Car Physics?
Does anyone know a good source for physics on a car, I''ve got the forward force (gas pedal) working, but need help with rotations and friction (both when turning and driving straight).
Thanks
Well I just had my final for my Engineering Physics course a coupla' days ago... I'll just offer my opinion on tire friction:
When your car is turning, the tires are providing a "central" or "radial" force, which prevents the car from just going in a straight line. This force *should* be: F(friction) = v^2/r (where v is the normal car velocity and r is the radius of the turn). However, the amount of friction force that the tires can provide is limited to: F(friction) = mu(static) * F(normal). Here, mu(static) is a constant, probably around 1.0, that represents the grip of the tire when it has traction, and F(normal) is the force down on the car, usually just mass * gravity.
Based on the equations, you should be able to tell if the tire has traction. If it does, then you could simply make the car go in the direction it is pointing. If there is no traction, the tire still provides *some* friction force, but not as much. This lesser force is: F(friction) = mu(kinetic) * F(normal). Mu(kinetic) is always less than mu(static). If the car is currently sliding, it will regain traction once v^2/r <= mu(kinetic) * F(normal).
To be accurate, you should actually calculate the net forces on the tire, which would include the car's foward acceleration/decceleration.
I don't know if you actually need to use these force equations to govern the car's acceleration; you could at least use them to determine the car's traction/slipping state.
Edited by - Eric on May 13, 2000 4:25:55 AM
When your car is turning, the tires are providing a "central" or "radial" force, which prevents the car from just going in a straight line. This force *should* be: F(friction) = v^2/r (where v is the normal car velocity and r is the radius of the turn). However, the amount of friction force that the tires can provide is limited to: F(friction) = mu(static) * F(normal). Here, mu(static) is a constant, probably around 1.0, that represents the grip of the tire when it has traction, and F(normal) is the force down on the car, usually just mass * gravity.
Based on the equations, you should be able to tell if the tire has traction. If it does, then you could simply make the car go in the direction it is pointing. If there is no traction, the tire still provides *some* friction force, but not as much. This lesser force is: F(friction) = mu(kinetic) * F(normal). Mu(kinetic) is always less than mu(static). If the car is currently sliding, it will regain traction once v^2/r <= mu(kinetic) * F(normal).
To be accurate, you should actually calculate the net forces on the tire, which would include the car's foward acceleration/decceleration.
I don't know if you actually need to use these force equations to govern the car's acceleration; you could at least use them to determine the car's traction/slipping state.
Edited by - Eric on May 13, 2000 4:25:55 AM
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