Traction in car physics

This topic is 3329 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

Recommended Posts

I've been reading several articles about car physics in order to code a car physics model as realistic as efficiently possible. I think I understand physics kinda easily and hold all the theoretical knowledge to hadle this, but now I think I've encountered a problem about the principle how to deal with traction. Marco Monster's car physics tutorial talks about slip ratio and how minor slipping of the driving wheels causes the traction:
Quote:
 In a typical situation where the car is cruising at constant speed, the rear wheels will be rotating slighty faster than the front wheels. -- 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. -- The traction depends on the amount of slip.
I can't understand how'd that be true. Please tell me if I'm wrong, but I think that's just bullsh*t. I'm simply using my common sense and it says that if a circular object is in accelerating on some surface, it's not necessarily slipping but likely rolling. The same applies to friction in general: the relative speed of two surfaces moving with respect to each other doesn't affect the magnitude of the kinetic friction force between the surfaces, right? What I'm trying to say is that I've always thought that the friction between a car's tire and surface of the road is actually static friction, not kinetic friction, as the tire is rolling (so the contact point isn't moving with respect to the road). So, if I do not use the method described above to determine the traction force, then what? Again I'm trying to be rational and to separate the pieces of the jigsaw puzzle: First consider a car wheel at rest. Assuming that our car is on flat ground no other forces exist but the gravitational weight of the tire and car combined and the normal force that prevents the tire from sinking into the ground: As the driver steps on throttle and releases clutch, the engine generates torque that is transmitted to the wheel. Now the wheel would experience high angular acceleration and start rotating if there were no friction. So we add a torque and a Friction force: Seeing the picture briefly could result in an imagination that everything is OK, but, in fact the traction force also causes a torque to the wheel that is not drawn in the picture. The approximate lw of static friction says that the friction force can't exceed magnitude F = μN, where μ is the friction coefficent and N is the force normal to the surface the object is on. So the maximum engine torque the friction force can resist is Tmax = μrN, where r is the radius of the wheel. Considering that our driver is about to make a smooth start we assume that the engine torque on the wheel is less than the frictional torque limit so the wheel doesn't spin aka slip. This causes the two torques to cancel each other so the total torque and therefore angular acceleration equals zero. However the friction force still exists and causes the wheel (and the whole car) to be in linearly accelerating motion although the wheel stays rotationally at rest. To me this seems kind of a paradox or then I've just forgotten to take something account. Any explanation would be appreciated!

Share on other sites
(1) I agree that the car physics tutorial seems incorrect. I looked up slip ratio and found it relates to anti-lock brakes!? You are right that traction to drive the car forward is primarily static friction. In fact, the dynamic friction limit is less than the static limit, so people put lots of effort into making sure that the wheel does not slip (especially important in racing).

(2). Your free-body diagrams are correct. I assume that you are talking about the driver accelerating from rest. In this case, the friction is defined to be F <= mu*N (which you have). Your forgot the kinematic relationship between linear and angular acceleration. In this case, the acceleration of the axle (the center of the wheel, and thus the car) is related to the angular acceleration of the wheel by a = -R*alpha, where a is the linear acceleration , R the radius of the wheel, and alpha the angular acceleration of the wheel. When the driver is accelerating, neither is zero. In this case, let the motor torque be T. then F = m*a (where m = mass of car, e.g.). Further, we know from your free body diagram that F*r-T=J*alpha (where alpha is the wheel angular acceleration). Solving:

m*a*r-T=-J*a/r => a = T / (m*r+J/r), and alpha = -T / (m*r^2+J). So neither is zero, even though the contact point is not moving.

Hope this helps

Share on other sites
That's awesome. The connection between linear and angular acceleration really was the missin link needed to solve this. Thanks a lot!

I went on to the case where the tire is slipping. So the relation breaks up. Have I done this right:
Total torque on the wheel is SUM T = F_mu * r - T_drive = mu*N*r - T_drive = J*alpha (same as before but mu now is the kinetic friction coefficient..) which gives alpha = (mu*N*r - τ_drive) / J. Linear acceleration is only created by the friction force F_mu so we get following equation for linear acceleration: F_mu = ma <=> mu*N = ma, resulting a = mu*N / m

Share on other sites
You'll find a lot more information by googling slip angle as opposed to slip ratio.

1. 1
2. 2
Rutin
24
3. 3
4. 4
JoeJ
17
5. 5

• 14
• 29
• 11
• 11
• 9
• Forum Statistics

• Total Topics
631773
• Total Posts
3002265
×

Important Information

We are the game development community.

Whether you are an indie, hobbyist, AAA developer, or just trying to learn, GameDev.net is the place for you to learn, share, and connect with the games industry. Learn more About Us or sign up!

Sign me up!