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On one hand, I think they should be equal since the external force and internal force are equal in equilibrium. On the other hand, I don't see anything related between them, the inside pressure is hold by rubber strength, while ground pressure is equal to gravity, e.g a lightweight steel wheel can hold very high internal pressure but with low ground pressure

I have no idea, does the pressure inside the tire equal to ground pressure? Please explain it in detail

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No, the pressure inside the tyre is slightly less than the pressure at the tyre/ground interface.

The pressure everywhere inside the tyre is the same - let's calls this $P_i$. If the area of the contact patch is $A$ then the air inside the tyre exerts a force on the ground of $F_i = P_iA$.

But the tyre itself has some elasticity. If you've ever handled a tyre that is off the wheel you'll know that it can support a considerable load even when uninflated (though obviously far less than the weight of a car). So the weight of the car deforms the tyre and this deformation of the tyre also creates an elastic force in a Hooke's law sort of way. Let's call this force $F_e$. Then the total force on the ground is:

$$ F_g = F_e + P_iA $$

and the pressure on the ground is:

$$ P_g = \frac{F_g}{A} = \frac{F_e}{A} + P_i $$

So the pressure at the tyre/ground contact is greater than the tyre pressure by $F_e/A$.

I think it would be very hard to predict the elastic force caused by the tyre deformation from first principles. I suspect the only way to get an idea of what $F_e$ is would be to measure it.

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  • $\begingroup$ I have a cart for my kayak that has tires that are nominally inflated with air. The elastic force of the tires is sufficient to support the load with 0 pisg inside the tires. $\endgroup$ Jun 18, 2014 at 15:06
  • $\begingroup$ @RossMillikan: I was wondering if there's a way to measure the contact patch area for my Ford Focus. I know the tyre pressure and kerb weight, so in principle I could calculate the elastic force. My best idea so far is to drive it onto a piece of paper and spray round the wheel. The contact patch will stay unpainted. I don't particularly want to get paint on my tyres though. $\endgroup$ Jun 18, 2014 at 15:23
  • $\begingroup$ @John: tyre paint $\endgroup$ Jun 18, 2014 at 15:34
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    $\begingroup$ :-) See, this is what happens when you spend a lifetime working as an experimental scientist! $\endgroup$ Jun 18, 2014 at 16:32
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    $\begingroup$ "spend 3 days writing a program to calculate area from scanned image," @RedGrittyBrick Or, you reach onto the shelf and pull down your handy planimeter. Then you congratulate yourself on the effort you put into learning twentieth century techniques. $\endgroup$ Jun 18, 2014 at 23:46
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I assume that by 'ground pressure' you mean the atmospheric pressure at ground level?

The answer is no. The tyre pressure (in cars) is usually 2.1 to 2.3 bar, whereas the atmospheric pressure at sea level is about 1 bar.

You need to realise that car tyres need to sustain the whole of the car weight, and must ensure that the rotor and the bracket do not touch the ground. On top of this, they also have to balance the weight of the air above them, so their pressure must be greater than 1 bar.

Pressure is defined as $\frac{force}{area}$, where the force in this case would be the weight of the car (divided by 4 if the wheels shafts are equally spaced from the centre of mass of the car, because the weight is shared between the 4 wheels) and the area is the cross sectional area of the wheel in contact with the road.

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  • $\begingroup$ You are misreading the question - "ground pressure" is intended to be "pressure the tire applies to the ground". $\endgroup$
    – Floris
    Jun 16, 2015 at 10:37
  • $\begingroup$ Yeah I had realised that, but couldn't be bothered to correct it since an answer has already been accepted $\endgroup$
    – SuperCiocia
    Jun 16, 2015 at 13:53
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Test data shows that pressure on ground is calculated by dividing the weight of car divided by 4 then divided by patch area which will always be much less that the interior tire pressure. Test data shows that when the load is doubled the patch area only increases by 25%. e.g. under a 1000 pound load on tire the patch area is 100 sq.in while for a 2000 pounds load the patch area is 125 sq. Inch. Therefore even under extreme loading the pressure on ground will not approach the interior tire pressure. Here is aqout from some data I was just reading

just what is immediately obvious is that doubling the load does not double the contact patch area. I.e., the contact patch pressure does not remain constant.

From 390 lb to 773lb (almost double the tire load) the contact patch size only increases by 25%.

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    $\begingroup$ I'd love to see a reference to this test data. $\endgroup$
    – jvriesem
    Dec 12, 2021 at 22:29
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The tires of a car suspended above the ground are round and have a certain pressure. When the car is lowered, the part of tire which touches ground is flattened. Thus, the volume of the tire is reduced, which in turn increases the internal pressure. In equilibrium, the internal pressure is exactly the pressure which the tire exerts on the ground.

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  • $\begingroup$ This is incorrect. Imagine a tire suspended in the air. It has some non-zero pressure. Now touch it to the ground, barely. The pressure exerted on the ground is vanishingly small, but the tire still has nearly the same non-zero pressure as before. $\endgroup$
    – Kyle Oman
    Jun 18, 2014 at 14:50
  • $\begingroup$ I must admit that I did no rigorous calculations, but I still think I am right. Is the pressure vanishingly small? The force certainly is, but so is the area. If you want to dent the tire in the slightest way, you need to exert a pressure higher than that inside. And as long as the outside pressure is higher, the tire will deform. The equilibrium where the tire stays in a flat form is reached, when the pressure equalizes. $\endgroup$
    – M.Herzkamp
    Jun 18, 2014 at 21:55
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    $\begingroup$ Do the same exercise with a square tire, now the area is not vanishingly small... and my reasoning holds. $\endgroup$
    – Kyle Oman
    Jun 18, 2014 at 23:58
  • $\begingroup$ A square tire would not be square when the pressure inside exceeds atmospheric pressure. Only if the material of the tire is exceptionally strong. But then the whole argument is off point anyways.As stated in John Rennie's answer, I forgot to take into account the elasticity of the tire. $\endgroup$
    – M.Herzkamp
    Jun 19, 2014 at 10:38
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The tyre pressure exert more pressure than the less frictionless force. So the tyre has the same deflection as when maximum load and pressure needed for that on tire

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Part of the load on a tire is supported by the construction of the tire at the actual deflection of it. So pressure on the ground is higher then the over-pressure inside the tire. I estimate that part to be about between 5 and 15 % of the maximum load of a normal car tire , but can go as low as 1% of that for a Truck tire with high pressure .

Then if you would determine the surface on the ground and multiply it with the over-pressure the load on it calculated is lower then the real load on tire. For truck tire difference is small ( probably that 1% construction load, as I babtised it), but for normal car tire difference will be about 5 to 15%.

Even the profile makes the pressure on ground higher , and if you would place the tire on a blunt nail bed the tire bends the same and by that the pressure on the nails will be much higher. So if you would then determine the surface on ground by the number of nails depressed and compare that with normal underground, the calculated surface will be the same .

So its possible that the total load on a tire is only carried by the construction of the tire. then the real load on tire is about that 1 to 15% of the maximum load given on tire. Then the tire has the same deflection as when maximum load and pressure needed for that on the tire .

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