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Air or water pressure can be calculated by the volume of air (or water) column and density. The column volume is usually calculated as the base area times the height. However since on a spherical earth, weight is directed towards the center of the earth, it should be more accurate to think of the column shape as a conical frustum, where the base is slightly smaller in area compared to the top.

Practically, probably the difference between this and say a cylinder is negligible. Theoretically it exist, but I couldn't find any mention of it. Am I missing a certain keyword or a terminology? Or simply a case of "We don't care" ?

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    $\begingroup$ You seem to have a hidden assumption that the pressure at the bottom of a fluid column depends on the shape of whatever contains that fluid. If that is correct, please directly state so in your question. $\endgroup$ Commented Jun 29, 2021 at 19:53
  • $\begingroup$ @DavidWhite, no I don't assume it depends on shape in general. Also am not talking about a physical container, but rather on the open air/water column above us. The volume of this column is not simply the volume of a cylinder, but similar to a cone due to the direction of gravity vectors towards the center of the earth (converging weight lines vs parallel lines) $\endgroup$ Commented Jun 29, 2021 at 19:59
  • $\begingroup$ for this question physics.stackexchange.com/questions/639047/… the highest voted answer and the second one down gave the same result, even thought the second one used the weight of the fluid directly above the area - so it doesn't seem to matter $\endgroup$ Commented Jun 29, 2021 at 21:52
  • $\begingroup$ @JohnHunter Votes can change, and users can change the order of how questions appear, so "highest voted" as well as "second one down" are neither constant nor universal qualifiers. Best to just link to the answers directly. $\endgroup$ Commented Jun 29, 2021 at 22:23

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In the standard explanation using a cylinder, we can calculate the pressure by first dividing the cylinder in small slices. For each slice the total force on the slice should be zero in equilibrium:

$p_{bottom}*A_{bottom} - p_{top} * A_{top} - A \rho * g * h = 0$,

where $p$ is pressure, $A$ the surface area, $rho$ the density, $g$ acceleration and $h$ the height of the slice. Then, if one knows the pressure difference per slice, one can calculate the total pressure.

One important property of the cylinder is that the top and bottom surface of each slice have the same size. Another important property of the cylinder is that the sides are parallel. This means that any forces on one side of the cylinder will be canceled by forces on the other side.

Your intuition is based on the fact that for slices of a cone, the top and bottom surface are not equal. This leads you to think that this means that therefore the pressure increase per slice should be higher than for the cylinder.

However, you do not take into account that for a cone, the sides are not parallel. Therefore, the forces on the sides of the cone do not cancel. The cone is effectively pushed up by the surrounding air pressure. This effect exactly cancels the effect of the bottom area being smaller than the top area.

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  • $\begingroup$ Thanks for the detailed answer. One point though, is that we use a cylinder under the assumption that the "force" lines are parallel to the walls. However in case of very long column the lines are slightly converging rather than truly parallel. The assumption of a cone also assumes force parallel to the cone boundary. I understand though that at the end if there is a difference it would be minuscule $\endgroup$ Commented Jun 30, 2021 at 1:28
  • $\begingroup$ @BasharGharaibeh, pressure depends ONLY on the depth of the fluid and not the shape of anything that contains that fluid. A "cylinder" or "cone" or any other shaped container is totally irrelevant. $\endgroup$ Commented Jun 30, 2021 at 2:34
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In a liquid, the depth is not likely to be large enough for the difference between calculated pressure and measured pressure to be noticeable. In the atmosphere, variations in density due to altitude and temperature would likely mask any such effect.

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  • $\begingroup$ Yes. Even in very long liquid columns (e.g. Mariana trench) the densities might increase with lower depth (colder fluid on the bottom) which further reduces the difference. $\endgroup$ Commented Jun 30, 2021 at 19:30

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