I am in the process of editing the picture I posted. But even before
that is finished, I will try to answer Carl Mungan's question about my
red arrows, because these are the key in my opinion.
The air pressure pushing outward on the sidewalls creates some amount of
outward bulge of the sidewall. The arrows indicating this in my drawing
are the horizontal blue arrows. Now, think of the sidewall like the
very common force-resolution problem found in physics textbooks in which
you tie one end of a rope to a stuck car, and tie the other end to a
tree, and then push the rope sideways in the middle. This allows you to
create strong tension in the rope that tries to pull the tree and car
together. Of course the tree does not move, so the car moves and gets
unstuck. In this problem, the more straight the rope is, the more force
you get on the car in the direction of the tree.
I see that traditional physics problem very much like the sidewall of
the tire. As the flexible sidewall is pushed outward, the outer
circumferential edge of the sidewall (connected to the tread) is pulled
radially inward. These are the red arrows at the top and bottom that
Carl suggested would be #1 and #4. In addition, the inner
circumferential edge (connected to the bead) is pulled radially outward.
These are the inner pair of red arrows that Carl suggested would be #2
and #3.
Since the bottom sidewall is more bulged, these red arrows (#3, #4)
representing the tension at the circumferential edges of the sidewall at
the bottom of the tire, are smaller than those at the top (#1, #2).
In short, as I mentioned in the text with the drawing, it is the
difference in the profile between the top half of the tire and the
bottom half of the tire that reduces the radially outward pull on the
bead (by the sidewall) on the bottom half compared to the top half.
Michael D. Edmiston, Ph.D.
Professor of Chemistry and Physics
Bluffton University
Bluffton, OH 45817
(419)-358-3270
edmiston@bluffton.edu