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<!DOCTYPE doctype PUBLIC "-//w3c//dtd html 4.0 transitional//en">
<html><head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<meta name="keywords" content="game,programming, car, physics, rigid body dynamics, acceleration,tutorial,c, djgpp, allegro">
<meta name="description" content="game programming tutorial"><title>Car Physics</title></head>
<body bgcolor="#ffffff" link="#0000ff" vlink="#800080">
<center><b><font face="verdana"><font color="#0000ff"><font size="+2">Car
Physics for Games</font></font></font></b> <br>
by <a href="http://home.planet.nl/%7Emonstrous">Marco Monster</a></center>
<p><b>Level:</b> Advanced </p>
<p><b>Abstract: </b><i>An introduction to car physics modelling for
games.</i> </p>
<p> </p>
<hr>
<p>version 1.9 <br>
November, 2003 <br>
<br>
</p>
<h4>Introduction</h4>
<p>This tutorial is about simulating cars in games, in other words
vehicle physics. </p>
<p>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. </p>
<h4>Notation and conventions</h4>
Vectors are shown in<b> bold,</b> we'll be using 2d vectors. So the
notation <b>a</b> = -<b>b </b>would translate to the following code
snippet: <br>
<tt> a.x = -b.x</tt> <br>
<tt> a.y = -b.y</tt>
<p>Throughout this tutorial I'll be assuming that the rear wheels
provide all the drive (for four wheel drives apply the neccesary
adaptations). </p>
<p>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.) <br>
</p>
<h4>Straight line physics</h4>
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.
<p> <b>F</b><sub>traction</sub> = <b>u</b> *
Engineforce, <br>
where <b>u</b> is a unit vector in the direction of
the car's heading.</p>
<p>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. </p>
<p> <b>F</b><sub>drag</sub> = - C<sub>drag</sub> * <b>v
* |v|</b> <br>
where C<sub>drag</sub> is a constant and <b>v</b> is
the velocity vector and the notation <strong>|v| </strong> refers to
the magnitude of vector <strong>v</strong><br>
</p>
<p> 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: </p>
<p><tt> speed = sqrt(v.x*v.x + v.y*v.y);</tt> <br>
<tt> fdrag.x = - Cdrag * v.x * speed;</tt> <br>
<tt> fdrag.y = - Cdrag * v.y * speed;</tt></p>
<p>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. </p>
<p> <b>F</b><sub>rr</sub> = - C<sub>rr</sub> * <b>v</b> <br>
where C<sub>rr</sub> is a constant and <b>v</b> is
the velocity vector. </p>
<p>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 C<sub>rr</sub> must be
approximately 30 times the value of C<sub>drag</sub> </p>
<p>The total longtitudinal force is the vector sum of these three
forces. </p>
<p> <b> F</b><sub>long</sub> = <b>F</b><sub>traction</sub>
+ <b>F</b><sub>drag</sub> + <b>F</b><sub>rr</sub> </p>
<p>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 <b>F</b><sub>long</sub>
is zero. </p>
<p>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: </p>
<p> <b>a</b> = <b>F</b> / M </p>
<p>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. </p>
<p> <b>v</b> = <b>v</b> + dt * <b>a</b>, <br>
where dt is the time increment in seconds between
subsequent calls of the physics engine. </p>
<p>The car's position is in turn determined by integrating the velocity
over time: </p>
<p> <b>p</b> = <b>p</b> + dt * <b>v</b> </p>
<p>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. </p>
<center>
<p><img alt="graph" src="Car Physics for Games_files/ctgraph.jpg" height="289" width="492"></p>
</center>
<p>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). <br>
</p>
<h4>Magic Constants</h4>
So far, we've introduced two magic constants in our equations: C<sub>drag</sub>
and C<sub>rr</sub> . 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.
<p>Air resistance is approximated by the following formula (<i>Fluid
Mechanics</i> by Landau and Lifshitz, [Beckham] chapter 6, [Zuvich]) </p>
<p> F<sub>drag</sub> = 0.5 * C<sub>d</sub> * A
* rho * v<sup>2</sup> </p>
<p> where C<sub>d</sub> = coefficient of friction <br>
A is frontal area of car <br>
rho (Greek symbol <img src="Car Physics for Games_files/ctrho.gif" height="14" width="11">)= density of air <br>
v = speed of the car </p>
<p>Air density (rho) is 1.29 kg/m<sup>3 </sup>(0.0801 lb-mass/ft<sup>3</sup>),
frontal area is approx. 2.2 m<sup>2</sup> (20 sq. feet), C<sub>d</sub>
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 C<sub>drag</sub>: <br>
C<sub>drag</sub> = 0.5 * 0.30 * 2.2 * 1.29 <br>
= 0.4257 </p>
<p>We've already found that C<sub>rr</sub> should be approx. 30 times C<sub>drag</sub>.
This gives us <br>
C<sub>rr</sub> = 30 * 0.4257 <br>
= 12.8 <br>
</p>
<p>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. </p>
<h4>Braking</h4>
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.
<p> <b> F</b><sub>long</sub> = <b>F</b><sub>braking</sub>
+ <b>F</b><sub>drag</sub> + <b>F</b><sub>rr</sub> </p>
<p>A simple model of braking: </p>
<p> <b> F</b><sub>braking</sub> = -<b>u * </b>C<sub>braking</sub> </p>
<p>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. <br>
</p>
<h4>Weight Transfer</h4>
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.
<p>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. </p>
<p>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: </p>
<p> <b> F</b><sub>max</sub> = mu<b> * </b>W <br>
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. </p>
<p>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): <br>
W<sub>f</sub> = (c/L)*W <br>
W<sub>r</sub> = (b/L)*W <br>
where b is the distance from CG to front axle, c the
distance from CG to rear axle and L is the wheelbase. <br>
<br>
<br>
</p>
<center>
<p><img src="Car Physics for Games_files/ctwd.jpg" height="260" width="400"></p>
</center>
<p>If the car is accelerating or decelerating at rate a, the weight on
front (W<sub>f</sub>) and rear axle (W<sub>r</sub>) can be calculated as
follows: <br>
W<sub>f</sub> = (c/L)*W - (h/L)*M*a <br>
W<sub>r</sub> = (b/L)*W + (h/L)*M*a, <br>
where h is the height of the CG, M is the car's mass
and a is the acceleration (negative in case of deceleration). </p>
<p>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?</p>
<p>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, W<sub>f</sub> = 0.5*W -
(h/L)*M*a and W<sub>r</sub> = 0.5*W +(h/L)*M*a; <br>
<br>
</p>
<h4>Engine Force</h4>
<p>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 <em>torque</em>. 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.</p>
<p>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 </p>
<center>
<p> <img alt="torque curve for the Corvette LS1 engine" src="Car Physics for Games_files/cttorq.gif" border="0" height="368" width="587"> </p>
</center>
<p>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.</p>
<p>Oh, and by the way, this is the <em>maximum</em> 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.</p>
<i></i><i></i>
<p>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: </p>
<blockquote>hp = torque * rpm / 5252</blockquote>
<p>Because of this relationship, the two curves will always cross at
5252 rpm. Check for yourself in the diagram above. </p>
<p>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). </p>
<center>
<p> <img alt="Same curves but now in SI units" src="Car Physics for Games_files/cttorqsi.gif" border="0" height="392" width="451"> </p>
</center>
<p>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. </p>
<p> 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. </p>
<p>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). </p>
<p>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 (<a href="http://www.carbibles.com/">http://www.carbibles.com</a>
). 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. </p>
<p>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. </p>
<p> <b>F</b><sub>drive</sub> = <b>u</b> * T<sub>engine</sub>
* x<sub>g</sub> * x<sub>d</sub> * n / R<sub>w</sub> <br>
where <br>
<b>u</b> is a unit vector which reflects the car's
orientation, <br>
T<sub>engine</sub> is the torque of the engine at a
given rpm,<br>
x<sub>g</sub> is the gear ratio,<br>
x<sub>d</sub> is the differential ratio,<br>
n is transmission efficiency and <br>
R<sub>w</sub> is wheel radius. <br>
<br>
An example: </p>
<p>Engine is running at 2500 rpm, looking this up on the curve gives
engine torque of 448 Nm (=330 ft lbs) <br>
Gear ratio (first gear): 2.66 <br>
Differential ratio: 3.42 <br>
Transmission efficiency: 0.7 (guess) <br>
Wheel radius: 0.34 m (=13.4 inch) <br>
Mass: 1500 kg (= 3300 lbs of weight) including the driver. </p>
<p>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. </p>
<p>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/s<sup>2</sup>
= 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: <br>
a = 7350 N / 1500 kg = 4.9 m/s<sup>2</sup>
(=0.5 G) </p>
<p>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. </p>
<p>This means W<sub>f</sub> = 7350 - 2940 N and W<sub>r</sub> =
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. <br>
</p>
<h4>Gear Ratios</h4>
<p>The following gear ratios apply to an Corvette C5 hardtop
(Source: <a href="http://www.idavette.net/facts/c5specs/">http://www.idavette.net/facts/c5specs/</a>
)</p>
<table border="0" cellspacing="0" width="320">
<tbody>
<tr>
<td valign="center" width="59%">First gear</td>
<td valign="center" width="25%">
<center>g1</center>
</td>
<td valign="center" width="16%">
<center>2.66</center>
</td>
</tr>
<tr>
<td valign="center" width="59%">Second gear</td>
<td valign="center" width="25%">
<center>g2</center>
</td>
<td valign="center" width="16%">
<center>1.78</center>
</td>
</tr>
<tr>
<td valign="center" width="59%">Third gear</td>
<td valign="center" width="25%">
<center>g3</center>
</td>
<td valign="center" width="16%">
<center>1.30</center>
</td>
</tr>
<tr>
<td valign="center" width="59%">Fourth gear</td>
<td valign="center" width="25%">
<center>g4</center>
</td>
<td valign="center" width="16%">
<center>1.0</center>
</td>
</tr>
<tr>
<td valign="center" width="59%">Fifth gear</td>
<td valign="center" width="25%">
<center>g5</center>
</td>
<td valign="center" width="16%">
<center>0.74</center>
</td>
</tr>
<tr>
<td valign="center" width="59%">Sixth (!) gear</td>
<td valign="center" width="25%">
<center>g6</center>
</td>
<td valign="center" width="16%">
<center>0.50</center>
</td>
</tr>
<tr>
<td valign="center" width="59%">Reverse</td>
<td valign="center" width="25%">
<center>gR</center>
</td>
<td valign="center" width="16%">
<center>2.90</center>
</td>
</tr>
<tr>
<td valign="center" width="59%">Differential ratio </td>
<td valign="center" width="25%">
<center>x<sub>d</sub></center>
</td>
<td valign="center" width="16%">
<center>3.42</center>
</td>
</tr>
</tbody>
</table>
<p>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/s<sup>2</sup> (a=F/m) which
equals 0.65 g. </p>
<p>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. </p>
<p>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. </p>
<center>
<p> <img alt="torque curves per gear" src="Car Physics for Games_files/ctgrcrvs.gif" border="0" height="416" width="638"> </p>
</center>
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 <i>of
the rear axle</i> is 36 m/s or 128 km/h car speed (80 mph). 2000 rpm
is 256 km/h (160 mph). Etcetera.
<h4> Drive wheel acceleration</h4>
<p>Now beware, the torque that we can look up in the torque curves
above for a given rpm, is the <em>maximum </em>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:</p>
<p> max torque = LookupTorqueCurve( rpm )<br>
engine torque = throttle position * max torque</p>
<p>You could implement the function LookupTorqueCurve by using an array
of torque/rpm value pairs and doing linear interpolation between the
closest two points.</p>
<p>This torque is delivered to the drive wheels via the gearbox and
results in what I'll call the drive torque:</p>
<p> drive torque = engine_torque * gear_ratio *
differential_ratio * transmission_efficiency</p>
<p>Or written more concisely as:</p>
<p> T<sub>drive</sub> = T<sub>engine</sub> * x<sub>g</sub>
* x<sub>d</sub> * n <br>
</p>
<p>Because F<sub>drive</sub> = T<sub>drive</sub> / R<sub>w </sub>,
this is the same as the equation for drive force we saw earlier.</p>
<p>Note that the gearbox <i>increases</i> the torque, but <i>reduces</i>
the rate of rotation, especially in low gears.</p>
<p> <b>How do we get the RPM?</b></p>
<p>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. </p>
<p>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!</p>
<p>rpm = wheel rotation rate * gear ratio * differential ratio *
60 / 2 pi</p>
<p> 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.</p>
<p>Wheel rotates at 17 rad/s.<br>
First gear ratio is 2.66, differential ratio is 3.42 so crankshaft is
rotating at 153 rad/s.<br>
That's 153*60 = 9170 rad/minute = 9170/2 pi = 1460 rpm at the
engine.</p>
<p> 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.</p>
<pre>if( rpm < 1000 )<br> rpm = 1000;<br></pre>
<p>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.</p>
<p>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.</p>
<p>The easy way is to pretend the wheel is rolling and derive the
rotation rate from the car speed and the wheel radius.</p>
<p>For example, let's say the car is moving at 20 km/h = 20,000 m /
3600 s = 5.6 m/s.<br>
wheel radius is 0.33 m, so wheel angular velocity is 5.6/0.33 = 17
rad/s</p>
<p>Plug this into the previous equations to get the 1460 rpm, from
which we can look up the engine torque on the torque curve.</p>
<p>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).</p>
<h4>Slip ratio and traction force</h4>
<p> 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.</p>
<p> In a typical situation where the car is cruising at constant speed, <em>the
rear wheels will be rotating slighty faster than the front wheels</em> .
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:</p>
<div align="center"><img alt="SR = (omega.R - Vlong) / abs(Vlong)" src="Car Physics for Games_files/cteq_sr.gif" border="0" height="69" width="174"></div>
<br>
where<br>
<img alt="" src="Car Physics for Games_files/omega.gif" border="0" height="12" width="14"> <sub>w</sub> (pronounced:
omega) is wheel angular velocity (in rad/s)<br>
R<sub>w</sub> is wheel radius (in m)<br>
v<sub>long</sub> is car speed a.k.a.
longtitudinal velocity (in m/s)
<p></p>
<p> Note: there are a number of slightly different definitions for
slip ratio in use. This particular definition also works for cars
driving in reverse.</p>
<p>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.</p>
<p> The relationship between longtitudinal (forward) force and slip
ratio can be described by a curve such as the following:</p>
<div align="center"><img alt="slip ratio curve" src="Car Physics for Games_files/ctsrcurve.gif" border="0" height="274" width="451"></div>
<div align="center"> </div>
<p>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.</p>
<p>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.</p>
<p>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.</p>
<p>To get from normalized longtitudinal force to actual longtitudinal
force, multiply by the load on the wheel. </p>
<blockquote>F<sub>long</sub> = F<sub>n, long</sub> * F<sub>z</sub><br>
where F<sub>n, long</sub> is the normalized longtitudinal force
for a given slip ratio and F<sub>z</sub> is the load on the tyre.</blockquote>
<p> For a simple simulation the longtitudinal force can be approximated
by a linear function: </p>
<blockquote>F<sub>long</sub> = C<sub>t</sub> * slip ratio</blockquote>
<p>where C<sub>t</sub> 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:</p>
<div align="center"><img alt="graph" src="Car Physics for Games_files/ct_srapprox.gif" border="0" height="170" width="224"></div>
<h4>Torque on the drive wheels</h4>
<p>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:</p>
<p> traction torque = traction force * wheel radius</p>
<p>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.</p>
<p>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.</p>
<p> <img alt="diagram of torques on drive wheel" src="Car Physics for Games_files/tc_torques.png" border="0" height="266" width="485"> </p>
<div align="center"></div>
<p> The net torque on the rear axle is the sum of following torques:</p>
<p> total torque = drive torque + traction torques from
both wheels + brake torque</p>
<p>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.</p>
<p>This torque will cause an angular acceleration of the drive wheels,
just like a force applied on a mass will cause it to accelerate.</p>
<p> angular acceleration = total torque / rear wheel
inertia.</p>
<p>I've found somewhere that the inertia of a solid cylinder around its
axis can be calculated as follows:</p>
<p> inertia of a cylinder = Mass * radius<sup>2</sup> / 2</p>
<p> 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.m<sup>2 </sup>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. </p>
<p>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:</p>
<p> rear wheel angular velocity += rear wheel angular
acceleration * time step</p>
<p>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.</p>
<p><strong>The Chicken and the Egg</strong></p>
<p>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?</p>
<p>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.</p>
<p>To solve differential equations in computer programs we use the
technique of numeric integration: if we know all the values at time <em>t</em>,
we can work out the values at time <em>t + delta. </em>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.</p>
<p>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
well-behaved 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:<a href="http://home.planet.nl/%7Emonstrous/tutstab.html">Achieving a Stable Simulator</a> </p>
<p><b>Curves</b></p>
<p>Okay, enough of driving in a straight line. How about some turning? </p>
<p>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. </p>
<p>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. </p>
<p>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 <i>delta</i> 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. </p>
<p>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. <br>
</p>
<blockquote>
<blockquote>
<blockquote><img src="Car Physics for Games_files/ctturn.jpg" height="205" width="255"></blockquote>
</blockquote>
</blockquote>
<blockquote>
<blockquote>
<blockquote><i><font size="-1">This is a screen shot from a car
physics demo by Rui Martins.</font></i> <br>
<i><font size="-1">More illustrations and a download can be
found at</font></i> <br>
<i><font size="-1"><a href="http://ruimartins.net/software/projects/racer/">http://ruimartins.net/software/projects/racer/</a></font></i></blockquote>
</blockquote>
</blockquote>
<p><br>
The radius of the circle can be determined via geometry as in the
following diagram: </p>
<center>
<p><img src="Car Physics for Games_files/ctangles.jpg" height="260" width="312"></p>
</center>
<p>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 <i>delta</i>. This means the angle at the
circle centre is also <i>delta</i> (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: </p>
<dir>
<dir><img src="Car Physics for Games_files/cteq_r.gif" height="70" width="222"></dir>
</dir>
Note that if the steering angle is zero, then the circle radius is
infinite, i.e. we're driving in a straight line.
<p>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 (<img alt="" src="Car Physics for Games_files/omega.gif" border="0" height="12" width="14">), 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: </p>
<dir>
<dir><img src="Car Physics for Games_files/cteq_av.gif"></dir>
</dir>
<p>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. </p>
<h4>High Speed Turning</h4>
<p>Of course, there are not many games involving cars that drive around
sedately (apart from the legendary <i>Trabant Granny Racer</i>
;-). 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. </p>
<p>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). </p>
<p>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 (<i>beta</i>). </p>
<p> </p>
<center><img alt="" src="Car Physics for Games_files/ctbeta.jpg" border="0" height="143" width="113"></center>
<p>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. </p>
<p>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 </p>
<blockquote>F<sub>lateral</sub> = C<sub>a</sub> * alpha</blockquote>
<p>where the constant C<sub>a </sub> is known as the cornering
stiffness. </p>
<p>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. </p>
<center>
<p><img src="Car Physics for Games_files/ctimage8.gif" height="232" width="213"></p>
</center>
<p>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. </p>
<p>The sideslip angle <font face="symbol">b</font> (b�ta) 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. </p>
<p align="center"><img alt="" src="Car Physics for Games_files/ctbeta.gif" border="0" height="60" width="140"></p>
<p>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/(2.pi.omega) 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. </p>
<center>
<p><img src="Car Physics for Games_files/ctav.gif" height="170" width="148"></p>
</center>
<p>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. </p>
<center>
<p><img src="Car Physics for Games_files/ctdeltapic.gif" height="170" width="148"></p>
</center>
<p> The slip angles for the front and rear wheels are given by the
following equations: </p>
<p> </p>
<p> </p>
<p align="center"><img alt="slip angle equations" src="Car Physics for Games_files/ct_alphas.gif" border="0" height="169" width="368"></p>
<p>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: </p>
<p align="center"><img alt="slip angle curve" src="Car Physics for Games_files/ctsacurve.gif" border="0" height="274" width="451"></p>
<p>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. </p>
<p>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. </p>
<p>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.</p>
<blockquote>F<sub>lateral</sub> = F<sub>n, lat</sub> * F<sub>z</sub><br>
where F<sub>n, lat</sub> is the normalized lateral force for a
given slip angle and F<sub>z</sub> is the load on the tyre.</blockquote>
<p>For very small angles (below the peak) the lateral force can be
approximated by a linear function: </p>
<blockquote>F<sub>lateral</sub> = C<sub>a</sub> * alpha</blockquote>
The constant C<sub>a </sub>goes by the name of the cornering
stiffness. This is the slope of the diagram at slip angle 0.
<p>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. </p>
<p>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. </p>
<blockquote>F<sub>cornering</sub> = F<sub>lat, rear</sub> + cos(delta)
* F<sub>lat, front</sub></blockquote>
As a point of interest, we can find the circle radius now that we know
the centripetal force using the following equation
<blockquote>F<sub>centripetal</sub> = M v<sup>2</sup> / radius</blockquote>
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.
<p>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 -F<sub>lat, rear</sub> * c
and for the front wheels it's cos(delta) * F<sub>lat, front </sub>*
b. Note that the sign differs. </p>
<p>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: </p>
<blockquote>Torque = Inertia * angular acceleration.</blockquote>
<p>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. <br>
</p>
<p><strong>Epilogue</strong></p>
<p>You can <a href="http://home.planet.nl/%7Emonstrous/download.html#cardem08">download</a> 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 <a href="http://www.cs.uni-magdeburg.de/%7Esodeike/java/CarPhysics/CarPhysics.html">Java
version</a> by Marcel "Bloemschneif" Sodeike! </p>
<p align="center"> <img alt="screen shot of demo" src="Car Physics for Games_files/cardem08.gif" border="0" height="122" width="162"></p>
<p> The following information is shown on screen: </p>
<ul>
<li>steering angle (also known as delta) </li>
<li>side slip angle (also known as beta) </li>
<li>"rotation angle" (due to yaw rate) </li>
<li>slip angle front wheels </li>
<li>slip angle rear wheels </li>
<li>velocity vector relative to car reference frame </li>
<li>velocity vector in world space</li>
</ul>
<br>
<img alt="legend" src="Car Physics for Games_files/legend.gif" border="0" height="326" width="433">
<p> See also my work in progress on the game <a href="http://home.planet.nl/%7Emonstrous/drivin.html">Downtown
Drivin'</a> for an example of some of the stuff that was discussed here.</p>
<p></p>
<p> If you spot mistakes in this tutorial or find sections that could do
with some clarification, let me know. </p>
<p>-Marco- </p>
<h4>References</h4>
[Beckham] <i>Physics of Racing</i> Series, Brian Beckham, <br>
<a href="http://phors.locost7.info/">http://phors.locost7.info</a>
<p>[Bower] Richard Bower, <br>
<a href="http://www.dur.ac.uk/%7Edph0rgb">http://www.dur.ac.uk/~dph0rgb</a> </p>
<p>[Hecker] Chris Hecker, series on Physics in Game Development
Magazine, <br>
<a href="http://www.d6.com/users/checker/dynamics.htm">http://www.d6.com/users/checker/dynamics.htm</a> </p>
<p>[Lander] Jeff Lander, The Trials and Tribulations of Tribology, <br>
<a href="http://www.gamasutra.com/features/20000510/lander_01.htm">http://www.gamasutra.com/features/20000510/lander_01.htm</a> </p>
<p>[RQRiley] <i>Automobile Ride, Handling and Suspension Design</i>,
R.Q. Riley Enterprises, <br>
<a href="http://www.rqriley.com/suspensn.html">http://www.rqriley.com/suspensn.html</a> </p>
<p>[Zuvich] <i>Vehicle Dynamics for Racing Games</i>, Ted Zuvich, <br>
<a href="http://www.gdconf.com/2000/library/homepage.htm">http://www.gdconf.com/2000/library/homepage.htm</a>,
note that you need to obtain a (free) Gamasutra id to access the game
development conference archives. <br>
<br>
</p>
<p><strong>Car specifications</strong> </p>
<p>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 <a href="http://home.planet.nl/%7Emonstrous/c5specs.html">Corvette
C5</a>. Be warned: some of the data is still guesswork. </p>
<p> </p>
<h4>Units and Conversions</h4>
The following S.I. units apply: <br>
<br>
<br>
<table border="1" cellpadding="3" cellspacing="3">
<tbody>
<tr>
<td valign="center">Force</td>
<td valign="center">N (Newton)</td>
<td valign="center">= m.kg/s<sup>2</sup></td>
</tr>
<tr>
<td valign="center">Power</td>
<td valign="center">W (Watt)</td>
<td valign="center">= N.m/s = J/s = m<sup>2</sup> kg / s<sup>3</sup></td>
</tr>
<tr>
<td valign="center">Torque</td>
<td valign="center">N.m (Newton meter)</td>
<td valign="center"> </td>
</tr>
<tr>
<td valign="center">Speed</td>
<td valign="center">m/s</td>
<td valign="center"> </td>
</tr>
<tr>
<td valign="center">Angular velocity</td>
<td valign="center">rad/s</td>
<td valign="center"> </td>
</tr>
<tr>
<td valign="center">Acceleration</td>
<td valign="center">m/s<sup>2</sup></td>
<td valign="center"> </td>
</tr>
<tr>
<td valign="center">Mass</td>
<td valign="center">kg</td>
<td valign="center"> </td>
</tr>
<tr>
<td valign="center">Distance</td>
<td valign="center">m</td>
<td valign="center"> </td>
</tr>
</tbody>
</table>
Angle rad =360 degrees/ 2 pi
<p>Here are some handy unit conversions to S.I. units. <br>
<br>
<br>
<table border="1" cellpadding="3" cellspacing="3">
<tbody>
<tr>
<td valign="center">1 mile</td>
<td valign="center">= 1.6093 km</td>
</tr>
<tr>
<td valign="center">1 ft (foot)</td>
<td valign="center">= 0.3048 m</td>
</tr>
<tr>
<td valign="center">1 in (inch)</td>
<td valign="center">= 0.0254 m = 2.54 cm</td>
</tr>
<tr>
<td valign="center">1 km/h</td>
<td valign="center">= 0.2778 m/s</td>
</tr>
<tr>
<td valign="center">1 mph</td>
<td valign="center">= 1.609 km/h = .447 m/s</td>
</tr>
<tr>
<td valign="center">1 rpm (revolution per minute)</td>
<td valign="center">= 0.105 rad/s</td>
</tr>
<tr>
<td valign="center">1 G</td>
<td valign="center">= 9.8 m/s<sup>2 </sup>= 32.1 lb/s<sup>2</sup></td>
</tr>
<tr>
<td valign="center">1 lb (pound)</td>
<td valign="center">= 4.45 N</td>
</tr>
<tr>
<td valign="center">1 lb (pound)</td>
<td valign="center">= 0.4536 kg <sup>1) </sup>= 1 lb/1G</td>
</tr>
<tr>
<td valign="center">1 lb.ft (foot pounds)</td>
<td valign="center">= 1.356 N.m</td>
</tr>
<tr>
<td valign="center">1 lb.ft/s (foot pound per second)</td>
<td valign="center">= 1.356 W</td>
</tr>
<tr>
<td valign="center">1 hp (horsepower) = 550 ft.lb/s</td>
<td valign="center">= 745.7 W </td>
</tr>
<tr>
<td valign="center">1 metric hp = 0.986 hp</td>
<td valign="center">= 735.5 W </td>
</tr>
</tbody>
</table>
<br>
<br>
</p>
<p>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 <i>assuming the
gravity is 9.8m/s<sup>2</sup></i>. On the moon, a kilogram will weigh
less but still have the same mass. </p>
<p>Unit converter on the web: <a href="http://lecture.lite.msu.edu/%7Emmp/conversions/intro.htm">http://lecture.lite.msu.edu/~mmp/conversions/intro.htm</a> </p>
<p> </p>
<hr>
<p><font size="-1">Marco Monster</font> <br>
<font size="-1">E-mail: <br><img src="Car Physics for Games_files/MonstrousSoftware-email-address.gif" alt="" border="0" height="24" width="193"></font> <br>
<font size="-1">Website: <a href="http://home.planet.nl/%7Emonstrous">http://home.planet.nl/~monstrous</a></font> </p>
<p><font size="-1">Copyright 2000-2003 Marco Monster. All rights
reserved. This tutorial may not be copied or redistributed without
permission.</font></p>
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