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I just noticed this whenever I park the car for a while, the tire pressure goes to a lower value on all tires, and whenever I start driving, it goes higher.

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The difference is not so big, but it is considerable, and should not be caused by measurement error. Whenever I park the car after driving, the values go back to the lower ones, and whenever I start driving again they go up and stay there until the car is parked for a while.

The answer should be simple, by heating up the tire because of friction while driving, however, the pressure does only raise a little, however fast or however long I drive. Should the pressure not rise when I drive faster?

Why is the pressure independent of the driving speed?

The question is, why is tire pressure lower when parking and higher while driving(and not rise more when driving faster)?

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    $\begingroup$ Did you consider that the tires (and the air in them) heat up while you're driving? $\endgroup$
    – hdhondt
    Commented Mar 16, 2020 at 22:17
  • $\begingroup$ @hdhondt yes, but the pressure does not go above these levels, whatever, however long or however fast I drive. And when I park, the pressure goes down. If the weather turns hot, and I park the car, the tire that gets the most sunlight will not have more pressure at all. Shouldn't the tire have more pressure when sunlight heats it up and the weather is hot too? Why is just simply driving at any speed for any time raise the pressure? $\endgroup$ Commented Mar 16, 2020 at 22:22
  • $\begingroup$ Why the downvote? $\endgroup$ Commented Mar 16, 2020 at 22:32
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    $\begingroup$ It is not polite to change the question after your original question has been answered by three people. $\endgroup$
    – G. Smith
    Commented Mar 16, 2020 at 22:46
  • $\begingroup$ @G.Smith somebody downvoted, because the answer seems obvious, thus, the question seems too simple. But this is because it is not clear what I am asking, then I have to make it clear I believe. I just wanted to make it more clear, and I believe it is a meaningful question, not just why the pressure rises because of friction. I believe there is more to it, why the pressure does not rise with the speed (rotation) of the tire. $\endgroup$ Commented Mar 16, 2020 at 22:54

2 Answers 2

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Due to friction the tires temperature increase. Now you apply the ideal gas law $$ pV = n R T $$ Since the volume remains approx. constant the pressure has to increase.

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  • $\begingroup$ thank you, but why does it not rise more when driving faster? $\endgroup$ Commented Mar 16, 2020 at 22:33
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It's very simple. When you drive the car the the tires heat up due to rolling friction. This increases the temperature of the air in the tires. Considering the air in the tire to be an ideal gas, the relationship between temperature, pressure and volume for a fixed amount of mass $m$ of air in the tire is given by

$$PV=mRT$$

Where $R$ is the specific gas constant for air. Now, assuming the volume of the tire does not change appreciably, if the temperature goes up the pressure goes up proportionally.

Hope this helps.

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  • $\begingroup$ thank you, "the compression of the tire during each revolution" can you please elaborate on this? $\endgroup$ Commented Mar 16, 2020 at 22:24
  • $\begingroup$ @ÁrpádSzendrei See my update. It's due to rolling friction that results from "squeezing' the rubber of the tire as it rolls. $\endgroup$
    – Bob D
    Commented Mar 16, 2020 at 22:30
  • $\begingroup$ thank you, I try to ask why the pressure does not rise when driving faster, so why does the pressure rise only a little, however fast I am driving? $\endgroup$ Commented Mar 16, 2020 at 22:31
  • $\begingroup$ @ÁrpádSzendrei Rolling friction, which is the major reason for the increase in temperature and pressure in the tire, is due to heat generated during the "squeezing" and "unsqueezing" of the tire synthetic material. Compared to other types of friction heating (such as kinetic friction when skidding) it is generally much less. So when you drive faster, there will be an increase in heating, but the amount of the increase may not be large enough to significantly change the pressure. Hope this helps. $\endgroup$
    – Bob D
    Commented Mar 16, 2020 at 22:49
  • $\begingroup$ thank you so much. $\endgroup$ Commented Mar 16, 2020 at 22:58

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