XXL [SMBC] (discuss.online)
submitted 2 weeks ago by to c/comicstrips@lemmy.world
 

cross-posted from: https://discuss.online/post/47437499

http://www.smbc-comics.com/comic/xxl

Alt textI haven't dated in a long time, so this may be unrealistic.

Bonus panelBonus panel

you are viewing a single comment's thread
view the rest of the comments
[–] 66 points 2 weeks ago* (last edited 1 week ago) (21 children)

Edit: accedentally used the wrong density. Should be the desity of the surounsing medium (air) at 1.23 kg/m^3^- density of helium making it slightly more fesable. I.e. i was wrong on almost an order of magnitude, but its still infeasable... thanks @exasperation, @noneofyourbuissness

So... hear me out. Yes i know its just a comic... im just autistic. Soooo...

Assuming the guy is average weight (70 kg) the total force of gravity on him is Fg=70Γ—g N. The force of boyancy would be Fboyancy=dheliumΓ—VdisplacemwntΓ—g N.

If thease are in equilibrium (the guy floats): Fboyancy=Fg

dheliumΓ—VdisplacemwntΓ—g =70Γ—g

If dhelium is 0.178 kg/m^3^ then 70/0.178 = Vdisplacemwnt= 393.5 m^3^β‰ˆ 1.9 m^3^ per condom.

Like i know condoms are super strechy (can hold about 40L air befor burating) but wow. Thats like a phone booth per condom.

Did you need this info?... no. Was i compelled to calculate this? ... no .... stay in school kids.

  • source
  • hideshow 21 child comments
  • [–] 35 points 2 weeks ago (2 children)

    Classic rookie mistake.

    Not every store has an unlimited supply of condoms, especially the extra large ones*.

    He brought 200 at her store, and emptied the shelves. He probably hit a dozen places and brought all they had on hand.

    *I shall modestly refrain from enumerating the many, many times I myself have needed to purchase a vast number of extra large condoms...

  • source
  • parent
  • hideshow 2 child comments
  • [–] 15 points 2 weeks ago (1 child)

    Density of air is about 1.23 kg/m^3 at standard pressure/temperature at sea level.

    Density of helium is about 0.16 kg/m^3.

    That's a delta of 1.07 kg/m^3. So for a 70kg man, they would need helium balloons equal to 65.4 m^3, or 65400 liters. Spread across 200 condoms, that's 327 L per condom.

    I'm not sure where my numbers come up differently from yours, but that's still not feasible.

  • source
  • parent
  • hideshow 1 child comment
  • [–] 6 points 2 weeks ago (12 children)

    What if he used Hydrogen though

  • source
  • parent
  • hideshow 12 child comments
  • [–] 9 points 2 weeks ago (10 children)

    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.

  • source
  • parent
  • hideshow 10 child comments
  • [–] 10 points 2 weeks ago* (last edited 2 weeks ago) (7 children)

    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.

  • source
  • parent
  • hideshow 7 child comments
  • [–] 3 points 2 weeks 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%.

  • source
  • parent
  • [–] 2 points 1 week ago (5 children)

    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.

  • source
  • parent
  • hideshow 5 child comments
  • [–] 3 points 1 week ago* (last edited 1 week ago) (4 children)

    Tell that to the guy who has to buy a hundred eight extra condoms!

  • source
  • parent
  • hideshow 4 child comments
  • [–] 1 point 1 week ago (1 child)

    I dunno, anything that applies to the guy who has to get 1600 condoms probably applies to the guy trying to get 1500 condoms. That doesn't seem like a significant difference to me.

  • source
  • parent
  • hideshow 1 child comment
  • [–] 5 points 2 weeks ago* (last edited 2 weeks ago)

    The other person who replied to you is deeply wrong. See my reply to them here for complete calculations. You can't simply ignore the mass of the gas with which you're replacing the air, despite their suggestion. If the mass of what's inside the balloon doesn't matter, you might as well fill it with lead. Hydrogen is something like 7% More efficient than helium for balloons. You have a good idea, and hydrogen balloons don't just explode for no reason. They're dangerous, but saying that they're necessarily going to flameout is ridiculous

  • source
  • parent