Am I the only one who ever thinks about countering g-forces by allowing a pilot to be submersed in liquid instead with neutral boyancy. This would completely eliminate passing out from g-forces at any acceleration factor.

Though of course it would be complicated so I understand why it isnt done. You would need a complicated system to maintain pressure in the tank so it wont fluctuate quickly. But since g-forces are usually transient if done correctly decompression might not be a big issue

@freemo that's the idea behind the g-suit: en.wikipedia.org/wiki/G-suit

It directs fluid to bladders around the lower body to counteract the ρgh gradient the pilot's blood is subjected to, which means that pooling blood in the legs doesn't represent a low-energy state, which in turn means that blood has less tendency to leave the brain and impair function.

@khird yes, im aware.. but if the pilot were completely emersed in water then there would be absolutely no effect from g-forces in that fashion. Any increase in g-forces would present as if the person just was deeper under the water than they are (like a diver)... So at any g-force the the pilot would be immune from passing out (though you'd have to deal with the pressure changes now)

@freemo well, not exactly. Going deeper underwater changes the gauge pressure a diver observes, but not the hydrostatic pressure difference between his head and his toes (ΔP=ρgΔh). ρ and g are effectively constant no matter where the diver is, and Δh is just a function of his posture - his height, if he stands upright.

But for the pilot in a non-inertial reference frame, g is not constant. So when he manouevres, the ΔP changes substantially, while the diver will always see about 17.8kPa (182cm diver standing upright in freshwater) higher pressure at his toes than at his head if standing upright.

Follow

@khird
Let me rethink this now, you have me aecond guessing

· · Fedilab · 1 · 0 · 0

@khird
Ok so yea there would be a pressure gradient that is steeper than usual in the water, your right. Meaning for it to work the tank would need a higher than normal pressure applied since fradient becomes less at depth.

@freemo sorry, what gradient becomes less at depth? dP/dh = ρg no matter what depth h you have.

@khird lets say your 6.6 feet tall and in a tank of equal height. At rest you have 1 ATM at the head and 1.2 ATM at your feet. If you accelerate at 10G that would be 1ATM and 3 ATM, so your feet have 3x the pressure at your feat relative to your head.

However if the tank was at 10 ATM then at res you have 10 ATM at the head and 10.2 ATM at rest at your feet but when accelerating at 10g you would expiernce 10 ATM at the head and 13 ATM at your feet, which is 1/1th the pressure difference at only 0.3x the difference between head and feet.

@khird So the more I think about it you might be right. while at pressure the ratio of pressure difference between head and feet would be much smaller the absolute difference would be the same. I think its the absolute difference that would matter.

With that said it only applies if you are standing verticle. If you are neutrally boyant the pressure gradient should cause you to be pushed horizontally. Thus the pressure gradient would be much less and the difference would be between the front of your body and the back.

I dont think that would be much better though and as you suggest it might not solve anything.

Sign in to participate in the conversation
Qoto Mastodon

QOTO: Question Others to Teach Ourselves
An inclusive, Academic Freedom, instance
All cultures welcome.
Hate speech and harassment strictly forbidden.