Theo, et Listers,
I see the Laifman has given some good engineering formulas for applying to
the Rolling Radius problem. But now to answer your question, which I think
is a good one. How is it that one revolution of the wheel doesn't give a
distance that is always equal to the circumference, assuming no stretch or
compression of the tread ( a pretty good assumption for steel belts I
think) and no slip of the tread on the pavement (only good a slow speeds
and very little applied torque). But let's assume these "ideal" conditions.
And, just for exaggeration, let's let a whole lot of air out of the tire,
so there's a large discrepancy between rolling radius and
Circumference/2pi. So, we slowly roll the car forward exactly one
revolution (make a mark on both the tread and axle if you like). After one
revolution, how far has the wheel traveled; the length of the circumference
or 2pi times the rolling radius? The answer is the latter, but how can this
be reconciled without something slipping? Think about one of those crazy
bikes in a circus with the off-center axle. If you peddle at constant
angular velocity (constant rpm), the bike speeds up and slows down during
each revolution according to the instantaneous rolling radius. Now, in the
case of a soft tire, the rolling radius and forward velocity stay constant,
but the circumferential velocity changes depending on its particular radius
from the center of rotation. In effect, the tread motion accelerates during
the time it isn't in contact with the road so as to make it back to the
point of contact in the distance given by the rolling radius times 2pi.
Here is a classic case of where "the rubber meets the road" is all that counts.
I hope this helps answer you question Theo.
TTFN,
Bob
At 09:40 AM 3/26/99 -0700, Theo Smit wrote:
>I guess you did :). On the weather: we had record highs earlier this week: 70
>degrees (with an overnight low of about 30). This means either the soft top
>gets
>a lot of exercise, or else it's a chilly drive in in the morning. I don't like
>to do anything with the top if it's close to freezing, so I'll suffer windburn
>instead.
>
>Back on the odometer/speedo topic: Bob suggested that the effective tire
>circumference (i.e. the distance covered in one revolution of the wheel) is a
>little less than the tire radius x 2pi, presumably due to sidewall flex.
How do
>you reconcile this with the fact that you do present the whole tire
>footprint to
>the pavement? Does the tread surface shrink where it contacts the road? Or
does
>it scrub where it gets planted on the ground, and then again where it leaves
>the
>ground? It would be interesting to know how much of either effect takes place,
>and how much the speedo / odo readings are actually affected.
>
>Theo Smit
>tsmit@novatel.ca
>B382002705
Robert L. Palmer
Dept. of AMES, Univ. of Calif., San Diego
rpalmer@ames.ucsd.edu
rpalmer@cts.com
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