Not exactly sure of the measurements of that particular metric equivalent tire but
T = Tire radius (diameter/2), in inches
W = tire section Width, in inches
R = Rim radius (diameter/2), in inches
Then, you have:
r = Radius of tire body in a radial direction = (T-R)/2
w = radius of tire body in the width direction = W/2
The volume of a torus is equal to the length around its center times the cross sectional area, or;
Length = (2 * pi * (R+r))
Area = pi * r * w
So for a 33x12.50R15 tire, you get:
T = 33/2 = 16.5
W = 12.5/2 = 6.25
R = 15/2 = 7.5
r = (16.5 - 7.5)/2 = 4.5
w = 12.5/2 = 6.25
So the volume in cu.in. is:
Length * Area = Volume
(2 * pi * (7.5 + 4.5)) * (pi * 4.5 * 6.25) = 6662 cu.in.
231 cu.in./gallon and 1728cu.in./cu.ft
so then you need to compute one more time
PV = MrT
where:
P = Pressure
V = Volume
M = Mass of the gas
r = Ideal gas constant (air is not an ideal gas, but who cares)
T = Temperature
So how would you apply this in practice? The simplest case is to assume you have a closed system (no leaks) and you have two volumes at different pressures. Then open the valve between them and let the pressure equalize and see what pressure you end up with. Lets put some real-world numbers in here and see what happens.
Assume you had a 20gal tank at 150psi and a 4 - 30gal tires at 10psi and you then hook the tank to the tire (assumes Mass and Temperature remain constant), you'll see something like:
Tank + Tire(s) = Combined
(20*150) + (4*30*10) = newP * (20 + 120)
solve for newP and get
(3000+1200)/140 = 30psi.
So a 20gal tank could air up four 33x12.50 tires from 10 to 30 psi.
Mind you this is not exact but close enough to know you need a compressor to kick in as you would need a 30 gallon tank to fill them up once
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