Question How does pressure increase with depth in water?

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Henry Palmer used a similar technique to measure stages of the Thames in the 1830s. A protected onshore standpipe, hydraulically connected to the river by a horizontal pipe. The level in the standpipe was always the same as the level of the river, but unaffected by waves, currents, boats, etc.
The ancient Egyptians -- long before Pascal and lowwall -- did the same thing to measure the height of the Nile....a horizontal tunnel cut over to a well that had height markings on the wall.
 
OK, don't have too much time to write a detailed answer, but to get back to @lowwall 's point and question with his 2 gauges x and y, the 2 gauges will read the same pressure regardless of their position under the roof or the thin tube, as long as they are at the same depth. What you need to visualize, is the fact that when the gauge is under the roof, and not the tube, there is a seemingly shorter water column directly above the gauge, which creates the confusion. The same pressure as at the bottom of the thin tube is applied by the bottom of your rock wall onto the water column directly above that gauge. The water at the contact point with the roof is pressurizing that roof upwards, and the rock roof is applying an equal reaction force downwards, otherwise, the water in the thin tube would just flow down into the chamber.
If you increase the water column height in the thin tube, you increase the pressure of the water at the roof level, the roof applies a corresponding increased reaction force, which will transfer to that gauge placed below it.

I hope that makes more sense in visualizing how increasing the water level in the thin tube increases the pressure of the water column which does not sit directly below it.
Your explanation helped me understand it a lot and got me to draw this diagram:
1699031532296.png

The skinny pipe is 10 meters long. At the 10 meter mark, the container opens up and at that point x, the pressure is 2 ATA. Now point Y is to the right, it also has to be at 2 ATA. How could that point have a different pressure? Water would have to flow from X to Y until pressure equilibrium was established (and it already is). If say the skinny pipe is empty and you add water, then the pressure at X and Y will change at the same rate and be based upon the amount depth of water added to the pipe.

I hope this helps.
 
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If say the skinny pipe is empty and you add water, then the pressure at X and Y will change at the same rate and be based upon the amount of water added to the pipe.
Not correct. It's based upon the vertical distance of the water column. Not the amount of water. On other words, if the skinny pipe was wider you would need more water to achieve the same height and the same pressure.
 
Not correct. It's based upon the vertical distance of the water column. Not the amount of water. On other words, if the skinny pipe was wider you would need more water to achieve the same height and the same pressure.
That is what I meant to say and changed amount with depth. Good catch! That is what I meant, but I obviously chose the wrong word.
 
Okay, I’m going to add a few complications, because it is applicable to all divers. First, about pressure. We need to understand the units of the pressure. In U.S. units, that is PSI, or Pounds Per Square Inch. It doesn’t matter how large the surface area is, the pressure in PSI will be the same at the same depth. BUT, if you measure the force against a specific area, then you need to multiply the pressure in PSI by the number of square inches of surface area. This is why submarines have a “crush depth,” that is a depth where the force against the sub’s walls becomes greater than the strength of those walls, and the sub is crushed. This recently happened to the OceanGate Titan submarine on its trip down to the Titanic.



Now a number of people have stated that one atmosphere in the water is 33 feet of depth, but neglected to say that this was for seawater, and not freshwater. For freshwater, that depth is 34 feet, due to the lesser density of freshwater verses saltwater. How does this affect divers? Well, depth gauges are normally set to depth in salt water, and in order to use them in freshwater, a correction factor needs to be applied. I will post a chart which details these corrections for different types of depth gauges when I got to my main computer (I’m typing on an iPad right now).

This brings up another assumption that has been going on here in this thread, and that has to do with the 14.7 psi at the surface (one atmosphere of pressure). But we as divers don’t always dive at sea level; we sometimes dive at an altitude, which means that our assumptions about absolute pressure need to be adjusted for the altitude. At 4000 feet altitude, the atmospheric pressure isn’t 14.7 PSI, but is reduced. So if we were diving in a cave with a very limited opening (which doesn’t matter for pressure, as described multiple times above), but that cave is at 4,000 feet altitude, then the absolute pressure will be different from the same depth at sea level in salt water. See how fun this physics can be? This altitude factor affects how decompression is computed, and again, I’ll provide a table from the 1970s about altitude diving and correction factors. Looking at the Cross Tables, a dive to 60 feet in Clear Lake, Oregon (altitude 4,000 feet) would be the theoretical dive to 69 feet, and if using decompression tables we would need to use the 70 foot level to calculate our decompression (or no decompression limit, NDL). Luckily, these corrections have now been built into our dive computers (if you set them up for altitude diving). So an understanding of these tables in not currently required, as long as the correct settings are set into the computer.

SeaRat
 

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...Fix your pressure gauge in position just under the roof somewhere away from the pipe. Take a reading. Now double the cross section of the pipe, so it's only half as high. Did your reading change?

Now excavate just enough of the roof for the entire column of water in the pipe to collapse into the pit. The gauge is now under a couple of cm/inches of water. Did the reading change?

I would expect the pressure reading to stay the same in all three scenarios. How could it be different? The all have the same weight of water above the gauge. Why would the shape of the column matter?
Answering myself as I try to understand this. Potential energy is different in all three scenarios.
 
Glanced through this thread. Long time since I did physics. Seems like everyone ended up agreeing with whoever originally derived

P = density *height*gravity

Which is a fascinating conclusion genuinely. The capillary action point is interesting. I guess wherever the threshold is for capillary action would be the point at which the law breaks down secondary to the opposing force? But I guess then you just have to calculate net force from the two vectors? Or did someone earlier say pressure from a fluid is not a vector force? Then how do you factor in the opposing force of capillary action? This is deeply difficult to fully grasp without the math.
 
Something interesting is that water pumps are often rated in feet of head. How far the pump can raise the elevation of the water. Nobody asks how big the pipe or tank is because that doesn't matter for feet of head.
 

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