400 pounds per half a square inch, 800 psi. And walking with a heel strike is generally considered as more than body weight impact. I’ve read that in running maybe twice body weight? Even walking has got to be 1.2 or more? So yeah hitting with something like a 1000 psi all over the new floors. And stilettos are often small than half square inch surface area.
Oh. As to the ancient OP - of course the weight of the car is being held up by way of the area of contact with ground, the pounds of the car held up spread across the number of square inches. Where else is that force going? Yes that is the pressure you’d measure against the ground and in the tire at points the tire is not touching the ground.
Beakman’s World has you all covered: 13:00 in
I think it could if the contact area, the footprint, is small enough and the weight of the tank is nonzero.
Yes, of course it’s possible to produce almost any desired pressure by decreasing the contact area to the appropriate degree. Anything balanced on some needle-like support will produce a huge pressure on the ground. But the question as I understand it is, is there a direct unavoidable relationship between the air pressure in a tire and the pressure which the tire exerts on the ground? I say there is not. You can’t, for instance, increase the ground pressure without limit simply by adding air to the tires.
No. Eventually the tire explodes.
But before that point? Add air and the tire pressures go up, the footprint of the tires decrease. The pressure per square inch increases. The total adding each tire’s footprint area times its pressure together will still equal the weight of the car. Take air out of the tires. The footprint increases. Same “unavoidable” math. Until the rim is carrying the load.
Are you saying that the tire’s footprint area is inversely proportional to the air pressure in the tire? That seems unlikely to me. I think, for example, that doubling the pressure would decrease the area by a modest amount, but not by a factor of 2.
I also think that the exact relationship between pressure and contact area would depend strongly on the construction of the tire, existence of steel belts, radial construction, rigidity of the sidewalls, etc. and would not be a simple linear relationship. As the stiffness of the tire increases, becoming more like a scuba tank and less like a balloon, the relationship between pressure and contact area has to change.
(Missed edit window)
My last sentence was worded poorly. I don’t mean “as the stiffness of [an individual tire] increases”; I mean that if you consider different tires with differing stiffnesses, their response to changing internal pressure must be different from each other.
Theoretically maybe? But in my experience most tire sidewalls do not have enough stiffness to support even the weight of the empty tire itself. Its contribution to supporting the weight of the tire can safely be assumed as inconsequential.
Weight and pressure are two different physical quantities. They are expressed in different units. Weight can be measured in pounds, while pressure can be expressed in pounds per square inch.
Most of the weight of a car is supported by the air in the tires. As stated above that amount is the air pressure multiplied by the contact area. A small amount of the weight is supported by the tire sidewalls.
In the case of scuba tanks discussed upthread, the distribution is different. In that case, most of the weight would be carried by the metal making up the tank.
My old Subaru, which weighed around 3400 lbs recommended 32 PSI for the tires.
My Chevy Equinox EV weighs 5100 lbs, and has tires at 42 PSI.
The tire size and contact patch are similar.
Tire pressure x area = weight is well known among racers. Tire sizes can be built to take advantage of it. A dragster tire footprint is long and narrow to allow better acceleration. A road racing tire patch is more square to allow high G forces in all directions.
In the days of bias ply tires the stiff sidewalls did indeed have more contact pressure along the edges of the tire patch. But radial tires are specifically designed to have flexible sidewalls and the car can lean a bit without lifting the outer edge of the tire off the ground.
There’s an obvious wear pattern to the tread of tires over- or under-inflated. Over-inflated tires, the tread bows outward, and the center wears more than the edges of the tread. under-inflated, the sidewalls bulge, and the “corner” constructed between the sidewall and the flat tread therefore causes the middle to raise up a bit. The edges of the tread wear more than the center. And, the lower the pressure, the longer the segment of tread in contact with the road, also increasing surface area.
So one issue is that the pressure is not necessarily uniform with a somewhat flexible container bearing a load. The shape and construction of the container affects to what extent the vessel will exert pressure in the area in contact and how it is distributed.
The difference with a scuba tank is it is specifically (for all purposes) not flexible, so internal pressure is irrelevant.
But a tire’s flexibility is variable, depending on its current pressure. If you start with a tire that is practically deflated, say at 5 psi, and double the pressure to 10 psi, the shape of the tire and the ground contact area will change a lot. It will get closer to cylindrical than it when it was at 5 psi. But if a tire is close to normal pressure, say 30 psi, and you double it to 60 psi, assuming it doesn’t explode, the shape of the tire will change very little. At 30 psi it’s already pretty much as cylindrical as it will ever get. At that point it behaves more like a scuba tank.
No. It doesn’t.
Assume you are correct that it is inflated to pressure that does not allow for any additional deformation and you place the car’s weight upon it. The footprint cannot change in this hypothetical. What then does? The pressure in the tire. It increases. Possibly past the level that blows out the tire.
I have read that the weight of the car is supported by the sidewalls.
The reasoning goes like this:
Consider a car supported with the axles on pistons. The car is supported by the air pressure in the piston, the piston is free to move.
Consider a car suspended by the axles on pistons suspend from above. The car is supported by the vacuum in the piston, the piston sidewalls are free to move up and down.
Consider a car suspended and supported by both pistons. Now connect the air from one piston to the other: the air from the supporting piston bleads into the suspending piston, the car goes down.
And that is what a car tyre is: the air above the axle is connected to the air below the axle, any air below the wheel supporting the wheel can just move up above the wheel.
You may also say: the air pressure pushing up on the wheel is the same as the air pressure pressing down on the wheel.
In which we see that either the pressure of the tyre is changing as the tyre goes around, or the stiffness of the tyre is different at different points.
Now consider @markn_1 's scuba tank. You could make a “tyre” out of a cylindrical tube. There would be a small amount of deformation due to the ‘give’ of steel (steel is a well characterised material with important springiness and deflection behavior). The tube could be pressurised, or hold a vacuum, the car would still be supported, it is supported on the sidewalls.
Is the car supported in compression on the lower sidewall, or in tension by the upper sidewall, or in bending by the transverse sidewall? We could consider an inner tube in chains, or by a special arrangement of loose pistons which can move up and down, but can’t be bent.
But the car goes down when the tyre gets flat! Yes, the supporting sidewall buckles when not supported by air pressure. Buckling is a well-understood mode of failure of thin-wall structures. This indicates that, at least in part, the car is supported by compression of the sidewall.
Of course the car is only supported by tyre wall compression!. Well, perhaps not. The tread could carry the load around to the top, with the tread in compression, not the tyre wall.
Does the air do anything more? I haven’t seen it discussed. The air pressure supports the outside tread in the same way that is supports the sidewalls: this allows the outside tread to help distribute load to the upper and transverse sidewalls. This might be important, or might not.
The footprint changes when the pressure changes! Yes, the pressure supports the shape of the outer shell. The outer shell transmits the road force to the sidewalls. You could make a tread of steel plate, and it wouldn’t deflect as much on tyre pressure, but it wouldn’t work as well with the road surface or car suspension.
It should be possible to determine empirically how much of the car’s weight is supported by the sidewalls, the remainder being supported by the air in the tyres. Take a wheel off a car, place it standing on the ground, puncture the tyre, and apply downward force on the centre of the wheel until the sidewall buckles.