• CannonFodder@lemmy.world
    2·
    1 day ago

    Hardly matters as the salient amount is the weight of the displaced air, not the weight of the helium or hydrogen. However using hydrogen might lead to a nice flameout.

    • wolframhydroxide@sh.itjust.works
      3·
      1 day ago

      No! What matters is the comparative density, and you must calculate the total mass of the air displaced, as well as the mass of the gas being used to displace it. Hydrogen is half as dense as helium. Is it a tiny mass difference? Actually, no, it’s quite significant.

      Let us be charitable to my ability to calculate and say that the condoms are, themselves, massless (please don’t make me use more complex equations in an internet comment). Given 70kg of displaced air, and using the ISO atmospheric density of air of 1.2250kg/m^3 at 15C, and that hydrogen and helium have densities of 0.0841kg/m^3 and 0.1662kg/m^3 respectively at that temp, each cubic metre of air displaced by hydrogen reduces the effective weight of the balloon by 82 grams more than helium, meaning that, given a fixed number of condoms, they need be filled only 93% as full with hydrogen (reducing strain) or, given a fixed volume, we require only 93% as many balloons. For a 70kg person with massless balloons each containing 40L of gas, this amounts to 1642 condoms for helium, but only 1534 of Hydrogen! That’s a difference of over a hundred condoms that this guy doesnt have to buy.

      If need be, I can provide work shown for this problem.

      • exasperation@lemmy.dbzer0.com
        1·
        13 hours ago

        Hydrogen is half as dense as helium. Is it a tiny mass difference? Actually, no, it’s quite significant.

        7.7% increased buoyancy per unit volume isn’t nothing, but I’m not sure it would qualify as “quite significant” in a colloquial sense.

      • CannonFodder@lemmy.world
        2·
        1 day ago

        Yeah I agree with your math (well I haven’t checked it but it sounds about what I’d expect). But 7% isn’t a big amount within the other approximations going on here. The point is that it doesn’t double using helium even though the density is double because the relevant change is the difference between air and the other gas - which as you say is only 7%. Like if you could somehow make a weightless container that had a perfect vacuum in it, you’d only gain another 7%.