Does tire patch area + tire air pressure = car weight?

My supervisor claims you can compute the weight of any vehicle by multiplying each tire’s patch area (the area the tire makes with the road surface) by its pressure (in psi), then adding up all the weights.

Is this true? Perhaps it is, but it just doesn’t seem right. Wouldn’t sidewall rigidity also have something to do with it?

Just wondering what the straight dope was…

Nonsense.

Example:

Car weighs 3500 pounds. Tyre pressure 25 psi. 4 tyres. Tread width 5 inches.

Contact area per tyre = 3500/25/4 = 35 square inches.

Length of the contact patch is therefore 7 inches. No way.

I recant my last post. Still thinking about it.

Howstuffworks:

It just sounded silly to me at first.

And note the comments in the above cite about sidewall rigidity and run flat tyres.

There has to be a balance of forces at work here, and yes, in general, you can say that the contact area times the pressure gives you the weight.

However, don’t sell short sidewall stiffness, even in ordinary radial tires. There is some level of force that goes down through the sidewall. How much this is depends on several things, but mostly, the profile of the tire and wheel width (well, to a small extent). It can be very small on, say, a 70 series tire, but larger on a 45 series tire.

Yeah, anyone who has sat in a tire swing knows that it can hold up some weight. Perhaps even a noticeable fraction of the car weight?

Hmmm, I’m not sure that sidewall stiffness changes the picture. After all the legs of a table are pretty stiff but the pressure times the area of the feet still equals the weight of the table. However, I’m willing to be convinced otherwise.

The only way the sidewall can exert a suspending force is is via the patch of tire in contact with the road.

Doesn’t the psi relate to the temperature at all?
Ex: What if you had been power braking where the rear tires were speeding and heating up as the front ones stayed still and constant temp?

Or you were driving in Canada vs. Death Valley.

Then moe I think about it, if that is the case, I guess the patch would alter as well right?

Well I would think it would have to. I once saw a tire advertisement on TV for “run flat” tires, and a tire appeared to support a vehicle despite the fact a hole was drilled in the tire’s sidewall. The air pressure in the tire would be 0 psig in this case…

I suppose it all depends upon how the tire deforms. The “run flat” tires would still have a “footprint.” It would just be bigger than if there were air. As I said, I’m not at all sure about this.

I think maybe it depends upon the angle at which the sidewall enters the footprint. Most radial sidewalls that I look at enter the footprint on the road pretty much parallel to the road. In that case I don’t see how they can transmit an up and down force. If the sidewall isn’t parallel to the road then it looks to me like it would transmit vertcal forces and help carry the weight. In which case the footprint area*tire pressure = weight wouldn’t be true.

I’ll go with that until a tire designer (or someone else) comes along and blows it out of the water.

Think of a table supported by legs which are hollow metal tubes. The air inside the legs is at 0 PSIG and the “walls” are suporting all the weight. Note the OP asks about the relation between internal air tire pressure and vehicle weight, not about the pressure exerted by the tire on the pavement.

The weight supported by the sidewall would be reflected by the internal forces/pressure in the rubber, which is not related to the tire PSI.

As far as I can see, the car’s weight is equal to the contact area multiplied by the contact pressure acting between the outer surface of the tyre and the ground. This is probably similar to the pressure acting between the air in the tyre and the inner surface of the tyre (i.e. the ‘tyre pressure’). However, for a stiffer and thicker tyre, these pressures might be slightly less similar, as the pressure acting on the ground includes both the pressure acting in the air in the tyre, and the load transmitted through the tyre walls. Equally, the tyre walls could also constrain the action of the tyre pressure and prevent it acting on the ground below.

As a more extreme example, consider a cylinder of compressed gas. Internal pressure * contact area could equal several tonnes for some (they’re heavy, but can be lifted usually). When empty (or even evacuated), however, they still have some weight.

Actually, the more I think about it, the more tired I become. But there’s probably a simple explanation.