The answer to your question lies in the fact that the sum of the forces equal zero...........

1) the air pressure in the tire goes up slightly.

2) the tire deforms to a shape where the pressure force normal to the inside the tube balances out the external load.

Fine. My bad.

Stoli anyone?

How is that significant?

Are you talking about the pressure on the contact patch? How does that keep the rim supported above the ground? Where, how, and why is force applied to the rim?

Air pressure doesn't just push on the tire. It pushes on the rim as well. Shall I draw you a picture?

It's like a star. Only in reverse.

That was an interesting thread -- I would have predicted that inflating a tire would increase tension in the spokes, not decrease it. (!) Hmm...

Yes, a picture would help!

Be careful, a sum of the forces in a closed section confining a constant pressure is going to be zero.

I can't see why you or anyone would think that. Inflate a tube, it pushes against the tread, which stretches very, very little. In response, the tube now pushes against the wheel strip and therefore, against the wheel. Go figger.

p&r

Yep.

I had simplified the tire+wheel in my mind - imagine, say, a constant-thickness metal tube rolled into a ring shape (toroid). If you pressurized that shape, I think you'd end up with tension stresses throughout the part. But obviously, that model is incorrect -- now I think your explanation is right.

I'm certain that a spoke tension meter with an extremely accurate capacity would take two different measurements from a wheel with an inflated tire installed and a wheel without a mounted tire. Hell, I don't know, now where did the mud go?

The area of the contact patch is simply A = F/P, which balances out the reaction force from the ground.

The rim is supported by the pressure integrated over the rim area plus the deformed section of tire which supports the bead.

There is a slight volume change in the tire due to the deformation of the tire near the contact patch which will increase the tire pressure, though it is pretty much negligible.

Isn't the pressure in the tire the same all the way around -- so the force on the rim from tire air pressure would equal zero?

I'm getting myself confused now! Or maybe just realizing my confusion.

both incorrect.

In a clincher tire, the spoke tension does not change a whit with tire pressure. As the tire pressure is increased, the increased force on the rim is exactly opposed by the force on the bead holding the tire in place. Otherwise the tire would blow off the rim.

With a tubular tire, increasing the tire pressure decreases spoke tension because the tire casing is self supporting, so the pressure inside the tire is wholly born by the tire casing. As the pressure is increased, the casing stretches and the inside diameter of the tire wants to decrease but is opposed by the rim (thus compressing the wheel) while the outside diameter wants to increase but is kept in check by the membrane tension on the tire casing.

Yes, the force of pressure is equal all the way around. I'm not sure how you are coming to the conclusion though that the force on the rim from the tire pressure is equal to zero. The tire can be abstracted to be a perfect membrane (holds tension but cannot support a bending force), but the rim most definitely cannot. The rim opposes the air pressure through its mechanical structure.

EDIT: I think I see what you are saying. I think if you include the load and the reaction force from the ground in your analysis, you'll find that all thing sum out to zero net force. But instead of just air pressure loads, you have external loads as well. The tire deforms to support the external loads.

When I was building for Tim's Bikes in Everett, WA. I always had to jack the tires up from the minimal pressure the factory would put in them, usually around 20 psi. Most bikes being box bikes. When I took them to 60, the spokes would always do a little popping. 26" X 1.75/2.0" wheels/tires. When I moved out to GA. and started building for some of the big box stores in the Chattanooga area, the same thing happened. Always some spokes popping. I do think my statement was right. YMMV.

Volume nor pressure change at all. Any air that was in the section of tire that is pushed against the ground gets relocated through the rest of the tire. Pressure does not change, nor does volume of air, unless you have a leak.

Basically speaking, the tire inflates outward from the rim, so the tire does nothing to support the rim other than give a cushion to protect from bending. The tire is merely to add cushion and traction. If you didnt have a tire on the rim, it wouldnt bend just from your weight, but it would bend when you hit bumps because there is no spring back to change the direction of pressure against the rim.

Tires dont structurally support rims at all. In fact, the rim supports the tire and keeps it from raising off the to of the wheel when weight is rolling on them.

basically speaking, the spokes compress because the tire tube expands, and pushes on all surfaces that it could expand to, so spokes compress inward. But the rim would stay in tact the same without a tire.

incorrect. The volume changes slightly because of the contact patch. The tire becomes oval-ized at the contact point; ovals have less cross sectional area than circles of the same perimeter. It is pretty negligible, but that's not the same as saying it's not there.

on a clincher tire, incorrect. Yes, the air exerts force on the rim, but this is exactly balanced out by the opposing force on the rim at the bead. Otherwise the tire would fly off the rim.

on a tubular tire, then correct. The rim does not support pressure loads, so it does get compressed by the tire.

No answers yet? BR was close.

OK, here goes-
the SHAPE of the cross section of the tire changes where it contacts the road. Look down while you are riding and see how the tire bulges out at the bottom

Now think about how that shape changes- everywhere except the bottom the tire is somewhat of a circle. Most bicycle tires really are close to a circle actually. Air pressure pushes on the entire inside surface of the tire equally, and the rim. Where the sidewall contacts the rim, the sidewall is at a certain angle, so the air pressure on the sidewall becomes a tension force in the sidewall, that tension force pulls on the rim at that certain angle.

Now think about the shape where the tire bulges at the bottom. The sidewall has the same air pressure pushing on it because the air pressure is equal everywhere inside of the tire. That air pressure creates a tension in the sidewall, but the tension in the sidewall is now a little bit different because the radius of the sidewall is reduced due to the bulge, but maybe more importantly the angle where the sidewall meets the rim is different so there is more upwards component of the force acting on the rim at the bottom. This increase in component is what supports the rider's weight.

All I know is I have measured a difference in spoke tension after inflating a clincher. There is a reduction.

That doesnt matter..... If you have a heavier person sit on the same tire, the oval would increase in size, then the volume of air inside the tire would stay the same. The air that is compressed rushes to the rest and top of the tire, expanding the distance between the top of the tire and the top of the rim. Volume of air in no way changes.