Does the weight of a vehicle affect tire pressure?

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Does the weight of a vehicle affect tire pressure?

Say you have 4 tires that read 25 pounds pounds per square inch while off of the car. (The tires have a maximum pressure of 50 psi, so there is room for more pressure.)

Once you put them on the car would the weight of the car increase the psi reading?

Likewise, if one end of the car is heavier than the other end (the front because of the engine weight, for instance), would those 2 tires read higher than the other 2?
 
Yup. Its all variable, but pressure is effected by volume, and a tire is not exempt to being deformed when additional force is applied to it. So if a tire were to become more oval then the volume would change and the pressure would increase.
 
OnlyTellsTruths said:
Does the weight of a vehicle affect tire pressure?
Click to expand...

Very little. While what CMacDady said is literally true, keep in mind that the actual volume change required to double the contact patch surface area is a pretty small fraction of the total tire volume. So we can double the contact patch area, and thus double the support force, with a very small relative change in pressure. To make a big difference in pressure, we need to make a bid change in the volume of the tire, but tires aren't intended to operate under such conditions. They're intended to stay fairly close to constant shape. If your tires deform enough to significantly change volume, then they're either very underinflated or very overloaded.

Say you have 4 tires that read 25 pounds pounds per square inch while off of the car. (The tires have a maximum pressure of 50 psi, so there is room for more pressure.)

Once you put them on the car would the weight of the car increase the psi reading?
Click to expand...

Not appreciably.

Likewise, if one end of the car is heavier than the other end (the front because of the engine weight, for instance), would those 2 tires read higher than the other 2?
Click to expand...

Not appreciably.

But it will make a difference in the wear each tire experiences.
 
OnlyTellsTruths said:
Does the weight of a vehicle affect tire pressure?

Say you have 4 tires that read 25 pounds pounds per square inch while off of the car. (The tires have a maximum pressure of 50 psi, so there is room for more pressure.)

Once you put them on the car would the weight of the car increase the psi reading?
Click to expand...

The equation you are looking for is the ideal gas law:

PV = nRT

P = Pressure
V = Volume of the container
n = The amount of gas
R = 8.314472
T = Temperature of the gas in the container

Solving for P you get:

P = (nRT)/V

In your scenario, you fill the tires and get a reading. You then put them on the car and read another reading. What has changed?

The amount of gas, R, and the temperature remain constant. (The temperature actually increases a little but that's not important right now.) However, the car slightly decreases the volume of the tire by deforming it. In terms of the equation, if V is smaller, P must be larger. Ergo, the pressure goes up.
 
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Temperature probably makes a bigger difference than load. Between winter and summer you might have a swing of 40 degrees Celsius, which would make roughly a 15% difference in pressure. And if the tire heats up as the car travels that could affect it too.
 
OnlyTellsTruths said:
While I understand you when you say 'not appreciably', I'm wondering if you know about what percentage an average car (idk, 1 ton perhaps?) would affect that 25 psi?
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I'd guess it's less than a 1% change, but that's just a guess. In practice it's going to depend a bit on the tire type and initial pressure.
 
OnlyTellsTruths said:
Once you put them on the car would the weight of the car increase the psi reading?
Click to expand...

What they said: you have to change the volume to change the pressure. Think about stepping on a balloon: it deforms and collapses under your weight, pressure builds up and the balloon bursts.

The same thing happens to tires on a car, but tires are really quite strong and won't deform much even under very low pressure. Some tires can even be driven on when "flat" because they are strong enough to not actually go flat when they lose air pressure.

So, when you put the full weight of the car on the tires they deform slightly and the pressure will change slightly, but not enough for your typical tire gauge to register.
 
Some calculations. Say you've got a 15" diameter, 6" wide, 35psi tire supporting 1/4 of a 2600 lb car. That requires, in principle, an 18 sq.in. contact patch, so 6" wide and 3" long. So, how much do you have to move in from the edge of a 15"d circle to find a 3" chord? Only about 0.15". The area outside this chord is about 0.17% of the area of the disk. Likewise, the volume "displaced" in crushing a 15" tire from round to a round-with-a-3"-contact-patch, is 0.17% the initial volume, so via PV = nRT, for a small perturbation, it'll increase the pressure by about 0.17% or 0.06 psi.
 
ben m said:
Some calculations. Say you've got a 15" diameter, 6" wide, 35psi tire supporting 1/4 of a 2600 lb car. That requires, in principle, an 18 sq.in. contact patch, so 6" wide and 3" long. So, how much do you have to move in from the edge of a 15"d circle to find a 3" chord? Only about 0.15". The area outside this chord is about 0.17% of the area of the disk. Likewise, the volume "displaced" in crushing a 15" tire from round to a round-with-a-3"-contact-patch, is 0.17% the initial volume, so via PV = nRT, for a small perturbation, it'll increase the pressure by about 0.17% or 0.06 psi.
Click to expand...

I would have guessed larger than that, but I knew it would be small. And even this is may be an overestimate, since tires generally bulge to the side when they flatten, which would make the volume change less. But even a ballpark figure is enough to show drivers don't need to worry about it.
 
On the other hand, don't confuse tire pressure with load pressure, which is effected extremely by the weight of the car.

Also, offroad tires, and many dual purpose street/at tires are meant to deform extremely under less than recommended pressure
 
ben m said:
Some calculations. Say you've got a 15" diameter, 6" wide, 35psi tire supporting 1/4 of a 2600 lb car. That requires, in principle, an 18 sq.in. contact patch, so 6" wide and 3" long. So, how much do you have to move in from the edge of a 15"d circle to find a 3" chord? Only about 0.15". The area outside this chord is about 0.17% of the area of the disk. Likewise, the volume "displaced" in crushing a 15" tire from round to a round-with-a-3"-contact-patch, is 0.17% the initial volume, so via PV = nRT, for a small perturbation, it'll increase the pressure by about 0.17% or 0.06 psi.
Click to expand...

I was going to say less than 1%, but now I feel totally humbled.

I used to be able to do **** like this when I was an engineer, but then I went to law school and that part of my brain is atrophied.
 
Easy enough to find out.
Jack one wheel off the ground.
Measure the pressure.
Lower the jack.
Measure again.

Best if you don't remove the gauge in the process, as that might let some air out.

Old desert driving trick is to run slick tyres and let the pressure down so low they deform to give a bigger contact area. Useful in soft sand.

Last time I changed a flat, the spare was at 36psig off the car. So far as I recall it was exactly the same (to the precision of my footpump gauge) once on.
 
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I've had someone describe it as the pressure in a tire is just there to hold the tire shape, and then the tire sidewalls are holding the load. I'm sure that's not completely true, but it makes a lot of sense to me.

A mechanically-inclined friend said he measured the pressure of the tires on a lift and couldn't tell any difference between on the lift and off the lift. I know if I'm using a standard piston gauge, I probably couldn't be more precise than about 1 or 2 psi. Don't know what he was using (probably same thing).
 
ben m said:
Some calculations. Say you've got a 15" diameter, 6" wide, 35psi tire supporting 1/4 of a 2600 lb car. That requires, in principle, an 18 sq.in. contact patch, so 6" wide and 3" long. So, how much do you have to move in from the edge of a 15"d circle to find a 3" chord? Only about 0.15". The area outside this chord is about 0.17% of the area of the disk. Likewise, the volume "displaced" in crushing a 15" tire from round to a round-with-a-3"-contact-patch, is 0.17% the initial volume, so via PV = nRT, for a small perturbation, it'll increase the pressure by about 0.17% or 0.06 psi.
Click to expand...


Thanks for getting down and dirty and doing the math Ben.

So the actual change in pressure should be a (significantly) less than one percent increase; which, when reading 25 to 35 psi on a normal (cheap) tire gauge, should not be noticeable at all.

That's just what I needed to know.
 
bowlofred said:
I've had someone describe it as the pressure in a tire is just there to hold the tire shape, and then the tire sidewalls are holding the load. I'm sure that's not completely true, but it makes a lot of sense to me.
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It's half-true. The sidewalls hold the load, but it's a stretching load from the top of the tire, not a compressive load from the bottom of the tire. And the pressure ALSO holds the load, since that's what's pushing up against the top of the tire to hold it up.
 
As soon as you mount the tires on the car, they will deform and form a flat spot where they contact the pavement.. Once the area of the flat spots multiplied by the pressure in the tires equals the weight of the car, the tire won't deform any more. At typical tire pressures, say 45 pounds per square inch, and assuming a weight of the car of 2000 pounds, the tires will stop deforming when the area is 44.4 square inches. That's 11.1 square inches per tire, which would be a square a little over three inches on a side, or assuming a wider tire, say 2 inches by six inches. Look at your tire, and visualize how little the volume has to decrease to make a two inch by six inch flat spot. The pressure will increase proportionally to that volume decrease, which is to say, not by much at all.
 
Andrew Wiggin said:
As soon as you mount the tires on the car, they will deform and form a flat spot where they contact the pavement.. Once the area of the flat spots multiplied by the pressure in the tires equals the weight of the car, the tire won't deform any more. At typical tire pressures, say 45 pounds per square inch, and assuming a weight of the car of 2000 pounds, ....
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Which basically means using only a ruler and a tire pressure guage, I can weigh my car.

But here's my question:
Suppose I know the tire pressure (and it is the same for all tires) and I can measure only the contact area under one FRONT tire; How can I tell the weight of the car if the center of gravity is not at the symmetrical center of the tire axles ?
Lets say the C.O.Gravity is only 2 feet behind the front axle and the distance between the axles is 8 feet.

How do we get the % weight distribution ? And thus determine the total car weight from only the measurement of the front tire ?

JW
 
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