At 09:40 AM 3/26/99 -0700, you wrote:
>...
>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
Theo, the circumfarance of my well worn 235-50-13 is
68.5". The (slow) rolling distance is 66.75" at 30 psi. This is a
considerable change.
I assume that the tread patch works just
like a tank tread. The tread bends relative to the tangent of the tire when
the rubber hits the road. For an instance the rubber is stopped relative to
the road and then it raises off the road. The lower side walls bulge out to
accomodate the tread bending. At high speed the effective rolling distance
may increase a little if the centrificual force causes the tire to expand.
The above assumes that one is not applying so much power that the
tread slips on the ground. I can make the effective rolling distance
equal almost nill if I really get on it. Hard to see at that point for
the smoke.
James Barrett Tiger II 351C and others
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