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The other day I inflated my tires up to 34psi and then when I took one of them off the car. I measured the pressure again to check how much the difference was, and I was surprised that it was still 34 psi.

What I suppose is that when the tire is put on the car, all the weight of the car is making pressure on the tires, so the pressure should "be higher" compared to when the tire is out. Is that wrong?

I tried this because I was wondering if the spare tire should have the same pressure or maybe less/more. So, should the spare tire have also the same pressure?

What am I missing here?

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    The physics description in all its mathematical detail can be found in the Physics Stack Exchange HERE
    – spicetraders
    Commented Sep 27, 2016 at 1:52
  • Nice! Didnt find that.. So that also answers the question about the spare tire.. It should have the same pressure.
    – Pablo Matias Gomez
    Commented Sep 27, 2016 at 1:57
  • @spicetraders should I close the question or what?
    – Pablo Matias Gomez
    Commented Sep 27, 2016 at 1:57
  • 2
    Better yet make the link and a brief summary an answer.
    – dlu
    Commented Sep 27, 2016 at 4:04
  • 1
    @spicetraders, like dlu suggests a summary of the physics involved in the tire situation with a citation of the PSE Q&A is worth a good answer here. I'd vote for it.
    – Bob Cross ♦
    Commented Sep 27, 2016 at 11:52

1 Answer 1

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The very short answer is this: for a gas at constant temperature, the product of pressure and volume is constant. This is the ideal gas law - close enough for this purpose.

When you load a tire, it flattens a tiny bit. Compared to the total volume, you lose just a sliver. Because the volume is virtually unchanged, the pressure stays the same.

Now when you start driving, friction between rubber and road results in heating of the air in the tire - and that will make a measurable difference in the pressure.

If you are interested in detailed math, I covered that in an answer on the physics site. LINK

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    In one line: volume and temperature does not change -> pressure does not change. Thanks!
    – Pablo Matias Gomez
    Commented Oct 6, 2016 at 2:33

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