Cave-diving and water pressure question

[ClayJar]

Okay, you're trying to wrap your head around the "Why?", so I'll give that a shot.

Fundamental Rule: At any point in the liquid, there is a pressure. The pressure at that point is exerted equally in all directions.​

(If you don't want to take this as a given, I'll have to type a lot more. :biggrin

Important Concept: The pressure of a certain bit of water comes from the pressure of the bit of water above it plus the extra pressure due to the weight of that bit of water above it.​

This is basically where the whole diving thing comes from. The weight of the water is on you, basically.

The Kicker: If the pressure on a given bit of fluid is not the same as the pressure pushing back, the fluid will move.​

If a football player is shoving you from behind, you fly forward. If two identical football players are shoving you, one from each side, you feel the pressure, but you don't go anywhere.

This bit of water here (*imaginary pointing*), 10 feet down from the surface, is pushing in all directions. The bit of water below it is pushing hard enough to hold up that bit of water and still have enough pressure to balance the pressure from the bit of water above it. The bit of water on the side, however, doesn't have anything extra to hold up, so it must be pushing back with the same pressure.

Now, when you look at the air pocket, for the water's surface inside the pocket not to be shooting upward into it, the air pocket must be pushing back with the same pressure as the surface of the water. Of course, since each bit of air in the air pocket weighs quite a bit less than an equivalently-sized bit of water, the pressure inside the air pocket decreases far less as you go up inside it -- we can even assume that it stays the same (assuming it's not several thousand feet tall).

Now, as far as the rock is concerned: The rock is (hopefully ) not a fluid. It can "push back" as gently or as hard as is necessary. If you lean your motorcycle against a concrete wall, it presses back gently. When you ride away, it no longer has to press against you at all. If you crash the motorcycle into the concrete wall, it presses back quite a bit harder (although you probably won't be around to notice).

Rigid solids don't matter, then, to pressure equations. It's only the fluids they contain that have any effect. The cave is just a container that holds in whatever pressure there is. (Think about two big refinery tanks with a pipe connecting them at the bottom. The pressure in the pipe is based on the depth of liquid in the tanks, not on the size of the pipe. If the tanks are almost empty, the pipe will be holding in much less pressure than if they were almost full.)

Anyway, I've glossed over this a bit, but it may help with some of the concepts (if someone hasn't posted better while I've been typing ).

 

[Gombessa]

Regarding the pocket of air: Remember basic scuba class. If you take a jar without a lid, hold it upside down to trap air, and bring it down into the water with you, that "bubble" of air in the jar will get smaller and smaller as you descend, right? That's the water pressure forcing up into the jar through the opening and compressing the air. The sides of the jar are your cave. Any pocket of air in a cave will be pushed upon by water pressure (regardless of the direction) until the air is at ambient pressure with the water around it.

 

[Blackwood]

C is 2ATA.

 

[b1gcountry]

Think about it this way, if the rock at 66' was putting more pressure on the water, it would push the water at the top high enough to offset that pressure. So there can never be any greater pressure than what the water and atmosphere is putting on you.

Tom

 

[perdidochas]

what about if there was hypothetically a pocket of water just a bubble under 60ft of rock completely enclosed and surrounding it was the sea. think of a pillar sticking out of the sea with a bubble of water trapped in the middle.... What would the pressure be?

If the bubble is contigious with the sea, it would be like in 60 ft of water. If it's totally disconnected, it would depend on the pressure of the bubble.

 

[Mudd]

Thanks everyone for posting. I understand it now.

I was being too simplistic - locking onto solely (and literally) the idea that the water above this point *pointing to a water molecule* as causing the current pressure.

 

[Thalassamania]

The rock supports itself, the only thing pressing on you is the water column.

 

[Mudd]

C is 2ATA.

LOL - sorry for the flashback there

 

[muddiver]

what about if there was hypothetically a pocket of water just a bubble under 60ft of rock completely enclosed and surrounding it was the sea. think of a pillar sticking out of the sea with a bubble of water trapped in the middle.... What would the pressure be?

The simplified answer is: the pressure on the bubble is the weight of the rock above the bubble that is not offset by the supporting effect of the rock around the bubble, plus the weight of the water above the rock column, if the rock is impervious to water intrusion.

If the rock is fractured in any way or is pourous like sand stone, then the pressure is the weight of the rock on the bubble minus the buoyancy of the water displaced by the volume of rock plus the weight of the water above the bubble.

 

[DA Aquamaster]

Atmospheric pressure is about 15 psi. If I go outside and look straight up I can see all the way up to the "surface" of the atmosphere (about 60 miles for most practical purposes) . If I then go in my house and look up, I see the ceiling about 2 feet above my head - but the pressure does not change because the air pressure at my "depth" under the "surface" pushes equally in all directions - up, down, sideways, etc.

Water does the same thing in a cave, so 60' of depth is 60' ft of depth even if the pressure has to be transmitted sideways through a half mile of cave passage.