rec.autos.simulators

  When looking at slip ratio vs. longitudinal tire force data, is the
longitudinal force shown always in the direction opposite the true velocity of
the tire, or opposite the tire's direction?

  For instance, if a tire moving north (0 degrees), but pointed to the right at
10 degrees (10 degree slip angle), has the brake applied enough to allow almost
no sideforce, will the direction of the braking force (shown in measured data)
be at 180 degrees, or 190 degrees?  If it's at 190 degrees, the front of the
car should turn left when the wheel is turned right under near-limit braking
conditions.  This surely can't be right, can it?  If not, then isn't all
combined slip angle and slip ratio vs. lateral and longitudinal force tire data
showing the longitudinal force to be opposite the direction of tire travel,
rather than having anything to do with the tire's orientation?  (Did that make
sense?)

  If a tire is locked, wouldn't the force be applied more or less in the
opposite direction the tire is moving, regardless of the direction it's
pointing?

  I need to make sure my braking forces are going the right direction here!
Obviously, tractive force works in the direction the tire is pointing, is this
how tire data is displayed?

  Thanks in advance for any clarification on this.

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

Hi !

Well, the "longitudinal" force is always in the direction of the wheel, BUT you
have always "lateral" forces if the real velocity's direction si not the same as
the
wheel direction ( you have a slip angle ).

Moreover, in high slip ration ( locked tire ) you have a *** friction envolve :
something like  friction = k * velocity + Cst,  in opposite direction of the
velocity.
This friction force is the dynamic friction + static friction.

And last don't forget that lateral force is dependant of slip angle AND slip
ration,
and that the same for longitudinal force.

In lot of books you'll read that if you make a vector of slip_ratio and slip_angle
where slip_ration is X component and slip_angle is Y component, then this vector
cover an ellipse.

regards,

Seb

"J. Todd Wasson" a crit :

--
Seb
Game Developer
GPLRank -37.10
http://www.racesimcentral.net/

you cannot really separate longitudinal and lateral forces, you have to
look at the combined effect. When a tire with a slip angle of 10 degrees
from your example is under a braking force, it also develops a slip
ratio. It's best to combine the two and think instead of the slip vector
where one component is the slip ratio and the other is the slip angle.
In a slightly idealized situation, the force is dependent only on the
magnitude of this slip vector and points in the (opposite) direction. Of
course, in reality things aren't quite as simple, but it's a good
starting point, at least in terms of how to imagine it.

The lateral and longitudinal directions are defined with respect to the
wheel orientation, probably to ensure that a free rolling tire would
have (approximately) only a lateral component of force present when put
at a slip angle with respect to the road, as the force in this example
points approximately in the direction of the wheel axis (no camber
situation).

-Gregor

The force required to slide a tire is called the adhesive limit of the tire,
or sometimes the "stiction", which is a slang combination of ``stick'' and
``friction.'' This law, can be expressed in a mathematical form,  where the
force with which the tire resists sliding;  is the coefficient of static
friction or coefficient of adhesion; and  is the weight or vertical load on
the tire contact patch. Both  have the units of force (remember that weight
is the force of gravity), so  is just a number, a proportionality constant.
Thus,  the maximum sideways force the tire can withstand and is equal to the
stiction. We often like to speak of the sideways acceleration the car can
achieve.

Actually, the transition from sticking `mode' to sliding mode should not be
very abrupt in a well-designed tire. When one speaks of a ``forgiving''
tire, one means a tire that breaks away slowly as it gets more and more
force or less and less weight, giving the driver time to correct. Old, hard
tires are, generally speaking, less forgiving than new, soft tires.
Low-profile tires are less forgiving than high-profile tires. Slicks are
less forgiving than DOT tires. But these are very broad generalities and
tires must be judged individually. Some tires are so unforgiving that they
break away virtually without warning, leading to driver dramatics usually
resulting in a spin.

On the other hand, you might be over-complicating things too - the best way
to understand a "picture" of the braking forces is provided courtesy of "for
every action there is an equal AND opposite reaction,"  combine that with
another physics law "a body in motion will always seek the path of least
resistance" - (hence the "opposite" you refer to in the slip angle).

While I think I might have added confusion, I hope this helped...

Michael
FastCo86

www.MichaelLathrop.com

  Yes, I know this :0)  However, if I have a set of slip angle and slip ratio
data to extrapolate approximate force values from, there is a point where a
large negative slip ratio will more or less disallow any side force from being
developed by slip angle (friction circle/ellipse).  Instead of looking at the
combined effect, I'm looking at the total force as though it's broken into
longitudinal and lateral components, as this is how published data appears to
be represented, and the fitting formulas seem to reflect this as well.  If the
slip ratio is constant, then a slip angle will cause a certain lateral force
that could be looked up from tire data.  There will also be a longitudinal
force because of the slip ratio as you pointed out.  

  My question is, in what direction is this longitudinal force applied?  At 180
degrees to the tire heading, or at 180 degrees to the tire's velocity vector?
If it's at 180 degrees to the tire heading, then locking the tire or causing
the tire to run at a large negative slip ratio would cause a car to steer the
wrong way under near limit-braking, because the lateral force from slip angle
could be very small compared to the lateral component of the longitudinal force
from slip ratio (depending on steer angle).  This doesn't seem right to me,
hence the question.  

  What is the SAE standard for this?  It IS broken into two components, from
what I've read so far.  My confusion is in the area concerning the direction of
the longitudinal component of the "slip vector."

  This won't do me any good in my simulation work :0)  If I'm calculating
lateral force from a Pacejka Magic Tire Model or something similar, I need to
know the direction (90 degrees to the tire's heading).  Longitudinal force
would be calculated through a different equation with different coefficients,
but it would be a seperate calculation.  

  For instance, I may have two functions that look like this:
  LateralForce = GetLateralForce (SlipAngle,SlipRatio,InclinationAngle,Load)
  LongitudinalForce =
GetLongitudinalForce(SlipAngle,SlipRatio,InclinationAngle,Load)

  See what I mean?  Pacejka's formulas appear to work this way.  Once I've got
the LateralForce, I can apply it at 90 degrees to the tire's forward direction.
 But what about the LongitudinalForce?  Is this at 180 degrees to the tire's
direction, or at 180 degrees to the velocity vector?  Of course, I could add
these two together to get the "slip vector" you described, but first I'd need
to know the direction of the longitudinal component.

  I'm having doubts about this.  If I crank the front wheels to a 50 degree
angle to the right and lock the brakes, or apply "just enough" braking to cause
"pure" longitudinal force (friction circle, there's no or very little force
available now for lateral force), the car won't turn to the right or go
straight, it'll turn to the left instead.  This is why it doesn't seem right to
me.  If I lock the front tires, won't the car go straight ahead, regardless of
steer angle?  It's producing a big longitudinal force, but it seems this acts
in a direction opposite the tire's velocity vector, not the tire's heading
vector.  What's the SAE standard here?  How do they do things?

  Thanks for your time :-)

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

  So what direction does the longitudinal force get applied, assuming the rest
of what you mentioned is already done?  Is it opposite the tire's heading
vector, or opposite the tire's velocity vector?

  Thanks for your help :-)
Todd Wasson
---
Performance Simulations
Drag Racing and Top Speed Prediction
Software
http://PerformanceSimulations.Com

  What if the braking torque is enough to cause the slip ratio to be such that
there is very little or no lateral force (friction circle/ellipse)?  Turning
the front wheels right 80 degrees would cause the car to turn left if the
longitudinal force is applied along the wheel direction.  What am I missing
here?>Moreover, in high slip ration ( locked tire ) you have a *** friction

  This sounds like a Coulomb friction type model.  I wouldn't touch that with a
ten foot pole in regard to tire modelling, personally.  Seems much better to
use slip angle and slip ratio data.  

  Done :-)  Now I just need to make sure the longitudinal force is being
applied in the correct direction.

  Thanks for your help, Seb

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

  Make it stop!  My head hurts trying to follow this...
dave henrie

Having said that, like you I have also noticed that some combined slip
"data expansion" methods don't very closely meet this limit condition.
I suspect that locked cases are expensive to measure (because they would
eat up the tires and because of the large heat input and consequent
temperature variation during the test). Locked cases with slip angles
are then presumably even more rare.

I'm more willing to believe in a limited amount of SAE tire axes lateral
force component in wheel spinning cases, because my own game models
tend to spin the car in yaw "too much" without it. However, no one
has yet stepped forward to let me burn doughnuts with their
high performance car, so for me this is just a feel thing.... ;)

- Matt

Well, its more complicated than that :o)

I really recommand you Race Car Vehicle Dynamic page 58, and you'll understand :o)

If we take the extrem :

let the front wheel be right 90 degrees. and braking .. what hapening >

slip angle is -90 degrees so you have a small lateral force that is in direction of
the rear of the car.
slip angle is maximum, slip ratio is NULL, no longitudinal forces.

and it's a realistic behavior, that's why if you apply too much steering the car
turn less.

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

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

Longitudinal force is ALWAYS applied in wheel direction ! you'll see that in all
the books talking about car dynamics :o)

--
Seb
Game Developer
GPLRank -37.10
http://www.racesimcentral.net/

SAE longitudinal is always aligned with the wheel plane (and lateral is
always perpendicular).  So if you could turn the front wheels suddenly to
90 degrees, then the tire lateral force would all be slowing (braking) the
motion of the car (wheels free rolling or locked).  You have to learn to
think in multiple different coordinate systems....

-- Doug

                Milliken Research Associates Inc.

  Got it.  I think I see my problem.  What I was doing previously was
calculating longitudinal force from slip ratio, then calculating lateral force
from slip angle.  Then, if the resultant force was too large, I cut the lateral
force to stay inside the friction circle.  From looking at the graph, if I had
a slip ratio of .048, as slip angle is increased, the longitudinal force
decreases as lateral force increases.  My problem is that these are not working
together properly.

  That's what happens when I use simple approximations instead of a good tire
model :0)  

  With this information, it's clear now that as slip angle increases, even if
the slip ratio remains constant, the longitudinal force will decrease, and the
car should go the way it's supposed to.  So the answer to my question is: The
longitudinal force indicated by all the graphs and formulas (as you and a
couple others mentioned) is measured in the tire coordinate system.  Just
because the slip ratio is constant, *does not* make the longitudinal force
constant (as it does now in my tire model),  but rather, the longitudinal force
will vary as slip angle varies.  

  Thanks much :0)

    >More over your braking force only exsit when your vehicle is moving, if
speed

  Right, I got the friction reversal problem in the brakes fixed.  It wasn't
too hard after all :0)  The car works fine when running in reverse.  It'll do
Rockford's/backwards 180's pretty easily.  Fun!

 >Well, remember that slip_ratio/slip_angle model is not right at high

  Ok, I'll keep this in mind.  BTW, if the tire's are locked, what direction is
the force?  Is this the only case where the force will act opposite the
velocity vector?  Matt Jessick pointed out that there probably isn't much data
for locked tires, as they'd heat up and get destroyed in a real hurry.  Right
now, if I lock the front tires, the car goes straight ahead no matter what I
do, so I'm not really worried about putting in a special case routine, but I'll
keep it in mind anyway :-)

  Yes, I thought so.  Thanks for confirming this.  My problem again was I
didn't realize that a constant slip ratio won't mean a constant longitudinal
force, but instead it will vary with slip angle as well.  

  The problem that led to these questions reared it's head when I attempted to
slide the car sideways with four locked tires.  The lateral force (because of
my make-shift , very simple tire model) became zero.  The slip ratio went to
-1, causing my algorithm to cancel the lateral force in order to keep within
the bounds of traction.  Bang, the car keeps going sideways.  Now, I see that a
-1 slip ratio and a 80-90 degree slip angle will result in real close to 0
longitudinal force, allowing the rest to go along the lateral direction, just
like the graph on page 58 shows.  

  Thanks again to everybody for your help!

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

  Yes, this is what I'm thinking too.  Right now when my front tires lock, I
can't turn the car anyway, but that's because my algo kills the lateral force
entirely long before the tire locks.  I think I know exactly what my problem
has been.

  >I'm more willing to believe in a limited amount of SAE tire axes lateral

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

  Thanks, Matt

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

"J. Todd Wasson" a crit :

Yes, the friction model's direction is opposite to velocity vector . The friction
is a force tha work against motion.

Well, when you lock all the tire the car must goes in the car velocity direction.
Because in locked tires case you have no more a car whith tire but a car with
*** block instead :o)

Your welcome !

--
Seb
Game Developer
GPLRank -37.10
http://www.racesimcentral.net/

-Gregor