Being a simple (and elderly, since I got my first pensioner haircut this week
(a few years before I'm entitled I might add)) soul I'll struggle, but the
flatter a tyre is the more it squirms against the road as it rotates, which
takes up the 'excess' circumference, and in extreme cases the bead will become
detached from the rim and the tyre turn relative to the rim. As an extreme
example think of a tracked vehicle - the rolling circumference is governed by
the radius of the wheels plus the thickness of the track, not the
circumference of the track.
PaulH.
----- Original Message -----
In a message dated 6/3/2005 3:18:10 A.M. Mountain Daylight Time,
paul.hunt1@blueyonder.co.uk writes:
The *effective* circumference is governed by the radius at the point of
contact with the road. The *actual* circumference is immaterial.
PaulH.
Ohhh, my brain hurts. Can you 'splain this in simple Americanese (southern
if possible) so this elderly fat boy can understand?
It would seem to me that even if you drove the tire almost flat, effectively
lowering the radius on the lower half of the tire, you would still roll 100%
of the tread across the road. If the tread is X inches long (or centimeters
if you insist), you travel X inches, whether the tire is round, oval, or
elliptical, no?
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