Amounts of pressure within 33 ft?
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UnderSeaBumbleBee
Contributor
Each inch you descend underwater the pressure increases. At 33 feet, you have two atmospheres of pressure. at about 16. feet you have 1.5 ATA's
w ripley
Contributor
Well think about it: Do you have to clear your ears once for every 33 feet you drop, or continuously as you go down?
I am not sure I understood your question correctly, but as I understand it, the answer is no.
I assume you are talking about the effect on air spaces.
33 feet of sea water is the same pressure as air pressure at sea level. The change in pressure is actually determined by a simple equation: V(1) * P(1) = V(2) * (P2)
In simple terms, if you are at 33 feet, you are at 2 atmospheres of pressure--one for the air and one for the water. If you have 1 liter of air in a flexible container at the surface and take it to 33 feet, the formula tells you what will happen:
1 * 1 = V(2) * 2, so 0.5 = V(2). You container will be half its original size.
If you want to figure out the effect at intermediate depths, you have to figure out the atmospheres at each depth. At 10 feet, it will be 10/33 +1 (for the air pressure), or 1.3 atmospheres. At 20 feet it will be 20/33 +1, or 1.6 atmospheres.
At 10 feet the formula will be 1 * 1 = V(2) * 1.3, so V(2) = 0.77. The container will be more than 3/4 its original size.
At 20 feet the formula will be 1 * 1 = V(2) * 1.6, so V(2) = 0.63. The container will be about 5/8 its original size.
Generally, the closer you are to the surface, the greater the effect changes in depth have on volume.
I assume you are talking about the effect on air spaces.
33 feet of sea water is the same pressure as air pressure at sea level. The change in pressure is actually determined by a simple equation: V(1) * P(1) = V(2) * (P2)
In simple terms, if you are at 33 feet, you are at 2 atmospheres of pressure--one for the air and one for the water. If you have 1 liter of air in a flexible container at the surface and take it to 33 feet, the formula tells you what will happen:
1 * 1 = V(2) * 2, so 0.5 = V(2). You container will be half its original size.
If you want to figure out the effect at intermediate depths, you have to figure out the atmospheres at each depth. At 10 feet, it will be 10/33 +1 (for the air pressure), or 1.3 atmospheres. At 20 feet it will be 20/33 +1, or 1.6 atmospheres.
At 10 feet the formula will be 1 * 1 = V(2) * 1.3, so V(2) = 0.77. The container will be more than 3/4 its original size.
At 20 feet the formula will be 1 * 1 = V(2) * 1.6, so V(2) = 0.63. The container will be about 5/8 its original size.
Generally, the closer you are to the surface, the greater the effect changes in depth have on volume.
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The more water there is above you, the more pressure there is on you.
billtodive
Contributor
Math shows you how to get the answer.
Common sense will also.
Using both is good for your diving. Enjoy!
Common sense will also.
Using both is good for your diving. Enjoy!
DevonDiver
N/A
As you are aware...water is heavy. It weighs approx 1kg per litre.
As you descend underwater.... that weight is on top of you. It exerts pressure on you. The more weight/water that is above you, the more pressure there is. The pressure continually grows, as the weight of water above you grows.
The weight of 33ft/10m of water above you exerts a pressure on you which is the same as atmospheric pressure (14.7psi). Every additional 33ft/10m of water adds more weight on you equating to another 14.7psi (1atm).
As you descend underwater.... that weight is on top of you. It exerts pressure on you. The more weight/water that is above you, the more pressure there is. The pressure continually grows, as the weight of water above you grows.
The weight of 33ft/10m of water above you exerts a pressure on you which is the same as atmospheric pressure (14.7psi). Every additional 33ft/10m of water adds more weight on you equating to another 14.7psi (1atm).
spoolin01
Contributor
Pressure increases linearly with depth - a 5 ft drop changes the pressure, in absolute terms (psi/bar/atm/mmHg/etc.), by the same amount when going from 20 to 25 ft as going from 120 to 125 ft. Change in pressure divided by change in depth is a physical constant.
Is that the question you were asking?
Is that the question you were asking?
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That's what I always thought. Sounds logical. But I posed a question years ago on SB: If you sank a tube (that was just wide enough to fit a diver) 100' into the ground and filled it with water, would the pressure on a diver at the bottom be the same as that on a diver down 100' in the ocean? The resounding answer from most was YES. It's only the water directly on top of you that affects you. Still hard to believe. Anyone want to rehash this?
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