Viewing 8 posts - 1 through 8 (of 8 total)
  • How many tyres will a 16g CO2 cannister do?
  • Militant_biker
    Full Member

    I had to use my CO2 inflator the other day for the first time – 23mm road tyre to ~110psi. How much of the 16g cartridge will that have used?

    MrSmith
    Free Member

    more than half. does you pump not leak the rest of the air out when you remove it from the valve?

    mattsccm
    Free Member

    All of it. Or as near as can be measured. Whats left in the cannister will be bugger all and probably won’t be worth faffing about with. Just use a new one each time. My 32mm cross tyres make about 80psi on 1 cannister.

    vdubber67
    Free Member

    It’s something like 130/140psi from one 16oz on a 23 road tyre, and 40psi on 2.1 MTB tyre (from memory and a CO2 supplier’s website)

    simonfbarnes
    Free Member

    the density of carbon dioxide is around 1.98 kg/m³,

    so 16g is 8 litres. Now you just need to know the tyre volume… if it were 2L then you’d get 4 bar or 1L would give 8 bar (120psi) The initial pressure will be slightly lower as the gas comes out cold

    Militant_biker
    Full Member

    more than half. does you pump not leak the rest of the air out when you remove it from the valve?

    Nope, Specialized Air Tool – has a push valve.

    Thanks for the info all – sounds like a new cartridge is required.

    simon – good maths – now all I need to work out the volume of a torus, with a odd shaped cross section!

    Volume of torus = volume of cylinder = (cross-section area)(length)

    Can’t be bothered to try working that out!

    simonfbarnes
    Free Member

    π * (r squared) * circumference 🙂
    circumference = 700mm * π
    which gives 700mL for a 2cm diameter tube or 1L for 3cm. I don’t know how big your tyres are

    Militant_biker
    Full Member

    So that (?(r^2)).(2?R) where r is the radius of the tyre (11.5mm – see my first post), and R is the radius from centre of wheel to centre of tyre. All assuming the tyre is a circle cross section.

    (3.141592654*(11.5^2))*(2*3.141592654*315)
    (415.5)*(1979.2)
    (822357)mm^3
    (822cm^3/ml

    8 bar*822ml=6.576l used = 82% used.

    Yes. I’m bored. And it’s still empty 🙂

Viewing 8 posts - 1 through 8 (of 8 total)

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