig . 8-12 . As t he roller G en ters i nt o t he r adial slo t , t he driving plate drives t he Geneva w heel by t he roller G an d forces it to rot ate . Af ter t he driving plate 1 and t he Geneva w heel 2 rotate to the position sho w n in Fig . 8-13 , t he

Fig . 8-12

Fig .8-13

・ 1 65 ・

roller G leaves t he r adial slot . D riving plate 1 t hen con tinues t o ro tate w h ile t he G eneva w heel 1 d wells . Therefore , t he continuous even ro tation of t he driving plate 1 is tr ansfor med in to an in termitt en t rotation of t he driven Geneva w heel in one direction . In order t o star t and stop t he Geneva w heel smoot hly , t he Geneva m echanism should be so designed t hat t he c ran k O 1 G must be perpendicular t o the cen t ral line of t he radial slo t w hen t he roller G begins to en ter in t o and t o leave t he r adial slot , as s ho w n in Fig . 8-12 and Fig . 8- 13 , respectively . In t h is way , t he angular velocit y of the driven Geneva w heel will be zero at t hese position s . N o rigid impulse will exist . In orde r to ensure t hat t he roller G can ente r exactly in to t he nex t r adial slot at t he nex t cycle , t he Geneva w heel must not rotate after t he roller G leaves t he r adial slot . For t his pur- pose , a convex lock ing a rc plate m m is fixed to t he driving plate 1 and concave locking arcs n n a re made bet ween neighbouring radial slo ts . T he r adii of t he convex and the concave locking a rcs ar e t he sa m e . A t t he mom en t w hen t he rolle r G begins to en ter i nt o the radial slo t , one end poin t m of t he convex locking a rc m m should arrive exactly at t he cen t re line O 1 O2 , as show n in Fig . 8-12 . A t t his mom en t , t he concave locking a rc nn begin to disengage from t he convex locki ng a rc m m and t he roller G can drive t he Geneva w heel . At t he momen t w hen t he rolle r G begi ns t o leave t he radial slo t , anot he r end poin t m of t he convex locki ng a rc m m should arrive exactly a t t he cen t re li ne O 1 O 2 , as show n in Fig . 8-13 . Aft er the roller G leaves t he r adial slo t , t he convex lock ing arc m m and t he concave locking arc n n are engaged over som e arc . H ence , t he Geneva w heel 2 is held stationa ry until t he rolle r G ente rs t he nex t radi- al slot and t he nex t cycle begins . Since some clea rance exists bet ween t he t wo locking arcs , t he positional accur acy is no t very h igh . In modern high accuracy index mechanisms , a special po- sitioni ng mechanism is added t o position t he Geneva w heel accurately and preven t it from rot at- ing . 8.2.2   The Types of Geneva Mechanism Depending on t he relative positions of t he axes of driving plate and Geneva w heel, Geneva m echanisms can be divided in to planar Geneva m echanisms and spherical Geneva m echanisms . T he re ar e also t wo sub-t ypes of planar Geneva m echanisms: one is t he ex ternal Geneva m echanism , as show n in Fig . 8-12 . The rotating directions of t he driving pla te and t he driven Geneva w heel a re opposite . T he ot he r is t he in ternal Geneva mechan ism , as show n in Fig . 8- 14 . The rot ati ng directions of t he driving plate and t he driven Geneva w heel ar e t he sam e . T hese t wo kinds of planar Geneva mechanism s ar e used t o t ransmit motion bet ween tw o par allel shaf ts . Sho wn in Fig . 8-15 is a spherical Geneva mechanism . T he axis of driving plate 1 is per- pendicula r to t hat of the driven Geneva w heel 2 w hich is in t he shape of a halfsphere . T he ro- tation ax is of driving pla te 1 , driven Geneva w heel 2 and rolle r 3 all pass t hrough t he sphere cen t re O . ・ 1 66 ・

Fig . 8-14

Fig . 8-15

    T he re m ay be one or more rollers on t he driving pla te . If more t han one roller is used , as show n in Fig . 8-16 , t hen t he driven Geneva w heel will rota te and d well m any ti mes w hile t he driving plate rotates once . If t he times for each d well of t he Geneva w heel a re required to be diffe ren t , t hen t he rolle rs should be distribu ted unevenly . If t he tim es for each rotation of t he Geneva w heel are required t o be differen t , t hen t he ro tating radii of t he rollers should be differ- ent . For the Geneva m echanism show n in Fig . 8-17 , t he tim es for each d well and for each ro- tation of t he Geneva w heel a re all differen t during one cycle of t he driving plate .

Fig .8-16

Fig . 8-17

・ 1 67 ・

8.2.3   The Ratio k between Motion Time and Dwell Time T he ra tio bet ween t he motion tim e t m and t he dwell time t d of t he Geneva w heel is denot ed by k . Usually , t he driving plate ro tates at a constan t speed . T hus , t his ratio k can be ex- pressed by t he r atio of correspondi ng angles . F or Geneva m echanisms wit h one roller ( Fig . 8- 12) , t m corresponds to angle φ 1 of t he drivi ng plate 1 w hile td corresponds to angle ( 2 π- φ 1) of t he driving plate 1 . Therefore , t he ratio k is k= tm φ1 π - 2π / z z -2 = = = t d 2π - φ 1 π+ 2π / z z +2 (8-1)

From t he above form ula , we can see t hat t he nu mber z of t he radial slots must be la rger t han 2′since t he r atio k must be greate r t han zero . Si nce t he num era tor is always less t han t he denominator in Eq . ( 8-1 ) , t he mo tion tim e of t he Geneva w heel is always less t han its dwell tim e in Geneva mechanism s wit h a si ngle roller . If n rollers ar e dist rbut ed evenly on t he drivin g plate ( Fig . 8-16 ) , t hen t he mo tion time t m of t he Geneva w heel during one circuit of t he driving plate corresponds t o angle nφ1 of t he driving plate . F or t he re maining angle ( 2π - nφ 1 ) , t he Geneva w heel dwells , i . e . td corre- sponds t o ( 2 π - nφ 1 ) . T herefore , t he ratio k of Geneva mechanism wit h n evenly dist ribut ed rolle rs is k= tm nφ 1 n (π - 2π / z) z - 2 = = = t d 2π - nφ 1 2π - n (π - 2 π / z ) 2 z/ n - ( z - 2 ) (8-2)

Since t he ratio k must be gr eate r t han zero , t he denomi nat or in Eq . (8-2) m ust be grea ter t han zero , i .e . 2z 2z - ( z - 2 ) > 0   or ,   n < n z -2 (8-3)

According to Eq . ( 8-3 ) , t he allowed number n of rollers G can be deter mined af ter t he number z of r adial slots is k no w n . F or exa mple , if z = 3 , t hen n = 1 ~ 5 . If z = 4 or 5 , t hen n = 1 ~ 3 . If z ≥ 6 , t hen n = 1 ~ 2 . For in ternal Geneva m echanisms ( Fig . 8-14) , t he angle of t he driving plate corresponding t o t he motion time of t he Geneva w heel during one cycle is always greater t han π . T herefore , in ternal Geneva m echanisms can have only one roller on t he driving plate . U si ng a m et hod sim- ila r t o t he above , t he ra tio k between t he motion time t m and t he dw ell tim e t d of t he in ternal Geneva w heel can be de rived as follo ws : k= tm z+2 = td z -2 (8-4)

Since k must be gr eate r t han ze ro , t he number z of r adial slo ts in an in ternal Geneva m echanism can not be less t han 3 . In addition , according to Eq . ( 8-4 ) , t he motion tim e t m of an in ternal Geneva w heel is always greater t han its d well ti me t d . ・ 1 68 ・

T he motion relationsh ip of Geneva m echanism in motion stroke is equivalen t to t hat of t he oscillating guide ba r mechan ism show n in Fig . 4-9 . I t can be show n t hat t he angular accele ra- tion of t he Geneva w heel changes abruptly w hen t he Geneva w heel sta rts and stops . T here- fore , a soft im pulse exists at t hese t wo instants . T he lowe r t he number of radial slots , t he grea ter is t he abrup t change in angular acceler ation and t he w orse t he motion smoo t hness . T he refore, t oo sm all a value of z s hould no t be chosen during design . In r ealit y , ex ternal Geneva m echanisms wit h 3 radial slots are seldom used . According to Eq . ( 8-1) , t he ra tio k of t he exte rnal G eneva m echanism incr eases wit h increase of z . Bu t t he incr ease of k is not grea t w hen z > 8 . T her efore , in practice , z should be chosen bet ween 4 ~ 8 . If a larger k is required , an in ternal Geneva mechanism is suitable . The dynam ic char acte ristics of t he in ter- nal Geneva m echanism are much bette r t han t hose of t he ex ternal one . If it is possible to dist ribute t he axes of the driving plat e and t he driven Geneva w heel per- pendicula rly , t hen a sphe rical Geneva mechanism ( Fig . 8-15 ) s hould be used . T he dyna mic char acte ristics of t he spherical Geneva m echanism a re much bett er t han t hose of t he ex ternal Geneva m echanism , bu t sligh tly infe rior to t hose of t he in ternal Geneva m echanism . The r atio k of t he sp herical Geneva m echanism is iden tically equal to 1 and independen t of t he number of radial slo ts . This char acte ristic is better than t hat of t he in ternal Geneva mechanism . T here- fore , t he sp herical Geneva m echanism is used more in combined m achine tools wit h multiple wor king position s . 8.2.4   The Characteristics and Applications of Geneva Mechanisms Geneva m echanisms have many advan tages , e.g . simple st ruct ure , small dim ensions , easy t o m an ufactur e, swift t o change position , high mechanical efficiency , etc . Com par ed to t he ra tchet mechanism , t he Geneva mechan ism has no rigid impulse and t he motion is smoot her . T he str eng t h of t he Geneva mechanism is higher t han t ha t of t he ra tchet mecha- nism . T her efore , Geneva mechanisms can be used to drive rotating tables wit h la rge mom en ts of iner tia . Ho wever , t he angular acceleration of t he Geneva w heel changes abrup tly at tw o instan ts w hen t he Geneva w heel star ts and stops and a soft impulse is cr ea ted . Therefore , t he angular velocit y of a Geneva m echanism should no t be too high . The rotation angle of a Geneva w heel for each cycle can not be adjusted . I ts adaptabilit y t o pa rticular applica tions is poore r t han t hat of a ratchet mechan ism . T herefore , Geneva mechanisms ar e mostly used for index m echanisms in w hich it is not necessary to adjust t he ro tation angle .

8.3   C am- T y p e In d e x M e c h a n i s m s
T he re a re tw o types of cam-t ype i ndex mechanisms : cylindrical-cam index m echanisms ( Fig . 8-18 ) and hourglass-cam i ndex mechanism s ( Fig . 8-19 ) . T hey are of ten used in m a- ・ 1 69 ・

chines w hich need to t ransmit index ing motion or in te rmitten t motion between t wo crossed shaf ts .

Fig . 8-18

Fig . 8-19

8.3.1   Cylindrical Cam Index Mechanisms T he driving cam 1 of t he cylindrical ca m in dex mechanis m ( Fig . 8-18 ) is a cyli ndrical ridge ca m . A number of rollers , 3′ are evenly dist ributed a round the face of t he driven pla te 2 . T he driving ca m rot ates at constan t speed . As show n in t he figur e, t h roug h an angle β of t he driving cam , t he ridge takes t he form of a spiral . F or t he rest of t he ca m rotation , t he ridge is perpendicular to t he axis of t he ca mshaf t . As t he cam ro tates t hrough t he angle β ′t he driven plate rota tes t h roug h angle equal to 2π / z ( w here z is t he numbe r of rolle rs ) . In orde r to ensure accurate indexing , t he widt h of t he ca m ridge should be designed such t hat bo t h faces ar e alw ays in con tact wit h tw o adjacen t rolle rs on t he driven plate . 8.3.2   Hourglass Cam Indexing Mechanisms In t he hourglass cam indexing m echanism ( also called t he Ferguson Mechanism ) , t he rolle rs a re moun ted radially aroun d t he circumference of t he driven plate ( Fig . 8-19 ) . In prin- ciple , t his ca m mechanism is simila r to a single t hread hourglass cam wit h t he driven plate per- formi ng as a w orm gear . As i n t he cyli ndrical mechanism , t he re is a por tion of t he driving ridge w her e t he pitch is zero and t he driven plate is held stationary . In t he re maining section , t he pitch of t he driving ridge causes t he driven plate t o rotat e t hrough 2π / z . As t he i nput shaft ro tates con tinuously , t he driven plate rota tes in termitt en tly in one dir ection . 8.3.3   Advantages, Disadvantages and Applications T he cam profile is designed accordi ng t o t he required motion curve . Bot h rigid and soft im- pulse can be avoided w hen t he driven plate sta rts and stops . T he refor e, t hese kinds of mecha- ・ 1 70 ・

nisms can transmit motion smoo t hly and t heir dyna mical char acte ristics are very good . N oise and vibration is ve ry low . Any ratio between motion ti me and d well tim e can be designed . In addition , t he positional accuracy is high and reliable . N o ex t ra positioni ng device is needed . T heir st ruct ures are compact . In order to minimize po wer consump tion , t he rolle rs of these t wo mechan isms often use needle bea rings . H owever , t hey a re com plicated t o manufact ure . T he accuracy for m anufacture , assembly and adjust m en t must be high . T hese t wo m echanisms a re used widely in high-speed , high-accuracy index m echanisms , e. g . high-speed punch press- es , m ulti-colour print ers , packe rs , etc . In t he hourglass-cam index m echanism , clea rance between t he roller and t he side face of t he curved ridge can be eliminated by adjusting t he cent re distance betw een t he cams haft and t he driven shaf t . Th us , t he roller and curved ridge re main pr est ressed during d well and in dex- ing . T he refore , compar ed wit h t he cylindrical-ca m index m echanism , t he hourglass-ca m in- dex mechanism has higher index accuracy ( t he i ndex accuracy m ay r each 5″ ) and better dy- na mical char acte ristics , etc .

8.4   U ni v e r s a l J o in t s
8.4.1   The Single Universal Joint T he single unive rsal join t show n in Fig . 8-20 is a spatial lo w-pair m echanism w hich can tr ansm it rotation between t wo in tersecting shafts . I t consists of a driving shaft 1 , a cross piece 2 , a driven shaft 3 , and t he fra m e 4 . T he ends of driv- ing shaf t 1 and t he driven shaft 3 a re in t he form of a yoke . All axes cross at t he cen tr e O of t he crosspiece 2 . T he acu te angle between t he driving shaf t 1 and t he driven shaft 3 is β 13 . W hen t he driving shaft 1 completes one revo- lu tion , t he driven shaf t 3 will also complete one revolu tion . Ho wever , it can be sho wn t ha t t he in- stan taneous angular velocities of t he tw o shaf ts ar e not equal at all tim es , i .e . t he driven shaf t 3 will t urn at a variable speed if t he drivi ng s haft 1 t urns at const an t speed . T he tr ansmission ratio i1 3 of t he t wo shafts is a fu nction of t he angle β 1 3 and t he ro tation angle φ1 of t he drivi ng shaft 1 . If β , t hen i1 3 is iden tically equal t o 1 . If 1 3 = 0° β 1 3 = 90° , t hen i 1 3 = 0 , i .e ., t he t wo s hafts can no t t ransm it rotation . If 0° <β 1 3 < 90° , t hen i1 3 will vary periodically . W hen t he driving shaft 1 rota tes at constan t an gular velocit y ω 1 , ・ 1 71 ・
Fig . 8-20

t hen t he angula r velocity ω 3 will fluct uate periodically wit hin ω 1 cos β 1 3 ~ω 1/ cos β 1 3 . Show n in Fig . 8-21 ar e t he curves of i 1 3 vs φ 1 (0° ~ 180° ) for differ en t β 1 3 . We can see t hat t he ampli- tude of t he fluct uation of ω3 increases wit h t he increase of β 1 3 . This fluct uation of ω 3 will cause additional dyna mical loads and s haft vibration . T he refor e , β 1 3 must not be too gr ea t . In prac- tical application s , β ~ 45°. 1 3 does no t exceed 35°

Fig .8-21

T he st ructur e of t he universal join t is simple . I t is easy t o assemble . T he unive rsal join t can adapt to poor working conditions . T he angle β 1 3 between t he t wo shafts of t he single uni- versal joint is also allowed to change duri ng transmission . O nly t he a mplitude of t he fluct ua tion of ω3 will change . 8.4.2   The Double Universal Joint Since t he ratio i 1 3 of t he single unive rsal join t is not constan t , vibration will occur during t ransmission . In orde r to m ake ω 1 ≡ω 3 , a double universal join t should be used , i . e ., t he driving shaf t 1 an d t he driven s haft 3 a re connected th rough an in term ediat e shaft 2 and t wo single unive rsal join ts , as s ho wn in Fig . 8-22 .

Fig .8-22

I t can be sho wn that , in order to m ake t he ra tio i1 3 betw een t he inpu t shaft 1 and t he ou t- put shaft 3 identically equal to 1 , t he following two conditions must be satisfied . (1 ) The angle β 1 2 bet ween t he driving shaf t 1 and t he in term ediat e shaf t 2 must be equal t o t he angle β 2 3 betw een t he in term ediate shaft 2 and t he driven shaft 3 . ・ 1 72 ・

(2 ) When t he left yoke of t he in term ediate shaft 2 lies on t he plane defined by t he axes of t he drivi ng shaf t 1 and t he in term ediate shaft 2 , t he righ t yoke of t he in te rmediate shaf t 2 must lie exactly on t he plane defined by t he axes of t he in term ediate shaft 2 and the driven shaf t 3 . In realit y , a double universal join t is seldom used to connect tw o s hafts locat ed on diffe ren t planes . O bviously , if t he driving shaft 1 and t he driven s haft 3 a re located on t he sam e plane , t hen t he second condition can be simplified as follows: (2 ) T he t wo yokes of t he in term ediate shaft 2 must lie on t he sa me plane . Alt hough t he angular velocit y of t he int ermediate shaft 2 fluctuates , its dyna mical load can be neglected since its mom en t of inertia is small . In orde r to adapt au tom atically to t he change i n t he distance between t he t wo single universal join ts , t he in te rmediate shaft 2 is often m ade of two pa rts w hich ar e connected by a sliding key (spline ) , as show n in Fig . 8-22 . T he un ive rsal join t is easy t o assemble and can be used in poor working conditions . I t is suit- able for applica tions w he re t he angle and t he dis- tance bet ween t he t wo shaf ts change con tinuous- ly . T herefore , it is used widely in aut omobiles , m achine tools , and tex tile machine ry , etc . For exa mple , i n t he veh icle show n i n Fig . 8-23 , t he gearbox 1 will be forced to vibrate up and dow n w hen t he vehicle runs on rough roads . T hus , t he distance and t he angle bet ween t he gea rbox 1 and t he diffe ren tial 2 will change con tin uously . W it h a double un ive rsal joi nt , t he tr ansmission is smoo t h and con tinuous and the vehicle can run norm ally .
Fig . 8-23


8.5   S c r e w M e c h an i sm s
8.5.1   Working Principle and Types of Screw Mechanisms A m echanism including screw pair ( s ) is called a screw m echa nism . In addition to sc rew pair ( s) , sc rew mechanism s in common use m ay also include revolu tes and/ or prismatic pairs . Sc rew m echanisms can be divided in to single-t hread screw mechanism s , double- t hread sc rew m echanisms , etc . T he single-t hread screw m echanism is t he most commonly used . It consists of a t hr eaded bar and a nu t . I ts kine ma tic relations hip is quite si mple so it is not necessary to explain it in detail . The lead screws in lat hes , scre w presses , screw jacks , etc . are examples of prim ary application s . Sho wn in Fig . 8-24 is a simple t hr ee-link double-t hread sc rew m echanism . T he scr ew bar ・ 1 73 ・

1 and the stationary nut ( fra m e ) 3 form a scre w pair A . T he screw bar 1 and li nk 2 form a screw pair B . T he link 2 and t he fra me 3 form a prismatic pair C . T he lin k 2 can not rotate but t ranslate relative to t he fr am e . Suppose t hat t he lead of t he t hr eads A and B are lA and lB , respectively , and bot h are right -handed t hreads . W hen t he screw bar 1 ro tates once in t he di- rection sho w n in Fig . 8-24 , t he sc rew bar 1 will t ranslate a distance l A t o the left relative to t he stationa ry nut ( or fra m e ) 3 along its cen t re line . Sim ilarly , t he sc rew bar 1 will t ranslate a distance lB to t he left relative to t he nu t 2 since t he nu t 2 can not rotate . In ot he r words , t he nu t 2 will tr ansla te a distance lB t o t he righ t relative to t he scr ew bar 1 . T he refore , t he ac- tual tr ansla tion of t he nut 2 is t he algebr aic dif- fe rence bet ween t he t ranslation of t he nu t 2 rela- tive t o t he screw ba r 1 ( l B to the righ t ) and t he t ranslation of t he screw bar 1 rela tive to t he fra me 3 ( l A t o t he left ) . If l A > lB , t hen t he nu t 2 will t ranslate ( l A - lB ) to t he left . If lA < l B , t hen it will t ranslate ( lB - l A ) t o t he right . When t he screw bar 1 rota tes t hrough an angle φ relative t o t he stationa ry nu t (fram e ) φ . If t he leads lA and lB of t he 2 π t hr eads A and B are quite closely si mila r , t hen a fine feed will results i .e . t he t ranslation of 3 , t he absolu te t ranslation of t he nu t 2 will be | l A - lB | t he nu t 2 is quite small alt hough screw bar 1 ro tates t hrough a large angle . This kind of mech- anism is often called a differ en tial sc rew mechan ism . W hen the leads of t he t hreads A and B in t he differ en tial screw m echanism are closely sim- ila r , not only can t he nu t 2 have a fi ne t ranslation , bu t also t he leads of t he t wo t hreads A and B do not need t o be too sm all (so t hat t he sc rews have enough st reng t h ) . T her efore , diffe ren- tial sc rew mechanisms are often used as fine m easuring inst rum en ts , index mechanism s , and m any fine adjust m en t devices . φ 2π t he spiral directions of t he t hreads A and B a re opposite . T his gives a quick t ranslation of t he I t can be show n sim ilarly t hat t he absolu te t ranslation of t he nu t 2 will be | l A + lB | nu t 2 . T his ki nd of m echanism is often called a compound screw m echanism . 8.5.2   The Characteristics and Applications of Screw Mechanisms T he st ruct ur e of scr ew m echanisms is quite simple . T he screw is easy to manufact ure . T he motion accuracy is quite hig h . T he speed r atio is quite large . Very large axial t hrust can be produced wit h a small driving torque . The transmission is smoot h and gene rates no noise . T he screw is self-locking . Ho wever , t he mechanical efficiency is quite low because of t he slid- ing betw een scr ew bar and nut . For convenience of m an ufactur e, t he righ t -hand t hr ead is ・ 1 74 ・
Fig . 8-24

most of ten used . Left-hand t hreads a re used only in special applications . Sho wn in Fig . 8-25 is a fine adjust m en t mechanism for adjusting t he borin g t ool in a bor- ing m achine . T he t ool sleeve 3 is fi xed on t he boring bar 6 by setsc rew 5 . The t ool sleeve 3 and t he boring tool 1 form a prism atic pair C by setscrew 2 . Befor e adjust ment , setscrew 2 should be loosened so t hat t he boring tool 1 can t ranslate axially inside t he tool sleeve 3 . T he spiral direction of t he screws A and B are the sam e . If l A = 2 mm and lB = 2 .25 mm , t hen t he boring t ool 1 t ranslates only 0. 25 m m for one r evolution of t he adjusti ng screw 7 . T hus , t he feed of t he boring tool 1 can be adjusted accur ately . Compoun d sc rew m echanisms are often used in devices w hich need to move componen ts quickly , e. g . connection of t rucks , fix t ures , etc . Show n in Fig . 8-26 is a self-cen tering bench vice . T he leads of t he t wo screws ar e t he sam e and t heir spir al directions a re opposite . W hen t he screw ba r 3 is turned , t he cen te rin g clips 1 and 2 can approach t he w ork piece 5 equally quickly an d t he workpiece 5 can be cen ter ed and clamped .

Fig . 8-25

Fig . 8-26

Problems and Exercises 8- 1   Explain t he difference between t he tw o terminologies in ratchet m echanisms: double function and changeable direction . 8-2   List t he advan tages and t he disadvan tages of t he t oot hed r atchet m echanism and t he silen t ratchet m echanism . 8-3   In order to avoid t he rigid impulse at t he t wo mom en ts w hen a Geneva w heel star ts and st ops , w hat a re t he design require men ts ? 8-4   What measures must be taken to pr even t a Geneva w heel from moving during d well tim e ? Ho w are t he cen t ral angle and positions of end poin ts determined in designing t he convex lock- ing a rc of t he driving pla te ? 8-5   Is t he mo tion ti me of a Geneva w heel alw ays grea ter t han t he d well tim e i n ( a ) an ex ternal Geneva mechan ism wit h single roller , ( c ) an in ternal Geneva m echanism ? ・ 1 75 ・ ( b ) an ex ternal Geneva m echanism wit h more rolle rs ,

8-6   Can t he in ternal Geneva mechanism be used wit h mor e than one roller ? 8-7   W hy is t he ex ternal Geneva m echanism wit h 3 r adial slo ts seldom used in practice ? 8-8   Why is t he in dex accuracy of t he hourglass-ca m index m echanism higher t han t hat of cylindrical-ca m index m echanism ? 8-9   W hich two conditions should be satisfied if a double universal join t is used t o connect tw o shaf ts locat ed on t he sam e plane ? 8-10   In a differen tial scr ew m echanism as sho w n in Fig . 8-24 , bot h spiral dir ections of t he t hr eads A and B a re rig ht -handed . T he lead of t he sc rew A is 3 m m . T he nu t 2 is r equir ed to t ranslate 0. 2 m m to t he righ t if t he scr ew bar 1 turns one r evolution i n t he direction show n in t he figure . W hat is the lead of t he screw B ?

・ 1 76 ・

C h ap t e r 9 C ombi n ed Mec han i s m s
9.1   In t r o d u c t i o n
    The plana r lin kages , cam mechanisms , fixed axis gear mechanism s , gea r t rai ns , Geneva m echanisms , ratchet m echanisms , etc . discussed earlie r a re t ypical basic m echanisms . Many kinem atic requirem en ts can be m et wit h t hese basic m echanisms . H owever , t he mo tion pat- terns (position , velocity , and acceleration ) ob tainable wit h t hese basic m echanisms are limited . W it h the developm en t of production m et hods , higher degrees of m echanization and au- tomation a re r equir ed . Machines are required to realize more complicated motion pat terns and to have better dynamical char acte ristics . Because of t heir lim itations , it is gener ally difficult , or even impossible , to realize t he more complica ted motion require men ts using basic m echanisms . T he refore , i n production practice , som e basic m echanisms a re of ten combined so t hat every ba- sic mechanism can not only bring its pa rticular kinem atic function in to full play bu t also over- come its limitations . M echanism syste ms combined in such a way are called combined mecha- nisms and t he basic m echanisms in t hese combined m echanisms are called sub-m echanisms . Combined mechanisms can t hus be bet ter in perform ance and can m eet m any higher require- m en ts .

9.2   M e t h o d s f o r t h e C l a s s i f i c a t i on o f C omb in e d M e c h a- nisms
    In order to clarify t he kinem atic t ran smission rou te between t he sub-m echanisms and/ or Assur groups in a combined m echanism , st ructural analysis must be complet ed on t he combined m echanism . T he detailed steps of st ructur al analysis for combined m echanisms ar e : Disconnect t he sub-m echanisms and/ or Assur groups from t he combined mechan ism accordi ng to the ki ne- m atic diagram of t he specific combined m echanism , iden tify t heir tr ansmission rou te , t hen show t he struct ure and t ransmission rou te in block diagra ms . L astly , decide on t he t ype of t he combined mechanis m according t o t he patt ern of t he t ransmission rou te . T he re a re various m et hods for t he classifica tion of combined mechanism s . T hey may be classified by t he nam es of sub-m echanisms , e. g .gear-lin kage , ca m-linkage , gea r-ca m mecha- nism , double-ca m m echanism , etc . T hey may also be classified by t he pat terns of t ransm ission ・ 1 77 ・

rout e , e .g .i n series , in pa rallel , by feedback , etc . The lat ter classification sho ws t he tr ansmis- sion rou te, w hich simplifies t he kinem atic analysis and syn t hesis of combined m echanisms . T he refore , t he la tter classification is used in t his book .

9.3   S e r i e s C o mb i n e d M e c h a ni sm s
    Som e basic m echanisms and/ or Assur groups are connected in series so t hat t he out pu t mo- tions of one basic m echanism ( or Assur group ) is t he in pu t motion of t he next basic mechan ism (or Assur group ) . The mechan ism syst em combined in t his way is called a series com bi ned m echan ism . Sho wn in Fig . 9-1a is a t ransmission mechan ism in a tex tile m achine . By use of a ca m m echanism , t he inpu t rotation of t he ca m 1 is t ransform ed in t o t he oscillation of follower 2 and t he motion of poin t B . Wit h t he help of an R R R Assur group consisting of link s 3 and CD , t he determ ined mo tion of t he poin ts B an d D gives link CD a det ermined motion . The lin k CD is t he inpu t gea r 4 of t he gea r-rack m echanism . T hus , t he ou tpu t r ack 5 is given t he r equir ed motion . According to t he above st ruct ural analysis , t he struct ural block diagra m of t he com- bined m echanism should be t hat show n in Fig . 9-1b . T he ou tpu t mo tion of one basic mecha- nism ( or Assur group ) is t he i npu t mo tion of t he nex t basic m echanism ( or Assur group ) . T he refore , t his is a series combined mechanism .

Fig . 9-1

    In t he planetary gea r t rain sho wn in Fig . 9-2 , differ en t poin ts on the planet gear 2 will t race differen t loci w hich ar e often r eferred to as cycloids . T he shapes of t hese cycloids depend upon t he m eshi ng patte rn ( in ternal m esh or ex ternal m esh ) , ratio of teet h and t he position of t he poin t on t he planet gear 2 . Suppose that , in itially , t he planet carrier H is located at hori- zon tal position A B 0 and t he poin t on t he planet gear 2 is located at poin t C0 on the cent re line A B 0 , as sho w n in Fig . 9-2 . When t he planet carrie r H ro tates t hrough angle φ H and arrives at t he position A B , t his poi nt will arrive at poin t C . Si nce ・ 1 78 ・

φ z3 2 - φ H = N φ 3 - φ H z2 t herefore , φ 1 - N 2 = z3 φ H z2

w her e N is t he coefficien t for m eshing . ( N = + 1 for in ternal mesh . N = - 1 for ex ternal m esh .) φ 2 and φ H a re positive coun ter-clock wise , negative clock wise . T he refore, for t he gea r-li nkage show n in Fig . 9-2 , t he x-y coordinates of t he poin t C a re x C = lA B cos φH - lB C cos φ 2 yC = l A B sin φ H - l B C si n φ 2 If t he ratio z 3/ z 2 is an i nt eger m , t hen t he poin t C
Fig . 9-2

will draw m complete cycloids . Some curve sections of som e cycloids can be approxi mated to an a rc or a st raigh t line . Using t hese proper ties ingeniously , in term it ten t mechan isms can be con- st ructed in w hich t he follower can d well approxim ately for long tim e periods .

Fig . 9-3

In t he gea r-li nkage with planet gea r show n in Fig . 9-3a , z 3/ z 2 = 3 and t he poin t C is lo- cated on the circu mfe rence of t he pitch circle of t he planet gea r 2 . T he locus of t he poin t C will be t hree gene ral h ypocycloids C1 C2 C3 C1 , as show n in t he figure . The hypocycloid C3 CC1 is a close approxi mation t o an a rc . An R R P Assur group consisting of coupler 4 and sliding block 5 is added to t he original mechan ism . T he leng t h of t he couple r 4 is set equal to t he r adius of t he a rc . The slidew ay of t he slidi ng block 5 is set t o pass t hrough t he cen t re D of t he arc . Thus , t he driven sliding block 5 will d well approxi mately w hen t he poin t C moves from t he position C3 to t he position C1 ( t he corresponding rota tion angle of t he drivin g planet ca rrier A B being ・ 1 79 ・

120° ) . At t he ot her positions of t he driving planet ca rrier A B , t he sliding block will r ecipro- cate at a variable speed . As analyzed above , t he st ruct ure of t his com bined m echanism is sho w n in Fig . 9-3b . The m echanism is a se ries combined mechanism . If t he sliding block 5 is r eplaced by a rocker 5 (i .e. its motion . U sing t he cycloid with a st raigh t-line sec- tion , a combined m echanism can also be cr eat ed t he follow er of w hich can d well approxima tely for a long tim e . For exam ple , in t he gear-linkage show n in Fig . 9-4 , z 3/ z2 = 4 . If lB C = 0. 36 r2 , t hen t he locus of t he poin t C on gear 2 is a squar e wit h rounded corners . If an R P P As- sur group consisting of sliding block 4 and slid- ing bar 5 is added to t he original m echanism by revolu te C and prismatic pair D , then t he slid- ing bar 5 will have a long approxim ate d well at each end of its r eciproca tion .
Fig . 9-4

group is r eplaced by an

R R R Assur group ) , t hen t he driven rocker 5 will oscillate wit h a long tim e dw ell at one end of

    In t he gea r-li nkage show n in Fig . 9-5a , crank A B is a driver . T he planet gea r 2 is fixed t o t he couple r 2 w hile t he sun gea r 5 is mounted on shaft D . T he gear 2 , gea r 5 and rocke r 3 constitu te a diffe ren tial gea r t rain wit h tw o DO F . T he angles φ 2 and φ 3 of t he coupler 2 and rocker 3 are t he t wo ou tpu ts of t he crank-rocker mechanis m A BCD . T hey are also t he inpu t motions of t he differential gea r tr ain wit h t wo DO F . T hey a re com bined in t he differen tial gear t rain and φ 5 is used as t he out put . Therefore , t he m echanism is a se ries combi ned mechan ism and its st ruct ural block diagra m is show n in Fig . 9-5b .

Fig . 9-5

・ 1 80 ・

9.4   P a r a ll e l C omb in e d M e c h a ni s ms
    In som e combined m echanisms , t he motion of t he driver is inpu t simultaneously to n single DO F basic mechanism s ( or kinem atic chains ) arranged in parallel . The n out put motion s of t he n single DO F basic m echanisms (or ki ne matic chains ) are inpu t simultaneously to a basic m echanism ( or Assur group ) wit h n DO F and t hen combined to produce an ou t put . T he m echanism syste m combined in such a way is called a par allel com bi ned m echan ism . In t he gea r-linkage show n i n Fig . 9-6a , gear 1 (or crank A B ) is a driver and rot ates at a constan t speed . O n t he one hand , t he constan t rota tion of t he c ran k A B makes the rocker 3 os- cillate at a va riable speed t hrough a cr ank-rocker mechanism A BCD . O n t he o t her hand , t he constan t rotation of t he gea r 1 makes t he gear 7 rot ate at a different con stan t speed clockwise t hrough a fixed axis gear tr ain . T he sun gear 5 , sun gear 7 , planet gea r 6 and planet ca rrier D E con stitu te a differential gear tr ain wit h t wo DO F . I t com bines t he con stan t rotation of t he gear 7 and t he va riable oscillation of t he planet carrier D E and t hen ou t pu ts a variable rota tion of t he int ernal gear 5 . F rom analysing t he st ructure as above , we can see t hat t he strea mlined diagr am of kinem atic analysis , or st ruct ural block diagr am , should be t hat show n i n Fig . 9-6b . T he refore , t he m echanism is a par allel combined mechanism .

Fig . 9-6

    Many k inds of loci , even very complicated and special shaped loci , a re often r equir ed in engi neering . I t is difficult t o gene rate t hese loci satisfact orily using t he basic m echanisms and so t hese complicated loci are frequen tly realized by gear li nkages . The five-ba r gear-lin kage sho w n in Fig . 9-7a is one of t he most frequen tly used gear-lin kages to produce complicated coupler loci . O n t he one hand , t he driver 1 gives t he poin t B a dete rmined motion t hrough link A B ; on t he o ther hand , t he drive r 1 gives t he gear 2 and t he poin t D determined motions t hrough t he fixed axis gea r t rain . T he revolu tes B and D are tw o oute r revolu tes of t he R R R Assur group . T he determ ined motions of t he r evolutes B and D ar e t he two inpu t motions of t his ・ 1 81 ・

R R R Assur group . Since t he mo tions of t he poin ts B and D are dete rmined , t he poin t F has a determ ined locus . T he st ruct ural block diagr am of t he m echanism is show n in Fig . 9-7b . From t h is , we can see t hat t he gear-lin kage is a par allel combined mechan ism .

                  a)                        

            b)

Fig . 9-7

    T he locus shape can be easily adjusted if one of following is changed : t he toot h ra tio , t he link lengt hs , t he r elative angle of t wo cr ank s , t he position of t he poin t F on t he couple r , etc . If t he toot h ratio i 1 2 = m/ n and t he m and n ar e tw o i ntegers w hich can no t be divided by a sa me i nteger , t hen t he point F will complete a cycle only if t he cr ank 1 rotates m cir cuits and t he c ran k 2 rotates n circuits . T hus , t he locus of t he poin t F can be quite complicated .

Fig . 9-8

Sho wn in Fig . 9-8a is a cam-li nkage w hich can realized any predetermi ned locus . T he ca m 1 is fixed on cr ank A B . On t he one hand , t he ro tation of cam 1 gives t he poin t B a dete rmined motion t hrough t he lin k A B ; on t he ot her hand , it gives a det ermined motion t o poin t D t hrough t he planar cam m echanism wit h oscillating rolle r follo wer 4 . The revolu tes B and D a re t he t wo oute r revolu tes of t he R R R Assur group consisti ng of lin ks 2 and 3 . As t he mo- tions of the points B and D a re determined , t he mo tions of t he two lin ks in t he Assur group can ・ 1 82 ・

be determined . T hus , t he poin t G is given a determ ined motion . F rom t he st ructur al analysis above, we can see t hat t he mechan ism is a pa rallel combined m echanism . I ts st ructur al block di- agr am is s ho wn in Fig . 9-8b . T he point G can move accurately along a pr edete rmined locus by designing t he cam profile correctly .     Show n in Fig . 9-9a is a double ca m m echanism consisting of t wo ca m m echanisms . Ca me 1 and 1′ a re fi xed to each ot her . The slots on t he ca ms will force the follo wers 3 and 4 to re- ciprocat e i n t he x and y directions , respectively . T he sliding blocks 5 and 6 form a P R P Assur group . T he movem en t of t he followe r 3 con t rols t he mo tion of t he revolu te cen t re M i n t he x direction . Similarly , t he move men t of t he followe r 4 con t rols t he motion of t he revolu te cen tre M in t he y dir ection . The inpu t rotation of t he ca ms produces x and y reciprocations t hrough t wo ca m m echanisms . T hese reciprocations a re combined in t he P R P Assur group w hich gener- at es t he locus of poin t M . As a result , t he st ruct ural block diagra m of t he mechanism should be t hat show n in Fig . 9-9b . T her efore , t he mechan ism is a parallel com bined mechanism .

Fig . 9-9

9.5   C omp o un d C o mb i n e d M e c h a ni sm s
    In a pa rallel combined m echanism , t he n ou tpu ts of t he drive r a re input si multaneously in- t o n single DO F basic m echanisms ( or kinem atic chains ) arranged in parallel . T he n ou tpu t motions of t he n single DOF basic m echanisms ( or kinem atic chains ) a re com bined in a basic m echanism wit h n DOF to produce a determined ou t pu t . I n some cases , n of ( n + 1 ) ou t puts of t he driver are in pu t simultaneously in to n si ngle DOF basic mechan isms ( or kine matic chai ns) arranged in pa rallel . The n out pu ts of t he n single DOF basic mechanis ms ( or ki ne- ・ 1 83 ・

m atic chains ) and an ou t put come from t he driver directly a re inpu t sim ultaneously in t o a basic m echanism ( or Assur group) wit h ( n + 1) DOF to produce a determi ned ou tpu t . T he mech- anism syste m combined in such a way is called a co m poun d co m bi ned mechan ism . Sho wn in Fig . 9-10a is an inver ted gea r-li nkage consisting of a r evolute four-ba r linkage A BCD and a pair of ex ternal m eshed gears . The planet gear 2 is fixed to coupler 2 . T he ou t- put gea r 5 is located on shaft A of t he driving crank A B . In a double-c ran k mechan ism A BCD , t he motion of t he lin k 2 can be dete rmined . T he gea r 5 , gea r 2 and planet ca rrier 1 constitu te a differen tial gear t rain wit h t wo DOF . After t he mo tions of t he planet ca rrier 1 and t he planet gear 2 are determ ined completely , t he motion of t he sun gear 5 can be determi ned . As a result , t he struct ural block diagra m of t he m echanism is t hat s ho wn in Fig . 9-10b . From t h is , we can see t hat t he m echanism is a compound combined m echanism .

Fig . 9-10

Fig . 9-11

    In t he cam-gea r combined mechanis m show n in Fig . 9-11a , t he planet carrier H is a driver w hile t he sun gear 1 is an out pu t gea r . T he rolle r on t he sector planet gea r 2 runs in t he slot of t he stationary ca m 3 . W hen t he planet ca rrier H rota tes at a constan t speed , t he roller will move along t he slo t . T hus , t he motion of link BC and sector gear 2 will be dete rmined . In t he diffe ren tial gea r t rain H -1-2-3 wit h tw o DOF , t he ou tpu t motion of t he sun gear 1 can be de- ・ 1 84 ・

termined if t he mo tions of t he planet ca rrier H and t he planet gear 2 a re know n . As analyzed above, t he st ructural block diagr am of the m echanism should be t ha t show n in Fig . 9-11b . O ne of t he inpu ts of t he differential gear t rai n wit h tw o DOF com es directly from t he driver w hile t he o t her i nput com es from t he stationary ca m mechanism . F rom t he st ructural block dia- gram , we can see t hat t he m echanism is a compound combi ned m echanism .

9.6   Mul t i p l e C omb in e d M e c h a ni sm s
    In som e combined m echanisms , t he basic m echanisms are connected t o each ot her so t hat t he relative fra me of one is t he ou t put lin k of t he form er . Every basic m echanism has its ow n in- dependen t pow er source . E ach basic m echanism execut es its ow n motion t ransmission and t he combina tion is t he out put of t he w hole mechan ism . T he mechan ism system combined in such a way is called a m u l tip le co m bi ned mechan ism . In t he excavator s ho wn in Fig . 9-12a, t he fr am e 4 of t he first basic mechan ism 1-2-3-4 is t he chassis of t he excavator . The second basic m echanism 5-6-7-3 is connected to t he ou tpu t link 3 of the first basic m echanism , i .e. pu t link 3 of t he first basic mechanism is used as the relative fra m e of t he second basic m echanism . Equally , t he t hird basic mechanis m 8-9- 10-7 uses t he ou tpu t lin k 7 of t he second basic m echanism as its relative fra me . E ach basic m echanism has its o w n independen t pow er so t her e ar e t hree inpu ts and t he m achine is a multi- ple com bined mechanism . I ts st ructural block diagr am is show n in Fig . 9-12b .

Fig . 9-12

    S ho w n in Fig . 9-13a is the schem atic diagra m of a funfair horse . T he first basic mechan ism consisting of t he fr am e 5 and lin k 4 is t he carriage m echanism ( t he m echanism Ⅰ in Fig .9- 13b) . I ts ou t put lin k 4 rota tes abou t t he fixed shaft O-O . T he cr an k-oscillating block mecha- nism consistin g of links 1 , 2 , 3 , and 4 is t he carried mechanism ( t he mechanis m Ⅱ in Fig . 9-13b) . T he relative fra me 4 of t his m echanism is t he ou tpu t lin k of t he first mechanism I . T he power inpu t of carriage m echanism Ⅰ causes t he horse to move a round in a circle . T he ・ 1 85 ・

po wer input of the ca rried m echanism Ⅱ gives t he rise and fall of the cent re of gravit y M of t he horse and t he pitching movem en t . W hen t he t wo basic mechan isms oper ate at t he sa me , t he funfair horse can execu te more complicated movem en ts , e.g . ro tati ng , up and dow n , and pitching at sa me tim e . T he st ructur al block diagram of t he m echanism is show n in Fig . 9-13c . T his is ano t her example of a multiple combi ned m echanism .

Fig . 9-13

9.7   F e e d b a c k C om b in e d M e c h a ni s ms
    As is well know n , a differ en tial m echanism has tw o DOF . T wo inpu ts ar e needed t o ou t- put a determined motion . In some differ en tial m echanism , one of t he inputs comes from t he drive r directly . The ot her in pu t is t he feedback from t he out pu t of t he differen tial mechan ism t hrough t he si ngle DOF k ine matic chain . T he mechanism syste m combi ned in such a way is called a feedback co m bi ned mechan ism .     Sho wn in Fig . 9-14a is a correction and compensation device used in gear cu tting m achine t ools . T he w orm and worm w heel m echanism ( the m echanism Ⅰ in Fig . 9-14b ) is a t wo DO F mechanism . The rotations of t he wor m and t he w orm w heel are , respectively , t he inpu t and ou tpu t of the w hole m echanism . T he ro tation of t he ca m ( or t he worm w heel ) can drive t he worm to r eciproca te axially . The st ructur al block diagr am is show n in Fig . 9-14c . O ne of inpu ts of t he two DOF w orm an d worm w heel mechanism , i .e. t he r eciprocation of t he w orm , ・ 1 86 ・

is t he feedback of the out pu t ( t he ro tation of t he worm w heel or cam ) th roug h the single DOF cam mechanism . T he refor e, t he w hole mechanis m is a feedback combined m echanism .

Fig . 9-14

    In gear cut ti ng m achine tools , because of errors in manufact ure and assembly , t he gear blank and t he cut ting t ool do not m ain tain t he correct tr ansmission ratio . T his error reduces t he accuracy of t he gear . How ever , wit h t his correction and compensation m echanism , t he error can be corr ected con tinuously . T he cam profile is designed according to t he actually m easur ed error . T he w orm is t hus forced to t ranslate axially by t he cam mechanis m w h ile t he worm w heel rota tes . T hus , t he w orm w heel is given an additional rotation t o correct t he k ine matic er- ror .

9.8   M i x e d C omb in e d M e c h a ni sm s
    A mechanis m syst em form ed t hrough mor e t han one com bination pat tern is called a m i xed co m bi ned mechan ism . Sho wn in Fig . 9-15a is a t ransmission m echanism in a prin ter . W hen t he driver A B ro- tates , t he li nk 3 is given a det ermined mo tion t hrough a double-crank mechanism A BCD . T he cam is fi xed to t he crank 3 and drives the t ranslating roller follo wer 6 (or rack 6 ) . W it h t he link subroutine, t he motion of t he poin t E can be dete rmined . T hen , t he motion of t he poin t F can be determined wit h t he R R P subroutine . T he rack 6 , gea r 7 and t he ou tpu t rack ( plat- form ) 8 constit ute a gea r-rack m echanism wit h t wo DO F . T he translations of bo t h t he rack 6 ・ 1 87 ・

and t he poin t F cause t he r ack 8 t o r eciprocate . F ollowing t his analysis , t he st ructur al block di- agr am of t he m echanism is as show n in Fig . 9-15b . T he combination patte rn is neit he r series nor parallel . Therefore , we have a m ixed com bined mechanism . The use of t he cam mecha- nism is to corr ect the velocit y of t he ou t put rack 8 so t hat t he velocity of t he plat form 8 is ap- proxim ately constan t duri ng t he work st roke .

Fig . 9-15

Fig . 9-16

    T he t hree-gear drive sho w n in Fig . 9-16a is a well-know n m echanism w hich can be used to produce a varying ou tpu t motion from a con tinuous ro tary inpu t . The inpu t shaf t at A drives an eccen t ric gear at a constant speed ω 1 . T he offset A B of t his gear forms t he input cr an k of a four-ba r linkage A BCD t o w h ich t wo gears are a ttached at C and D . T he motions of coupler 2 and rocker 3 can be determined in t he rocker-c ran k mechanism A BCD . T he gear 1 , gear 6 and coupler 2 constit ute a differen tial gear t rain wit h tw o DOF . The motion of t he gea r 6 can be determined according t o t he motions of t he coupler 2 and t he gea r 1 . T he gear 6 , t he gear 5 and t he rocker 3 constit u te anot her differen tial gea r train wit h tw o DOF . T he mo tion of gear 5 is a com bination of those of t he rocke r 3 and the gea r 6 . A pplying t his analysis , t he st ruc- ・ 1 88 ・

tural block diagra m of t he m echanism is as show n in Fig . 9-16b . F rom t his , we can see t hat t h is is a mi xed combined m echanism .

9.9   M a t t e r s N e e d i n g A t t e n t i o n
    Lastly , t he re a re t wo poin ts for atten tion . ( 1) For t he sam e k ine matic chai n , if differ en t links are chosen as fra me or drive r , t hen t he combina tion pat tern of t he combined mechanism s will be differen t . For exa mple , t he kine matic chai ns i n Fig . 9-5a and Fig . 9-10a are t he sa me . Ho wever , since different lin ks are chosen as drive r , t he com bi ned m echanism in Fig . 9-5a is a series t ype w hile t hat i n Fig . 9-10a is a com- pound type . (2 ) T he m ain purpose of t he st ructural analysis and t he st ructural block diagr am is to clari- fy t he t ransmission rou te , not t he classification . For t he sa m e mechanism , if differ en t m et hods a re used for kinem atic analysis , t hen t he st reamli ne diagram s for ki nem atic analysis , t he st ruc- tural block diagra ms , and t he combination pat terns corresponding to t he differ en t kine matic analysis met hods will be differen t , even if t he fra me and t he drive r a re unchanged . T his is quite norm al .

Fig . 9-17

For exa mple , show n in Fig . 9-17a is a combined gear t rain with t hree sun gears 1 , 3 and 4 . The sun gear 1 is an input gea r w h ile t he sun gea r 4 is an ou tpu t gear . According t o differ- ・ 1 89 ・

ent kinem atic analysis routes , t he com bination patte rn of t his combined mechanism may be se- ries ( show n in Fig . 9-17b ) or compound (s ho wn i n Fig . 9-17c ) or feedback ( show n in Fig . 9-17d ) . Problems and Exercises 9-1   In t he com bined m echanism sho wn in Fig . 9-18 , lin k 1 is a driver w hile t he sliding block 6 is t he ou tpu t link . A nalyze t he motion t ransmission route and dr aw its struct ural block diagram . What is t he combina tion pat - tern of t he combined mechan ism ? 9-2   T he cam-linkage show n in Fig . 9-19 is t he holding m echanism i n a chocolate pack ing m achine . T he link A B is a driver w hile sliding block 4 is t he ou tpu t link . A nalyze t he motion t ransmission route and dr aw its struct ural
Fig . 9-18

block diagr am . W ha t is t he com bination patte rn of t he com bi ned m echanism ?

Fig . 9-19

Fig . 9-20

9-3   In t he planetary gea r t rain sho wn in Fig . 9-2 , z 3 = 50 , z 2 = 20 , module m = 3 mm and lB C = 0. 69 r2 . Dra w t he locus of t he poin t C . 9-4   In t he gea r-linkage show n in Fig . 9-3a , r3 = 150 m m and r 2 = 50 m m . T he drivi ng carri- ・ 1 90 ・

er A B runs at a constan t speed ωH = 10 rad/ s . (1 ) Draw t he locus of t he poin t C . (2 ) S uppose l 4 = 7. 11 r2 . W rite a program to calculate t he displace men t sD , velocity v D and accelera tion a D of t he slidi ng block . Dra w t he motion curves sD - φH , v D - φH , and aD - φH . 9-5   In t he gea r-linkage show n in Fig . 9-4 , r3 = 200 mm , r2 = 50 mm and lBC = 0 .36 r2 . T he driving carrie r A B runs at a const an t speed ωH = 10 r ad/ s . (1 ) Draw t he locus of t he poin t C . (2 ) W rite a progr am t o calculate t he displace ment s5 , velocit y v 5 and accele ration a 5 of t he followe r 5 . D raw t he mo tion curves s5 - φH , v5 - φ H , and a 5- φ H . 9-6   In t he gea r-linkage sho wn in Fig . 9-20 , lin k A B is a driver w hile t he oscillating bar 5 is t he ou tpu t link . A t t he position show n , t he t hree poin ts A , B , and C a re on t he sam e hori- zon tal line . l A B = 150 m m , lB C = 100 m m , l A D = 500mm , z 3 = 90 and z 2 = 30 . The driving car- rier A B runs a t a constan t speed ωH = 10 r ad/ s . (1 ) Analyze t he mo tion transmission rou te and dr aw its st ructur al block diagra m . W hat is t he combination pa ttern of t he combined m echanism ? (2 ) Draw t he locus of t he poin t C . (3 ) Write a progr am to calculate t he an gular displacem en t φ5 , angular velocit y ω 5 and an- gular acceleration ε 5 of t he driven rocker 5 . D raw t he motion curves φ 5 -φ H , ω 5 -φ H , and ε 5- φ H . 9- 7   In t he cam-linkage show n i n Fig . 9- 21 , t he driving rota ry block 1 rot ates con tin- uously . By designing t he con tour of t he fixed cam correctly , t he sliding block 4 can get a predet ermined motion . Analyze t he motion t ransmission rou te and dr aw its st ruct ural block diagra m . What is t he combination pat - tern of t he combined mechan ism ? 9- 8   In t he combined mechanism show n in Fig . 9-22 , t he driving crank A B runs at a constan t speed and drives guide ba r 2 and oscillating block 3 . T he guide bar 2 will r eciprocate relative to t he oscillating block 3 . A rack is fixed t o guide ba r 2 and gea r 4 is moun ted on shaft C . T he rack m eshes wit h t he gea r so t hat t he gear can ou tpu t a ve ry large angula r oscillation . A nalyze t he motion t ransmission rout e and draw its st ructur al block diagra m . W hat is t he com- bina tion pat tern of t he combined mechanism ? 9-9   I n t he gear-lin kage show n in Fig . 9-7a, suppose t hat r1 = 28 mm , r2 = 42 mm , l A B = lC D = 24 mm an d lB F = l D F = 110 mm . T he relative position bet ween t he cr anks A B and CD is show n in t he figur e . Dra w t he locus of t he poin t F . 9-10   S how n Fig . 9-23 is a ca m-lin kage w hich can t race any complicated locus . T he driving ・ 1 91 ・
Fig . 9-21

cr ank A B ro tates toget he r wit h t he cam about t he shaft A . By designi ng t he cam profile cor- rectly , t he revolu te C bet ween t he couplers 2 and 3 can move along a pr edeterm ined locus . A nalyze t he motion t ransmission rout e and draw its st ructur al block diagra m . W hat is t he com- bina tion pat tern of t he combined mechanism ?

Fig . 9-22

Fig . 9-23

9-11   S how n in Fig . 9-24 is a cam-linkage used in a prin ter t o tr anspor t pape r . T he t wo ca ms a re fixed to each ot her . When t he double cam ro tates , t he sucke r J fixed to coupler 2 will move along t he locus m m so t hat paper can be lifted and t ranspor ted . Analyze t he motion tr ansmis- sion rou te and draw its st ructural block diagr am . W hat is t he combination patte rn of t he combined m echa- nism ? 9- 12   In t he gear-linkage show n in Fig . 9-25 , t he cr ank A B is a driver . The ou tpu t pinion 5 is located on t he shaf t A . A nalyze t he motion t ransmission rout e and draw its st ructur al block diagr am . What is t he combination patte rn of t he combined m echa- nism ? 9-13   In t he cam-w orm combined mechan ism show n in Fig . 9-26 , t he worm 2 and t he cylindrical ca m 2′ a re fixed toget he r . T hey are connected to t he input shaf t 1 by a sliding key , i .e. ω 2 = ω 1 . T he sta- tiona ry pin 4 runs in the slot of t he cylindrical ca m 2′and causes reciprocation of ca m 2′and worm 2 .
Fig . 9-24

T he t ranslation of t he worm 2 will cause an additional rotation of worm gea r 3 . Analyze t he motion t ransm ission rout e and draw its struct ural block diagra m . What is t he combi nation pat- tern of t he combined mechan ism ? 9-14   In t he cam-gear-linkage sho w n in Fig . 9-27 , t he cam is fi xed to gear 1 . They a re bot h inpu t lin ks . Gear 3 is t he ou tpu t . A nalyze t he motion t ransmission rou te and dr aw its struct ural ・ 1 92 ・

block diagr am . W ha t is t he com bination patte rn of t he com bi ned m echanism ? 9- 15   Show n in Fig . 9-28 is a schem atic diagra m of an indust rial robo t . Analyze t he mo tion t ransmission rou te and draw its struct ural block diagra m . W hat is t he combination pa ttern of t he combined m echanism ?

Fig . 9-25

Fig . 9-26

Fig . 9-27

Fig . 9-28

9-16   In t he gea r-linkage show n in Fig . 9-29 , gea r 1 is a drive r w hile gea r 6 is t he out pu t . A nalyze t he motion t ransmission rou te and draw its st ruct ural block diagram . W hat is t he com- bina tion pat tern of t he combined mechanism ? 9-17   In t he gea r-linkage s ho wn in Fig . 9-30 , t he crank A B is a driver w hile gea r 4 is t he ou t- put . A nalyze t he motion t ran smission route and draw its st ruct ural block diagr am . W hat is t he combina tion pat tern of t he combined mechanism ? 9-18   S how n in Fig . 9-31 is anot he r kind of gea r-li nkage w hich can t race complicated loci . In t h is combined m echanism , t he t wo m eshed gears 1 and 5 are moun ted on coupler 2 . W hen t he driving cr ank A B rotates , t he poi nt M fixed to gea r 5 will t race a complicated locus . Analyze ・ 1 93 ・

t he motion t ransmission rou te and dr aw its st ruct ur al block diagram . W hat is t he combina tion pat tern of t he combined mechan ism ?

Fig . 9-29

Fig . 9-30

9-19   In t he kine matic chain show n in Fig . 9-16a , if t he driver is t he link 3 instead of t he gear 1 , w hile t he gea r 5 is still an out put gear . Analyze t he motion tr ansmission rou te and draw its st ruct ural block diagra m . What is t he combination pat tern of t he combined m echanism ?

Fig . 9-31

・ 1 94 ・

C h a p t e r 10 B ala n ci n g o f Mac h i n e r y

10.1   P u r p o s e s a n d M e t h o d s o f B a l a n c i n g
10.1.1   Purposes T he cen t re of mass of som e machine ele ments , e. g .ca ms , m ay no t coincide wit h t heir ro- tating cent res because of t he asymm et ry of t he st ruct ure . Even for symm et rical m achine ele- m en ts , t he cent re of m ass m ay still be eccen t ric because of uneven dist ribu tion of m aterials , er- rors in m achin ing and also in castin g and forging . O t her errors m ay be caused by im proper bor- ing , by keys or by assembly . Sho wn in Fig .10-1 is a disk t he mass cen t re of w hich is displaced by an eccen tricity e from its ro- tating cen tr e . W hen t he disk ru ns at constan t an- gular speed , ω, t he cent rifugal for ce F c rea ted
2 by t he mass m is F = me ω dir ected radially ou t-

wards . Since t he magnitude of t he cen trifugal force F is propor tional t o t he squar e of angular speed ω, t he magnit ude of cen t rifugal force F of high speed pa rts m ay reach quite high levels even for a small eccen t ricit y e . For exa mple , t he
Fig . 10-1

weigh t W of a round plane cu tter is W = 500 N and t he ro tati ng speed n = 2 000 r/ m in . Al- t hough t he eccen tricit y e is only 1m m , t he cen t rifugal inertia force F cr ea ted is F = m e ω =
2 ( W/ g ) e ω = 2 238 N , w hich is much larger t han its weight . F urt hermore , t he direction of 2

t he cen t rifugal force F exe rted on t he fram e is time varying , and it will t he refor e i mpa rt vibra- tion to t he fr am e . T his vibration can adversely affect t he st ruct ural in tegrit y of t he m achine foundation . It also decr eases t he machining precision and efficiency , inc reases t he st ress in componen ts , causes wea r an d subjects bearings t o r epeated loads w hich cause par ts t o fail pre- m atur ely by fatigue . If t he rot atin g frequency of a pa rt is nea r to its resonant frequency , t he a mplitudes of vibr ation m ay reach dan gerous levels . T her efore , w e should try t o eliminate t he un wan ted cent rifugal forces in m achines , especially in high-speed m achinery and precision m a- ・ 1 95 ・

chine ry . H oweve r , som e machi nes , tion . 10.1.2   Methods

g . vibrat ors , shock drillers , etc . do work by vibra-

T he process of designin g or modifying m achine ry in order to r educe un wan ted vibration to an accept able level, and possibly to eliminate it en tir ely , is called bal ancin g . T he balancing of m achine ry can be divided in t o balancing of ro tors and balanci ng of m echanisms . (1 ) Balancing of rotors Par ts constr ained t o rotate about a fixed axis ar e called rotors . Such rot ors can be divided furt her in to rigid rotors and flexible rotors . ( a ) Rigid ro tors If t he rot ati ng frequency of t he rotor is less t han ( 0. 6 ~ 0 .7 ) n C 1 ( w he re : n C 1 is t he first resonan t frequency of t he ro tor) , t hen t he ro tor is supposed t o have no deforma tion duri ng ro- tation and is called a ri gi d rotor . According t o t he ratio between t he axial di men sion and dia m- eter , rigid ro tors can be divided in t o disk-li ke rot ors and non-disk rigid ro tors . Ro tors w hose ax ial dimensions B a re sm all compared to t heir dia m eters D (usually B/ D < 0. 2) , such as gea rs , pulleys , ca ms , flyw heels , fan s , grinding w heels , and impellers , are called d isk-like rotors . T he m asses of such ro tors are assum ed practically to lie in a common t ransverse plane . If B/ D ≥ 0. 2 , such as c ran kshaf ts , spindles in machi ne tools , ax les of elect ric mo tors , etc ., t he rotor is called a non-d isk ri gi d rotor . (b ) Flexible rotors In som e m achines , g . steam t urbines , t he mass and axial dim ensions of t he ro tor are quite large , w hile t he diam eter is quite small . T herefore , its t he first r esonan t frequency n C 1 is low . If t he working rotating frequency of t he rot or is larger t han (0 .6 ~ 0. 7) n C 1 , t hen t he ro tor will have large elastic deforma tion due t o imbalance during ro tation . T he elastic deform a- tion m akes t he eccent ricit y la rger t han t he original one so t hat a new imbalance factor is added and t he balancing problem becomes mor e complica ted . S uch a rot or is called a f le xible rotor . T he balancing t heory of flexible rot ors is beyond t he scope of t his book . In t his book , empha- sis is pu t on t he balance of rigid rotors . (2 ) Balancing of m echanisms T he coupler of a lin kage has a complex motion . The acceleration of its m ass cen t re and its angular accelera tion vary t hroughou t t he motion cycle . The couple r will t herefore create a va ry- ing inertia force and iner tia momen t of force for any m ass distribu tion . So t he balance of link- ages must be considered as a w hole . The resultan t i ner tia force of all moving pa rts is equal to t he net unbalanced force acting on t he fr am e of a m achine , w hich is refe rr ed t o as t he shaki ng f orce . Li kewise , a r esultan t unbalanced moment acting on t he fra me , caused by t he ine rtia forces and iner tia mom en ts of all moving pa rts , is called t he shak in g m om en t . The s haking ・ 1 96 ・

force and t he s haking momen t will cause t he fra me t o vibrate . Alt hough t her e exist m any m et hods t o m ake a lin kage m echanism fully shak ing force and/ or fully shaking mom en t balanced , t he results a re usually unr easonable , g ., t he in ternal bearing forces m ay be too gr eat , t he st ruct ure m ay be too cumbersom e , t he driving mom en t required m ay be too la rge , etc . T hus , a par tial balance may represen t t he best compromise be- tween many dynam ic c rite ria . Op timization m et hods can be used to reach a reasonable result .

10.2   B a l a n c i n g o f Di s k- li k e R o t o r s
10.2.1   Conditions for the Balancing of a Disk-like Rotor Conside r t he rigid disk-like ro tor show n in Fig . 10-2 w h ich is assum ed t o rotate wit h con- stan t an gular velocity ω . U nbalanced m asses ar e depicted as poin t masses m i at radial distances ri . In t his case , t her e a re t hree masses , m 1 , m 2 , and m 3 , but t her e could be any n umber . Each of t he unbalanced m asses in Fig . 10-2 produces a cen trifugal for ce Fi ( i = 1 , 2 , 3 ) acting radially out - wards from t he axis of ro tation wit h a magn it ude equal to m i riω2 ( i = 1 , 2 , 3 ) . All cen t rifugal forces Fi in t his disk-like rot or a re plana r and concurren t . If t he vector sum of t hese forces is zero , t hen t he mass cen t re of t he system coi ncides wit h t he shaft cen t re and t he rotor is bal- anced . Ot her wise , it is called i m ba lance . Since t he im- balance can be show n statically , such im balance is called st atic i m ba la nce . I n t his case , a four t h m ass m C with ro tati ng radius of rC is added to t he system so t hat t he vec- torial sum of t he four cen trifugal forces is zero and balance is achieved . T he added mass is called a cou n ter wei gh t . Such balance is called st atic bal ance . T he refor e , t he condition for t he balancing of a disk-like ro tor is F C + ∑ Fi = 0 w her e F C is t he cen t rifugal force produced by t he coun te rweight m C . or m C rC ω2 + ∑ m i riω2 = 0
2

Fig . 10-2

( 10 -1 ) ( 10-2) ( 10-3)

T he quan tit y ω can be elim inated from Eq . (10-2 ) , yielding t he following relations hip : m C rC + ∑ m i ri = 0 w her e m i ri is called t he m ass-rad i us p rod uct , w hich has a simila r char acte ristic t o t hat of t he cen t rifugal force Fi . N ote t hat m i ri is a vector , not a scalar quan tit y . Therefore , t he condi- tion for t he balancing of disk-like rotor can be desc ribed as: t he vect or sum of all ine rtia forces ・ 1 97 ・

or t he vect or sum of all m ass-radius products must be zero . T he st atic balance is equivalen t to bringing t he syste m cen t re of gravit y t o t he axis of ro tation . 10.2.2   Calculation for the Balancing of a Disk-like Rotor Eq .(10-3 ) lows: m C r C cos θ C + ∑ m i r i cos θ i = 0 m C r C sin θ C + ∑ m i ri sin θ i = 0 ( 10-4) y by resolving it in t o x and y com ponen ts , as fol-

w her e θ i represents t he location angle of t he m ass i . Solving the tw o equations simultaneously for ( m C rC ) and θ C , we have m C rC = ∑ m i ri cos θ i θ C = a rctan
2 i + ∑ m iri sin θ 2

( 10-5) ( 10-6)

- ∑ m i ri sin θ i - ∑ m i ri cos θ i

    No te t hat t he proper quadran t for t he angle θ C must be dete rmined by t he signs of bot h t he num erat or and denominat or of t he arctan function in Eq . (10-6 ) . Because only t he value of t he product is required , eit he r m C or r C can be selected arbitr arily . U sually , coun ter weigh ts are placed at as large a rotating radius as is practicable to mi nimize the amount of t he added mass . In pr actice , rotors a re oft en balanced by removing mass in t he direction of im balance ( g . drilling a hole as t he open circle m C′show n in Fig . 10-2 ) , rat her t han by adding coun ter- weigh ts to t he periphery 180° from t he i mbalance . Removing m ass is not only easy t o achieve but also dec reases t he m ass and t he mom en t of iner tia so t hat t he iner tia mom en t of force can be decreased w hen t he rot or rotates at an uneven speed . From t he above , we can conclude t hat an y number of m asses in a disk-like ro tor can be balanced by adding a single m ass or re moving a mass at an appropria te position . 10.2.3   Static Balancing Machines Imbalance t ha t arises from t he uneven densit y of mate rials , manufact uring errors , etc ., cannot be predicted purely by compu tation duri ng design . T he refor e, experim en tal procedur es and equipmen t are necessary t o detect t he imbalance in a given rotor even if t he ro tor has been designed t o be balanced . In m any instances , it is mor e econom- ical t o allow an im balance caused by manufact ur e and t hen to balance t he rot or by adding or re moving ma- terial as indica ted by a balancing machine . T he static balancing mach ine show n in Fig . 10- 3 has t wo hard horizon tal pa rallel r ails . The rigid ・ 1 98 ・
Fig . 10-3

ro tor for test is laid on t he r ails . If t he cen tr e of m ass of t he system coincides wit h t he axis of ro tation , t he rotor will not roll regardless of t he an gular position of the ro tor . Si mple experi- m en ts to dete rmine w het he r t he disk is statically unbalanced can be conducted as follows . Roll t he disk gently by hand and pe rmit it t o rock unde r t he action of gravit y un til it com es to r est . T he r est position of t he disk indicates t he lowest angula r location of t he mass cent re , bu t not t he a moun t . So a t rial weigh t ( e. g . a lump of plasticine ) can be at tached to t he hig hest poin t of t he ro tor . Since t he a moun t of unbalance is un know n , t hese corrections m ust be made by t rial and e rror un til t he ro tor does not roll along t he r ails from any initial location . T his will de- termine t he amount and t he location of m ass-radius product required for balance . Then balanc- ing m ay be achieved by drilli ng out m aterial at t he location of t he imbalance or by adding m ass t o t he periphery 180° from th is location according t o t he unbalanced m ass-radius product .

Fig . 10-4

Since corr ection by t he static balancing machine in Fig . 10-3 m ust be m ade by trial and er- ror , it will t ake much ti me t o balance a rot or . The static balancing m achine show n in Fig . 10- 4 can in dica te bot h t he magn it ude and t he location of imbalance at t he sa me tim e . There is a spherical pair bet ween t he fra m e and t he plat form . T he m achine is essen tially a pendulu m w hich can tilt in any direction . W hen an unbalanced rot or is moun ted on t he platform of t he m achine , t he pendulum tilts . T he dir ection of t he tilt gives t he angular position of t he im bal- ance w hile t he tilt angle θ indicates t he amoun t of t he imbalance .

10.3   B a l a n c i n g o f N o n-d i s k R i g i d R o t o r s
10.3.1   Conditions for the Balancing of a Non-disk Rigid Rotor All cen t rifugal forces on a disk-like rotor are planar and concurren t . Afte r a disk-like rotor is sta tically balanced on a static balancing m achine , t her e will be no vibr ation w hen it runs at a constan t speed . H oweve r , even if a non-disk rigid ro tor is statically balanced , it may still ex- hibit un wan ted vibration w hen rot ated abou t its axis . For example , in t he non-disk rigid rotor show n in Fig . 10-5 , tw o equal masses m 1 and m 2 lie in a com mon axial plane and at equal dis- ・ 1 99 ・

t ances r 1 and r 2 from t he axis of rotation . T her efore , t he mass cen t re of t he rotor is located at poin t S on t he ax is of rotation and t he rotor will rest on t he rails of a sta tic balancing machi ne in any angular position , as sho wn in Fig . 10-5b . T her efor e it is statically balanced .

Fig . 10-5

Ho wever , if t he non-disk rigid rotor of Fig . 10-5 is placed in bearings and caused t o rotate
2 2 at an angular velocit y ω, t he cen trifugal forces F1 = m 1 r1 ω and F2 = m 2 r2 ω act ou twa rds .

Alt hough t he resultan t of t he two cen t rifugal forces is zero , t he forces are not collinea r and a resultan t couple will exist . T he direction of t he r esultan t couple changes during rotation . T he resultan t couple will act on t he fra me and tend to produce rot ational vibration of t he fram e . Such an imbalance can only be detected by mean s of a dyna mic test in w hich t he rotor is spin- ning . Therefore , t his is r eferred t o as dyna mic im balance . T he rotor in Fig . 10-5 is t her efore sta tically balanced and dyna mically un balanced . From t he above , we can see t hat t he conditions for t he balancing of a non-disk rigid rot or a re : Bo t h t he vect or sum of all ine rtia forces an d t he vector sum of all mom en ts of iner tia forces about any poin t must be zero , 10.3.2   Resolution of Forces
2 In Fig . 10-6 , an eccen t ric mass m creates a cent rifugal force F = m rω . F rom T heoreti-

Fi = 0 and ∑ M i = 0 .

cal M echanics , we know t hat t he cen t rifugal force F can be re- placed dynamically by a pair of forces F A and FB parallel to F and acti ng i n t wo a rbit rarily chosen tr ansverse planes A and B , as sho wn in Fig . 10-6 . T he necessa ry conditions t o replace F by F A and F B dynamically ar e: (1 ) T he resultan t force of F A and F B must be equal t o F . (2) T he resultan t of t he mom en ts of F A and FB abou t any poin t must be equal t o t he mom en t of F abou t t he sa m e poin t . ・ 2 00 ・
Fig . 10-6

T aking poin t C as t he cent re of momen t , we have FA + F B = F FA l A = FB lB Solving t he t wo equations simultaneously will give F A = Fl B/ l and   F B = Fl A/ l planes A and B , r espectively . Therefore ,
2 2 FA = m A rA ω   and   FB = m B rB ω

( 10-7)

T he forces F A and FB can be seen as t he cen trifugal forces created by m A and m B on t he ( 10-8) ( 10-9)

    Substit uting Eq . (10-7 ) in to Eq . ( 10-8) will give m A r A = m rlB/ l     and     m B rB = m rlA/ l From t he above , we can see t hat a mass-radius product m r can be replaced dynam ically by a pair of m ass-radius products ( m A r A and m B rB ) par allel t o t he original one and acting in t wo a rbit rarily chosen transverse planes . 10.3.3   Calculation for the Balancing of a Non-disk Rigid Rotor Sho wn in Fig . 10-7 is a non-disk rigid rot or on w hich t hr ee m asses ( m 1 , m 2 and m 3 )

exist in t hree t ransverse planes . The cen t rifugal forces of t he t hr ee m asses will act in radial di- rections as show n . T he t h ree masses have mass-r adius products F1 , F2 , an d F3 , respective- ly . F1 = m 1 r1 , F2 = m 2 r2 , and F3 = m 3 r3 , . We can i magine t hem t o form a spa tial force system ( w here w e have divided t he forces by ω ) .
2

Fig . 10-7

・ 2 01 ・

According t o Eq . ( 10-9 ) , F1 can be replaced dyna mically by tw o m ass-radius products , F1 A and F1 B , on tw o arbitr arily chosen t ransverse planes A and B , respectively , as sho w n in Fig . 10-7 . F1 A = m 1 r1 ( l - l1 )/ l   and   F1 B = m 1 r1 l1/ l Sim ilarly , F2 and F3 can be replaced dynamically by F2 A and F2B , and F3 A and F3 B , on t he planes A and B , r espectively . F2 A = m 2 r2 ( l - l2 )/ l   and   F2 B = m 2 r2 l2/ l F3 A = m 3 r3 ( l - l3 )/ l   and   F3 B = m 3 r3 l3/ l No w , F1 , F2 , and F3 have been r eplaced dyna mically by F1 A , F2 A , and F3 A on t he plane A , and F1B , F2 B , and F3 B on t he plane B .In t his way , t he complicat ed spatial for ce syste m has been convert ed in to tw o simpler planar concurren t for ce syst ems on tw o planes .In orde r to balance F1 A , F2 A , and F3 A on t he plane A , a coun ter weigh t m A is added .Accordi ng to Eq .( 10-3 ) , t he m agnitude and t he location of t he coun ter weigh t m A must sa tisfy t he following condition . m A r A + m 1 r1 ( l - l 1 )/ l + m 2 r2 ( l - l 2 )/ l + m 3 r3 ( l - l3 )/ l = 0 W it h t he sa me met hod as in Sec . 10 .2 .2 , t he above vect or equation can be solved by di- viding in to x and y components and t hen ( m A rA ) and θ A can be calculated by solvi ng t he t wo equations sim ultaneously . Sim ilarly , in orde r to balance F1 B , F2 B , and F3 B , on t he plane B , a coun ter weigh t m B can be used . T he m agnit ude and t he location of t he counte rw eigh t m B must satisfy following condition . m B rB + m 1 r1 l1/ l + m 2 r2 l2/ l + m 3 r3 l3/ l = 0 T he values of ( m B r B ) and θ B can be calculated by dividing t he above vector equation in to x and y componen ts and t hen solvi ng t he two equations si mult aneously . Af ter choosi ng as large rot ati ng radii rA and rB as is practicable , t he m agnit udes of m A and m B can be calculated . Since t he vector su m of t he four mass-r adius products on the plane A is ze ro an d t hey ar e plana r an d concurren t , bo t h resultant force and resultan t mom en t abou t any poin t ar e ze ro . Similarly , t he vector sum an d resultant mom en t abou t any poin t of t he four m ass-radius products on t he plane B is also ze ro . T he refore , t he non-dis k rigid ro tor becom es dyna mically balanced . T he above met hods can be ex tended to any rot or wit h any number of im balances on any number of tr ansverse planes . The conclusion is t ha t any number of masses on any numbe r of t ransverse planes of a non-disk rigid rotor can be balanced dyna mically by a mini mum of t wo m asses placed in any t wo arbitr arily selected transverse planes . T he selected tr ansverse planes a re called bala nce pl anes . In pr actice , t hose planes on w hich coun ter weigh ts can be mount ed easily , or mass can be re moved easily , may be chosen as t he balance planes . Usually , t hey are near bea ri ng suppor ts in order t o min imize t he bending mom en ts and t he r esulting shaft deflec- tion . ・ 2 02 ・

For t he disk-like ro tor sho wn i n Fig . 10-2 , mom en t equilibrium is inheren t if ∑ Fi = 0 or ∑ m iri = 0 , since t he i ner tia force vectors ar e plana r concurren t . For a non-dis k rigid rot or , howeve r , t he ine rtia forces are not concurren t . T hus , for t he balance of a non-disk rigid ro- tor , bot h r esultan t force and resultant mom en t of for ce must be ze ro . T her efore , static bal- anci ng is sufficien t for disk-like rot ors , w hile non-disk rigid rot ors must also be dynam ically balanced . How ever , if t he unbalance t olerance of a high speed disk-like rot or is very sm all , it is also necessa ry to balance t he disk-li ke ro tor in a dyna mic balancing machi ne , as mentioned in t he nex t section , since t he precision in dyna mic balancin g is higher t han t hat in static balanc- ing . A dyna mically balanced ro tor is also statically balanced , but , in general, t he converse is not t rue . 10.3.4   Dynamic Balancing Machines We have seen t hat any dyna mically unbalanced rot or wit h any numbe r of imbalances on any t ransverse planes is equivalen t to a non-disk rigid rot or consisting of t wo lumped m asses locat ed on t wo balancing planes . T he task of a dynamic balancing machi ne is to locate t he magnitude and t he angular position of t hese tw o equivalen t masses on t he tw o use r-specified balancing planes . According to t he differen t principles and m et hods of m easurin g t he unbalanced forces , a va riety of dyna mic balanci ng m achines has been evolved ove r t he yea rs . A detailed descrip tion of t hem is beyond t he scope of t his book . Most bal- anci ng m achines consist of a fra mew ork t hat car- ries t wo bea ri ngs C and D in t o w hich t he rotor journals are placed , usually wit h t he rot or axis horizont al , as sho w n in Fig . 10-8 . The fra me is at tached t o flexible pedest als t hat permit t he bea r- ing to oscillate in one direction ( usually horizon- ta l) norm al t o t he ro tor axis . T he rotor is
Fig . 10-8

brought up to a desired angula r speed by means of a belt drive , or a universal join t at one end . T he a mplitudes of t he resulting lateral vibra tions of t he bearings at C and D caused by the im- balances are m easur ed wit h highly sensitive elect ronic pickups . Modern balancing m achines perform all of t he required calculations au tom atically after t he use r has en ter ed t he lengt hs be- tween balancing planes ( A and B ) and bea ri ng planes ( C and D ) . After t hat , t he angular positions and t he values of t he unbalanced m ass-radius products on the two balancing planes are show n on r eadout devices . I t is impor tan t to no te he re t hat if sever al w heels are to be moun ted upon a shaft , t he w heels should be balanced individually on a static balancing m achine before moun ti ng . O t her- wise , additional bendi ng momen ts inevitably exist after t he assem bly is balanced dyna mically in ・ 2 03 ・

t wo planes .

10.4   U n b a l a n c i n g A l l o w a n c e o f R o t o r
T he unbalancing effect will be r educed sharply afte r t he rotor is balanced on a balancing m achine . H owever , we can no t expect absolut ely perfect balancing , because t he resolution of any m easure men t instrum en t is limited . Som e im balance will re main in t he ro tor aft er balanc- ing . Alt hough a bet ter balanci ng can be achieved by a bet ter balanci ng m achine wit h higher precision , it i nvolves highe r cost . T he amoun t of residual eccen t ricit y t hat can be tole rated is called tolerable i m bal ance . I t depends upon the speed at w hich a rotor is to oper ate and t he wor king requirem en t . Usually , t he higher t he angular speed ω, t he lower t he allowed eccen- t ricit y e . The product of e and ω is often used t o repr esent t he balance pr ecision of a rot or . T he Comm it tee of t he In ternational Standards O rganization (I SO ) has formulated r ecomm en- dations for t he allowable qualit y grade G = [ e ω] for diffe ren t classes of m achine ry . T ypical recom men ded values of G are listed in T able 10-1 , w he re t he dim ensions of e and ω a re m m and rad/ s , respectively . Table 10-1   Recommended values of G
G/ ( mm/ s ) 4 000 ro tor types   Cr anksh aft assem bly of slow ma rine diesel e ngi nes wit h uneven nu mbe r of cylin- ders . 1 600 630 100 40   Cr anksh aft assem bly of large two-cycle engi nes   Cr anksh aft assem bly of large four-cycle engines   Com plete e ngin es for cars , truck s, locomo tives   Car wh eels , drive sh afts , cr anksh aft assem bly for engines of cars , t rucks , loco- motives 16 6 .3 2 .5 1 0 .4   Individu al engine com pon en ts   Fa ns , flywh eels , m ach ine t ools   Gas and steam t urbines, small elect ric a rmat ures   T ape record er a nd phonogra ph drives , gri nding machine driver   Gyroscopes , pr ecision grinder com pon en ts

    First , choose t he balance precision G from T able 10-1 according t o t he t ype and working condition of machine . T hen calculat e t he allowed eccen t ricit y [ e] from G/ ω . T he product of [ e ] an d t he m ass m of t he ro tor is t he allowed unbalanced mass-radius product . For a disk-like ro tor , t he calculated m [ e] is t he allo wed unbalanced m ass-radius product [ me ] . F or a non- disk rigid rot or , m [ e] should be resolved in t o tw o allowed unbalanced mass-r adius products ・ 2 04 ・

on tw o balancing planes . For exa mple , show n in Fig . 10-9 is a m achine tool spindle . T he w ork ing rotational speed is n = 1 000 r/ mi n , the m ass m = 10 kg , lA = 100 mm and lB = 200 mm . The balance pr ecision is G = 6 .3 mm/ s from T able 10-1 . T her efore , angular speed ω = 1 000 × 2 × π / 60 rad/ s≈ 100 rad/ s . T he allowed eccen t ricit y [ e ] = G/ ω = ( 6 .3/ 100 ) mm = 0 .063 mm . T he to- tal allow ed re maining unbalanced mass-radius product [ m e ]
Fig . 10-9

10 × 0 .063 kg・m m = 0 .63 kg・mm . T he allo wed unbalanced mass-r adius products on t he t wo balancing planes A and B can be calcula ted by Eq . ( 10-9) as follow : [ me]A = [ me] l B/ l = 0. 42 kg・mm and [ me] B = [ me] l A/ l = 0 .21 kg・ mm . Problems and Exercises 10-1   W hat is static balancing ? W hat is t he condition for sta tic balancing ? What is t he least number of coun te rweights to be used to balance a disk-like rot or if t here are five concent rat ed m asses on t he ro tor ? 10-2   What is dyna mic balancing ? W hat ar e t he conditions for dyna mic balancing ? W hat is t he least numbe r of cou nte rweigh ts to balance a non-disk rigid ro tor if t here ar e five imbalances on five differen t transverse planes ? What is a balancing plane ? Can t he balancing planes be chosen a rbit rarily ? W hat kind of plane can be chosen as balancing plane in pr actice ? 10-3   Is static balancing sufficien t for a disk-li ke rotor ? Is static balancin g sufficien t for a non- disk rigid rotor ? Is it cer tai n t hat a ro tor is dyna mically balanced if it has been st atically bal- anced ? Is it ce rtain t hat a rot or is statically balanced if it has been dynam ically balanced ? 10-4   T he disk-like rot or in Fig . 10-2 has unbalanced m asses m 1 = 10 kg , m 2 = 21. 4 kg and m 3 = 36 kg at rotating radii r 1 = 100 mm , r2 = 112 m m and r3 = 110 m m , and angula r orien ta- tions θ 1 = 30° , θ 2 = 130° and θ 3 = 270° . Determine the m agnitude m C and t he location angle θ C of t he coun ter weig ht if it is t o be placed at a rot atin g r adius r C of 115 mm . 10- 5   S how n in Fig . 10-10 is a disk-li ke rot or on w hich t her e are four imbalances . The masses , rot ati ng radii , and angula r orien tations a re : m 1 = 5 kg , m 2 = 10 kg , m 3 = 8 kg and m 4 = 7 kg , r1 = 10 mm , r 2 = 10 mm , r3 = 15 m m and r4 = 20 mm . T he rot or is to be balanced by remov- ing a mass m C at a ro tating radius of 25 m m . Find t he m agnit ude m C and its location angle θ C . 10- 6   The m asses , rotating radii , and angula r locations of t he u nbal- anced m asses ( m 1 , m 2 , and m 3 ) show n i n Fig . 10-7 ar e: m 1 = 2 kg , m 2 = 3 kg an d m 3 = 1 k g , r 1 = 20 mm , r2 = 15 m m and r 3 = 25 mm , and θ 1 = 240° , θ 2 = 150° and θ 3 = 330° . T he distances between planes are : l 1 = 200 m m , l2 = 350m m , l3 = 550 m m , and l = 900 mm . T he system is t o be balanced dynam ically by adding a m ass m A on t he balancing plane A at a ro tati ng radius of 50 mm and a mass m B on t he balancing plane B at a rot atin g radius of ・ 2 05 ・
Fig . 10-10

45 mm . Determine t he magn it udes ( m A and m B ) and angular location s ( θ A and θ B ) of t he required coun ter weigh ts . 10-7   An unbalanced m ass m is moun ted out board of bal- anci ng planes A and B , as show n in Fig . 10-11 . m = 2 k g , r = 10 mm . T he rotor is to be balanced dynam ically by adding tw o coun ter weigh ts m A and m B on t he balancing planes A and B , respectively , at a ro tating r adius of 50 mm . l 1 = 100 m m , l2 = 200 mm . Dete rmine t he a moun ts ( m A and m B ) and angular locations of t he t wo coun ter weights . 10-8   Th ree unbalanced masses m 1 , m 2 r 2 = 6 kg ・m m , m 2 and m 3 exist on t hr ee t ransve rse planes 1 , 2 and 3 , m 1 r 1 = 2 kg・mm , respectively, as s ho w n in Fig . 10-12 . T heir m ass-radius products are :
Fig . 10-11

m 3 r3 = 3 kg・mm , respectively . T he locations are as show n . Suppose t he

system is to be balanced fully by t wo balancing mass-r adius products , Fb 1 and F b 3 , on t he planes 1 and 3 , respectively . Dete rmine t he a moun ts and angula r locations of t he tw o balanc- ing m ass-radius products .

Fig . 10-12

10-9   T wo unbalanced masses m 1 and m 2 ex ist on t wo t ransverse planes 1 and 2 of a non-disk rigid rotor , as sho wn in Fig . 10-13 . U nbalanced mass m 1 is moun ted out board of balancing planes A and B . m 1 = 2 kg , m 2 = 5 kg , r1 = 100 mm , r2 = 50 mm , l 1 = 200 mm , l2 = 180 mm , l 3 = 300 mm . S uppose t he system is t o be balanced dyna mically by sub t racting masses m A and m B on t he balancing planes A and B , respectively , at a rota ting radius of 150 mm . Determ ine t he a moun ts ( m A and m B ) and angula r locations (θ A and θ B ) of t he m asses to be removed .

Fig . 10-13

10-10   O n t he non-disk rigid rotor show n in Fig . 10-14 , t he re exist four u nbalanced m asses . ・ 2 06 ・

T heir masses , rotating radii, and angula r locations ar e : m 1 = 10 kg , m 2 = 15 kg ,

m 3 = 20 kg

and m 4 = 10 kg , r 1 = 40 mm , r2 = 30 mm , r3 = 20 mm and r4 = 30 mm and θ 1 = 120° , θ 2 = 240° , θ . l 1 2 = l2 3 = l 3 4 . T he system is to be balanced dyna mically by 3 = 300°and θ 4 = 30° addi ng a m ass m A on t he balancing plane A at a rotating r adius rA of 50 mm and removing a m ass m B on t he balanci ng plane B at a ro tating radius r B of 60 m m . Determ ine t he m agnitudes ( m A and m B ) and angular locations ( θ A and θ B ) of t he r equir ed m asses .

Fig . 10-14

・ 2 07 ・

C h a p t e r 11 M o t io n o f Mec h a n ical S y s t em s a n d I t s R e g u la t i o n
11.1   In t r o d u c t i o n
In t he kinem atic analysis of a m echanism , t he motion of t he input li nk should be given . In most cases , t he input link is supposed t o run a t a constan t speed . T his is only an approxim a- tion t o realit y . T he act ual mo tion of t he inpu t lin k depends on t he m ass dist ribution of t he m echanism and t he ex te rnal forces acting on t he m echanism . In t his chap ter , w e will st udy t he act ual motion of t he m echanism accordi ng t o t he m ass dist ribu tion of t he m echanism and t he exte rnal forces acting on it . T he w hole ope rating process of a machine can be divided in to th ree phases : t he star ting phase, t he steady working phase, and t he stopping phase . In t he star ti ng phase , t he work done by t he drivi ng forces ( inpu t w or k ) is greater t han t he sum of t he w ork done by resis- tances ( ou t pu t work ) and t he frictional loss . Therefore , t he speed of t he machi ne inc reases . In t he st opping phase , t he sit ua tion is reve rsed and t he speed decr eases . In a w orki ng cycle of t he steady w ork ing p hase , t he i nput work is equal t o t he su m of t he ou t put work and t he fric- tional loss . Bu t t his m ay not be t rue in ot her int ervals and t he speed of t he machi ne is no t nec- essa rily constant . There may exist som e speed fluct uation w hich is not desirable and should be limited t o below som e level to ensur e t he qualit y of perform ance . T hat is t he tas k of t he r egula- tion of t he motion .

11.2   M o t i o n E q u a t i o n o f a M e c h a n i c a l S y s t em
11.2.1   General Expression of the Equation of Motion As we have seen in previous chap ters , most m echanisms in use have only one degree of freedom . I n ot he r words , one independen t pa ra mete r or generalized coordi nate is enough to desc ribe t he kinem atic state of such m achines . In t his case , t he principle of work and energy is useful in deriving t he equation of motion . Suppose t ha t t he kinetic energy of t he system in- cr eases by d E in an in te rval d t and t he elem en tary work done by all t he ex te rnal forces acting on t he system in t he sa me in terval is d w . According to t he principle of work and energy , we ・ 2 08 ・

have d E = d w = Nd t w her e N is t he sum of t he i nstan taneous power of all t he ex te rnal forces . For a planar mechanism consisting of n movi ng links , t he kinetic energy of t he syste m can be expressed as
n n

( 11-1)

E =

∑E
i= 1

i

=


i= 1

1 1 2 2 m i vC i + JC i ωi 2 2

( 11 -2 )

w her e E i is t he k inetic energy of lin k i , m i is t he m ass of lin k i , Ci is t he m ass cent re of lin k i , JC i is t he momen t of ine rtia of link i about its m ass cen tr e Ci , cen t re Ci , and ωi is the angula r velocit y of lin k i . If t her e a re k non-conservative forces deno ted as F j ( j = 1 , … , k ) acting on the mechan ism and t he sum of mom en ts of forces acting on link i is M i , t hen t he sum of instan taneous po wers N can be w rit ten as
k n

v C i is t he velocit y of m ass

N =

∑ F v cos α + ∑ M ω
j j j i i
j= 1 i =1

( 11 -3 )

w her e v j is t he velocity of t he poin t at w hich t he force F j acts and α j is t he angle bet ween F j and v j . If t he angle α j is acu te , t hen cos α j is positive and it m eans t hat t he force F j is a driv- ing force . If t he angle α j is obt use , t hen cos α j is negative and it m eans t hat t he force F j is a resistance . Bot h M i an d ωi are defined positive if t hey a re coun ter-clock wise an d negative if clock wise . If t he directions of M i and ωi are t he sa m e, it m eans t he M i is a driving mom en t and t he po wer M iωi is positive . If t he directions of M i and ωi a re opposite , it m eans t he M i is a resist ance mom en t an d t he power M i ω i is nega tive . Inse rting Eqs . ( 11-2) and (11-3 ) in to Eq . ( 11-1) , w e get
n

d


i =1

1 1 2 2 m i vC i + JC iωi 2 2

k

n

=

∑ F v cos α + ∑ M ω
j j j i i
j= 1 i =1

dt

( 11 -4 )

    T his is t he general equation of motion of a planar m echanism wit h t he degree of freedom F = 1 in diffe ren tial form . 11.2.2   Dynamically Equivalent Model of a Mechanical System In Eq . ( 11-4 ) , t he re are still m any unkno wn velocities . For a mechanical system wit h one degr ee of freedom , suppose link 1 ro tates about a fixed pivot and its angle of rotation φ 1 is chosen as t he generalized coordinate of t he syste m , t hen Eq . (11-4 ) can be rew ritten as . 1 d 2∑ i =1
n

mi

vC i ω1

2

+ JC i

ωi ω 1

2

k

ω

2 1

=


j =1

vj Fj cos α j + ω1

n


i= 1

Mi

ω i ω1

ω 1 d t ( 11 -5 )

    T his can fur t he r be simplified as d 1 Je ω2 1 = Me ω1 d t 2 ( 11 -6 )

w her e Je is called t he equivalent momen t of iner tia and M e t he equivalent mom en t of force wit h ・ 2 09 ・

n

Je = Me =


i= 1 k j= 1

mi

vC i ω 1

2

+ JC i
n

ω i ω1

2

( 11 -7 ) ( 11 -8 )



vj ω i Fj cos α j + ∑ Mi ω1 ω 1 i= 1

    If t he di men sions of t he m echanism are given , t he speed ratios v C i/ ω 1 , ω i/ ω 1 and v j/ ω 1 a re functions of φ 1 on ly . Bu t F j and M j t he mselves may be functions of v j , ω i , and t as well . We have , t herefore , Je = Je ( φ 1) M e = M e (φ 1 ,ω 1 , t) ( 11 -9 ) ( 11 -10 )

In t his w ay , a m echanical syste m with one degree of freedom can be assu med as an im aginary lin k rotating abou t a fixed pivot , as show n in Fig . 11-1 . I ts m ass cent re is located at t he fixed pivot and its mom en t of iner tia abou t its m ass cent re is Je ( φ 1 ) . T he moment of force M e ( φ 1 , ω 1, t ) acts on it and m akes it rotat e wit h an angular velocity ω 1 . The mo tion equation of link 1 in Fig . 11-1 is t he sa me as Eq . (11-6 ) . ω 1 calculated according to Eq . ( 11-6) for li nk 1 in Fig . 11-1 is t he sam e as t ha t calculated accordin g t o Eq . ( 11-4)
Fig . 11-1

for t he w hole m echanism . H oweve r , t he calculation of Eq . (11-6 ) is sim- pler t han t hat of Eq . (11-4 ) . S uch an i maginary li nk is called t he equivalen t

link , or t he dyna mically equivalen t model , of a m echanical system . Sim ilarly , if we choose a t ranslating m ember as an equivalen t link and take its displace- m en t s1 as t he generalized coordinate , t hen we can define t he equivalen t mass m e ( s1 ) and t he equivalen t force F e ( s1 , v 1 , t ) as follo ws .
n

m e ( s1 ) =


i =1 k j= 1

mi

vC i v1

2

+ JC i

ω i v1
n

2

( 11 -11 ) ( 11 -12 )

F e ( s1 , v 1 , t ) =

∑F

j

vj ωi cos α j + ∑ Mi v1 v1 i= 1

w her e v1 = d s1/ d t . The correspondi ng equation of motion is then d 1 2 m e v1 2 = Fe v 1 d t ( 11 -13 ) y , t he equivalent

T his is ano t her k ind of equivalen t link , as show n in Fig . li nk should have t he sam e dynam ical response as t he chosen link i n t he act ual mechanis m . T his is ensured by t he e- quivalen t moment of iner tia Je ( or equivalen t mass m e ) , and t he equivalen t moment of force M e ( or equivalen t force Fe ) , as t he k inetic energy of t he equivalen t lin k is e-

Fig . 11-2

qual to t he sum of t he kinetic ene rgies of all links in t he original m echanism and t he powe r M e ω 1 ( or Fe v 1 ) is equal to t he sum of pow ers of all t he ex ternal forces and ex ternal momen ts of force acting on t he origi nal mechanism . T hese a re t he principles used in calculati ng Je ( ・ 2 10 ・ me )

and M e (

Fe ) . In addition , w hen we choose one lin k as t he equivalen t lin k , t his m eans t hat

we take its displacem en t ( angular or linear ) as t he displacem en t of t he equivalent lin k or t he generalized coordi nate of t he syst em . The dynamically equivalen t model of a m echanical syste m as given above simplifies gr eatly the dynamical problem of t he mechanical system . Example 11-1 In t he mechanism show n in Fig . 11-3 , cr ank O2 A is fixed with gea r 2 . The numbers of t eet h of gear 1 and gea r 2 a re z 1 and z2 , r espectively . T he m ass cen tr es of gea r 1 and gear 2 (including cr ank O 2 A ) ar e located at O 1 and O 2 , respectively . T he moment of i ner tia of t he gear 1 abou t O1 is J1 . The momen t of ine rtia of t he gear 2 ( including cr an k O2 A ) abou t O 2 is J2 . T he lengt h of t he cr an k O 2 A is l . T he m ass of lin k 3 is m 3 and t he m ass of lin k 4 is m 4 . T he driving mom en t M 1 acts on t he gear 1 an d t he resistance F 4 acts on t he lin k 4 , as show n in t he figur e . T ake t he gear 2 as t he equivalent lin k . Calculat e the equivalen t mom en t of iner tia Je and t he equivalen t mom en t of for ce M e . Solution: (1 ) Ki nem atic analysis of t he mechanism ω1 = z2 ω 2 z1

v A 3 = lω 2 v 4 = lω2 sin φ 2 (2 ) Calcula tion of Je T he links 3 and 4 are tr ansla ting . T hey have no rotating kinetic energies . T he links 1 and 2
Fig . 11-3

have no t ranslating kinetic ene rgies . Since t he kinetic energy of the equivalen t link 2 s hould be equal t o t he sum of t he kinetic energies of all links in t he original mechan ism , we have 1 2 1 2 1 2 1 2 1 2 ω 2 Je = ω1 J1 + ω2 J2 + v A 3 m 3 + v 4 m 4 2 2 2 2 2 ω 1                 Je = ω 2 z2 = z1 (3 ) Calcula tion of M e
2

J1 + J2 +
2 2

vA 3 ω2

2

v4 m3 + ω 2
2 2

2

m4

J1 + J2 + l m 3 + l m 4 sin φ2

M 1 is a driving mom en t of force and its work is positive . F4 is a resistance and its work is negative . Since t he po wer M e should be equal t o the sum of t he powers of all t he ex ternal forces and ext ernal mom en ts of for ce acting on t he original m echanism , we have Meω 2 = M1 ω 1 - F4 v 4 Me = M1 z2 - F 4 l si n φ 2 z1 ・ 2 11 ・

We can see here t hat t her e are two par ts i n t he equivalen t mom en t of ine rtia Je . O ne is t he constan t ( first t hree term s i n the example ) and anot he r is variable ( wit h respect t o φ 2 ) . If t he t ran smission ratio of one link t o t he equivalen t link is constan t , its con t ribu tion t o t he equivalen t mom en t of i ner tia is also const an t . Ot herwise, it is va riable . T he equivalen t mo- m en t of force can also be divided in to tw o par ts . O ne is the driving momen t (positive ) and an- o t her is t he r esist an t mom en t ( negative ) , i .e . M e = Me d - M e r ( 11 -14 ) For more complicated mechanisms , Je can be calculated num erically . If t he equivalen t lin k runs at an a rbit rary speed , we calcula te t he corr esponding velocities and put t he m i nt o t he for- mulae (11-7 ) and get t he num erical results of Je for t he given position . M e can be calculat ed simila rly if it is independen t of ω1 and t . o t her wise Me should be determined step by step dur- ing t he solu tion of t he equa tion of motion . 11.2.3   Other Forms of Motion Equation Eqs . ( 11-6) and (11-13) are t he equations of motion in differ en tial form . Ot her forms of motion equation a re available for diffe ren t purposes . F or convenience , we delete t he subsc ripts of all t he va riables and consider only t he rota ting equivalen t link . Suppose φ = φ0 , ω = ω 0 and J = J0 w hen t = t0 . Int egrating Eq. ( 11-6) , we get 1 1 2 2 J ( φ) ω ( φ) J0 ω0 = 2 2

∫ M ( φ, ω, t ) dφ
φ
0

φ

( 11 -15 )

    T he left hand side of t he equation is t he incr em en t of k inetic energy and t he righ t hand side is t he work done by all t he forces in t he sam e in terval . Eq . (11-15) is t her efor e called t he en- ergy form of motion equation . By differen tiation , Eq . ( 11-6) can be rew rit ten as d ω( φ) ω ( φ) d J ( φ) J ( φ) + = M ( φ, ω, t ) dt 2 dφ T his is t he equation of mo tion in t he form of mom en t of force .
2

( 11 -16 )

11.3   S o l u t i o n o f t h e M o t i o n E q u a t i o n o f a M e c h a ni c a l Sys t em
    Gener ally , t he equivalen t momen t of i ner tia J is a function of position and t he equivalen t momen t of force M is a function of position φ, velocity ω and tim e t , as described above . In cert ain situations , t hese functions can be w rit ten in differen t forms so t hat t he solu tion process can be simplified . We discuss a few such cases below . 11.3.1   Solution for Constant M and J T his is t he simplest case and is suitable for mechan ical syste ms wit h constan t t ransm ission ratios and constan t forces . F rom Eq . ( 11-16 ) , we have ・ 2 12 ・

M= J

dω = J α dt M J

( 11 -17 )

w her e α is t he angular acceler ation . F ur t he rmore , w e get α=

ω = ω0 + α t φ= φ 0 + ω 0 t + 1 2 α t 2

( 11 -18 )

w her e φ 0 , ω 0 a re initial an gular position and initial angula r velocit y , respectively . Eliminating t in Eq . ( 11-18 ) , we ob tain ω - ω 0 = 2α( φ - φ 0)
2 2

( 11 -19 )

Example 11-2 T he spindle of a m achine ro tates at t he speed n = 700 r/ min . T he equivalen t mom en t of ine rtia is J = 0 .05 kgm , assuming t he spindle is taken as t he equivalen t link . If t he machine is t o be stopped wit hin t hree t urns of t he spindle , find t he mini mal braking mom en t of force on t he spindle and t he corresponding time t t o stop . Solution: ω 0 = π n π× 700 - 1 = s = 73 .3 s 30 30 ω= 0 φ- φ 0 ≤6 π From Eq . ( 11-19 ) , we have 73 .3 α ≤ s - 2 = - 142 .5 s 2× 6 π From Eq . ( 11-18 ) , we get M= J α= 0 .05 × ( - 142 .5 ) N m = - 7. 13 N m t= ω- ω 0 0 - 73 .3 = s = 0. 51 s α - 142 .5
2 2 1 2

Since t his is in t he st opping phase , t he direction of t he braking momen t M should be op- posite to n , as t he minus sign implies . 11.3.2   Solution for M = M ( ω ) J

T he driving moment of an elect ric mot or is a function of speed . Resist ances of come kinds of mach ines , such as cen trifugal pu mps and blowers , ar e also functions of speed . If t hey are driven by elect ric mot ors wit h constan t tr ansmission ratios , t hey onstit u te a system wit h M = M ( ω) J. ・ 2 13 ・

Since J is constant , Eq . (11-17) is still valid . By separ ati ng va riables , we get dt dω = J M ( ω) or d φ ωd ω = J M ( ω) In tegration of t hese equations r esults in t = t0 + J and φ= φ 0 + J Example 11-3 A mechanical syste m has an equivalen t momen t of force M = M 0 - kω and a constan t equivalen t mom en t of inertia J .Find t he motion of t he equivalen t link ω = ω( t ) Solution: According to Eq .( 11-20 ) , t = t0 + J



ω

ω
0

dω M ( ω) ωd ω M ( ω)

( 11 -20 )



ω

ω
0

( 11 -21 )



ω

ω
0

dω J = t0 ln ( M 0 - kω) M 0 - kω k
M0 - ω0 e k k ( t - t ) J 0

ω ω
0

    Af ter som e tr ansform ations , w e get M0 ω( t ) = k

(11-22)

11.3.3   Solution for M = M ( φ) and J = J( φ) T he driving mom en t of an in ternal combustion engine is a function of t he position of t he cr anks haft . Resistances of som e m achines , such as shaping m achines and punching machines , a re also functions of position . T hese r esult in t he position-varian t equivalent moment of force M ( φ) .J = J ( φ) is t he general case if t he tr ansmission r atio of a given m achine is not con stan t . From Eq . ( 11-15 ) , we have 1 1 2 2 J ( φ) ω ( φ) = J0 ω 0 + 2 2     Fur t her more , ω( φ) can be solved as ω( φ) = J0 2 2 ω 0 + J ( φ) J ( φ)
φ



φ

φ
0

M ( φ) d φ



φ

φ
0

M ( φ) d φ

( 11 -23 )

    Since ω( φ) = dφ / d t , by in tegr ation we ob tai n t = t0 +

∫ ω( φ)
φ
0



( 11 -24 )

・ 2 14 ・

    Eqs .(11-23) and (11-24) toget her provide a par am etric form of t he function ω = ω( t ) , wit h φ as a para m eter . 11.3.4   Solution for M = M ( φ, ω ) and J = J( φ)     A shaping m achine driven by an elect ric mo tor is an exa mple of t his model . Generally , t he equation of motion i n t his case can not be solved analy tically . We provide her e a m ethod of nu- m erical solu tion . Eq . ( 11-6) can be rew rit ten as 1 2 ω d J ( φ) + J ( φ) ωd ω = M ( φ, ω) d φ 2 (11-25)

Dividing t he angle φ in to small segm en ts , we get disc rete poin ts of φ as φi ( i = 0 , 1 , 2 , … ) wit h φi + 1 - φ i = Δφ . T he diffe rences of J and ω a re t hen Δ J = Ji + 1 - J i and Δω i = ωi + 1 - ωi . Turning t he differen tial Eq . (11-25) in t o a difference equation , we have 1 2 ωi ( Ji + 1 - Ji ) + Jiωi ( ωi + 1 - ωi ) = M ( φ i ,ω i )Δφ 2 Solving for ωi + 1 , we get ωi + 1 = M (φ i ,ω i )Δφ ( 3 J i - Ji + 1 ) ωi + Jiωi 2 Ji ( 11 -26 )

From t he initial condition ω 0 , t he velocit y ω can be calcula ted , poin t by poin t , wit h Eq . (11-26) .

11.4   P e r i o d i c S p e e d F lu c t u a t i o n a n d i t s R e g ul a t i o n
11.4.1   Condition for a Periodic Steady Working State Constan t speed wit hout any fluct uation is nat urally an ideal state in t he steady working phase . I n t his case , from Eq . ( 11-6) , we have 1 2 dJ ω = M 2 dφ Practically , t his is i mpossible if M ≠ 0 . T her efore , t he conditions for t he con stant speed wor king stat e must be dJ =0 dφ M = Md - M r = 0 If t he above conditions can no t be m et at all tim es , t he steady w or king state is at most pe- riodic . T he periodic steady working sta te m eans ω( φ) = ω( φ + φT ) w her e φT is t he period of t he syste m , w hich is t he lo west common multiple of periods of J and M , i .e. J ( φ) = J ( φ + φT ) ・ 2 15 ・

M ( φ) = M ( φ + φT ) From Eq .(11-15) , we have


from w hich

φ+ φ φ

T

M dφ =

1 1 2 2 J ( φ + φT ) ω ( φ + φT ) J ( φ) ω ( φ) = 0 2 2



φ+ φ φ

T

M d dφ =



φ+ φ φ

T

M r dφ

( 11 -27 )

    T his m eans t hat t he work done by driving forces is equal t o t he work done by resistan t forces wit hin a period . T his is t he condition for a pe riodic steady w ork ing state . 11.4.2   Coefficient of Speed Fluctuation T he speed fluctuation of a m achine may cause an ex t ra dyna mic load and vibration of t he system and t herefore should be con t rolled wit hin som e limits to en sur e good working quality . T he coefficien t of speed fluct uation is t hen in troduced and defined as δ= ωm a x - ωm i n ωm ( 11 -28 )

w her e ωm a x and ωmi n are t he m axi mum and minimum angula r speed . ωm is t he ave rage angular speed and is approxi mated as ωm = ωm a x + ωm i n 2 ( 11 -29 )

Multiplying t he above t wo equations will give
2 2 2 ω ω max - ω mi n = 2δ m

Recom mended allowable coefficients of speed fluct uation for som e kinds of m achines are listed in T able 11-1 . Table 11-1   Allowable Coefficients of Speed Fluctuation [ δ]
Mac hines Rock c rush er P un ching mac hine , Shear ing machin e Rolli ng mill A uto mobile , t ractor Mach ine t ools [δ] 1/ 5 ~1/ 20 1/ 7 ~1/ 10 1/ 10 ~ 1/ 25 1/ 20 ~ 1/ 60 1/ 30 ~ 1/ 40 Mac hines Pu mp , blower Pap er mac hine , loom Spinning mach ine DC gen erat or AC ge nera tor [δ] 1/ 30~ 1/ 50 1/ 40~ 1/ 50 1/ 60~ 1/ 100 1/ 100~ 1/ 200 1/ 200~ 1/ 300

    From Eq . ( 11-15 ) , we have
φ

1 J ω2 = 2


φ

M dφ +

0

1 J0 ω2 0 2

    T he w or k W is t he i ntegr al of t he moment of force, w hich can be defined as ・ 2 16 ・

W =


φ

φ

M dφ =

∫( M
φ
0

φ

d

- M r ) dφ

( 11 -30 )

0

    If J is constan t , t hen ω r eaches ex t re me values w hen W reaches ex tr em e values . We have 1 1 2 Jω max = W max + J ω2 0 2 2 1 2 1 2 Jωmi n = W mi n + J ω0 2 2     T he diffe rence between t he above tw o equations gives 1 2 2 J(ω max - ω mi n ) = W m a x - W mi n 2 Pu tting Eq . (11-29) in to t he above equation will give
2 J δ ωm = W m a x - W mi n

Defi ning Δ W m a x = W m a x - W m i n as t he m axi mum incre ment of w ork , we get δ= Δ W max 2 ωm J ( 11 -31 )

11.4.3   Calculation of moment of inertia of a Flywheel T he regulation of pe riodic speed fluct uation means limitation of t he coefficien t of speed fluct uation so t hat δ ≤ [δ] ( 11 -32 ) As we can see from Eq . ( 11-31) , if we increase J , δ will be dec reased . T herefore , if δ> [δ] , we can add a fly w heel wit h enough mom en t of i ner tia t o r educe t he value of δ so t hat δ ≤ [δ] . T he equivalen t mom en t of inertia of t he flyw heel J F can be calulated by JF = Δ W m ax - JC 2 ωm [δ] ( 11 -33 )

w her e JC is t he constan t pa rt of t he equivalent mom en t of i ner tia of t he original system wit hou t t he flyw heel . T he va riable par t of t he equivalen t mom en t of inertia is neglect ed ( if no t zero ) t o simplify t he proble m and t his approxi mation leads t o a more reliable result . If JC n J F , even JC can be neglected . Essen tially , t he fly w heel is an energy storage device . In som e par t of a period φT , if W d > W r , or t he wor k W is postive and t he kinetic energy is inc reasing , t he fly w heel absorbs ki- netic energy wit h inc rease of speed . H owever , such an increase in speed is less compared t o t he case wit hou t t he fly w heel . On t he ot her hand , w hen t he w or k W is negative and t he kinetic energy is dec reasing , t he fly w heel r eleases kinetic energy and t he speed decreases less . In t his way , t he speed fluct uation is r educed . Ho wever , t he speed fluctuation can not be eli minat ed completely wit h t he help of t he fly w heel only , as we can see from Eq . (11-31 ) . Besides , if possible , t he fly w heel is bette r moun ted on a shaft wit h higher speed . The higher t he speed of t he fly w heel , t he grea ter t he con t ribu tion of t he flyw heel t o t he equivalent moment of iner tia . ・ 2 17 ・

Example 11-4 T he equivalent resistant mom en t of force of a m achine is as sho w n i n Fig . 11-4 wit h t he period φT = 2π . T he equivalen t driving momen t of for ce M d is con stant . The average speed of t he equivalen t link is n m = 1 000 r/ min . T he equivalen t momen t of inertia of all t he li nks ex- cep t t he flyw heel is negligible . The allowable coefficien t of speed fluct uation is [δ] = 0 .05 . Fin d t he mini mum momen t of ine rtia JF of t he flyw heel on t he equivalen t link .

Fig . 11-4

Solution: (1 ) Fin d M d From Eq . ( 11-27 ) , we get 2π M d =


0



M r d φ = 0 .25 π× 60 N m + 0 .75 π× 10 N m + 0. 5π× ( 10 + 60 ) N m M d = 28. 75 N m

    ( positions w here ωm a x and ωm i n appear According to calculus and Eq . ( 11-17 ) , ωm a x and ω m i n can appear only w her e M = 0 or M is discon tinuous or at t he ending poin ts of t he in terval . Therefore w e need t o calculate t he work W defined by Eq . ( 11-30 ) only at positions A , B , C and D as in Fig . 11-4 . Af ter some simple calculations , we have t he following results:
position φ / rad W/ N m A 0 0 B 0 .25π - 24 .54 C 1 .375π 30 .68 D 2π 0

Hence , ωm a x appears at φ = 1 .375π and ωm i n appears at φ = 0 .25 π. (3 ) Calcula te t he maximum i ncre men t of w ork Δ W m a x = W m a x - W mi n = 30. 68 N m - ( - 24 .54 ) N m = 55 .22 N m (4 ) Calcula te t he minimum mom en t of iner tia of t he fly w heel Δ W max 55 .22 2 2 JF ≥ 2 = kgm = 0 .1 kgm 2 ωm [δ] 1 000 × 2 π × 0 .05 60 ・ 2 18 ・

11.5   In t r o d u c t i o n t o A p e r i o d i c s p e e d F lu c t u a t i o n a n d i t s R e gu l a t i o n
    D uri ng t he working process of a m echanical syste m , t he resistance may change irregularly or randomly . T his results in t he so-called aperiodic speed fluct uation . Suppose t hat t he nomi- nal equivalen t driving mom en t M d and resistan t moment M r of a m echanical system ar e as show n in Fig . 11-5 . T he curves of M d an d M r in tersect at t he working poin t s wit h working Md Mr < as show n i n Fig . 11-5a . If t he speed is dec reased to ω ω ωa for som e r eason , as M d > M r in th is region , t he speed will ret urn t o ωs . If t he speed is in- speed ωs . Con sider t he case cr eased to ωb for som e ot he r reason , as M d < M r now , t he speed will come dow n to wards ωs . Md Mr > ω ω as show n in Fig . 11-5b , t he speed will no t come back to ward ωs if there is a dist urbance . T he T he ope ration of t he machine is st able in t his case . In t he opposit e sit uation w he re system is u nstable in t his case .

Fig . 11-5

if t he syst em is unstable, a speed regulat or or governor is needed to r egulate t he input en- ergy and make t he syste m stable . O ne kind of cen- trifugal governor is sketched in Fig . 11-6 . The wor king m achine 2 is connected t o an i nte rnal-com- bustion engine 1 . T he spindle of t he gove rnor is also connected t o t he engine . If t he load is reduced , t he speed goes up . Th is increases t he cen trifugal forces on w eigh ts A , lifting t he slide B and pu tting dow n t he valve C . The gas in take int o t he engine is de- cr eased . This reduces t he inpu t po wer and t he ene rgy balance is achieved . The aut omatic r egulation w orks
Fig . 11-6

also in t he opposite dir ection and t his m akes t he syst em stable . T he design of speed r egulat ors ・ 2 19 ・

is beyond t he scope of t his book an d is not discussed furt her . Problems and Exercises 11-1   What is t he equivalen t force or equivalen t moment of force of a m echanical syst em ? Are t hey t he r esultan t force and t he resultan t mom en t of t he syst em ? H ow can t hey be calculat ed ? 11-2   What is t he equivalen t m ass or equivalent mom en t of iner tia of a mechan ical syst em ? Are t hey the sums of t he masses and mom en ts of inertia of t he syste m ? H ow can t hey be calculated ? 11-3   What is t he dynamically equivalen t model of a mechan ical system and w hy is it set up ? W hat does it m ean if one link is chosen as t he equivalen t link ? 11-4   W rite do wn t he motion equations of a m echanical syste m in differ en t forms . H ow can t hey be solved ? 11-5   W hat is t he periodic speed fluct ua tion ? W ha t is t he coefficient of speed fluct uation ? W he re can t he m axi mum and min imum speeds appear ? H ow can t hey be checked ? 11-6   Wh y and how should t he periodic speed fluct uation be regula ted ? W ha t is t he function of a flyw heel and w here should it be mou nted ? How can its momen t of iner tia be calculat ed ? 11-7   What is t he aperiodic speed fluct uation ? W hy and ho w s hould it be regulat ed ? W hat is t he function of a speed regulator ? 11-8   A planetary gear t rain wit h tw o planets is show n in Fig . 11-7 . T he module of all t he gears is m . T he nu mbers of t eet h a re z 1 , z2 and z 3 respectively . T he mom en ts of iner tia of t he lin ks are J1 , J 2 , and J H . T he m ass of a planet is m 2 . The resistant momen t of force M H acts on t he planet ca rrier H . T ake gear 1 as t he equivalen t lin k . Find t he equiv- alen t resistan t momen t of force M r of M H and t he equivalen t mom en t of ine rtia J of t he w hole gea r t rai n .
Fig . 11-7

11-9   A planar linkage is sho w n in Fig . 11-8 wit h dimensions lA B = 3 l , lC D = 2 l , lD E = 6 l ,

Fig . 11-8

l A C = l . The mom en t of iner tia of c ran k 1 abou t its cen tr e of m ass A is J1 . T he momen t of in- er tia of rota ting guide 3 abou t its cen t re of m ass C is J3 . The m ass of slider 2 is m 2 and its mo- m en t of iner tia abou t its cen tr e of m ass B is J2 . The m ass of coupler 4 is m 4 . I ts cen t re of m ass s4 is located at the middle of DE . I ts momen t of iner tia abou t s4 is J4 . T he mass of slider ・ 2 20 ・

5 is m 5 . T he r esistan t force F 5 acts on slider 5 . T ake crank 1 as the equivalen t link . Find t he equivalen t r esistan t moment of force M r of F5 and t he equivalen t mom en t of ine rtia J of t he w hole linkage for φ . 1 = 90° 11-10   In a lif tin g device sho w n in Fig . 11-9 , t he nu mbers of teet h of t he gea rs are z 1 = 20 and z2 = 80 . T he r adius of t he reel is R = 150 m m . T he moment of inertia of gear 1 is J1 = 0. 01 kgm 2 . T he mom en t of inertia of gear 2 t oget her wit h t he reel is J2 = 2 k gm 2 . T he lift ed weigh t is G = 2 000 N . The driving mom en t on gear 1 is M 1 = 100 N m . Find t he angula r ac- celer ation α 1 , t he angular speed ω 1 and t he rotating angle φ 1 of gear 1 , 0. 5 second after star t- ing from standstill .

Fig . 11-9

Fig . 11-10

11-11   A m echanical syste m has an equivalen t moment of ine rtia J = 0 .5 kgm 2 and equivalen t momen t of for ce { M } N・m = - 70 - 0 .3 { ω} r a d/ s . T he star ting angular speed is ω0 = 150 r ad/ s . Fin d t he ti me t for ω = 0 . 11-12   A m echanical syste m wit h a con stan t equivalent mom en t of iner tia J = 1 kgm 2 is periodic and stable . T he equivalen t r esistan t momen t is M r = 70 φ and t he equivalen t driving mom en t π M d is constan t . T he angular pe riod of t he motion is φT = 2π . T he initial speed for φ 0 = 0 is

ω 0 = 100 rad/ s . Find t he angula r speed ω and angular accelera tion α of t he equivalen t link as functions of t he rotating angle φ . 11-13   T he ou tpu t mom en t of force M d of an in ternal combustion engine on t he crankshaft is a given function of t he rotating angle φ as show n in Fig . 11-10 . T he angula r pe riod of motion is φT = π . T he average speed is nm = 620 r/ min . T he engine is used t o drive a m achine wit h con- stan t r esistan t mom en t . T he allow able coefficient of speed fluctuation is [δ] = 0 .01 . Masses and mom en ts of inertia of all t he links a re neglected . Find t he needed mom en t of iner tia JF of t he fly w heel on t he c ran kshaf t .

・ 2 21 ・

C h a p t e r 12 C r eat iv e D e s ign o f Mec han i sm S y s t em s
12.1   In t r o d u c t i o n t o D e s i g n o f Me c h a ni s m S y s t em s
12.1.1   Mechanisms systems In t he previous chap ters , we have discussed some of t he most commonly used mecha- nisms . While som e simple m achines con sist of only one kind of such m echanisms , in most cas- es , using only one simple m echanism is not enough to perform t he required mechanical actions in a mach ine . Take t he shaping machine show n i n Fig . 12-1 as an exa mple . T w o working links ( or ou t put lin ks) are needed t o shape a flat surface . T hey ar e t he sliding block wit h t he shaping t ool ( cu tter ) and the work table holding t he workpiece . Carrying t he cu tte r , t he slid- ing block moves back and for t h to pe rform t he cu tting motion and t he stroke of t his motion is adjustable to fit t he size of t he w orkpiece . T he work table moves int ermittently t o provide t he feeding action w hile t he sliding block moves back . The a moun t of feed is also adjustable . Such a w orki ng process needs several simple m echanisms w or king t oget her to fulfill t he w hole func-

Fig . 12-1   Shaping machine

・ 2 22 ・

tion .T hese mechanism s opera te toget her in a m achine and form a mechanism syste m .Anot her exa mple of a mechanism system is t he well-know n in te rnal combustion engine , show n in Fig . 1-1 , w hich consists of a c ran k and slider mechan ism , cam m echanisms an d a gea r m echanism . T he crank and slider mechan ism converts the back and fort h movem en t of t he pist on in t o rota- tion of t he crankshaft . T he gea r m echanism and ca m m echanisms con t rol the movem en ts of t he valves exactly and ensure t he synchronised operation of t he w hole engine . According t o syste m t heory , a machi ne can be seen as a syste m of m echanisms and a m echanism is a sub-syste m of a m achine . Hence , t he design of a mach ine is t he design of a m echanism system . T he qualit y , performance and compa tibilit y of a mechanical product depend m ainly on its design . Any error , defect or car elessness in design may result i n considerable ex t ra cost in m anufact ure or even t he failure of t he product . T he impor tance of design is obvious here . 12.1.2   Routine design and creative design T he re are differ en t levels of design . W hen you design a m achine w hich uses sim ilar mod- els , t hen you can design by im ita tion . T ake one si mila r m achine as a model . t hen , by keep- ing t he m ai n st ruct ure unchanged bu t changing some of t he dim ension s or sizes of t he m achine or replacing som e par ts with new ones , you can carry ou t a design quickly . Such a design is called a rou tine design . If you design a totally new machine or apply a new working principle in a machi ne , you have t o c reate a new st ructure , not just im ita te t he existing one . T his is cre- ative design . Of course, creative design is more difficult t han t he rou tine one . Creative de- sign s play an im portan t pa rt in developing new products to m eet t he growing de m ands of cus- tome rs . T he kinem atic function of a m echanism system is to conver t the motion of a prime mover in to t he required motion of the ou tpu t lin ks of a mechanism syste m . To carry ou t t he design of a m echanism system , an en gineer should be fa miliar with t he analysis , syn t hesis , and design m et hods of various m echanisms as discussed in t he pr evious chap ters . F urt hermore , he or she should be able t o choose t he most suitable mechanis ms and combine t hem in to an in tegrat ed sys- tem . T his requir es som e experience and technique about w hich various t heories and m et hods a re curr en tly being developed . T his chap ter includes som e of t hese t heories and met hods . 12.1.3   The four phases of the design process Gene rally , t he design process of a m echanism syste m can be divided in to t he follo wi ng four phases : (1 ) Planni ng of t he product In t his p hase , t he w hole function of t he product should be dete rmined and t he design task should be defined . Som e investigation of t he ma rket and a feasibility analysis should be done he re . (2 ) Designing t he kine ma tic diagr am of t he m echanism syst em ・ 2 23 ・

T his phase can be divided furt her in to t he following steps: ( a ) Dete rmina tion of t he w orki ng principle of t he machine Designe rs are required to have ext ensive kno wledge of science and technology . Special at- ten tion should be paid to t he so-called high technologies and t heir recen t developm en ts . Ad- vanced w orking principles will produce excellen t products if applied properly . F ur t hermore , t he function of all working link s should be prescribed . (b ) Type syn t hesis of t he m echanisms Suitable types of m echanisms should be chosen and combined i nt o a mechanism system ac- cordi ng to t he technical requirem en ts of t he mach ine . T he t ypes of m echanisms chosen s hould all perform t he required motions . T his is t he most cr eative step an d effort should be concent rat - ed her e . ( c ) D rawing t he diagra m of t he w orki ng cycle T his diagram is also called t he diagram of t he motion cycle of t he m echanism syste m . I t is actually t he ti me table for t he actions of all t he w ork ing link s . Such a diagra m ensures t he syn- chronisation of all m echanisms in a machine . (d ) Dim ensional syn t hesis of mechan isms According to t he actions of t he working links and the w ork ing cycle diagram of t he mecha- nism system , t he kinem atic dim ensions of t he m echanisms can be dete rmined based on t he m et hods given in previous chap ters . ( e ) D rawing t he kine matic diagra m of t he m echanism syste m T his is t he final step in t his phase . We shall concent rat e m ainly on t his phase accordi ng to t he scope of t his book . (3 ) Conc rete design Based on t he kine ma tic diagra m of t he mechanism syste m an d t he for ce analysis of t he sys- tem , t he st ruct ural design can be done . T he technical drawings of all machine elem en ts and t he assem bly dr awi ng of the m achine should be complet ed . (4 ) Improve men t of t he design Af ter a prot otype m achine is made , a series of tests should be carried out on it . Corr ec- tions can t hen be m ade to improve its perform ance . In t he design process described above , som e feedback or iteration m ay be necessary to achieve a bet ter result . We shall discuss t he design process in more det ail la ter .

12.2   C r e a t i v e T hin k in g a n d C r e a t i v e M e t h o d s
As mentioned above , c rea tive design is very impor tant in t he developm en t of new prod- ucts . Cr eative t hin king and creative met hods are helpful t hroughou t t he w hole process of de- sign . It is t her efore desirable for engineers t o lea rn somet h ing abou t cr eative t hink ing and cre- ative m et hods . ・ 2 24 ・

12.2.1   Creative thinking Some ways of cr eative t hink ing ar e in troduced i n pairs as follows . (1 ) T hin king in im ages and abst ract t h inking T he elem en ts of t hin king in images are concr ete im ages . A n i mage r eflects t he gene ral fea- tures of one kind of object . During t he design of a machine for example , t he shape , t he colour and ot her ex ternal feat ures appear in t he br ain of t he designer . H e or she can im agine ho w to assemble, disassemble or run t he m achine . S uch a t hinki ng activit y is th inking in i mages . A bst ract t hinkin g takes concep ts as t he t hink ing ele men ts . I nfer ence is its main feat ure . For exam ple , w hen a force is t o be enla rged , t he speed of t he working link should be decreased according t o t he pri nciple of conse rvation of m echanical ene rgy . O f t hese t wo , t hinking i n im ages is more flexible w hile abst ract t hin king or reasoning t h inking is rigorous . T hey should complem en t each ot her in cr ea tive design . (2 ) Divergen t t hi nking and convergen t t hinking Divergen t t hinki ng sea rches for differen t ways t o solve a proble m . The infor mation from t he given proble m is r ea rr anged in differen t directions wit hou t rou ti ne const rain ts to produce as m any ways as possible to solve t he proble m . For exa mple , in order to r educe t he frictional loss in a m achine, we can t ry to use mat erials wit h lower coefficien t of friction , to r educe t he r eac- tions or t he rela tive velocities betw een kinem atic pair ele men ts , to decrease t he numbe r of ki ne- m atic pairs , to lubricate t he bearings , or t o r eplace sliding wit h rolli ng , and so on . A good designer should have ex tensive knowledge and wide in terests in m an y differ en t fields . More- over , he or she should be able to apply t he knowledge and experience to produce new ideas . O n t he ot her hand , convergen t t hin king derives t he best idea from many possible ones a rising in dive rgen t t hinkin g . Dive rgen t t hinki ng and conve rgen t t hinking a re comple men tary t o each ot he r in a creative activit y . Convergence is of little significance wit hou t sufficien t diver- gence w hereas divergence is t he basis of convergence . The purpose of divergence is t o provide enough choices for conve rgence . A cr ea tive process often needs several cycles of convergence and divergence and finally stops at convergence . (3 ) Logical t hinki ng and nonlogical t hin king Logical t hink ing is a rigorous way of t hink ing . It reflects t he act ual process and reveals t he essence of a subject based on scien tific concepts . T he m ain forms of logical t hink ing are analy- sis , judgm en t , induction and inference . Logical t hin king is widely used in c rea tion . The in- ven tion of t he comput er can be seen as t he c reative achievem en t of logical t hin king . In cont rast t o logical t hink ing , nonlogical t hin king does not obey logical formulae st rictly . I t is a flexible free way of t hi nking and can lead for tuitously to some unusual and novel results . T he main form s of nonlogical t hinking are association , imagination , in tuition an d inspiration . Association is a m en tal activit y relating one t hing t o anot her . I magi nation c rea tes a new im age on t he basis of association . In t uition is a quick dir ect judgm en t based on vast experience . In tu- ・ 2 25 ・

ition is not so st rict , bu t quite valuable in creating new models and concep tions . Inspiration is t he st rong excitation of in telligence . I t follows long hard work on a difficult problem and bursts ou t suddenly . T he brain t hen becomes ve ry active and cr eative . M any new ingen ious ideas sprin g up and t he problem is solved quickly in th is period . Nonlogical t hinki ng plays an impor tan t par t in creative design . We should try t o m ake full use of it . 12.2.2   Creative methods I t is said t hat t here ar e hundreds of cr eative m et hods in t he world . In fact , everyone can cr eate some new m et hods wit h a lit tle effort . This m eans t hat t he re is no myste ry surrounding cr eation . O f course , cr eation is also not easy and it is t her efore helpful to know som et hing about c rea tive met hods . (1 ) Br ainstorm A sm all group of people ( not more t han 10) is called t oget her to hold a meeting . T he sub- ject of t he m eeting should be announced in advance . Everyone is encouraged to expr ess his opinion s freely no mat ter ho w str ange t hey a re . I n fact , unusual or even curious ideas are wel- come and should not be criticized . I n such a free at mosphere , people activate and inspire each o t her . New concep ts , new met hods and new assumptions can easily well up . This is a diver- gen t process and proves t o be an effective way t o collect t he wisdom of t he m asses . (2 ) Q uestionnaire A series of question s abou t a problem or invention a re listed first . T hey a re t hen checked and discussed carefully . Solu tions or new inventions m ay arise from such checks and discus- sions . T ypical questions asked a re as follows : ( a ) Are t here an y ot her applica tions of t he presen t inven tion ? (b ) Can t he met hod of m anufact ure , s hape , colour or o t her aspects of t he pr esen t inven- tion be changed ? ( c ) Can t he presen t inven tion be m ade s maller and ligh ter ? (d ) Can t he pr esent invention be replaced by som et hing else ? ( e ) Is it possible t o apply the presen t inven tion inve rsely ? (f) Is it possible to combine som e of t he present inventions ? A good questionnaire leads people to sea rch for a bette r solu tion or make new inven tions in diffe ren t direction s . It can be done pe rsonally or i n a group . Q uestions should be checked one by one or even be repeated to en sur e effectiveness . (3 ) Imitation By imitating or simulating som e natur al process or living bei ngs , new machi nes or tech- niques can be developed . Various ki nds of robo ts and manipulat ors a re exam ples . Som e popu- la r techniques in compute r science , such as t he simulat ed annealing algorit hm , genetic algo- rit h ms an d neural net works can be seen as t he achievem en ts of imitation . ・ 2 26 ・

(4 ) Listing draw backs Listing all t he defects or weakness of a product , we may find som e ways to improve its performance or even to inven t som et hing new . Si mila rly , w e can list the hopes for a product and find a way to improve it . (5 ) T ran splan tation T ran splan tation of an advanced technique from one field or product t o ano t her will proba- bly r esult in a new inven tion . Some milit ary tech niques have been successfully used i n civil products in t his way . Mor e creative m et hods can be found in t he literat ure . T he key poin t he re is to get t he spir- it of t hem , t hat is changing and associatin g somet hing som ew here .

12.3   K in e ma t i c P a r am e t e r s o f t h e S y s t em
12.3.1   Prime kinematic parameters of the system Af ter determ ination of t he wor king principle of a m achine , t he mo tion forms and t heir kinem atic pa ra mete rs of working lin ks should be determined . T he main motion forms of w ork- ing links a re rotation , r ectilinear t ranslation , curvilinea r motion ( of a poin t ) and com pound motion . We now discuss t hese one by one . T he re a re t hree types of rotation . ( a ) Continuous rotation , the ki nem atic para mete r of w hich is its angular velocit y or r evolution pe r mi nute ( rpm ) , e. g . t he spindle of a lat he , a drilling m achine or a m illing m achine and so on . ( b ) Step ( or in term it ten t ) rotation ( for som e rotating tables in au tomatic machines ) , t he kinem atic para met ers of w hich are t he fre- quency , stepping angle , kinem atic coefficien t and so on . ( c ) R ocking motion , t he kinem atic par am eters of w hich a re t he frequency , angle of ro tation and the time ratio . Rectili near t ranslation s also have differen t forms . ( a ) Reciprocating t ranslation ( e. g. as in t he sliding block in a shapi ng m achine ) , t he kinem atic pa ram eters of w hich are t he frequen- cy , t he lift and t he ti me ratio . (b ) Rectilinea r tr anslation wit h dwells , t he kine matic pa ram e- ters of w hich are t he number of d wells in a cycle , t he sequence of in tervals of mo tion and dw ell, t he lift , m ean velocity and so on . T he kinem atic para met ers of a curvilinea r motion ( e. g. for a mixer ) a re involved in t he pat h w hich t he end poin t of t he w ork ing li nk should follow . They can be given as functions of t he coordinates of t he movi ng point ( wit h respect t o ti me ) , or at several disc rete positions of t he poin t . T he pat h can be planar or spatial , fixed or adjustable . A compound motion is a combination of t he above forms of motion . F or example , t he mo- tion of a drill is a combi nation of rot ation ( cu tting motion ) an d a r ectilinear t ranslation ( feed- ing) . T he kine ma tic para m eters of a compound motion depend on its components and t heir combina tion . ・ 2 27 ・

T he above kinem atic pa ram eters of t he w orki ng link s a re determ ined in t he planning of a product . Car eful planning and calculation a re needed to dete rmine t he pa ra mete rs wit h proper accuracy . We can t hen choose prime movers and t heir kine matic par am eters . 12.3.2   Selection of prime mover Motion forms of prime movers can be rotation , rock ing and reciprocating motion . Electric motors provide rota tion . H ydraulic motors and pneumatic mo tors provide bot h rotations and rocking mo tion . H ydraulic cylinde rs , gas cylinders and electric li near motors provide r ecipro- cating motion . The kinem atic pa ra meters of prim e movers a re velocity ( angula r or linear ) , rocking angle and frequency , lift and so on . T he most widely used prime mover in machi nery is t he AC asynchronous electric mot ors wit h diffe ren t synchronous speeds . The highe r t he synchronous speed , t he sm aller t he size and t he weigh t of t he mo tor , and t he lower t he price . Bu t h igher velocit y of t he motor will in- cr ease t he size and price of t he tr ansmission system if t he velocity of t he working link is low . Modern indust ry provides m any new t ypes of mo tors , e.g . servo motors , step mot ors and frequency modulating motors . Also , spri ngs and weigh ts can serve as prim e movers occasional- ly . Differen t pri me move rs have differ en t char acte ristics and a re suitable for different cases . Selection of prim e movers is one of t he im portan t steps in machi ne design .

12.4   S e l e c t i o n o f M e c h a ni s ms
According t o t he motion forms and kinem atic pa ra mete rs of t he working links and prime movers described above , mechanis ms can be chosen t o conver t t he motions of pri me movers in- to t hose actions of t he working lin ks . O ne way to do t his is by decomposition and combina tion of t he functions of t he m achine . Functions of t he syste m ar e first resolved int o seve ral sub- functions or function ele men ts . For each function ele ment , all t he possible function ca rriers (m echanisms w h ich carry ou t t he function ) ar e listed . The combination of all t he function car- riers in each group provide a great many solutions among w hich some feasible ones can be cho- sen . F urt her evaluation of t hese feasible sche mes will produce the best one . A not he r effective way t o select mechanism s is im it ation and re-formation . First , deter- mine t he key t echniques of t he m achine to be designed an d t hen search for t he corresponding devices as models . According t o t he functions and design de mands of t he machine , t he models can be reformed or combined and t he m echanism syste m determined . This m et hod is often ap- plied w hen relevan t r efe rences can be found . I t is now helpful to review differen t m echanisms wit h t heir kine ma tic functions . 12.4.1   Rotation transmitting mechanisms T hese k inds of m echanisms a re used t o change t he velocity and t he dir ection of rota tion . ・ 2 28 ・

T hey can be divided furt her in to t hr ee t ypes . (1 ) Frictional mechanism s T his type of mechan ism includes belt m echanisms and frictional w heel drives . T hey are simple i n const ruction and smoot h in oper ation . Sliding in t he m echanisms can provide prot ec- tion against overload . T hey can be designed as stepless speed regula tors . T he draw backs are t he inaccuracy of t ransmission ratio , low er po wer capacit y and lo wer efficiency . (2 ) Engagi ng mechanism s T his type of mechanism includes gear mechanisms , worm gea r mechan isms , chain mecha- nisms , gea r t rains , and so on . A chain mechanism is of ten used betw een tw o pa rallel shafts wit h longe r cen t re distances and lowe r accuracy compa red to a gea r m echanism . I t ope rates less smoot hly t han gear m echanisms and should not be used at high speed . Gear m echanisms and wor m m echanisms can t ransmit rotation bet ween t wo shafts wit h any relative orien tation . T hey have h ig her powe r capacit y and efficiency an d oper ate smoo t hly . T hey are used widely in a range of m achinery . (3 ) Linkage mechanism s Exa mples of t his kind of mechanis m are t he double-cr ank m echanism ( Fig . 4-1b ) , pa ral- lel-crank m echanism ( Fig. 4-22) and rota ting guide-bar m echanism ( Fig. 4-2b ) . S uch mecha- nisms con tain only lo wer pairs and provide va rious of transmission fu nctions . They are easy to m anufact ure bu t no t easy t o design t o suit t he given tr ansmission function . 12.4.2   Step mechanisms T he out pu t li nk of a step m echanism performs un idirectional motion with periodic d wells . T he Geneva m echanism ( Fig .8-12 ) , ra tchet mechan ism ( Fig . 8-1 ) and index ca m mechan ism ( Fig. 8-18 and Fig. 8-19) ar e t ypical st ep mechanisms . As mentioned in Chapt er 8 , a ratchet m echanism conver ts a rockin g motion in to a step (or in te rmitten t ) motion . T he rotating angle of t he out pu t link ( r atchet w heel ) can be adjusted . T his m echanism should be used only at lo w speeds and loads because of t he impact between t he r atchet w heel and t he jaw . A Geneva m echanism operates more smoot hly t han a ratchet m echanism , bu t t he rotating angle of its ou t- put link can not be adjusted . Indexing cam mechanisms a re ideal step mechanism s for high speed applications . Some geared linkages ( Fig . 9-16a ) have been designed as pilgrim-step m echanisms in w hich t he tr ansmission function has t he cha racteristic of an overall unidir ectional motion bu t wit h pe riodically r ecurring reversals . T hey have found applications in ligh t indust ry m achin- ery . 12.4.3   Reciprocating and rocking motion generating mechanisms     Reciprocating and rock ing motions ar e frequently used form s in machinery . Since rota tion is t he usual mo tion form of a prim e move r , t his t ype of mechanism should be used t o conver t a ・ 2 29 ・

ro tation in to a reciprocatin g or rocking mo tion . T he slider-crank mechan ism ( Fig . 4-2a ) , c ran k and tr anslati ng guide-ba r mechan ism (Fig . 4-6 ) an d ca m m echanism with t ranslating followe r ( Fig . 5-18 and Fig . 5-30 ) conver t ro tation int o r eciprocating motion . The c ran k-rocker mechanism ( Fig . 4-1a ) , oscillating guide-bar m echanism ( Fig . 4-9 ) and ca m m echanism wit h oscillati ng followe r ( Fig . 5-28 and Fig . 5-34 ) conver t rotation int o rocki ng motion . Ca m m echanisms can carry out almost any t ransmission function exactly wit h lo wer load capacit y because of the higher pair . T he lif t of cam mechanism s is also limited wit hin a small range to m ain tain a favorable pressure angle . O n t he con tra ry , linkage mechanisms can not carry ou t exactly t he given t ransmission function in most cases . Bu t t heir load capacit y and lift of t he ou t pu t lin k are greater t han t hose of ca m m echanisms . T he screw m echanism conver ts rotation in to t ranslation wit h high accuracy and a gr eat de- cr ease in speed . It se rves oft en as a fine t uning m echanism . I t can produce great force wit h self-locking in t he reve rse direction . The efficiency of a t raditional screw m echanism is not high but nowadays , rolling sc rew mechanis ms wit h high efficiency a re available . Sim ilarly , the rack and pinion mechan ism conver ts ro tation in t o t ranslation wit h a higher velocit y , but it oper ates less smoo t hly t han t he screw m echanism . 12.4.4   Path-generating mechanisms Linkage mechanism s ( Fig . 4-41) , geared linkages ( Fig . 9-7) and som e ot her combined m echanisms ( Fig . 9-8 ) can serve as pat h-gene rating m echanisms . F our-ba r linkages a re t he simplest ones and easy to build . Bu t t hey can gener ally follo w t he given pat h only approxim ate- ly . Linkage mechanisms wit h more ba rs or gea red linkages can sometim es do better , bu t still can not follow t he pat h exactly . Combined mechanisms wit h a t least one ca m ( Fig . 9-8) pro- duce almost an y given pat h exactly bu t wit h higher cost due t o ca m m anufact ure .

12.5   C o o p e r a t i o n o f W o r k in g L i n k s
12.5.1   Types of cooperation Af ter m echanisms for t he w orki ng li nks have been chosen , t hey must be connected in to a m echanism syste m . T he key problem her e is the cooperation of t he working link s . (1 ) Independen t kinem atic chain for each w orking lin k In som e cases , motions of w orkin g li nks a re independen t of each ot he r . F or exam ple , in t he grindin g m achine sho wn in Fig . 12-2 , t he w or king lin ks are t he grinding w heel , t he grin ding w heel carrier and t he spindle . T here is no cooper ation problem and no accurate t rans- mission r atio is needed betw een t hem . In t his case , we can design an independent ki nem atic chai n wit h its prim e mover for each working lin k . In t his way , t he m echanism syste m can be ・ 2 30 ・

Fig . 12-2   Grinding machine

simplified . Excep t in t he above cases , cooperation a mong working lin ks must be consider ed . There a re tw o k inds of cooper ating r ela tionships , velocit y and displace ment . (2 ) Velocity cooperation Velocity coope ration r equires accurate t ransmission ra tios among working lin ks . F or exam- ple , in a gea r cu tting m achine, the t ransm ission ra tio bet ween t he cut ting tool and t he w ork- piece should re main constant t o ensur e t he correct gener ating motion . Similarly , t he tr ansmis- sion ra tio bet ween t he spindle and t he ca rriage of a lat he should be constan t i n cut ting a screw . In such a m echanism system , all w ork ing links should be driven by t he sa m e prim e mover . Belt mechanism s should not be used bet ween t hese w orki ng links . The corresponding tr ansmis- sion ratios are determined according to t he required velocit y cooperation . (3 ) Displace ment coope ration Displace ment cooper ation r equir es t ha t mo- tions of all t he w orkin g links should be timed and spaced cooperatively , e . g . in a shaping m a- chine and an int ernal com bustion engine . T he biscuit packi ng machine show n in Fig . 12-3 is ano t her example . T he tw o folding bars ar e t he wor king links in t he folding m echanism . T hey rock and fold t he packing paper a round t he bis- cuit periodically . T he pat hs of t heir ends in ter- sect at poin t M . Therefore , t heir motions
Fig . 12-3   Biscuit packing machine

should be ti med and spaced to avoid any i nte rfer-

ence . F or example , if t he left bar moves first , t hen before it ret reats t o poin t M , t he righ t bar should no t en ter t he region to t he left of t he point M . 12.5.2   Working cycle diagrams To ensure t he displace men t cooperation of working lin ks , t he w orking cycle diagram of t he m echanism syste m s hould be draw n . Her e one link should be chosen as refer ence link t he posi- tions of w hich ar e taken as ti me ma rke rs to indicate t he sequence of actions . T he re are t hr ee forms of working cycle diagra ms as follows: ・ 2 31 ・

Fig . 12-4   Working cycle diagram of a shaping machine

(1 ) Rectili near working cycle diagra m Fig . 12-4 is t he w ork ing cycle diagram of a shaping m achine ( Fig . 12-1 ) . T he c ran k in t he guiding ba r mechanism is t he reference link wit h a period of 360°. E ach st rip indicates t he sequence of actions of one w ork ing lin k . I t can be seen from t he diagram t hat t he sliding block has a quick-r et urn motion . T he feeding of t he table is carried out duri ng t he return stroke of t he sliding block . (2 ) Circula r w or king cycle diagra m Fig . 12-5 is t he w ork ing cycle diagra m of an in ternal combustion engine ( Fig . 1-1 ) . T he cr ank shaft is t aken as refer ence link wit h a pe- riod of 720°. Each ring in t he diagr am indicates t he sequence of actions of one working link . (3 ) Rectangular coordinate working cycle diagr am Fig . 12-6 is t he w ork ing cycle diagra m of t he foldi ng mechanism ( Fig . 12-3 ) in t he bis- cuit packing m achine men tioned above . T he spindle of t he machine is t he refer ence link with a period of 360°. T he rot atin g angle φof t he spindle is taken as t he abscissa and t he angles of t he foldi ng bars a re t he ordinates . T he diagra m shows not only t he sequences of actions bu t also t he motion la ws of t he working links and t heir kinem atic relation . T his kind of w orking cycle diagr am is superior to t he pr evious t wo .
Fig . 12-5   Working cycle diagram of an internal combustion engine

Fig . 12-6   Working cycle diagram of the folding mechanism

・ 2 32 ・

A working cycle diagra m is act ually a tim etable of w orkin g links . It provides a basis for designing , connecting and adjusting t he m echanism syst em to ensur e t he cooperation of t he w hole system . Af ter t he working cycle diagra m is fi nished , t he kine matic dim ensions of t he m echanisms can be dete rmined . T hen t he ki ne matic diagra m of t he m echanism syste m can be draw n .

12.6   E v a lu a t i o n o f t h e M e c h a n i s m S y s t e m
12.6.1   Indexes of evaluation D urin g t he syn t hesis of a m echanism syste m , we gener ally get more t han one solu tions as t he result of divergence . No w we t ry to get t he best of all from t hese feasible solu tions by m eans of evaluation . A m echanism syste m can be evaluat ed in five aspects as follows: (1 ) System functions: Motion laws , Tr ansmission accuracy . (2) Work ing perform ance : Applicable range , Adjust ability , O per ating speed , Load ca- pacit y . (3) Dynam ic pe rform ance : Peak value of acceleration , N oise , Resist ance t o wea r , Reli- abilit y . (4) Economics: M anufact urabilit y , Sensitivit y t o error , Convenience of adjust m en t , Ef- ficiency . (5 ) St ructure : Size , Weight , Complex it y . Altoget her we have 17 indexes as above , deno ted by A i ( i = 1 , 2 , … , 17 ) . Of course , t he num ber of i ndexes can be adjusted ( neglected or added) in a par ticula r design . According t o t he design dem ands and t he experiences of expe rts , w e dist ribute weigh ting fact ors k i ( i =
17

1 , 2 , … , 17 ) t o all t he indexes wit h ∑ k i = 100 .Nex t , for each index A i , we choose a value
i =1

gi ( gi ∈ [ 0 , 1 ] ) to evaluate t he corresponding fit ness . We can t hen calculate the t otal score H for t he m echanism syste m as follows:
17

H =

∑k
i =1

i

gi

( 12 -1 )

T he maximu m of H he re is 100 , indicati ng t he ideal system . Comparing t he to tal scor es of different m echanism syste ms , we can iden tify the best one . 12.6.2   Hints for improvement If none of t he m echanism system s is good enough , som e modification must be made to im- prove t he system . Seve ral hin ts for i mprovem en t ar e listed below . ・ 2 33 ・

(1 ) Choose t he kine ma tic chai n t o be as short as possible A shorte r kinem atic chain her e m eans fewer link s an d kinem atic pairs . T his can reduce t he cost of m anufacture and t he kinem atic error . Efficiency can also be inc reased . (2 ) Use t he m echanism with highe r efficiency T he efficiency of a kinem atic chain is t he product of t he efficiencies of t he componen t m echanisms if t hey are connected in series . T her efore , t he kinem atic chain t ransmitting t he m ain power should no t i nclude a mechanism wit h lower efficiency . (3 ) A rrange t he mechanisms in proper orde r Gene rally , t he speed of w ork ing links a re lowe r t han t he ot hers and variable . T herefore , t he m echanisms w hich conver t t he forms of motion ( e . g . cam m echanisms and lin kage mech- anisms) are placed at t he end of t he kine ma tic chain , i . e . near t he w orkin g link . I n t his way , t he vibr ation of t he mechanism can be reduced . In addition , t he frictional t ransmitting m echanisms such as belt drives should be arranged nea r t he motors w here t he speed is higher and t he torque is lower . Such an arr ange ment can r educe t he size of t he driving system wit h overload pro tection and t he pri me mover can be loca ted wit h more flexibilit y . (4 ) Choose reasonable tr ansmission ratios Every m echanism should w or k wit hin its applicable r ange of t ransmission ra tio t o ensure proper perform ance . F urt hermore , careful choice of t he tr ansmission ratios can consider ably reduce t he size of t he syste m . O bviously , t he above hin ts can also be used in selection of the m echanisms or rough initial evaluation of t he m echanism syste m .

12.7   D e s i g n E x am p l e o f a M e c h a ni sm S y s t em
We will now t ake t he shaping m achine as an exa mple t o explain t he process of design of a m echanism syste m . 12.7.1   Design requirements and kinematic decomposition A shaping m achine is gene rally used for cu tting planes on t he workpiece by t he relative re- ciprocating motion betw een t he cut ter and t he workpiece . T he st roke of t he reciprocating mo- tion of t he cu tte r s hould be adjustable t o fit t he leng th of t he plane to be cu t and t he velocit y of t he cu tt er motion should be uniform ( no t necessa rily constan t ) to ensure t he cu tting quality . A feeding in t he dir ection pe rpendicular to t he cu tting mo tion is necessa ry to form a complete plane . Such a feeding mo tion should be adjust able to fit differ en t m aterials and cut ting require- m en ts . The feedi ng mo tion should not in te rfer e wit h t he cut ti ng motion . T he supple men tary tim e should be reduced t o increase productivity . T he re a re tw o possible decompositions of t he cu t ting process . In t he first decomposition , t he workpiece reciprocates and t he cu tter feeds in termit ten tly . T his fits a la rge workpiece, as ・ 2 34 ・

in a double housing plane r . In t he second decomposition , t he cu tte r reciprocates and t he w ork- piece feeds in termitt en tly as m en tioned at t he beginning of t his chap ter . Th is fits t he shaping m achine cu tting workpieces of small or middle size . According to t he kinem atic decomposition of t he tech nical process , t her e ar e two working links in t he machi ne . For such a common m achine tool , we choose t he electric mot or as its prime mover . Nex t , we discuss t he mechanisms for cu tting and feedi ng sepa rately . 12.7.2   Selection of the mechanisms for cutting T he mechanism for t he cut ting motion must conve rt rotation in to r eciprocation wit h a quick-ret urn to incr ease productivit y . Som e possible choices a re listed as follows . (1) Sc rew mechan ism wit h t he cut ter fixed to t he nu t . T his mechanism can provide t he cu tte r wit h constant velocity and con tains only lo wer pairs . But t he efficiency of t he sc rew m echanism is lo w and ot her m echanisms ar e needed t o give a quick-ret urn stroke . T his would complicate t he system . ( 2) Rack and pinion m echanism wit h t he cut ter fixed to t he rack . This mechanism can al- so give t he cut ter a constan t velocit y wit h high efficiency . But it can not provide the quick-re- turn itself as can t he scr ew mechanism . (3) Offset slider-c ran k m echanism wit h t he cut ter fixed to t he slide r . The offset of t he slider can produce a quick-r eturn st roke . Th is mechanism w orks reliably and is easy to make since it is a lowe r pair m echanism . The m ain dr awback is t hat t he velocit y of t he slide r in t he cu tting st roke is not un iform enough . Also , t he lengt h of coupler must be la rge t o ensure a practical pressure angle . (4 ) Ca m m echanism wit h t he cu tte r fixed t o t he t ranslating follo wer . W it h a properly designed ca m profile , t his mechan ism can provide t he cut ter wit h constan t velocit y in t he working stroke and the requir ed quick-ret urn st roke . But t he lengt h of the st roke in t his m echanism is not adjustable and t he higher pair elem en ts ar e not suitable for t he la rge cu tting force . (5 ) Ca m m echanism wit h oscillating followe r con nected in se ries wit h a rocker and slider m echanism show n in Fig . 12-7 . The oscillating followe r in t he cam m echanism is bet ter in t ransmitting force and it is easy t o ex tend the st roke , but t he proble m of t he previous case rem ain s . ( g guide-bar mechanism show n in
Fig . 12-7   Cutting mechanism for shaping machine ( No . 1)

Fig . 12-8 wit h t he cu tte r fixed t o t he slide r . T his m echanism m ay provide t he cu tter wit h quite uniform move ment in t he

wor king st roke and a quick-ret urn . By adjusting t he lengt h of t he crank , t he st roke of the slid- er can be changed easily . T his m echanism con tai ns only lo wer pairs w hich is t herefore benefi- ・ 2 35 ・

cial for t he t ransmission of force . Similar mechanism s ar e show n in Fig . 12-9 , Fig . 12-10 and Fig . 12-11 .

Fig . 12-8   Cutting mechanism for shaping machine ( No . 2)

Fig . 12-9   Cutting mechanism for shaping machine ( No . 3)

Fig . 12-10   Cutting mechanism for shaping machine ( No . 4)

Fig . 12-11   Cutting mechanism for shaping machine ( No . 5)

    Evaluating t he above mechanisms , we find t hat t he m echanism in Fig. 12-9 satisfies all t he requirem en ts quite well and is simple and compact in st ruct ure . T his m echanism is finally cho- sen t o drive t he sliding block with t he cu t ter . 12.7.3   Selection of the feeding mechanisms T he mechanism for t he feeding motion should conver t ro tation in to in term it ten t movem en t of t he table wit h self-locking to keep t he workpiece st eady w h ile cu tting . Some possible choices a re listed as follows . (1) Rectilinear Geneva m echanism and ra tchet mechanism show n in Fig . 12-12 and Fig . 12-13 . T hese t wo m echanisms evolve from t he general Geneva w heel and ratchet w heel mecha- nisms . T hey can provide in termitt en t feeding of t he workpiece and are simple in st ruct ure . ・ 2 36 ・

Bu t t hey need an ex t ra locking m echanism since t hey ar e not self-locking in t he reverse dir ec- tion . T he a moun t of feed cannot be adjusted in t he case of t he Geneva m echanism .

Fig . 12-12   Rectilinear Geneva mechanism

Fig . 12-13   Rectilinear ratchet mechanism

(2 ) Pinion and double-r ack mechan ism con nected in series wit h ratchet mechan ism sho w n in Fig . 12-14 . This m echanism can change t he movi ng direction of t he table au t om atically at t he end of t he st roke , bu t it is no t self-locking . (3 ) Screw mechanism con nected in series wit h ratchet m echanism and crank and rocker m echa- nism show n in Fig . 12-15 . Here t he scr ew m echa- nism is self-locking . T he tw o-way r atchet m echa- nism can change t he rotating direction of t he screw and also t he moving direction of t he t able . By changi ng t he len gt h of t he crank , t he swing angle of t he rocker can be adjusted and t he refor e t he feed is adjustable . T his mechanis m m eets all t he re- quire men ts and is t her efore chosen for feeding motion .
Fig . 12-14   Pinion and double-rack m echanism

Fig . 12-15   Feeding mechanism for shaping machine

12.7.4   Kinematic diagram of the shaping machine Displace ment coope ration bet ween t he cut ter and t he table holding t he workpiece is re- ・ 2 37 ・

quired . T he w ork ing cycle diagra m of t he mechanism system is t hen dra wn as in Fig . 12-4 . Based on t his diagra m and t he funda ment al r equir em en ts of t he machi ne , t he k ine matic dim en- sions of t he m echanism syste m can be determi ned and finally t he kine ma tic diagram of t he m echanism syste m can be draw n as in Fig . 12-16 .

Fig . 12-16   Kinematic diagram of shaping machine

・ 2 38 ・

V ocab u la r y
abscissa acceler ation addendu m allowan ce amplitude anneal an ti-cloc kwise apex Arch ime des spiral Arcta n ar mat ur e Assur group atlas axis axle backlash bear ing bevel gear bisect or block diagram blower bold line bolt bori ng bot tom cleara nce cam ca ntilever ca rdioid ca rrier Ca rtesia n coordi nate system categorize cen t rifugal gove rnor chain do tte d lin e chassis chord 横坐标 加速度 齿顶 ( 高 ) 许用值 , 公差 振幅 , 幅度 退火 逆时针 顶, 尖 , 脊 阿基米德螺旋线 反正切 盔甲 , 电枢 , 转子 阿苏尔组 ( 杆组 ) 图册 , 图谱 轴 ( 线 , 心) ( 轮, 车 ) 轴 间隙 轴承 锥齿轮 平分线 框图 鼓风机 粗线 螺栓 镗, 钻 顶隙 凸轮 悬臂 心形线 系杆 , 托架 , 搬运器 直角坐标系 分类 离心式调速器 点划线 底座 弦 circum fere nce clam p clar it y classification classify cleara nce clip clock wise clutch coefficien t coincide nt com pact com pensat e com plemen t com ponen t com position com pound com pression com pressor con cavit y con crete con jugate conseque nce consist ence const rain t consu mp tion con tact ratio con t our con t rast converse convex coordi nate corr ected gear corr espondence cosine 圆周 夹紧 , 夹具 透明 ( 度 ) , 清晰 ( 度 ) 分类 ( 法 ) 分类 清除 , 间隙 夹子 , 夹紧 顺时针 ( 的 ) 联轴器 , 离合器 系数 重合 紧凑的 , 压实 补偿 余角 , 补充 分量 , 部件 组成 , 合成 ( 物 ) 复合 ( 的 ) , 复合物 压缩 压缩机 凹面 , 凹度 具体的 , 混凝土 共轭 后果 , 推论 一致性 , 稠度 约束 , 强迫 消耗 ( 量 ) 重合度 轮廓 ( 线 ) , 画轮廓 对比 逆的 , 转换 凸的 , 凸面体 坐标 ( 的 ) , 协调 , 一致 变位齿轮 对应 , 通信 余弦

・ 2 39 ・

counte r-clock wise counte rweigh t couple coupler cra nk cra nk shaft criterion cross-h atch crosspiece cross-section crossed h elical gear crow n gea r crush er curta te epicycloid curvat ure curvilinear cusp cycloid cylinder dash ( and ) dot line dash line decele ration decomposition ded endu m deflection defor mation de nomin ator de nsity de rivative de rive det ect diagram diesel diffe ren tiat e dimension disconnect disengage displace DO F ( degree of freedom ) double Haas plane r double-helical g ea r drag-link mecha nism

逆时针 配重 双 , 连 (联 ) 接 连杆 曲柄 曲轴 指标 , 规范 , 标准 网状线 十字叉 , 十字架 横截面 交错轴斜齿轮 冠轮 ( 颚式 ) 破碎机 短幅外摆线 曲率 曲线 ( 的 ) 尖点 , 歧点 摆线 圆柱 , 汽 ( 气 ) 缸 点划线 虚线 减速 ( 度 ) 分解 齿根 偏移 变形 分母 密度 导数 , 导出的 推导 , 派生 发觉 , 检测 (简 ) 图 ( 狄塞尔 ) 内燃机 区分 , ( 求 ) 微分 , 求导 尺寸 拆开 , 拆卸 解开 , 分离 , 脱开 位移 , 置换 自由度 龙门刨床 人字齿轮 双曲柄机构

drill driller driven link duplicat e dur ation dwell dya d eccen tr ic eccen tr icit y elastic elasticity engage envelope enveloping worm equilibriu m evaluate evolu tion ex cavat or ex er t ex te nsion facewidt h feed feedback fit ness fi xt ure flexsplin e fluct uate flyw heel forge for m cu tti ng for m ula for t h for t uity fr ame fr amework friction frustu m fulc ru m gea r gea r t rai n gea rbox gea red linkage

钻孔 , 训练 , 钻头 , 钻床 钻床 , 钻探工 从动件 重复 ( 的 ) , 复制品 延续 ( 性 , 时间 ) , 期间 小停歇 Ⅱ级杆组 偏心的 , 偏心轮 偏心 ( 距 ) 橡皮带 弹性 , 灵活性 从事 , 啮合 包络线 , 封皮 环面蜗杆 平衡 , 均衡 求值 , 评价 , 测定 展开 , 渐进 , 演化 挖掘机 发挥 , 施加 延长 ( 部分 ) 轮齿宽 进给 反馈 适当 , 适应性 夹具 柔性轮 ( 使 ) 波动 飞轮 锻造 成形切削 公式 向前 偶然事件 , 偶然性 机架 , 框架 机架 , 框架 摩擦 ( 力 ) 锥台 , 截锥体 支点、支轴 齿轮 轮系 齿轮箱 齿轮连杆机构

・ 2 40 ・

gen erati ng cut ting gen erat ri x Ge neva mecha nism gravit y grind grinder groove group dividing guidance guide ba r gyroscope Har monic h elicoid h elix h erringbone h ing e hob hoist horizon tal hour-glass hydraulic hypocycloid iden tifier idle illust rate im bala nce imita tion im pelle r incline inconsiste ncy increme nt index induction inequ alit y iner tia infe rence infinite inlet inspiration inst an t cent r e inst ru men t int eger

范成法切削 发生线 槽轮机构 重力 磨 , 研磨 磨床 , 砂轮 凹槽 , 开槽 拆杆组 引导 , 操纵 导杆 陀螺仪 谐波 ( 的 ) , 调和 ( 的 ) 螺旋面 , 螺状的 螺旋线 人字形 ( 的 ) 铰链 , 枢纽 滚刀 , 滚铣 卷扬 ( 机 ) , 起重 ( 机 ) 水平线 砂漏 液压装置 内摆线 标志符 , 鉴别器 空转 ( 的 ) 图解 不平衡 模仿 , 仿造品 推进器 , ( 水泵 ) 转子 ( 使 ) 倾斜 不一致 ( 性 ) 增加 , 增量 分度 ( 头 ) , 指标 , 目录 感应 ( 现象 ) , 归纳 不等式 , 不相等 惯性 ( 物 ) , 惯量 推理 , 推论 无限的 , 无穷大 入口 , 引进 启发 , 鼓舞 , 灵感 瞬心 仪器 , 手段 整数 , 整体

in tegrate in tegrity in terc ha ngeability in terfere nce in ter mediate in tersect in terval inverse inversion involut e jack jam journal key kin em atic diagram kin em atics label lat he lead lead screw lif t li ne of action li nk li nkage local view loco motive locus loom loosen lubrican t lubricate mach inery magnitude manipulat or marin e match maxim um mech anics mech anism mech anization mesh milling

( 求 ) 积分 , 集成 完整性 , 完善 可互换性 干涉 , 阻碍物 中间的 , 中间体 相交 , 交叉 区间 , 间隔 相反的 , 倒数 倒置 , 机架变换 渐开线 , 渐伸的 千斤顶 挤住 , 堵塞 轴颈 , 期刊 键 运动简图 运动学 标签 , 做记号 车床 引导 , 导程 , 螺距 , 铅 导杆 , 丝杆 举起 , 升降机 , 行程 啮合线 构件 , 杆 , 连接 连杆机构 局部图 火车头 轨迹 织布机 松开 , 放宽 润滑剂 , 润滑的 润滑 机械 大小 , 数量 机械手 , 操纵者 航海的 , 船舶 比赛 , 匹配 最大的 , 最大值 力学 机构 ( 学 ) , 机理 ( 制 ) 机械化 网孔 , 啮合 铣削

・ 2 41 ・

minimize minim um mobile modulate module mo men t of iner tia m ultiply n etwork nor mal notch nu merat or nu t offset omit op timiza tion orbit ordi nate orien tation oscillat e overrun Pack er pa rallel pa rallelogram pa ramete r passive D O F pa th way pawl ped al ped estal pe ndulu m pe riphery pe rp endicular phonograph pin pin ion pist on pitch pitch ci rcle pitch curve pivo t plasticine polar

最小化 最小的 , 最小值 可动的 , 汽车 , 运动物体 调节 模数 转动惯量 乘 , 增加 , 倍增 网络 法向 ( 的 ) , 法线 槽口 , 凹口 , 开槽 分子 螺母 偏移量 , 偏置 ( 量 ) 省略 , 忽略 最优化 轨道 , 沿轨道运行 纵坐标 方向 , 倾向性 振荡 超出限度 , 超过 打包机 平行 ( 的 ) , 平行线 平行四边形 参数 , 参量 局部自由度 导路 , 通道 棘爪 ( 踩 ) 踏板 底座 钟摆 , 摇锤 周边 , 圆周 , 外围 垂直的 , 垂线 留声机 钉, 销 小齿轮 活塞 齿距 , 周节 节圆 理论廓线 , 节线 枢轴 , 支点 橡皮泥 极坐标的 , 极性 ( 的 )

polygon polynomial portion positive-drive precision predete rmine preload prime circle prime move r prismatic pair profile proper ty pulley punc h punc h press qua dr an t qua dr ilat eral rack radius rail ratc het ratio readout rear reciprocate rectangle rectify reel refer ence cen t re dista nce refer ence circle regulat or relia bilit y represen tation reproduce rest rict resulta nt reversal righ t t r iangle rigidity rivet rock er rod

多边形 多项式 ( 的 ) 部分 强制驱动 精确 , 精密度 , 精度 预定 , 预先确定 预加载 基圆 原动机 移动副 外形 , 轮廓 性质 滑轮 冲床 , 冲孔 冲床 象限 四边形 , 四边 ( 的 ) 齿条 , 架子 半径 横杆 , 围栏 , 铁轨 棘齿 比率 读出器 , 读出 后面 ( 的 ) ( 使 ) 往复 ( 运动 ) 长方形 , 矩形 矫正 , 调整 卷轴 标准中心距 分度圆 调速器 , 调整器 可靠性 表示法 , 代表 再生 , 复制 限制 , 约束 , 限定 合成的 , 合力 , 合量 反转 , 反向 , 逆转 直角三角形 坚硬 , 刚性 铆钉 , 固定 摇杆 杆、竿

・ 2 42 ・

rolle r rot or saw too th scala r scope screw sect or segmen t self-tipping vehicle sense sensitivity seque nce servo setscrew seven-piece puzzle sh ape sh aper sh ea r shock side link side-t hrust sine sinusoid slan t sleeve slider slide-way slo t socket spacewidt h spindle spin spir al spline spring spur gea r stepping mot or step wise st reamlin e st roke st ruct ur e st rut

滚子 , 滚筒 , 辊子 转子 , 回转轴 锯齿 标量的 , 标量 范围 螺丝钉 , 螺旋 扇形 节 , 片断 , 线段 自卸车 感觉 , 方向 敏感 , 灵敏 ( 度 ) 次序 , 顺序 , 序列 伺服 , 伺服系统 紧定螺钉 七巧板 刨削 , 形状 , 定形 牛头刨床 剪 , 修剪 , 剪切 震动 , 冲突 连架杆 侧压力 正弦 正弦曲线 ( 使 ) 倾斜 , 倾斜 套筒 , 衬套 滑块 滑道 狭槽 , 开槽 窝 齿槽宽 锭子 , 主轴 旋转 螺旋形的 , 螺旋 ( 线 ) , 盘旋 花键 弹簧 直齿轮 步进电机 逐步的 流线型 ( 的 ) , 流水线 行程 , 冲程 , 敲 , 打击 结构 , 建筑物 支柱 , 支承

subrou tine subsc ript substit ute sub t ract sucke r swing tangen t ter minology th eor em th ickn ess tip ci rcle tole rance tole rate toot h space toot h t r ace torque t raject ory t rammel t ransfer t ransformation t ranslate t ransmission t ransmit t ranspor tation t rial t ria ngle turbine unde rcu t unfold unfolding a ngle universal join t valve vane vectorial velocity vice viewe r vi rt ual gear wear wit hdr aw workpiece wor m

子程序 下标 代用品 , 替代 减去 , 减 进气管 , 吸管 摇摆 切线 ( 的 ) , 相切的 , 正切 术语 定理 , 法则 厚度 , 浓度 齿顶圆 公差 , 容许量 忍受 , 容忍 齿间 , 齿槽 齿线 扭矩 , 转矩 轨道 , 弹道 , 轨线 拘束 , 阻碍 ( 物 ) 传递 , 转移 变化 , 转化 平移 , 转化 传动 , 传送 传输 , 转送 , 传达 运输 试验 , 考验 三角形 涡轮 根切 打开 展角 万向联轴节 阀 叶片 向量 ( 的 ) 速度 台钳 , 虎钳 阅读器 当量齿轮 磨损 , 穿 收回 , 撤消 工件 蜗杆

・ 2 43 ・

wor m gea r wor m w heel wr ap

蜗轮 蜗轮 包裹 , 包装 , 缠绕

yield yoke

产生 , 屈服 轭 , 轭状物

・ 2 44 ・

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32   唐晓腾 , 叶仲和 . 平底摆动推杆盘形凸轮实际廓线曲率半径 . 机械设计 , 1999 (5) : 10~ 13 33   叶仲和 . 按杆组装配顺序分解机构中的杆组 . 机械设计与研究 , 1987 (1 ) : 36 ~ 39 34   叶仲和 . 三齿轮连杆机构临界尺寸计算公式 . 机械设计与研究 , 1990 (2 ) : 38 ~ 39 35   叶仲和 , 刘鹰潭 . 凸轮廓线的通用公式 . 机械科学与技术 , 1991 (2 ) : 44 ~ 47 . 36   叶仲和 , 陈桂芳 . 滚子摆动推杆盘形凸轮机构的凸轮最 小基圆半 径和最 佳匹配 摆杆长 度 . 机械科 学与技 术 , 1995 ( 5) : 57~ 60 37   叶仲和 , 刘鹰潭 . 滚子摆动从动件盘形凸轮机构优化设计线图 . 机械科学与技术 , 1995 ( 6) : 37~ 116 . 38   叶仲和 , 唐晓腾 . 滚子摆动盘形凸轮机构优化设计方法的比较 . 机械设计 , 1999 ( 3) : 42~ 43 . 39   张世民主编 . 机械原理 . 北京 : 中央广播电视大学出版社 , 1993 40   郑文纬 , 吴克坚主编 . 机械原理 . 第 7 版 . 北京 : 高等教育出版社 , 1997 41   邹慧君主编 . 机械运动方案设计手册 . 上海 : 上海交通大学出版社 , 1994 42   邹慧君等 . 机械原理 . 北京 : 高等教育出版社 , 1999 43   王玉新编著 . 机构创新设计方法学 . 天津 : 天津大学出版社 , 1997

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