2d car physics question

235b371d8fea8dbf60a687b31f8b5f67
0
staticVoid2 101 Apr 02, 2010 at 19:43

hi there, I’m creating a 2d car simulator and I’m trying to get turning to be as physically correct as possible.

The first thing I done was to create a rigid body structure for the car body that has some properties: mass, inertia etc. The main function of this structure is to be able to apply forces at specific points and for it to rotate and translate itself accordingly. All this works fine.

The problem I’m having is with the wheels… I know the velocity of the wheels and the direction they are heading, The angle between these two vectors is commonly refered to as ‘slip angle’ I think. Currently I split the velocity into two components - side velocity and forward velocity (side velocity is how many m/s the tyre is moving on its side and same with forward) and then I apply a force in the opposite direction of the side velocity to the car, this force is proportional to the velocity. This method seems to work, I just set a coefficient that gets multiplied with the inverse side velocity and use this vector as a force that I apply to the car. Can someone tell me velocity is linearly proportional to force in this case (seems to work).

I’ve tryed doing it the proper way by storing the previous frames side velocity and then differentiating over a a certain time (1/50th second (50fps)) to give me the acceleration which I then plug into (F=ma).(I say ‘proper way’ but I really dont know if it is) If the tyre pushes the ground and the ground pushes back with an equal amount of force then why doesn;t this work. (do I also need to take inertia into account).

p.s. I dont take weight transfer into account, in the calculations mass is just car.mass / 4.0

thanks:)

6 Replies

Please log in or register to post a reply.

A8433b04cb41dd57113740b779f61acb
0
Reedbeta 167 Apr 02, 2010 at 20:00

Car tires sliding sideways on pavement are subject to friction. The simplest model for friction force, which most games use, is for the force to be in the opposite direction to the velocity (in this case the side velocity) and proportional to the force with which the two surfaces are pressed together - in this case 1/4 the car’s weight. The coefficient of proportionality depends on whether the surfaces are currently sliding (kinetic friction) or not (static friction), the static friction coefficient being larger - it takes more force to get an object sliding from a stop than to keep it sliding once it’s already moving.

So basically you’re already doing the right thing except for the friction force being proportional to the velocity - friction is substantially independent of velocity. (If you were talking about fluid drag, such as from a boat motoring through the ocean, then the drag would be proportional to velocity, at slow speeds - and velocity squared at high speeds.)

235b371d8fea8dbf60a687b31f8b5f67
0
staticVoid2 101 Apr 02, 2010 at 20:23

I think I understand now, if you are going really fast and suddenly turn your tyres 90 degrees the car won’t suddenly stop but rather slide because friction force is not enough to overcome such a velocity.

So the magnitude of the friction force stays the same (assuming no weight transfer) its just the direction of the force changes. And this also means I don’t need to calculate the net force from the current and previous velocity… I just apply friction force and its gets integrated in on the next update. thanks, I’ll try implemnting it and see what happens.

235b371d8fea8dbf60a687b31f8b5f67
0
staticVoid2 101 Apr 02, 2010 at 23:09

Thanks a lot, Its working perfect now.

235b371d8fea8dbf60a687b31f8b5f67
0
staticVoid2 101 Apr 03, 2010 at 19:35

I have another problem now.

basically the car body has an angular velocity as soon as I turn the wheels when driving - this velocity will go up (or down) clockwise/anti-clockwise.

The problem is that I get the velocity of the tyre using this angular velocity and the relative offset of the tyre from the cars center of mass. The velocity of the tyre is then used to determine the direction of friction, therefore I apply the friction force in the opposite direction of velocity, which , in the next frame, causes the velocity of that tyre to be in the other direction. This is an infinite loop where the velocity of the tyre switches between negative and positive every 2 frames, the forces of friction balance each other out but not in the same frame. therefore the image jitters about quite noticably. Also when I turn the wheels when the car is stationary it starts to slowly rotate about its center and then back again. I think this is because a force is being applied one frame, the wheels turn (between frames) and then a new set of friction forces applied in the next frame causing an imbalanced force.

Is there any way I can solve this?

A8433b04cb41dd57113740b779f61acb
0
Reedbeta 167 Apr 03, 2010 at 20:10

The friction force should be capped at the force necessary to bring any sideways motion of the wheel to a halt. The friction force should never cause the wheel to start sliding in the opposite direction. (This is physically correct; it’s the way real-world friction behaves, to a very good approximation.) This should help with the first problem and maybe the second problem as well.

235b371d8fea8dbf60a687b31f8b5f67
0
staticVoid2 101 Apr 03, 2010 at 20:49

Thanks again, its working. I just calculated the amount of force required from the current frames velocity. i.e.

F = ma
a = (vf - vi)/t
a = (0 - vi)/t
F = -vi/t * (carMass/4);

I then just check if friction force is greater than this, ifso use this instead.