rec.autos.simulators

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.racesimcentral.net/#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

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...

  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

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

|   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

  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.  

  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.  

  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 thir*** 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.  

  I know the feeling!

Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://www.racesimcentral.net/

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/

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.

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
***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

                Milliken Research Associates Inc.

Great to see you around here,

-Gregor

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

|

| >   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
|***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

|                 Milliken Research Associates Inc.
|
|

...

With no camber or suspension angles? That seems strange. Perhaps some
toe in/out?

...

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).

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/

| On Wed, 17 Jan 2001 12:40:15 -0000, "Ashley McConnell"
|
| ...
| >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?

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 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).

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 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.

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

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

                Milliken Research Associates Inc.

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/

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

| On Fri, 12 Jan 2001 16:08:16 -0000, "Ashley McConnell"
|
| >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/

...

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/