Re: Archery test

Bad Idea said:

Carl, maybe you should ask Jobst to draw a free body diagram for you.
Oh, wait....nevermind.

Dear Bad Idea,

If Jobst becomes curious about the matter, I expect that he'll quite
sensibly support one horizontal spoke with a ceiling rope, hang a few
5-lb weights off the spoke above it, and find out what happens to the
spoke's tension by using a tool that he designed.

If he does, I predict that he'll see one spoke's tension not changing
appreciably, while the other spoke first loses tension and then gains
tension back to its original level by about 20 pounds of squeeze
force.

If my prediction is wrong and tension just rises for both spokes, then
I'm confident that Jobst will say so. He and I have our differences,
but this is pretty much something between anyone who cares to measure
the tension and the spokes themselves.

If my prediction is right, then Jobst may reason that the result is
odd, but misleading and try to explain why.

Or he might decide that the measurements make sense and try explain
how.

Or he might say that the results are odd, but that he doesn't know
what's going on.

That's the beauty of a simple, repeatable test. Other people can try
it, if they're interested, and then try to explain what it means.

Consider the familiar claim that the area of the contact patch will be
equal to the load on the tire divided by the air pressure. At 30 psi,
a 700 x 25 tire with a 100-lb load should have a contact patch area of
about 100/30 in^2, or about 3.3 square inches.

Conventional wisdom admits that the contact patch could be a little
larger than 3.3 square inches, with a central area at 30 psi of force
against the ground and wide edges with the force tapering off to 0.

Conventional wisdom also admits that the flimsy bicycle sidewall might
add a little support, but even a few pounds will flatten it, so it can
be ignored. But the contact patch certainly won't be significantly
smaller than 3.3 square inches--how could it?

But if you inflate a tire to 30 psi, hang weights until it presses
down on a scale with 100 lbs, ink the tread, and measure the contact
patch . . .

Darn. The dumb contact patch is much too small.

Multiply the length by the width as a generous rectangle, and it's
only about 2.5 square inches. Treat it as the ellipse that it
resembles, and it's only about 2.1 square inches.

How can only 60 to 75 pounds of air pressure (30 psi x 2.1~2.5 square
inches) support a 100 pound load?

The uninflated sidewalls aren't stiff enough to support the 25 to 40
pounds of unaccounted-for load, so something is wrong.

Most likely, the contact patch at low pressure consists of a central
30 psi area and a ring of high pressure around its edges.

The air pressure pushes the tread down with 30 psi everywhere that it
touches the ground.

The extra force needed to support the tire probably is probably
resistance added by the sidewall, where it meets the ground.

Look at those bulging sidewalls, deforming against the 30 psi of air
pressure that's trying to push them back into a more circular curve.

Anyone with a bicycle, tire pump, air gauge, bathroom scale, weights,
$2 ink pad, paper, and ruler can test the size of the contact patch at
low pressure.

Cheers,

Carl Fogel