Quick question

[trux]

I'd say not: the air bubbles are compressed by the pressure differential between the water and air,

Not sure what you mean, but the pressure is proportional to depth. If there were theoretically some bubbles of gas in the great depth, they would have just the same ambient pressure as the water, hence roughly the 1000 bars at 10km depth as Siku wrote.

The problem is the solubility of gases in liquids described by Henry's law. The deeper you go, the more the gas disolves in liquids. Hence at such great depth the bubbles have little chance to exist.

But even if they existed, the density of the gas in bubbles would not be proportional to the pressure, since the compressibility factor changes with pressure. So at the pressure of 1000 bars the compressibility of air would be roughly 2.3 smaller than that of ideal gas (it may be influenced a bit by the presence of other molecules such as CO2). It means that at 1000 bars, the specific weight of air (assuming it would be contained to prevent it dissolves in water) would be 515 kg/m3, still much lighter than water.

 

[FreeRestriction]

So since air is compressable thats why we free fall after x depth? And thus why every foot deeper makes us that much less bouyant?

 

[trux]

So since air is compressable thats why we free fall after x depth? And thus why every foot deeper makes us that much less bouyant?

Yes, of course

 

[Siku]

Damn! I knew I shouldn't have exhaled at 10,000m !

 

[baiyoke]

Ya thats the main point im trying to make to him. Once something is not bouyant enough and sinks at the surface theres no saving the descent all the way to the bottom with exsisting bouyancy.

Yes, it is an unstable system, that will either float more and more, or sink more and more...

From this page: http://en.wikipedia.org/wiki/Cartesian_diver

It might be though that if the weight of displaced water exactly matched the weight of the diver, it would neither rise nor sink, but float in the middle of the container, however, this does not occur in practice. Assuming such a state were to exist at some point, any departure of the diver from its current depth, however small, will alter the pressure exerted on the bubble in the diver due to the change in the weight of the water above it in the vessel. If the diver rises, by even the most minuscule amount, the pressure on the bubble will decrease, it will expand, it will displace more water, and the diver will become more positively buoyant, rising still more quickly. Conversely, should the diver drop by the smallest amount, the pressure will increase, the bubble contract, additional water enter, the diver will become less buoyant, and the rate of the drop will accelerate as the pressure from the water rises still further. This positive reinforcement will amplify any departure from equilibrium, even that due to random thermal fluctuatuns in the system. A range of constant applied pressures exists that will allow the diver either to float at the surface, or sink to the bottom, but to have it float within the body of the liquid for an extended period would require continuous manipulation of the applied pressure

On this page is a stable system using saltwater and freshwater (bottom of page). But it is not a real "on-board" stable system:

http://physics.stackexchange.com/questions/15678/is-mid-water-bouyancy-a-classic-example-of-a-balanced-but-unstable-system

The links are not active, try google the words,I'm having great trouble posting in this thread for 24 hours now...

 

[baiyoke]

Perhaps this works...

Cartesian diver:

http://www.google.co.uk/url?sa=t&rc...GqlN0z&usg=AFQjCNEJjGlfKUiJlbD4phgyWdHyaTmX_Q


Unstable system:

http://www.google.co.uk/url?sa=t&rct=j&q=mid-water-bouyancy&source=web&cd=4&sqi=2&ved=0CGAQFjAD&url=http%3A%2F%2Fphysics.stackexchange.com%2Fquestions%2F15678%2Fis-mid-water-bouyancy-a-classic-example-of-a-balanced-but-unstable-system&ei=N1C-T4q1PNGK6QGYtKlK&usg=AFQjCNHmPDPR510WRmkgjpWGhR69zTG_9Q

 

[Felipe M]

Don't have time to think about it right now but my hunch is that even if you could compress air that much, you'd end up turning it into a liquid anyway! (liquid air lol)

I think in reality many other weird things will happen to it before it is compressed this much as air is obviously a mixture of many individual gases that will liquify etc at different pressures etc...

Yes. gases with the right amount of presure can become liquid.

Other thing that might happen is that some of the gases in the air might just be diluted into water (as nitrogen get diluted in blood with pressure)

 

[trux]

No, you would need much more pressure than that in Mariana Trench to turn the air liquid at the temperatures there are in that depth. I found (not verified) that it would need around 35,000 bars for liquifying air without supercooling it. You would need to cool it down considerably for liquifying it at 1000 bars, but in that case the water would freeze into ice anyway, so there would be no point of speaking about the buoyancy anymore

And the solubility of gas in liquids (Henry's law) is exactly what I wrote in my earlier post

 

[Felipe M]

No, you would need much more pressure than that in Mariana Trench to turn the air liquid at the temperatures there are in that depth. I saw that it would need around 35,000 bars for liquifying air without supercooling it. You would need to cool it down considerably for liquifying it at 1000 bars, but in that case the water would freeze into ice anyway, so there would be no point of speaking about the buoyancy anymore

And the solubility of gas in liquids (Henry's law) is exactly what I wrote in my earlier post

Oh...true...sorry xD

 

[Bill]

I finally found it. Galileo thermometer.

As the temperature of the fluid in the cylinder increases, the pressure drops and the liquid becomes less dense
When the precisely measured floats reach their specific calibrated temperature they drop
As the temperature of the liquid cools, the process is reversed and the floats rise from the bottom of the cylinder

If I read this right, at a specific temperature, the sphere sinks. When the water gets cooler, it rises. Does this meet the requirements?

 

[sai]

So since air is compressable thats why we free fall after x depth? And thus why every foot deeper makes us that much less bouyant?

It may be a little offtopic, but how can you calculate for example how much more weight you have to push up at -40m compared to -20m?

 

[Simos]

It may be a little offtopic, but how can you calculate for example how much more weight you have to push up at -40m compared to -20m?

I doubt you can calculate this reliably - don't forget that apart from the volume of air in your lungs, you also have the air in the suit etc. You can probably measure it in practice though.

If you were a round, air-filled balloon then I could have probably helped you lol

 

[sai]

As a round air-filled balloon, that's offensive... rofl
Guess I'll have to find out it some time then. :t

 

[Bill]

If you take the suit buoyancy at the surface (about the same as the thickness in mm) and add your total lung volume = B take B/3 and subtract B/5 you should be close.
My numbers are (7 +3) B = 10 10/3 - 10/5 = 1.33 kilo or about 1.45 kilo of lead. Like Simos said too many variables so give me + or minus 25%.

 

[Simos]

As a round air-filled balloon, that's offensive... rofl
Guess I'll have to find out it some time then. :t

I think it should be easy to get an approximation if you can get in the water (similar idea to what Bill said) - find out how much weight you need to be neutral at the surface (wearing you suit and full lung, assuming you dive full lung) and the do the same at 10m say. Eg might be 5kg at surface and 3kg at 10m.

Then you'd know that you lose 2kg of buoyancy every time the pressure doubles so when you go from 20m to 40m you'd lose 1.67kg of buoyancy.

Then take the average of 1.67kg and 1.33kg (that Bill calculated) and you get the answer to your question which is, 1.5kg - nice and round number, there you go lol

 

[Antoine]

I doubt you can calculate this reliably - don't forget that apart from the volume of air in your lungs, you also have the air in the suit etc. You can probably measure it in practice though.

If you were a round, air-filled balloon then I could have probably helped you lol

As simos said it would be unreliable to calculate this mathematically for several reasons the main one being that there are multiple air-spaces in the body and in the equipment which cannot be accurately measured. Also the density of the water would have a minor effect.
However, if measured in practise and recorded, there would certainly be a clear pattern from which a equation could be deduced.
The problem is that this equation would only be valid for that person with a precise volume of air in the lungs at a precise location. Therefore, it is completely impractical to attemp to come up with an equation for the net weight(weight-buoyancy) of a person at any given depth.