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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/

Justin H
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    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. – 10ppb Apr 19 '21 at 04:53

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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.

silverrahul
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  • 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. – Justin H Apr 19 '21 at 05:00
  • 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 – silverrahul Apr 19 '21 at 05:05
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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.

Adrian Howard
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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.

geshel
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    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. – Justin H Apr 19 '21 at 06:06
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    @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 – silverrahul Apr 19 '21 at 06:28
  • @geshel the weight of the car is supported by the friction force between the tire and the wheel rim. It is not supported by air pressure on the wheel. – Roger Wood Apr 19 '21 at 23:16
  • What if there were no friction? – geshel Apr 20 '21 at 00:40
  • @RogerWood It is the normal forces that keep a car from sinking into the road, not friction. – Adrian Howard Apr 20 '21 at 04:12
  • @geshel I'm sorry but your answer is incorrect, it is the inner and outer pressure difference that matters. Say you add or subtract air so you always have 32 psi more pressure inside than outside, then your tire will always have the same stiffness. – Adrian Howard Apr 20 '21 at 04:18
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    @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. – Roger Wood Apr 20 '21 at 05:44
  • @geshel I don't know the answer to the question, but I would like to reinforce the point that the weight of the vehicle is transmitted through the sidewalls of the tires. The air pressure serves only to maintain the shape of the tire so it can transmit the weight. A moderate pressure is chosen so that the tires are not completely rigid but have some desired degree of compliance to the road. – Roger Wood Apr 20 '21 at 05:58
  • OK I was incorrect about the pressure on the wheel providing support, but I am still inclined to believe that the absolute pressure is what determines the deformation. So the weight is supported initially by the sidewall. Is it strong/rigid enough? If so, there's no deformation. If not, then the tire will deform outwards. When will it stop deforming? When the absolute pressure times the downward normal area of the deformed bulge is enough to offset the sidewall's lack of support. The outside air pressure plays no role in this equation. – geshel Apr 20 '21 at 18:39
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    @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... – Roger Wood Apr 20 '21 at 19:25
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    @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. – Roger Wood Apr 20 '21 at 19:33
  • @geshel I still don’t think the absolute pressure determines the tire footprint / firmness even when the tire has its shape. Consider a cube, with 5 sides fixed. The top side is detached, allowing it to depress inwards with an airtight seal. It is intuitive to see that if the pressure inside the cube equals the pressure outside the cube, (0 gauge pressure, could be any absolute pressure) it is easy to press the top of the cube inwards, which equates to a soft tire. If you increase the pressure in the cube (increasing gauge pressure) it is harder to press the top in, equating to a firm tire. – Justin H Apr 20 '21 at 22:43
  • Well, at least in regards to the interface between the tire and the ground, the only force that can push the middle of the tread into the ground is the air pressure inside the tire, and it will be the absolute pressure since there is no atmosphere on the other side (at that moment) pushing back. – geshel Apr 21 '21 at 00:38