rec.autos.simulators

A slide is distinct from slip. Slip is OK and normal; slide is not good. A
clear definition might help. The tire is sliding when no part of its contact
patch is in static contact with the road surface. In a slip, lateral forces
deform the tire carcass to the side. Each point on the contact patch in
static contact is slightly offset from the next point to come down as the
tire rolls. Thus, the wheel track is offset from the plane of rotation by an
amount known as the slip angle. So long as some part of the contact patch is
holding the road, some part of the tire is acting as a tire and not as a
*** eraser.

A tire locked under braking is sliding just as surely as one overwhelmed by
lateral force, or spinning from power. What little directionality remains is
miniscule, and due more to the shape of the contact patch and weathervaning
than virtue of its being donut shaped and mounted on a wheel. The lateral
force available is very close to that in fore-aft direction, and is equal by
definition to the sliding coefficient of friction integrated over the
sliding contact area.

Michael.

  That's true of a ski on a person but can it be true of a ski on a
skidoo/  I'm not sure.

: A tire locked under braking is sliding just as surely as one overwhelmed by
: lateral force, or spinning from power. What little directionality remains is
: miniscule, and due more to the shape of the contact patch and weathervaning
: than virtue of its being donut shaped and mounted on a wheel. The lateral
: force available is very close to that in fore-aft direction, and is equal by
: definition to the sliding coefficient of friction integrated over the
: sliding contact area.

  Sure - I agree that the directionality of a sliding tyre has nothing
to do with its "tyre nature" and you would get a similar directionality
from (say) a huge pencil eraser mounted on our front axle...  But I'm
just interested in how miniscule that is - is it significant and do
sims model it?

Richard G. Clegg       Only the mind is waving
    Networks and Non-Linear Dynamics Group
      Dept. of Mathematics, Uni. of York
    www:  http://manor.york.ac.uk/top.html

: The contact patch is somewhat of a malformed ellipse, blunter in front than
: the rear; the major axis typically running across the width of the tire,

  That's kinda what my pencil eraser looks like... well it was the
nearest everyday object I could think of.

: depending mostly on the tire radius to width ratio, and also influenced by
: inflation pressure. Pressure distribution varies across the contact patch. I
: am not aware of any quantitative studies of how the scrub loads vary on a
: sliding contact patch. My guess, based on nothing more than conjecture, is
: less than 10% variance from strongest to weakest angle of presentation. I
: further expect that the small difference wouldn't be apparent at all from
: casual observation (as when reviewing a game replay). I would also guess
: that, in a proper sim, the sliding condition is not treated much differently
: from a more "normal" condition. Thus, it's entirely likely that a reasonably
: sophisticated physics model would use different values for lateral and
: longitudinal forces acting on a sliding contact patch. For the definitive
: answer, however, you'll have to ask the game authors if they do indeed
: account for this.

  Hmm..,  I was just wondering really how accurate the GPL "lock the
wheels and off you go" model was...  

: Aside from the obvious academic interest, why do you ask?

  I have a sick and perverse interest in the physics of racing cars.
:-) (True, you can't expect me to have anything other than an
academic interest - I'm a professional academic).

--
Richard G. Clegg       Only the mind is waving
    Networks and Non-Linear Dynamics Group
      Dept. of Mathematics, Uni. of York
    www:  http://manor.york.ac.uk/top.html

From casual observation, it's pretty close. :-) The transition from almost
sliding to fully sliding is gradual; this is reflected in the gradation of
skid marks. Without more rigorous verification, I'm pretty satisfied that
they have a good understanding of the forces acting at the contact patch.

Milliken & Milliken's "Race Car Vehicle Dynamics", SAE, ISBN 1-56091-526-9.
GPL is a wonderful workshop for these explorations.

Michael.

I assume that what you mean is that you have
trouble recovering from locking the wheels.
Once you are able to read the cues from GPL,
that's not a problem

--
Pat Dotson
IMPACT Motorsports
http://www.impactmotorsports.com/pd.html

-- Doug

                Milliken Research Associates Inc.

First of all, the truth is that a LOCKED tyre has no directional
integrity. A locked tyre is just a piece of *** in a fixed position
sliding along the ground, and it doesn't matter much in which direction
it slides, as the friciton force is always pointing in (approximately)
the oposite direction to the velocity of the tyre relative to the
ground.

Consider the example of a tyre that is freely rolling, but is pointing
(sliding) at an angle (say 'alpha'-the slip angle) relative to its
velocity with respect the ground. There can be no force in the direction
of the orientation of the tyre, as the tyre is freely rolling, there is
just a (friction, F) force  perpendicular to it (in the direction of the
tyre axis). The component of the force in the direction of the velocity
is then -|F|*Sin(alpha), and the perpendicular component is
|F|*Cos(alpha). If alpha is not too big but still large enough so that
the tyre is sliding (really a hard to define term, though), MOST of the
force is still in the lateral direction, so a sliding tyre (but not a
locked one) still has lots of directional integrity.

Gregor Veble

  Thanks - much appreciated.

: First of all, the truth is that a LOCKED tyre has no directional
: integrity. A locked tyre is just a piece of *** in a fixed position
: sliding along the ground, and it doesn't matter much in which direction
: it slides, as the friciton force is always pointing in (approximately)
: the oposite direction to the velocity of the tyre relative to the
: ground.

  That's one way of looking at it - but the other way of looking at it
is that it's a fixed "slab" of *** at an angle being driven across
the ground.  I'd have thought this would have some boat rudder like
turning effect - experiments with pushing my ***ised mouse-mat
across a desk (my favourite "locked tyre" contact patch experiment) seem
to show that it does try to turn around a drawing pin pushed through
its centre.

  Anyway, this is all getting waaaay off topic.  I had just come to
wonder about how accurate the statement "a locked tyre has no
directional integrity" was....  I guess the conclusion from this group
is "very accurate indeed"

--
Richard G. Clegg       Only the mind is waving
    Networks and Non-Linear Dynamics Group
      Dept. of Mathematics, Uni. of York
    www:  http://www.racesimcentral.net/

:   That's one way of looking at it - but the other way of looking at it
: is that it's a fixed "slab" of *** at an angle being driven across
: the ground.  I'd have thought this would have some boat rudder like
: turning effect - experiments with pushing my ***ised mouse-mat
: across a desk (my favourite "locked tyre" contact patch experiment) seem
: to show that it does try to turn around a drawing pin pushed through
: its centre.

I believe it turns because your pushpin is not perfectly centered on the
pad.  If it is in the center, then integrating the friction force on
either side of the pushpin shows that the forces are balanced and there is
no net turning moment trying to twist the pad.  If you move the pushipn
off center, then there is a greater friction force to one side of the pin
that the other (area x contact pressure x mu) and there is a turning
moment which realigns the pad until the area on either side of the pin
("sides" here defined by the direction of travel) is equal, and then it
stops turning.  Of course, you could probably set up some good
oscillations depending on how fast you can "drive" your eccentric
mousepad.

As for a tire, that's an interesting question.  I'm still puzzling over
it.  If an individual wheel is loaded through its hub, which is centered
w.r.t. the tire contact patch, then the orientation of the tire w.r.t. the
direction of travel shouldn't matter, as there will be no net turning
moment trying to twist the tire.  But not all wheels are centered.

Now my brain is puzzling about weight distribution in a car, and polar
moments of inertia.  I'll get back to you.  It's a nice challenge for
myself to see if I can dust out the cobwebs and get back to some old
textbook stuff.

Stephen

Now, if the total contact patch would be sliding uniformly all the time,
no lateral force is possible as the direction of the force on any little
part of the patch is exactly in the opposite direction to the velocity
vector. But, since *** doesn't exactly behave this way as it tends to
stick and then elastically slip with a high frequency which is percieved
as the tyre squeal, the time average of the lateral force when sticking
and then slipping can be nonzero and an overall lateral force can occur.

A nonzero torque (the turning of your mousepad) around a certain point
can occur even if no stick-slip scenario occurs, it really depends on
where you put that point and on the manner of how you apply the vertical
forces on the patch.

In the limit when the surface of the contact patch goes to zero,
however, all lateral forces and twisting torques cease so no directional
integrity exists anymore.

Of course, only a scientist can ask such a question, any engineer or
driver will tell you that when you lock, you plow ;). The fact remains
that this can only be a minor effect which can probably not be felt when
driving.

My assertion is that a sliding tire has very little directional integrity.
Any variation is due only to the assymetry of the contact patch. I realize
this is counter-intuitive, given the discussion above. Try the following
thought experiment.

The stated condition for a slide is that no part of the contact patch is in
static contact with the road surface. Driving on normal dry pavement, the
sliding coefficient of friction is such that a true slide is difficult to
achieve, let alone maintain. If we reduce the coefficient of friction,
however, we begin to see that a slide truly doesn't match our expectations.
It's easier to imagine how a polished steel drum, instead of sticky ***,
will slide freely once we apply a large enough side force to interrupt its
free rolling state. The same is true when driving on an oil slick, or on
ice. We've all witnessed the effects of oil on the race track, and probably
experienced the terror of driving on ice.

Let me know if these examples don't hold up to your scrutiny.

Michael.

I like your definiton of a sliding tire, namely that all (or at least
most) of the contact patch is sliding, so only the coefficient of
friction (and not stiction applies). As you stated this can be better
observed not with *** but a metal drum. My analysis is, however,
still valid for this case, in fact, it is easier to show its point here.

When a metal drum acquires a nonzero slip angle, the stiction
immediately ceases and the force on it becomes purely frictional, thus
the wheel develops a proper slide in accord with the definition above.
The frictional force always shows in the opposite direction to the
relative velocity between the rubbing surfaces.

The main point is that, with a free rolling drum in a sliding
equlibrium, the relative velocity of the contact point on the bottom of
the drum with respect to the ground is always in the direction of the
wheel axis, i.e. perpendicular to the plane of rotation.

This happens because, unless there is rotational acceleration involved
which must be 0 in equlibrium anyway, there can be no torque along the
wheel axis and consequently no friction force component in the wheel
plane of rotation. The friction force in equlibrium thus always points
along the wheel axis and so does the relative slip direction.

The magnitude of this force is just the coefficient of friction times
the normal force, and is the magnitude of the force |F| in my previous
post (see below). With a metal wheel, the slip angle alpha can be set to
almost 0 and still only friction is occuring. The only loss of lateral
force in this case is the difference between the coefficient of friction
and the coefficient of stiction. But the force is still purely lateral
at small angles, so directional integrity is indeed preserved.

What is lost is something more subtle. While the lateral forces are
still preserved, although a bit smaller than when the wheel is still
sticking, the steering looses any practical effect. While the car may
still be executing a turn due to the preservation of the lateral forces,
when the slide of the steered tyres occurs the input of the steering
wheel does not have any (linear) effect. So what is lost is the control,
but not (a great part of) the turning ability.

I know from first hand experience, I did a lot of wonderful long turning
drifts with my little Renault Clio in the snow this winter. Only when
you aplly the brakes to lock the wheels or the accelerator to spin them
does the directional integrity gradually cease relative to the amount of
the longitudinal slip of the corresponding tires.

I really enjoy these kinds of discussions, so please, anyone, be
critical about what is written here!

Greetings,

Gregor Veble

: :   That's one way of looking at it - but the other way of looking at it
: : is that it's a fixed "slab" of *** at an angle being driven across
: : the ground.  I'd have thought this would have some boat rudder like
: : turning effect - experiments with pushing my ***ised mouse-mat
: : across a desk (my favourite "locked tyre" contact patch experiment) seem
: : to show that it does try to turn around a drawing pin pushed through
: : its centre.

: I believe it turns because your pushpin is not perfectly centered on the
: pad.  If it is in the center, then integrating the friction force on
: either side of the pushpin shows that the forces are balanced and there is
: no net turning moment trying to twist the pad.  If you move the pushipn
: off center, then there is a greater friction force to one side of the pin
: that the other (area x contact pressure x mu) and there is a turning
: moment which realigns the pad until the area on either side of the pin
: ("sides" here defined by the direction of travel) is equal, and then it
: stops turning.

  Being a thorough experimenter I tried to compensate for that by
(without moving the pin) making my mousemat execute a left then a right
turn.  However, retrying this I think my results were not very
reproducible.  (I think there was a significant amount of seeing "what
I expected" in the experiment).

  Hmm... point taken about integrated frictional forces though.  I
must rethink.  Instinctively I had thought of a locked tyre across a
road being "like" a rudder on a boat going through water.  

: Of course, you could probably set up some good
: oscillations depending on how fast you can "drive" your eccentric
: mousepad.

  It's much better now

: As for a tire, that's an interesting question.  I'm still puzzling over
: it.  If an individual wheel is loaded through its hub, which is centered
: w.r.t. the tire contact patch, then the orientation of the tire w.r.t. the
: direction of travel shouldn't matter, as there will be no net turning
: moment trying to twist the tire.  But not all wheels are centered.

  I think in my earlier experiment with the mousemat, my drawing pin was
slightly to the rear of the "tyre".  This would lead to a "tyre" which
when turned and locked would have greater friction on the side it
turned to.  

  (Actually, if you designed a race car wheel like that, you might get a
wheel which would turn the car while locked up - a positive boon for
***poor drivers who like to go into a corner with the front wheels locked).

  Are real wheels not pivotted more or less about the centre of the
contact patch then?  I'd always assumed that they were - wouldn't
turning be a bit peculiar if they weren't.

  (grin) and, of course, if we have a force trying to twist the tyre and
we hold the tyre steady (using the wheel) then we will have a force
trying to twist the car (which is what I thought might happen).

: Now my brain is puzzling about weight distribution in a car, and polar
: moments of inertia.  I'll get back to you.  It's a nice challenge for
: myself to see if I can dust out the cobwebs and get back to some old
: textbook stuff.

  Heh... well, I'm glad to see that at least I've interested someone
else in the problem....  let me know your conclusions.

  Now, of course, having thought about the problem deeply, I've no
idea why a rolling tyre produces a turning effect (or rather don't
understand it in the depth I'd like to).

--
Richard G. Clegg       Only the mind is waving
    Networks and Non-Linear Dynamics Group
      Dept. of Mathematics, Uni. of York
    www:  http://www.racesimcentral.net/

[snip]

My interpretation is that what make the car steer is really not changing
the direction of the forces generated by the front wheels, but changing
the magnitude of the force component lateral in car coordinates
(i.e. the amount of sideways force).

This means that when you are not sliding and increase the steering input
the total increase in force will be large enough to increase the lateral
force component even though turning the wheels is rotating the total
force vector backwards. But when you are near the maximum force, increasing
steering input will only rotate the force vector backwards leaving the
lateral component constant or even smaller.

      _
Mats Lofkvist

On the contrary, turning would be quite a bit peculiar (i.e. unstable)
if caster angle didn't put the pivot point ahead of the contact patch :-)

      _
Mats Lofkvist