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Does tire pressure measured by a meter on tire gauge change with load? (I am not interested in pressure produced by car tires onto the road).

Car spec usually says "inflate to 220kPa normal load, 300kPa full load". Does this mean the measured pressure should be 300kPa only after the car was loaded, or can one inflate the tires to the recommended 300kPa while empty and then load the car with 500kg without needing to re-measure the pressure?

There are basically both answers to be found when researching the non-physics forums on the internet:

  • One explanation says that the tire will compress under bigger load, making the volume inside the tire smaller, hence pressure higher. It also states this is the reason why the spec plate in the car has two values - you should simply expect to measure higher pressure on the tires while the car is loaded.

  • There are also explanations stating that the air does not escape from the tire, hence the amount of air inside the tire is constant and the pressure measured on the tires is constant even if the car is loaded (and based on observations, tires normally compress when the car is loaded). This would mean one needs to put more air into the tires when the car is loaded to ensure higher pressure.

To a layman, both sound feasible.

Is one of the explanations simplifying based on normal load assumptions? (i.e. tire pressure does not change if car is loaded with at most 1000kg, but would change if there was an enormous load put on the car?)

Where is the truth?

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Isn't this a "no brainer"? Tires are necessarily flexible, that is why we have them around the wheels. If I supply sufficient load, the tires will burst. – Jodrell Aug 28 '14 at 15:13
........yes........... – Anixx Aug 28 '14 at 16:05
@Jodrell: The vast majority of the air in a tire is above the horizontal plane marked by the bottom of the rim. Forcing all the air below that level to move above that level would thus no more than double the pressure of the tire (for many tires, it probably wouldn't even increase it by 50%). Applying enough load to squish the bottom of the tire flat would probably cause the seal to fail (perhaps in spectacular fashion) because it's not designed to deform that much, but I don't think loading a car could put tire pressure much over design limits. – supercat Aug 28 '14 at 16:44
It's already been worked out here:… – Hot Licks Aug 28 '14 at 19:17
"air does not escape from the tire, hence the amount of air inside the tire is constant and the pressure measured on the tires is constant". This is false (unless your meter is broken). To anyone who has studied fluid dynamics this will sound a bit like "Socrates was a man. All men are mortal. Therefore all men are Socrates". – quant Aug 29 '14 at 5:46

10 Answers 10

up vote 16 down vote accepted

TL;DR: The load does not significantly increase the pressure in the tire, but not inflating the tire more will increase friction. This will heat up the tire. Correct pressure ensures correct contact area - preventing wear on the tire, and keeping rolling friction low.

Full answer: Going to use simple math, round numbers (no calculator): 1000 kilo car, 4 tires, 2.2 bar pressure. Contact area for each tire approximately $250 / 2.2 = 110 ~\rm{cm^2}$. With the tire 15 cm wide, the contact patch is 6 cm long.

Now "load" the car with 50% more weight (500 kg). The additional contact area needed is $55~\rm{cm}^2$ per tire. If you assume that the side walls don't deform, the contact length increases to 9 cm.

The change in volume from this additional flattening of the tire is quite small. Looking at the diagram below, you can compute the volume change (assuming all deformation happens in this plane)

enter image description here

The volume of the air in the undeformed tube:

$$\begin{align}V &= \pi (r_o^2 - r_i^2) w\\ V &= \rm{volume}\\ w &= \rm{width\ of\ tire}\end{align}$$ The angle subtended by the flat region: $$\theta = 2\sin^{-1}(\frac{L}{2r_o})$$ when $L<<r_o$ this approximates to $\theta = L/r_o$

The area of the flattened region is $$A_{flat}=\frac12r_o^2\theta - \frac{L}{2} r_o \cos\frac{\theta}{2}$$

Small angle approximation:

$$\begin{align} A_{flat}&=\frac12r_oL(1-(1-\left(\frac{L}{2r_o}\right)^2))\\ &=\frac{L^3}{8r_o}\end{align}$$

For a constant width $W$ of the tire, the flattened volume is of course $Aw$.

If we assume to first order that friction is proportional to the volume that is being distorted, you can see that a slightly flat tire (larger contact area) will significantly affect fuel consumption.

How big is the effect? With the numbers I used above, the fractional volume change is only 0.03% (for $r_i = 30~\rm{cm}, r_o = 40~\rm{cm}, w = 15~\rm{cm}$). That means that the pressure will not increase due to the deformation of the tire / the additional mass.

And that in turn means that the reason to inflate the tire more is precisely to prevent the increased contact area, which would lead to higher friction and potentially higher temperature.

Note that there is a feedback loop - if the tire is underinflated and heavily loaded, it will get hot which will increase the pressure somewhat. But it is better just to start with a bit more air in it...

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This is a great answer with very good combination of physics and practicality. Thanks! – Marek Sep 5 '14 at 18:35

The data manufacturers provide assume the tyre pressure is adjusted with tyres cold (ambient temperature) prior to loading. Though the small error introduced by making adjustments after loading will not be significant, the difference between cold and hot tyres is greater.

Recommended pressure is higher for a 'full laden' car. That is why a spare tyre has to be kept at higher pressure.

I am looking for the ultimate truth to this question and while your answer provides very good practical advice, the answer seems rather vague and still mentions some "small error". Could you please clarify in physics terms and provide an answer whether tire pressure increases under load or not supported by physics (even if the difference under would be insignificant?

I ignore if there is a specific concrete measurement that answers your question, my reply implied a logical deduction: if pressure increased in a non negligible way, the manufacturers would not recommend to increase the pressure.

Moreover, it is impossible to give a general answer that will be valid for all tyres. I am afraid an ultimate truth is not there because it cannot exist. The increase of pressure takes place only if the tyre is deformable, and if it is, then you must take into account if the difference of volume is meaningful. If an object is spherical or cylindrical, any deformation will automatically produce a decrease of volume. No tyre is such.

enter image description here

Since the lateral walls are never circular, it is even possible that vertical pressure on the thread will produce a deformation of the lateral wall that will give it a more circular shape, increasing the volume (in a negligible way).

The physical answer can be only theoretical and is quite obvious and simple: if there is such a load that produces such a deformation that produces a certain reduction of volume, the pressure increases proportionally to the latter.

... you should simply expect to measure higher pressure on the tires while the car is loaded.

Probably what you wrote in the 'first explanation' above is what baffled and misled you, is it not so: you must inflate your tyre up to the higher recommended pressure before the car is loaded anyway, (and I am not sure you if should expect to measure higher pressure when it is loaded). This is because higher pressure prevents deformation. That's all I can figure out.

You suggest that in practice the second explanation should be observed and one should inflate tires when expecting load, but I have not found this in my car manual. Do you have any references?

Probably they take it for granted, but it really makes no difference if you measure it before or after you loaded the car: pressure at full load must and will be 300kPa anyway. The web is full of references here is NSCEP "test laboratories make the pressure check prior to loading the tire". Any attendant at a filling station will confirm that.

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thank you for your quick answer, but I am looking for the ultimate truth to this question and while your answer provides very good practical advice, the answer seems rather vague and still mentions some "small error". Could you please clarify in physics terms and provide an answer whether tire pressure increases under load or not supported by physics (even if the difference under would be insignificant)? – Marek Aug 28 '14 at 11:28
thank you for the clarification, this put things into practical perspective as well. It looks like the first is true (pressure increases with load given that tire is deformed, but never enough to approach the "pressure when loaded" figure). You suggest that in practice the second explanation should be observed and one should inflate tires when expecting load, but I have not found this in my car manual. Do you have any references? – Marek Aug 28 '14 at 12:46
You have found that, that is what the spec you quoted means. They don't expect people to overthink it. :) For faster driving or heavier loads you need to put in more air. – JamesRyan Aug 28 '14 at 14:20
On most cars, the spare is at a higher pressure, because its a donut tire. If you had a full sized spare, it would be kept at the same pressure as the regular tires (or maybe slightly different so that it stays inflated between uses or due to a different lighter tire construction). – Batman Aug 29 '14 at 3:08
@Batman, ".. If you had a full sized spare, it would be kept at the same pressure as the regular tires" No,it's not true. The tyre is kept at higher pressure because in this way you can use it properly both with full load and, by deflating it a little, with normal load – abc Aug 29 '14 at 4:34

The question: Does the pressure inside a tire equal to its average ground pressure? is related.

If we can ignore the rigidity of the tyres then the air pressure in the tyres multiplied by the four tyre contact patches must be equal to the car weight, so the pressure would be given by:

$$ P = \frac{Mg}{A} $$

where $M$ is the car mass and $A$ is the total contact patch area.

At first glance it looks as if $P \propto M$, and therefore increasing the weight of the car (and its contents) will increase the tyre pressure in proportion. However the increased weight will also flatten the contact region and increase the contact patch area $A$ and this effect will reduce the tyre pressure.

I guess you could model the deformation of the tyres to find out how $A$ depends on $M$, but this seems to me to be a complicated task, and it's not obvious to me which effect would dominate. However any deformation of the tyre will reduce its volume and therefore tend to increase the pressure inside, so it seems likely that the increase in contact patch area will not offset the increased mass. In other words the tyre pressure will increase with load but it probably wouldn't be simply proportional to the load.

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"and increase the contact patch area A and this effect will reduce the tyre pressure". Actually, as discussed in other answers, as the tyre deforms it is likely that the volume decreases and the pressure goes up. Maybe "reduce the required tyre pressure". – Rob Jeffries Aug 28 '14 at 16:51
@RobJeffries: well my answer does say: However any deformation of the tyre will reduce its volume and therefore tend to increase the pressure inside. My point is simply that for constant $M$ we have $P \propto A^{-1}$. I'll look at the wording and see if I can make it clearer. – John Rennie Aug 28 '14 at 16:58

I will attempt to go more in-depth into the physics here. In practice, the unloaded weight of the car is non-zero, but I'll consider two states where the first state refers to zero weight on the tire. In order to make a practical calculation for the car, then, you'll use the numbers for the second state, subtracting one vehicle weight from another.

There are two governing equations that I see playing out, and these have been laid out by other answers, although not in so-many variables. Again, I'll use "state 1" to denote no loading and "state 2" the loaded state.

$$ P_1 V_1 = P_2 V_2 $$

A note about justification: this assumes the temperature is set. There is an infinite temperature sink available in the form of the ambient environment. So you'll have some temperature change if you load a vehicle quickly, but over time the tire temperature is dictated entirely by operating conditions. Moving on to the second equation I have in mind, we equate the footprint and the pressure on that footprint to the weight being supported. I'm assuming the tire has no weight.

$$ A_{ft,2} P_2 = M g $$

For state 1, this is trivial equation, so I do not write it. The A, area, here is the footprint area. This is different from the total geometric tire area. This is only the area of contact between the tire and the ground. I'll have to neglect any contributions from the rigidity of the tire itself.

It would be a straightforward matter to equate the footprint area of a torus to the remaining volume above-ground. However, the torus equation is complicated. I'll assume a spherical tire. This is still more detailed than any other attempt so-far, although the door is open if someone wants to do more work.

$$ A_{ft} = \frac{M g }{ P} = \pi \left( R^2 - (h-R)^2 \right) = \pi ( 2 R h - h^2 ) \\ V = \int_{h-R}^{R} \pi \left( R^2 - z^2 \right) dz = \left[ z R^2 - \frac{1}{3} z^3 \right]_{h-R}^{R} \\ = \pi \left( \frac{4}{3} R^3 - h^2 ( R - \frac{1}{3} h ) \right) $$

To apply this, for the unloaded state observe that h=0. For the loaded state, the first equation above can be solved for the height displacement. We will imagine that we know the unloaded tire pressure, so everything in that equation is solved for. Now we can return to the P1 V1 = P2 V2 equation, and solve for P2. The V1 term is simply the volume of a perfect sphere. Then, we have V2 in terms of h, and we have h in terms of P2. This yields an equation that you can solve for P2, but it's a high order polynomial.

Here are some numbers, which vaguely represent a Corolla.

  • P1 = 30 psi = 206842.7 Pa
  • R = 28.85/2 inches = 0.3155 meters

Plugging these in, we can solve it numerically. However, in doing this, we need to correct for the egregious assumption of a spherical tire, as well as account for the 4 wheels. A typical tire volume is on the order of 10 L, and the sphere in my case is 131.5 L. So for realism, I'm correcting the loading by the ratio between these two, 131.5 L to 40 L.

With that, here are the numbers I get. This is the pressure in Pa versus the car loading in Newtons.


EDIT: This is the corrected, second version of the graph. It still uses all the above parameters, but take note this assumes 40 L tire total volume. I had missed a factor of pi.

Now this was assuming a sphere, and also note the y-axis doesn't start at zero. The non-linearity is due to the changes in the footprint as the depression height first goes above 0. Since real tire geometry start with a much flatter contact area to begin with, it follows that a lot of the initial curve here would not be observed in practice.

Nonetheless, when we're at roughly 1 metric ton of loading, the depression height is about half the radius. Geometrically, I think this point would somewhat resemble the footprint to volume ratio of normal tires, and the above curve's 2nd derivative near that point is thus similar to what we'd see in practice. Furthermore, the adjusted contact area in this example comes out to about 0.06 m^2, which is very similar to what I expect running some simple numbers on the Corolla.

This is much more non-linearity than what I expected, and also a much greater effect from the weight of the car on tire pressure. I would certainly believe that you could cause some problems by failing to account for the impact of loading on tire pressure.


I really should have assumed a cylinder model. Basically, you imagine all tires are horizontally oriented cylinders, with a hole in the middle for the hub. This is realistic to an extent, and if you have the dimensions, then no correction factor is needed like the above example.

Tire measurements I assumed were:

  • R = 0.292 m
  • Rin = 0.206 m
  • L = 0.711 m

Here, the "L" is the width of the 4 tires, if they were theoretically lined up all next to each other. So this is just multiplying the width of one tire times 4. "Rin" is the radius of the rim. This volume is subtracted out of the total cylinder volume.

These measurements, which are based on outer dimensions, come out to about 95 L. A major difference between this and the above example is thus the volume. However, the fundamental difference in shape is still apparent.


To put this in real terms, if the load is 3 metric tons, tire pressure increases 0.5 psi relative to the unloaded state. It's normal weight causes only about 0.05 psi above the unloaded tire pressure.

The Corolla weighs about 1.3 tons. The maximum legal load is 0.45 tons. Assuming a 0.5 psi error on a pressure gauge seems reasonable.

So my conclusion is that there is no legal way to measure this effect in your garage.

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awesome! This is the kind of answer I have been hoping for. Your conclusion is that a significant (nonlinear) portion of pressure increase would be from weight increase without adding air to the tires. Therefore tires should be probably inflated to target pressure only after the car was loaded, otherwise we will get significantly more pressure after loading. Would it make sense to do estimate this ratio for typical car load? (x:pressure to which we inflated the tires before car was loaded, y:pressure after car was loaded with 500kg). I would love to verify this with an experiment afterwards. – Marek Aug 28 '14 at 14:39
@Marek I now realize my answer was over-complicated. I can produce a usable number for exactly what you speak of, and possibly with less computation. The proper assumption is that of a cylinder lying down on the ground. The 10 L figure also looks about 50% too low, which isn't surprising since it came from Yahoo answers. You can easily do this test yourself and compare to the number from this method. Just make sure to carefully read your car's instructions on changing a flat if you're using the crank which comes with it. – Alan Rominger Aug 28 '14 at 15:40
Nice answer, +1. A Corolla isn't made for hauling stuff. Now try this analysis on a regular cab Ford F-550 with dualies (four rear tires). Empty weight: 6700 lbs. Max payload: 12300 lbs, almost twice the empty weight of the vehicle. – David Hammen Sep 24 '14 at 13:07
This answer is wrong and not realstic as it assumes a cylinder model which is not the model of any existing or possible tyre – user59485 Jan 24 '15 at 9:10

The first explanation in the question is correct: tire pressure increases with an increasing load.

The second explanation has a problem in that the phrase "the amount of air inside the tire is constant" is being too vague as to what "the amount of air" means. The number of molecules of air inside the tire remains constant. But the volume of air does not remain constant; the volume decreases with an increasing load. This means that the pressure increases due to Boyle's law, which says that for a constant number of molecules of gas at a constant temperature, the pressure of the gas is inversely proportional to its volume, $$P \propto 1/V \ \ .$$

It may be counterintuitive that pressure and volume are so closely related for air, because most other things that are encountered in everyday life don't behave that way. Solid objects in general have very little compressibility. And although liquids are more similar to a gas than a solid is in that liquids are fluid like gasses are, liquids and gasses are different in that liquids are an incompressible fluid, but gasses are a compressible fluid.

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This is actually hinted at in an existing answer (on re-reading) but (and this is a rather handwaving take on the question as we don't have much data on tyre deformation, expanded from an intended comment):

There's a further explanation which I believe to be closer to the correct one, as the pressure change shown can be large compared to the mass change, and the tyre is neither rigid nor completely free to deform. We know this because an uninflated tyre collapses under the weight of (1/4 of) a car, but not under the weight of a wheel.

The aim (I propose) is that the pressure should be adjusted to maintain something close to the desired contact area on the road, taking into account tyre deformation as a function of load. As the tyre pressure specs are a feature of the car not the tyres, these will have to be typical tyre deformations. The effect of the cushioning from the tyre will require a change in the same direction (i.e. to support a heavier mass, use a stiffer spring); this is a not-insigificant part of the suspension of a car.

The deformation won't suddenly increase as more load is added (your last paragraph), but there may become a point at which it is worth topping up the tyres.

I'm tempted to experiment (on a bike rather than a car, where I can change the loading significantly by sitting on it).

Edit it looks like your answer from Red Act (+1) now goes into more detail on this, but I'll leave mine here as it approaches from a different perspective.

Edit2:* I tried it with a bike tyre which I was pumping up anyway. Basically the stiction in a normal pressure gauge prevents a result from being observed in a realistic situation, and I didn't have the kit to attach a U-tube manometer to a schrader (car style) tyre valve pressing in the centre pin. By running the tyre at about 1 bar and leaning a fair proportion of my weight on it I could get the needle to twitch by around its own width - detectable but not really measurable.

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Thanks, I would love to see the results of your bike experiment! – Marek Aug 28 '14 at 14:40

My answer is that you can fill up to the needed pressure when no load on tire so off the ground or to the spare tire for instance , sy to 2.5 bar . Then placed under the car and fully loaded you still measure 2.5 bar ( if temperature in tire and outside pressure is the same).

Did a test myself with front tire of my car. Mesured 3 times on the ground , then yacked up the wheel off the ground and again measured 3 times , on the ground 3 times , off the ground 3 times and again on the ground 3 times. outside and inside tire temp where the same during the test and off the ground zero load and on the ground estimated about 450 kg on tire. Every measurement was within the range you would expect for inacuracy of this cheap digital device and at the end about 0,03 kg/cm2 lower and I blame that on the little air that escaped by every measurement ( 15 measurements )

So fill up to the pressure needed for the situation and then fill up the car , no problemm.

Mayby if you go to extremes there will be a small diffence but in the range cartires are filled no measurable difference.

The explanation for this is that what goes of the vollume by deflection of tire , comes to it by the belly of the sidewall and the flexing of the tire on places of the ground. Once made a paint picture of it to give an idea of how a tire bends when on the ground. how a tire bends

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Did the tires have time to cool down or warm up after changing load? I would think the temperature change caused by a pressure change in the inside air would not be noticable on the outside surface for some time; And I think the heat capacity of the tire material should be in the range of or above the heat capacity of the air. – Volker Siegel Apr 30 '15 at 11:47

Take an empirical approach; go to a oil-change place. Put the car up on a rack/lift (the kind that supports the frame, not the tires - allows tire rotation during oil change). Have them adjust tire pressure while car is up in the air; have them remeasure when the car is back on the ground (tell them why - give them "scientist for a day" certificate). Because the tire structure provides a significant resistance to deformation (otherwise you'd always be riding on the rims) I'm betting the tire pressure will be higher.

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Obviously the pressure increases with load. If it didn't then you could have an infinite load! If the load is increased, eventually the pressure will increase to the point where the tire bursts (or the load is resting on the wheels, not the tires).

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layman, It takes more PSI to suspend a heavier load with equal square inches. Or the same PSI with more square inches. Rubber tires will mostly just flex to give more square inches. Foot print of tires x psi = weight of car

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This doesn't exactly answer the question – Sean Mar 28 '15 at 2:25

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