# Bushman's Hole diving - does the geometry affect pressure?

This thread on diving Bushman’s Hole (fascinating and tragic story) got me wondering, does the geometry of the cave affect the water pressure? There is a narrow chimney that opens into a large chamber below 150 feet. Is the water pressure at, say, 160 feet in the chamber the same as it would be if the cave were a uniform diameter all the way up?

My first way of thinking about it is, at 140 feet in the chimney, it feels like 140 feet of water above me. I wouldn’t think the pressure would decrease when I go down into the chamber, so the pressure should be independent of the shape.

But then I think, what if the chimney were very narrow, like 1 inch in diameter? I could fill the entire 150 feet with just over 6 gallons of water. It doesn’t make sense that the pressure in the chamber could change by ~4.5 atmospheres by adding that amount of water to the chimney.

I’m obviously wrong somewhere. Where is the flaw?

Pressure is independent of volume and acts equally in all directions. Basically the way to think about it is in 1 square inch tubes for each foot of fresh water in the tube .433 pounds are added. While the total weight added increases per tube you count your pounds per tube is fixed and pressure only measures the second.

It doesnt matter, not one iota.

Wanna be more confused ?

Imagine a 6 foot diameter tube, big enough to just scuba dive in. Imagine it has a very shallow slope. You swim down it for thousands of feet at which point you are, say a 100 feet lower than when you started.

Guess what the water pressure is ? Its not the 3 feet above your head, its the 100 foot.

Some trained rescue types got killed on at least one occasion misunderstanding this.

This part of it makes sense to me. I’m having trouble reconciling the huge difference in volumes when you have a large tank and narrow chimney.

I’m floating in a million gallon tank 10 feet under the surface, with roughly 1/3 atm water pressure. I have a very narrow tube extending 150 feet from the top, fill that tube with 6 gallons of water, and suddenly I’m experiencing ~4.7 atm? How does that work?

In a quiescent body of water with a free surface - that is, a surface exposed to atmosphere - the pressure (above atmospheric) is proportional to the vertical distance below that free surface, and the density of the water (seawater is more dense than fresh water). Nothing else matters (OK fine, gravity matters, if you want to count that as a variable).

Oredigger’s onto it: think of a column of water above you, whose weight you have to bear. You only have to bear the weight of the water that is directly above you, and the weight of that water will be determined by the distance you are below the surface. If the column is really wide, then the water off to the sides (not directly above you) has its weight born by the cavern floor.

If you’ve got a chimney with a diameter of 1 inch, that’s a cross-sectional area of 0.785 square inches. Pour six gallons (50 pounds) of water onto that cross section, and you’ve got yourself about 64 psi. The water depth is 115.5 feet.

If you’ve got a chimney with a diameter of 50 feet (600 inches), that’s a cross-sectional area of 282,743 square inches. It takes 2,161,092 gallons of water ( 18,001,904 pounds) to fill it up to the same depth (115.5) feet. That many pounds of water, borne on that much area, works out to:

18008904/282,743 = 64 psi.

So the width of the body of water doesn’t matter: as it gets wider, you’ve added more water weight, but you’ve added a proportional amount of cross-sectional area, so the pressure at a given depth remains constant.

Look at this way: What keeps the water in the tube from coming into the massive tank below it?

Suppose instead of a closed tank, you had an infinite amount of water below you. What happens when you pour water into the tube? It would all flow down into the infinite depths below you.

In the tank, though, the water doesn’t flow down. Why not? Because it’s being supported by the water in the tank. The tank keeps the water inside it from flowing out. It’s liquid sitting on a liquid. In order to support a 150’ column of water, the water in the tank needs to push up with a weight equal to the weight of the water on top of it. And water is quite heavy. (edit: as Joe Friday showed)

Additionally all the rest of the water in the tank must be pushing back with the proper force, otherwise the water under or in the column wouldn’t stay where it is.

With respect to the ‘suddenly’ portion, think about what happens if this did occur as if by magic. The water below initially can’t support the weight of the column, as it’s not at the right pressure. There’s a pressure wave that needs to reach the walls of the tank (and ‘reflect’) and even out before the whole tank pressure stabilizes.

There is a limit to this, but it wouldn’t be a tube big enough to swim in. Get small enough, and other forces can support the water in the tube aside from the water below it.

OK, ya got your big-ass chamber filled to its ceiling, with a small hole at its top open to atmosphere. The pressure at the ceiling is zero psi.

Now you screw a 150-foot pipe into that small hole, and fill it up to the top of the pipe with water. Pressure at bottom of pipe is 83 psi. The water in the pipe is desperately trying to get into the big-ass chamber, but all the water in the chamber is pushing back. At this point the ceiling of the big-ass chamber, which is at the same depth as the bottom of the pipe is actually pushing down on the water in the chamber with - you guessed it - 83 psi, and the walls of the big-ass chamber at any given depth are pushing inward on the water with 83 psi more than they were pushing before we added the 150-foot pipeful of water.

Double the cross-sectional area at the bottom of the pipe, and the pressure at its bottom will be the same 83 psi (see my previous post), and you will see that same 83 psi at the ceiling of the big-ass chamber. Triple the cross sectional area - hell, make the cross-sectional area of the pipe the same as that of the big-ass chamber - and the pressure at the bottom of the pipe will still be the same 83 psi, as will the ceiling of the big-ass chamber (if there’s any ceiling area left - how big is that pipe now?).

I’m with you through that, but if I swim to the side of the cavern away from the chimney, now I only have 10 feet of water above me. Above that is the cavern ceiling, which we’re assuming is self-sustaining (not held up by the water). How is the weight of the water in the chimney affecting the pressure throughout the cavern to the same extent?

ETA: never mind, the follow-up posts from panamajack and Joe Frickin Friday make more sense to me now. It’s one of those cases for me where the physics don’t always match what we think is common sense.