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Recently had new tires installed on my vehicle, and checked the tire pressure. I found that tire gauges read about 3 PSI higher than the TPMS system in the vehicle. I did some more research, and found that this is because TPMS system in the vehicle gives you a type of "absolute pressure" referenced to sea level, while a tire gauge is reporting the relative difference between the tire pressure and atmospheric pressure, known as "gauge pressure". Since I live at altitude, this creates that 3 PSI discrepancy. That lead me to the question which pressure, absolute or gauge, defines the tire characteristics, i.e. how firm or soft it will be? Intuitively I believe that it is gauge pressure, however the below link contradicts that.

https://www.coyoteents.com/compensating-tire-pressure-readings-for-elevation/

EDIT: I did some more research, and these links agree with my intuition that gauge pressure is what matters.

https://www.flight-mechanic.com/types-of-pressure/

https://www.fullbay.com/blog/psi-psia-psig-air-pressure-measurements/

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    $\begingroup$ I agree with you, pressure determines the net forces on the walls of the tire, so that is what matters. The writer of the article you cite is confused when he says that the ‘amount’ of air determines the footprint. $\endgroup$
    – 10ppb
    Apr 19, 2021 at 4:53

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Gauge pressure is what matters. this tells you the pressure difference between the inside and outside of the tire. As you increase in elevation the outer air pressure lowers, so the pressure difference between the inside and outside of the tire becomes greater and the tire becomes over inflated. It is not addressed in your question but I would like to add that temperature also affects tire pressure, as air expands as it becomes warmer. So you should recheck the tire pressure with season changes.

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I am not sure, how that link contradicts that. The link says, that you have to change the pressure in your tire according to elevation. Because for the same amount of air in your tyre, the gauge pressure changes according to change in elevation . So, you have to add or remove air so that the gauge pressure remains constant. Hence, it is consistent with your intuition that gauge pressure is what matters.

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  • $\begingroup$ Here is the quote that contradicts that: "I’d like to also add that the footprint is unchanged despite what your gauge reads! This is due to the fact that footprint is determine by the amount of air in the tire, literally the number and temperature of the air molecules within the tire." So the author is implying that absolute pressure is what actually matters for the tire. $\endgroup$
    – Justin H
    Apr 19, 2021 at 5:00
  • $\begingroup$ I cannot help but believe that the author is mistaken. The footprint cannot be independent of the outside atmospheric pressure. Think of how balloons or bags of chips expand when put in lower pressure. How much the tire expands or contracts depends on the difference between internal and external pressure i.e. the gauge pressure $\endgroup$ Apr 19, 2021 at 5:05
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OK after many good comments and a bunch of edits, here's what I believe. It's a mixture of both.

The gauge pressure will affect the shape of the tire without a load on it. The deformation of the tire once load is applied will depend on both the gauge and absolute pressure. And the distribution of force across the footprint will depend more heavily on the absolute pressure.

Begin by inflating the tire to a given gauge pressure, but without load from the car. The sidewall will attain a certain stiffness which is due in part to the gauge pressure.

Next, lower the car down so that the tires start supporting the weight of the car. Initially, this support will come from the sidewall of the tire, as the wheel presses it down and the ground supports that.

If the sidewall is strong (due to the material) and rigid (due to the gauge pressure) enough to support the car, then it won't deform and we're almost done. All that is left is to analyze the force of the middle of the tread on the ground -- this will be due to absolute pressure.

But if it is not, which is frequently the case, the tire will deform. It will only stop deforming when something has changed such that the sidewall can now support the full weight of the car.

There are two routes to this: the sidewall becomes more rigid, and some other force augments the force the sidewall's internal force applied upward to the rim.

The former will be true if the gauge pressure increases when the tire deforms. This is likely to be true for most tires that have a cross-sectional aspect ratio which is already squashed compared to a circle, since squashing it more will reduce the internal volume (and tires aren't balloons, they generally do not stretch). So this will play a part.

The latter will also be true because the deformation allows for some of the internal air pressure to push upwards on the sidewall surface that is under the rim. This will mean that the downward-facing surface area becomes larger than the upward-facing surface area on the top of the inside of the wheel, so there will be a net upward force from the air pressure. It will also change the stresses on the sidewall so they are no longer purely compression, so I feel like the full solution could be complicated.

So to recap: I believe the full answer to what shape the tire takes is complicated. However, once it has taken that shape, if you want to look at the contact patch and how evenly the pressure on the road is distributed, I believe the absolute pressure plays the largest role in that.

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    $\begingroup$ If you had an “empty” tire at sea level and brought it up to some very high elevation like 20,000 feet, it would essentially appear to fill up as you gained elevation. The absolute pressure in the tire hasn’t changed, but the gauge pressure would. That is why it seems to me that it has to be the gauge pressure that matters. $\endgroup$
    – Justin H
    Apr 19, 2021 at 6:06
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    $\begingroup$ @JustinH Geshel's answer is not inconsistent with what you are saying. He is agreeing that for an empty tire that would indeed happen. But he is saying, that , for a tire under load , gauge pressure does not matter . I do not think , he is correct though. I think , even for a tire under load, gauge pressure would STILL affect the shape of the tire $\endgroup$ Apr 19, 2021 at 6:28
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    $\begingroup$ @geshel if there were no friction, the entire weight would be carried by bead against the short horizontal lower portion of the wheel surface and could well lose contact at the upper horizontal surface. In practice most of the load is transmitted through the frictional contact between the vertical lip on the wheel and the vertical surface on the tire. The force is present through all 360 degrees of the contact region. The air pressure is uniform inside the tire and cannot transmit any net force to the wheel or the tire. $\endgroup$
    – Roger Wood
    Apr 20, 2021 at 5:44
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    $\begingroup$ @geshel (I see the edit). I wasn't thinking clearly either. I think there's some analogy with a bicycle wheel. All the spokes in a bicycle wheel are in tension. The ones at the top have more tension and the ones at the bottom have less. It's that difference that supports the weight of bicycle+rider. It must be the same for the car tire. The entire tire is in strong tension because of the internal air pressure. The tension in the sidewall is greater at the top than the bottom and that is what supports the weight. It's obvious in hindsight that the sidewalls cannot support compression... $\endgroup$
    – Roger Wood
    Apr 20, 2021 at 19:25
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    $\begingroup$ @geshel ... The difference in tension between top and bottom is associated with 'hoop stress'. The sidewall radius of curvature is greater on the bottom where the bulge occurs and so the tensile hoop stress is lower there. Hoop stress is directly proportional to the radius of curvature. It all seems quite couterinuitive but I think it's correct. $\endgroup$
    – Roger Wood
    Apr 20, 2021 at 19:33

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