Faking car physics

Dear readers,

Today I'm going to write about faking car physics, in particular adding a sense of weight to the car. What I mean by this is as follows: when you see a car breaking you'll notice how the car tilts forward as the center of mass moves forward. In the same way when a car accelerates you'll see how it tilts backwards. Likewise when a car turns left, it banks rights and vice versa. See these images:

Car tilting forward

Car tilting sideways

To fake this effect I'm applying two extra forces on the car: one force that lifts the car and one force that pushes the car down. The trick is to find the proper points.
See the following image:

Applying a lifting force (arrow up) and pushing force (arrow down).

Here we have a lifting and pushing force that are moved along the red circle. The lifting force is always directly on the opposite side of the pushing force on the circle. In this particular situation we are pushing the nose down and lifting the back which corresponds to a car that is decelerating. Other situations can be done in the same manner.

So now, given the car speed in global coordinates, how do we find where we position the lifting (and pushing force)? First we store the difference in speed between the last frame and the current frame: this is our deceleration/acceleration vector and determines where we are going to apply our force. This vector should be in car space (not global space) and determines where we apply the lifting force. The larger the length of the deceleration/acceleration vector the stronger the lifting force. At exactly the other side (simply multiply the position of the lifting force by -1) we apply pushing force with the same strength.

This should already work nicely, but a problem with the circle is that the car is longer than it is wide, as a result if we apply the same forces the car will bank too much and even flip the car. So we have to correct for this by either changing the forces when they're applied near the side of the car or change the circle to an ellipsoid. Either should work fine.

-- Stijn