Car Physics : Pacekja Revisted!

Hi Folks, I now have the longitudnal Pacekja formula working fairly well, with an acceleration and top-speed that is reasonable for my example cars, thanks everyone for the help! I am now returning to the lateral model. I am beginning to think that when the slip angle of a tyre is zero it does not necessarily mean that the force produced is zero. I have found the following webpage:- http://www.rrci.com/pacejka96.html#Pacejka%2096 Can anyone tell me why the graph does not centre on zero? I just cant understand why, perhaps they have got it wrong too? Thanks again for the help Ashley

If the tire is not stricteley vertical there is a lateral force for a slide angle of 0. I've made Car Physics for my job and i can tell you that the LateralForce vs Slip Angle is more complicated in reallity. Normaly there is a pick at 6degres then the Force decreased and Stablised at 15degre. MoreOver, the simpliest thing is to make a Graph K = F( SlipAngle ) and then LateralForce = K * TireLoad. But if you want a real thing K = F( SlipAngle , SlipRatio ). I really recommand you all the S.I.A articles, in French, SIA means Society of Automotive Engineer. http://www.sia.fr I use all there articles to make my car physics... But it's all in French ... sorry... Ashley McConnell wrote:

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> Hi Folks, > > I now have the longitudnal Pacekja formula working fairly well, with an > acceleration and top-speed that is reasonable for my example cars, thanks > everyone for the help! > > I am now returning to the lateral model. I am beginning to think that when > the slip angle of a tyre is zero it does not necessarily mean that the force > produced is zero. > > I have found the following webpage:- > > http://www.rrci.com/pacejka96.html#Pacejka%2096 > > Can anyone tell me why the graph does not centre on zero? I just cant > understand why, perhaps they have got it wrong too? > > Thanks again for the help > Ashley

-- Seb Game Developer GPLRank -36.24 http://magicfr.multimania.com

Hi Ash, The graphs on the site you show are for a tire set at a 3 degree camber angle. The graph doesn't center on zero primarily because of camber thrust, although if the Pacejka Magic Formula has coefficients to approximate conicity and ply steer, those may also be a (much smaller) contributor. Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

Todd, Sebastien Aha - I see, didnt notice the camber. One thing that is confusing me though :- I can see that camber thrust would produce such a graph, but would the graph not be different for the left and right wheels. I mean from my R/C car days - 3 degrees -ve Camber meant that each of the wheels were "bent inwards" by 3 degrees each, which would produce an equal and OPPOSITE force by each wheel. Should the -ve camber on the left wheel be an equal +ve camber on the right? What I mean is, if there is equal camber on each (left and right) wheel the resultant Lateral force will be off to one side, but what we want is for them to cancel eachother out. Another thing is my own data is producing a lateral force at zero slip and zero camber. My brain is a bit fried (work at the minute) Thanks for the help Ash "J. Todd Wasson" <[email hidden]> wrote in message news:[email hidden]... | Hi Ash, | | The graphs on the site you show are for a tire set at a 3 degree camber | angle. The graph doesn't center on zero primarily because of camberthrust, | although if the Pacejka Magic Formula has coefficients to approximateconicity

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| and ply steer, those may also be a (much smaller) contributor. | Todd Wasson | --- | Performance Simulations | Drag Racing and Top Speed Prediction | Software | http://PerformanceSimulations.Com

>I can see that camber thrust would produce such a graph, but would the graph >not be different for the left and right wheels. I mean from my R/C car >days - 3 degrees -ve Camber meant that each of the wheels were "bent >inwards" by 3 degrees each, which would produce an equal and OPPOSITE force >by each wheel. Should the -ve camber on the left wheel be an equal +ve >camber on the right? I think the camber angle here is really "inclination angle", which is the same angle, only a negative or positive value on both of the front/rear tires means the tires tilt in the same direction. So, yes, on my first impression, the -ve camber on the left wheel would be an equal +ve camber on the right, only technically, this is "inclination angle" at this point. >What I mean is, if there is equal camber on each (left and right) wheel the >resultant Lateral force will be off to one side, but what we want is for >them to cancel eachother out. Right. However, according to the chart, this will actually mean you've got positive inclination on one tire, with negative on the other. I guess you'd need another data set for the other angle. Perhaps the graph can be flipped around to reflect this somehow. >Another thing is my own data is producing a lateral force at zero slip and >zero camber. I haven't worked much with Pacejka's formulas yet, so I'm not sure. I did some experimenting with the non-dimensionalization technique in Milliken's book, but didn't see this effect happen, possibly because there were only five constants for the lateral force behavior, rather than the thirteen you listed in an earlier post. Perhaps Dr. Pacejka's other constants reflect conicity and ply steer. These two factors supposedly can make a tire oriented at 0 camber/slip angle produce some lateral force. Apparently, many real tires indeed exhibit this behavior, according to Milliken & Milliken's "Race Car Vehicle Dynamics." I'm far from being an expert in this area, so I couldn't tell you what in the world conicity or ply steer actually is! It must have something to do with how tires torque themselves when a vertical load is applied during rotation, even at 0 camber/slip, or something like that. I wouldn't immediately jump to the conclusion that the data you're seeing is incorrect, however. Perhaps setting one or more of the constants to 0 may get rid of the effect, who knows? If you're like me, you'd have already tried that a million times by now out of shear frustration, so I'm at the end of my rope here otherwise. If you'll post, email, or show a link to the full 13 constant equation for lateral force you're using, I'll try to duplicate the effect using the constants you posted. Maybe we can figure it out that way. >My brain is a bit fried (work at the minute) I know the feeling! Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

Hi Todd, Sorry this is so short, I am shattered (was ill on Saturday and havent recovered) - I will formulate a response in the morning :) This is the formula for Fy from the book Fy = D sin(C arctan {B(1 - E)(Sideslip + Sh) + E arctan [B(sideslip + Sh)]}) + Sv where C = a0 D = u(yp)Fz -> u(yp) = a1Fz + a2 E = a6Fz + a7 BCD = a3 sin[2 arctan(Fz/a4)](1 - a5 |CamberAng|) Sh = a8 CamberAng + a9Fz + a10 Sv = a11 CamberAng Fz + a12Fz + a13 a11 = a111Fz + a112 Here are the coefficients 1.6929 //a0 -55.2084 //a1 1271.28 //a2 1601.8 //a3 6.4946 //a4 0.0047966 //a5 -0.3875 //a6 1.0 //a7 -0.045399 //a8 0.0042832 //a9 0.086536 //a10 -7.9730 //a111 -0.2231 //a112 7.6680 //a12 45.8764 //a13 Thanks for the help, hope this is helping you too (at least a bit!) Ash Sent via Deja.com http://www.deja.com/

On 15 Jan 2001, J. Todd Wasson wrote: > I think the camber angle here is really "inclination angle", which is the > same angle, only a negative or positive value on both of the front/rear tires > means the tires tilt in the same direction. So, yes, on my first impression, > the -ve camber on the left wheel would be an equal +ve camber on the right, > only technically, this is "inclination angle" at this point. Tire data is normally given in terms of inclination angle (IA), because when a tire is measured (test rig) there is usually no way to know if it will be mounted on the left or right side of the car. There is a little sketch in RCVD that shows the difference between camber and inclination. We normally use IA all the way through our models to avoid confusion when inputting new tire data, etc -- lean to the right is + in the SAE axis system. > Perhaps Dr. Pacejka's other constants reflect conicity and ply steer. These > two factors supposedly can make a tire oriented at 0 camber/slip angle produce > some lateral force. Apparently, many real tires indeed exhibit this behavior, > according to Milliken & Milliken's "Race Car Vehicle Dynamics." I'm far from Basically all tires are asymmetric, so they make sideforce when tested at 0 SA and 0 IA. Conicity refers to behavior like a cone that rolls in a circle, as if the sidewalls were different height. Plysteer is like a screw thread, possibly due to the fact that the outer belt (usually steel wire) is wound onto the tire in a helix. The test for these is to set the tire at 0 SA & IA then slowly run the test rig forward _and_ backward, then look at the sum and difference of the lateral force measured (and the aligning torque too). Both of these effects are most prominent at low SA (slip angles), and tend to wash out at high SA -- so for racecar sims they can probably be ignored. -- Doug Douglas Milliken <[email hidden]> Milliken Research Associates Inc.

The sacrilege! Ignoring any effect in a racing sim! ;) Great to see you around here, -Gregor Doug Milliken wrote:

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> > Both of these effects are most prominent at low SA (slip angles), and tend > to wash out at high SA -- so for racecar sims they can probably be ignored. > > -- Doug > Douglas Milliken <[email hidden]> > Milliken Research Associates Inc.

Doug, Thanks for the reply, I have been thinking this over and come up with what I think could be a solution. My graph is different from the graph on the website. For a -ve slip I get a -ve Force, for a +ve slip I get a positive force. I have a force of -190N (if I recall) when there is no sideslip. What I think is that the result from the formula gives an INWARD force of 190N when there is no sideslip. Therefore the righthand wheel gives a "left force" of 190N and the lefthand wheel gives a "right force" of 190N. This means they cancel each other out and the car goes straight....wahey! :) So what I can do is:- --- Say we have a +20deg slip angle --- get the slip angle for each wheel (I am not entirely sure of how exactly to do this yet - probably because I dont have a representation of "the world" yet). --- For my graph a +ve sideslip yields a +ve force (i.e. to the right) - so the right hand wheel's force does not need to be modified --- *** In order to get the correct left wheel force we should put in the slipangle * -1 (i.e. -20 degrees). --- Run the slip angle through the Fy Pacekja equation to get a positive or negative force --- Then take the result force and multiply it by -1 to get it facing in the right direction (for the Left wheels) Have I gone nuts? Is this the right way to go about things? It all makes sense using my cunning a4 page of diagrams :) Thanks for the help Ashley "Doug Milliken" <[email hidden]> wrote in message news:[email hidden]... | | On 15 Jan 2001, J. Todd Wasson wrote: | > I think the camber angle here is really "inclination angle", which isthe | > same angle, only a negative or positive value on both of the front/reartires | > means the tires tilt in the same direction. So, yes, on my firstimpression, | > the -ve camber on the left wheel would be an equal +ve camber on theright,

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| > only technically, this is "inclination angle" at this point. | | Tire data is normally given in terms of inclination angle (IA), because | when a tire is measured (test rig) there is usually no way to know if it | will be mounted on the left or right side of the car. There is a little | sketch in RCVD that shows the difference between camber and inclination. | We normally use IA all the way through our models to avoid confusion when | inputting new tire data, etc -- lean to the right is + in the SAE axis

system. | | > Perhaps Dr. Pacejka's other constants reflect conicity and ply steer.These | > two factors supposedly can make a tire oriented at 0 camber/slip angleproduce | > some lateral force. Apparently, many real tires indeed exhibit thisbehavior, | > according to Milliken & Milliken's "Race Car Vehicle Dynamics." I'm farfrom

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| | Basically all tires are asymmetric, so they make sideforce when tested at | 0 SA and 0 IA. Conicity refers to behavior like a cone that rolls in a | circle, as if the sidewalls were different height. Plysteer is like a | screw thread, possibly due to the fact that the outer belt (usually steel | wire) is wound onto the tire in a helix. The test for these is to set the | tire at 0 SA & IA then slowly run the test rig forward _and_ backward,

then | look at the sum and difference of the lateral force measured (and the | aligning torque too). | Both of these effects are most prominent at low SA (slip angles), andtend | to wash out at high SA -- so for racecar sims they can probably beignored. | | -- Doug | Douglas Milliken <[email hidden]> | Milliken Research Associates Inc. | |

On Wed, 17 Jan 2001 12:40:15 -0000, "Ashley McConnell" <[email hidden]> wrote: ... >My graph is different from the graph on the website. For a -ve slip I get >a -ve Force, for a +ve slip I get a positive force. I have a force of -190N >(if I recall) when there is no sideslip. With no camber or suspension angles? That seems strange. Perhaps some toe in/out? ...

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>--- Say we have a +20deg slip angle > >--- get the slip angle for each wheel (I am not entirely sure of how exactly >to do this yet - probably because I dont have a representation of "the >world" yet). > >--- For my graph a +ve sideslip yields a +ve force (i.e. to the right) - so >the right hand wheel's force does not need to be modified > >--- *** In order to get the correct left wheel force we should put in the >slipangle * -1 (i.e. -20 degrees).

That would not seem right. When you look at it, one wheel doesn't know where the other's are. The slipangle for both wheels is +20 degrees. Consider a normal situation, where slipangle for example doesn't exceed 6 degrees (above that, most tires will behave worse in turning force than from 0..6 degrees). If you're turning left, both front wheels will create a leftdirected force of 'x' Newton. If both slipangles are 6, both forces will be 'x' Newton. Not -x for any wheel. Not even when you take the back wheels (it's just the rotational velocity of the car and the geometry to match that with the general velocity of the car will make changes in the slip angle for the back wheels). >--- Run the slip angle through the Fy Pacekja equation to get a positive or >negative force I really must get into Pacekja (Pacejka?). Seems like a very nice formula. :) Fortunately, I can finally drive on a track, so I've picked up a book on car physics again yesterday; had been a long time. As Todd indicated, perhaps you're mixing camber with inclination angle, which have different signs depending on which side of the car you are. As I don't limit myself to 4 wheels, I will only use inclination angle I guess. I could have a wheel right at the center. Now what would that generate for camber sign? ;-) Ruud van Gaal, GPL Rank +53.25 Pencil art : http://www.marketgraph.nl/gallery/ Car simulation: http://www.marketgraph.nl/gallery/racer/

"Ruud van Gaal" <[email hidden]> wrote in message news:[email hidden]... | On Wed, 17 Jan 2001 12:40:15 -0000, "Ashley McConnell" | <[email hidden]> wrote: | | ... | >My graph is different from the graph on the website. For a -ve slip Iget | >a -ve Force, for a +ve slip I get a positive force. I have a forceof -190N | >(if I recall) when there is no sideslip. | | With no camber or suspension angles? That seems strange. Perhaps some | toe in/out? Yes, nothing at all.....no mention of toe-in though. There is provision for camber angle - but it is set at zero. | | ... | >--- Say we have a +20deg slip angle | > | >--- get the slip angle for each wheel (I am not entirely sure of howexactly | >to do this yet - probably because I dont have a representation of "the | >world" yet). | > | >--- For my graph a +ve sideslip yields a +ve force (i.e. to the right) -so

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| >the right hand wheel's force does not need to be modified | > | >--- *** In order to get the correct left wheel force we should put in the | >slipangle * -1 (i.e. -20 degrees). | | That would not seem right. When you look at it, one wheel doesn't know | where the other's are. The slipangle for both wheels is +20 degrees. | Consider a normal situation, where slipangle for example doesn't | exceed 6 degrees (above that, most tires will behave worse in turning | force than from 0..6 degrees). If you're turning left, both front | wheels will create a leftdirected force of 'x' Newton. If both | slipangles are 6, both forces will be 'x' Newton. Not -x for any | wheel. Not even when you take the back wheels (it's just the | rotational velocity of the car and the geometry to match that with the | general velocity of the car will make changes in the slip angle for | the back wheels).

How I was thinking about it was - if its an -190N at zero degrees slip, then the left wheel must be under a force of 190N towards the centre of the car and the right wheel must also be under 190N force in the opposite direction. If the left wheel is under a positive slip angle (i.e. to the right) - eg. 6degrees, then the force would be MORE on the left wheel than a negative slip of 6 degrees (due to the force that is already acting right at zero degrees). Sorry thats a bit incoherent, but its getting late in the working day ;) | | >--- Run the slip angle through the Fy Pacekja equation to get a positiveor | >negative force | | I really must get into Pacekja (Pacejka?). Seems like a very nice | formula. :) Fortunately, I can finally drive on a track, so I've | picked up a book on car physics again yesterday; had been a long time. I think he is from around your parts - Tudelft universiry in holland (i think) | | As Todd indicated, perhaps you're mixing camber with inclination | angle, which have different signs depending on which side of the car | you are. As I don't limit myself to 4 wheels, I will only use | inclination angle I guess. I could have a wheel right at the center. | Now what would that generate for camber sign? ;-) Problem is there is neither inclination or camber .....it is quite strange. | | | Ruud van Gaal, GPL Rank +53.25 | Pencil art : http://www.marketgraph.nl/gallery/ | Car simulation: http://www.marketgraph.nl/gallery/racer/ Thanks for your help, had a good look at your website, some good stuff on there! Ash

On Wed, 17 Jan 2001, Ashley McConnell wrote:

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> How I was thinking about it was - if its an -190N at zero degrees slip, then > the left wheel must be under a force of 190N towards the centre of the car > and the right wheel must also be under 190N force in the opposite direction. > If the left wheel is under a positive slip angle (i.e. to the right) - eg. > 6degrees, then the force would be MORE on the left wheel than a negative > slip of 6 degrees (due to the force that is already acting right at zero > degrees).

_If_ your Magic Formula expansion is correct and one front tire is producing some side force at 0 slip angle (SA) and 0 inclination angle (IA), then the other front tire (same Magic Formula coefficients) is also producing the same sideforce, in the same absolute direction (assume that the tires are under equal load). The result is that your car will turn... So, to go straight down the road, you will need to add some steer angle (to both wheels), so that the sideforce produced by the slip angle counters the sideforce that the tire is making "all by itself", due to plysteer and/or conicity (or whatever asymmetry is built into that set of Magic Formula coefficients). This is perfectly normal -- most real cars don't drive straight with the steering wheel centered, because the tires are asymmetric. Even if the car was carefully aligned with the wheel planes parallel to the center plane of the car and the steering wheel carefully adjusted to point straight, small tire forces mean that some "non-zero" steering is required to move in a straight line. A much more interesting problem is to figure out why some cars run straight with no hands on the steering wheel -- even when the tires produce some small lateral forces at 0 SA and 0 IA. And it gets even more interesting to analyze when you have four different tires, each built slightly differently so that they all have different amounts of asymmetry -- the real world is actually _very_ messy. -- Doug Douglas Milliken <[email hidden]> Milliken Research Associates Inc.

On Fri, 12 Jan 2001 16:08:16 -0000, "Ashley McConnell" <[email hidden]> wrote: >I now have the longitudnal Pacekja formula working fairly well, with an >acceleration and top-speed that is reasonable for my example cars, thanks >everyone for the help! Hi Ash, I've looked around several times now on the net, but haven't found any real good information on Pacejka's Magic formula yet. Do you happen to have a URL that will provide some info, or perhaps a book where you got the actual info from (to implement it in software?) Thanks, Ruud van Gaal, GPL Rank +53.25 Pencil art : http://www.marketgraph.nl/gallery/ Car simulation: http://www.marketgraph.nl/gallery/racer/

Hi Ruud, Nice to hear from you! I have only got the book "Motor Vehicle Dynamics by Giancarlo Genta" - it has a full set of values for (inc. Pacekja for tyres) a Fiat Tipo, Saab 9000 and a Ferrari F348 (also some large truck and a motorbike). I am a bit suspicious on how correct and consistant they are due to the large lateral forces at zero degrees camber and sideslip. There is a problem with one of the inertia values in the F348 though - it should be .7kg per something instead of 7 - this really messed with my head for a while ;) I have recently bought Doug Millikens book (he will be relieved to hear ;)) which is great (and I think I heard you mention you had bought), it is excellently written and alot easier to follow compared with G.Genta's, however I was a little disappointed in the level of detail in the tyre section (I am only on chapter 5 at the minute), but it may be only a simple model to be expanded later. The book is essential to gain a proper understanding in a well layed out manor, however G.Genta's book has alot of the formulas needed for actually implementing a model, which is handy. Sorry I am rambling ;) - If you find any pacekja coefficients anywhere please let me know Ash "Ruud van Gaal" <[email hidden]> wrote in message news:[email hidden]...

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| On Fri, 12 Jan 2001 16:08:16 -0000, "Ashley McConnell" | <[email hidden]> wrote: | | >I now have the longitudnal Pacekja formula working fairly well, with an | >acceleration and top-speed that is reasonable for my example cars, thanks | >everyone for the help! | | Hi Ash, | | I've looked around several times now on the net, but haven't found any | real good information on Pacejka's Magic formula yet. Do you happen to | have a URL that will provide some info, or perhaps a book where you | got the actual info from (to implement it in software?) | | Thanks, | | Ruud van Gaal, GPL Rank +53.25 | Pencil art : http://www.marketgraph.nl/gallery/ | Car simulation: http://www.marketgraph.nl/gallery/racer/

On Wed, 31 Jan 2001 15:50:26 -0000, "Ashley McConnell" <[email hidden]> wrote:

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>Hi Ruud, > >Nice to hear from you! I have only got the book "Motor Vehicle Dynamics by >Giancarlo Genta" - it has a full set of values for (inc. Pacekja for tyres) >a Fiat Tipo, Saab 9000 and a Ferrari F348 (also some large truck and a >motorbike). I am a bit suspicious on how correct and consistant they are >due to the large lateral forces at zero degrees camber and sideslip. There >is a problem with one of the inertia values in the F348 though - it should >be .7kg per something instead of 7 - this really messed with my head for a >while ;)

I'll keep that in mind, eh, archive! Always nice to stay sane. :) ... >model to be expanded later. The book is essential to gain a proper >understanding in a well layed out manor, however G.Genta's book has alot of >the formulas needed for actually implementing a model, which is handy. Sounds good, just ordered it (@B&N@$59). >Sorry I am rambling ;) - If you find any pacekja coefficients anywhere >please let me know Perhaps I'll find some near where I left my socks, eh, no, probably not. ;-) Thanks for the reply. Ruud van Gaal, GPL Rank +53.25 Pencil art : http://www.marketgraph.nl/gallery/ Car simulation: http://www.marketgraph.nl/gallery/racer/

>I was a little disappointed in the level of detail in the tyre >section (I am only on chapter 5 at the minute), but it may be only a simple >model to be expanded later. I'll rear my ugly head in here for a second. Flip to chapter 14 for the Radt nondimensional method. BTW, I still have the Pacejka coefficient data you supplied, and will have a look shortly. Thanks! Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

Todd, Much appreciated - I am still reading and learning :). I am having a break for a couple of months to finish off my old Radio Controlled car timing system ( http://www.race-timing.com ) if anyone is interested (*plug plug*). But it will be great to lay back and do some reading without the urge to code all the time - might help get a bit of structure to it all! :) I am still a bit confused about why I am getting a massive sideways force at zero slip angle and camber angle. I have visions of my car sitting sliding to one side off the start line ;) - I am not sure how to calculate the counter friction force (or more specifically the Coeff of Friction) when the car is in the "Friction Zone" as Doug puts it, I know it must be linked to some pacekja values, but I dunno how. The product BCD yields the cornering stiffness and the maximum sideforce coefficient is also defined as (a1 x Fz + a2). Perhaps the max coefficient of friction is just a1? Trying to work things out while typing a post is probably a bad idea, especially at work ;) I better go and earn my keep :) Thanks for the help Ash "J. Todd Wasson" <[email hidden]> wrote in message news:[email hidden]... | >I was a little disappointed in the level of detail in the tyre | >section (I am only on chapter 5 at the minute), but it may be only asimple | >model to be expanded later. | | I'll rear my ugly head in here for a second. Flip to chapter 14 for theRadt

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| nondimensional method. BTW, I still have the Pacejka coefficient data you | supplied, and will have a look shortly. Thanks! | | Todd Wasson | --- | Performance Simulations | Drag Racing and Top Speed Prediction | Software | http://PerformanceSimulations.Com

> I am having a break >for a couple of months to finish off my old Radio Controlled car timing >system ( http://www.race-timing.com ) if anyone is interested (*plug plug*). Very cool program! I've saved it and will check in occasionally to see what's new :0) >I am still a bit confused about why I am getting a massive sideways force at >zero slip angle and camber angle. I still haven't run your #'s to see what results you're getting, but I think the numbers might be correct. As others have pointed out, conicity and plysteer can cause a lateral force, even at 0 camber and slip angle. >I have visions of my car sitting sliding >to one side off the start line ;) I wouldn't worry about that too much. Right now, I have a seperate low speed tire model for the rotational direction, but use the same high speed version (slip angle based) for lateral forces. The car usually oscillates slightly when sitting still, with +/-90 degree slip angle. However, since the main frequency is about 300Hz usually, it's barely noticable. >Perhaps the max coefficient of friction is just a1? From the info you posted a couple weeks ago, I'd say that a0 is probably peak lateral coefficient of friction, while b0 is peak longitudinal. I still haven't tried it though :-) Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

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>This is the formula for Fy from the book > >Fy = D sin(C arctan {B(1 - E)(Sideslip + Sh) + E arctan [B(sideslip + >Sh)]}) + Sv > >where > >C = a0 > >D = u(yp)Fz -> u(yp) = a1Fz + a2 > >E = a6Fz + a7 > >BCD = a3 sin[2 arctan(Fz/a4)](1 - a5 |CamberAng|) > >Sh = a8 CamberAng + a9Fz + a10 > >Sv = a11 CamberAng Fz + a12Fz + a13 > >a11 = a111Fz + a112 > > >Here are the coefficients >1.6929 //a0 >-55.2084 //a1 >1271.28 //a2 >1601.8 //a3 >6.4946 //a4 >0.0047966 //a5 >-0.3875 //a6 >1.0 //a7 >-0.045399 //a8 >0.0042832 //a9 >0.086536 //a10 >-7.9730 //a111 >-0.2231 //a112 >7.6680 //a12 >45.8764 //a13

Ash, I'm playing with these now and have a couple questions for you, if you don't mind :0). Where is BCD used? Also, where is B defined? Sideslip is in degrees or radians? Will these coefficients only work in the metric system? It's tricky figuring out where to put 180*PI, and PI/180 in the fy equation!! lol Thanks, Todd Wasson --- Performance Simulations Drag Racing and Top Speed Prediction Software http://PerformanceSimulations.Com

On Thu, 1 Feb 2001 09:52:57 -0000, "Ashley McConnell" <[email hidden]> wrote: ... >I am still a bit confused about why I am getting a massive sideways force at >zero slip angle and camber angle. I have visions of my car sitting sliding >to one side off the start line ;)... >the maximum sideforce coefficient is also defined as (a1 x Fz >+ a2). Perhaps the max coefficient of friction is just a1? Hi Ash, I've emailed you as well since you might otherwise miss this post from something you wrote a long time ago. :) I've started using Pacejka numbers from Genta's book, and it is working fine here. Now side slipping. I'm using the Ferrari numbers btw. The weird thing was the units; Fz in kN, slipAngle in degrees, and slipRatio in percentage (what on Earth was he thinking! ;-)). But anyway, the car works better than before, although it generates a lot of longitudinal forces, so I have the flipflop effect with the tires again a bit (the wheel spinning up too fast, then the slipratio->Fx force stops the wheel from rotating, then SR=0 and the wheels spins up again; all until you get a bit of speed, and even then I can hear the RPM beating as velGround and velFreeRollingWheel pass eachother. Not to worry too much; I'm going to add RPM inertia, which probably damps the RPM sweeps quite a bit). As Fz is in kN and a2 is something like 1668, I'd say the friction coeff. is 1.668 in this case. 'a1' must be the loading sensitivity. So a1*Fz+a2 in total makes the maximum available force for your friction circle. Well, that's what I use since yesterday, so testing should be done still. I still can't do forward driving donuts though. :( Was just wondering, I think you got it to work because of the percentage thing or something, but in any case, those are my findings upto now. Cheers, Ruud van Gaal, GPL Rank +53.25 Pencil art : http://www.marketgraph.nl/gallery/ Car simulation: http://www.marketgraph.nl/gallery/racer/