Chapter 06 - Vector Conversion

LECTURE 6

REPRESENTATION OF VECTORS IN DIFFERENT COORDINATE SYSTEM
6.01 VECTOR REPRESENTATION IN DIFFERENT COORDINATE SYSTEMS a) Rectangular Coordinate System r A = Ax a x + Ay a y + Az a z b) Cylindrical Coordinate System r A = Aρ a ρ + Aφ aφ + Az a z c) Spherical Coordinate System r A = Ar a r + Aθ aθ + Aφ aφ

6.02 DIRECTION OF UNIT VECTORS a) Rectangular Coordinate System Directions of unit vectors are invariant of the location.

Fig 6.1 Direction of unit vectors in rectangular coordinate system b) Cylindrical Coordinate System Directions of unit vectors a ρ and aφ are dependent of the value of φ . The direction of a z is invariant of the location.

Fig 6.2 Direction of unit vectors in cylindrical coordinate system c) Spherical Coordinate System Directions of unit vectors in Spherical coordinate are dependent of the value of θ and φ .

Fig 6.3 Direction of unit vectors in spherical coordinate system 6.03 VECTOR CONVERSION The conversion of vector components from one coordinate system to another coordinate system can be written in the form, r r A new = TA old r where: A new = vector in the new coordinate system r A old = vector in the old coordinate system

T = transformation matrix

VECTOR CONVERSION FORMULAS a) Rectangular to Cylindrical r r A cyl = Trec−to−cyl A rec

 Aρ   cos φ  A  = − sin φ  φ    0  Az   
b) Cylindrical to Rectangular r r A rec = Tcyl −to−rec A cyl

sin φ cos φ 0

0  Ax    0   Ay  1   Az  

 Ax  cos φ  A  =  sin φ  y    0  Az   
c) Rectangular to Spherical r r A sph = Trec−to− sph A rec

− sin φ cos φ 0

0  Aρ    0   Aφ  1   Az  

 Ar   sin θ cos φ     Aθ  = cosθ cos φ  Aφ      − sin φ
d) Spherical to Rectangular r r A rec = Tsph−to−rec A sph

sin θ sin φ cosθ sin φ cos φ

cosθ   Ax    − sin θ    Ay  0    Az  

 Ax  sin θ cos φ  A  =  sin θ sin φ  y    Az     cosθ

cosθ cosφ cosθ sin φ − sin θ

− sin φ   Ar    cos φ    Aθ   0    Aφ 

e) Cylindrical to Spherical r r A sph = Tcyl −to−sph A cyl

 Ar   sin θ     Aθ  = cos θ  Aφ      0

0 cos θ   Aρ    0 − sin θ    Aφ  1 0    Az  

f) Spherical to Cylindrical r r A sph = Tcyl −to−sph A cyl

 Aρ   sin θ A  =  0  φ   cos θ    Az 

cos θ 0 − sin θ

0  Ar    1   Aθ   0   Aφ 

6.04 EXAMPLES:

r Example 6.01 Convert vector B = 3a x + 0.50a y + 0.25a z at
R(3.052, 85.301°, 170.538°) in Cylindrical coordinates.
Solution:

 Bρ   cos(170.538°) sin(170.538°)  B  = − sin(170.538°) cos(170.538°)  φ   0 0  Bz     r In Cylindrical coordinates, B = −2.877a ρ − 0.986aφ

0  3    0   0.5  1  0.25  + 0.25a z

r Example 6.02 Convert vector B = 3a x + 0.50a y + 0.25a z at

R(3.052, 85.301°, 170.538°) in Spherical coordinates.
Solution: θ = 85.301°, φ = 170.538°

 Br   sin θ cos φ sin θ sin φ cosθ   3        Bθ  = cosθ cos φ cosθ sin φ − sin θ   0.5   Bφ   cos φ 0   0.25     − sin φ r In Spherical coordinates, B = −2.847a r − 0.485aθ − 0.986aφ

r Example 6.03 Convert vector C = 5a ρ − 0.30aφ + 0.50a z at
S (6.403, 128.66°, 25°) in Rectangular coordinates.
Solution:

Cx  cos(25°) − sin(25°) 0  5  C  =  sin(25°) cos(25°) 0 − 0.30    y   0 1   0.50    0   C z  r In Rectangular coordinates, C = 4.658a x + 1.841a y + 0.50a z

r Example 6.04 Convert vector C = 5a ρ − 0.30aφ + 0.50a z at
S (6.403, 128.66°, 25°) in Spherical coordinates.
Solution:

Cr   sin(128.66°) 0 cos(128.66°)   5       Cθ  = cos(128.66°) 0 − sin(128.66°) − 0.30 Cφ    0 1 0   0.50      r In Spherical coordinates, C = 3.592a r − 3.514aθ − 0.30aφ

r Example 6.05 Convert vector D = 10a r + 3aθ − 4aφ at T (7, 25°, 45°)
in Rectangular coordinates. Solution: θ = 25°, φ = 45°

 D x  sin θ cos φ  D  =  sin θ sin φ  y    Dz     cos θ

cos θ cos φ cos θ sin φ − sin θ

− sin φ   10    cos φ   3  0   − 4 

r In Rectangular coordinates, D = 7.739a x + 2.083a y + 7.795a z r Example 6.06 Convert vector D = 10a r + 3aθ − 4aφ at T (7, 25°, 45°)
in Cylindrical coordinates. Solution:

 Dρ   sin(25°) cos(25°) 0  10  D  =  0   0 1  φ   3    Dz    cos(25°) − sin(25°) 0   − 4  r In Cylindrical coordinates, D = 6.945a ρ − 4aφ + 7.795a z

6.05 QUESTIONS: 6.06 SUPPLEMENTARY PROBLEMS: