I recently saw an assertion that the pressure exerted on the road* by a tire is equal (roughly) to the tires air pressure. At first I dismissed it as insane ramblings, but the more I think about it the more sense it makes.
Some quick calculations show that this theory is at least in the ballpark. Suppose I have a truck that weighs 3000 lbs and the tires are at 35 PSI.
21.4 in^2 of surface contact between each tire and the road sounds plausible to me.
A friend and I got into a discussion about it and he thinks that the pressure exerted on the road would be greater than the air pressure of the tire. He thinks that the sidewalls of a tire transfer a significant amount of the vehicles weight to the road. That may be, I don’t know, what do you guys think?
*I’m defining pressure exerted on the road as the weight of the vehicle divided by the total tire surface that is in contact with the road at any given time.
Unless I’m misunderstanding you, I think you’re wrong. You’re suggesting that the pressure exerted on the road is independent of the weight of the vehicle. If that’s true, then a car running over your foot would feel the same as a tire (minus the car) being rolled over your foot. I’ve experienced the latter, and it didn’t really hurt that much. If I ever experience the former I’ll know for sure, but my common sense tells me it would hurt plenty.
You’re missing the fact that the contact area increases as the weight of the car increases. With just the tire sitting on the ground, the contact area is just a thin line. If you put the weight of the car on it, the tire would deform and the contact area becomes much wider.
The internal pressure goes up a bit too, which helps.
As for the sidewalls - I’ve never sat on an uninflated tire, but surely someone on here has? How much weight can an uninflated tire support? My WAG would be <10% of the car weight.
By the way I have had a car roll over my foot when I was little, while trying to ‘help’ my father park… It was only a little Honda Civic though (and they were much smaller back then in the 70s). Didn’t hurt that much - didn’t scream, just waited for the car to move forward. No injuries or lasting pain.
Geoff, I believe you are correct, although your truck must be pretty small if it only weighs 3000 pounds. 4 cylinder S-10?
The sidewalls are relatively thin and don’t really have a significant effect on the weight distribution (other than the shape of the tire). If you deflate your tire, the contact patch will become larger to compensate for the lower pressure until it’s so low that the rim hits the ground. Obviously a heavier vehicle with the same size tires will require a higher pressure to inflate properly.
The weight per tire contact area is a function of the tire pressure since contact area is a function of tire pressure and vehicle weight, but the total weight on the road is equal to the total weight of the vehicle.
A friend and I got into a discussion about it and he thinks that the pressure exerted on
the road would be greater than the air pressure of the tire. He thinks that the sidewalls
of a tire transfer a significant amount of the vehicles weight to the road. That may be, I
don’t know, what do you guys think?
If you reduce the amount of air pressure in the tire until the pressure is zero, you can get an idea how much weight the sidewalls actually bear, which is basically “none”.
You might also note that different cars have different weight distributions (ie. front engine v. rear engine), so there’s at least one other variable to consider…
One of my first real jobs was in the tire testing machine industry and I remember a little of it. We used to chuck (mount) a tire horizontally and inflate it to 35 PSI or so. Then we drove a “load wheel” into contact with the tire until it hit a certain load, say 500 pounds. The load wheel was also spun horizontally and looked like a tire but it was about 4 feet in diameter and about 1 foot thick. As the tire was spinning in contact with the load wheel, the load wheel would see a variation in the force exerted by the tire, say between 490 and 510 pounds. This was caused by differences in tread and sidewall thickness around the tire. By having grinding wheels actually grind off the tire a bit in certain spots, this variation was minimized, hence a smoother ride. I hope this addresses the OP but I’m not sure.
It seems we are looking at two separate questions here: Individual tire loading vs vehicle weight.
A vehicle can’t exert anything less than its’ gross weight on the ground (unless you’ve delveloped anti-grav? ).
The more tires you have, the less weight per contact patch, regardless of area. This is why you see semis and mixer trucks with override axles, but the total weight distribution on the road will always equal gross vehicle weight.
And, just in case anyone wasn’t aware, most (if not all) state highway codes specify both a maximum GVW (gross vehicle weight) and maximum load per axle, and drivers can be fined for being over either of these. The maximum load per axle times the number of axles in a typical semi is considerably larger than maximum GVW.
My high school physics teacher was once called as an expert witness in a case where the driver claimed that his rig was within the permissible GVW range, though due to the method of weighing used (one axle at a time on a portable scale; add up all the individual axle weights) it was logged as being overweight because the cargo (livestock) moved in such a way as to cause the axle being weighed at the time to always be carrying more than its share of the load.
We calculated it out in class as an exercise, and determined that it was theoretically possible but not very likely given that cattle are not point masses and don’t generally like to have other cattle stand on them.
Late in the game (26 years), but I recall a guy called Issac Newton who insisted that every action has an equal and opposite reaction. In general, the scientific community believes him.
Based on Mr. Newton’s law, yes, tire pressure will equal the pressure on the road. As an aside, my road bike at 110 psi will make ruts in a dirt track that happily accommodates my MTB (50 psi).
Of course pressure on the road is equal to the weight of the vehicle (divided by area of tires in contact with the road). If tire pressure is low, the tire will deform until the volume of the tire with the same amount off air results in the approriate pressure.
So, the obvious question (as mentioned) is how much of the weight of the vehicle i supported by the sidewall, and I will agree with scr4 that the fact sidewalls deform easilly indicates while they are good at containing pressure (tension) they suck at supporting weight (compression) and so deform fairly easily with the weight of a vehicle without support of tire pressure.
I suppose the siplest test is to take the valve stem out and put progressive loads on a tire to see what the sidewalls hold up before the rim is resting on the ground. But since flat tires are notably “flat” I would suggest sidewall strength is minimal. I’ve never tried it, but if you sit on the top (tread) of an unmounted tire will it collapse?
(I have a Tesla, 4100lb and a BMW, 3800lb, so a rough rule of thumb would be about 1,000lb per tire is the typical load from a medium-large sized car.)
Yes, the pressure of the tires on the road is the vehicle weight divided by the contact area. But this is not necessarily related to the air pressure of the air inside the tires. A fully charged scuba tank is pressurized to about 3000 psi. That doesn’t mean that a scuba tank sitting on the ground is exerting a force of 3000 psi on the ground. If it were, it would impossible for a human to pick up a scuba tank.
You cannot equate an inflexible scuba tank with a flexible tire because the crux of the matter is that the footprint of a tire changes directly in response to the weight it’s carrying.
To better understand why tires exert (on road surfaces) pretty much the pressure they are pumped to, blow up a balloon. Now place it on one of your palms (flat on a table). Imagine that it’s a spherical tire. Balance a half-full bottle of water on the balloon; note the footprint and, subjectively, the pressure on your palm. Now do the same with the bottle topped up. You should find that the pressure on your palm stays more or less the same but the footprint of the balloon has increased.
To a small extent. But a bike tire at any pressure, even fully deflated, won’t have the contact area of a car tire. The geometry of the tire is more important than the internal pressure.
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