# How exactly does a tire hold up its internal metal wheel against gravity?

This question isn't about how the tire holds weight up against the ground based on its contact patch, I understand that. It's about what's happening inside the tire such that the wheel (the metal hub in the middle, which serves as the inner wall of the pressure chamber) doesn't fall down to rest against the ground (like it does when the tire is flat.)

(This is a rephrasing of a question I asked on Physics.SE which was closed as off-topic and more suitable for Engineering.)

The intuitive answer is "air pressure" but my confusion stems from the fact that, since pressure is roughly equal throughout the tire, the net pressure force on the wheel should cancel out. It feels the same pressure radially inwards all around its circumference, it's just being squeezed in place.

A commenter on the now-closed Physics questions mentioned that the air pressure pre-stresses the tire rubber, which I assume gives it some form of rigidity that allows it to resist the tendency of the heavy wheel to deform the rubber? I.e. the asymmetric force opposing gravity would be applied by the tire rubber pressed against the wheel, not the air. But I don't understand enough about pre-stressing to know if this is accurate/how it works, and it was just a comment, not a full answer. I'd like to know more about the forces at work.

Included is a diagram of my understanding of the forces involved in a tire vs. a simple airbag holding up a mass. Obviously something is missing in the tire diagram, because it doesn't balance.

Also included is a diagram of my current understanding of the pre-stress explanation. What I don't understand is why Fr=(Fr2-Fr1) is larger in the presence of air pressure. I understanding why Fr2>Fr1, the tire is more deformed at the bottom and is acting like a spring, I just don't get why the air pressure so drastically increases the force with which it resists deformation.

• @DKNguyen yeah, I understand some applications of pre-stressing - that cannon example makes sense to me, and pre-stressed concrete makes sense to me. In both those examples, the pre-stress force is directly opposed to the load force. But I don't get how the concept transfers to the geometry of a tire, where the weight force is downward but the pre-stress force is radially outward on the walls of a torus. Commented Sep 21, 2022 at 20:15
• What is Ft supposed to be?????? Commented Sep 21, 2022 at 20:16
• @DKNguyen Ft = Force of tension, i.e. inward force from the rubber holding itself together with tensile strength. Commented Sep 21, 2022 at 20:17
• To be honest, I don't think the pre-stress analogy helps that much here. Commented Sep 21, 2022 at 20:26
• @DKNguyen I've added a FBD of the tire walls in cross section, including which forces are confusing me. Commented Sep 21, 2022 at 20:29

After puzzling it out and discussing here, I think I've talked myself around to the answer. The key is that when the bottom of the tire is deformed, the tension forces created by the rubber holding itself together with tensile strength are reduced - the tire in that region is not being pulled apart, it's being compressed, and so the tensile force resisting being pulled apart disappears.

As a result, the tensile forces cease canceling the pressure forces near the bottom of the wheel. This creates a net pressure force upwards at the bottom of the wheel, with no corresponding net pressure force downwards at the top of the wheel, because the top of the tire is not deformed.

The diagram below shows the situation when a tire is at atmospheric pressure, vs pressurized. At atmospheric pressure, the tire creates only a normal force upwards into the wheel to resist deformation, but isn't strong enough. Fg>Fn and the wheel falls to the ground.

When the tire is pressurized, the tire also creates tension forces inwards, cancelling the pressure force in most places. But there are less tension forces near the bottom of the wheel, which is not experiencing tension, and Fp>Ft. The tire transfers this force to the wheel (I represented this as a large Fp arrow but you could equivalently just make Fn larger.) Fn+Fp = Fg, and the system is in equilibrium.

I believe this does count as a form of pre-stress, with the pressure "converting" tensile forces in one direction into normal forces in the other direction. But that's semantics on a term that I don't fully understand.

The tire has maximum volume inside when the hub is perfectly centered. This also means that in this configuration, the pressure inside the tire is lowest possible for a given amount of air (given the condition that this amount of air can still exceed atmospheric pressure in this configuration) which, not surprisingly, is also when there is no external forces on the tire other than atmospheric pressure.

The configuration where the hub is sitting on the ground is much lower which means that for the same mass of air inside, the pressure is much higher. Thus the total force this pressure can exert is increased.

If there is enough air in the tire such that there is enough pressure to produce a collective force enough to overcome the weight of the hub and the car connected to it, it will lift up the hub and car and move towards a configuration where there is more volume inside the tire, and more volume means reduced pressure.

The hub will continue to be raised towards center while the volume inside increases as a result until equilibrium is reached which is when the pressure inside produces a force equal to the weight of the car.

In other words, it's no difference than a balloon being spherical and maximum volume versus deforming it to be anything other than a sphere. Except that with a car wheel you're distracted by the hub forming part of the inner wall, and the arrangement lower and higher volume configurations are not as obvious. But nothing has changed from an airbag or balloon.