• starman2112@sh.itjust.works
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    25 days ago

    Not at 15 feet. I don’t know enough to say how fast the water would be leaving that hole, but it’s maybe a couple hundred pounds of pressure. If he even got caught, it would be super uncomfortable, but he ain’t about to get ∆p’d

    If you wanna see a real crab-in-a-pipe situation, look up that Byford Dolphin everyone’s talking about

    • AnIndefiniteArticle@sh.itjust.works
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      25 days ago

      Let’s convert to metric so we can tell.

      15 ft is about 5 m.

      Water pressure increases by 10,000 pa per meter (rhogh, rho=1000 kg/m^3, g~10m/s^2), so total pressure is 50 kpa, or 1/2 earth atmospheric pressure.

      One side of that hole has ambient pressure of 1 atm. The other side has that plus water pressure totalling 1.5 atm.

      A pressure is just an energy density. Multiply by the cross-sectional area of the interface to get the energy gradient across the interface. An energy gradient is a force. We don’t have a measure of the cross-sectional area of the hole, but if we expect a person to fit through let’s call it 1m^2.

      50 kpa = 50 kJ/m^3, so total force felt across this opening is 50kN which is the equivalent weight of five metric tons.

      Size of the hole absolutely matters. If it’s only the size of a fist (10cm x 10cm) then instead of 5 metric tons it’s only 50 kg of equivalent weight, or about the weight of a person and easily survivable.

    • Spiritsong@lemmy.world
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      25 days ago

      Yeah I read the entire Wikipedia entry on the Byford Dolphin and I almost threw up because how vivid the description is. I think this would be my third time saying this but that’s not a nice way to go (to die) at all.