# Pressure , gauge pressure and absolute pressure

I was given the following question a while back :

An empty utility vehicle weighs 16.5 kN. Each of its tires has a gauge pressure of 205 kPa. What is the total contact area of the four tires with the pavement (assume that the tire walls are flexible so that the pressure exerted by the tire on the pavement equals the air pressure inside).

Of course, we can use $A = \frac{F}{P}$ (equation 1; A is area, F is force and P is pressure).

However, I was told to use PTotal = PGauge + PAtmosphere , where PAtmosphere is 101 kPa, and then sub PTotal into equation 1.

I am convinced this is wrong, and that we only need to take into consideration the gauge pressure i.e. the pressure difference between the tire and the atmosphere.

If you do need to use the total pressure in equation 1, then why?

What is the difference between a tire being pumped to 205 kPa (gauge pressure) in a vacuum where the atmospheric pressure is 0 Pa, and a tire pumped to the same gauge pressure where the atmospheric pressure is 101 kPa. Surely then, its the gauge pressure that matters, why would the whole thing be determined by its value relative to 0 Pa?

You should be fine as long as you are consistent. If you use total pressure, as they do, then you also need to include the effects of the air pressure on top of the tire pushing down on it with 1 atmosphere of force!

As a thought experiment let's use some really fancy spherical-cow materials. Let's make a gigantic tire out of material that has negligible mass and put 100kN of force on it. If we pump the tires up to slightly above atmospheric pressure, say $P_{gague} = 0.00000001 kPa$. These are the flattest tires on the planet!

By your method, using just gauge pressures, you would see that these tires would flatten out like pancakes until their area was massive.

By using their method, converting to absolute pressures, $P_{total}=101.00000001 kPa$, so by $A=\frac{F}{P}$ you would get an area of 1 square meter or so. This would be an absolute minimum. No matter how flat your tires are, they would never take up more than a square meter to support 100kN. Something's wrong!

On the other hand, if you use all absolute pressures and you account for the fact that there's atmospheric pressure pushing down on the tires as well as the weight of the vehicle, then you would once again be able to show that the world's flattest tires are, indeed, flat.

• so , with the information given, one would have to use gauge pressure Mar 19, 2017 at 22:02
• That's the way I would do it. You could do it both ways, but to do it with absolute pressures you'd have to go through a lot of extra arithmetic which all cancels out in the end. Mar 19, 2017 at 22:04
• but to use absolute pressure, would you need to know the area of the car and tires to find the force the atmosphere is pushing down with Mar 19, 2017 at 22:05
• or the volume of the whole vehicle and apply Archimedes principle Mar 19, 2017 at 22:06
• Yeah, if you account for the difference in pressure from the top of the car to the bottom, Achimedes principle would be needed too. If you neglected that difference (which is small on the scales we are talking about), what you'd find is that there is air pressure pushing down from above, and up from below, and that the differences in those areas is exactly the footprint of the tires, which is the only part of the vehicle not surrounded by air. Mar 19, 2017 at 22:08

An open tube mercury manometer is used to measure the pressure in an oxygen tank. When the atmospheric pressure is 1040 mbar, what is the absolute pressure in Pascal in the tank if the height of the mercury in the tube is

A) 28.0 cm higher

B) 4.2 cm lower, than the mercury in the tube connected to the tank.