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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.
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.
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.
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!:)
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.
