On Sun, 21 Aug 2005 02:10:50 -0400, 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circumference of a circle (C = pi *2 r or C = pi * D) is no longer
valid.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.

If the radius is less at the flat spot, then it must be greater
elsewhere. No length of tread is lost, it just isn't all at the same
distance from the axel.

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.

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.

This is true if you use the r that the tire had when it was circular
in shape.

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.

These formulae were derived for circular geometry. The tire covers
once circumference per rev regardless of its shape, so work output =
work input (minus losses that go to heat the tire). Torque is
exerted on all partsof the periphery, not just the part that touches
the road. The parts of the tire not touching the road still have
torque due to "pulling" the tread around with circumferential tension.

The total torque is the sum of the various moments (at various radii)
around the axel.

It is true that the *average* radius is always r, which is the
(constant) radius of the tire when it is circular in shape. If you
use that r in your assertion, then your assertion is correct. If r
varies with theta, then the average radius is
(1/(2*pi)) * integral ( r(theta) d theta) integrated over 2pi
radians. If r is constant, as in a circle, this comes out to r,
fancy that!