Car Physics for Games

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Car Physics for Games
by Marco Monster Level: Advanced Abstract: An introduction to car physics modelling for games. version 1.9 November, 2003

Introduction This tutorial is about simulating cars in games, in other words vehicle physics. One of the key points in simplifying vehicle physics is to handle the longtitudinal and lateral forces separately. Longtitudinal forces operate in the direction of the car body (or in the exact opposite direction). These are wheel force, braking force, rolling resistance and drag (= airresistance). Together these forces control the acceleration or deceleration of the car and therefore the speed of the car. Lateral forces allow the car to turn. These forces are caused by sideways friction on the wheels. We'll also have a look at the angular moment of the car and the torque caused by lateral forces. Notation and conventions Vectors are shown in bold, we'll be using 2d vectors. So the notation a = -b would translate to the following code snippet:
a.x = -b.x a.y = -b.y

Throughout this tutorial I'll be assuming that the rear wheels provide all the drive (for four wheel drives apply the neccesary adaptations). I'll be mainly using S.I. units (meters, kilograms, Newtons, etc.), but I've included a handy conversion table at the end for those readers more familiar with imperial measures (pounds, feet, miles, etc.) Straight line physics First let's consider a car driving in a straight line. Which forces are at play here? First of all there's what the tractive force, i.e. the force delivered by the engine via the rear wheels. The engine turns the wheels forward (actually it applies a torque on the wheel), the wheels push backwards on the road surface and, in reaction, the road surface pushes back in a forward direction. For now, we'll just say that the tractive force is equivalent in magnitude to the variable Engineforce, which is controlled directly by the user. Ftraction = u * Engineforce, where u is a unit vector in the direction of the car's heading. If this were the only force, the car would accelerate to infinite speeds. Clearly, this is not the

case in real life. Enter the resistance forces. The first and usually the most important one is air resistance, a.k.a. aerodynamic drag. This force is so important because it is proportional to the square of the velocity. When we're driving fast (and which game doesn't involve high speeds?) this becomes the most important resistance force. Fdrag = - Cdrag * v * |v| where Cdrag is a constant and v is the velocity vector and the notation |v| refers to the magnitude of vector v The magnitude of the velocity vector is more commonly known as the speed. Note the difference of data type: speed is a scalar, velocity is a vector. Use something like the following code:
speed = sqrt(v.x*v.x + v.y*v.y); fdrag.x = - Cdrag * v.x * speed; fdrag.y = - Cdrag * v.y * speed;

Then there is the rolling resistance. This is caused by friction between the rubber and road surface as the wheels roll along and friction in the axles, etc.etc.. We'll approximate this with a force that's proportional to the velocity using another constant. Frr = - Crr * v where Crr is a constant and v is the velocity vector. At low speeds the rolling resistance is the main resistance force, at high speeds the drag takes over in magnitude. At approx. 100 km/h (60 mph, 30 m/s) they are equal ([Zuvich]). This means Crr must be approximately 30 times the value of Cdrag The total longtitudinal force is the vector sum of these three forces. Flong = Ftraction + Fdrag + Frr Note that if you're driving in a straight line the drag and rolling resistance forces will be in the opposite direction from the traction force. So in terms of magnitude, you're subtracting the resistance force from the traction force. When the car is cruising at a constant speed the forces are in equilibrium and Flong is zero. The acceleration (a) of the car (in meters per second squared) is determined by the net force on the car (in Newton) and the car's mass M (in kilogram) via Newton's second law: a=F/M The car's velocity (in meters per second) is determined by integrating the acceleration over time. This sounds more complicated than it is, usually the following equation does the trick. This is known as the Euler method for numerical integration. v = v + dt * a, where dt is the time increment in seconds between subsequent calls of the physics engine. The car's position is in turn determined by integrating the velocity over time:

p = p + dt * v With these three forces we can simulate car acceleration fairly accurately. Together they also determine the top speed of the car for a given engine power. There is no need to put a maximum speed anywhere in the code, it's just something that follows from the equations. This is because the equations form a kind of negative feedback loop. If the traction force exceeds all other forces, the car accelerates. This means the velocity increases which causes the resistance forces to increase. The net force decreases and therefore the acceleration decreases. At some point the resistance forces and the engine force cancel each other out and the car has reached its top speed for that engine power.

In this diagram the X-axis denotes car velocity in meters per second and force values are set out along the Y-axis. The traction force (dark blue) is set at an arbitrary value, it does not depend on the car velocity. The rolling resistance (purple line) is a linear function of velocity and the drag (yellow curve) is a quadratic function of velocity. At low speed the rolling resistance exceeds the drag. At 30 m/s these two functions cross. At higher speeds the drag is the larger resistance force. The sum of the two resistance forces is shown as a light blue curve. At 37 m/s this curve crosses the horizontal traction force line. This is the top speed for this particular value of the engine power (37 m/s = 133 km/h = 83 mph). Magic Constants So far, we've introduced two magic constants in our equations: Cdrag and Crr . If you're not too concerned about realism you can give these any value that looks and feels good in your game. For example, in an arcade racer you may want to have a car that accelerates faster than any car in real life. On the other hand, if you're really serious about realistic simulation, you'll want to get these constants exactly right. Air resistance is approximated by the following formula (Fluid Mechanics by Landau and Lifshitz, [Beckham] chapter 6, [Zuvich])

Fdrag = 0.5 * Cd * A * rho * v2 where Cd = coefficient of friction A is frontal area of car rho (Greek symbol )= density of air v = speed of the car Air density (rho) is 1.29 kg/m3 (0.0801 lb-mass/ft3), frontal area is approx. 2.2 m2 (20 sq. feet), Cd depends on the shape of the car and determined via wind tunnel tests. Approximate value for a Corvette: 0.30. This gives us a value for Cdrag: Cdrag = 0.5 * 0.30 * 2.2 * 1.29 = 0.4257 We've already found that Crr should be approx. 30 times Cdrag. This gives us Crr = 30 * 0.4257 = 12.8 To be honest, I have my doubts about this last constant. I couldn't confirm its value anywhere. Be prepared to finetune this one to get realistic behaviour. Braking When braking, the traction force is replaced by a braking force which is oriented in the opposite direction. The total longtitudinal force is then the vector sum of these three forces. Flong = Fbraking + Fdrag + Frr A simple model of braking: Fbraking = -u * Cbraking In this model the braking force is a constant. Keep in mind to stop applying the braking force as soon as the speed is reduced to zero otherwise the car will end up going in reverse. Weight Transfer An important effect when accelerating or braking is the effect of dynamic weight transfer. When braking hard the car will nosedive. During accelerating, the car leans back. This is because just like the driver is pushed back in his seat when the pedal hits the metal, so is the car's centre of mass. The effect of this is that the weight on the rear wheels increases during acceleration and the front wheels conversely have less weight to bear. The effect of weight transfer is important for driving games for two reasons. First of all the visual effect of the car pitching in response to driver actions adds a lot of realism to the game. Suddenly, the simulation becomes a lot more lifelike in the user's experience. Second of all, the weight distribution dramatically affects the maximum traction force per wheel. This is because there is a friction limit for a wheel that is proportional to the load on that wheel:

Fmax = mu * W where mu is the friction coefficient of the tyre. For street tyres this may be 1.0, for racing car tyres this can get as high as 1.5. For a stationary vehicle the total weight of the car (W, which equals M *g) is distributed over the front and rear wheels according to the distance of the rear and front axle to the CM (c and b respectively): Wf = (c/L)*W Wr = (b/L)*W where b is the distance from CG to front axle, c the distance from CG to rear axle and L is the wheelbase.

If the car is accelerating or decelerating at rate a, the weight on front (Wf) and rear axle (Wr) can be calculated as follows: Wf = (c/L)*W - (h/L)*M*a Wr = (b/L)*W + (h/L)*M*a, where h is the height of the CG, M is the car's mass and a is the acceleration (negative in case of deceleration). Note that if the CG is further to the rear (c < b), then more weight falls on the rear axle and vice versa. Makes sense, doesn't it? If you want to simplify this, you could assume that the static weight distribution is 50-50 over the front and rear axle. In other words, assume b = c = L/2. In that case, Wf = 0.5*W - (h/L)*M*a and Wr = 0.5*W +(h/L)*M*a;

Engine Force

When I said earlier that the engine delivers a certain amount of force, this was a a bit of a simplification. An engine delivers an amount of torque. Torque is like a rotational equivalent of force. Torque is force times distance. If you apply a 10 Newton force at 0.3 meters of the axis of rotation, you've got a torque of 10 x 0.3 = 3 N.m (Newton meter). That's the same torque as when you apply a 1 N force at 3 m from the axis. In both cases the leverage is the same. The torque that an engine can deliver depends on the speed at which the engine is turning, commonly expressed as rpm (revolutions per minute). The relationship torque versus rpm is not a linear relationship, but is usually provided as a curve known as a torque curve (the exact shape and height of the curve is specific for each engine type, it is determined by engine tests). Here's an example for the 5.7 liter V8 engine found in Corvettes from 1997 to 2000: the LS1

Note that the torque curve peaks at about 4400 rpm with a torque of 350 lb-ft (475 N.m) and horsepower peaks at 5600 rpm at 345 hp (257 kW). The curves are only defined in the range from, in this particular case, about 1000 to 6000 rpm, because that is the operating range of the engine. Any lower, and the engine will stall. Any higher (above the so-called "redline"), and you'll damage it. Oh, and by the way, this is the maximum torque the engine can deliver at a given rpm. The actual torque that the engine delivers depends on your throttle position and is a fraction between 0 and 1 of this maximum. We're mostly interested in the torque curve, but some people find the power curve also interesting. You can derive the horsepower from the torque in foot-pounds using the following equation: hp = torque * rpm / 5252

Because of this relationship, the two curves will always cross at 5252 rpm. Check for yourself in the diagram above. And here's the same curves in SI units: Newton meter for torque and kiloWatt for power. The curves are the same shape, but the relative scale is different (and because of that they don't cross either). .

Now, the torque from the engine (i.e. at the crankshaft) is converted via the gear and differential before it's applied to the rear wheels. The gearing multiplies the torque from the engine by a factor depending on the gear ratios. Unfortunately, quite some energy is lost in the process. As much as 30% of the energy could be lost in the form of heat. This gives a so-called transmission efficiency of 70%. Let me just point out that I've seen this mentioned as a typical value, I don't have actual values for any particular car. The torque on the rear axle can be converted to a force of the wheel on the road surface by dividing by the wheel radius. (Force is torque divided by distance). By the way, if you want to work out the radius of a tyre from those cryptic tyre identfication codes, have a look at the The Wheel and Tyre Bible ( ). It even provides a handy little calculator. For example, this tells us the P275/40ZR-18 rear tyres of a Corvette have an unloaded radius of 34 cm. Here's an equation to get from engine torque to drive force: the longtitudinal force that the two

rear wheels exert on the road surface. Fdrive = u * Tengine * xg * xd * n / Rw where u is a unit vector which reflects the car's orientation, Tengine is the torque of the engine at a given rpm, xg is the gear ratio, xd is the differential ratio, n is transmission efficiency and Rw is wheel radius. An example: Engine is running at 2500 rpm, looking this up on the curve gives engine torque of 448 Nm (=330 ft lbs) Gear ratio (first gear): 2.66 Differential ratio: 3.42 Transmission efficiency: 0.7 (guess) Wheel radius: 0.34 m (=13.4 inch) Mass: 1500 kg (= 3300 lbs of weight) including the driver. This gives us a potential drive force of (448*2.66*3.42*0.7/0.34 = ) 8391 N if the driver puts his foot down. Meanwhile, in the static situation, the weight on the rear wheels is half the weight of the car and driver: (1500 kg / 2 ) * 9.8 m/s2 = 7350 N (=1650 lbs). This means the maximum amount of traction the rear wheels can provide if mu = 1.0 is 7350 N. Push the pedal down further than that and the wheels will start spinning and lose grip and the traction actually drops below the maximum amount. So, for maximum acceleration the driver must exert an amount of force just below the friction threshold. The subsequent acceleration causes a weight shift to the rear wheels. The acceleration is: a = 7350 N / 1500 kg = 4.9 m/s2 (=0.5 G) Let's say that b = c = 1.25m and L is therefore 2.50 m, the CG is 1.0 m above ground level. After a brief moment the amount of shifted weight is then (h/L)*M*a, that is (1.0/2.50)*1500*4.9 = 2940 N. This means Wf = 7350 - 2940 N and Wr = 7350 + 2940 N. The rear wheels now have extra weight which in this case is sufficient to allows the driver to put his foot all the way down. Gear Ratios The following gear ratios apply to an Corvette C5 hardtop (Source: ) First gear g1 2.66 1.78 Second gear g2 Third gear Fourth gear Fifth gear g3 g4 g5 1.30 1.0 0.74

Sixth (!) gear Reverse Differential ratio

g6 gR xd

0.50 2.90 3.42

Max torque 475 N.m (350 lb ft) at 4400 rpm, mass = 1439 kg (ignoring the driver for now). In first gear at max torque this gives us a whopping 475*2.66*3.42*0.7/0.33 = 9166 N of force. This will accelerate a mass of 1439 kg with 6.4 m/s2 (a=F/m) which equals 0.65 g. The combination of gear and differential acts as a multiplier from the torque on the crankshaft to the torque on the rear wheels. For example, the Corvette in first gear has a multiplier of 2.66 * 3.42 = 9.1. This means each Newton meter of torque on the crankshaft results in 9.1 Nm of torque on the rear axle. Accounting for 30% loss of energy, this leaves 6.4 N.m. Divide this by the wheel radius to get the force exerted by the wheels on the road (and conversely by the road back to the wheels). Let's take a 34 cm wheel radius, that gives us 6.4 N.m/0.34 m = 2.2 N of force per N.m of engine torque. Of course, there's no such thing as a free lunch. You can't just multiply torque and not have to pay something in return. What you gain in torque, you have to pay back in angular velocity. You trade off strength for speed. For each rpm of the wheel, the engine has to do 9.1 rpm. The rotational speed of the wheel is directly related to the speed of the car (unless we're skidding). One rpm (revolution per minute) is 1/60th of a revolution per second. Each revolution takes the wheel 2 pi R further, i.e. 2 * 3.14 * 0.34 = 2.14 m. So when the engine is doing 4400 rpm in first gear, that's 483 rpm of the wheels, is 8.05 rotations per second is 17.2 m/s, about 62 km/h. In low gears the gear ratio is high, so you get lots of torque but no so much speed. In high gears, you get more speed but less torque. You can put this in a graph as a set of curves, one for each gear, as in the following example.

Note that these curves assume a 100% efficient gearing. The engine's torque curve is shown as well for reference, it's the bottom one in black. The other curves show the torque on the rear axle for a given rpm of the axle (!), rather than the engine. As we've already seen, the rotation rate of the wheels can be related to car speed (disregarding slip for the moment) if we know the wheel radius. This means 1000 rpm of the rear axle is 36 m/s or 128 km/h car speed (80 mph). 2000 rpm is 256 km/h (160 mph). Etcetera. Drive wheel acceleration Now beware, the torque that we can look up in the torque curves above for a given rpm, is the maximum torque at that rpm. How much torque is actually applied to the drive wheels depends on the throttle position. This position is determined by user input (the accelerator pedal) and varies from 0 to 100%. So, in pseudo-code, this looks something like this: max torque = LookupTorqueCurve( rpm ) engine torque = throttle position * max torque You could implement the function LookupTorqueCurve by using an array of torque/rpm value pairs and doing linear interpolation between the closest two points. This torque is delivered to the drive wheels via the gearbox and results in what I'll call the drive torque: drive torque = engine_torque * gear_ratio * differential_ratio * transmission_efficiency Or written more concisely as:

Tdrive = Tengine * xg * xd * n Because Fdrive = Tdrive / Rw , this is the same as the equation for drive force we saw earlier. Note that the gearbox increases the torque, but reduces the rate of rotation, especially in low gears. How do we get the RPM? So we need the rpm to calculate the engine's max torque and from there the engine's actual applied torque. In other words, now we need to know has fast the engine's crankshaft is turning. The way I do it is to calculate this back from the drive wheel rotation speed. After all, if the engine's not declutched, the cranckshaft and the drive wheels are physically connected through a set of gears. If we know the rpm, we can calculate the rotation speed of the drive wheels, and vice versa! rpm = wheel rotation rate * gear ratio * differential ratio * 60 / 2 pi The 60 / 2 pi is a conversion factor to get from rad/s to revolutions per minute. There are 60 seconds in a minute and 2 pi radians per revolution. According to this equation, the cranckshaft rotates faster than the drive wheels. For example, let's say the wheel is rotating at 17 rad/s. Wheel rotates at 17 rad/s. First gear ratio is 2.66, differential ratio is 3.42 so crankshaft is rotating at 153 rad/s. That's 153*60 = 9170 rad/minute = 9170/2 pi = 1460 rpm at the engine. Because the torque curve isn't defined below a certain rpm, you may need to make sure the rpm is at least at some minimum value. E.g.
if( rpm < 1000 ) rpm = 1000;

This is needed to get the car into motion from a standstill. The wheels aren't turning so the rpm calculation would provide zero. At zero rpm, the engine torque is either undefined or zero, depending how your torque curve lookup is implemented. That would mean you'd never be able to get the car moving. In real life, you'd be using the clutch in this case, gently declutching while the car starts moving. So wheel rotation and engine rpm are more or less decoupled in this situation. There are two ways to get the wheel rotation rate. The first one is the easiest, but a bit of a quick hack. The second one involves some more values to keep track of over time, but is more accurate and will allow for wheel spins, etcetera. The easy way is to pretend the wheel is rolling and derive the rotation rate from the car speed and the wheel radius. For example, let's say the car is moving at 20 km/h = 20,000 m / 3600 s = 5.6 m/s. wheel radius is 0.33 m, so wheel angular velocity is 5.6/0.33 = 17 rad/s Plug this into the previous equations to get the 1460 rpm, from which we can look up the engine

torque on the torque curve. The more advanced way is to let the simulation keep track of the rear wheel rotation rate and how it changes in time due to the torques that act on the rear wheels. In other words, we find the rotation rate by integrating the rotational acceleration over time. The rotational acceleration at any particular instant depends on the sum of all the torques on the axle and equals the net torque divided by the rear axles inertia (just like linear accelaration is force divided by mass). The net torque is the drive torque we saw earlier minus the friction torques that counteract it (braking torque if you're braking and traction torque from the contact with the road surface). Slip ratio and traction force Calculating the wheel angular velocity from the car speed is only allowed if the wheel is rolling, in other words if there is no lateral slip between the tyre surface and the road. This is true for the front wheels, but for drive wheels this is typically not true. After all, if a wheel is just rolling along it is not transfering energy to keep the car in motion. In a typical situation where the car is cruising at constant speed, the rear wheels will be rotating slighty faster than the front wheels . The front wheels are rolling and therefore have zero slip. You can calculate their angular velocity by just dividing the car speed by 2 pi times the wheel radius. The rear wheels however are rotating faster and that means the surface of the tyre is slipping with regard to the road surface. This slip causes a friction force in the direction opposing the slip. The friction force will therefore be pointing to the front of the car. In fact, this friction force, this reaction to the wheel slipping, is what pushes the car forwards. This friction force is known as traction or as the longtitudinal force. The traction depends on the amount of slip. The standardised way of expressing the amount of slip is as the so-called slip ratio:

where (pronounced: omega) is wheel angular velocity (in rad/s) Rw is wheel radius (in m) vlong is car speed a.k.a. longtitudinal velocity (in m/s)

Note: there are a number of slightly different definitions for slip ratio in use. This particular definition also works for cars driving in reverse. Slip ratio is zero for a free rolling wheel. For a car braking with locked wheels the slip ratio is -1, and a car accelerating forward has a positive slip ratio. It can even be larger than 1 of there's a lot of slip. The relationship between longtitudinal (forward) force and slip ratio can be described by a curve such as the following:

Note how the force is zero if the wheel rolls, i.e. slip ratio = 0, and the force is at a peak for a slip ratio of approximately 6 % where the longtitudinal force slightly exceeds the wheel load. The exact curve shape may vary per tyre, per road surface, per temperature, etc. etc. That means a wheel grips best with a little bit of slip. Beyond that optimum, the grip decreases. That's why a wheel spin, impressive as it may seem, doesn't actually give you the best possible acceleration. There is so much slip that the longtitudinal force is below its peak value. Decreasing the slip would give you more traction and better acceleration. The longtitudinal force has (as good as) a linear dependency on load, we saw that earlier when we dicussed weight transfer. So instead of plotting a curve for each particular load value, we can just create one normalized curve by dividing the force by the load. To get from normalized longtitudinal force to actual longtitudinal force, multiply by the load on the wheel. Flong = Fn, long * Fz where Fn, long is the normalized longtitudinal force for a given slip ratio and Fz is the load on the tyre. For a simple simulation the longtitudinal force can be approximated by a linear function: Flong = Ct * slip ratio where Ct is known as the traction constant, which is the slope of the slip ratio curve at the origin. Cap the force to a maximum value so that the force doesn't increase after the slip ratio passes the peak value. If you plot that in a curve, you get the following:

Torque on the drive wheels To recap, the traction force is the friction force that the road surface applies on the wheel surface. Obviously, this force will cause a torque on the axis of each drive wheel: traction torque = traction force * wheel radius This torque will oppose the torque delivered by the engine to that wheel (which we called the drive torque). If the brake is applied this will cause a torque as well. For the brake, I'll assume it delivers a constant torque in the direction opposite to the wheel' s rotation. Take care of the direction, or you won't be able to brake when you're reversing. The following diagram illustrates this for an accelerating car. The engine torque is magnified by the gear ratio and the differential ratio and provides the drive torque on the rear wheels. The angular velocity of the wheel is high enough that it causes slip between the tyre surface and the road, which can be expressed as a positive slip ratio. This results in a reactive friction force, known as the traction force, which is what pushed the car forward. The traction force also results in a traction torque on the rear wheels which opposes the drive torque. In this case the net torque is still positive and will result in an acceleration of the rear wheel rotation rate. This will increase the rpm and increase the slip ratio.

The net torque on the rear axle is the sum of following torques:

total torque = drive torque + traction torques from both wheels + brake torque Remember that torques are signed quantities, the drive torque wil generally have a different sign than the traction and brake torque. If the driver is not braking, the brake torque is zero. This torque will cause an angular acceleration of the drive wheels, just like a force applied on a mass will cause it to accelerate. angular acceleration = total torque / rear wheel inertia. I've found somewhere that the inertia of a solid cylinder around its axis can be calculated as follows: inertia of a cylinder = Mass * radius2 / 2 So for a 75 kg wheel with a 33 cm radius that's an inertia of 75 * 0.33* 0.33 / 2 = 4.1 kg.m2 You must double that to get the total inertia of both wheels on the rear axle and perhaps add something for the inertia of the axle itself, the inertia of the gears and the inertia of the engine. A positive angular acceleration will increase the angular velocity of the rear wheels over time. As for the car velocity which depends on the linear acceleration, we simulate this by doing one step of numerical integration each time we go through the physics calculation: rear wheel angular velocity += rear wheel angular acceleration * time step where time step is the amount of simulated time between two calls of this function. This way we can determine how fast the drive wheels are turning and therefore we can calculate the engine's rpm. The Chicken and the Egg Some people get confused at this point. We need the rpm to calculate the torque, but the rpm depends on the rear wheel rotation rate which depends on the torque. Surely, this is a circular definition, a chicken and egg situation? This is an example of a Differential Equation: we have equations for different variables that mutually depend on each other. But we've already seen another example of this before: air resistance depends on speed, yet the speed depends on air resistance because that influences the acceleration. To solve differential equations in computer programs we use the technique of numeric integration: if we know all the values at time t, we can work out the values at time t + delta. In other words, rather than trying to solve these mutually dependent equations by going into an infinite loop, we take snapshots in time and plug in values from the previous snapshot to work out the values of the new snapshot. Just use the old values from the previous iteration to calculate the new values for the current iteration. If the time step is small enough, it will give you the results you need. There's a lot of theory on differential equations and numeric integration. One of the problems is that a numeric integrator can "blow up" if the time step isn't small enough. Instead of wellbehaved values, they suddenly shoot to infinity because small errors are multiplied quickly into

larger and larger errors. Decreasing the time step can help, but costs more cpu cycles. An alternative is to use a smarter integrator, e.g. RK4. For more information on this topic, see my article on numerical stability:Achieving a Stable Simulator Curves Okay, enough of driving in a straight line. How about some turning? One thing to keep in mind is that simulating the physics of turning at low speed is different from turning at high speed. At low speeds (parking lot manoeuvres), the wheels pretty much roll in the direction they're pointed at. To simulate this, you need some geometry and some kinetics. You don't really need to consider forces and mass. In other words, it is a kinetics problem not a dynamics problem. At higher speeds, it becomes noticeable that the wheels can be heading in one direction while their movement is still in another direction. In other words, the wheels can sometimes have a velocity that is is not aligned with the wheel orientation. This means there is a velocity component that is at a right angle to the wheel. This of course causes a lot of friction. After all a wheel is designed to roll in a particular direction and that without too much effort. Pushing a wheel sideways is very difficult and causes a lot of friction force. In high speed turns, wheels are being pushed sideways and we need to take these forces into account. Let's first look at low speed cornering. In this situation we can assume that the wheels are moving in the direction they're pointing. The wheels are rolling, but not slipping sideways. If the front wheels are turned at an angle delta and the car is moving at a constant speed, then the car will describe a circular path. Imagine lines projecting from the centre of the hubcabs of the front and rear wheel at the inside of the curve. Where these two lines cross, that's the centre of the circle. This is nicely illustrated in the following screenshot. Note how the green lines all intersect in one point, the centre round which the car is turning. You may also notice that the front wheels aren't turned at the same angle, the outside wheel is turned slightly less than the inside wheel. This is also what happens in real life, the steering mechanism of a car is designed to turn the wheels at a different angle. For a car simulation, this subtlety is probably not so important. I'll just concentrate on the steering angle of the front wheel at the inside of the curve and ignore the wheel at the other side.

This is a screen shot from a car physics demo by Rui Martins. More illustrations and a download can be found at

The radius of the circle can be determined via geometry as in the following diagram:

The distance between front and rear axle is known at the wheel base and denoted as L. The radius of the circle that the car describes (to be precise the circle that the front wheel describes) is called R. The diagram shows a triangle with one vertex in the circle centre and one at the centre of each wheel. The angle at the rear wheel is 90 degrees per definition. The angle at the front wheel is 90 degrees minus delta. This means the angle at the circle centre is also delta (the sum of the angles of a triangle is always 180 degrees). The sine of this angle is the wheel base divided by the circle radius, therefore:

Note that if the steering angle is zero, then the circle radius is infinite, i.e. we're driving in a straight line. Okay, so we can derive the circle radius from the steering angle, now what? Well, the next step is to calculated the angular velocity, i.e. the rate at which the car turns. Angular velocity is usually represented using the Greek letter omega ( ), and is expressed in radians per second. (A radian is a full circle divided by 2 pi). It is fairly simple to determine: if we're driving circles at a constant speed v and the radius of the circle is R, how long does it take to complete one circle? That's the circumference divided by the speed. In the time the car has described a circular path it has also rotated around its up-axis exactly once. In other words:

By using these last two equations, we know how fast the car must turn for a given steering angle at a specific speed. For low speed cornering, that's all we need. The steering angle is determined from user input. The car speed is determined as for the straight line cases (the velocity vector always points in the car direction). From this we calculate the circle radius and the angular velocity. The angular velocity is used to change the car orientation at a specific rate. The car's speed is unaffected by the turn, the velocity vector just rotates to match the car's orientation.

High Speed Turning Of course, there are not many games involving cars that drive around sedately (apart from the legendary Trabant Granny Racer ;-). Gamers are an impatient lot and usually want to get somewhere in a hurry, preferably involving some squealing of tires, grinding of gearboxes and collateral damage to the surrounding environment. Our goal is to find a physics model that will allow understeer,oversteer, skidding, handbrake turns, etc. At high speeds, we can no longer assume that wheels are moving in the direction they're pointing. They're attached to the car body which has a certain mass and takes time to react to steering forces. The car body can also have an angular velocity. Just like linear velocity, this takes time to build up or slow down. This is determined by angular acceleration which is in turn dependent on torque and inertia (which are the rotational equivalents of force and mass). Also, the car itself will not always be moving in the direction it's heading. The car may be pointing one way but moving another way. Think of rally drivers going through a curve. The angle between the car's orientation and the car's velocity vector is known as the sideslip angle (beta).

Now let's look at high speed cornering from the wheel's point of view. In this situation we need to calculate the sideways speed of the tires. Because wheels roll, they have relatively low resistance to motion in forward or rearward direction. In the perpendicular direction, however, wheels have great resistance to motion. Try pushing a car tire sideways. This is very hard because you need to overcome the maximum static friction force to get the wheel to slip. In high speed cornering, the tires develop lateral forces also known as the cornering force. This force depends on the slip angle (alpha), which is the angle between the tire's heading and its direction of travel. As the slip angle grows, so does the cornering force. The cornering force per tire also depends on the weight on that tire. At low slip angles, the relationship between slip angle and cornering force is linear, in other words Flateral = Ca * alpha where the constant Ca is known as the cornering stiffness. If you'd like to see this explained in a picture, consider the following one. The velocity vector of the wheel has angle alpha relative to the direction in which the wheel can roll. We can split the velocity vector v up into two component vectors. The longtitudinal vector has magnitude cos(alpha) * v. Movement in this direction corresponds to the rolling motion of the wheel. The lateral vector has magnitude sin(alpha) * v and causes a resistance force in the opposite direction: the cornering force.

There are three contributors to the slip angle of the wheels: the sideslip angle of the car, the angular rotation of the car around the up axis (yaw rate) and, for the front wheels, the steering angle. The sideslip angle β (bèta) is the difference between the car orientation and the direction of movement. In other words, it's the angle between the longtitudinal axis and the actual direction of travel. So it's similar in concept to what the slip angle is for the tyres. Because the car may be moving in a different direction than where it's pointing at, it experiences a sideways motion. This is equivalent to the perpendicular component of the velocity vector.

If the car is turning round the centre of geometry (CG) at a rate omega (in rad/s!), this means the front wheels are describing a circular path around CG at that same rate. If the car turns full circle, the front wheel describes a circular path of distance 2.pi.b around CG in 1/( seconds where b is the distance from the front axle to the CG. This means a lateral velocity of omega * b. For the rear wheels, this is -omega * c. Note the sign reversal. To express this as an angle, take the arctangent of the lateral velocity divided by the longtitudinal velocity (just like we did for beta). For small angles we can approximate arctan(vy/vx) by vx/vy.

The steering angle (delta) is the angle that the front wheels make relative to the orientation of the car. There is no steering angle for the rear wheels, these are always in line with the car body

orientation. If the car is reversing, the effect of the steering is also reversed.

The slip angles for the front and rear wheels are given by the following equations:

The lateral force exercised by the tyre is a function of the slip angle. In fact, for real tyres it's quite a complex function once again best described by curve diagrams, such as the following:

What this diagram shows is how much lateral (sideways) force is created for any particular value of the slip angle. This type of diagram is specific to a particular type of tyre, this is a fictitious but realistic example. The peak is at about 3 degrees. At that point the lateral force even slightly exceeds the 5KN load on the tyre.

This diagram is similar to the slip ratio curve we saw earlier, but don't confuse them. The slip ratio curve gives us the forward force depending on amount of longtitudinal slip. This curve gives us the sideways (lateral) force depending on slip angle. The lateral force not only depends on the slip angle, but also on the load on the tyre. This is a plot for a load of 5000 N, i.e. the weight exerted by about 500 kg of mass pushing down on one tyre. Different curves apply to different loads because the weight changes the tyre shape and therefore its properties. But the shape of the curve is very similar apart from the scaling, so we can approximate that the lateral force is linear with load and create a normalized lateral force diagram by dividing the lateral force by the 5KN of load. Flateral = Fn, lat * Fz where Fn, lat is the normalized lateral force for a given slip angle and Fz is the load on the tyre. For very small angles (below the peak) the lateral force can be approximated by a linear function: Flateral = Ca * alpha The constant Ca goes by the name of the cornering stiffness. This is the slope of the diagram at slip angle 0. If you want a better approximation of the the relationship between slip angle and lateral force, search for information on the so-called Pacejka Magic Formula, names after professor Pacejka who developed this at Delft University. That's what tyre physicists use to model tyre behaviour. It's a set of equations with lots of "magic" constants. By choosing the right constants these equations provide a very good approximation of curves found through tyre tests. The problem is that tyre manufactors are very secretive about the values of these constants for actual tyres. So on the one hand, it's a very accurate modeling technique. On the other hand, you'll have a hard time finding good tyre data to make any use whatsoever of that accuracy. The lateral forces of the four tyres have two results: a net cornering force and a torque around the yaw axis. The cornering force is the force on the CG at a right angle to the car orientation and serves as the centripetal force which is required to describe a circular path. The contribution of the rear wheels to the cornering force is the same as the lateral force. For the front wheels, multiply the lateral force with cos(delta) to allow for the steering angle. Fcornering = Flat, rear + cos(delta) * Flat, front As a point of interest, we can find the circle radius now that we know the centripetal force using the following equation Fcentripetal = M v2 / radius The lateral force also introduce a torque which causes the car body to turn. After all, it would look very silly if the car is describing a circle but keeps pointing in the same direction. The cornering force makes sure the CG describes a circle, but since it operates on a point mass it does nothing about the car orientation. That's what we need the torque around the yaw axis for. Torque is force multiplied by the perpendicular distance between the point where the force is applied and the pivot point. So for the rear wheels the contribution to the torque is -Flat, rear * c and for the front wheels it's cos(delta) * Flat, front * b. Note that the sign differs.

Applying torque on the car body introduces angular acceleration. Just like Newton's second law F = M.a, there is a law for torque and angular acceleration: Torque = Inertia * angular acceleration. The inertia for a rigid body is a constant which depends on its mass and geometry (and the distribution of the mass within its geometry). Engineering handbooks provide formulas for the inertia of common shapes such as spheres, cubes, etc. Epilogue You can download a demo (car demo v0.8) and the source code which demonstrates high speed cornering using the method just described. Or easier yet, try the Java version by Marcel "Bloemschneif" Sodeike!

The following information is shown on screen:
• • • • • • •

steering angle (also known as delta) side slip angle (also known as beta) "rotation angle" (due to yaw rate) slip angle front wheels slip angle rear wheels velocity vector relative to car reference frame velocity vector in world space

See also my work in progress on the game Downtown Drivin' for an example of some of the stuff that was discussed here. If you spot mistakes in this tutorial or find sections that could do with some clarification, let me know. -MarcoReferences [Beckham] Physics of Racing Series, Brian Beckham, [Bower] Richard Bower, [Hecker] Chris Hecker, series on Physics in Game Development Magazine, [Lander] Jeff Lander, The Trials and Tribulations of Tribology, [RQRiley] Automobile Ride, Handling and Suspension Design, R.Q. Riley Enterprises, [Zuvich] Vehicle Dynamics for Racing Games, Ted Zuvich,, note that you need to obtain a (free) Gamasutra id to access the game development conference archives. Car specifications It's handy to know what sort of values to use for all these different constants we've seen. Preferably some real values from some real cars. I've collected what I could for a Corvette C5. Be warned: some of the data is still guesswork. Units and Conversions The following S.I. units apply:

Force Power Torque Speed

N (Newton) W (Watt) N.m (Newton meter) m/s

= = N.m/s = J/s = m2 kg / s3

Angular velocity rad/s Acceleration Mass Distance m/s2 kg m

Angle rad =360 degrees/ 2 pi Here are some handy unit conversions to S.I. units.

1 mile 1 ft (foot) 1 in (inch) 1 km/h 1 mph 1 rpm (revolution per minute) 1G 1 lb (pound) 1 lb (pound) 1 lb.ft (foot pounds)

= 1.6093 km = 0.3048 m = 0.0254 m = 2.54 cm = 0.2778 m/s = 1.609 km/h = .447 m/s = 0.105 rad/s = 9.8 m/s2 = 32.1 lb/s2 = 4.45 N = 0.4536 kg 1) = 1 lb/1G = 1.356 N.m

1 lb.ft/s (foot pound per second) = 1.356 W 1 hp (horsepower) = 550 1 metric hp = 0.986 hp = 745.7 W = 735.5 W

1) To say a pound equals so and so much kilograms is actually nonsense. A pound is a unit of force and a kilogram is a unit of mass. What we mean by this "conversion" is that one pound of weight (which is a force) equals the weight exerted by 0.4536 kg of mass assuming the gravity is 9.8m/s2. On the moon, a kilogram will weigh less but still have the same mass. Unit converter on the web:
Marco Monster E-mail: Website: Copyright 2000-2003 Marco Monster. All rights reserved. This tutorial may not be copied or redistributed without permission.

The Physics of Racing, Part 26: The Driving Wheel, Chapter I Brian Beckman, PhD ©Copyright July 2001
Imagine 400 ft-lbs of torque measured on a chassis dynamometer like a DynoJet (see references at the end). This is a very nice number to have in any car, street or racing. In dyno-speak, however, the interpretation of this number is a little tricky. To start a dyno session, you strap your car down with the driving wheels over a big, heavy drum, then you run up the gears gently and smoothly until you're in fourth at the lowest usable engine RPM, then you floor it, let the engine run all the way to redline, then shut down. The dyno continuously measures the time and speed of the drum and the engine RPM through a little remote radio receiver that picks up spark-plug noise. The only things resisting the motion of the driving wheels are the inertia of the driveline in the car and the inertia of the drum. The dyno 'knows' the latter, but not the former. Without these inertias loading the engine, it would run up very quickly and probably blow up. Test-stand dynos, which run engines out of the car, load them in different ways to prevent them from free running to annihilation. Some systems use water resistance, others use electromagnetic in any case, the resistance must be easy to calibrate and measure. We are only concerned with chassis dynos in this article, however. What is the equation of motion for the car + dyno system? It is a simple variation on the theme of the old, familiar second law of Newton. For linear motion, that law has the form F = ma, where F is the net force on an object, m is its mass, and a is its acceleration, or time rate of change of velocity. For rotational motion, like that of the driveline and dyno, Newton's second law takes the form where T is the net torque on an object, J is its moment of inertia, and is its angular acceleration, or time rate of change of angular velocity. The purpose of this instalment of the Physics of Racing is to explain everything here and to run a few numbers. To get the rotational equation of motion, we assume that the dyno drum is strong enough that it will never fly apart, no matter how fast it spins. We model it, therefore, as a bunch of point masses held to the centre of rotation by infinitely strong, massless cables. With enough point masses, we can approximate the smooth (but grippy) surface of the dyno drum as closely as we would like.

Assume each particle receives a force of F / N in the tangential direction. Tangential, of course, means the same as circumferential or longitudinal, as clarified in recent instalments about slip and grip of tyres. So, each particle accelerates according to F / N = ma / N. The N cancels out, leaving a = F / m. Now, a is the rate of change of the velocity, and the velocity is defined as , where r is the constant radius of the circle and is the angular velocity in radians per second. The circumference of the circle is , by definition, so a drum of 3-foot radius has a circumference of about 6.28 x 3 = 18.8 feet. At 60 RPM, which is one rev per second, each particle goes 18.8 feet per second, which is about 15 x 18.8/22 = 12.8 mph. RPM is one measurement of angular velocity, but it's more convenient to measure it such that angular units go by every second. Such units save us from having to track factors of all over the math. So, there are radians per revolution, and the equation is seen as a general expression of the example that x revolutions per second = 18.8 feet per second = velocity. Since r is constant, it has no rate of change. Only has one, measured in radians per second per second, or radians per second squared, or , and denoted with an overdot: . The equation of motion, so far, looks like . Now, we know that torque is just force times the lever arm over which the force is applied. So, a force of F at the surface of the drum translates into a torque of T = Fr applied to the shaft-or by the shaft, depending on point of view. So we write , which we can rearrange to . We can make this resemble the linear form of Newton's law if we define , the moment of inertia of the drum, yielding < , which looks just like>F = ma if we analogise as follows: . This value for J only works for this particular model of the drum, with all the mass elements at

distance r from the centre. Suffice it to say that a moment of inertia for any other model of the drum could be computed in like manner. It turns out that the moment of inertia of a solid cylindrical drum is half as much, namely . Moments of inertia for common shapes can be looked up all over the place, for instance at So, now, the dyno has a known, fixed value for J, and it measures very accurately. This enables it to calculate trivially just how much torque is being applied by the driving wheels of the car to the drum. But it does not know the moment of inertia of the driveline of the car, let alone the radius of the wheels, the gear selected by the driver, the final-drive ratio in the differential, and so on. In other words, it knows nothing about the driveline other than engine RPM. Everyone knows that the transmission and final-drive on a car multiply the engine torque. The torque at the driving wheels is almost always much larger than the flywheel torque, and it's larger in lower gears than in higher gears. So, if you run up the dyno in third gear, it will accelerate faster than if you run it up in fourth gear. Yet, the dyno reports will be comparable. Somehow, without knowing any details about the car, not even drastic things like gear choice, the dyno can figure out flywheel torque. Well, yes and no. It turns out that all the dyno needs to know is engine RPM. It does not matter whether the dyno is run up quickly with a relatively large drive-wheel torque (DWT) or run up slowly with a relatively small DWT. Furthermore, the radius of the driving wheels and tyres also does not matter. Here's why. Wheel RPM is directly proportional to drum RPM, assuming the longitudinal slip of the tyres is within a small range. The reason is that at the point of contact, the drum and wheel have the same circumferential (longitudinal, tangential) speed, so Let's write , where . Engine RPM is related to wheel RPM by a . We .

factor that depends on the final-drive gear ratio f and the selected gear ratio write expect

. Usually, engine RPM is much larger than wheel RPM, so we can to be larger than 1 most of the time. So, we get

We also know that, by Newton's Third Law, that the force applied to the drum by the tyre is the same as the force applied to the tyre by the drum. Therefore the torques applied are in proportion to the radii of the wheel + tyre and the drum, namely that


. Recalling that the transmission gear and final drive multiply engine torque, , so . But we already know , or, more

we also know that : it's the ratio of the RPMs, so usefully,

Every term on the right-hand side of this equation is measured or known by the dyno, so we can measure engine torque independently of car details! We can even plot effectively taking the run-up time and the drum data out of the report. versus ,

Almost. There is a small gotcha. The engine applies torque indirectly to the drum, spinning it up. But the engine is also spinning up the clutch, transmission, drive shaft, differential, axles, and wheels, which, all together, have an unknown moment of inertia that varies from car-to-car, though it's usually considerably smaller than J, the moment of inertia of the drum. But, in the equations of motion, above, we have not accounted for them. More properly, we should write

This doesn't help us much because we don't know the equation:

, so we pull a fast one and rearrange

This is why chassis dyno numbers are always lower than test-stand dyno numbers for the same engine. The chassis dyno measures , and the test-stand measures . Of course, those trying to sell engines often report the best-sounding numbers: the test-stand numbers. So, don't be disappointed when you take your hot, new engine to the chassis dyno after installation and get numbers 15% to 20% lower than the advertised 'at the crankshaft' numbers in the brochure. It's to be expected. Typically, however, you simply do not know you take on faith. : it's a number

Let's run a quick sample. The following numbers are pulled out of thin air, so don't hang me on them. Suppose the drum has 3-foot radius, is solid, and weighs 6,400 lbs, which is about 200 slugs (remember slugs? One slug of mass weighs about 32 pounds at the Earth's surface). So, the moment of inertia of the drum is about . Let's say that the engine takes about 15 seconds to run from 1,500 RPM to 6,000 RPM in fourth gear, with a time profile like the following: t e RPM V MPH v FPS drum 0 1,500 1 1,800 2 2,100 3 2,400 4 2,700 5 3,000 6 3,300 7 3,600 8 3,900 9 4,200 10 4,500 11 4,800 12 5,100 13 5,400 14 5,700 15 6,000 35 51.33 42 61.60 49 71.87 56 82.13 63 92.40 70 102.67 77 112.93 84 123.20 91 133.47 98 143.73 105 154.00 112 164.27 119 174.53 126 184.80 133 195.07 140 205.33 17.11 20.53 23.96 27.38 30.80 34.22 37.64 41.07 44.49 47.91 51.33 54.76 58.18 61.60 65.02 68.44 drum drum RPM RPM ratio Torque 0.00 163.40 0.1089 0.00 3.42 3.42 3.42 3.42 3.42 3.42 3.42 3.42 3.42 3.42 3.42 3.42 3.42 3.42 3.42 196.08 228.76 261.44 294.12 326.80 359.48 392.16 424.84 457.52 490.20 522.88 555.56 588.24 620.92 653.60 0.1089 335.51 0.1089 335.51 0.1089 335.51 0.1089 335.51 0.1089 335.51 0.1089 335.51 0.1089 335.51 0.1089 335.51 0.1089 335.51 0.1089 335.51 0.1089 335.51 0.1089 335.51 0.1089 335.51 0.1089 335.51 0.1089 335.51

The " v MPH" column is just a straight linear ramp from 35 MPH to 140 MPH, which are approximately right in my Corvette. The " v FPS" column is just 22 / 15 the v MPH. The drum is in radians per second and is just v FPS divided by 3 ft, the drum radius. The drum is just the stepwise difference of the drum numbers. It's constant, as we would expect from a run-up of the dyno at constant acceleration. The drum RPM is times the drum . The RPM ratio is just drum RPM divided by engine RPM, and it must be strictly constant, so this is a nice sanity check on our math. Finally, the torque column is the RPM ratio times J = 900 slug - ft2 times drum . We see a constant torque output of about 335 ft-lbs. Not bad. It implies a test-stand

number of between 394 and 418, corresponding to 15% and 20% driveline loss, respectively. Looks like we nailed it without 'cooking the books' too badly. Of course, we have a totally flat torque curve in this little sample, but that's only because we have a completely smooth ramp-up of velocity. Dyno reports often will be labelled 'Rear-wheel torque' (RWT) or, less prejudicially, 'drive-wheel torque' (DWT) to remind the user that there is an unknown component to the measurement. These are well intentioned misnomers: do not be mislead! What they mean is 'engine torque as if the engine were connected to the drive wheels by a massless driveline', or 'engine torque as measured at the drive wheels with an unknown but relatively small inertial loss component'. It should be clear from the above that the actual drive-wheel torque cannot be measured without knowing A, the ratio of the drum radius to the wheel + tyre radius. It's slightly ironic that an attempt to clear up the confusion risks introducing more confusion. In the next instalment, we relate the equations of motion for the driving wheel to the longitudinal magic formula to compute reaction forces and get equations of motion for the whole car. References: [sic] Attachments: I've included the little spreadsheet I used to simulate the dyno run. It can be downloaded here. ERRATA: * Part 14, yet again, the numbers for frequency are actually in radians per second, not in cycles per second. There are cycles per radian, so the 4 Hz natural suspension frequency I calculated and then tried to rationalize was really 4 / 6.28 Hz, which is quite reasonable and not requiring any rationalization. Oh, what tangled webs we weave… * Physical interpretations of slip on page 2 of part 24: " Car (hub) moving forward, CP moving slowly forward w.r.t. ground, resisting car motion." Should be " Car (hub) moving forward, CP moving slowly forward w.r.t. HUB, resisting car motion." * Part 21, in the back-of-the-envelope numerical calculation just before the 3-D plot at the end of the paper, I correctly calculated tan-1(0.8222) = 0.688, but then incorrectly calculated tan-1(SB) SB as -0.266. Of course, it's 0.688 - 0.822 = -0.134. One of the hazards of doing math in one's head all the time is the occasional slip up. Normally, I check results with a calculator just to be really sure, but some are so trivial it just seems unnecessary. Naturally, those are the ones that bite me. More info

Drivetrain Layout
When I open the front hood of my Mazda Protege sedan, I find its engine and many other goodies. Same for my S2000. I also saw many people opening their front hood to inspect the engine. This gave me an impression that all cars have engine stored under their front hood. I used to think this is the truth. But after going to my first car show, I realized that I was like a toad that was living in the bottom of a well. I am always an admirer of Porsche. They made a lot of uncompromising performance car while maintain the exquisite styling. [Lately they disappointed me with the introduction of automatic 911 Turbo and Porsche SUV(!)] So naturally the first car I wanted to check out at the car show is the legendary Porsche 911 Turbo. After watching it inside out, I lifted the front hood to look for the engine that powers this Godly machine. Heck! There was nothing there, under the hood it is like a front trunk for people to store things. I found the rear trunk too but still no engine. I asked people around and finally someone answered me. That person says the 911 Turbo is a rear-engine car. The engine is mounted after the rear driving axle. So it is located under the rear trunk. I finally find the engine by unlatching a lock. It is under the smallish rear trunk. Well, why do Porsche put their engines at the back of a car? There was another thing about cars that bugged me. I am an avid player of the Gran Turismo racing game. Initially, I was baffled by the many designations like FF, FR, MR, RR and so on. I later learned that the first letter of the designation is used to designate the engine location and the second letter is to designate the which pair of wheels the engine drives.

Drivetrain Layout
The car body, the front panel, the rear trunk, the seats and other amenities that come with the car are just distractions to a performance enthusiast. What makes a car move are its engine, its transmission and its wheels. Collectively speaking, these three components comprise the drivetrain.

Engine Positioning
There are two major characteristics of a drivetrain that impacts the performance of a car: the engine placement and the driving wheels location. The engine placement is a big factor to determine the moment of inertia and the weight distribution of car because many other mechanical/electrical components of a car are usually located close to the engine. The driving wheels location determines which wheels the transmission to send the engine-generated torque to. Due to weight transfer, you will soon find out that driving wheels location is a very important factor in car handling. For simplicity purposes, car nuts classified different engine placements into three types: frontengined, mid-engined and rear-engined. Front-engined cars have their engine placed in front of the passenger seats. Mid-engined cars have their engine placed behind the passenger seats but in front of the rear driving axle. Rear-engined cars have their engine placed behind the rear driving

axle. There is also a subclass of front-engined cars called front-mid-engined. In this subclass, the engine is behind the front drving axle but in front of the passenger seats. An example is the Honda S2000.

Driving Wheels
As most people know, the engine isn't necessarily driving all the four wheels. There are actually four types of driving wheels. The first is All Wheel Drive (AWD) in which all four wheels are driven by the engine all the time. Then there is Front Wheel Drive (FWD) in which only the front wheels are driven by the engine. It then follows by Rear Wheel Drive (RWD) in which only the rear wheels are driven. The final one is an odd one. It is called Four Wheel Drive (4WD). It is usually employed by Sports Utility Vehicles (SUV). There is a switch in the car that allows you switch between AWD and FWD mode. Since 4WD doesn't really introduce anything new, so by reading materials related to AWD and FWD should help you completely understand 4WD.

Some common Drivetrain Layouts
In this section, I will give you examples of cars that represents the seven common drivetrain layouts. There will also be some simple description of their handling and acceleration characteristics. Acceleration characteristic will be explained in details at the bottom of this page. However, to understand the handling characteristics, you need to understanding the concepts described in the Cornering Chapter.
1. Front Engine, Front Wheel Drive (FF)

This is most common drivetrain layout. It is used for all the low cost economy car like Toyota Camry/Corolla, Honda Civic/Accord, Mazda Protege/Millenium, etc. FF cars are more front heavy. It can counteract with the understeer characteristics exhibited by front wheel drive cars. The overall effect is that it has slight understeer in all acceleration situations. This actually makes the car more stable in city driving. The main reason for building FF cars is that it is cheaper to build. Steering, engine, transmission, wheels and so on are all very close by, there is no need to build long axles to transmit the engine power to the other end of the car.



2. Front Egnine, Rear Wheel Drive (FR)

This is most common drivetrain layout for luxury sedans and low end sports car. Examples are Mercedes sedans, BMW sedans, Mazda Miata, Honda S2000. One of the characteristics of these cars is that they usually have a almost neutral weight distribution due to the driving axle that traverses from the front to the rear. Since the weight distribution is neutral, to attain higher acceleration potential, the car needs to be RWD or AWD (see the bottom for math-



oriented people). For not powerful enough engines in passenger cars, RWD is good enough to exploit the potential.

RWD cars exhibit oversteer under mild acceleration and understeer under heavey acceleration. The oversteer characteristic allows an RWD car to accelerate after exits from the apex and hence attain higher speed when it enters the straight. For details, please refer to the Understeer and Oversteer section and the Cornering Line section. Front-mid-engined car like S2000 can turn faster than normal FR cars because it has a smaller moment of inertia.


3. Mid Engine, Rear Wheel Drive (MR)

This drivetrain layout is usually employed by high end sports car and most of the formula one race cars. Notable examples are Porsche Boxster, Ferrari Modena. Mid-engined is the configuration that has the lowest moment of inertia and hence it turns the fastest. Weight Distribution is a little bit biased to the rear and hence more prone to oversteer under mild acceleration.

o o

4. Rear Engine, Rear Wheel Drive (RR)
o o

This is one of the rare drivetrain layouts. Notable examples are Porsche 911 Carrera and the original Volkswagen Beetle. Rear-engined cars are similar to mid-engined cars but they have higher moment of inertia and are even more prone to oversteer under mild acceleration.

5. Front Engine, All Wheel Drive (FA)

There are two types of cars that employ this drivetrain layout. The first type includes cars that want to provide traction on all four tires such that you can move the car around in snow or unfavorable terrain. Examples are Subarus, Audis, BMW 330xi. The second type includes high power sports car. Examples are Nissan Skyline GTR, Mistubishi Lancer Evolution and so on. For high power sports cars, the reason for AWD is to exploit all the traction of the four tires to attain the greatest acceleration possible (see bottom for the gory math details). Most AWD cars are rear biased in which they allocate more torque to the rear than the front. Therefore they all have mild oversteer under mild acceleration.



6. Mid Engine, All Wheel Drive (MA)

MA car was built in the same spirit as FA cars but the mid engine

configuration reduces moment of inertia and hence makes the car turn more quickly. An example is Lamborghini Murcielago. 7. Rear Engine, All Wheel Drive (RA)

RA cars are an extension of RR cars. They take advantage of the AWD to exploit full acceleration potential. An example is Porsche 911 Carrera 4.

Quantitative Effects of Drivetrain Layout during acceleration
Here we would like to use a simple example to quantify the effects of different driving wheels. We will still use our Skyline GT-R as an example. This time in addition to the default all wheel drive (AWD) setting, we will also study the cases when it is front wheel drive (FWD) and rear wheel drive (RWD). To compare, we will check to see what is the maximum force that can be applied to the ground in each case.

The Math
First, let's assume for a while that our Skyline is FWD. For this kind of car, the maximum force that can be applied to the ground is limited by the limiting friction on the front tires. From the Weight Transfer section, we know that
Wf* = (mgLr - hFf - hFr) / (Lf + Lr)

Since this is FWD, Fr is 0. Note that the maximum of Ff is μWf*. Beyond that, the front tires will slip and hence no force can be transmitted to the ground. By substituting Wf* with Ff/μ, we can solve for maximum Ff:
Ff/μ =(mgLr - hFf) / (Lf + Lr) Ff =μmgLr / (Lf + Lr + μh) 0.94×1550×9.8×1.519 / (1.146 + 1.519 = + 0.94×0.34) =7267.04N

Next we consider the case when it is RWD. In this case, the maximum force is limited by the limited friction on the rear tires. Again, from the Weight Transfer section, we know that
Wr* = (mgLf + hFf + hFr) / (Lf + Lr)

This time, Ff is 0. The maximum force will be limited by μFr. Beyond that, no force can be transmitted to the ground. Again, by subsituting Wr* with μFr, we can solve for maximum Fr:
Fr/μ =(mgLf + hFr) / (Lf + Lr)


=μmgLf / (Lf + Lr - μh) 0.94×1550×9.8×1.146 / (1.146 + 1.519 = - 0.94×0.34) =6976.75N

For the AWD case, thanks to all the all powerful technology called Limited Slip Differential (LSD), the math is way simpler. LSD can distribute the torque generated by the engine to not yet slipped tires. So during acceleration, when the rear tires start to slip, it can transfer the torque to the front tires that still has traction. Therefore, the car basically won't slip until all four tires slip. To make all four tires slip, you have to apply a force greater than μWf* + μWr* = μmg = 0.94×1550×9.8 = 14278.6N.

As we can see AWD allows the most torque to be transferred to the ground. Then what about RWD and FWD? In our example, FWD allows for great acceleration. But if you look at the formulas more carefully, you can discover that there is a condition for this to happen. Suppose we want to know what is the condition such that RWD works better than FWD, we look at the condition Fr >= Ff:
μmgLf μmgLr

----------------- > -----------------= --Lf + Lr - μh μh Lf + Lr + μh > L - Lf = r

Note that Lr and Lf is determined by the weight distribution of the car. The higher the weight is at the front, the longer Lr is and the shorter Lf is (Note that Lf+Lr is a constant). The converse holds true for rear-heavy cars. For a perfectly balanced (ie 50:50) car, Lr equals Lf and hence this condition holds. But as we can see if the front is heavy enough such that Lr is way longer than Lf, then FWD will prevail. Now we can check whether AWD is the correct configuration for our Skyline GT-R. Recall that our car gives out 392.27Nm of torque at 4400rpm. When it is transmitted to the rear tires at the first gear, it is Γ×g1×G/r = 392.27×3.827×3.545/0.3266 = 16,294.6N. In reality, there should be about 20% loss, so only 13,035.68N can be transmitted to the tires. Note that this is beyond what an RWD GT-R can handle but below what an AWD GT-R can handle. Therefore we can conclude that the engineers at Nissan did the right thing to make GT-R an AWD car. After all these analysis, a question naturally arises: so if AWD is better than FWD and RWD for powerful engines, why the race cars still use MR configuration? The reason is that installing an AWD system put more weight on the car. For an example, compare the curb weight of BMW

325i and 325xi sedans: 1,463kg vs 1,573kg. There is about 110kg difference. For BMW 325 sedan, there is an increase of 7.5% in weight and hence 7% reduction in acceleration. However, for a 500kg race car, the weight increases by 22% and the acceleartion decreases by 18%! Discussions related to understeer and oversteer in relation to the drivetrain layout will be discussed in the Understeer and Oversteer section. First Draft: January 18th, 2004 1.0 Published: February 29th, 2004 Yee Man Chan

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Frequently we hear riders talk about how many horsepower their motorcycle makes. These claims are often not believed, and sometimes not believable. Also, we hear some riders state that they like torque more than horsepower. In this article, I'll explain exactly what torque and horsepower are, and why some motorcycles are dramatically faster than others. Your motorcycle moves forward because the tire pushes against the road. This is the basic fact we'll use to understand everything else. Push is a non-technical word, and engineers, like doctors and lawyers, love to use more complicated words. The engineering word for push is Force, and a force applied by something round like a wheel is called torque. So, torque is a fancy word for a push by a wheel. More torque equals more push, which equals more acceleration, which wins races.

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In chart 1, there's a graph of torque vs. rpm for two Harley-Davidson big twins. Rpm is, of course, revolutions per minute - the number of times your engine crankshaft turns over each minute, and the number which is reported by your tachometer. One of the Harleys on the chart is a stock California model, and the other is modified "for off-road use only". A surprising fraction of all Harley big twins are apparently purchased for dirt riding and closed course competition. This particular modified bike has a Crane Fireball 310 cam, a Mikuni carburetor, and Cycle Shack exhaust pipes. As you can see, the modified engine produces about 25% to 50% more torque, and therefore 25% to 50% more acceleration. Since velocity is acceleration times time, we would expect the modified machine to take about 1/3 less time to go 0 to 60. A popular performance test is a 50 mph, 200 yard roll-on: full throttle for 200 yards, starting at 50 mph. Let's see what our modified Harley will do in these circumstances. At 50 mph in 5th gear, a big twin is doing about 2200 rpm. At 70 mph, it's doing about 3000 rpm. Over the range of 2000 to 3000 rpm, the modified engine is only producing about 5-10 more foot- pounds of torque than the stock engine; a gain of about 10%. The final speed in a 200 yard roll-on improves from about 72 to about 73. This is not so impressive as the 30% reduction in 0 to 60 times. The calculations used to figure this out are in the "mathematics" sidebar. The problem is that the modified engine is producing a lot more power, but at a higher rpm. In 0 to 60 tests, we run the engine up to the red line, but in the roll-on, we never get much beyond 3000 rpm. If we did the same roll on test in 3rd gear, we would expect the stock engine to get to about 74, and the modified engine to get to about 80. Now, we get our complication: the torque is applied at the rear wheel, but it is not produced by the wheel, it's produced by the engine. If the wheel and the engine were directly connected, as they are in a speedway bike, we'd be done. But, on almost all bikes, there's a transmission. The

gear ratios in the transmission divide the engine rpm and the torque, with higher gears producing less rpm and less rear wheel torque at a given speed. The best way to understand this is to make a picture. We'll graph the torque available at the rear wheel at various road speeds. Of course, this graph is going to look a little complicated, because, for example, at 30 mph you could be in any of the five gears, which means you can choose between five available rear-wheel torques and five engine rpm's. To make this graph, first we need to know how fast the bike goes in its various gears. Harley Davidson supplies the overall gear ratios in their product literature, so all we need is one calibration point. A motorcycle magazine recently reported than in 5th gear at 60 mph a '94 Harley is doing 2570 rpm. Since all '94 Harley big twins use the same rear tire, transmission, and belt pulleys, they all share the same overall gear ratios. Using this information, I filled in the numbers in table 1. For example, at 4000 rpm in 5th gear, the Harley is going 93 mph (60 mph * 4000 / 2570). At 3000 rpm in 3rd gear, the Harley is going 44 mph (60 mph * 3000 / 2570 * 3.15 for fifth gear / 5.05 for third gear). If you own an earlier evolution big twin, then here are the rules: in '94 all big twins use 29 tooth and 70 tooth pulleys. In earlier years, some used 27 tooth and 70 tooth pulleys, and some used 27 tooth and 61 tooth pulleys. If you have a 27-70 (FXR's and glides), multiply all speeds in this article by 27/29, about .93. If you have a 27-60 (most softtails), multiply all speeds in this article by 70/61*27/29, about 1.07. Table 1: Road speed at various rpm in each gear of a '94 Harley big twin
RPM 1st gear (10.212) 2nd gear (6.96) 3rd gear (5.05) 4th gear (3.86) 5th gear (3.15) 2000 3000 4000 5000 6000 14 mph 22 mph 29 mph 36 mph 43 mph 21 mph 32 mph 42 mph 52 mph 63 mph 29 mph 44 mph 58 mph 72 mph 87 mph 38 mph 57 mph 75 mph 94 mph 113 mph 47 mph 70 mph 93 mph 117 mph 140 mph

Now, this is not to say a Harley will necessarily go 140. This just says that if a '94 big twin were ever in 5th gear doing 6000 rpm, it would be doing 140 mph. In fact, the stock Harley rev limiter kicks in at 5200 rpm, and a California big twin will actually only do about 95-105. Also, different model years of Harleys have different gear ratios. Now, using this table, we can construct our real-wheel torque vs. road speed graph. To do this, we look at the torque curve in chart 1, and at the road speeds in table 1. At 2000 rpm in 1st gear, the bike is going 14 mph. At 2000 rpm, the engine is making 65 foot-pounds of torque. At the rear wheel, this is 65*10.212 = 664 foot-pounds of torque. So, at 14 mph, we make a dot at 664 foot-pounds for 1st gear. At 4000 rpm in 3rd gear, the bike is going 58 mph and the engine is making 55 foot-pounds, which is 55*5.05 = 278 foot-pounds at the rear wheel, so we make a dot at 58 mph and 278 foot-pounds for 3rd gear. I'm sure you see how this goes now, so, here's the resulting graph:

In chart 2, we see a graph of the rear wheel torque at various road speeds in each gear. If you look at 50 mph, you see that there is 200 foot-pounds of torque available in 5th gear, about 240 ft-lbs in 4th gear, about 270 ft-lbs in 3rd gear, and 320 ft-lbs in 2nd gear. Thus, down-shifting this engine from 5th to 3rd gear at 50 mph only gets us about 35% more torque; similarly, upshifting from 3rd gear to 5th gear at 50 mph only costs us about 28% of our torque. Thus we see that stock Harleys don't change their performance very much when you shift gears. In fact, at 80 mph, 5th and 3rd are almost identical in performance, and 4th is only about 10% better. The arrows in the graph point at the places where the rear wheel torque curves cross going from one gear to the next. These are the optimum shift points: at about 40, 56, 73, and 95 mph. However, we also see that with this engine, there is very little difference if we shift directly from 2nd to 4th at about 55, or from 3rd to 5th at about 70. We must remember that these shift points are for a '94 Harley; different years have different gearing and different shift points. Since we put in the dots at 2000, 3000, 4000, 5000, and 6000 rpm, we can estimate the optimum shifting rpm: about 5500, 5500, 5000, and 5000 rpm on a 94 Harley. Actually, these rpm estimates are not too bad for nearly any nearly stock Harley big twin.

In chart 3, we see overlaid on the stock torque curves the torque curves for the modified engine. I've left off the dots for the stock CA Harley torque curves to make the chart more readable. From chart 1, we see that the modified engine torque peak is at 4000 rpm. Thus, if we have the modified engine, and we are willing to run it at 3500 to 5500 rpm, we can see from chart 3 that the performance will be much better - there's more available torque, and therefore more available push and acceleration. On the other hand, if we shift at 3000 rpm, as many Harley riders do, there will be very little improvement. For example, at 3000 rpm in first gear, the bike is doing about 22. At 22, there is very little difference between the stock and modified engines in first gear, and there is also very little difference between them in second gear at 22 mph. So, our modified engine will pull much harder, and go a lot faster, but only at a much higher rpm. In chart 3, the shift points are indicated for both the stock and the modified engines. We see that they are not very different: 42, 57, 78, and 102 compared to 40, 56, 73, and 95 mph. However, we see that shifting gears in the modified engine produces a large effect, while shifting gears in the stock motor produces relatively little effect. At 50, when we down-shifted the stock motor from 5th to 3rd, the torque increased from 200 to 270 - about 30%. With the modified motor, the torque increases from about 220 to 400 - almost double. However, as long as we stay in 5th gear, there's not much difference between the two motors. Again, these numbers are all for a '94 Harley; different years with different gearing will have different shift points in mph. The modified engine has shift points we can estimate at 5700, 5500, 5400, and 5300 rpm. These numbers are consistently a bit higher than for the stock engine, and above the stock 5200 rpm rev limiter, but within reasonable limits for the engine. Evolution engines are pretty well engineered and can take a reasonable amount of abuse.

Now we see a general rule about making power. In order to make more power, mostly you have to spin the engine faster. Also, each time you shift gears, the torque gets divided by the same ratio that the rpm gets divided. So, to go fast, you want a motor where you don't have to shift gears as often. In fact, if two bikes make the same peak torque - say, 80 foot pounds - the motor which makes the torque at a higher rpm can stay in lower gears longer and therefore go faster. This makes us think it would be useful to have a single number which reported both the peak torque and the peak rpm. In fact, there is just such as number: horsepower. Horsepower equals torque times rpm; thus an engine which makes 80 foot pounds of torque at 8000 rpm makes twice as many horsepower as another engine which makes the same 80 foot pounds at 4000 rpm. Although the two engines deliver the same amount of torque, and therefore the same amount of push, the first engine can go twice as fast in each gear. While I'm doing 60 topped out in second gear on my 80 foot-pound Harley, some other guy can be doing 60 not quite topped out in first gear on his 80 foot-pound Kawasaki ZX-11. Of course, he's winning the race.

It's about time we considered some different bikes. In chart 4, we see the torque curves for several different engines overlaid. The bikes are all about the same in terms of engine size - a bit over one liter. Interestingly, with the exception of the California Harley and the BMW, we see that they all produce about the same amount of peak torque - about 80 foot- pounds. We all know that Harley engines can make significantly more power than what the factory supplies. Based on these curves, we suspect the same is true for the BMW.

In chart 5, we see the horsepower curves for the same five bikes. In spite of the similarity of the peak torque, the horsepower curves are very different. Notice that the modified Harley, which makes the most torque up to almost 5000 rpm, also makes the most horsepower up to 5000 rpm. The Honda ST1100 makes the most torque and horsepower from 5000 to 6500 rpm, and the Kawasaki makes the most from there up. Let's see how to calculate horsepower. There's no big trick at all. From chart 1, we see that a stock California Harley is making 65 foot-pounds at 2000 rpm, 64 ft-lbs at 3000, 55 at 4000, and about 45 at 5000. Horsepower is torque times rpm divided by 5252; so this Harley is making 25 horsepower at 2000 rpm, 37 hp at 3000 rpm, 42 at 4000, and 43 at 5000. (Yes, it's true: stock California big twins make about 43 horsepower. 49 state big twins make about 50 horsepower.) Here's something interesting: the torque peaks at about 2000 rpm, but the horsepower peaks at about 5000 rpm. Even though the torque is going down, the rpm is going up faster than the

torque is going down, and therefore the horsepower is increasing. Also, we see from the formula that at 5252 rpm, horsepower is equal to torque. This is always true. If your dyno tells you that the horsepower and torque curves cross anywhere other than 5252 rpm, your dyno is broken. Finally, we see that at any given rpm, the engine which makes the most torque at that rpm also makes the most horsepower at that rpm. To make more horsepower, an engine has to either make more torque or run at higher rpm. The ZX-11 makes the same 80 foot-pounds of torque as the modified Harley, but it makes this torque at 8000 rpm - exactly twice the 4000 rpm peak of the modified Harley. Because of this, the ZX-11 makes twice the horsepower - 120 hp - at 8000 rpm that the modified Harley makes at 4000 rpm - 60 hp - even though both are making 80 footpounds of torque. Since the Harley torque curve drops off a bit gradually from 4000 to 5000 rpm, and the ZX-11 torque curve drops more quickly from 8000 to 10000 rpm, the ZX-11 doesn't quite make twice as much horsepower as the modified Harley - "only" 130 hp compared to 71 hp. Kawasaki riders need not totally despair - the ZX-11 ram air system is not active on a dyno. It is believed that at 150 mph, the Kawasaki is making about 145 hp due to the ram air effect. We see that if all you care about is engine peak torque, you would be equally happy with the Harley and the ZX-11. If all you care about is Horsepower, you definitely want the ZX-11. If what you're really interested in is power at low rpm, now you definitely want the Harley. So, those riders who claim to prefer torque over horsepower really mean they prefer engines with power at low rpm. Let's see how this horsepower advantage translates into a torque advantage. We'll make the same picture for the ZX-11 rear wheel torque that we made for the Harleys. In table 2 below, there are the gear ratios and road speed worked out for the ZX-11 (the gear ratio data was supplied by U.S.Kawasaki). The ZX-11 has a six speed transmission. I've worked out the road speed in 2,000 rpm increments from 2,000 rpm to 10,000 rpm. The ZX-11 actually has a red line of 11,000 rpm, so each gear actually goes a little faster than the table shows. For those of you who aren't familiar with ZX-11's, this bike has been clocked by radar guns at 172 mph, stock. It does 135 mph in the quarter mile in about 10.3 seconds. These days, the ZX11 pretty much defines fast. The ZX-11 engine torque curve is illustrated in chart 4 above. We do the same cross-reference look up exercise as before, and we get the rear wheel torque curves. Interestingly, these curves seem to indicate that, for racing purposes, fourth gear could be eliminated with no effects, and fifth gear is only useful from 120 mph to 135 mph. It looks like removing fourth and fifth gear and replacing them with one gear with the ratio 5.372 (31/25 internally) would be an improvement.

In chart 6, we see a direct comparison of the rear wheel torque of a modified Harley, and a Kawasaki ZX-11. As you can see, the ZX-11 has a much larger advantage over the modified Harley than the modified Harley has over the stock Harley. Not only does the ZX-11 make more rear wheel torque, it can also go faster in first gear than the Harley can go in second gear, and it can go about as fast in third gear as the Harley can go in fifth. And, in fact, it is now well established that a ZX-11 can pretty much beat any Harley, any where, any time, in any race. So, how come Harleys outsell ZX-11's about a hundred to one?
Table 2: Road speed at various rpm in each gear of a '94 Kawasaki ZX-11
RPM 2000 4000 6000 8000 10000 1st gear (12.133) 12 mph 25 mph 37 mph 49 mph 62 mph 2nd gear (8.905) 17 mph 34 mph 50 mph 67 mph 84 mph 3rd gear (6.890) 22 mph 44 mph 65 mph 87 mph 109 mph 4th gear (4.996) 26 mph 52 mph 78 mph 104 mph 130 mph 5th gear (3.15) 30 mph 60 mph 90 mph 120 mph 150 mph 6th gear (4.485) 33 mph 67 mph 100 mph 134 mph 167 mph

In the Fall 1994 issue of American Rider, Joe Minton presents a graph of horsepower for a Harley 1340. His article did not specify the modifications made, but in the context of his article, we can safely assume they include substantial head work. His engine's torque curve can be easily deduced from the horsepower curve: at each rpm, we multiply the horsepower by 5252 and divide by the rpm. The result indicates an engine making between 87 and 93 foot-pounds of torque from 2500 to 6250 rpm. This is an engine with the same torque per cubic inch as a ZX-11, and a power band which is 25% wider. If true, this is quite an accomplishment.

In chart 7, I have graphed the rear wheel torque curves for Mr. Minton's bike. As you can see, the performance would be substantially better than either the stock or the lightly modified Harley. Also, we see that this engine is poorly mated to its gear box: optimum shifting points would occur at 7,500 to 6,500 rpm. This level of rpm severely over stresses an evolution big twin. It appears that a change of camshafts to move the torque curve down 500 to 1000 rpm would give this bike performance better suited to its gear box and engine design parameters. With Mr. Minton's bike, we would expect the 0 to 60 times to be cut roughly in half compared to a stock CA Harley, and the 50 mph roll-on speed to be about 75 mph in fifth, and 86 mph in third. We would also expect that when shifting from first gear to second at 6000 rpm, the bike's acceleration would drop dramatically. Mark Lawrence has a degree in Engineering. He has been riding motorcycles since 1969. In this time he has owned 21 bikes, ridden 305,000 miles through 28 states, accumulated 27 traffic tickets, and set a California state record for traffic school attendance. He currently rides a Road

King with a modified "closed course competition use only" motor. He has not yet raced his Road King in a Grand Prix, but expects to do well if he ever does. Sidebar 1: The Mathematics For those who are interested, here's a brief summary of the physics and mathematics used in this article.
1. force = mass * acceleration 2. torque = force * moment arm length 3. distance = 1/2 * acceleration * time * time 4. velocity = acceleration * time = force / mass * time 5. time = velocity * mass / force 6. time = square root( 2 * distance / acceleration ) 7. velocity = square root( 2 * distance * acceleration ) 8. velocity = square root( 2 * distance * force / mass ) 9. horsepower = torque * rpm / 5252 10.rear wheel torque = engine torque * primary gear ratio * transmission gear ratio * final gear ratio 11.rear wheel force = rear wheel torque * rear wheel radius 12.rear wheel radius is almost always 1 foot, +/- about 1 inch, for any motorcycle

To calculate the roll-on performance, here's what I did: published final speed = 72 mph. Modified bike has 10% greater torque, average, over 50-80 mph. By equation 5, velocity = K * square root( force ) new velocity / old velocity = square root( new force / old force ) = square root( 1.1 ) = 1.05 old velocity = 72 - 50 = 22 mph. 5% of 22 mph = 1 mph. new velocity = 50 + 22 + 1 = 73 mph. For third gear modified bike, I noted that the modified bike is making about 370 foot-pounds of rear-wheel torque in third gear from 50 to about 80, and it's making about 220 foot-pounds from 50 to 80 in fifth gear, so the math is the same up to: new velocity / old velocity = square root( new force / old force ) = square root( 380 / 220 ) = square root( 1.7 ) = 1.3 30% of 23 is 7, so the modified bike hits about 50 + 23 + 7 = 80 mph in third gear.

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Engine torque to Wheel torque calculation

Is there a way to calculate engine torque to wheel torque? What I am looking for is a way to calculate (if possible) the amount of torque at the wheels taking into account the engine torque at a specific RPM and gear. (This is to show some unbeliever why a car like the NSX with such 'low' torque can & will outrun a diesel-engined car which (officially) has much more engine-torque) Answer: Yes, just multiply the engine torque at RPM by the gear ratio and final drive ratio. Or multiply the Horsepower by 5252 and divide by the RPM of the tire. Or multiply the Horsepower by 5252, by final drive ratio, by trans gearing, and divide by the RPM of the engine. All three will give the same result, but did not subtract for frictional losses. Bob

By Rick Johnson of TOO TECH RACING

Rick graduated from General Motors Institute in 1977. He has worked with Dirt bike Suspension since 1982. He raced the local Pro class in the 80’s and still rides and competes regularly in the over 40 Master A and Pro classes. Rick continues to perform shock-

tuning trackside at a variety of Southern California Tracks. BACKGROUND There seems to be a lot of confusion regarding how tension in the chain affects a motorcycle suspension. Contrary to many track side opinions, the tension in the chain during acceleration does not cause the rear end to sag, but instead, actually reduces the frame’s tendency to squat. To prove this, perform this simple experiment: Put your bike in 1st gear directly in front of some immovable object such as a tree. Rev up the motor and slowly let out the clutch loading the motor. Notice how the rear fender rises under this load. The net squat that one sees between the rear wheel and the fender during acceleration is caused from the vehicle’s weight being transferred to the rear wheel and the rear spring deflecting under this increased load. The chain tension force is actually trying to raise the rear end! Riders use chain tension to their advantage. For instance, if you gas it as you go up a jump ramp the bike will fly higher and farther. If you chop the gas as you go off a jump the bike will stay lower and not go as far in the air. This is all because of tension in the chain. Read on for the technical explanation of chain tension and how it affects bike action. What effects I will explain: To better understand the effects of chain tension, we will look at the forces the chain exerts on the rear wheel and frame, how these forces affect suspension performance, and then discuss how the rider can use these forces to his advantage on the track. Then we will discuss the pros and cons of the ‘Torque Eliminator” product’s ability to eliminate these chain reactions. We will choose some numbers for engine torque, bike & rider weight, and calculate the approximate lifting force generated by the tension in the chain. We will ignore some other forces acting during acceleration so we can focus on chain tension and the associated lifting forces. This is not completely accurate, but close enough to make the point of this analysis. Sketch Explanations: Straight lines with arrows indicate forces and their direction of application. Circular lines indicate torque with the arrow showing the direction of application. The numbers indicate magnitude. To understand the effects of a force (straight arrows) drawn at some angle other than vertical or horizontal, we “resolve” the angled force into its horizontal and vertical components. In our example, ‘X’ will indicate horizontal and ‘Y’ will indicate vertical forces. All forces must balance at some force equilibrium condition. When we twist the throttle, the tension in the chain is added to the force equilibrium analysis. This added force causes a new equilibrium condition and a short-term vertical acceleration (rear end lift) to get to this point! CHAIN TENSION CALCULATIONS AND HOW THEY AFFECT THE FRAME We will assume a total weight of 400 lbs for bike and rider. We will look at suspension reactions with all the weight shifted to the rear wheel modeling maximum acceleration. We will place the rear swing arm in the horizontal position to keep things simple. First we must approximate the tension in the chain. To accelerate, the engine produces torque at the drive sprocket; which applies tension to the chain; which then

drives the rear sprocket. We will choose a 250cc engine that produces about 25 ft/lbs of torque. The second gear ratio is about 2 to 1 and the primary reduction ratio is about 2.5 to 1. Total engine torque multiplication in second gear equals 5 to 1; thus drive sprocket torque is 125 ft/lbs (25x5=125). For any torque, the force exerted is dependent on the leverage arm length. In our example the front sprocket diameter is 3.5”. The radius is 1.75” or .145 ft. Dividing 125ft/lb of torque by the front sprocket radius of .145ft. yields approximately 850 lbs of potential chain tension (125/.145 = 850). Since the rear tire typically slips in second gear, it is apparent that tire traction cannot contain all the available engine torque. The engine starts to ‘free wheel’ at some point below its maximum torque potential, thus the tension in the chain doesn’t reach its full potential. Tire slippage is dependent on terrain and rider technique. A reasonable ‘SWAG’ (scientific wild ass guess) for actual chain tension in 700 lbs. We’ll use this number for our analysis.

Note Sketch ‘A’. In this sketch we first see a diagram of the two sprockets with a chain drawn between them. Note the actual rise (vertical component) is 3.25”.

Note Sketch ‘B’. The second diagram resolves the chain angle into X and Y components (distances) so we can calculate the angle of the chain. The angle calculates to 8 degrees thus the 700 lbs of chain tension is at an 8-degree angle with respect to the horizontal swing arm.

Note Sketch ‘C’. To determine the effects of 700 lbs of chain tension at the rear wheel and at the frame, we must resolve the chain tension into its perpendicular components with respects to the swing arm. Note from our example that the horizontal force ‘Fx’ calculates to 693 lbs. The vertical force ‘Fy’ calculates to 100 lbs.

Note Sketch ‘D’. This sketch shows the forces acting on the rear wheel and the frame that hold the system in equilibrium. First note the horizontal force labeled Fx at the rear wheel. It is held in equilibrium by an equal and opposite Fx1 force exerted by the frame at the front swing arm pivot. These two forces cancel each other and combine to compress the swing arm some negligible amount. The vertical forces are much more interesting. The Fy force acts at the rear wheel and its associated equilibrium force Fy1 is applied at the frame. In the standard ‘rigid body’ analysis, Fy1 and Fy would combine to form a ‘couple’, or torque, which would exert a torque or twisting force on the entire bike. However in our case, the swing arm is attached with bearings at both ends, so our two vertical forces are separated by a pivoting link system. This link allows the two opposite vertical forces to cause vertical displacement instead of a torque. The rear wheel Fy tries to make the swing arm go down in the rear. The ground prevents this motion. The opposite vertical force at the front sprocket Fy1 tends to lift the frame, which unloads the rear shock spring. By applying 100 lbs of lifting force directly to the frame, the rear spring is relieved of 100 of the total 400 lbs of rider and bike weight. This causes the bike to rise.

Out on the track, when the throttle is applied quickly, chain tension occurs instantly causing the bike to lift as if the rider had suddenly lost 100 lbs. This sudden loss of a 100-pound load from the rear spring causes the bike to accelerate upward to its new equilibrium point. DISCUSSION AND APPLICATION Remember, we chose 400 lbs for the total bike weight and we said all the weight would be transferred to the rear wheel during acceleration. Without chain tension forces, all 400 lbs is transferred to the rear wheel through the rear shock spring. When we consider the chain tension and analyze the resulting vertical forces at the rear wheel and frame, we realize that 100 of the 400 lbs of bike weight is applied directly to the rear wheel via the tension in the chain. The remaining 300 lbs are supported by the rear shock spring. The advantage of the conventional chain layout is the rider’s ability to use this upwards force to his advantage for pre-jumping obstacles and clearing double jumps. As discussed earlier, this 100 lbs upward force reduces the weight carried by the spring. When you relieve the spring of part of its load, it accelerates upwards to a new equilibrium position. When the rider coordinates this upward frame acceleration with a jump takeoff, the result is more upward motion and more distance. Chain tension also tends to reduce bottoming when landing from large jumps. Landing with the throttle on reduces suspension bottoming because the shock spring only has to support the impact forces from 300 lbs rather than the 400 lbs combined weight of the bike and rider. The remaining 100 lbs is supported by the tension of the chain. Note that the ‘Torque Eliminator’ system eliminates these forces so the rider cannot ‘dial in’ frame lift or anti-bottoming with the throttle. The disadvantage of the traditional chain alignment is that 100 lbs of the bike weight is carried by the “rigid” chain connection between the frame and rear wheel, thus the rear wheel behaves as if it had 100 extra pounds of ‘unsprung’ weight during acceleration. Any ‘unsprung’ weight has trouble following the changes in terrain. Adding another 100 lbs to this ‘unsprung’ weight during acceleration adversely affects the rear tire’s ability to follow the terrain. (This adds another reason to exit a turn one gear higher. The higher the gear, the less the engine torque is multiplies; thus, the rigid connection between the frame and rear wheel is minimized!) The ‘Torque Eliminator” product effectively eliminates the rigid connection between the frame and rear wheel. By aligning the chain parallel to the swing arm, the chain tension becomes parallel to the swing arm. Without an angle between the chain and the swing arm, the forces are purely horizontal with no vertical components. No vertical forces mean no frame lift and no rigid connection between the frame and rear wheel. All 400 lbs of bike and rider weight act as ‘sprung’ weight transmitted to the ground through the rear shock spring. This also explains why the ‘Torque Eliminator’ product requires a stiffer shock spring. Without the 100 lbs vertical force at the frame to counter rear end squat during acceleration, the bike requires a stiffer spring rate to eliminate excessive squat. Most riders experimenting with this ‘Torque Eliminator’ system found that they preferred the potential for frame lift over a smoother ride. It should be noted that a 500 cc bike has about 1.5 times more torque than

a 250 cc bike and proportionally greater chain tension. Conversely a 125 cc bike would have about half the potential chain tension of a 250 cc bike. Additionally, there is more engine torque multiplication / chain tension in the lower gears and less in the higher gears. This explains why reactions at the rear wheel and frame will be more noticeable in the lower gears with large displacement bikes. Happy trails and don’t forget to ‘Gas It’ to clear that next obstacle or double jump! Back to My Home Page

Horsepower and Torque: A Practical Explanation
This may be the most highly debated question in all automotive internet forums. It's not that the definitions are in doubt. They're obviously objective. The controversy centers around which is more important.

Force is the pressure of one mass against another, and is one of the primary units in all of physics. In the metric system, force is calculated in "Newtons". Gravity is an easy example of a natural force and is written in the English system as "pounds". So we also use pounds as the basic unit of force.

Work is defined as force over distance and is calculated as Work= Force * Distance. In other words, work is achieved when force causes an object to move. The force placed on the object and the distance it moves are calculated as the work done.

Power was originally defined by the engineer James Watt as the amount of work that can be done in a certain amount of time. So its function is Power = Work / Time.

Torque is defined as the force at any one point on the edge of a circle in the exact direction of the rotation multiplied by the radius (distance from the center). This comes from the calculus/geometry concept of a "tangent", a line which touches exactly one point of the edge of a circle.

In the metric system, force is calculated in newtons, and distance is in meters, so the standard torque unit is Newton-Meters or N-M. In the Standard/English system, force is calculated in pounds and distance in feet. So the torque unit is lb-ft, usually pronounced as "Foot-pounds" and sometimes written as "ft/lb".

Horsepower is a unit of power. It can be defined in many ways. In its basic sense, it's defined as work done in a straight line as described above under "Power". But when the work is not done in a straight line, it must be defined in a different way: torque.
Horsepower = Torque X RPM


Now although horsepower in this instance is defined by rotational forces, it is no different than straight-line horsepower. For instance, if you wrapped a rope around the circle and allowed the torque to pull the rope, the force on the rope would now be exactly as defined above.

Gearing and Towing
Now when it comes to just about any type of racing known to mankind, besides engine output, gearing is the single most contributor to acceleration. It will make or break any car and the right gear selection can and will mean the difference between winning and losing a race. How important is gearing? Gearing nearly makes torque obsolete. Yes, it's that important. In a perfect environment with no limiting factors such as size and weight, the actual peak torque output of an engine would be totally meaningless because of gearing. How's that possible? It's simple. Gearing multiplies peak torque to the wheels to any

amount desired. Increasing the ratio increases torque. The limiting factors are the biggest problem with this ideal setup. Torque is multiplied through gear ratios, but the higher the gear ratio, the larger the gear and the more space it takes up. Unfortunately, in the real world, there's only so much space for a gear to occupy. It's this space limit that contributes to the "torque = towing capacity" philosophy. If space were unlimited and we could make the ratios anything we wanted, then towing capacity would be limitless since we could easily just utilize a higher ratio gear. But since the car world doesn't operate like that, there is generally a maximum amount of torque that can be generated at the wheels. It's this maximum which defines the towing capacity of the vehicle. If you start with an engine that already generates a great deal of torque, then the towing capacity will be easier to manipulate to higher levels.

Horsepower and Torque "At the Wheels"
Now when we're talking about automobiles, the amount of horsepower or torque generated at the flywheel is not very useful when determining acceleration. What is useful, however, is horsepower and torque "at the wheels". The problem here is that drivetrains cannot be perfectly efficient and pass 100% of the power of the engine through its components to the wheels. Some of the power is lost for several reasons. Generally 15-25% of engine power never makes it to the wheels. Different types of drivetrains will have different levels of efficiency. Generally, drivetrains with more weight and those with more components will be less efficient. Let's use my own car for some sample calculations. In stock form, it has 165 hp @ 5600 RPM and 166 lb-ft @ 4000 RPM. Dyno results have shown that the car has around 127 peak hp at the wheels. That's a 23.03% loss. Note that this is higher than most cars because of the heavy and sophisticated all-wheel-drive system. Here's a chart to show how the power and torque change before they reach the wheels. Although, the efficiency loss is difference for each gear, we'll assume that 127 peak hp is attainable in every gear. At 5600 RPM, the flywheel torque calculates as 154.7 lb-ft. Calculating the same efficiency loss (23.03%) as horsepower, this would come out to 119.1 lb-ft.

Engine RPM = 5600
Gear 1 2 Gear Ratio 3.545 2.110 Axle Ratio 4.11 4.11 Total Ratio 14.57 8.67 Flywheel Horsepower 165 165 Wheel Horsepower 127 127 Flywheel Torque 154.7 154.7 Wheel Torque after loss 1735 1033

3 4 5

1.448 1.088 0.780

4.11 4.11 4.11

5.95 4.47 3.21

165 165 165

127 127 127

154.7 154.7 154.7

707 532 382

To prove the accuracy of the wheel torque numbers, let's look at the example of 1st gear. Using the Speed/RPM Calculator, we can determine that the vehicle will be traveling at approximately 27.5 mph in 1st gear at 5600 RPM. Using the Tire Size Calculator, we can determine that the circumference of the tire is approximately 78.16 inches. Let's calculate the RPM of the tire: 27.5 mph = 145,200 ft/hour 145,200 ft/hour = 1,742,400 in/hour 1,742,400 in/hour = 22,294 revs/hour 22,294 revs/hour = 372 RPM Now we know that there is 127 hp generated at the wheels. If we use the horsepower formula above: 127 HP = (Torque * 372 RPM) / 5252 667,004 = Torque * 372 Torque = 1793 lb-ft Notice the difference between 1793 and 1735. This is caused by the reduction of the tire's size when fitted onto the vehicle. To help explain this, please read

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