Does car tire pressure change with weight of car load?

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)

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.

As @Tom pointed out, under load a tire sidewall will also deform, and this deformation will cause additional wear on the tire. This is another reason why tire pressure needs to be adjusted to the load.

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...

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.

http://www.sturgeontire.com/images/glossary/crosssection.jpg

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.

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.