Pacejka magic formula, longitudinal force and slip ratio

Hello, I am trying to code my own car simulator, and have some difficulties to understand how slip ratio is to be interpreted and how I should interpret it to generate proper wheel angular velocities. I've read this carefully: http://groups.google.com/group/rec.autos.simulators/tree/browse_frm/thread/1e4d0a2b2ce69ad7/580a3b90f0f630e3?rnum=1&q=pacejka+slip+ratio&_done=%2Fgroup%2Frec.autos.simulators%2Fbrowse_frm%2Fthread%2F1e4d0a2b2ce69ad7%2F566abcb8cf36deb2%3Flnk%3Dgst%26q%3Dpacejka%2Bslip%2Bratio%26#doc_580a3b90f0f630e3 but there are still some contradictions I don't grasp. From what I understand, in real life, slip ratio is the proportion of contact patch that is sliding (usually the aft part as the wheel tread is lifted as the wheel rolls) over the part that adheres on the ground (fore area of the patch, pressed into adherence by the tread being pushed down on the ground as the wheel rolls). In other words, in my wheel model, as long as slip ratio is not 100%, at contact patch wheel tangential velocity (radius x angular velocity x 2PI) more or less matches ground velocity. If slip ratio is 100%, then the wheel slides, and the generated force is constant: whatever the velocity delta, the resulting force is the same, and wheel angular velocity change depends on the resultant of all torques, and the wheel inertia (plus connexions to the drive train). But Pacejka magic formula takes as input a totally different slip ratio: it is the ratio of the two velocities I mentioned previously, measured at the previous simulation step. Using these formulae, typical road tyre generates maximum force at slip ratios between 10 and 20%, and doesn't generate anything significant when slip ratio is 0. This means that for the tyre to generate maximum force the totality of the contact patch slides, because I can't really well see how 80% of the patch could be in static friction state when I have a tangential velocity of 1.2m.s-1 over a ground that travels at 1m.s-1. And I see that even less when the tyre has a tangential velocity of 60m.s-1 with ground at 50m.s-1 (same slip ratio, but 10m.s-1 of relative velocity!). This means that when I want to render the wheel angular position, using the velocity ratio definition, it constantly slides over the ground, which is hardly acceptable. So, how can I use Pacejka magic formula, slip ratio as it is said to be defined, and still have realistic wheel angular position rendering ? Thanks for any insights on that matter, Benoit Germain

I can't provide the math that describes this, but you can find it in Wong's 'Theory of Ground Vehicles'. Wong refers to "Juliens theory" on slip deformation, and provides the equations. I can describe the mechanical/physical reason that the circumferential velocity of the wheel/tire doesn't match the ground speed of the vehicle during longitudinal acceleration and braking. The reason is that as a tractive force is applied to the tire, the tire tread immediately in front of the contact patch is compressed. As these compressed tread elements enter the contact patch the distance that the tire travels when subject to driving torque is less than a free- rolling wheel. I suppose that as tractive force increases the compression of the tread increases which increases the slip ratio. At low slip ratios adhesion remains constant through the contact patch. The relationship between drive torque and tractive effort remains relatively linear until approaching peak slip ratio. Then the tractive efficiency tails off as more and more of the back side of the contact patch starts to slide relative to the road. Pat Dotson UltraForce Simulations LLC http://www.ultraforcesim.com On Mar 18, 5:39 am, [email hidden] wrote:

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> Hello, > > I am trying to code my own car simulator, and have some difficulties > to understand how slip ratio is to be interpreted and how I should > interpret it to generate proper wheel angular velocities. I've read > this carefully:http://groups.google.com/group/rec.autos.simulators/tree/browse_frm/t... > but there are still some contradictions I don't grasp. > > From what I understand, in real life, slip ratio is the proportion of > contact patch that is sliding (usually the aft part as the wheel tread > is lifted as the wheel rolls) over the part that adheres on the ground > (fore area of the patch, pressed into adherence by the tread being > pushed down on the ground as the wheel rolls). In other words, in my > wheel model, as long as slip ratio is not 100%, at contact patch wheel > tangential velocity (radius x angular velocity x 2PI) more or less > matches ground velocity. If slip ratio is 100%, then the wheel slides, > and the generated force is constant: whatever the velocity delta, the > resulting force is the same, and wheel angular velocity change depends > on the resultant of all torques, and the wheel inertia (plus > connexions to the drive train). > > But Pacejka magic formula takes as input a totally different slip > ratio: it is the ratio of the two velocities I mentioned previously, > measured at the previous simulation step. Using these formulae, > typical road tyre generates maximum force at slip ratios between 10 > and 20%, and doesn't generate anything significant when slip ratio is > 0. This means that for the tyre to generate maximum force the totality > of the contact patch slides, because I can't really well see how 80% > of the patch could be in static friction state when I have a > tangential velocity of 1.2m.s-1 over a ground that travels at 1m.s-1. > And I see that even less when the tyre has a tangential velocity of > 60m.s-1 with ground at 50m.s-1 (same slip ratio, but 10m.s-1 of > relative velocity!). > > This means that when I want to render the wheel angular position, > using the velocity ratio definition, it constantly slides over the > ground, which is hardly acceptable. > > So, how can I use Pacejka magic formula, slip ratio as it is said to > be defined, and still have realistic wheel angular position > rendering ? > > Thanks for any insights on that matter, > > Benoit Germain

On 18 mar, 21:42, [email hidden] wrote: > The reason is that as a tractive force is applied to the tire, the tire tread > immediately in front of the contact patch is compressed. As these > compressed tread elements enter the contact patch the distance that > the tire travels when subject to driving torque is less than a free- > rolling wheel. Thanks for you reply. Let's see if I get this right: Because of engine induced torque, a particular tread point compresses tangentially in front of the patch. This compressed point enters the patch, is kept compressed because of friction, and then is released when the tread leaves the ground.In wheel hub reference frame, this point travels a circle of unloaded radius sligthly deformed toward wheel hub when traversing the contact patch and its neighbourhood. As a result, because of tangential compression and relaxation, tangential velocity decreases in the compression phase, is more or less constant while the point is on the ground, then increases again when it starts sliding as load diminishes because the wheel rolls. I suppose that if we measure tangential distance between the tread point rest position and its position in fully compressed state it should be on the order of a few millimeters, which is at the very most 1% of the wheel's circumference. If a tread point moves more than that, I would expect the sidewalls to deform considerably to the point of generating significant ripples, and I don't think it happens in real life even with a powerful vehicle. I suspect the wheel would spin before it can happen. But more important for me, during one wheel revolution the considered tread point travels less than the unloaded circumference, but more than the loaded circumference, the exact amount being closer to the former (this is my gut feeling here). In the mean time, the vehicle has moved forward by that exact same distance, since there is always a part of the contact patch that does not slide. Am I correct, or is there something I miss here ? > Then the tractive efficiency tails > off as more and more of the back side of the contact patch starts to > slide relative to the road. Now back to my original question: Assume that peak performance is indeed reached at 15% slip ratio on real life everyday tyres on an everyday road, and we have a vehicle in steady state (constant velocity) at that slip ratio. In practice, it is constantly accelerating (bad choice of words, but I don't know how to state it better) to oppose frictions, but as you said, the major part of the contact patch *does not slide*. Now, in my simulation I don't do actual contact patch simulation, but get the generated force from the Pacejka magic formulas. Because of this, slip ratio is not a consequence of patch simulation and the various forces acting on the tyre, but a cause of those forces. I therefore have to measure slip ratio, then feed it to the formula, that yields a force. According to what I have seen so far, slip ratio in that context is the ratio of vehicle forward travel velocity and tread point tangential velocity, computed as a fraction of the unloaded velocity (this fraction being of the order of 99%, or even 100% if the loaded radius is neglected). And here comes my mind dislocation: On one side during a full wheel revolution, the distance travelled by the vehicle should be near equivalent to the wheel's circumference (unloaded or not, as stated above) since the major part of the contact patch is not sliding. On the other side, for Pacejka formula computed force to oppose friction and be able to maintain current velocity, I should feed 15% of slip ratio, and therefore measure 15% of velocity difference. Replace 15% with 3% everywhere in the above text, and it still would be visible and not acceptable, as they represent respectively 53° and 10° of excess angular travel each time the vehicle move forward the equivalent of a wheel circumference. Which means that I should render my vehicle with wheels rotating that way, and therefore show them spinning much faster than they should compared to the vehicle velocity. Because in real life, if I try to move forward a wheel the equivalent of its circumference while I rotate it by one revolution plus 53 degrees, I just can't see how I can do this while having the majority of the contact patch in static friction with the ground at the same time. So, I'm afraid I still don't understand the true reason why Pacejka magic formulas take such a velocity difference as input, and how it can mix well with vehicle rendering. Benoit.

I think you understand the basics of the explanation that I gave. The tread is compressed circumferentially prior to entering the contact patch. That means that the portion of the contact patch that is in constant adhesion to the road is shorter circumferentially than it is at any other point in the rotation of the tire. Slip is the ratio of the difference in length between compressed and uncompressed sections of tire tread. You have to integrate(?) that difference over a complete rotation of the tire to realize the potential magnitude of the difference. The main thing to keep in mind is that, even though it's called 'slip', the tire is really not necessarily sliding at all. At slip ratios less than 100% some part of the tire is in constant contact with the road. I hadn't thought it through enough to consider the magnitude of angular difference. It is amazing that at 15% slip ratio the wheel is really turning an extra 53° every time the vehicle travels a linear distance equaling the circumference of the tire. Per my understanding of slip ratio this is exactly what is happening! And to consider that the tire does that while in constant contact with the road is equally amazing. It might help to imagine taking a cross section of the tire, laying it down on a flat surface, and then applying a weight to the tread portion of the cross section. This would be akin to tangential force applied to a tire via drive torque. Imagine a 1" thick cross section of tire laying on a table. Place a 1000 pound weight on the tread part of the tire section. How much will the rubber tread be compressed? It will probably be significant, and I imagine you've gone well past 15% compression. This is essentially what is happening with regard to slip ratio. > Now back to my original question: Assume that peak performance is > indeed reached at 15% slip ratio on real life everyday tyres on an > everyday road, and we have a vehicle in steady state (constant > velocity) at that slip ratio. If you assume steady state (constant velocity) then there should be no appreciable slip ratio. It takes very little throttle input to maintain a constant speed, which means there is very little force working to compress the tire tread at the contact patch. You would have to have an unreasonable amount of drag on the car to reach 15% slip at constant velocity. Do I get to beta test? :) Pat Dotson UltraForce Simulations LLC http://www.ultraforcesim.com On Mar 20, 6:00 am, Maximilien <[email hidden]> wrote:

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> On 18 mar, 21:42, [email hidden] wrote: > > > The reason is that as a tractive force is applied to the tire, the tire tread > > immediately in front of the contact patch is compressed.  As these > > compressed tread elements enter the contact patch the distance that > > the tire travels when subject to driving torque is less than a free- > > rolling wheel. > > Thanks for you reply. > Let's see if I get this right: > > Because of engine induced torque, a particular tread point compresses > tangentially in front of the patch. This compressed point enters the > patch, is kept compressed because of friction, and then is released > when the tread leaves the ground.In wheel hub reference frame, this > point travels a circle of unloaded radius sligthly deformed toward > wheel hub when traversing the contact patch and its neighbourhood. > > As a result, because of tangential compression and relaxation, > tangential velocity decreases in the compression phase, is more or > less constant while the point is on the ground, then increases again > when it starts sliding as load diminishes because the wheel rolls. I > suppose that if we measure tangential distance between the tread point > rest position and its position in fully compressed state it should be > on the order of a few millimeters, which is at the very most 1% of the > wheel's circumference. If a tread point moves more than that, I would > expect the sidewalls to deform considerably to the point of generating > significant ripples, and I don't think it happens in real life even > with a powerful vehicle. I suspect the wheel would spin before it can > happen. > > But more important for me, during one wheel revolution the considered > tread point travels less than the unloaded circumference, but more > than the loaded circumference, the exact amount being closer to the > former (this is my gut feeling here). In the mean time, the vehicle > has moved forward by that exact same distance, since there is always a > part of the contact patch that does not slide. Am I correct, or is > there something I miss here ? > > > Then the tractive efficiency tails > > off as more and more of the back side of the contact patch starts to > > slide relative to the road. > > Now back to my original question: Assume that peak performance is > indeed reached at 15% slip ratio on real life everyday tyres on an > everyday road, and we have a vehicle in steady state (constant > velocity) at that slip ratio. In practice, it is constantly > accelerating (bad choice of words, but I don't know how to state it > better) to oppose frictions, but as you said, the major part of the > contact patch *does not slide*. > > Now, in my simulation I don't do actual contact patch simulation, but > get the generated force from the Pacejka magic formulas. Because of > this, slip ratio is not a consequence of patch simulation and the > various forces acting on the tyre, but a cause of those forces. I > therefore have to measure slip ratio, then feed it to the formula, > that yields a force. According to what I have seen so far, slip ratio > in that context is the ratio of vehicle forward travel velocity and > tread point tangential velocity, computed as a fraction of the > unloaded velocity (this fraction being of the order of 99%, or even > 100% if the loaded radius is neglected). And here comes my mind > dislocation: > > On one side during a full wheel revolution, the distance travelled by > the vehicle should be near equivalent to the wheel's circumference > (unloaded or not, as stated above) since the major part of the contact > patch is not sliding. > > On the other side, for Pacejka formula computed force to oppose > friction and be able to maintain current velocity, I should feed 15% > of slip ratio, and therefore measure 15% of velocity difference. > Replace 15% with 3% everywhere in the above text, and it still would > be visible and not acceptable, as they represent respectively 53° and > 10° of excess angular travel each time the vehicle move forward the > equivalent of a wheel circumference. > > Which means that I should render my vehicle with wheels rotating that > way, and therefore show them spinning much faster than they should > compared to the vehicle velocity. Because in real life, if I try to > move forward a wheel the equivalent of its circumference while I > rotate it by one revolution plus 53 degrees, I just can't see how I > can do this while having the majority of the contact patch in static > friction with the ground at the same time. > > So, I'm afraid I still don't understand the true reason why Pacejka > magic formulas take such a velocity difference as input, and how it > can mix well with vehicle rendering. > > Benoit.

On 20 mar, 14:50, [email hidden] wrote: > Slip is the ratio of > the difference in length between compressed and uncompressed sections > of tire tread.  You have to integrate(?) that difference over a > complete rotation of the tire to realize the potential magnitude of > the difference. OK, I understand what you mean. > I hadn't thought it through enough to consider the magnitude of > angular difference.  It is amazing that at 15% slip ratio the wheel is > really turning an extra 53° every time the vehicle travels a linear > distance equaling the circumference of the tire.  Per my understanding > of slip ratio this is exactly what is happening! In fact, I didn't reach the figure that way, but it happens that 53° is actually 15% of 360°, so that's consistent. > And to consider that the tire does that while in constant contact with > the road is equally amazing. Yes, to me as well, hence my dubiousness. But it looks like I have to make myself a reason and accept is as reality. > If you assume steady state (constant velocity) then there should be no > appreciable slip ratio.  It takes very little throttle input to > maintain a constant speed, which means there is very little force > working to compress the tire tread at the contact patch.  You would > have to have an unreasonable amount of drag on the car to reach 15% > slip at constant velocity. Yeah, for example 350km/h as in a racing game. Aerodynamics drag is quite enourmous, and tractive force should compress the tread notably. I'd love to see a high frequency video capture of a F1 racing car wheel at that velocity to see if angular velocity is perceivably higher than ground velocity... Not that it would be visible when rendering of course, because the wheel spins so fast we would use a different material to render the spinning rim, or else we would just get a horrible stroboscopic effect. > Do I get to beta test?  :) Given the stage of the project, I fear you'd in for a few years wait. Benoit.

On Mar 20, 12:59 pm, Maximilien <[email hidden]> wrote:

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> On 20 mar, 14:50, [email hidden] wrote: > > > And to consider that the tire does that while in constant contact with > > the road is equally amazing. > > Yes, to me as well, hence my dubiousness. But it looks like I have to > make myself a reason and accept is as reality.

Wong's book describes exactly the idea that I relayed to you. I trust that source so I assume its accurate. I'll confess that I've never looked that closely at the concept of slip ratio before. My only real racing experience is with relatively underpowered karts, so I've never seen a need to apply the concept. All I knew was the basic idea that its the difference between angular wheel speed and linear travel. I never knew the mechanics behind it. This has been a good learning experience. > Yeah, for example 350km/h as in a racing game. Aerodynamics drag is > quite enourmous, and tractive force should compress the tread notably. I'd never considered that before, but you are absolutely right. At terminal velocity with 100% throttle all available engine torque is working through the tires to overcome aero drag. Can you put a number to the drag force in terms of pounds or newtons? I think, though, that available torque at the tires is going to be greatly reduced by gearing, such that in 7th gear there may not be as much torque on the wheels as one might imagine. > > Do I get to beta test?  :) > Given the stage of the project, I fear you'd in for a few years wait. It was really only a joke. I noticed your email domain after a couple of replies and responses. I had previously thought you were doing this for fun. :) Hope this discussion helped in some way. Pat Dotson

> This means that when I want to render the wheel angular position, > using the velocity ratio definition, it constantly slides over the > ground, which is hardly acceptable. > Benoit/Maximilien: Man, I think it's time to get a grip/smell the coffee/stop missing the boat here. All mixed metaphors aside, please be advised that what you deem "hardly acceptable" is the de facto standard in the automotive engineering world. Many, if not most, custom tire models are proprietary enhancements to the Pacejka tire model. >in real life, slip ratio is the proportion of >contact patch that is >sliding (usually the aft part as the wheel tread is lifted as the >wheel rolls) over the part that adheres on the ground Semantics. In my "real life," the slip ratio is as defined by the Pacejka model. It is almost amusing that you're bemoaning what you consider an unsatisfactory formulation utilized by the rest of the automotive engineering world because you can't get your mind around the concept. But in short, you are getting tire design issues and vehicle dynamics issues confused, as well as missing a few key vehicle dynamics considerations. I'll cover both levels. The mechanics of how a given tire generates forces (what is going on in the contact patch) is studied in the world of tire design. The net force(s) a tire DOES generate that act upon a vehicle are studied in the field of vehicle dynamics. Your perceived problem is one of level. Consider the following comparison: Materials develop their strength based on molecular slip; i.e., under load a material deforms based on whether atoms/molecules are slipping past one another, or if the space between them is merely stretched. Applying your point of view to the strength of materials, the only valid view is some explicit measure of molecular slip and displacement. Net properties such as yield strength, ultimate strength, percent elongation, and fatigue properties would be "hardly acceptable." What causes slip and distortion in the contact patch? The only thing that CAN cause it is a difference in the states of motion of the contact patch and the axle. Which means what? It means that tire properties that are described as a function of net motions such as slip ratio and slip angle are both mathematically and philosophically valid. >it constantly slides over the ground, which is hardly acceptable. Maybe you haven't thought about this, but if the contact patch sliding over the ground is truly "hardly acceptable" to you, then you should have already been having heartburn over lateral force generation and the concept of slip angle. I can tell you for a certainty that tire testing done for the purposes of characterizing a tire's handling characteristics for vehicle dynamics purposes, concerns itself with net slip ratio, net slip angle, camber, radius, load, etc. - all net properties, and you get six net forces and moment in your data. >So, how can I use Pacejka magic formula, slip ratio as it is said to be >defined, and still have realistic wheel angular position rendering ? If your tire performance data is right, the wheel speed that comes from a Pacejka model will be right, because the force generated is related to the speed of the axle and the speed of the contact patch. Advanced Pacejka models are speed-sensitive, and can take into account the speed-related concerns you have. Contact-patch modeling for force generation is done largely in FEA, as I recall. You can incorporate that into a full-vehicle simulation if you like, but go raise a family or something to while away the time once you hit the "solve" button, because it will move with glacial speed. If someone (with an understanding of tires and data for a given tire) tells you a given tire generates maximum force at around 15% slip (the definition of slip you cite in conjunction with the Pacejka model is not limited solely to Pacejka formulations, BTW), please be advised that said tire was MEASURED to perform that way on a tire test machine, therefore any any wheelspeed "mismatch" per se, IS a REAL attribute of that tire, if the tire test was carried out properly and the data processed properly. You cite the case of F1 cars at 350 kph. It's a good case. I've got a news flash though: An F1 car at 350 kph is NOWHERE NEAR delivering maximum longitudinal force to the tire. Notice I said longitudinal force, and not power. Q: When is a car capable of deliviering maximum longitudinal force to the tires? A: Regardles of the power band of the engine, THE CAR DELIVERS MAXIMUM FORCE TO THE DRIVEN WHEELS WITH THE TRANSMISSION FIRMLY IN FIRST GEAR!!! I hope I don't need to repeat that. Think about it, the principle is simple: Torque muliplication. F1 tires, as well as NASCAR tires, as well as street tires, have to have longitudinal force capacity to permit the car to get its power/ torque to the road WHEN THE CAR IS IN LOW GEAR. Delivering the power (longitudinal force, really) in top gear becomes a piece of cake. So, in reality, The F1 car at 350 mph is nowhere near pushing the tires to their longitudinal force limits, nor is a Nextel/Spring Cup car at 205 mph doing so, either. In top gear, those cars likely have integer multiples of longitudinal force capability in the tires above and beyond what the motor and gear combination can generate. I really hope this helps, and that it doesn't come across as merely a good b*tch-slapping. Pat recommended Wong's "Theory of Ground Vehicles" to you. I'd like to recommend a couple as well. I have Wong's book, but I typically do not like it. I suggest Gillepsie's "Fundamentals of Vehicle Dynamics". I don't know if it covers Julien's theory, but where it and Wong do cover the same material, I like Gillespie's treatment much better. I would also heartily recommend Milliken & Milliken's "Race Car Vehicle Dynamics" for tire data treatment and much, much more. speedmd

On 20 mar, 22:50, speedmd <[email hidden]> wrote: > All mixed metaphors aside, please be advised that what you deem > "hardly acceptable" is the de facto standard in the automotive > engineering world. Many, if not most, custom tire models are > proprietary enhancements to the Pacejka tire model. This is exactly why I came here in the first place. I am by no way a physicist, and my own reality sensors (eyes, namely), told me that when a vehicle moves, the wheels rotate so that vehicle velocity and tread tangential velocity match with no perceptible difference. Because of this, figures saying that maximum force occurs with a 15% difference or more seemed strange. It happens that I got things wrong and mixed transitional and steady states (more on that below) > Semantics. In my "real life," the slip ratio is as defined by the > Pacejka model. > > It is almost amusing that you're bemoaning what you consider an > unsatisfactory formulation utilized by the rest of the automotive > engineering world because you can't get your mind around the concept. I am just bemoaning that fact that I don't understand it, nothing more. I know very well that all this doesn't come out of thin air. I just wanted to know where I was wrong in my "understanding" of the tyre dynamics. > Maybe you haven't thought about this, but if the contact patch sliding > over the ground is truly "hardly acceptable" to you, then you should > have already been having heartburn over lateral force generation and > the concept of slip angle. I wasn't clear enough. The only unacceptable thing (to me) here is not the fact that it happens, but the visual discrepancy between what I see in real life, as described above, and what I think I will have to render. But it seems I got the englightenment I needed (see below). Regarding slip angle, I had no preconception, true or false (which I admit were what caused my misunderstanding of longitudinal force generation). And just visually speaking, a few degrees difference between car heading and wheel heading aren't as visible as in the longitudinal case, so it simply didn't bother me as much. > Contact-patch modeling for force generation is done largely in FEA, as > I recall. You can incorporate that into a full-vehicle simulation if > you like, but go raise a family or something to while away the time > once you hit the "solve" button, because it will move with glacial > speed. Well, the family is already being raised, so I'd have to find something else. For lack of other ideas, I guess I'll take the easy route and forget contact patch modeling :-)

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> You cite the case of F1 cars at 350 kph. It's a good case. I've got a > news flash though: An F1 car at 350 kph is NOWHERE NEAR delivering > maximum longitudinal force to the tire. > ... > So, in reality, The F1 car at 350 mph is nowhere near pushing the > tires to their longitudinal force limits, nor is a Nextel/Spring Cup > car at 205 mph doing so, either. In top gear, those cars likely have > integer multiples of longitudinal force capability in the tires above > and beyond what the motor and gear combination can generate.

Here is exactly what I've needed to know: slip ratio is very low even at high velocities, which was not self-evident with my background. Just out of curiosity, if we assume that the car has the necessary drive train capacity, at what velocity would the tyre peak ratio be reached ? Anyway, if I had understood/known this in the first place, then all would have been much clearer. So I assume that if a skilled driver accelerates just short of starting wheelspin, and I can somehow watch the wheel rotation compared to ground movement, I'll actually observe this 15% difference, but it will be much less when he reaches a constant velocity however high it might be. > I really hope this helps, and that it doesn't come across as merely a > good b*tch-slapping. It does help a lot, and it was worth the understanding I gained in the process :-)

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> Pat recommended Wong's "Theory of Ground Vehicles" to you. I'd like to > recommend a couple as well. I have Wong's book, but I typically do not > like it. I suggest Gillepsie's "Fundamentals of Vehicle Dynamics". I > don't know if it covers Julien's theory, but where it and Wong do > cover the same material, I like Gillespie's treatment much better. I > would also heartily recommend Milliken & Milliken's "Race Car Vehicle > Dynamics" for tire data treatment and much, much more.

I am already aware of Gillespie's and Milliken&Milliken's books, but haven't had time to get them yet. But once I do I'll have to read them too. > > speedmd Thanks to you and Pat for the explanations, they were really necessary. Maybe I should have started by explaining that I can't really work on any given subject and apply whatever models it without having the necessary understanding. If I don't have it, I can't see the difference between what I do wrong because I didn't do it right (and know this is the case), or because I don't understand what I am doing (and therefore can't tell if I do it right or not, which gets on my nerves). I was just seeking that understanding. Benoit.

I've observed something in everyday driving that has always puzzled me, and it may have its roots in this discussion... Have you ever noticed that when you apply throttle in a car the engine RPM will increase immediately, and any increase in velocity will lag behind RPM increase. Then, when you release the throttle, RPM will drop with no corresponding drop in velocity. The point being that in real life the relationship between engine RPM and velocity is not entirely linear. I always figured maybe it was just mechanical 'slop', or something else in the drive train that causes this observable disconnect between engine RPM and vehicle speed. Slip ratio seems to explain what I've observed. Maybe Manuel can confirm this? Pat Dotson

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> You cite the case of F1 cars at 350 kph. It's a good case. I've got a > news flash though: An F1 car at 350 kph is NOWHERE NEAR delivering > maximum longitudinal force to the tire. Notice I said longitudinal > force, and not power. Q: When is a car capable of deliviering maximum > longitudinal force to the tires? A: Regardles of the power band of the > engine, THE CAR DELIVERS MAXIMUM FORCE TO THE DRIVEN WHEELS WITH THE > TRANSMISSION FIRMLY IN FIRST GEAR!!!

Not with a high downforce car like a F1 car. With 800hp, 1300 lb car, and downforce up to 2 g's. Even though first gear on F1 cars redline around 100mph, there's downforce isn't high enough and the car still has enough power to spin the tires in 2nd gear until enough downforce has been achieved. 1g of downforce occurs around 120mph (depending on wing settings), and my guess is that maximum slip ratio occurs about that speed. Link below to actual downforce numbers for 2001 F1 cars which had the more powerful 3 liter V10s. The left side list the tracks, the car setups vary from track to track. Minimum weight of a 2001 car plus driver, oil, basically everything but fuel is 605kg. This is a French web site. In the table Cx = Cd (coefficient of drag), Eff = efficiently (overall lift to drag ratio), V = velocity, the speed at which all the other parameters were measured or calculated, A = aerodynamic downforce (in kg). http://www.one-pablo.com/technique/tablaero.gif Regarding slip angle, the Indy Racing League cars have tires with the stiffest sidewalls, and their "optimal" slip angle is only about 2 degrees. Formula 1 cars are closer to 3 degrees. Typical radial tires are around 3 to 5 degrees, and bias ply racing slicks are even higher. Going back to the old days of pre-downforce Formula 1 cars, 1967, the bias ply racing tires reached "optimal" grip at over 10 degrees slip angle.

Pat: You're right. You can see it in collected data, too: a sstep increase in RPM (or a speed signal coming from a driven wheelsped sensor) when hitting the gas, and a step decrease when going to the brakes. As a simulation engineer a few years ago, one of my greatst difficulties was getting "true" car speed from a wheelspeed sensor, which was usually one of the rears. I KNEW the car was going faster than the indicated speed under braking, and slower than indicated when driving, it was just hard to quantify it. So, you tried to match RPMs. Then along came manageable (read: small, accurate, quick logging) GPS logging units, and speed measurment has become more direct, more accurate. (Except big grandstands causing reflection of satellite signals or tall catch fences causing the temporary losss of one or more satellites) Speaking of matching RPMs, my simulated RPMs typically match real RPMs within about half a percent, accounting for noise in both speed and RPM in the real car data. Jeff: >Not with a high downforce car like a F1 car. I respectfully (and I mean that) beg to differ. Downforce doesn't enter into the particular question posed and answered. Here are the question and answer again: >Q: When is a car capable of deliviering maximum longitudinal force to the tires? >A: Regardles of the power band of the engine, THE CAR DELIVERS MAXIMUM FORCE >TO THE DRIVEN WHEELS WITH THE TRANSMISSION FIRMLY IN FIRST GEAR!!! Notice both the question and answer pertain to when the engine (driveline, really, I suppose) can deliver the most force to the rear tires, which will always be in low gear. You're addressing when the tires can successfully transmit the most longitudinal force to the road. You're right in what you stated, but you've answered a different question than the one I asked. But your answer is still useful to this discussion, because it is one more reason that Benoit can be confident that the rear tires aren't smoking on the road with 15% slip at 350kph. And some credit has to go to Benoit; not everyone is willing to speak up in a public forum about what he doesn't know and risk getting the rocks thrown at them... speedmd

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> Jeff: > >>Not with a high downforce car like a F1 car. > > I respectfully (and I mean that) beg to differ. Downforce doesn't > enter into the particular question posed and answered. Here are the > question and answer again: > >>Q: When is a car capable of deliviering maximum longitudinal force to the tires? >>A: Regardles of the power band of the engine, THE CAR DELIVERS MAXIMUM FORCE >>TO THE DRIVEN WHEELS WITH THE TRANSMISSION FIRMLY IN FIRST GEAR!!!

You're not taking into account the energy consumed by angular acceleration of the engine, clutch, transmission, drive shaft, differential, and half shafts to each tire. When the tires are slipping badly, a lot of power is being consumed by this brief angular acceleration that occurs before the engine hits the rev-limiter, and the reduced torque delivered to the wheels is well below the peak torque possible. Maximum torque applied to the rear tires requires sufficient grip that power losses to the drivetrain aren't as significant, and the grip only increases with speed and downforce. The difference between 1st and 2nd gear on a F1 car is pretty small, and I don't know for sure that maximum torque applied to the rear tires occurs in 1st or 2nd gear on a F1 race car, since the gear ratios between 1st and 2nd are fairly close, and the engine rpms may correspond with the torque peak in 2nd gear at the speed required to generate enough downforce to prevent tire spin.

> You're not taking into account the energy consumed by angular acceleration of > the engine, clutch, [yada, yada, yada] I didn't have to. The simple question posed was when is a car CAPABLE of delivering maximum longitudinal force to the tires? And the answer is STILL when the torque multiplication is greatest; i.e, in lowest gear. The ENTIRE ORIGINAL CONTEXT of this thread was concerning high longitudinal slip/wheelspin - my reply was to give him the proper perspective for when a tire COULD be subjected to maximum longitudinal force. And it STILL is the proper perspective. >When the tires are slipping badly... ...they are past generating maximum longitudinal force, so it really isn't the disproof of my case, is it? I am well aware, thank you, that delivered torque is acceleration-rate- based. Inertia dynos with different inertias will yield different torque numbers simply because the engines accelerate at different rates. You can make an engine generate any number you want to on a water brake dyno. You could generate different numbers on a Mustang dyno by presenting the motor with a different load. If you had your half shafts strain gauged and you accelerated uphill and downhill, you would get different numbers, wheelspin or not. And you know what? My answer would STILL be sufficient and right. This really reminds me of what was so unpleasant about the Papyrus forums - there always seemed to be somebody with something to prove who had to get into a pissing contest with a racing engineer who was willing to put his name and knowledge on the line. That's probably why I lurked this forum for five years before ever posting here. One thread, one post, one attempt to help, and it's all back again. Lovely... speedmd

"speedmd" <[email hidden]> wrote in message news:[email hidden]...

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>> You're not taking into account the energy consumed by angular >> acceleration of >> the engine, clutch, [yada, yada, yada] > > This really reminds me of what was so unpleasant about the Papyrus > forums - there always seemed to be somebody with something to prove > who had to get into a pissing contest with a racing engineer who was > willing to put his name and knowledge on the line. That's probably why > I lurked this forum for five years before ever posting here. One > thread, one post, one attempt to help, and it's all back again. > Lovely... > > speedmd

You're a bit soft there - usenet has always been and always will be a process of sorting the shit from the good stuff = like panning for gold. It's a simple matter of ceasing conversation with those who are too much attitude and not enough information. If all the cluey people were so easily intimidated - well, there'd be none posting at all!

Byron Forbes wrote: > You're a bit soft there - usenet has always been and always will be a > process of sorting the shit from the good stuff = like panning for gold. > It's a simple matter of ceasing conversation with those who are too much > attitude and not enough information. If all the cluey people were so easily > intimidated - well, there'd be none posting at all! Quite. Head off to RASF1 and discover what a bunch of softies we are :)

> I didn't have to. The simple question posed was when is a car CAPABLE > of delivering maximum longitudinal force to the tires? Ok, I was getting a bit carried away with the specific details of F1 race cars. In some cases, 1st gear is "undergeared" so the cars can exit pits, but 2nd gear is "better" once the car is moving.

Pacejka magic formula is just an approximation. It's not perfect, doesn't take hysteresis effects into account (once a tire is in full slip, it's slip ratio has to decrease significanly below the optimal slip ratio before it regains static friction). Regarding the formula itself, as mentioned in the web site below: "don't try to find any physics in here: it's just a convenient formula that fits the data reasonably well." http://www.miata.net/sport/Physics/Part21.html

On Mar 18, 4:39am, [email hidden] wrote: > Hello, ><snip> > > Benoit Germain Slip ratio generating force does seem strange at first, but that's how it works. If you imagine 4 elements in a contact patch starting at the front and progressing toward the rear, here's essentially what happens: If the contact patch is 1 unit long, under a free rolling condition our evenly spaced elements are located at these positions: 0.2 0.4 0.6 0.8 If we add in 10% slip at the rim and assume there's enough friction to keep all the elements stuck to the ground so there's no sliding, the position that each element wants to be will move 10%. The "deformed positions" have now stretched away from the 0.2, 0.4, etc., to this: 0.22 0.44 0.66 0.88 Each of those elements is like a spring. Granted, in a real tire this is generating shear, but it's fundamentally the same thing as far as you need to be concerned. The elements have deformed the following amounts 0.22 - 0.20 = 0.02 0.44 - 0.40 = 0.04 0.66 - 0.60 = 0.06 0.88 - 0.80 = 0.08 The force is getting larger and larger toward the rear and at some point will slip. However, the percentage of the contact patch that is actually slipping has nothing to do with the slip ratio and is not important for you. "20% slip" does not mean that 20% of the contact patch is sliding. It just means the rim is rotating 20% faster than it would be if it were free rolling just as the slip ratio equation says. Conceptually, however, once these elements have left the contact patch they eventually return to their (perhaps nearly) undeformed position, so it's perhaps not too unreasonable to accept that this occurs with something deformable like a tire. It's indeed what happens. In my simulator I am modelling the contact patch directly and the end result is quite similar to the Magic Model as far as force vs slip ratio or slip angle go (although the low speed/stopping issues are solved much more easily with my approach; you'll have to cross that bridge when you come to it: Search for "relaxation length" when it falls apart at low speed). One other thing: The slip ratio here has nothing to do with the vertical tire squash. It's free rolling angular velocity versus current angular velocity *at the current vertical displacement.* 0 slip ratio ought to give you 0 or very near 0 force. That's what free rolling means. The tire is rotating just as you imagine it. Don't worry about 10-15% slip ratio and the effects on rendering. It's imperceptible unless you want to look really closely under slow motion. But hey, it looks pretty wierd in reality too: ;-) http://youtube.com/watch?v=V3yj_OGezWc

Never thought I'd see you post here Todd. Thanks for the info. Getting back on topic, I would assume that low profile tires, like a 345/30/19 tire (the 30 being the profile percentage) would have a lower optimal slip ratio than a higher profile tire. Also, what about the hysteresis effect mentioned earlier (the transition from dynamic friction back to static firction)? Lastly there's the classic case of car stopped on a hill or banked track where the Pacejka model because actual velocity is zero. Has your approached solved this problem? Nice video link to those wrinkle wall tires.

On Apr 1, 2:47am, "jeffareid" <[email hidden]> wrote:

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> Never thought I'd see you post here Todd. Thanks for the info. > > Getting back on topic, I would assume that low profile tires, like a 345/30/19 > tire (the 30 being the profile percentage) would have a lower optimal > slip ratio than a higher profile tire. > > Also, what about the hysteresis effect mentioned earlier (the transition > from dynamic friction back to static firction)? > > Lastly there's the classic case of car stopped on a hill or banked track > where the Pacejka model because actual velocity is zero. Has your > approached solved this problem? > > Nice video link to those wrinkle wall tires.

Hi Jeff, RAS used to be about the only place I ever posted. Really it was my first spot on the internet I think. I don't come around here too often nowadays. Low profile tires: Generally speaking it's probably a safe rule of thumb to assume that that's indeed the tendency if you look at all tires overall. The cord angles and construction (# of plies, etc) have a lot to do with this too though, so it's not a hard and fast rule. You can make 2 tires that are identical dimensions in this regard that vary considerably in stiffnesses in different directions. Hysteresis: I'm not sure I'm following what you mean on that one. Static versus dynamic friction go right out the window when it comes to rubber and I'm not sure of the connection you're implying with hysteresis there. Classic low speed problem/hill stopping: Yes, my approach solves this (finally) and there's a perfectly smooth transition between all states. There is no separation between low and high speed (it's all just 1 model; no switching equations around at x speed). I have this running in the RC simulator (Virtual RC Racing), but not in my big car sim thingy but it should work just fine. A relaxation length addition to Pacejka's Magic Model solves it nicely enough too. When I was playing with those models I didn't really know all that much about tires and never got the approach to work. Since I've not used it forever there isn't any reason to go back and make it work, but anyone using the Magic Model should definitely look into it. It's one of those big problems every sim guy runs into, often unexpectedly.

Jeff: I think most of our locking of horns stems from you addressing things from a roadholding (what the tires can transmit to the road) issue, whereas I was presenting it from a driving torque issue (i.e. when are the tires least likely to develop the high rates of slip that caused Benoit's orignal concern). >Ok, I was getting a bit carried away with the specific details of F1 >race cars. In some cases, 1st gear is "undergeared" so the cars can >exit pits, but 2nd gear is "better" once the car is moving Brings to mind a story from Jackie Stewart's old book ("Faster!"?) where he and Ken Tyrrel hit upon using a high first gear that they thought would be ideal for the chicanes installed at Monza, I think, but didn't practice a start with it on full tanks, and ended up being done at the drop of the green flag... I suppose I was a bit quick to have flashbacks of bygone days at the Papy forums. Could be worse; I could tell folks I won't share my ARCA sim setups with them and REALLY get the **** kicked out of me. speedmd

On Apr 3, 11:21 pm, speedmd <[email hidden]> wrote: > I suppose I was a bit quick to have flashbacks of bygone days at the > Papy forums.  Could be worse;  I could tell folks I won't share my > ARCA sim setups with them and REALLY get the **** kicked out of me. I really don't mind being kicked around if it brings some action to RAS :) Pat Dotson

Todd Wasson a écrit :

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> A relaxation length addition to Pacejka's Magic Model solves it nicely > enough too. When I was playing with those models I didn't really know > all that much about tires and never got the approach to work. Since > I've not used it forever there isn't any reason to go back and make it > work, but anyone using the Magic Model should definitely look into > it. It's one of those big problems every sim guy runs into, often > unexpectedly.

Yeah, I just got myself a copy of SAE950311 on the subject. Still missing Milliken&Milliken's bible though. So all in all, I'm convinced now, and I have learned something, so I'm happy :-) Thanks, Benoit.

> > Yeah, I just got myself a copy of SAE950311 on the subject. Still > missing Milliken&Milliken's bible though. > amazon.com, or direct from the SAE. It ain't cheap no matter where you buy it, though. speedmd

On Apr 10, 6:44pm, [email hidden] wrote:

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> > Yeah, I just got myself a copy of SAE950311 on the subject. Still > > missing Milliken&Milliken's bible though. > > amazon.com, or direct from the SAE. It ain't cheap no matter where you > buy it, though. > > speedmd

My copy is signed by Mario Andretti and Doug Milliken. Whoot! :-)

> My copy is signed by Mario Andretti and Doug Milliken.  Whoot! :-) We had Doug in the shop a few years back, and he signed the team owner's copy, but I didn't think to have him sign mine. speedmd