A Question On Pressure

Maybe try to search for: "how to calculate atmospheric pressure with depth"

The short answer is Density of Air versus depth.

Check these out:

https://appmeas.co.uk/resources/pressure-measurement-notes/how-is-pressure-related-to-altitude-and-depth/

http://theory.uwinnipeg.ca/physics/fluids/node8.html

Let's just pretend that "atmospheric pressure" adds 1 bar of pressure.

At the deepest point on Earth, the pressure would be 3.66 bar.

Could you survive at the deepest point of the ocean, if there was no ...

A lot will depend on the temperature of the air column....

http://nopr.niscair.res.in/bitstream/123456789/2506/1/IJRSP%2037%281%29%2064-67.pdf

https://en.wikipedia.org/wiki/Kola_Superdeep_Borehole

https://en.wikipedia.org/wiki/Density_of_air

Deep mine barometric concerns have been studied. That paper concluded that nitrogen narcosis and oxygen toxicity would only become a concern at about twice the depth of Challenger Deep.

If the water were removed from the oceans and replaced with the air currently available ( with water in the oceans ) . Would your hypothesis still hold true ?

It seems now that we have more air and less space, if you remove the water, then you would have less air and more space.

Actually, I was taking into consideration the volume of the oceans. If the average depth of the oceans is 10000 ft and the oceans cover 3/4 of the globe, then I estimated the atmospheric pressure at the current shoreline (sea level) would be what the current pressure is at 7500 ft (3/4 x 10000), based on a volume measurement.

Actually, 10000 ft pressure would be a more accurate estimate for the shoreline pressure, but even this is not exact. The ocean basin is a container with an irregular bottom contour, whereas the current ocean surface is very flat.

Gas concentrates at the bottom of a container, so to get a good estimate would require getting a contour map of the world (oceans and land), say with resolution 100 ft. With a contour map, you can compute the mass distribution of the atmosphere by multiplying the area at each elevation by 100 (layer thickness) by the density at that elevation. Doing this integration (with oceans filled) to a very high altitude where the density is negligible would give a value of the total mass of the atmosphere.

If you do this contour integration again, this time with the oceans empty using the ocean bottom contour and assuming a density value, you could come up with a new atmosphere total mass. Adjust the density value until the new total atmosphere value is the same as the current value.

You still have the problem that the density is inversely proportional to temperature, so you have to assume a temperature profile. It really is very difficult to come up with an accurate estimate.

Good explanations.

In Death valley, it is cooler at the top of the valley, as you descend( below sea level ) the temperature increases. Would it be possible to use the differential as a base in approximating the temperature at lowest and highest elevation ?

That's a very good point. Death Valley might be a good small-scale model of how dry ocean beds might behave.

"Death Valley has a subtropical, hot desert climate (Köppen: BWh), with long, extremely hot summers and short, mild winters, as well as little rainfall. As a general rule, lower altitudes tend to have higher temperatures. When the sun heats the ground, that heat is then radiated upward, but the dense below-sea-level air absorbs some of this radiation and radiates some of it back towards the ground. In addition, the high valley walls trap rising hot air and recycle it back down to the valley floor, where it is heated by compression.^[14]"

https://en.wikipedia.org/wiki/Death_Valley#Climate

The temperature gradient in the lower atmosphere is primarily caused by solar radiation heating the ground, which in turn heats the air.

As air rises it expands and cools due to the decreased pressure and when it descends it is heated by the compression, a phenomenon known as the adiabatic lapse rate.

As long as the temperature gradient caused by the sunshine is less than the adiabatic lapse rate the air will be stable. When the temperature gradient exceeds the adiabatic lapse rate, the air begins to circulate, limiting the temperature gradient.

In an oceanless earth, the air pressure would be greater in the ocean basins, the rate of change with of pressure with elevation (the slope) would be greater and therefore the adiabatic lapse rate and maximum temperature gradient would be greater.

(An extreme example of this effect is Venus, where the surface pressure of 90 bar can support a very large temperature gradient.)

The oceanless earth would have hot dense air in the ocean basins that would make Death Valley seem like a nice place.

"I do think a significant correction might be needed that would further reduce expected pressure in most places on the waterless globe due to the greater amount of atmosphere stored at the deepest points due to the depth dependent density."

Good point, I agree. My initial estimate was a bit of a fudge.

There's another calculator here

https://www.mide.com/pages/air-pressure-at-altitude-calculator

Which works if you're boring a dry hole, but as Rixter and TINAC have pointed out removing the water reduces the total "height" of the air. Here's a picture which shows how much water and air (at STP _{Standard Temperature and Pressure}) there is on earth:

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The water sphere (blue) in this computer visualization measures 1390 kilometres across and has a volume of 1.4 billion cubic kilometres. This includes all the water in the oceans, seas, ice caps, lakes and rivers as well as ground water, and that in the atmosphere. The air sphere (pink) measures 1999 kilometres across and weighs 5140 trillion tonnes. As the atmosphere extends from Earth it becomes less dense. Half of the air lies within the first 5 kilometres of the atmosphere. The spheres show how finite water and air supplies are.

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its interesting on the size comparison when you group the water like that.... going off topic here...

when comparing the earth to size... if the earth was proportionally (Highest mountain deepest trench kept in tacked proportionally) shrunk down to the size of a pool cue ball, the earth would be smoother then an actual cue ball.

They have misinterpreted the tollerance in the link you provided. The tollerance is for diameter. It is unlikely to be a tollerance for smoothness/roughness as typical pool balls are far far more smooth.

The link claims the +0.005" is for smoothness, but that is pretty coarse. Far more course than a pool ball. 120 grit sandpaper is more smooth with a grit size around 0.0045".

Pretty funny it's called 'bad astronomy'.

Thank you all for your answers. They were all helpful. There was a vital flaw in the original experiment which was that no allowance was made for the air having replaced the removed water. This was well explained by truth is not a compromise in reply 8. I think I will have to accept that it would be very difficult to calculate the pressure in the trenches due to too many variable factors. My main concern was the well being of any explorers that descended to these dry depths. I will go with the assumption that the air pressure, density and temperature will still be adequate for survival. Again, thanks every one. Time to welcome the dolphins back and give them some fish.