Tire pressure

Here is what i came up with, and sent in an email to whoever it is that got it. They pointed me here…

Tire pressure: the math and the answer…

The question posed on 11/3/07 got me
thinking, so i figured i would do the math
and get a real answer.

Why doesn’t the tire pressure change when the
car is put on a lift?

It does, but not much.

I drive a 2007 subaru impreza wagon. It
weighs just over 3000 lbs, and has four tires
which are about 7" wide and about 25" around.
I run them at 35 PSI.

The load on each tire is, of course, 1/4th of
the total weight of the car, or about 750
lbs. At 35 psi, to support the car, the car
needs 21 square inches of contact patch per
wheel (pressure times area equals force).
This is a patch roughly 3 inches long by 7
inches wide. I went outside and measured the
patch, and it is indeed about 3 inches long.

The length of the contact patch has a name,
in mathematics: It’s called the chord, which
is a line connecting any two points on the
circumference of a circle. The largest chord
of any circle, of course, is the diameter.

Anyway, if you know the diameter of a circle
and a chord length, you can figure out the
area of the section the chord sets aside. In
the case of a cylinder, like a tire, you can
find the volume of that section. In my
vehicles case, the volume of the section set
aside by a 3" chord is about 130 inches
cubed. The volume of the entire tire is about
2114 inches cubed (the volume of the tire
minus the volume of the wheel or rim). 130
inches is a mere 6 percent of the total tire
volume. So, when i put my car down, i would
expect a change in tire pressure of six
percent of 35 psi, or 2.1 PSI.

It doesn’t change nearly that much, of
course. My calculation assumes that the tire
simply collapses in on itself, which is
doesn’t. In fact, the sidewalls will expand
outwards (very obvious when the tire is
extremely low on air) and the top of the
tire will expand out in all directions
slightly. Also, the tire will become out of
round, with the extra volume moving to the
front and rear of the contact patch. If those
three things combined accounted for only 4%
of the volume, the pressure change would drop
to about 0.6 PSI . The friction from the seal
in a normal pen-type pressure gauge is
probably enough to throw your reading by that
much.

Cheers.

As you can see, my results matched your measured change fairly well, so i think i’m right. I’d also like to point out that the pressure MUST go up to some degree, because the belting keeps the tire from expanding any more than it already has. Therefore, the only direction volume can go is down.

I did, got no change, but I didn’t follow “scientific protocols” either. To do this correctly, you need to design an experiment that would eliminate all sources of variation except the one you are measuring. You would need a very accurate gauge with small increments. Then do what I described in my original post above.

The car I did a quick test on was an 86 Tercel 4wd with oversized (185/65-14 vs, 175/70-13) tires at 32 psi.

I have read the answers below. I sent the following comment to the general e-mail, and was directed here. I think it says the same thing, but maybe it will be clearer to some the way I say it.

Tire pressure question

The question was why the pressure isn’t higher with the car on the ground compared with it suspended. The
physicist you consulted said that the pressure doesn’t change because the volume and temperature don’t
change, and that is certainly true, but why is it still unsatisfying?

You would expect that, if you applied a 3000 lb force to a piston against a 25-35 lb per square
inch pressure, you would move the piston, compress the air in the container, and so the volume would go down,
and the pressure would go up, right?

The key is the “per sq in” part. A 3000 lb car is putting 750 lbs of force on each tire. If the tire pressure is 30 lb/
sq. in. (oh for the metric system!) the tire only needs to have 25 sq in. of contact with the ground to support the
weight of the car. That’s just 5x5 in., or 3x8 in.–quite compatible with what the tire looks like with the car
sitting on the ground.

Another way of looking at it, if you had a piston with an area of 100 sq. in. , enclosing a chamber that is pressured
to 30 lb/sq.in, (the piston can be compressed but has a stop that keeps the volume from expanding) you could
put a 3000 lb car on the piston and still not move the piston.

I think this explains it better than just invoking Boyle’s Law. [That is, PV=nRT.]

Addendum–I spoke too soon, and sounded pretty conceited at that. The answer immediately ABOVE says the same thing, and is correct, plus addresses the issue of deformation and the tiny increase in pressure that actually occurs. Kudos.

Good news: The pressure gauge arrived. It?s an Accutire Model MS-5510B. It measures to the nearest 0.1 psi.

I used a 1995 Ford Mustang 2.3L (4 cyl) equipped with P205/65R15 Continental CH95?s.

Results: Loaded / Unloaded
LF ? 29.7 / 29.5
RF ? 28.5 / 28.3
LR ? 27.0 / 26.8
RR ? 29.4 / 29.2

Notice the consistency!

What should happen now is that others should also report their tests to verify the principle.

Most tire gauges only read in .5 psi increments so that would explain why no one else go any difference. But I have a question or two. Does your gauge screw onto the valve stem or do you push it on like most of the store bought brands? If it screws on, did you leave it on while lifting the vehicle. If you did not, or its a push on type, did you take a reading after putting the car back down on the ground as you lose a little air every time you use the gauge?

If you don’t mind, where did you get this gauge from?

The gauge is available through Tire Rack. It’s a digital and has a clip on hose and I tried using that, but it didn’t seem to like the inactivity - It shut itself off while I was jacking the car up. So I took individual readings.

I discovered that technique affected the reading, but I didn’t have enough time to find out which method worked best nor try to see if I could get the gauge to read continuously.

I commend CapriRacer for his/her experimental method and for not being put off by the pontifications of so many people. His result has the “ring of truth” to me. I hope anybody inquiring about his method will keep it civil and do so in the spirit of learning.

I got in on the tail end of this discussion and only heard the radio program because we taped it last weekend and I finally got around to listening to it.

Some comments:

There’s nothing wrong with “using math” to calculate an estimated solution–what you need to do, though, is make sure you are modeling “the entire system” and not making any erroneous assumptions Math, by its definition, is a way of describing reality. And if you get an answer that doesn’t match reality, then you must have something wrong in your model or in the equations you’re applying to the model. One approach in the “scientific method” is to do the math based on a model, and then do an experiment to see if you get the same answer. If the two results are close, you have SOME reason to believe that your model is close to representing reality–though it could still be wrong! Anyway, this is why I offer kudos to CapriRacer for doing the calculations well AND doing the experiment and comparing results.

Other comments:

It would seem to me like the Ideal Gas Law (PV=nRT) ought to apply and be accurate enough in this instance. Yes, a real gas does turn non-linear at high pressures, but unless I’m not remembering rightly, in the neighborhood of 30 psi (plus 15 psi atmosphere) it still compresses pretty linearly.

The initial physics professor appears to have made an assumption that the tire will not change in volume when you set the car down on it. It may not LOOK like it expands all over, but keep in mind that any bulging of the tire will be distributed over the whole thing and thus might be pretty slight and hard to notice by eye. Anyway, if the tire did not change in volume, then his analysis using PV=nRT would be correct. If variables V,n,R,and T didn’t change, then Pressure couldn’t change either. He might have had a clue though, about the assumption that the tire doesn’t expand: anybody who has inadvertently put too much air in a tire and then suddenly found that the tire started rubbing on the side of the wheelwell…Also, didn’t you ever change a flat tire, and find that you have to jack the car up higher to get the pressurized tire on. Yes, a tire does change volume when you put more air in it–when you increase the pressure.

The ideal gas law still is valid, though. If you recognize that the volume of the tire does change with pressure, if you measured (or somehow knew) that change in volume, then that number should lead you the change in pressure (by calculation).

One thing to note about the above method is the definition of “the system.” In that instance the system is defined as just the air inside the tire. If you were to draw a picture of the system (a FREE BODY DIAGRAM!) you would draw a dotted line around the air in the tire and define it as the system. No other components involved. And if your equation pertains only to that system, then that’s just fine. So PV=nRT should be a valid mathematical approach. The difficulty is in knowing what the change in volume is.

Other calculation methods were suggested in all the above discussion. Some of them were valid, too. Using a “balance of forces” equation is certainly valid–when/where it applies. So putting the car down on the pavement should produce an equal force in the opposite direction–somewhere. Note that in this first bit of the chain of logic, you define the system as the car (draw a dotted line around it)–as a rigid body-- being let down onto the pavement. If you model this in terms of force vectors, then there has to be some vector pointing upward, which increases in magnitude by, say, 800 pounds at each tire… This force comes from the pavement. Fine. Then you have to recognize that in your next logical step you are then selecting a different system (you have to be a little systematic): Now draw a dotted line (your system definition) about the tire, instead. The pavement is pushing up on it with 800 lbs. Correspondngly, by balance of forces, there has to be something about the tire which is pushing down on the pavement with 800 lbs. The difficulty in this approach comes from the mental model of what happens inside the tire during all this… that the pavement pushes on the tire, and the tire in turn pushes on the air. If it were that simple it would be great. But this model neglects the stiffness of the rubber. I can sit on an unpressurized tire and hardly deflect it–the rubber is pretty rigid. So the 800 lbs pushing on down on the pavement comes partly from the rubber and partly from the air pushing back from the inside. How much comes from each? Dunno. So it seems pretty hard to guess how much the air pressure will have had to increase–and how much downward force comes from stiffness in the rubber. This is another instance in which the model was OK and the equations were applied properly–but there is some unknown (force from rubber stiffness) that keeps you from deducing the answer.

If you have a Finite Element Analysis package in your CAD system, and you input the material properties of the rubber (and the air), you can model the whole tire and get a very accurate estimate of how much force comes from the rubber and how much the air volume changes, etc. You just can’t do it with a simple balance of forces equation.

Not to put too fine a point on it… a Balance of Vertical Pressures method doesn’t hold up in this situation. Pressure upward from the pavement and pressure downward from the air do not balance. I won’t bore everybody further by talking about when balance of vertical pressures would work–but you can see that the difference is in effects from the tire.

Another mathematical method is also valid: Balance of Strain Energies. This a “before and after” balance. Energy is conserved (or, close enough for our purpose). You have the energy of setting the car down on the pavement. That is calculate-able. It’s just force times distance. That number has to match the energy that goes into straining (distorting) the rubber and the air. You could get a good estimate of the energy required to strain the rubber–again if you used an FEA package on your computer model of the tire. It would also give you the energy of compressing the air. (you could also calculate it from the gas law) The two strain energies together have to add up to equal the energy of dropping the car to the pavement.

Any other engineering types out there have other valid methods in mind?

Anyway, do I know the right answer? No. My GUT FEEL (even engineers have gut feel) is that when you drop the car to the pavement there is a little expansion of the tire as a whole (like a stiff balloon), and that the bottom of the tire flattens, and that there is a net small decrease in volume, and a net small increase in air pressure. But I’d have to do a test to verify that. But CapriRacer already did the test! And thanks!

In the above, I made two misrepresentations: If you wanted to use a Strain Energy Balance approach, you could get at the energy of dropping the car to the pavement. I said that you’re treating the car as a rigid body. Never mind that–it has springs, etc.–the point is that you initially think of the forces of the pavement as pushing upward on each tire. I also said the energy would be just force times distance. Actually it is force INTEGRATED over distance as the car lowers to the pavement.

The other mistake was one of omission. In prgh 10 I said that upward pressure from the pavement and downward pressure from the air don’t balance. Of course, there was pressure in the tire before it was lowered to the pavement. What I meant was that the INCREASE of pressure from the tire doesn’t balance the pressure from the pavement. There are effects from the rigidity of the rubber that need to be included.

My take on this tire pressure question is a little different from those considering changes in tire contact area or internal volume. I think it might help to consider the total force due to the pressure inside a tire. If we estimate the total surface area inside a tire and multiply that by say 30 psi, we get an estimate of the total force inside the tire. Of course, unless the tire is in the process of exploding, this internal force is exactly balanced by the outside forces provided by the tire, the tire bead, the wheel, and the weight of the car! We can discount the outside force due to atmospheric pressure because we are measuring our 30 psi as gauge pressure, not absolute pressure. Anyway, if we figure our tire is 27" in diameter, is mounted to a 15" wheel, is 9" wide, and has a rectangular cross-section (this makes our calculations easier), we can figure the area of the tread (763 in^2), the rim (424 in^2) and the two side walls (393 in^2 each) for a total inside area of 1,979 in^2. Multiply by 30 psi for a total internal force of 59,376 pounds! Adding the 1,000 pound corner of a 4,000 pound car to that will change the internal pressure at most by 1.7% or 0.5 psi, not enough difference to see on most commonly available pressure gauges. Now this is certainly not a scientific answer. I only offer it to suggest that this question is a matter of scale, i.e., pressure (a relatively small number) vs. total force (a large number).

Why is the tire pressure the same whether on the car or off? Ah! The eternal question. This is the question that separates the mechanics from the automotive technician. You have quite an imagination - so does that physicist. Physicists are not mechanics - those guys are used to dealing with absolutes. They know nothing about the punishment that the modern pneumatic tire is subjected to. Back to the question… I submit to you that the tire has nothing to do with keeping the car off the ground. Let me explain… The tire pressure measurement is the difference in the air density and pressure inside the tire relative to the atmospheric pressure and air density on the outside of the tire. The tire itself acts as a vessel to contain the air under pressure. It is the air inside of the tire under pressure that does the work of holding the car up. Another example of air at work is your air tools that you use in the shop. It’s all relative, man. Rodd

That’s too much math. This is a simple question. The volume doesn’t change. Think about it. Does the volume change when you hit a pothole?

Take your tire gauge and your car to the top of Mt. Kilamonjaro. See what happens then.

It’s OK to use math to solve the problem. You just have to keep track of where the air went. The volume still doesn’t change.

How heavy is your car? Let me tell you about that. You know those big tractors? John Deere, etc. 3 - 10 psi in those big tires. Your bicycle? If it’s a mountain bike - 65 psi. 10-speed racer - 100 psi.

Have you ever tied to stretch a tire? How do you explain a fully loaded cement truck?

The professor is right. Clear and concise. I am sure you are an excellent teacher.

Keep adding weight to the car. You will see how insignificant it is.

Don’t get mad. You are the one that overloaded the tire.
You don’t need an MIT grad to figure this out. Ask Jay Lenno. He might know the answer. He’s a real comedian. He’s also a gearhead.

How much does it cost to buy a tire gauge to measure to the nearest 0.1 psi ?