rec.autos.simulators

Never. The pivot axis around which the front wheels are turned to steer a car is
determined by the caster angle (longitudinal plane) and the king pin angle
(lateral plane).

This is necessary to create forces that make the car stable and provide feed back
through the steering wheel...the car wants to go straight and the force required
to turn it is proportional to the steering angle.

Marty

Yes, non-zero slip angles occur in *** tires due to their deformation.
Steel drums have a much smaller range of deformation. :-)

This is where intuition departs for me. To demonstrate your point, say I
drag a steel drum behind me as I cruise down the road. The harness allows me
to "steer" the drum. The edges of the drum are rounded to prevent edge
effect, which would negate the conditions under discussion. Thus, only
sliding friction of the contact patch and the pinned connection at the axle
acts on the wheel.

The frictional force acts to directly oppose the direction of travel. This
force can be decomposed to components perpendicular and parallel to the
plane of wheel rotation. Intuitively, I expect the wheel will spin up to
road speed (unless this force is reduced to zero by steering it
perpendicular to the direction of travel). The component of force in the
plane of rotation is entirely dissipated by accelerating the wheel's
rotation about its axis. The perpendicular component is the resultant
lateral force. No other result is possible.

The difficulty with this explanation is what happens when the wheel is
already rotating at its steady state speed? The frictional force no longer
accelerates the wheel's rotation, and must be dissipated somewhere else.

So, refining the mental picture of what's happening: As the wheel
accelerates (spins faster), the apparent slide angle differs from the actual
steer angle to account for the motion of the face of the wheel across the
road. At steady state, each infinitesimal element on the face of the drum
has a component of velocity equal to the road speed. Thus, incredibly, the
drum is actually turning faster than it would if the steering angle was held
at zero.

My head hurts from trying to reconcile this, so I'll just offer a guess on
the nature of the new resultant force. The rotation speed is such that the
perpendicular component just cancels the lateral force arising from the
steering angle. Thus, the net resultant frictional force is still directly
behind the direction of travel, regardless of steering angle, meaning there
is no directional integrity.

I expect I'm more surprised than you at this result. :-)

I drive a late model Cobra all year round; its fat tires have almost zero
traction in snow. Forgive me for presuming; I believe I am better acquainted
with true slides versus simply extreme slip (which I also play with from
time to time). So long as you still have directional control, you are not
yet sliding. :-)

Michael.

...err, I think I meant to say longitudinal AXIS &  lateral AXIS.

Let us define the experiment a bit further. All measurements are done at
a constant velocity of the tyre so no translational acceleration accurs
(i.e. accelerating the center of gravity of the tyre), but accelerations
around the axis of the wheel are possible. All velocities considered
here will be measured with respect to the ground (this is important).

To be more exact, the frictional force acts to directly oppose the
direction of the slipping, and not the direction of travel of the center
of the wheel. If the tyre would not be rotating, the direction of the
travel would be the same as the direction of the slipping and the
frictional force would then be opposing the direcion of the travel of
the wheel (i.e. its velocity vector). As soon as the wheel rotation is
present, the direction of slipping does not match the direction of
travel anymore and things get a bit more involved. For example, when the
rotation of the wheel is faster than the corresponding road travel
velocity but the wheel still points in the direction of travel, the
friction force does not oppose the direction of travel but actually
points in the same direction (this happens in real cars when you have,
for example, wheel spin on acceleration -  bit let's forget about real
cars for now).

However, the forces do not get dissipated in the way, for example,
energy does. The component of the frictional force that is used to
accelerate the rotation of the wheel through the torque it produces on
the rotational axis when the wheel is pointing at a slip angle, is also
in effect when considering the overall force on the wheel. The fact that
it is used for rotational acceleration does not negate ('dissipate') its
effect when translational effects are in question.

So if you press a stationary yet freely rotating (i.e. no brakes on the
axis) wheel onto a moving road surface, the frictional force - at that
exact instant - points exactly in the opposite direction to the road
travel as the slipping is done in that direction, and there is NO
translational component at this moment. Still, the component of the
frictional force in the plane of rotation produces a torque which starts
accelerating the wheel, but this does not have any effect on the total
force felt by the wheel, i.e. the wheel stil feels that component of
force.

As soon as the wheel is rotating, the velocity vector of the point on
the wheel at its bottom (at the contact point) is not the same as the
one at the center of the wheel, but gets a component that is
proportional to the angular velocity of the wheel and is pointing in the
plane of rotation of the wheel.

To put it in mathematical terms, the components of the total velocity of
the point at the bottom in the plane of the wheel are (after a little
bit of geometry)

V_in_plane      = V_travel * Cos(alpha)-Omega * R (1)

V_perpendicular = V_travel * Sin(alpha)           (2)

where alpha is the slip angle, Omega the angular velocity and R the
radius of the wheel. As this is the velocity as measured with respect to
the ground, this are exactly the components of the slipping velocity,
because it is the point that is in contact with the ground.

The friction force points exactly opposite to this velocity but its
magnitude is always the same, namely the coefficient of friction times
the normal force. Now, as long as V_in_plane is not 0, there is a
component of the friction force opposing it which in turn produces the
torque accelerates the rotation of the wheel (changes the Omega) until
V_in_plane=0. In this case,

Omega = (V_travel/R) * Cos(alpha).                (3)

This means that the slipping velocity in equilibrium always points
perpendicular to the plane of rotation (as its component in the plane of
rotation is zero), and from this it follows that the friction force is
also perpendicular to the plane of rotation as it points exactly the
opposite way.

Not true, see equation (3). The rotation is maximal when the wheel is
pointing in the direction of the travel, and gradually goes to zero when
alpha goes towards 90 degrees.

Due to the points presented above I cannot agree. The friction force in
equilibrium is, written in the components perpendicular and in the wheel
plane,

F_in_plane      = 0
F_perpendicular = F_total_friction,

which means that in another coordinate system,

F_in_travel               = -F_total_friction * Sin(alpha)
F_perpendicular_to_travel =  F_total_friction * Cos(alpha)

So when alpha goes to 0, only the component perpendicular to the
direction of travel exists, and the one opposing the travel direction in
fact tends to 0. This may sound very surprising, but is a subtle
consequence of the fact that the friction force is ideally independent
of the slipping velocity.

So I stand by my original point, that is that the free rolling and
purely frictioanally slipping wheel preserves most of the directional
integrity, only degraded by the ratio of the coefficient of friction and
the coefficient of stiction and the cosine of the slip angle (which is
close to 1 for small angles).

I apologize if the sign conventions are not obvious, but I would need a
sketch to present them better. But I am sure the general idea can be
extracted from the above equations.

I forgive you for presuming ;), if you alow me to presume that, although
the practical advantage is no doubt yours, the theoretical advantage is
on my side :-). I do have a rather good understaning of a slide versus
an extreme slip, although the boundary is fuzzy (and sometimes hairy, as
any enthusiastic driver should know ;).

Never did I say I have directional control in the snow slides, turning
the wheel doesn't do diddly-squat then. Control can be had by pulling on
the handbrake which locks the rear wheels and changes the direction of
the force on the wheel from perpendicular to the wheel towards the
direction opposite to travel, thus diminishing the lateral force and
sending the tail further out. Or, in a front wheel drive car like mine,
flooring it puts the direction of the friction force directly in the
forward direction of the orientation of the tyres, which diminishes the
lateral force on the fronts. Note that when in a slide, only the
direction of the friction force can be changed, and not much can be done
about its ...

read more »

Gregor, I am a structural engineer by training and early vocation, but now
write (non-engineering) software for a living. This background predisposes
me to initially think in terms of non-moving systems, so please bear with
me.

It's clear to me now that I had presumed the proof when I thought this
through. The free-body ignored a lateral force that might be present at the
point where the harness attaches to the tow vehicle. This being the whole
discussion, we would naturally come to quite different conclusions. :-)

Let's try this again.

[Clarifying: each element on the drum has a component of velocity equal to
the road speed ( plus a perpendicular component). ]

Ignoring everything else for the time being, an infinitesimal tire element
in contact with the road is acted on by an impulse caused by the sliding
friction. At this point, we only know the velocity of the road surface, and
the wheel's steer angle. For a wheel not already rotating, the condition I'm
most comfortable with, the relative velocity is precisely the same as the
road speed. The component of the frictional force in the rotation plane adds
to the tire's rotation. The component parallel to the wheel axis is the
disputed lateral force. This defines one boundary condition. It appears that
we're done at this point, as there clearly is a non-zero lateral force, and
your point has already been made.

The other boundary condition, steady state rotation, is not so clear in my
mind, and I have a need to bring it back to first principles. Simplistically
speaking, the force resulting from sliding friction does not depend on
speed. That is, we expect it to be the same at 10 ft/sec as at 100 ft/sec.
However, frictional force is zero when the velocity is zero. Also,
superposition applies. At steady state, wheel rotation is neither increasing
nor decreasing, implying that the force in this direction is zero. Thus,
steady state rotation occurs when the velocity component on the face of the
wheel, in the plane of rotation, matches the road speed. There is also a
velocity component parallel to the wheel axis. There are two points to be
made about lateral velocity component. First, the magnitude of the resultant
of these vectors is clearly larger than that of the road speed alone. In
fact, it should become infinite when the wheel is steered perpendicular to
the direction of travel, the direct opposite of what you state. Second, I'm
having trouble this morning mapping across reference frames. I suddenly
can't distinguish if this resultant lateral force directly opposes that in
the first case above, or if it is the same.

Gregor, since we're so far apart on this one point, I feel it pointless to
continue without coming to terms with the discrepancy. Help me, if you can,
find the flaw in the thinking.

If you ask me, that sounds suspiciously like directional control. :-) All
kidding aside on this one; the Cobra just goes in whatever direction it was
already going, no matter what I do with my hands or feet, even at very slow
speeds. It has so far always recovered in time, but I try to avoid those
situations now. The situations being having to change speed or direction of
travel on snow covered roads. The only saving grace is that it takes a lot
of cussing and begging just to get it going in those conditions.

Michael.

I thought I had an engineer on the other side of the net :). It brings
back all those engineer/scientist jokes, in which the engineer is
invaribly always better off. One wonders why.

In the case of the nonrolling wheel, my point was actually that no
lateral force exists. If you reread my previous post on this topic, the
fact that the force component of the rotation plane produces a torque on
the wheel axis thus starting to rotate the wheel does not mean this
force is not felt by the wheel when considering its translational (as
opposed to rotational) effects (i.e. the force felt by the harness). To
use your choice of words, the force is not 'dissipated' for rotation. As
the element at the bottom of the wheel that is in the contact with the
road slides in the direction of the travel, the friction force is in
parallel with the travel and no lateral force exists.

It does not matter whether the tyre is forced to a nonspinning condition
(i.e. locked) or is freely rolling but at the moment of contact its
rotation is zero, in both cases the (initial) friction force has no
lateral component.

Correct, the component of the friction force in the plane of rotation is
zero in the steady state.

This is incorrect. This is only true when the wheel is oriented in the
direction of travel. Once the wheel is turned at an angle to the
direction of travel, this then in fact contradicts the previous
statement of the zero component of force in the rotation plane.

Let me try to demonstrate this with the equations I posted earlier,

V_in_plane      = V_travel * Cos(alpha)-Omega * R , (1)

V_perpendicular = V_travel * Sin(alpha)   .         (2)

These equations show the velocity of the point at the bottom of the
wheel with respect to the ground (i.e. the slip velocity) in the
components of the wheel coordinate system (in and perpendicular to the
plane of the wheel). Both parameters V_travel and alpha are controlled
from the vehicle onto which the wheel is connected, the only free
variable is Omega, the angular velocity of the wheel with radius R.

I hope you agree that the above equations are always valid for the
bottom point of the wheel under the conditions we put it to, no matter
what the friction or any other force on the wheel, as they purely state
the wheel dynamics without any kinematic effects.

When you mention the velocity component on the face of the wheel, I
guess you refer to the velocity of the point at the bottom of the wheel
due to the rotation of the wheel, expressed in the system of the wheel
(i.e. Omega * R in the above equations).

The main point is, in order to get the overall zero velocity component
relative to the road in the plane of the wheel (for a nonslip condition
in that direction, and therefore no corresponding friction force
component), the wheel doesn't need to rotate so as to fully match the
road speed, but only the component of the road velocity in the plane of
the wheel (i.e. V_travel * Cos(alpha)). This is where our discrepancy
comes from, I believe.

These equations also show what you stated earlier, namely that each part
on the wheel has the component of velocity that is equal to the road
speed (the part proportional to the V_travel), plus a NONperpendicular
component proportional to the angular velocity of the wheel. The fact
that these contributions are not perpendicular (in fact, when the wheel
is parallel to the ground, they are also parallel for the point at the
bottom of the wheel) means that the absolute value of the total velocity
can actually be smaller than the road speed (is actualy 0 for alpha=0,
as is expected for a fully nonslipping tyre).

I will not admit it, but I am a bit envious about your Cobra. The fact
that you cannot do much about the slide is that it is a rear wheel drive
thingy. Have you ever seen any footage from any touring car championship
with front wheel drive cars? They can get into up to 90 degrees slides
(especially after bumps with other cars), and when in any RWD racing
series a spin is inevitable, these guys and gals just floor the gas and
beautifully recover. Looks almost impossible, but is quite obvious from
just simple physics.

Greetings,

Gregor

:-) It isn't so much what we know or don't know; it's our proximity to the
guy paying for services. I presume you meant "paid more" when you say better
off. I can honestly say an academic's stress level appears to be far lower
than an engineer's. What price sanity?

I believe V_in_plane is where we differ. The cos(alpha) term is not right;
it should be V_travel/cos(alpha). Steady state rolling behind the tow
vehicle with 0 alpha, V_in_plane is precisely V_travel. At alpha approaching
pi/2, V_in_plane approaches infinity to maintain a component of velocity
equal to V_travel.

Force F_in_plane diminishes with larger alpha, while V_in_plane increases.

Yes it is; I'm fairly confident that V_in_plane has to increase with alpha.

I was talking about something far more dull. In 2" of slushy snow at walking
speed, every change in speed or direction is a challenge. Brakes lock
instantly, the rears spin at the slightest provocation, and the car just
plows straight ahead when I turn the wheel. I've had trouble in flat parking
lots where the surface was just packed snow, not even really ice. The tires,
Firestone SZ50's, simply don't work in the snow. It might be a temperature
thing; the tread is plenty aggressive, and work ok when the road is simply
wet.

They are spectacles to behold. :-) I've been saying for three winters now
that I need a second car for winter. I truly miss thrashing the parking lots
in the early morning after a good snow.

Michael.

As a side question, the forces generated through the tire, how are they
transmtted to the vehicle. Is it through the roll center of the suspension?
And if so what if only one tire is on the ground, is the roll center
calculated as the intersection of the CG center of the car, or do the forces
go through the intersection point of the aarms (if it has a arms). If anyone
has any insight on this I would love to know.
Regards
Carol

At the point where only one tire is on the ground, the car is a unicycle. At
this and all other times, the loads are transferred through the spindle arm.
The spindle arm is in turn connected to the suspension, which moves in
accordance with its geometry and the applied loading. Roll center might be a
useful concept in calculating roll coupling and distribution, but is not a
fundamental concept in itself. Again, Milliken's and Gillespie's books are
good sources of information.

Michael.

        The "Zero Gee News" is running a 12 part series on "The Physics of
Racing" by Brian Beckman.  This month is the section on Tires, Check it
out.. http://bovineracing.com/zgnews.htm in the Editorial section.
Section one was on Weight Transfer and relates directly to the forces
upon each tire, so..

        The ZGN is a comic and serious paper for everything GPL and I
understand that Real Racing may be added when the real racing season
starts, but don't quote me.  And I'd like to see more articles wrote up
so if you have something worthy to be posted I'm sure the editors would
be more then happy to post it for you.

Mike Barlow