Car physics: multiple tire contacts?

Suppose I have a tire making contact with the ground at two different points (imagine a tire pressed up against a steep curb). How does one go about figuring out the tire forces? The only option that seems to make sense to me would be to use the contact closest to the "front" of the wheel, depending on which way the wheel is rotating. Has anyone else considered this situation? Any ideas? Jason 379

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 My little car sim screenshots: http://performancesimulations.com/scnshot4.htm

hi, If you have two contact on your tire so the total load on the tire must be distributed on those 2 contacts, so you have two contact with 2 differents load, compute the lateral/longitudinal forces as there were only 1 contact and just add them. I don't know how you calculate your tire load, but the sum of the load of all contact on the tires must be equal of the load if there were only 1 contact. >The only option that seems to make sense to me would be to >use the contact closest to the "front" of the wheel, >depending on which way the wheel is rotating. Has anyone >else considered this situation? Any ideas? But to be honest i din't implement this in our game. but i will implement it in my personall simulation ;o) Sebastien TIXIER - Game Developer Dynamics and Car Physics http://www.eden-studios.fr GPLRank Normal:-44.24 Monster:-124.44

> I don't know how you calculate your tire load, but the sum of the load > of all contact on the tires must be equal of the load if there were > only 1 contact. There is a problem with that assumption, in that a load say due to a kerb can is not the same as a normal tyre load, you need to take into account how the tyre is being driven up the kerb as this will increase the load, and a tyre contact cuase by a kerb or other non flat surface dosnt follow the same rules as a normal tyre contact so formulas like Pacejka no longer work correctly. To model a situation such as that you need to different approach to tyre modelling, something like a brush method may be off more help. Its a complicated area.

>There is a problem with that assumption, in that a load say due to a kerb >can is not the same as a normal tyre load, you need to take into account how >the tyre is being driven up the kerb as this will increase the load, and a >tyre contact cuase by a kerb or other non flat surface dosnt follow the same >rules as a normal tyre contact so formulas like Pacejka no longer work >correctly. I think a good way to handle this is to modify the Pacejak coefficient according to the number and position of the contacts. In fact the pacejka coef are already variables according to tire pressure, tire temperature, etc ... The problems is to find how to modify this ... i think there's some information in Giancarlo Genta's book to help but i still didn't read entirely his book ;o) >To model a situation such as that you need to different approach to tyre >modelling, something like a brush method may be off more help. Its a >complicated area. > > Sebastien TIXIER - Game Developer Dynamics and Car Physics http://www.eden-studios.fr GPLRank Normal:-44.24 Monster:-124.44

>There is a problem with that assumption, in that a load say due to a kerb Carol, check your mail please :-) Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com My little car sim screenshots: http://performancesimulations.com/scnshot4.htm

ok try to come up with theroy engine is not good stuff. Please consider the physics.Apply then the theory engine. mm weight transfer verses spring rate!! mm NO please choose a tire manufacturer "J. Todd Wasson" <[email hidden]> wrote in message news:[email hidden]... > 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 pointsto > find more than one "compression distance," maybe along a centerline "disc" > aligned with the tire. Then, maybe it would be fine to apply more thanone > force at the tire based on these contacts, making sure to aim them throughthe > 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 theycould > 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 ifyou > 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 surfaceyou > 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 takethe > average of the two. > > With a curb compressing the tire .02 meters at an angle of 30 degrees tothe

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> 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 flatand > indistinguishable from one big triangle). Both compressions might be .1,but

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> 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 intersectthe > wheel at the very bottom, the "compressions" might be lower, giving thesame

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> 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 > > My little car sim screenshots: > http://performancesimulations.com/scnshot4.htm >

Umm, wha? [email hidden] wrote: > ok try to come up with theroy engine is not good stuff. > Please consider the physics.Apply then the theory engine. > mm weight transfer verses spring rate!! mm > NO please choose a tire manufacturer

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>Umm, wha? > >[email hidden] wrote: > >> ok try to come up with theroy engine is not good stuff. >> Please consider the physics.Apply then the theory engine. >> mm weight transfer verses spring rate!! mm >> NO please choose a tire manufacturer >

My thoughts exactly.... ? Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com My little car sim screenshots: http://performancesimulations.com/scnshot4.htm

On Mon, 21 Jan 2002 10:43:58 +0000, Ashley McConnell <[email hidden]> wrote: >Umm, wha? Seems the Martians have *just* landed. This is all way beyond our heads. >[email hidden] wrote: > >> ok try to come up with theroy engine is not good stuff. >> Please consider the physics.Apply then the theory engine. >> mm weight transfer verses spring rate!! mm >> NO please choose a tire manufacturer Ruud van Gaal Free car sim : http://www.marketgraph.nl/gallery/racer/ Pencil art : http://www.marketgraph.nl/gallery/

Phew, I am not going mad! All your physics belongs to us! :D Ash Ruud van Gaal wrote:

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> On Mon, 21 Jan 2002 10:43:58 +0000, Ashley McConnell > <[email hidden]> wrote: > > >>Umm, wha? >> > > Seems the Martians have *just* landed. > This is all way beyond our heads. > >

Phew, I am not going mad! All your physics belongs to us! :D Ash

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> On Mon, 21 Jan 2002 10:43:58 +0000, Ashley McConnell >>Umm, wha? > Seems the Martians have *just* landed. > This is all way beyond our heads.