# How much does tire pressure affect the weight of the wheels

Kind of inspired by this question, but something I've been thinking about for a while.

How much does the air in a bike tire weigh? Is it an appreciable amount? Is there a point where using a wider tire, like at 28c at 80 psi would be lighter than a 25c tire at 100psi? Obviously, this depends on the specific tires used. I don't have a scale precise enough to measure, and I don't have the math/physics knowledge to figure this out.

• I seriously doubt that it's an appreciable amount, but this is a fascinating question. I hope someone has the knowledge and/or equipment to come up with an answer. – jimchristie Apr 19 '14 at 14:12
• Once you've got your head around the eloquent and informative answers below, there's also the schools of thought that say you should inflate tyres from bottled gases rather than air. Not only will this affect the mass of the wheel it will also affect how quickly the tyre deflates. Or possibly not. – PeteH Apr 19 '14 at 20:38

## 7 Answers

The ideal gas law (which is a good approximation in this case) says PV=nRT where P is pressure, V is volume, n is mols of gas, R is the ideal gas law constant, and T is temperature in Kelvin.

Thus, solving for n, we see n = (PV)/(RT). Then, assuming air is made up of {gas1, gas2,...} with fractions {p1,p2,...} (so p1+p2+...=1) and corresponding molar masses {m1,m2,...}, the mass of air in a tire is (PV/(RT))(p1*m1+p2*m2+...). So, what we see is that the mass of air in a tire is directly proportional to the volume of the tire and directly proportional to the pressure in the tire, and inversely proportional to the temperature of the air in the tire.

We will make the following (reasonable) assumptions: Assume the temperature is around room temperature (293 Kelvin) and the volume of the tire regardless of pressure is the same (primarily determined by the shape of the rubber, assuming not severely under/over inflated). For convenience, air is about {nitrogen, oxygen} with {p1,p2}= {0.8,0.2} and molar masses {28 g/mol,32 g/mol}. Thus, under these assumptions (V is fixed, and T is fixed), the mass of the air in tire grows linearly with pressure.

So, the mass of air in a tire of volume V and pressure P and temperature T is about (PV/RT)(0.8*28+0.2*32) grams. It may be better to write it as "P ((V/(RT)) (0.8*28+0.2*32)) grams" noting that V/(RT) is a constant for us.

Since I don't want to put the units into wolfram alpha carefully, you can put in the entry "(7 bar* 10 gallons)/(ideal gas constant*293 Kelvin)*(0.8*28+0.2*32)" and read the result off in grams (ignoring the unit it says there) to get an estimate for the weight of air in a 7 bar (~100 psi), 10 gallon volume tire as around 313 grams. Is 10 gallons reasonable? No.

Lets be crude about estimating the volume of a tube using a torus. The volume of a torus is V=(pi*r^2)(2*pi*R) where R is the major radius and r is the minor radius. Google will calculate it for you (and has a picture of what major and minor radius is).

I can't be bothered to actually go outside and measure these things, but lets be crude and use a massive tire. Say the minor radius is 2 inches, and the major radius is 15 inches (this is probably larger the size of the tire on something like a Surly Moonlander). This has a volume of about 5 gallons. If you were a nutcase and running this at 7 bar, it would be around 150 grams of air. At a more reasonable 1 bar or 2 bar, youd be at 45 or 90 grams.

What about a thin road bike tire? Lets also assume the major radius is around 15 inches, and the minor radius is around a half inch. Thats around 0.3 gallons of volume. Plugging into our formula, at 7 bar, we see that this is about 9 grams. At 10 bar, a whopping 13.5 grams.

• For reference, according to the bag of potato chips i have next to me (Lays Wavy Hickory Barbacue), one chip is about 2 grams. So, if you're a road biker and you're worried about the weight of air in your tires, note that one serving of potato chips (28 grams) is more than the air in both of your tires. All of which are much lower than even a light tire (lightest I can find is 130 grams). – Batman Apr 19 '14 at 16:22
• +up for the examples calculated, very interesting. – olee22 Jul 12 '14 at 8:20
• What are these gallons and inches of which you speak? – andy256 Jul 13 '14 at 5:31
• I'm American =) – Batman Jul 13 '14 at 5:41
• @Batman - But what is the drag coefficient of that potato chip? – Rider_X Aug 13 '14 at 16:46

To calculate the weight of a gas you need the volume, pressure and temperature.

A bike tyre is a torus (doughnut) with volume given by the formula:

V=(πr^2)(2πR)

where R is the radius of the wheel and r is the radius of the tyre. For a 700c25 tyre, R will be 311mm and r will be 12.5mm that gives a volume of 9.59×10^5 cubic millimetres or 0.000959 cubic metres.

Pressure is 100 PSI, which is 689475 Pascals.

Room temperature is about 295 Kelvin.

Using the Ideal Gas Law:

n = PV / RT

where R is the gas constant, gives n as 0.27 moles of gas.

To keep things easy, lets assume the tyres are filled with 100% nitrogen. 1 mole of nitrogen weighs 28g so the gas in the tyre weighs 7.56g.

Just in case you prefer general knowledge over physics: the density of air at a reasonable temperature is around 1.2 kgm-3.

The volume of your tire (accepting Tom77's answer) is 0.000959m3.

So the mass of air in it at 15°C and atmospheric pressure is around 1.1g.

Then we do need one bit of physics, the relationship between mass and pressure for a given gas in a given volume and temperature is linear. This comes from just Boyle's law provided we're prepared to believe that twice as much gas at the same temperature and pressure has twice the mass. Which is a lot like saying that two buckets of water weighs twice as much as one bucket of water, so hopefully not controversial ;-) So I've cleverly(?) avoided needing to know the ideal gas law and the value of the universal gas constant in favour of a direct crib off Wikipedia measurement of air.

Atmospheric pressure is 15 psi (ish), so when you measure 80psi that's really 95, so it's 95/15 = 6.3 times as dense as the external air. So the answer is 6.3 * 1.1.

7g (0.2 ounce), at the 15°C stated by the Wikipedia article for my estimate of the density of air.

If you change the temperature from there then the pressure changes linearly, according to the combined gas law (or "Gay-Lussac's law" apparently is the name for this component of it, I had to look this up), provided you measure temperature in Kelvins not Celsius. 0°C is 273.15K. So to consider variations in temperature and pressure starting from my value, just multiply the 7g in proportion. Adding 3°C is about 1%, so the difference is smaller than my margins of error. Adding 20psi to the pressure is about 20%, or another 1g. The mass of air is already far smaller than the weight of the wheels. So pressure has more effect than temperature for the examples you give but no, it does not appreciably affect the weight of the wheels.

There's also another small confounding factor, which is that inner tubes are stretchy and so the volume does increase a bit as the pressure changes, requiring a little bit more gas. But not much.

• Well now, is it 7.56g or 7 even?? You guys make up your mind!!! – Daniel R Hicks Apr 19 '14 at 17:29
• @DanielRHicks: Right, we don't even agree to 1sf! – Steve Jessop Apr 19 '14 at 17:30
• @DanielRHicks: A value to withing 10% is often useful. You didn't specify the conditions in your original question, so the responders had to guess. In fact, nobody has specified whether the pressures are absolute or gauge (relative to atmosphere). This makes more than 10% difference at usual bike tire pressures. Just the fact that we are talking single digits of grams says the air mass difference will be dominated by tire/rim mass difference-a useful fact. – Ross Millikan Apr 20 '14 at 17:00
• @RossMillikan - I was being facetious. A value within a factor of 10 is good enough here. – Daniel R Hicks Apr 20 '14 at 18:26

actually it affects more than has been suggested. I tested the theoretical derivations. I have a super single (huge) truck tire. At 115psi it weighed 219lbs. At 0psi it weighed 214lbs. Using V=(πr^2)(2πR) and n=PV/RT (r=0.178m and R=0.15m) i got 1.65lbs of air weight. But the actual difference was 5lbs. I eyeballed the r and R so those are major estimates, but i didn't expect to be off by 4lbs!:) I had to lift the tire to mount it on the truck as a spare and I appreciated the 5lbs off its weight!:)

• Welcome to Bicycles SE. We ask that you write to the best of your ability on this site. This means proper capitalization, proper punctuation, and complete sentences. You can edit your answer with the "edit" button at the bottom of the answer. If you do not do so, it is likely to be downvoted, flagged for moderator intervention, and possibly deleted. – jimchristie Oct 29 '15 at 23:08
• A truck tire isn't a torus -- its closer to a washer (i.e. a cylinder with a concentric cylinder removed). If the tire is w wide, inner radius r and outer radius R, you should use pi (R^2-r^2) w to estimate the volume. A bicycle tire or motor cycle tire is closer to a torus than to a washer, which is why I used it in my calculation. Also, this tire has a near 11-12 foot diameter, which seems enormous! – Batman Nov 18 '15 at 17:22
• I don't know what kind of scale you have been using, but some scales are notoriously imprecise. I used to have a bathroom scale that could be off by a full percent of my body weight, and each time I stood on it, it would show a different weight, even when done straight in a row. So, I guess that the 4lbs discrepancy may have more to do with the scale than with the actual weight of the tire. – cmaster - reinstate monica Aug 27 '20 at 10:20

Even though this (actually, these, as there are three) question(s) has(have) been answered, like, a year & a half ago, it's early (well, it was when I started typing this). And raining. So I'm not riding. So here I am...

Anyway, my answer is really crude (as in rough, not precise, inexact, approximate, but close enough for government work), but should be well within the indicated parameter (noted in one of the comments) of "A value within a factor of 10 is good enough here".

Q1: "How much does the air in a bike tire weigh?"

A1: In short: less than 12 to 16 grams (for a 700cx23 tire at 105psi).

The "12 to 16" values are based on CO2, which is, I believe, somewhat heavier than air. However, the difference is well within the "good enough" factor of 10.

The "12 to 16" values were determined via experimentation. That is, a 12g CO2 cartridge fills a common 700c x 23mm tire to about 80psi. A 16g CO2 will fill the same tire to about 105psi. (The unknown precision of my pressure gauge notwithstanding.)

Q2: "Is it an appreciable amount?"

A2: That depends: how much do you appreciate a few grams of air? :)

Q3: "Is there a point where using a wider tire, like at 28c at 80 psi would be lighter than a 25c tire at 100psi?"

A3: No.

This is because 80psi of air is only a few grams (2 to 4?) lighter than that at 100psi (in a 700c X 23mm tire), and I'd guess that a 28mm tire is more than those same few grams heavier than either a 23mm or 25mm tire, and the larger tires will contain more air, somewhat offsetting the reduced amount of air due to lower pressure.

No one has really addressed the size versus pressure part of the question.

Nominally different sized tires will have about the same mass of air. As the size of the tire goes up the design pressure goes down. The contact patch must support the weight of the rider. Assume bike with rider is 100 lbs on the rear wheel. At 100 psi the size of the contact patch is 1 square inch. On a bigger tire you can drop the pressure down to get a bigger contact patch. At 80 psi the same rider would have a contact patch of 1.25 square inches. You cannot just reduce the pressure on a small tire to get a bigger contact patch without without banging the rim.

Lets assume the n in PV=nRT the same in all diameter tires. If so what would the relationship of diameter to pressure be? S for small and B for big

nS = Pb * Vb / (R * T)
nB = Ps * Vs / (R * T)
the assertion (test) is the nS = nB
Pb * Vb / (R * T) = Ps * Vs / (R * T)
R * T drops out
Pb * Vb = Ps * Vs
Pb / Ps = Vs / Vb
Pb / Ps = (πrS^2)(2πR) / (πrB^2)(2πR)
Pb / Ps = rS^2) / rB^2
Pb / Ps = (rS/rB)^2

If Pb / Ps = (rS/rB)^2 then the two tires will have the same mass of air.
If the pressure is inversely proportional to diameter squared the two tires have the same mass of air.

So let's test at 25mm 100psi and see what pressure at 28mm is the same weight
Pb = (25/28)^2 * 100
Pb = 79.7 PSI

So in your example of 28c at 80 psi versus 25c tire at 100psi
The answer is almost exactly the same mass

Not the question but if you assume the same mass how does contact patch size scale with diameter. Contact patch is load / pressure So that ratio is
(Lb / Pb) / (Ls / Ps)
but Lb = Ls so
Ps / Pb
sub in for Pb from above
Ps / Ps * (rS/rB)^2
1 / (rS/rB)^2
(rB/rS)^2

So if you keep the mass in the tire constant then the contact patch goes up with the square of the diameter. And that makes sense since area is proportional to diameter square.

Why would you keep the mass the same? Because it makes sense. Consider the force the beads must withstand. If the mass is the same then the total force on beads is the same. Same number of molecules will produce the same force. The Force is proportional to pressure * area. Force is proportional to r^2 * P.
Consider the ratio of the force on beads from big diameter to small at constant air mass.
Fb / Fs
Pb * rB^2 / Ps * rS^2
sub for Ps again with constant mass assumption
Ps * (rS/rB)^2 * rB^2 / (Ps * rS^2)
1
If you keep the number of molecules constant then the total force on the beads is constant regardless of tire diameter.

I know a lot of you are going to think I am full of BS. But various sized diameters have about the same number of molecules in them. As the diameter goes up the contact patch size goes up with the square of the diameter. So a 2" tire will nominally have 1/2 the pressure and 4 x the contact size of a 1".

Even at the lower pressure a larger diameter is less susceptible to pinch flats because it has further to travel to the rim and it builds area faster relative to deflection. I know even more of you are not going to believe me on this but even at the lower pressure the pinch resistance is proportional to the diameter squared.

Wait, what? The above answers comment on the mass of the air inside a tire (which I assume is what is being asked). However, what is the weight difference from an empty to an inflated tire? Buoyancy says zero!

Only measure from this point on is the change of moment of inertia of the tire i.e. how easy it is to accelerate.

• I'm not so sure about that. The air in the tire is compressed. It is denser than the air surrounding it (outside of the tire), I don't think that buoyancy will make up for all that is lost as you cram more and more air into the tire. Am I missing something here? – dlu Oct 27 '15 at 16:25
• @dlu, I didn't consider that (so may answer is wrong). Still wouldn't weigh as much as in vacuum. – Vorac Oct 27 '15 at 16:31
• :-) depends, I suppose, on where you find the vacuum (and being pedantic about the difference between weight and mass). – dlu Oct 27 '15 at 16:50
• The weight of the air will be zero only for zero gauge pressure. – Vladimir F Aug 27 '20 at 9:57