In article ,
Ned Simmons wrote:
In article ,
says...
On Fri, 19 Aug 2005 20:27:08 -0400, Ned Simmons
wrote:
It seems to me no matter what the shape the tire is forced into, the
distance traveled will be 2*pi*r per rev, where r is the distance from
the ground to the axle. Think about it from the standpoint of torque. If
the tire is driving the auto, the torque at the axle is clearly equal to
F/r, where F is the force required to move the car. If the car travelled
^^^^^F*r
more than 2*pi*r per rev you'd have the basis for a perpetual motion
machine.
r is not constant because the tire has a flat patch. C = pi *2 r
comes from integrating dC = r(theta) d(theta) thru 2 pi radians with r
constant. If r(theta) is not constant, then the formula for
circumference of a circle (C = pi *2 r or C = pi * D) is no longer
valid.
Even a properly inflated tire has a flat patch. An underinflated tire
just has a bigger flat patch. Circumference can remain unchanged, so
revs/rolled_distance also can remain unchanged.
Then where is the length of tread that compensates for the length that
is lost to the flat spot? It seems to me that it either must be in an
inward wave in the middle of the contact patch, or causing an outward
bulge just outside the contact patch, or possibly both.
An inward wave doesn't seem likely on a tire that's anywhere near to
normal inflation: that would mean the tire tread lifting off the road;
but the air pressure inside is pressing the tire tread down onto the
road. Even if the tire is only at 15 psi, with a contact patch of
several tens of square inches that's still a lot of force. For the tire
to lift off the road in the middle is plausible for a completely flat
tire, which is where someone who posted to this thread observed it
happening; but it's different if the tire is just low, not flat.
Note that we all seem to be accepting the fact that the tread length is
fixed. Do we really know this to be the case? I'm sure the belts are
pretty effective at limiting the length of the tread in tension, but how
do they really behave in compression? If the tread can compress slightly
as it rotates into contact with the road that would resolve the entire
controversy.
The nitpicker in me says that the tread will certainly compress, even if
the belt doesn't, since the tread is outside the belt and is normally of
a larger radius, yet as the two go over the contact patch, they are
flattened out to the same length, or rather, to be more precise, to the
same (infinite) radius. But this happens even with a tire at normal
pressure, and happens about the same amount, since the initial radius is
almost the same, and so is the final (infinite) radius.
To address your real question, the steel belt is prestressed by the
tire's air pressure. For it to compress much, it'd have to lose all that
prestress. Then it'd have to buckle -- which the individual wires in the
belt are restrained from doing by the rubber surrounding them and binding
them together. Either the rubber would have to disintegrate, or the
tread as a whole would have to buckle. It can't buckle downward into the
road, because the road is hard, so it would have to buckle up off the
road, which gets us back to the question of whether the tire lifts up off
the road in the middle of the contact patch; as mentioned above, I don't
think this is likely for the sorts of tire pressures which you'd like to
catch and report to the driver via an automated system of the type that
started this thread.
In any case, I just can't accept that the car travels anything other
than 2*pi*r per rev (as before, r is the distance between the axle and
the road), regardless of what the tread does. To carry the torque and
work argument further, consider that the horizontal reaction of the
driving tire on the road is equal in magnitude to the horizontal force
at the axle pushing the car forward, call it F. Work is equal to F * d,
d being the distance the car travels. Work is also equal to Torque *
angular displacement in radians, T * theta. The torque at the axle is F
* r. So...
F * d = T * theta = F * r * theta
But if d per rev is greater than 2*pi*r, the work moving the car forward
is greater than the work input to the system by the torque turning the
axle.
Much as I like conservation-of-energy type arguments, they are made
inapplicable here by the energy loss that heats up the tire. The actual
torque at the axle is going to be greater than the number yielded by your
above formula, because the torque at the axle must also provide the
energy that goes into heating the tire.
The basic situation seems to be this:
-- The tire can't slip on the rim; one revolution of the tire
always means one revolution of the rim.
-- At the middle of the tire contact patch, the tire wants to
move the wheel at the angular speed omega given by the formula
omega = v/r,
where r is the distance from the axle to the road, and v is
the car's speed.
-- At the very front of the tire contact patch, the tire wants to
move the wheel at approximately
omega = v/r0,
where r0 is the distance from the axle to the front of the tire
contact patch, which is about the normal radius of the wheel.
The two parts of the tire are thus fighting each other: each wants to
move the wheel at a different angular velocity. The resulting angular
velocity will be between the two figures, and there'll be scrubbing of
the tire against the road, with the front of the contact patch trying to
go faster over the ground, but slipping backwards, and the middle of the
contact patch trying to go slower over the ground, but slipping forwards.
(Some of the people in this thread have been arguing for omega=v/r;
others have been arguing for omega=v/r0; well, you're not the only ones;
the tire is arguing with itself too.)
Sharp eyes will have noticed that there's an "approximately" in my above
argument for omega=v/r0. That's largely because the tire tread is
hitting the ground at an angle; that's really a formula for the velocity
of a point on the tread just before it hits the ground, not just after.
The correction factor for this is roughly cos(theta), where theta is the
angle between vertical and the line between the axle and the front of the
contact point: really the formula should be more like
omega = v*cos(theta) / r0,
except even that isn't correct, since it's based on the approximation
that the tread makes a perfect circle interrupted only where it touches
the ground, where it's perfectly flat. Still, it's not too far off; and
for small theta, cos(theta) is approximately equal to one, yielding the
approximation-to-an-approximation formula omega=v/r0.
--
Norman Yarvin
http://yarchive.net