rec.autos.simulators

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 ***.

-= mike =-

  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

F = mA.  If the vehicle is slowing down, there is a net force acting on
it.

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

....

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

  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

- Matt

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 ***ed if I can figure it
out.

Colin

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

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).

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.

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/

Ruud van Gaal, GPL Rank +53.25
Pencil art    : http://www.marketgraph.nl/gallery/
Free car sim  : http://www.marketgraph.nl/gallery/racer/

Yes, only when the car is stationary :-)

Colin

  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.

  >The other thing that I am having trouble with is this :-

  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.

   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

...

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'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

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.

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 ***
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.racesimcentral.net/
Free car sim  : http://www.racesimcentral.net/