Car physics -- Braking forces

When looking at slip ratio vs. longitudinal tire force data, is the longitudinal force shown always in the direction opposite the true velocity of the tire, or opposite the tire's direction? For instance, if a tire moving north (0 degrees), but pointed to the right at 10 degrees (10 degree slip angle), has the brake applied enough to allow almost no sideforce, will the direction of the braking force (shown in measured data) be at 180 degrees, or 190 degrees? If it's at 190 degrees, the front of the car should turn left when the wheel is turned right under near-limit braking conditions. This surely can't be right, can it? If not, then isn't all combined slip angle and slip ratio vs. lateral and longitudinal force tire data showing the longitudinal force to be opposite the direction of tire travel, rather than having anything to do with the tire's orientation? (Did that make sense?) If a tire is locked, wouldn't the force be applied more or less in the opposite direction the tire is moving, regardless of the direction it's pointing? I need to make sure my braking forces are going the right direction here! Obviously, tractive force works in the direction the tire is pointing, is this how tire data is displayed? Thanks in advance for any clarification on this. Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

Hi ! Well, the "longitudinal" force is always in the direction of the wheel, BUT you have always "lateral" forces if the real velocity's direction si not the same as the wheel direction ( you have a slip angle ). Moreover, in high slip ration ( locked tire ) you have a rubber friction envolve : something like friction = k * velocity + Cst, in opposite direction of the velocity. This friction force is the dynamic friction + static friction. And last don't forget that lateral force is dependant of slip angle AND slip ration, and that the same for longitudinal force. In lot of books you'll read that if you make a vector of slip_ratio and slip_angle where slip_ration is X component and slip_angle is Y component, then this vector cover an ellipse. regards, Seb "J. Todd Wasson" a �crit :

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> When looking at slip ratio vs. longitudinal tire force data, is the > longitudinal force shown always in the direction opposite the true velocity of > the tire, or opposite the tire's direction? > For instance, if a tire moving north (0 degrees), but pointed to the right at > 10 degrees (10 degree slip angle), has the brake applied enough to allow almost > no sideforce, will the direction of the braking force (shown in measured data) > be at 180 degrees, or 190 degrees? If it's at 190 degrees, the front of the > car should turn left when the wheel is turned right under near-limit braking > conditions. This surely can't be right, can it? If not, then isn't all > combined slip angle and slip ratio vs. lateral and longitudinal force tire data > showing the longitudinal force to be opposite the direction of tire travel, > rather than having anything to do with the tire's orientation? (Did that make > sense?) > > If a tire is locked, wouldn't the force be applied more or less in the > opposite direction the tire is moving, regardless of the direction it's > pointing? > > I need to make sure my braking forces are going the right direction here! > Obviously, tractive force works in the direction the tire is pointing, is this > how tire data is displayed? > > Thanks in advance for any clarification on this. > > Todd Wasson > --- > Performance Simulations > Drag Racing and Top Speed Prediction > Software > http://PerformanceSimulations.Com

-- Seb Game Developer GPLRank -37.10 http://magicfr.multimania.com

Hi, you cannot really separate longitudinal and lateral forces, you have to look at the combined effect. When a tire with a slip angle of 10 degrees from your example is under a braking force, it also develops a slip ratio. It's best to combine the two and think instead of the slip vector where one component is the slip ratio and the other is the slip angle. In a slightly idealized situation, the force is dependent only on the magnitude of this slip vector and points in the (opposite) direction. Of course, in reality things aren't quite as simple, but it's a good starting point, at least in terms of how to imagine it. The lateral and longitudinal directions are defined with respect to the wheel orientation, probably to ensure that a free rolling tire would have (approximately) only a lateral component of force present when put at a slip angle with respect to the road, as the force in this example points approximately in the direction of the wheel axis (no camber situation). -Gregor "J. Todd Wasson" wrote:

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> > When looking at slip ratio vs. longitudinal tire force data, is the > longitudinal force shown always in the direction opposite the true velocity of > the tire, or opposite the tire's direction? > > For instance, if a tire moving north (0 degrees), but pointed to the right at > 10 degrees (10 degree slip angle), has the brake applied enough to allow almost > no sideforce, will the direction of the braking force (shown in measured data) > be at 180 degrees, or 190 degrees? If it's at 190 degrees, the front of the > car should turn left when the wheel is turned right under near-limit braking > conditions. This surely can't be right, can it? If not, then isn't all > combined slip angle and slip ratio vs. lateral and longitudinal force tire data > showing the longitudinal force to be opposite the direction of tire travel, > rather than having anything to do with the tire's orientation? (Did that make > sense?) > > If a tire is locked, wouldn't the force be applied more or less in the > opposite direction the tire is moving, regardless of the direction it's > pointing? > > I need to make sure my braking forces are going the right direction here! > Obviously, tractive force works in the direction the tire is pointing, is this > how tire data is displayed? > > Thanks in advance for any clarification on this. > > Todd Wasson > --- > Performance Simulations > Drag Racing and Top Speed Prediction > Software > http://PerformanceSimulations.Com

Todd, It would seem as you are over-simplifying a bit, the coefficient of static friction is not exactly a constant. Under driving conditions, many effects come into play that reduce the "stiction" of a good tire. These effects are deflection of the tire, suspension movement, spring rates, schock valving, tire-surface temperature, inflation pressure, and so on. But the proportionality law still holds reasonably true under these conditions. Now you can see that if you are cornering, braking, or accelerating at the limit, which means at the adhesive limit of the tires, any weight transfer will cause the tires unloaded by the weight transfer to pass from sticking into sliding. The force required to slide a tire is called the adhesive limit of the tire, or sometimes the "stiction", which is a slang combination of ``stick'' and ``friction.'' This law, can be expressed in a mathematical form, where the force with which the tire resists sliding; is the coefficient of static friction or coefficient of adhesion; and is the weight or vertical load on the tire contact patch. Both have the units of force (remember that weight is the force of gravity), so is just a number, a proportionality constant. Thus, the maximum sideways force the tire can withstand and is equal to the stiction. We often like to speak of the sideways acceleration the car can achieve. Actually, the transition from sticking `mode' to sliding mode should not be very abrupt in a well-designed tire. When one speaks of a ``forgiving'' tire, one means a tire that breaks away slowly as it gets more and more force or less and less weight, giving the driver time to correct. Old, hard tires are, generally speaking, less forgiving than new, soft tires. Low-profile tires are less forgiving than high-profile tires. Slicks are less forgiving than DOT tires. But these are very broad generalities and tires must be judged individually. Some tires are so unforgiving that they break away virtually without warning, leading to driver dramatics usually resulting in a spin. On the other hand, you might be over-complicating things too - the best way to understand a "picture" of the braking forces is provided courtesy of "for every action there is an equal AND opposite reaction," combine that with another physics law "a body in motion will always seek the path of least resistance" - (hence the "opposite" you refer to in the slip angle). While I think I might have added confusion, I hope this helped... Michael FastCo86 www.MichaelLathrop.com

>you cannot really separate longitudinal and lateral forces, you have to >look at the combined effect. When a tire with a slip angle of 10 degrees >from your example is under a braking force, it also develops a slip >ratio. Yes, I know this :0) However, if I have a set of slip angle and slip ratio data to extrapolate approximate force values from, there is a point where a large negative slip ratio will more or less disallow any side force from being developed by slip angle (friction circle/ellipse). Instead of looking at the combined effect, I'm looking at the total force as though it's broken into longitudinal and lateral components, as this is how published data appears to be represented, and the fitting formulas seem to reflect this as well. If the slip ratio is constant, then a slip angle will cause a certain lateral force that could be looked up from tire data. There will also be a longitudinal force because of the slip ratio as you pointed out. My question is, in what direction is this longitudinal force applied? At 180 degrees to the tire heading, or at 180 degrees to the tire's velocity vector? If it's at 180 degrees to the tire heading, then locking the tire or causing the tire to run at a large negative slip ratio would cause a car to steer the wrong way under near limit-braking, because the lateral force from slip angle could be very small compared to the lateral component of the longitudinal force from slip ratio (depending on steer angle). This doesn't seem right to me, hence the question. What is the SAE standard for this? It IS broken into two components, from what I've read so far. My confusion is in the area concerning the direction of the longitudinal component of the "slip vector." >It's best to combine the two and think instead of the slip vector >where one component is the slip ratio and the other is the slip angle. This won't do me any good in my simulation work :0) If I'm calculating lateral force from a Pacejka Magic Tire Model or something similar, I need to know the direction (90 degrees to the tire's heading). Longitudinal force would be calculated through a different equation with different coefficients, but it would be a seperate calculation. For instance, I may have two functions that look like this: LateralForce = GetLateralForce (SlipAngle,SlipRatio,InclinationAngle,Load) LongitudinalForce = GetLongitudinalForce(SlipAngle,SlipRatio,InclinationAngle,Load) See what I mean? Pacejka's formulas appear to work this way. Once I've got the LateralForce, I can apply it at 90 degrees to the tire's forward direction. But what about the LongitudinalForce? Is this at 180 degrees to the tire's direction, or at 180 degrees to the velocity vector? Of course, I could add these two together to get the "slip vector" you described, but first I'd need to know the direction of the longitudinal component. >In a slightly idealized situation, the force is dependent only on the >magnitude of this slip vector and points in the (opposite) direction. I'm having doubts about this. If I crank the front wheels to a 50 degree angle to the right and lock the brakes, or apply "just enough" braking to cause "pure" longitudinal force (friction circle, there's no or very little force available now for lateral force), the car won't turn to the right or go straight, it'll turn to the left instead. This is why it doesn't seem right to me. If I lock the front tires, won't the car go straight ahead, regardless of steer angle? It's producing a big longitudinal force, but it seems this acts in a direction opposite the tire's velocity vector, not the tire's heading vector. What's the SAE standard here? How do they do things? Thanks for your time :-) Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

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>Todd, >It would seem as you are over-simplifying a bit, the coefficient of static >friction is not exactly a constant. Under driving conditions, many effects >come into play that reduce the "stiction" of a good tire. These effects are >deflection of the tire, suspension movement, spring rates, schock valving, >tire-surface temperature, inflation pressure, and so on. But the >proportionality law still holds reasonably true under these conditions. Now >you can see that if you are cornering, braking, or accelerating at the >limit, which means at the adhesive limit of the tires, any weight transfer >will cause the tires unloaded by the weight transfer to pass from sticking >into sliding. > > >The force required to slide a tire is called the adhesive limit of the tire, >or sometimes the "stiction", which is a slang combination of ``stick'' and >``friction.'' This law, can be expressed in a mathematical form, where the >force with which the tire resists sliding; is the coefficient of static >friction or coefficient of adhesion; and is the weight or vertical load on >the tire contact patch. Both have the units of force (remember that weight >is the force of gravity), so is just a number, a proportionality constant. >Thus, the maximum sideways force the tire can withstand and is equal to the >stiction. We often like to speak of the sideways acceleration the car can >achieve. > >Actually, the transition from sticking `mode' to sliding mode should not be >very abrupt in a well-designed tire. When one speaks of a ``forgiving'' >tire, one means a tire that breaks away slowly as it gets more and more >force or less and less weight, giving the driver time to correct. Old, hard >tires are, generally speaking, less forgiving than new, soft tires. >Low-profile tires are less forgiving than high-profile tires. Slicks are >less forgiving than DOT tires. But these are very broad generalities and >tires must be judged individually. Some tires are so unforgiving that they >break away virtually without warning, leading to driver dramatics usually >resulting in a spin. > >On the other hand, you might be over-complicating things too - the best way >to understand a "picture" of the braking forces is provided courtesy of "for >every action there is an equal AND opposite reaction," combine that with >another physics law "a body in motion will always seek the path of least >resistance" - (hence the "opposite" you refer to in the slip angle). > >While I think I might have added confusion, I hope this helped...

So what direction does the longitudinal force get applied, assuming the rest of what you mentioned is already done? Is it opposite the tire's heading vector, or opposite the tire's velocity vector? Thanks for your help :-) Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

>Well, the "longitudinal" force is always in the direction of the wheel, BUT >you >have always "lateral" forces if the real velocity's direction si not the same >as >the >wheel direction ( you have a slip angle ). What if the braking torque is enough to cause the slip ratio to be such that there is very little or no lateral force (friction circle/ellipse)? Turning the front wheels right 80 degrees would cause the car to turn left if the longitudinal force is applied along the wheel direction. What am I missing here?>Moreover, in high slip ration ( locked tire ) you have a rubber friction >envolve : >something like friction = k * velocity + Cst, in opposite direction of the >velocity. >This friction force is the dynamic friction + static friction. This sounds like a Coulomb friction type model. I wouldn't touch that with a ten foot pole in regard to tire modelling, personally. Seems much better to use slip angle and slip ratio data. >And last don't forget that lateral force is dependant of slip angle AND slip >ration, >and that the same for longitudinal force. > > Done :-) Now I just need to make sure the longitudinal force is being applied in the correct direction. Thanks for your help, Seb Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

"J. Todd Wasson" <[email hidden]> wrote in message news:[email hidden]... > >Todd, > >It would seem as you are over-simplifying a bit, the coefficient ofstatic > >friction is not exactly a constant Make it stop! My head hurts trying to follow this... dave henrie

"J. Todd Wasson" wrote: > So what direction does the longitudinal force get applied, assuming the rest > of what you mentioned is already done? Is it opposite the tire's heading > vector, or opposite the tire's velocity vector? I like the resultant force in the direction opposite to the velocity vector when the tire is locked because I don't much like the idea of being able to do much steering of the car by twisting locked wheels back and forth. Having said that, like you I have also noticed that some combined slip "data expansion" methods don't very closely meet this limit condition. I suspect that locked cases are expensive to measure (because they would eat up the tires and because of the large heat input and consequent temperature variation during the test). Locked cases with slip angles are then presumably even more rare. I'm more willing to believe in a limited amount of SAE tire axes lateral force component in wheel spinning cases, because my own game models tend to spin the car in yaw "too much" without it. However, no one has yet stepped forward to let me burn doughnuts with their high performance car, so for me this is just a feel thing.... ;) - Matt

On 24 Jan 2001, J. Todd Wasson wrote: > What is the SAE standard for this? It IS broken into two components, from > what I've read so far. My confusion is in the area concerning the direction of > the longitudinal component of the "slip vector." SAE longitudinal is always aligned with the wheel plane (and lateral is always perpendicular). So if you could turn the front wheels suddenly to 90 degrees, then the tire lateral force would all be slowing (braking) the motion of the car (wheels free rolling or locked). You have to learn to think in multiple different coordinate systems.... -- Doug Douglas Milliken <[email hidden]> Milliken Research Associates Inc.

> What if the braking torque is enough to cause the slip ratio to be such that > there is very little or no lateral force (friction circle/ellipse)? Turning > the front wheels right 80 degrees would cause the car to turn left if the > longitudinal force is applied along the wheel direction. What am I missing > here? Well, its more complicated than that :o) I really recommand you Race Car Vehicle Dynamic page 58, and you'll understand :o) If we take the extrem : let the front wheel be right 90 degrees. and braking .. what hapening > slip angle is -90 degrees so you have a small lateral force that is in direction of the rear of the car. slip angle is maximum, slip ratio is NULL, no longitudinal forces. and it's a realistic behavior, that's why if you apply too much steering the car turn less. More over your braking force only exsit when your vehicle is moving, if speed = 0, then your braking force is NULL. Wheen turning front wheel right, you move right, and the braking force will decrease this force. But will be never greater otherwise you car would go backward when braking. Speaking of backward, do you car turn left or right when turning front left right and driving backward ? :o)

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> >Moreover, in high slip ration ( locked tire ) you have a rubber friction > >envolve : > >something like friction = k * velocity + Cst, in opposite direction of the > >velocity. > >This friction force is the dynamic friction + static friction. > > This sounds like a Coulomb friction type model. I wouldn't touch that with a > ten foot pole in regard to tire modelling, personally. Seems much better to > use slip angle and slip ratio data.

Well, remember that slip_ratio/slip_angle model is not right at high slip_angle slip_ratio value, at this level the rubber friction model is more right ( especially when tires are totally locked ).

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> >And last don't forget that lateral force is dependant of slip angle AND slip > >ration, > >and that the same for longitudinal force. > > > > > > Done :-) Now I just need to make sure the longitudinal force is being > applied in the correct direction.

Longitudinal force is ALWAYS applied in wheel direction ! you'll see that in all the books talking about car dynamics :o) > Thanks for your help, Seb > you'r welcome ! -- Seb Game Developer GPLRank -37.10 http://magicfr.multimania.com

>I really recommand you Race Car Vehicle Dynamic page 58, and you'll >understand :o) > Got it. I think I see my problem. What I was doing previously was calculating longitudinal force from slip ratio, then calculating lateral force from slip angle. Then, if the resultant force was too large, I cut the lateral force to stay inside the friction circle. From looking at the graph, if I had a slip ratio of .048, as slip angle is increased, the longitudinal force decreases as lateral force increases. My problem is that these are not working together properly. That's what happens when I use simple approximations instead of a good tire model :0) With this information, it's clear now that as slip angle increases, even if the slip ratio remains constant, the longitudinal force will decrease, and the car should go the way it's supposed to. So the answer to my question is: The longitudinal force indicated by all the graphs and formulas (as you and a couple others mentioned) is measured in the tire coordinate system. Just because the slip ratio is constant, *does not* make the longitudinal force constant (as it does now in my tire model), but rather, the longitudinal force will vary as slip angle varies. Thanks much :0) >More over your braking force only exsit when your vehicle is moving, ifspeed

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>= 0, >then >your braking force is NULL. Wheen turning front wheel right, you move right, >and >the braking force >will decrease this force. But will be never greater otherwise you car would >go >backward when braking. >Speaking of backward, do you car turn left or right when turning front left >right >and driving backward ? :o) >

Right, I got the friction reversal problem in the brakes fixed. It wasn't too hard after all :0) The car works fine when running in reverse. It'll do Rockford's/backwards 180's pretty easily. Fun! >Well, remember that slip_ratio/slip_angle model is not right at high >slip_angle >slip_ratio value, at this level the rubber friction model is more right ( >especially when tires are totally locked ). Ok, I'll keep this in mind. BTW, if the tire's are locked, what direction is the force? Is this the only case where the force will act opposite the velocity vector? Matt Jessick pointed out that there probably isn't much data for locked tires, as they'd heat up and get destroyed in a real hurry. Right now, if I lock the front tires, the car goes straight ahead no matter what I do, so I'm not really worried about putting in a special case routine, but I'll keep it in mind anyway :-) >Longitudinal force is ALWAYS applied in wheel direction ! you'll see that in >all >the books talking about car dynamics :o) Yes, I thought so. Thanks for confirming this. My problem again was I didn't realize that a constant slip ratio won't mean a constant longitudinal force, but instead it will vary with slip angle as well. The problem that led to these questions reared it's head when I attempted to slide the car sideways with four locked tires. The lateral force (because of my make-shift , very simple tire model) became zero. The slip ratio went to -1, causing my algorithm to cancel the lateral force in order to keep within the bounds of traction. Bang, the car keeps going sideways. Now, I see that a -1 slip ratio and a 80-90 degree slip angle will result in real close to 0 longitudinal force, allowing the rest to go along the lateral direction, just like the graph on page 58 shows. Thanks again to everybody for your help! Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

>I like the resultant force in the direction opposite to the velocity vector >when the tire is locked because I don't much like the idea of being able >to do much steering of the car by twisting locked wheels back and forth. Yes, this is what I'm thinking too. Right now when my front tires lock, I can't turn the car anyway, but that's because my algo kills the lateral force entirely long before the tire locks. I think I know exactly what my problem has been. >I'm more willing to believe in a limited amount of SAE tire axes lateral >force component in wheel spinning cases, because my own game models >tend to spin the car in yaw "too much" without it. Yes, mine is very hard to control when you spin the tires too (unless you do it in reverse, of course :-)). It's hard to believe that a tire would not distort laterally when spinning while at some non-zero slip angle. I think there's still got to be some lateral force there. Friction circle theory isn't perfect after all :0) Thanks, Matt Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

Tire model, is the most important and difficult simulation in car physics :o) As you can see in my signature i'm a game developer, if you check www.eden-studios.fr , i'm the bold man on the go-kart photo :o) Anyway .... me and my co-worker on car physics have spend lot of time and we will spend again lot of time on tire model because it's the only link with the ground and must be very accurate and tweakable ... a very good tire model imply that you can do a good sim but also a good less-sim game :o) "J. Todd Wasson" a �crit :

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> > >Well, remember that slip_ratio/slip_angle model is not right at high > >slip_angle > >slip_ratio value, at this level the rubber friction model is more right ( > >especially when tires are totally locked ). > > Ok, I'll keep this in mind. BTW, if the tire's are locked, what direction is > the force? Is this the only case where the force will act opposite the > velocity vector?

Yes, the friction model's direction is opposite to velocity vector . The friction is a force tha work against motion. > Matt Jessick pointed out that there probably isn't much data > for locked tires, as they'd heat up and get destroyed in a real hurry. Right > now, if I lock the front tires, the car goes straight ahead no matter what I > do, so I'm not really worried about putting in a special case routine, but I'll > keep it in mind anyway :-) Well, when you lock all the tire the car must goes in the car velocity direction. Because in locked tires case you have no more a car whith tire but a car with rubber block instead :o) > Thanks again to everybody for your help! Your welcome ! -- Seb Game Developer GPLRank -37.10 http://magicfr.multimania.com

"J. Todd Wasson" wrote:

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> > Yes, mine is very hard to control when you spin the tires too (unless you do > it in reverse, of course :-)). It's hard to believe that a tire would not > distort laterally when spinning while at some non-zero slip angle. I think > there's still got to be some lateral force there. Friction circle theory isn't > perfect after all :0) > > Thanks, Matt > > Todd Wasson > --- > Performance Simulations > Drag Racing and Top Speed Prediction > Software > http://PerformanceSimulations.Com

Of course, the tire distorts laterally even when spinning if travelling sideways, but this is already inherent in the friction circle equations, as the force will in fact oppose the slip velocity, which in this example doesn't point in the direction of the wheel plane but also has a corresponding lateral component. Another factor to consider is if you take into account the velocity component on the rear tires that comes from the yawing velocity of the whole vehicle. This means that the slip velocity vector at the rear tires will acquire a component that opposes the yawing tendency, and will have the more relative effect the less the tires are spinning. A similar thing also happens at the front, and it all causes damping in that axis, making the transition less sudden. -Gregor

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>> Yes, mine is very hard to control when you spin the tires too (unless you >do >> it in reverse, of course :-)). It's hard to believe that a tire would not >> distort laterally when spinning while at some non-zero slip angle. I think >> there's still got to be some lateral force there. Friction circle theory >isn't >> perfect after all :0) >Of course, the tire distorts laterally even when spinning if travelling >sideways, but this is already inherent in the friction circle equations,

I guess what I was really trying to say here was that according to friction circle theory (from what I thought I got out of it, anyway), if the tire's maximum force is being applied along the tire's longitudinal axis, there is none left for lateral force. Matt indicated that he liked to include some lateral force anyway, because otherwise there can be very strong yawing tendencies. With 0 lateral force at the rears, you can spin around rather quickly. For instance, if the tire's spinning wildly, at 100 slip ratio, all the available force should be applied along the longitudinal axis according to what little I've read on friction circle use. I agreed with his approach, saying it would probably be best for me to go ahead and allow a small amount of lateral force anyway. >Another factor to consider is if you >take into account the velocity component on the rear tires that comes >from the yawing velocity of the whole vehicle. That's already included in my model when calculating the instantaneous direction of each tire, before finding the tire heading, and before subtracting the two to find the slip angle. >This means that the slip >velocity vector at the rear tires will acquire a component that opposes >the yawing tendency, and will have the more relative effect the less the >tires are spinning. By "a component", you mean the lateral component. I think I see what you're doing. You're letting the velocity vector define the total available force (and possibly limiting it if it gets too big), then braking it into lateral and longitudinal components if you want to analyze those seperately. This way, even a wildly spinning tire that has some slip angle/lateral movement will produce a lateral force. That's interesting. Different from what I'm doing. I'll ponder that awhile :0) Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

Thanks, Seb. Yes, the tire model is very important indeed. Unfortunately, I've not included much of one yet. Still working on getting other things working first. Out of curiousity, did you have and solve the low speed slip ratio instability problem? If so, how? Thanks, Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

"J. Todd Wasson" a �crit : > Thanks, Seb. Yes, the tire model is very important indeed. Unfortunately, > I've not included much of one yet. Still working on getting other things > working first. Out of curiousity, did you have and solve the low speed slip > ratio instability problem? If so, how? Well, the main problem was that at high slip angles there is a static lateral forces ... if we look the factor = f( slip_angle), if slip_angle > 15� factor = 0.75 and Lateral force = tire_load * factor. So at low speed ,in theory there is an oscilation because the lateral force tend to opposite the slip_angle. the slip_angle could switch from -90� to 90�. But as we compute dynamic by interuption at 200hz, the oscilation is not perceptible and the model works well at low speed . -- Seb Game Developer GPLRank -37.10 http://magicfr.multimania.com

S�bastien Tixier wrote:

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> > Well, the main problem was that at high slip angles there is a static lateral > forces ... > > if we look the factor = f( slip_angle), if slip_angle > 15� factor = 0.75 and > Lateral force = tire_load * factor. > > So at low speed ,in theory there is an oscilation because the lateral force tend > to > opposite the slip_angle. the slip_angle could switch from -90� to 90�. > > But as we compute dynamic by interuption at 200hz, the oscilation is not > perceptible and > the model works well at low speed . > -- > Seb > Game Developer > GPLRank -37.10 > http://magicfr.multimania.com

I was thinking about the low speed model as well a while ago, and it's obvious that the slip angle model is just a high speed approximation. The reason is that, when the slip angle is changed, the tire footprint requires some rolling length to adapt to it, and if you take this into account the low speed behaviour is regularized, and the car behaviour at low speeds should become more natural (this is one of the few areas where GPL seems to fail). I believe WSC will take the adaptation length into account. -Gregor

On Mon, 29 Jan 2001, Gregor Veble wrote: ... > requires some rolling length to adapt to it, and if you take this into > account the low speed behaviour is regularized, and the car behaviour at > low speeds should become more natural (this is one of the few areas > where GPL seems to fail). > > I believe WSC will take the adaptation length into account. Terminology quibble: Normal term for this in the USA is "relaxation length".

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>Well, the main problem was that at high slip angles there is a static lateral >forces ... > >if we look the factor = f( slip_angle), if slip_angle > 15� factor = 0.75 >and >Lateral force = tire_load * factor. > >So at low speed ,in theory there is an oscilation because the lateral force >tend >to >opposite the slip_angle. the slip_angle could switch from -90� to 90�. > >But as we compute dynamic by interuption at 200hz, the oscilation is not >perceptible and >the model works well at low speed .

Ok, this is what I'm doing as well. I was really referring to slip ratio (longitudinal, or rotational direction), where the denominator approaches zero. I've written a low speed model based on a torsion spring/damper that seems to work well, eliminating the instability at low speeds, and also allowing the car to stop on hills, as well as "pop back" a little when you stop. Any tips on this area? Also, I've looked and for the life of me can't figure out how to include the relaxation length in a slip angle based model. Any ideas? Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com