Car Physics: Engine torque to wheel RPM

Hi, I've recently began writing a vehicle simulator, and I currently trying to get to grips with engine/wheel stuff. I think I have a model which is correct, but I just wanted to run it by you guys to see if I'm completely on the wrong track. Here's the way I think things work, feel free to poke holes in it :) I have an engine which for a given RPM produces an amount of Torque Te (for now I'm ignoring resistive/frictional forces). For a given gear, we can calculate the torque on the axle Ta by: Ta = Te*Rg*Rd where Rg is the gear ratio and Rd is the differential ratio. Now here's the bit I'm not convinced about. The torque on the axle(s) is used to accelerate the rotation of the wheels, ie. from the torque, the angular acceleration of each wheel can be calculated by Aw = Ta / Iw where Iw is the MOI for the wheel. This angular acceleration can then be integrated to give a new angular velocity for the wheel. Once we have an angular velocity for each wheel, the slip ratios can be calculated. These can then be used to calculate a traction force using the longitudinal version of the magic formula. This traction force (ignoring friction limits) then accelerates the car forwards, as well as again applying a torque to the wheel in the opposite direction (by Newton's 3rd law). OK, so thats what I think happens so far. The thing I'm not very clear on is rolling resistance. In the litrature I've seen, rolling resistance is always treated as a force which is applied to the vehicle as a whole. To my mind, rolling resistance should be applied to the wheels' rotation, which would indirectly cause the vehicle to slow down using the above. Am I right? and if so how do I calculate rolling resistance. What I've read seems to be very vague on this, and just says it's some constant multiplied by the velocity of the vehicle. This seems like a cheap approximation to me ... Any help appreciated ... Cheers, Mike.

I'm not sure if you're making an open source sim but you might want to talk to Ruud van Gaal who is working on Racer, an open source driving sim. http://www.marketgraph.nl/gallery/racer/ "Mike Stanley" <[email hidden]> wrote in message news:G9Z47.58470$[email hidden]... > Hi, > > I've recently began writing a vehicle simulator, and I currently trying to > get to grips with engine/wheel stuff. I think I have a model which is > correct, but I just wanted to run it by you guys to see if I'm completelyon > the wrong track. Here's the way I think things work, feel free to pokeholes > in it :)

I'm afraid you lost me after "Hi," :-) Would love to help Mike... but this just isnt my field. ed_ Mike Stanley wrote: > Hi, > > I've recently began writing a vehicle simulator, and I currently trying to > get to grips with engine/wheel stuff. I think I have a model which is > correct, but I just wanted to run it by you guys to see if I'm completelyon > the wrong track. Here's the way I think things work, feel free to pokeholes > in it :)

> OK, so thats what I think happens so far. The thing I'm not very clear onis > rolling resistance. In the litrature I've seen, rolling resistance isalways > treated as a force which is applied to the vehicle as a whole. To my mind, > rolling resistance should be applied to the wheels' rotation, which would > indirectly cause the vehicle to slow down using the above. Am I right? and > if so how do I calculate rolling resistance. What I've read seems to bevery

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> vague on this, and just says it's some constant multiplied by the velocity > of the vehicle. This seems like a cheap approximation to me ... > > Any help appreciated ... > > Cheers, > > Mike. >

I am in very much the same situation. What I was taught in A-Level mechanics is that the friction of a sliding object is = uR, where u is the coefficient of friction 0<=u<=1, and R is the reaction force from the ground. The reaction force from the ground is equal to the weight of the car on that suspension point + the weight of the wheel, brake and suspension units. The weight on each suspension unit can be calculated from the Centre of Mass using simple moments. We always assumed rolling resistance to be negligible when compared to air resistance, and treated air resistance as the total car resistance. The formula for air resistance (from a technical aeroplane book) is proportional to the surface area of the object, the velocity of the object and the density of the air. There is a great explanation in the Air Pilot's Manual Vol 4: The Aeroplane: Technical available at http://www.amazon.co.uk/exec/obidos/ASIN/1840371552/qid%3D995399866/202-8008 782-1771049 for �16. However, I cannot help you in your quest for the rolling resistance. Nick.

I agree that air resistance is important, but I'm not too concerned with that at the moment. From what I've seen in the litrature, air resistance makes a large contribution to the resistive forces at high speeds, but at low speeds rolling resistance is dominant. The main reason I'm interested in rolling resistance is to slow down the rotation of the wheels. If you have a vehicle suspended in the air with it's wheels rotating, the wheels will gradual slow down then stop. This is the behaviour that I want to model. This will also make a contribution to slowing the vehicle down if the wheel is in contact with the ground. Thanks for the link, I'm sure it will become useful soon. Mike. "Nick G" <[email hidden]> wrote in message news:[email hidden]... > We always assumed rolling resistance to be negligible when compared to air > resistance, and treated air resistance as the total car resistance. The > formula for air resistance (from a technical aeroplane book) isproportional > to the surface area of the object, the velocity of the object and the > density of the air. There is a great explanation in the Air Pilot's Manual > Vol 4: The Aeroplane: Technical available at >http://www.amazon.co.uk/exec/obidos/ASIN/1840371552/qid%3D995399866/202-8008

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> 782-1771049 for �16. > > However, I cannot help you in your quest for the rolling resistance. > > Nick. > >

"Kirk House" <[email hidden]> wrote in message news:3b548636$0$73969$[email hidden]... > I'm not sure if you're making an open source sim but you might want totalk > to Ruud van Gaal > who is working on Racer, an open source driving sim. > http://www.marketgraph.nl/gallery/racer/ It was from that web page that I found out about this newsgroup :) At the moment I'm just trying to learn this stuff, and I don't expect the first iteration to be very useful other than as a learning aid to me. Once I'm sure I have a suitable model, then I'll start to write a sim for real, and the physics part will probably be open source. I'll post a link to the group when that happens.

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>I've recently began writing a vehicle simulator, and I currently trying to >get to grips with engine/wheel stuff. I think I have a model which is >correct, but I just wanted to run it by you guys to see if I'm completely on >the wrong track. Here's the way I think things work, feel free to poke holes >in it :) > >I have an engine which for a given RPM produces an amount of Torque Te (for >now I'm ignoring resistive/frictional forces). For a given gear, we can >calculate the torque on the axle Ta by: > >Ta = Te*Rg*Rd > >where Rg is the gear ratio and Rd is the differential ratio.

Right. If you decide later to model a differential gear itself, this would be torque applied to the ring gear.

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>Now here's the bit I'm not convinced about. The torque on the axle(s) is >used to accelerate the rotation of the wheels, ie. from the torque, the >angular acceleration of each wheel can be calculated by > >Aw = Ta / Iw > >where Iw is the MOI for the wheel. This angular acceleration can then be >integrated to give a new angular velocity for the wheel. >

Yes, although if you want to go a bit further, remember that (at least) the half-shaft MOI also should be included with this. This will help stability in your system somewhat, as well as keep overall acceleration closer to real life results.

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>Once we have an angular velocity for each wheel, the slip ratios can be >calculated. These can then be used to calculate a traction force using the >longitudinal version of the magic formula. This traction force (ignoring >friction limits) then accelerates the car forwards, as well as again >applying a torque to the wheel in the opposite direction (by Newton's 3rd >law). > >OK, so thats what I think happens so far.

That's how I do it too.

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>The thing I'm not very clear on is >rolling resistance. In the litrature I've seen, rolling resistance is always >treated as a force which is applied to the vehicle as a whole. To my mind, >rolling resistance should be applied to the wheels' rotation, which would >indirectly cause the vehicle to slow down using the above. Am I right? and >if so how do I calculate rolling resistance. What I've read seems to be very >vague on this, and just says it's some constant multiplied by the velocity >of the vehicle. This seems like a cheap approximation to me ...

Currently, I'm using a cheap force approximation. However, looking through Milliken's "Race Car Vehicle Dynamics" on pages 71 and 72 (tire behavior), there is a good treatment on rolling resistance. If you don't have the book yet, get it :-) Hopefully, Doug himself will pop in here with the best link to use. The section describes how, due to the tire elements' compression and re-extension as they move through the contact patch, there is an offset where this force is applied (rather than being dead-center in the patch.) The example shows a .22 inch offset for one particular tire, so that might be a good ballpark figure. I'm not sure if this stays constant or not, but is probably a decent approximation for starters. Given the effective tire radius (squashed tire from spring type behavior), vertical load, and the offset described above, the rolling resistance force is given in the example by: rolling_resistance_force = offset * vertical_load / effective_tire_radius In the example, it was: rolling_resistance_force = .22inch * 1200lbs / 13.7inches = 19.2 lbs Anyway, the way I was originally handling rolling resistance was by adding a braking torque to the tire directly that would give the same results as the approximation you spoke of in the last post. However, this doesn't appear to be the correct way to handle it. Since the rolling resistance force is applied in the direction opposite to vehicle travel (if wheels are free rolling), the resulting torque would actually speed up the tire and slow down the car simultaneously, rather than slow down both at the same time as my braking torque idea would. In short, it looks as though the original approximation you talked about is probably right. Now, try finding the resulting torque on the tire and add it in while also applying the force directly to the car. The wheels should accelerate slightly (keeping non-negative slip ratios, unlike a braking method would produce), yet the car should decelerate, *I think* :-) Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

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>I agree that air resistance is important, but I'm not too concerned with >that at the moment. From what I've seen in the litrature, air resistance >makes a large contribution to the resistive forces at high speeds, but at >low speeds rolling resistance is dominant. > >The main reason I'm interested in rolling resistance is to slow down the >rotation of the wheels. If you have a vehicle suspended in the air with it's >wheels rotating, the wheels will gradual slow down then stop. This is the >behaviour that I want to model. This will also make a contribution to >slowing the vehicle down if the wheel is in contact with the ground. >

Oh, that's not rolling resistance, it's bearing friction combined with compression losses with closed throttles (if the clutch is engaged.) To do this, add a negative torque to the driveline/wheels, as though a small brake was at work somewhere. Not sure on real values for this though. The Straightline Acceleration Simulator program at my site works well with real life values of about 75 lbs-feet for driveline loss. That seems like an awful lot of friction to me, but the results are consistent with real car tests so far. Rolling resistance is a seperate effect, and is due to tire flex in the contact patch. If the car's suspended in mid air, there is 0 "rolling resistance," if my definition is correct. Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

I know almost nothing in this field but I would like to make a small point which has oozed from my feeble mind. <g> This is only being applied to wind resistance. If it takes 50hp to push a 340lb motorcycle w/ 145lb rider 120mph through the air.. at 60mph it takes 1/8 the hp which would be 6.25hp. Doesnt rolling resistance have almost no effect on the vehicle above a certain speed when wind resistance very quickly becomes such a huge factor? BTW although not exact, I used myself on my motorcycle as a model. Also I found a page where they apply rolling resistance, wind resistance etc to bicycle riding.. both faired and unfaird, formulas included, quite interesting.. http://www.wisil.recumbents.com/wisil/hein/hpvpaper.htm Lowering flame retardent face shield... Charlie "J. Todd Wasson" <[email hidden]> wrote in message news:[email hidden]... > >I've recently began writing a vehicle simulator, and I currently tryingto > >get to grips with engine/wheel stuff. I think I have a model which is > >correct, but I just wanted to run it by you guys to see if I'm completelyon > >the wrong track. Here's the way I think things work, feel free to pokeholes > >in it :) > > > >I have an engine which for a given RPM produces an amount of Torque Te(for

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> >now I'm ignoring resistive/frictional forces). For a given gear, we can > >calculate the torque on the axle Ta by: > > > >Ta = Te*Rg*Rd > > > >where Rg is the gear ratio and Rd is the differential ratio. > > > Right. If you decide later to model a differential gear itself, this

would > be torque applied to the ring gear. > > > >Now here's the bit I'm not convinced about. The torque on the axle(s)is > >used to accelerate the rotation of the wheels, ie. from the torque, the > >angular acceleration of each wheel can be calculated by > > > >Aw = Ta / Iw > > > >where Iw is the MOI for the wheel. This angular <snip>

Good grief, that's quite an intensive site! Put down your flame retardent face shield, you don't need it :-) For automobiles, a common formula for figuring how much power is required to move at a given speed is: Horsepower = ((Coefficient_rolling_resistance * Weight + air_density*frontal_area*miles_per_hour^2) * miles_per_hour) / 375 With a 3500 lb car, using a supposedly high typical rolling resistance coefficient of .015, a frontal area of 24 sq feet, and a coefficient of drag (the wind resistance part) of .42, to travel at these speeds would require this much power: (air density is .0025 in this example) Rolling resistance --Wind resistance -Both effects -% rolling resistance only only 10 mph---> 1.4 hp .06 hp 1.46 hp 96% 20 mph---> 2.8 hp .53 hp 3.33 hp 84% 40 mph---> 5.6 hp 4.30 p 9.90 hp 56% 45 mph---> 6.3 hp 6.12 h 12.42 hp 50% At 45 mph (for this particular car), the wind and rolling resistance both contribute about the same amount. Continuing on, the wind resistance really takes over: Rolling resistance --Wind resistance -Both effects -% rolling resistance only only 60 mph---> 8.4 hp 14.51 hp 22.91 hp 36.6% 80 mph---> 11.2 hp 34.40 hp 45.60 hp 24.6% 100 mph---> 14.0 hp 67.20 hp 81.20 hp 17.2% 120 mph---> 16.8 hp 116.12 hp 132.92 hp 12.6% 140 mph---> 19.6 hp 184.39 hp 204.00 hp 9.6% 160 mph---> 22.4 hp 275.25 hp 297.65 hp 7.5% 180 mph---> 25.2 hp 391.91 hp 417.11 hp 6.0% 200 mph---> 28.0 hp 537.60 hp 565.60 hp 4.9% 220 mph---> 30.8 hp 715.00 hp 746.34 hp 4.1% 240 mph---> 33.6 hp 928.97 hp 962.57 hp 3.5% 260 mph---> 36.4 hp 1181.11 hp 1217.50 hp 3.0% 280 mph---> 39.2 hp 1475.17 hp 1514.37 hp 2.6% 300 mph---> 42.0 hp 1814.40 hp 1856.40 hp 2.3% I've never layed this out this way before, but it looks as though rolling resistance is a larger factor than I previously believed, but yeah, it does look like it becomes somewhat negligable after a point. The percentage column is "rolling resistance only" divided by "both effects", giving the percentage contribution of rolling resistance. The interesting part (to me, anyway :-)) here is, it looks as though the rolling resistance force is constant and independent of speed, while the wind resistance force (not power, like shown above) quadruples when the speed doubles (squared relationship.) Power is cubed, of course. Note that racing bicycles with hard, skinny tires should have far, far less rolling resistance than this car example does. They're designed probably with that sole purpose in mind, so the percentages would drop off extremely quickly, leaving aerodynamics (even at typical bicycle racing speeds) as the dominant force. Well, that's what I'd bet anyway :-)

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> I know almost nothing in this field but I would like to make a small >point which has oozed from my feeble mind. <g> > This is only being applied to wind resistance. If it takes 50hp to push >a 340lb motorcycle w/ 145lb rider 120mph through the air.. at 60mph it takes >1/8 the hp which would be 6.25hp. Doesnt rolling resistance have almost no >effect on the vehicle above a certain speed when wind resistance very >quickly becomes such a huge factor? BTW although not exact, I used myself on >my motorcycle as a model. Also I found a page where they apply rolling >resistance, wind resistance etc to bicycle riding.. both faired and unfaird, >formulas included, quite interesting.. >http://www.wisil.recumbents.com/wisil/hein/hpvpaper.htm > > Lowering flame retardent face shield... > Charlie

Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

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> The thing I'm not very clear on is >rolling resistance. In the litrature I've seen, rolling resistance is always >treated as a force which is applied to the vehicle as a whole. To my mind, >rolling resistance should be applied to the wheels' rotation, which would >indirectly cause the vehicle to slow down using the above. Am I right? and >if so how do I calculate rolling resistance. What I've read seems to be very >vague on this, and just says it's some constant multiplied by the velocity >of the vehicle. This seems like a cheap approximation to me ...

One more thing. If you do seperate the driveline/bearing and engine friction from true rolling resistance at the tires, note that the rolling resistance force pushes in the opposite direction of the car itself. This would give the same negative acceleration to the car as it would if you applied the force directly to the center of mass of the car, so the way you've read is correct. The thing that intrigues me though, is that since the force is at the contact patch of the tire, it will torque the body (pitch) as well as torque the tire itself, and unless I'm missing something here, this torque would actually serve to *accelerate* the tire, while simultaneously slowing the vehicle. Kinda wierd. Anyone know for sure if this is true, or have I been drinking too much lately? :-) Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

Get your bathroom scales and hold them vertically up against the back of your car. Make sure it's in neutral, handbrake off, on flat ground, and then shove the car along with the scales jammed between your palms and the back of the car. That way you can measure the force you're applying. It doesn't matter if you're moving at a slow walk or an all-out sprint, the force is the same. At least, it was for my Vauxhall Nova down our drive.... by the way it's a good idea to have someone in the front seat who can apply the brakes when you're done :0) As for aerodynamic drag, it's 0.5 * Cd * rho * A * V^2 where Cd is the drag coefficient (approx 0.3 to 0.5 for road cars), rho is density of air, A is frontal area, and V is the velocity w.r.t the air. Not sure if any of this stuff helps in your conversation as I sort of lost my way in your maths, but hopefully it's slightly relevant! Cheerio Jim Seamus "J. Todd Wasson" <[email hidden]> wrote in message news:[email hidden]... > Good grief, that's quite an intensive site! Put down your flameretardent

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> face shield, you don't need it :-) For automobiles, a common formula for > figuring how much power is required to move at a given speed is: > > Horsepower = ((Coefficient_rolling_resistance * Weight + > air_density*frontal_area*miles_per_hour^2) * miles_per_hour) / 375 > > With a 3500 lb car, using a supposedly high typical rolling resistance > coefficient of .015, a frontal area of 24 sq feet, and a coefficient of

drag > (the wind resistance part) of .42, to travel at these speeds would requirethis > much power: (air density is .0025 in this example) > Rolling resistance --Wind resistance -Both effects -%rolling

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> resistance > only only > 10 mph---> 1.4 hp .06 hp 1.46 hp > 96% > 20 mph---> 2.8 hp .53 hp 3.33 hp > 84% > 40 mph---> 5.6 hp 4.30 p 9.90 hp

56% > 45 mph---> 6.3 hp 6.12 h 12.42 hp50% > > > At 45 mph (for this particular car), the wind and rolling resistanceboth > contribute about the same amount. Continuing on, the wind resistancereally > takes over: > > Rolling resistance --Wind resistance -Both effects -%rolling > resistance > only only > 60 mph---> 8.4 hp 14.51 hp 22.91 hp > 36.6% > 80 mph---> 11.2 hp 34.40 hp 45.60 hp24.6% > 100 mph---> 14.0 hp 67.20 hp 81.20 hp17.2% > 120 mph---> 16.8 hp 116.12 hp 132.92 hp12.6% > 140 mph---> 19.6 hp 184.39 hp 204.00 hp9.6% > 160 mph---> 22.4 hp 275.25 hp 297.65 hp7.5% > 180 mph---> 25.2 hp 391.91 hp 417.11 hp6.0% > 200 mph---> 28.0 hp 537.60 hp 565.60 hp4.9% > 220 mph---> 30.8 hp 715.00 hp 746.34 hp4.1% > 240 mph---> 33.6 hp 928.97 hp 962.57 hp3.5% > 260 mph---> 36.4 hp 1181.11 hp 1217.50 hp3.0% > 280 mph---> 39.2 hp 1475.17 hp 1514.37 hp2.6% > 300 mph---> 42.0 hp 1814.40 hp 1856.40 hp2.3% > > I've never layed this out this way before, but it looks as thoughrolling > resistance is a larger factor than I previously believed, but yeah, itdoes > look like it becomes somewhat negligable after a point. The percentagecolumn > is "rolling resistance only" divided by "both effects", giving thepercentage > contribution of rolling resistance. > > The interesting part (to me, anyway :-)) here is, it looks as though the > rolling resistance force is constant and independent of speed, while thewind > resistance force (not power, like shown above) quadruples when the speed > doubles (squared relationship.) Power is cubed, of course. > > Note that racing bicycles with hard, skinny tires should have far, farless > rolling resistance than this car example does. They're designed probablywith > that sole purpose in mind, so the percentages would drop off extremelyquickly, > leaving aerodynamics (even at typical bicycle racing speeds) as thedominant

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> force. Well, that's what I'd bet anyway :-) > > > > > > > > > > I know almost nothing in this field but I would like to make a small > >point which has oozed from my feeble mind. <g> > > This is only being applied to wind resistance. If it takes 50hp to

push > >a 340lb motorcycle w/ 145lb rider 120mph through the air.. at 60mph ittakes > >1/8 the hp which would be 6.25hp. Doesnt rolling resistance have almostno > >effect on the vehicle above a certain speed when wind resistance very > >quickly becomes such a huge factor? BTW although not exact, I used myselfon > >my motorcycle as a model. Also I found a page where they apply rolling > >resistance, wind resistance etc to bicycle riding.. both faired andunfaird,

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> >formulas included, quite interesting.. > >http://www.wisil.recumbents.com/wisil/hein/hpvpaper.htm > > > > Lowering flame retardent face shield... > > Charlie > > > Todd Wasson > --- > Performance Simulations > Drag Racing and Top Speed Prediction > Software > http://PerformanceSimulations.Com

Mike, Have you read the physics of racing series by Brian Beckman? Its all great stuff, but the earlier parts are keyed towards the race-driver wanting to know "how things work", it is great background reading. The later parts are just awesome for the budding Sim-writer talking about the pacekja "magic" formula as a way to get the Force generated by a particular tyre at a given slip-ratio and slip-angle. He is currently writing part 26 and 27, which cover slip-ratio near zero (a problem with the normal formulas for calculating slip ratio). Sorry thats a bit illegible, I haven't woken up properly yet :) Oh, here is the link - http://www.esbconsult.com.au/ogden/locost/phors/phors.htm All the best, Ash - budding sim writer member #131151232 ;)

Hi Todd, the question is really interesting. If you only applied the rolling resistance as a force pointing backwards to the direction of travel at the contact point of the wheel, when the system settles a slip ratio develops with another force being produced that exactly counters this force, and, as amazing as it may seem, there would be absolutely no braking effect from such a force. The rolling resistance, however, is defined as happening in the already settled situation. The total torque on the wheel is actually very close to 0 in this situation (*), but a net braking force exists. The explanation for this is that there is an additional torque produced by the vertical force due to the offset in vertical load that you mentioned, and in a settled situation (*) this exactly cancels the torque on the wheel as produced by the rolling resistance. The overall result is that rolling resistance as a total force must effectively be applied at the center of the wheel (*). Of course, a very good question remains whether slip ratio is nonzero at this point or not (meaning that the tyre may indeed rotate faster or slower). I do think that the effect should be pretty negligible, though, at least with low rolling resistance. -Gregor (*) as long as the effective MOI/R^2 is much lower than the mass of the car "J. Todd Wasson" wrote:

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> > Anyway, the way I was originally handling rolling resistance was by adding a > braking torque to the tire directly that would give the same results as the > approximation you spoke of in the last post. However, this doesn't appear to > be the correct way to handle it. Since the rolling resistance force is applied > in the direction opposite to vehicle travel (if wheels are free rolling), the > resulting torque would actually speed up the tire and slow down the car > simultaneously, rather than slow down both at the same time as my braking > torque idea would. >

Hi Todd, "J. Todd Wasson" <[email hidden]> wrote in message news:[email hidden]... > Oh, that's not rolling resistance, it's bearing friction combined with > compression losses with closed throttles (if the clutch is engaged.) Todo > this, add a negative torque to the driveline/wheels, as though a smallbrake > was at work somewhere. Not sure on real values for this though. The > Straightline Acceleration Simulator program at my site works well withreal > life values of about 75 lbs-feet for driveline loss. That seems like anawful > lot of friction to me, but the results are consistent with real car testsso

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> far. > > Rolling resistance is a seperate effect, and is due to tire flex in the > contact patch. If the car's suspended in mid air, there is 0 "rolling > resistance," if my definition is correct. > Todd Wasson > --- > Performance Simulations > Drag Racing and Top Speed Prediction > Software > http://PerformanceSimulations.Com

That's where I've been getting confused. The stuff I've read tends to group all these effects together under the name rolling resistance. Thanks for clearing that up.

"J. Todd Wasson" <[email hidden]> wrote in message news:[email hidden]... > Yes, although if you want to go a bit further, remember that (at least)the > half-shaft MOI also should be included with this. This will helpstability in > your system somewhat, as well as keep overall acceleration closer to reallife > results. > Sounds sensible. Any idea of approximate values for this combined MOI in a real life situation? > > That's how I do it too. > Good to see I'm going in the right direction! > Currently, I'm using a cheap force approximation. However, lookingthrough > Milliken's "Race Car Vehicle Dynamics" on pages 71 and 72 (tire behavior), > there is a good treatment on rolling resistance. If you don't have thebook > yet, get it :-) Hopefully, Doug himself will pop in here with the bestlink to > use. > > The section describes how, due to the tire elements' compression and > re-extension as they move through the contact patch, there is an offsetwhere > this force is applied (rather than being dead-center in the patch.) The > example shows a .22 inch offset for one particular tire, so that might bea > good ballpark figure. I'm not sure if this stays constant or not, but is > probably a decent approximation for starters. Given the effective tireradius > (squashed tire from spring type behavior), vertical load, and the offset > described above, the rolling resistance force is given in the example by: > > rolling_resistance_force = offset * vertical_load /effective_tire_radius

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> > In the example, it was: > > rolling_resistance_force = .22inch * 1200lbs / 13.7inches > = 19.2 lbs > > Anyway, the way I was originally handling rolling resistance was by

adding a > braking torque to the tire directly that would give the same results asthe > approximation you spoke of in the last post. However, this doesn'tappear to > be the correct way to handle it. Since the rolling resistance force isapplied > in the direction opposite to vehicle travel (if wheels are free rolling),the > resulting torque would actually speed up the tire and slow down the car > simultaneously, rather than slow down both at the same time as my braking > torque idea would. > > In short, it looks as though the original approximation you talked aboutis > probably right. Now, try finding the resulting torque on the tire and addit > in while also applying the force directly to the car. The wheels should > accelerate slightly (keeping non-negative slip ratios, unlike a brakingmethod > would produce), yet the car should decelerate, *I think* :-) > Hmmm .... this is confusing ... have to think about this one. In my current model applying a torque in this direction will accelerate the vehicle. Cheers, Mike.

On Wed, 18 Jul 2001, Mike Stanley wrote: > "J. Todd Wasson" <[email hidden]> wrote in message > news:[email hidden]... > Sounds sensible. Any idea of approximate values for this combined MOI in a > real life situation? Not too hard to estimate -- weigh the various parts separately, then guess at a radius of gyration (place where the inertia would be the same if all the mass was concentrated in a ring at that radius) for each, then add up all the parts that are spinning at the same speed. There was a previous thread on how to deal with parts that are geared together... > > Currently, I'm using a cheap force approximation. However, looking through > > Milliken's "Race Car Vehicle Dynamics" on pages 71 and 72 (tire behavior), > > there is a good treatment on rolling resistance. If you don't have the book > > yet, get it :-) Hopefully, Doug himself will pop in here with the best link to > > use. Best place to order is right from SAE, our publishers -- go to <www.sae.org> and search for Milliken in the bookstore, order number is R-146. They have the book in stock and give a decent price on shipping. They have a lot of other books on racing too -- support small specialty publishers! -- Doug Milliken Milliken Research Associates Inc.

I suspect Mike dosnt have the parts to weigh :O) Any hints then as to the MOI Doug?

Depends totally on the tire sizes, wheel type, etc... Also, I'm used to working in what we call "English" units -- will you know what to do if I give you mumbers in slug-ft^2? <evil grin>! -- Doug Douglas Milliken <[email hidden]> Milliken Research Associates Inc. On Wed, 18 Jul 2001, Carol Bekker wrote: > I suspect Mike dosnt have the parts to weigh :O) Any hints then as to the > MOI Doug? > > > >

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>Get your bathroom scales and hold them vertically up against the back of >your car. Make sure it's in neutral, handbrake off, on flat ground, and then >shove the car along with the scales jammed between your palms and the back >of the car. That way you can measure the force you're applying. It doesn't >matter if you're moving at a slow walk or an all-out sprint, the force is >the same. At least, it was for my Vauxhall Nova down our drive.... by the >way it's a good idea to have someone in the front seat who can apply the >brakes when you're done :0)

Good, then I was right for once :-P This tripped me up when trying to put in rolling resistance at first. When letting the force increase with speed, the coefficient had to be so low that it wouldn't stop for a long time. If it was high enough to slow down realistically at low speeds, the car was seriously deprived of top end speed. Luckily, the force remains more or less constant, so just picking a low number worked well.

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>As for aerodynamic drag, it's 0.5 * Cd * rho * A * V^2 where Cd is the drag >coefficient (approx 0.3 to 0.5 for road cars), rho is density of air, A is >frontal area, and V is the velocity w.r.t the air. > >Not sure if any of this stuff helps in your conversation as I sort of lost >my way in your maths, but hopefully it's slightly relevant! >

It's absolutely relevant! The formula you're showing gives the force from aero drag, while the one I showed gave the horsepower. It's better to use the one you're showing in simulation. Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

>Depends totally on the tire sizes, wheel type, etc... > >Also, I'm used to working in what we call "English" units -- will >you know what to do if I give you mumbers in slug-ft^2? <evil grin>! Hey! I could use some of these numbers! I use "English" units too. Send 'em my way if you get a few extra minutes :-) Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

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>Hi Todd, > >the question is really interesting. If you only applied the rolling >resistance as a force pointing backwards to the direction of travel at >the contact point of the wheel, when the system settles a slip ratio >develops with another force being produced that exactly counters this >force, and, as amazing as it may seem, there would be absolutely no >braking effect from such a force. > >The rolling resistance, however, is defined as happening in the already >settled situation. The total torque on the wheel is actually very close >to 0 in this situation (*), but a net braking force exists. The >explanation for this is that there is an additional torque produced by >the vertical force due to the offset in vertical load that you >mentioned, and in a settled situation (*) this exactly cancels the >torque on the wheel as produced by the rolling resistance. The overall >result is that rolling resistance as a total force must effectively be >applied at the center of the wheel (*). > >Of course, a very good question remains whether slip ratio is nonzero at >this point or not (meaning that the tyre may indeed rotate faster or >slower). I do think that the effect should be pretty negligible, though, >at least with low rolling resistance. > >-Gregor >

Well, we've got vehicle models, so let's add a forward force, kill wind resistance, put in an offset and try it. You go first ;-) Interesting though, I hadn't thought of the vertical force torque effect that would develop. Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

Could someone possibly give a quick explaination of 'tire load' as required by Pacejka as in my head suspension forces should be taken into condition, but I have seen in other places that suspensions forces are not taken into consideration but the 'tire load' is calculated from the amount the tire is sinking into the ground (tire compression). I am still waiting for my copy of RCVD to arrive from Amazon (cost me a whopping $101)!!! Anyone? TIA, Colin

Doug, you... engineer :) -Gregor Doug Milliken wrote:

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> > Also, I'm used to working in what we call "English" units -- will > you know what to do if I give you mumbers in slug-ft^2? <evil grin>! > > -- Doug > Douglas Milliken <[email hidden]> > Milliken Research Associates Inc. >

Oh no the 'name-calling' has started <g> Ash

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| Doug, | | you... engineer :) | | -Gregor | | Doug Milliken wrote: | > | > Also, I'm used to working in what we call "English" units -- will | > you know what to do if I give you mumbers in slug-ft^2? <evil grin>! | > | > -- Doug | > Douglas Milliken <[email hidden]> | > Milliken Research Associates Inc. | >

Oh, to be young again and have the time for creative abuse...<g> If anyone does have access to the parts that they need inertias for, I seem to remember that there was a nice write up on this cycling site, but I haven't looked lately to see if it's still there. <http://www.analyticcycling.com> (a very cool site, by an ex-DuPont engineer) On Thu, 19 Jul 2001, Ashley McConnell wrote:

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> Oh no the 'name-calling' has started <g> > > Ash > > | Doug, > | > | you... engineer :) > | > | -Gregor > | > | Doug Milliken wrote: > | > > | > Also, I'm used to working in what we call "English" units -- will > | > you know what to do if I give you mumbers in slug-ft^2? <evil grin>! > | > > | > -- Doug > | > Douglas Milliken <[email hidden]> > | > Milliken Research Associates Inc. > | > > > > >

Well, no problem, we always have this site handy: http://www.digitaldutch.com/unitconverter/ As long as someone manages to explain me the difference between all the pounds, I'll be just fine :). -Gregor Doug Milliken wrote:

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> > Depends totally on the tire sizes, wheel type, etc... > > Also, I'm used to working in what we call "English" units -- will > you know what to do if I give you mumbers in slug-ft^2? <evil grin>! > > -- Doug

On Fri, 20 Jul 2001 09:58:33 +0200, Gregor Veble <[email hidden]> wrote: >Well, no problem, we always have this site handy: > >http://www.digitaldutch.com/unitconverter/ > >As long as someone manages to explain me the difference between all the >pounds, I'll be just fine :). Indeed! :) When do I use pound-force (0.45?) and when to use pound (0.44?)? All these nice numbers that sometimes trickle in are sometimes hard to convert correctly. Probably something with gravity, although it would seem odd why there are so many lbf units while these change I think depending on where you are on the planet, right? Ruud van Gaal, GPL Rank +53.25 Pencil art : http://www.marketgraph.nl/gallery/ Free car sim : http://www.marketgraph.nl/gallery/racer/

Still some good stuff there. Bicycles only though :( Looks like I'll be off to the breaker's yard with a set of bathroom scales then :-) "Doug Milliken" <[email hidden]> wrote in message news:[email hidden]...

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> > Oh, to be young again and have the time for creative abuse...<g> > > If anyone does have access to the parts that they need inertias for, > I seem to remember that there was a nice write up on this cycling > site, but I haven't looked lately to see if it's still there. > <http://www.analyticcycling.com> (a very cool site, by an ex-DuPont > engineer) > > > On Thu, 19 Jul 2001, Ashley McConnell wrote: > > > Oh no the 'name-calling' has started <g> > > > > Ash > > > > | Doug, > > | > > | you... engineer :) > > | > > | -Gregor > > | > > | Doug Milliken wrote: > > | > > > | > Also, I'm used to working in what we call "English" units -- will > > | > you know what to do if I give you mumbers in slug-ft^2? <evil grin>! > > | > > > | > -- Doug > > | > Douglas Milliken <[email hidden]> > > | > Milliken Research Associates Inc. > > | > > > > > > > > > >

"J. Todd Wasson" <[email hidden]> wrote in message news:[email hidden]... > The thing that intrigues me though, is that since the force is at the > contact patch of the tire, it will torque the body (pitch) as well astorque > the tire itself, and unless I'm missing something here, this torque would > actually serve to *accelerate* the tire, while simultaneously slowing the > vehicle. Kinda wierd. Anyone know for sure if this is true, or have Ibeen

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> drinking too much lately? :-) > Todd Wasson > --- > Performance Simulations > Drag Racing and Top Speed Prediction > Software > http://PerformanceSimulations.Com

Thinking about this, wouldn't the rolling resistance actually act against the direction of the wheel's roll? Effectively this would slow down the wheels rotation causing a deceleration of the vehicle.

"Mike Stanley" <[email hidden]> wrote in message news:<CuT57.54813$[email hidden]>...

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> "J. Todd Wasson" <[email hidden]> wrote in message > news:[email hidden]... > > The thing that intrigues me though, is that since the force is at the > > contact patch of the tire, it will torque the body (pitch) as well as > torque > > the tire itself, and unless I'm missing something here, this torque would > > actually serve to *accelerate* the tire, while simultaneously slowing the > > vehicle. Kinda wierd. Anyone know for sure if this is true, or have I > been > > drinking too much lately? :-) > > Todd Wasson > > --- > > Performance Simulations > > Drag Racing and Top Speed Prediction > > Software > > http://PerformanceSimulations.Com > > Thinking about this, wouldn't the rolling resistance actually act against > the direction of the wheel's roll? Effectively this would slow down the > wheels rotation causing a deceleration of the vehicle.

I agree, if you imagine a situation of a car coasting from a smooth road (low rolling resistance) onto, say, gravel, then the wheels will be slowed exactly as if you applied outboard brakes. Todd, I'm not sure it's valid to regard rolling resistance as a force, since it's a representation of work done in the contact patch. The various forces at play between rubber and road are converting kinetic energy into heat and sound/vibration. I.e., the reaction forces in the suspension components are transmitting the stored kinetic energy of the coasting vehicle to the footprints, where it is getting dissipated. Thinking about it you can calculate an "effective force" for rolling resistance by using the relationship between work and force times distance so maybe they're both valid viewpoints, but in any event the effective force will be acting rearward on the rubber. -= mike =-

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>> The thing that intrigues me though, is that since the force is at the >> contact patch of the tire, it will torque the body (pitch) as well as >torque >> the tire itself, and unless I'm missing something here, this torque would >> actually serve to *accelerate* the tire, while simultaneously slowing the >> vehicle. Kinda wierd. Anyone know for sure if this is true, or have I >been >> drinking too much lately? :-) >> Todd Wasson >> --- >> Performance Simulations >> Drag Racing and Top Speed Prediction >> Software >> http://PerformanceSimulations.Com > >Thinking about this, wouldn't the rolling resistance actually act against >the direction of the wheel's roll? Effectively this would slow down the >wheels rotation causing a deceleration of the vehicle. >

It seems it would at first thought since the force slows the car, but I'm not really sure. The force acts on a spot that would accelerate the wheel rather than slow it down. If you read Gregor's thoughts in the other post on this he points out something I hadn't thought of. The force is acting at a point slightly forward of the contact patch center, so the vertical component (gravity reaction, or tire "spring" force) would act to slow the wheel. With both things happening at once, the vertical component might win and slow the wheel as you've said. It's an interesting thing to ponder. One of these days I'll try it in my simulator and see what it does. Gregor was talking a bit on how the forces might be measured after the slip ratio stabilized, sort of compensating for the effect. But the question then becomes, is the slip ratio positive, negative, or zero when the wheel is rotating at a constant rate, with a forward force acting directly at the wheel hub equal to the rolling resistance? If it's negative, I'd think this would mean the rolling resistance force is actually acting to slow the wheel (probably the positive gravity force reaction). If it's positive, then the first quote in this post would probably be right. If it's zero, then it all equals out some how. Either way though, the force still acts on the center of gravity of the car (if the suspension can't move forward or backwards) and would act to slow the car. What I'm curious to know is what is the slip ratio sign in this situation! It might switch depending on the location of the offset, but since the rolling resistance itself is dependent on the offset.... Hmmm.... :-) ? I don't know! Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

"Mike Nunan" <[email hidden]> wrote in message news:[email hidden]... > "Mike Stanley" <[email hidden]> wrote in messagenews:<CuT57.54813$[email hidden]>... > > I agree, if you imagine a situation of a car coasting from a smooth > road (low rolling resistance) onto, say, gravel, then the wheels will > be slowed exactly as if you applied outboard brakes. Todd, I'm not > sure it's valid to regard rolling resistance as a force, since it's a > representation of work done in the contact patch. F = mA. If the vehicle is slowing down, there is a net force acting on it.

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> The various forces > at play between rubber and road are converting kinetic energy into > heat and sound/vibration. I.e., the reaction forces in the suspension > components are transmitting the stored kinetic energy of the coasting > vehicle to the footprints, where it is getting dissipated. > > Thinking about it you can calculate an "effective force" for rolling > resistance by using the relationship between work and force times > distance so maybe they're both valid viewpoints, but in any event the > effective force will be acting rearward on the rubber. > > -= mike =-

I would suggest drawing a free-body diagram of the wheel. If we know that the wheel is slowing down, there must be moment on it counter to its direction of rotation. If we know that the vehicle is slowing down, there must also be a rearward force on the tire. Both of these forces must act through the contact patch. We can solve it by drawing two perpendicular forces, one horizontal rearward force which represents the decelerative force on the vehicle, and one vertical upward force which must equal the load on the wheel, and which acts at a center that creates the correct moment on the wheel, i.e. it must be forward of the center of axle. Conceptually the reason the vertical center of force is forward of the center of the axle is because the tire is not a perfect spring, i.e. you lose energy as you compress it. Therefore the tire is not pushing down as strongly at the rear as it is at the front, meaning the road isn't pushing up as strongly at the rear as it is at the front. HTH, Hal

On Sat, 21 Jul 2001, Haqsau wrote: .... > > be slowed exactly as if you applied outboard brakes. Todd, I'm not > > sure it's valid to regard rolling resistance as a force, since it's a > > representation of work done in the contact patch. > I would suggest drawing a free-body diagram of the wheel. If we know..... > center of axle. Conceptually the reason the vertical center of force is > forward of the center of the axle is because the tire is not a perfect > spring, i.e. you lose energy as you compress it. Therefore the tire is > not pushing down as strongly at the rear as it is at the front, meaning > the road isn't pushing up as strongly at the rear as it is at the front. And just to bring that nice visualization back to modeling, which definition of slip ratio are you using (or planning to use)? There are quite a few (see RCVD, pp 39-40) and they all have been used for one reason or another... -- Doug Milliken

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>Could someone possibly give a quick explaination of 'tire load' as >required by Pacejka as in my head suspension forces should be taken >into condition, but I have seen in other places that suspensions >forces are not taken into consideration but the 'tire load' is >calculated from the amount the tire is sinking into the ground (tire >compression). I am still waiting for my copy of RCVD to arrive from >Amazon (cost me a whopping $101)!!! > >

There are two basic ways to do this that I'm aware of. 1. Use the tire "sinking into the ground." In short, the tire in this model acts as a spring, which is compressed a given amount (calculated from distance from center of wheel to plane on contact). The tire spring rate is multiplied by this compression distance, and that gives the tire load directly. The tire itself then moves up or down using this force added to the suspension forces. Put simply, you could use the suspension spring and damper forces as a start, although the suspension geometry itself would have an effect. This is a bit complicated and not necessary unless you want to accurately get anti-dive, anti-squat, and jacking forces included. 2. (The method I'm using now.) Find the distance from some arbitrary point above a wheel to the ground (for example, the point directly above the wheel that crosses the horizontal plane through the center of mass of the sprung weight of the car). When the suspension is at full droop, there would be a set distance (you'd just pick one.) Find the difference between these two distances and that gives the amount of bump travel at the wheel. This could be multiplied by the "wheel rate" (another spring). If you calculate the velocity towards/away from the ground (using a 3-D velocity vector, the normal vector, and a cross product operation), you can multiply this by the shock absorber/damper rate to get another force. Add this to the "spring" force and you've got wheel load. This force also acts upwards on the car at this point. The drawback to method one is that you need high sampling rates. The cool thing about it is the tires themselves will be bouncy, so a perfectly rigid suspension would allow some body roll and pitch that could vary with tire pressure. IMO, it's overkill when doing "stock" type cars, as the suspension is usually 5-10 times softer than the tires anyway. It is, however, more correct and accurate than method two. If you're doing Champ cars or F-1 cars, method one should really be used, as the suspension system's stiffness approaches tire stiffness. Method two's benefits are primarily simplicity and speed. It also makes it easier to include the physical, geometric details of the suspension system itself later on if you want to add it in (IMO). Still, someday I'll probably switch to method one. Basically, if you're treating the wheels as seperate masses that move due to forces acting on them, method one, the "tire sinking into the ground" method, would give wheel load. Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

It's no fun to change the slip ratio convention once you have it implemented. We just changed for several reasons and it included a change in sign. I was mildly surprised that the car didn't run off backward the first time after the change! ;) - Matt Doug Milliken wrote: > And just to bring that nice visualization back to modeling, which > definition of slip ratio are you using (or planning to use)? There are > quite a few (see RCVD, pp 39-40) and they all have been used for one reason > or another...

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> Method two's benefits are primarily simplicity and speed. It also makes > it easier to include the physical, geometric details of the suspension system > itself later on if you want to add it in (IMO). Still, someday I'll probably > switch to method one. > > Basically, if you're treating the wheels as seperate masses that move due > to forces acting on them, method one, the "tire sinking into the ground" > method, would give wheel load. >

Thanks, I actually sat down and thought about it for a while and came up with my own solution (I usually find this works quite well :-)). I actually find the proportion that the suspension has been compressed as a ratio from 0.0 - 1.0, we already know the rate of the springs so just multiply the ratio and spring rate and voila, weight load per wheel in Newtons, if I add all of the weights for each of the wheels I get EXACTLY the weight of the car (in Newtons) just divide by 9.8 to get Kg. I do like the idea of having the tire as an extra spring, but currently my wheels are 'extremely' virtual, having no velocity, vertically in relation to the car which would make simulating the compression of the 'tire springs' impossible to do, for the while anyway, I'll be upgrading to slightly less virtual wheels at a later date. The other thing that I am having trouble with is this :- 1) My engine produces a torque through to the wheels (via gears, drive shaft, diff etc.) which spins the wheels. 2) I can correctly calculate the tire weight load, and slip ratio and hence am able to calculate the longitudinal force due to Pacejka's algorithm. What I don't understand is what do I do with this force that is calculated, obviously most of it is used to propel the car forward, but surely some of it is applied to the wheel because of the friction involved (if no force was applied the wheel would just be free to spin from the torques from the engine and would never slow down). For example, if the car is moving forward and the wheels are locked under braking, there is obviously a backward force which slows the car. But, and here is where I am stuck, there is either no force applied to the wheel (because if there was it would start spinning again) or, the force that is applied is less than the force that is braking the wheel and hence no wheel rotation. If it is the second case, the magnitudes of the forces would be greatly different. But as you can see there is obviously two results from the force calculated from Pacejka's formula and I'm buggered if I can figure it out. Colin

"Colin Reed" <[email hidden]> wrote in message news:[email hidden]... > > Method two's benefits are primarily simplicity and speed. It alsomakes > > it easier to include the physical, geometric details of the suspensionsystem > > itself later on if you want to add it in (IMO). Still, someday I'llprobably > > switch to method one. > > > > Basically, if you're treating the wheels as seperate masses thatmove due

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> > to forces acting on them, method one, the "tire sinking into the ground" > > method, would give wheel load. > > > > Thanks, I actually sat down and thought about it for a while and came > up with my own solution (I usually find this works quite well :-)). I > actually find the proportion that the suspension has been compressed > as a ratio from 0.0 - 1.0, we already know the rate of the springs so > just multiply the ratio and spring rate and voila, weight load per > wheel in Newtons, if I add all of the weights for each of the wheels I > get EXACTLY the weight of the car (in Newtons) just divide by 9.8 to > get Kg.

Hopefully you only get EXACTLY the weight of the car when it's perfectly stationary, cos if the car is "bouncing" at all on its suspension (over bumps) then you ought to get a different value. I think :0) Jim

On 22 Jul 2001 17:48:19 -0700, [email hidden] (Colin Reed) wrote: >Thanks, I actually sat down and thought about it for a while and came >up with my own solution (I usually find this works quite well :-)). I >actually find the proportion that the suspension has been compressed >as a ratio from 0.0 - 1.0, we already know the rate of the springs so >just multiply the ratio and spring rate and voila, weight load per >wheel in Newtons, Well, that's the suspension load, but when the wheel's not on the ground, the load is 0 still (I use method #1 btw, without tire flex damping, that's all done through the suspension's dampers, which may be off a little as indeed the tires are quite stiff).

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>1) My engine produces a torque through to the wheels (via gears, drive >shaft, diff etc.) which spins the wheels. > >2) I can correctly calculate the tire weight load, and slip ratio and >hence am able to calculate the longitudinal force due to Pacejka's >algorithm. > >What I don't understand is what do I do with this force that is >calculated, obviously most of it is used to propel the car forward, >but surely some of it is applied to the wheel because of the friction >involved (if no force was applied the wheel would just be free to spin >from the torques from the engine and would never slow down).

I use wheel rotational inertia, though some people don't and may add other methods. What I do is take the engine torque, subtract the Pacejka longitudinal force, and that makes the force which rotates the wheel (the road fighting the engine's wish to accelerate the tire). So you apply the resulting force as a torque (T=F*r) at the contact patch, then calculate angular acceleration as T=I*w there T=torque, I=inertia of the wheel and w=angular acc. Note however I do this for the whole drivetrain, not just the tire, so I add all (effective) inertiae and accelerate the lot, including driveshaft, gearbox (see my article on geared effective inertia on my site) and such. Seems like you can do it without inertia though somehow I think.

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>For example, if the car is moving forward and the wheels are locked >under braking, there is obviously a backward force which slows the >car. But, and here is where I am stuck, there is either no force >applied to the wheel (because if there was it would start spinning >again) or, the force that is applied is less than the force that is >braking the wheel and hence no wheel rotation. If it is the second >case, the magnitudes of the forces would be greatly different.

Braking is not your regular force, except when moving. It is more a potential force, and can keep a wheel locked when it's potential is bigger than the applied force (from the road and differential/engine). This is where you get an everpresent velocity reversal thing; when applying the brakes hard, you will get a point where the wheel start spinning the other direction (in your sim). At this point, lock the wheel (keep ang.vel=0) and don't let go until the brakeForce<otherForces. This same concept can be used for the clutch too, for example. Hope this helps, Ruud van Gaal, GPL Rank +53.25 Pencil art : http://www.marketgraph.nl/gallery/ Free car sim : http://www.marketgraph.nl/gallery/racer/

On Mon, 23 Jul 2001 14:09:37 +0100, "Jim Seamus" <[email hidden]> wrote:

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>> I >> actually find the proportion that the suspension has been compressed >> as a ratio from 0.0 - 1.0, we already know the rate of the springs so >> just multiply the ratio and spring rate and voila, weight load per >> wheel in Newtons, if I add all of the weights for each of the wheels I >> get EXACTLY the weight of the car (in Newtons) just divide by 9.8 to >> get Kg. > >Hopefully you only get EXACTLY the weight of the car when it's perfectly >stationary, cos if the car is "bouncing" at all on its suspension (over >bumps) then you ought to get a different value. > >I think :0)

Confirmed. A bouncing car doesn't have all the weight at the contact patches. Some of it is, uhm, flying! Ruud van Gaal, GPL Rank +53.25 Pencil art : http://www.marketgraph.nl/gallery/ Free car sim : http://www.marketgraph.nl/gallery/racer/

> Hopefully you only get EXACTLY the weight of the car when it's perfectly > stationary, cos if the car is "bouncing" at all on its suspension (over > bumps) then you ought to get a different value. Yes, only when the car is stationary :-) Colin

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>Thanks, I actually sat down and thought about it for a while and came >up with my own solution (I usually find this works quite well :-)). I >actually find the proportion that the suspension has been compressed >as a ratio from 0.0 - 1.0, we already know the rate of the springs so >just multiply the ratio and spring rate and voila, weight load per >wheel in Newtons, if I add all of the weights for each of the wheels I >get EXACTLY the weight of the car (in Newtons) just divide by 9.8 to >get Kg. > >I do like the idea of having the tire as an extra spring, but >currently my wheels are 'extremely' virtual, having no velocity, >vertically in relation to the car which would make simulating the >compression of the 'tire springs' impossible to do, for the while >anyway, I'll be upgrading to slightly less virtual wheels at a later >date.

That'll probably work just fine. Even if you combined a tire spring rate with the suspension spring rate as a sequential thing so people could adjust air pressure and such, no one would probably know see the difference (well, maybe Gregor or Ruud ;-)) Anyway, that's basically how I'm doing it right now too.

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>The other thing that I am having trouble with is this :- > >1) My engine produces a torque through to the wheels (via gears, drive >shaft, diff etc.) which spins the wheels. > >2) I can correctly calculate the tire weight load, and slip ratio and >hence am able to calculate the longitudinal force due to Pacejka's >algorithm. > >What I don't understand is what do I do with this force that is >calculated, obviously most of it is used to propel the car forward, >but surely some of it is applied to the wheel because of the friction >involved (if no force was applied the wheel would just be free to spin >from the torques from the engine and would never slow down).

I think I'm just echoing exactly what Ruud said, but more than one explanation couldn't hurt ;-) The force you're talking about here causes a torque around the wheel, according to the radius of the wheel. This torque, when added to the torque from the engine (through all the gears and whatever differential stuff you might have going on), keeps the tire from accelerating the way you're worried about. In fact, the tire will end up sitting at "just the right" slip ratio most of the time, and also get the little transient stuff going on automatically too when you adjust the throttle or brakes.

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>For example, if the car is moving forward and the wheels are locked >under braking, there is obviously a backward force which slows the >car. But, and here is where I am stuck, there is either no force >applied to the wheel (because if there was it would start spinning >again) or, the force that is applied is less than the force that is >braking the wheel and hence no wheel rotation. If it is the second >case, the magnitudes of the forces would be greatly different. >

The force that gets applied to the wheel here is the same thing as above. Take the radius of the tire, the force, and then calculate the torque acting on the tire *by the road* from that. The reason the wheel stays locked is because the brake itself won't allow the wheel to rotate unless the incoming torque, from both the road and the drivetrain, is great enough. In other words, if your brake is pressed together with enough force to stay locked at up to 10000 Nm torque, the wheel will stay locked until the force from the locked tire (actually, the torque you'd get from that), is greater than 10000Nm. If it goes any less than this, you'll need to write code that keeps the wheel locked, otherwise, the wheel will shake forward and backwards rapidly instead of locking. This causes wierd things to happen. As Ruud mentioned, you can do the same thing with the clutch. Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

On 24 Jul 2001 22:10:41 GMT, [email hidden] (J. Todd Wasson) wrote: ...

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>In other words, if >your brake is pressed together with enough force to stay locked at up to 10000 >Nm torque, the wheel will stay locked until the force from the locked tire >(actually, the torque you'd get from that), is greater than 10000Nm. > > If it goes any less than this, you'll need to write code that keeps the wheel >locked, otherwise, the wheel will shake forward and backwards rapidly instead >of locking. This causes wierd things to happen.

Indeed, what I once tried (but everybody seems to use the lock state) is to detect the velocity reversal, and then keep it *just* over the edge. So if your braking makes it go from +5rpm to -2rpm, I just top things off at -0.0001rpm. Next step, it tries to brake from -0.0001rpm to +6rpm for example. Then I again see the sign change, and keep it at +0.0001rpm. Now this is all moot ofcourse here, as the lock code looks cleaner, but I don't think the 'just over the sign' method gets you different car behavior. The point where I'm going to is that I think this might be useful perhaps too in the differential. For now, Gregor and I have this diff which has different equations depending on combinations of locked wheels, so 4 in all for 1 rear diff. If somehow this reversal thing might be used there, it could be easier to implement a 4WD car, which has 3 differentials. Normally, you would end up with 4x4x4=64 differents sets of equations, quite a pain to debug, to say the least. My intuitive feeling is that perhaps keeping one end jittering if locked but thereby using just 1 set of equations can help here. But my brain's cracking up now, so I can't think straight and figure this out right now. Been seeing too much network ideas for the last 24h. Ruud van Gaal, GPL Rank +53.25 Pencil art : http://www.marketgraph.nl/gallery/ Free car sim : http://www.marketgraph.nl/gallery/racer/

I would just like to thank all of you that have taken the time to reply to my questions so far, I am still in the early stages so will have many more questions no doubt. I wish you all could have seen the size of the smile on my face when I got all of the longitudinal slip code working, power slides, skidding to a halt, all working nicely. I'm gonna work on doing some better braking code soon but for the while if you brake it just locks the wheels, the car slides nicely to a stop :-) I have noticed some annoying jittering on the wheels when at low speeds ( <= 20 kph), I have done some code to counter this but there must be a more elegant way of accomplishing the same result. Does anyone have an answer to why the amount of longitudinal force is the same for 10% slip at 10kph as it is for 10% slip at 100kph. Is this right? As a tip to anyone else beginning to learn the Pacejka stuff, SlipRatio = ((w * Re) - V) / fabsf(V) This is from part 24 of the 'The Physics of Racing' by Brian Beckman, this ensures that even when the car has a -ve forward velocity the slip ratio is correct as opposed to your wheels exploding with excessive -ve forces :-( Colin

On 25 Jul 2001 04:18:32 -0700, [email hidden] (Colin Reed) wrote: >I have noticed some annoying jittering on the wheels when at low >speeds ( <= 20 kph), I have done some code to counter this but there >must be a more elegant way of accomplishing the same result. That's low speed problems and there are a lot of approaches. Some do kinematic. For slip angle jittering, I just let it jitter, and at high enough speed (frequencies) it doesn't show and let's you stop sideways on a hill without sliding down. Try to get your hands on SAE950311 (or info on it) from sae.org. It contains a method to make SR and SA differential equations instead of deriving them anew each step. Have some problems with it though, see another thread. > Does >anyone have an answer to why the amount of longitudinal force is the >same for 10% slip at 10kph as it is for 10% slip at 100kph. Is this >right? That's right. Thought it was weird at first too, but it's not the amount of slipvelocity that matters, it's the ratio as it's the rubber compressing and extending speed that I think makes it a ratio and not really absolute. Still, it's not really intuitive I agree, but right nonetheless. Ruud van Gaal, GPL Rank +53.25 Pencil art : http://www.marketgraph.nl/gallery/ Free car sim : http://www.marketgraph.nl/gallery/racer/

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>I would just like to thank all of you that have taken the time to >reply to my questions so far, I am still in the early stages so will >have many more questions no doubt. I wish you all could have seen the >size of the smile on my face when I got all of the longitudinal slip >code working, power slides, skidding to a halt, all working nicely. > >I'm gonna work on doing some better braking code soon but for the >while if you brake it just locks the wheels, the car slides nicely to >a stop :-) >

Yes, it's very exciting the first time, isn't it ? You'll experience both bouts of extreme frustration and big smiling (even evil, devilish sounding laughter) attacks in turn in the future :-) >I have noticed some annoying jittering on the wheels when at low >speeds ( <= 20 kph), I have done some code to counter this but there >must be a more elegant way of accomplishing the same result. This is an age old problem that even today engineers are still developing new methods to eliminate. Ruud pointed out a paper that has an excellent solution, probably the best one any of us here have seen, although you might need to drop him or Gregor a note for help (sorry guys ;-)). It's confusing to say the least, although rather brilliant. Thanks can go to Chris West for pointing it our way. The problem you're having is (we all have this problem), the tires tend to overshoot their "ideal" slip ratio by a small amount, then fall back the next step, and continue oscillating. At higher speeds, since the rotational speed of the wheel and theoretical free rolling speed are high, the oscillation in slip ratio isn't enough to notice or cause an instability. However, since slip ratio is essentially (TrueRotVel / TheoreticalRotVel), once the car slows down enough, even a small change in true rotational velocity causes a large change in slip ratio, which causes a really big torque back the other direction. This goes back and forth, and gets worse the slower you go. This is why I sometimes cringe when people here at r.a.s. occasionally get frustrated at some low speed behavioral "quirk" in one of their various racing sims, and immediately dub it "arcade" instead of "hardcore super realistic simulation." They don't understand how the situation is completely different when it moves to high speed racing modelling. The paper Ruud talks about treats slip ratio (and slip angle) as a state variable. Meaning you calculate what it "should" be the same way you're doing it now, but then you only let it increase/decrease a little bit towards this value each time step, depending on the speed of wheel rotation and linear motion. It slowly gains on the slip ratio (and angle) while the wheel rotates according to a somewhat arbitrarily defined "relaxation length" the wheel moves/rotates through, and eventually arrives at the slip ratio you're currently calculating. This lets the slip ratio and slip angle slowly build up, and when stopping on a hill (a problem you'll soon discover that will rear it's ugly head, causing a massive hair pulling episode), the slip ratio and angle settle at some non-zero value, even though the wheel isn't rotating, keeping the forces right where they should be. It's elegent and beautiful, IMO :-) I haven't included this in my system yet, although I've played around a little with the gain effect in a seperate program as an experiment to see how/why/if it worked. Anyway, since I didn't know about this relaxation approach when I first encountered the problem, I did two things: 1. I decoupled the rotational velocity/acceleration of the tires from the main system and let it operate at its own frequency. This lets me set the sampling frequency of the tires and differential to a really high value without doing every single body motion calculation and ground intersection 10-100 times as often (actually, it's adjustable on the fly, I should experiment with that come to think of it.) To see what I mean, increase the main frequency (decrease your timestep) and you'll notice that the speed where instabilities start to occur drops. This lessens the problem, but you'd need an infinite frequency to completely eliminate it this way. Perhaps there will be an Infinite HZ holographic laser/time warp operated CPU someday :-) (Inside joke: Would Papy make a CPU patch for this as quickly as they did for the 1.4 GHz if it didn't work right away? WOW, was that fast! ;-)) 2. I wrote something similar to a "rigid-ring" tire model for low speed/stationary movement. This treats the tire and wheel as two seperate things that can rotate around each other like a torsion spring. The angular difference between the two "rings" is multiplied by a spring rate, which provides a forward/aft force and torque, and is damped by the difference in rotational velocity. It works great for slow motion stuff, although it's tricky to adjust it so you don't notice the sudden change between the low and regular (high speed) model. It also doesn't deal well in certain situations with my differential gear code if the diff unlocks, so it isn't perfect, but it lets you stop on hills or move slowly up and down them with no stability problems and without resorting to absurdly high sampling frequencies. As far as the lateral (slip angle) stuff goes, my experience is the same as what Ruud mentioned in that letting the slip angle oscillate quickly between high negative and positive slip angles is usually unnoticeable, even at low speeds, and lets the car stop sideways on embankments. I haven't put anything special in my system to handle this, and you can only see the oscillations when I've got extremely high grip tires on a really light car, then it shakes just like it has a big engine, so maybe it's not all bad ;-) Still, if you've got a few bucks to pay SAE for the paper Ruud mentioned and are mathematically inclined enough to work it, it's the best solution I've seen for all the low speed problems. I'm planning on using it as a basis when it's time to get back into my tire model. Thanks for the slip ratio equation, I've currently got an IF block to handle the different forward/aft movement vs. rotation combinations to make sure the slip ratio is calculated correctly. That's ugly and slow compared to Brian Beckman's equation :-) > Does >anyone have an answer to why the amount of longitudinal force is the >same for 10% slip at 10kph as it is for 10% slip at 100kph. Is this >right? > I believe it's because the amount of longitudinal stretching in the tire is dependent on slip ratio, rather than slip velocity (that's kind of circular reasoning, but what the heck). Imagine a piece of rubber moving through the contact patch from front to rear (ignoring distortion from tire radius change.) Imagine it's moving rearwards 1 m/s quicker than it would be in a free rolling tire: 1. In a tire that had a free rolling velocity of only 1 m/s, the piece of rubber would leave the contact patch twice as quickly as it would if it was rolling freely. If the contact patch was 1 meter long (big tire!), it would have stretched out of place .5 meters from where it should be in a free rolling situation. If a longitudinal tire spring rate could be estimated at 1000N/meter, you'd now have 500 N of force. The tire acts like a big rubber band, so the amount of stretching (.5 meters in this case) is what causes the force. 2. In a tire that had a free rolling velocity of 100 m/s, the piece of rubber would leave the contact patch only 1/100th more quickly than it should. That would only amount to a tiny amount of stretching in the print (.01 * .5 meter print = .05 meters ahead, instead of .5 meters ahead.) Multiply that by our 1000N/meter spring rate and you'd only have 50 N of force. Anyway, I don't know if my math is quite right there, but maybe the reason is apparant now anyway. The slip "velocity" in both cases is identical at 1 m/s slip. If you tried to find the amount of stretching using different velocities calculated from a constant slip ratio, the amount of stretching would probably be constant. Therefore, since the "rubberband" is stretched out of shape the same amount, it'd make the same force. Make sense? Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

On 25 Jul 2001 22:31:47 GMT, [email hidden] (J. Todd Wasson) wrote:

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> The paper Ruud talks about treats slip ratio (and slip angle) as a state >variable. Meaning you calculate what it "should" be the same way you're doing >it now, but then you only let it increase/decrease a little bit towards this >value each time step, depending on the speed of wheel rotation and linear >motion. It slowly gains on the slip ratio (and angle) while the wheel rotates >according to a somewhat arbitrarily defined "relaxation length" the wheel >moves/rotates through, and eventually arrives at the slip ratio you're >currently calculating.

For the slip angle this is not entirely true, since sliding sideways will never get you at the exact 90 degree slipangle (or was it the 180 degree case). May be good enough just the same. > 1. I decoupled the rotational velocity/acceleration of the tires from the >main system and let it operate at its own frequency. This lets me set the >sampling frequency of the tires and differential to a really high value without >doing every single body motion calculation and ground intersection 10-100 times >as often (actually, it's adjustable on the fly, I should experiment with that >come to think of it.) Actually, Chris does that when things get stiffer (or you get to the bumpstops), IIRC. Well, perhaps the whole frequency goes up in that case, but it's the same idea, based on the situation at hand. > Would Papy make a CPU patch for this as quickly as they did for the 1.4 GHz if >it didn't work right away? WOW, was that fast! ;-)) Actually, they had seen the problem in N4, and it was a 32-bit counter that needed to be 64-bit (or 16 vs. 32). But still, yes, the fix was there fast. :)

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> 2. I wrote something similar to a "rigid-ring" tire model for low >speed/stationary movement. This treats the tire and wheel as two seperate >things that can rotate around each other like a torsion spring. The angular >difference between the two "rings" is multiplied by a spring rate, which >provides a forward/aft force and torque, and is damped by the difference in >rotational velocity. It works great for slow motion stuff, although it's >tricky to adjust it so you don't notice the sudden change between the low and >regular (high speed) model. It also doesn't deal well in certain situations >with my differential gear code if the diff unlocks, so it isn't perfect, but it >lets you stop on hills or move slowly up and down them with no stability >problems and without resorting to absurdly high sampling frequencies.

Sounds good; I'd have thought the longitudinal spring coefficient are so stiff that you STILL need high frequencies. This is the method somewhat in SAE980453 (may need to shuffle the 453 ;-) ), although that one uses a real 3D spring to fix things. Has its problems (read: things it doesn't support like camber), so I recommend SAE950311. Problem also is to get the 3 stiffness coefficients. Pacejka numbers are easier to guess almost. :)

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> As far as the lateral (slip angle) stuff goes, my experience is the same >as what Ruud mentioned in that letting the slip angle oscillate quickly between >high negative and positive slip angles is usually unnoticeable, even at low >speeds, and lets the car stop sideways on embankments. I haven't put anything >special in my system to handle this, and you can only see the oscillations when >I've got extremely high grip tires on a really light car, then it shakes just >like it has a big engine, so maybe it's not all bad ;-)

You can add that to your list of features! ;-) > Still, if you've got a few bucks to pay SAE for the paper Ruud mentioned >and are mathematically inclined enough to work it, it's the best solution I've >seen for all the low speed problems. I'm planning on using it as a basis when >it's time to get back into my tire model. I'm just still confused about the damping, but should bring that back after I've gotten energy back to work on that part again... > Thanks for the slip ratio equation, I've currently got an IF block to handle >the different forward/aft movement vs. rotation combinations to make sure the >slip ratio is calculated correctly. That's ugly and slow compared to Brian >Beckman's equation :-) It's also explained in that 200+ page PDF thesis btw. Including the implications for SAE950311 (another abs()). > Anyway, I don't know if my math is quite right there, but maybe the reason >is apparant now anyway. The slip "velocity" in both cases is identical at 1 >m/s slip. If you tried to find the amount of stretching using different >velocities calculated from a constant slip ratio, the amount of stretching >would probably be constant. Therefore, since the "rubberband" is stretched out >of shape the same amount, it'd make the same force. Make sense? Uhm, nice weather here today. ;-) Ruud van Gaal, GPL Rank +53.25 Pencil art : http://www.marketgraph.nl/gallery/ Free car sim : http://www.marketgraph.nl/gallery/racer/

Colin Reed <[email hidden]> wrote in message news:[email hidden]... > What I don't understand is what do I do with this force that is > calculated, obviously most of it is used to propel the car forward, > but surely some of it is applied to the wheel because of thefriction > involved (if no force was applied the wheel would just be free tospin > from the torques from the engine and would never slow down). Right... erm... <carefully ensures brain is in gear>

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> For example, if the car is moving forward and the wheels are locked > under braking, there is obviously a backward force which slows the > car. But, and here is where I am stuck, there is either no force > applied to the wheel (because if there was it would start spinning > again) or, the force that is applied is less than the force that is > braking the wheel and hence no wheel rotation. If it is the second > case, the magnitudes of the forces would be greatly different.

Okay, bug 1: the *torque* applied by the brake, to the wheel is *exactly* equal (and opposite) to the torque applied by the tyre (due to the tractive force produced). If they were different the wheel would accelerate (either +vely or -vely, i.e. decelerate). Remember that for friction "F <= mu N", not "F = mu N". Now, back to the original question. We have two free bodies (the car and the wheel/axle) plus the ground. (Not for nothing is my nickname "first principles" on the polo shirt I'm wearing ;-) - Engine (i.e. car) applies a torque to the wheel, which applies the same torque back to the car (see Newton, law 3) (causing a small amount of weight transfer - but I'll ignore that unless you /want/ me to go into it). - Tyre (i.e. wheel) accelerates (rotates), acquires a slip ratio and so generates a tractive force. Wheel pushes back on ground, ground pushes forward on wheel. - Wheel pushes forward on car (at axle bearings). Car applies equal and opposite reaction force through bearings. We now have two equal[1] forces on the wheel: the ground pushing it forwards, and the car pushing it back. However, these are offset by the tyre's rolling radius and so although there's no overall *force* on the wheel (it's all transferred to the car) there is a *torque*, which in equilibrium is equal to that applied by the drivetrain. Does that help? Jonny [1] They're not *quite* equal since some force is used in accelerating the wheel *forwards* (as opp. rotationally). But we'll conveniently ignore that for now.

> I'm gonna work on doing some better braking code soon but for the > while if you brake it just locks the wheels, the car slides nicely to > a stop :-) I've just found one big problem with locking the wheels when pressing the brake button, as we all know due to the traction circle, if the wheels are locked the longitudinal force is the maximum -ve force possible so there is not place on the 'circle' left for lateral forces. Result, I just go straight on and slam into the tires :-( This all means that I'll have to do some proper braking code sooner rather than later ;-) Colin

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>I've just found one big problem with locking the wheels when pressing >the brake button, as we all know due to the traction circle, if the >wheels are locked the longitudinal force is the maximum -ve force >possible so there is not place on the 'circle' left for lateral >forces. Result, I just go straight on and slam into the tires :-( > >This all means that I'll have to do some proper braking code sooner >rather than later ;-) >

This is still an area most of us are not clear on yet. Ruud allows two different types of friction circle activity to occur. I'm doing it the way you are right now, where once a wheel locks, the longitudinal force goes to maximum, and there is zero force left for lateral control. In other words, I'm calculating slip ratio and longitudinal force first, then, "just enough" force is applied in the lateral direction to keep inside the friction circle (if slip angle indicates the force is too large.) I don't think this is correct though. If a tire was running at a 10 degree slip angle, for example, and was locked, there would still be lateral distortion in the print, so there would still be lateral force. Ruud talked recently about and allows a "vector trim" approach, where the entire slip velocity is trimmed down to contact the edge of the friction circle, thus allowing for some lateral forces to be developed. This probably is more correct, as a car sliding sideways at 90 degrees with locked tires will certainly develop a lateral force! 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]... > I don't think this is correct though. If a tire was running at a 10degree > slip angle, for example, and was locked, there would still be lateral > distortion in the print, so there would still be lateral force. Ruudtalked > recently about and allows a "vector trim" approach, where the entire slip > velocity is trimmed down to contact the edge of the friction circle, thus > allowing for some lateral forces to be developed. This probably is more > correct, as a car sliding sideways at 90 degrees with locked tires will > certainly develop a lateral force! > This sounds similar to what I do. I first define the vector that results from lateral and longitudinal forces (without clipping to friction). I then calculate the length of this vector. If this exceeds the maximum force allowable due to friction, I clip to the outside of the friction circle (keeping the same vector direction), so the situation where you can't get any longitudinal or lateral force never arises. This seems to work quite well for me, and I'm sure it's the correct way to do things. All that's happening is that you're making sure that the force applied to the tyres doesn't exceed the friction limit. The force is a combination of lateral and longitudinal forces, and there's not really any reason for clipping each one seperately - the two directions are not independent as far as friction goes.

The only problem with the below is, when the wheels are locked, the force should be lower than the maximum one, no matter what the attitude of the car. With the below, assume the tyre is locked and the slip angle is past optimal as well. Each of these forces as calculated separately will give less than the maximum force, but when you combine them so that the length of the vector may in some cases be higher than the available force, you then clip them and end at the edge of the friction circle with a force that is higher than it should be. A good way of doing things properly is explained here: http://www.esbconsult.com.au/ogden/locost/phors/phors.htm Check chapters 24 & 25. Basically, you need to first calculate the combined slip and its components, and use these to explicitly (without clipping) calculate both the lateral and longitudinal forces from the equations for separate longitudinal and lateral response. This approach returns the same results as the separate operation when there is only slip ratio or slip angle present without the other. -Gregor Mike Stanley wrote:

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> > This sounds similar to what I do. I first define the vector that results > from lateral and longitudinal forces (without clipping to friction). I then > calculate the length of this vector. If this exceeds the maximum force > allowable due to friction, I clip to the outside of the friction circle > (keeping the same vector direction), so the situation where you can't get > any longitudinal or lateral force never arises. > > This seems to work quite well for me, and I'm sure it's the correct way to > do things. All that's happening is that you're making sure that the force > applied to the tyres doesn't exceed the friction limit. The force is a > combination of lateral and longitudinal forces, and there's not really any > reason for clipping each one seperately - the two directions are not > independent as far as friction goes.

On Mon, 30 Jul 2001 11:44:08 +0200, Gregor Veble <[email hidden]> wrote: >A good way of doing things properly is explained here: > >http://www.esbconsult.com.au/ogden/locost/phors/phors.htm > >Check chapters 24 & 25. And this method will be in Racer v0.4.7 btw. If I don't cling on car window transparency too much, it should be out, uhm, tonight, hopefully. Otherwise somewhere this week. Method #3, with thanks to Gregor & Brian for coming up with the same method. :) Ruud van Gaal, GPL Rank +53.25 Pencil art : http://www.marketgraph.nl/gallery/ Free car sim : http://www.marketgraph.nl/gallery/racer/

On Sun, 29 Jul 2001 20:59:30 +0100, "Jonny Hodgson" <[email hidden]> wrote:

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>Okay, bug 1: the *torque* applied by the brake, to the wheel is >*exactly* equal (and opposite) to the torque applied by the tyre (due >to the tractive force produced). If they were different the wheel >would accelerate (either +vely or -vely, i.e. decelerate). > >Remember that for friction "F <= mu N", not "F = mu N". > >Now, back to the original question. We have two free bodies (the car >and the wheel/axle) plus the ground. (Not for nothing is my nickname >"first principles" on the polo shirt I'm wearing ;-) > >- Engine (i.e. car) applies a torque to the wheel, which applies the >same torque back to the car (see Newton, law 3) (causing a small >amount of weight transfer - but I'll ignore that unless you /want/ me >to go into it).

You can play with the effects of the weight transfer in GPL; hold the clutch button and throttle away. The body moves a bit. You can get an exaggerated example of this in Racer v0.4.7, setting engine.torque_reaction to 1 (or beyond 1 to make it unrealistic as can be). Looks nice on low-sprung cars, but is a bit much, a setting of 0.2 may be better (but just making that up here). >- Tyre (i.e. wheel) accelerates (rotates), acquires a slip ratio and >so generates a tractive force. Wheel pushes back on ground, ground >pushes forward on wheel. > >- Wheel pushes forward on car (at axle bearings). Car applies equal >and opposite reaction force through bearings. I think I apply the forward forces at the wheel's roll center, but I'm not really simulating rod links very well so this is the best I could do perhaps. Feels ok-ish, but have no idea whether the rollcenters are on the mark. :) Ruud van Gaal, GPL Rank +53.25 Pencil art : http://www.marketgraph.nl/gallery/ Free car sim : http://www.marketgraph.nl/gallery/racer/

I feel that the proper boundary condition of tire forces when locked is to have the resulting force be anti-velocity. This makes it not a function of the direction the SAE tire axes are pointing, unless your combined slip model is sophisticated enough to impose this boundary condition. -- Matthew V. Jessick Motorsims [email hidden] http://www.moto1.net Vehicle Dynamics Engineer (972)910-8866 Ext.125, Fax: (972)910-8216

Hi Matthew, agreed. That's why I use directly the components of the slip velocity (velocity_size*sin(slip_angle) and Omega*R-velocity_size*cos(slip_angle), and in my program I explicity calculate the slip velocity vector to get these two components) divided by the size of the velocity to do my calculations instead of slip angle, slip ratio. For small slip angles, the definitions are equivalent anyway, but for larger ones this one is more natural and gives the correct locked wheel behaviour. -Gregor mjessick-Motorsims wrote:

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> > I feel that the proper boundary condition of tire forces > when locked is to have the resulting force be anti-velocity. > > This makes it not a function of the direction the SAE > tire axes are pointing, unless your combined slip model > is sophisticated enough to impose this boundary condition. > > -- > Matthew V. Jessick Motorsims > [email hidden] http://www.moto1.net > Vehicle Dynamics Engineer (972)910-8866 Ext.125, Fax: (972)910-8216

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>I feel that the proper boundary condition of tire forces >when locked is to have the resulting force be anti-velocity. > >This makes it not a function of the direction the SAE >tire axes are pointing, unless your combined slip model >is sophisticated enough to impose this boundary condition. >

That's what I was thinking too. Otherwise, front tires that are locked can generate steering forces, like they do sometimes in my stuff ;-) Glad to know you feel this way about it. Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

On 03 Aug 2001 00:32:56 GMT, [email hidden] (J. Todd Wasson) wrote:

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>>I feel that the proper boundary condition of tire forces >>when locked is to have the resulting force be anti-velocity. >> >>This makes it not a function of the direction the SAE >>tire axes are pointing, unless your combined slip model >>is sophisticated enough to impose this boundary condition. >> > > That's what I was thinking too. Otherwise, front tires that are locked can >generate steering forces, like they do sometimes in my stuff ;-) Glad to know >you feel this way about it.

Just have 1 NDE (Near Death Experience) and you'll know for all time. In fact, getting of the brakes at one time saved me from hitting a sudden line of cars, and I could steer onto the exit lane (that I was supposed to take, mind you :) ). Just didn't have the time to test how soon it started to loose all steering control, hehe. (in other words, at what point does the tire really begin to act like a block of rubber instead of the normal flexing and all that). Ruud van Gaal, GPL Rank +53.25 Pencil art : http://www.marketgraph.nl/gallery/ Free car sim : http://www.marketgraph.nl/gallery/racer/