"But the road cannot push up on what is not there."
Correct, but you don't have a measure of the pressure exerted by the
road on the tread. Rather, you have a measure of the pressure exerted
by the air on the inside of the tire, and the inside of the tire is
smooth.
"The area of the patch would change if the tread was missing and a slick
tire was used."
I don't think so. The contact area between the rubber and asphalt would
go up, and the pressure at the rubber/asphalt boundary would go down
accordingly. The air pressure wouldn't change.
"The correction is not small and is needed to get agreeable results."
Maybe so, but that doesn't make it right. It means something else in
your procedure or assumptions is not right.
Here is a more extreme example to demonstrate my point. Take a rubber
balloon, inflate it, and tie it off. Clamp or otherwise hold a regular
wood pencil vertically with the eraser end up. Hold the balloon over
and in contact with the eraser. Have someone lower a 1-pound box of
butter on top of the balloon so the butter rests on the top of the
balloon while the 1/4-inch eraser is pushed into the bottom of the
balloon. Your hands don't lift anything; they just keep everything
aligned.
The area of the 1/4-inch diameter eraser is about 0.049 sq in and it is
supporting one pound. Does that mean the pressure in the balloon is
1/0.049 = 20.4 psig? If so, your lungs are *considerably* stronger than
mine, and your balloon is *considerably* stronger than balloons I buy.
But how can the much smaller air pressure in the balloon push down on
area that is not there? There is only 0.049 sq in of balloon in contact
with the eraser. Isn't the eraser feeling a pressure of 20.4 psi?
(yes) Doesn't that imply 20.4 psig in the balloon? (no) Look at the
shape of the balloon over the eraser and figure it out.
Michael D. Edmiston, Ph.D.
Professor of Physics and Chemistry
Bluffton University
Bluffton, OH 45817
(419)-358-3270
edmiston@bluffton.edu
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