The easiest way would be to ignore it :-) If you really wanted to do it
though, I suppose if you were modelling the tire as a spring, you could
calculate the distance from the center axis to any and all contact points to
find more than one "compression distance," maybe along a centerline "disc"
aligned with the tire. Then, maybe it would be fine to apply more than one
force at the tire based on these contacts, making sure to aim them through the
center of the wheel. So you'd have a distance to the ground as normal, but
also a second distance to the curb at some angle of contact. Maybe they could
just be added together in their respective directions then? Just an idea, I
haven't done it so can't say for sure.
It seems that could eat up a lot of collision detection cycles though if you
happened to be sitting in the middle of a triangle fan or something.
Actually, you might need to take the average of the two forces and their
direction vectors. With a system like this it seems that your tire could
actually be contacting two or more ground triangles (or whatever surface you
use) at a time. Then, if you had two triangles on a ground plane (ignoring
curbs for now), the force would suddenly double, so maybe you could take the
average of the two.
With a curb compressing the tire .02 meters at an angle of 30 degrees to the
car/suspension's vertical axis, and the ground below compressing it at .1
meters, assuming a tire spring rate of 10000 N/M (I don't have a feel for
metric units well, bear with me), you'd have:
Fx = Force longitudinal
Fy = Force vertical
Curb Ground
Fy = [.02 * Cos(30) + .1] * 10,000
Fy = 1173 N
Fx = [.02 * Sin(30)] * 10,000//Ground has no longitudinal effect here
Fx = 100 N
Something like this maybe? But then the problem is, what if you have two
contact "points", both at ground orientation (because there might be a
collision at two ground triangles/polys, even though the surface is flat and
indistinguishable from one big triangle). Both compressions might be .1, but
you'll have a sum of .2 for compression.
You'd wind up with:
Fy = [.1 * Cos(0) + .1] * 10,000
Fy = (.1 + .1) * 10,000
Fy = 2000 N
When really, you should only have 1000 N.. So maybe the forces from more
than one contact point could be averaged together/combined somehow?
Actually, since the two flat ground triangles are not going to intersect the
wheel at the very bottom, the "compressions" might be lower, giving the same
effect as only having one triangle...
Sorry for the confusing rambling, maybe it will help anyway :0)
Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com
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