Buoyancy and the math required

[Diver0001]

Hello
I am having some issues with the following question
I have an object 5' square and 5' high. It is floating in saltwater with only 6" showing. What line strength would I need to lift the object out of the water.
Here is how I did the calculation.
5 x 5 x 4.5= 112.5 cubic feet.
112.5 x 64 = 7200 {cubic feet x the weight of one cubic foot salt water) = {the total lb of test line I would require}

Is this correct?

You can't answer that question without knowing the density of the material from which the object is made. Unless you assume that it is neutrally buoyant with whatever portion of it sticking out of the water and you know the density of the water (the salt content, for example).

The second part depends upon altitude. You would have to assume a certain altitude (say sea level) in order to answer that.

R..

 

[rongoodman]

Isn't there also an issue here of what color swallow is carrying the coconut?

 

[Buoy_A]

You can't answer that question without knowing the density of the material from which the object is made. Unless you assume that it is neutrally buoyant with whatever portion of it sticking out of the water and you know the density of the water (the salt content, for example).

Can’t we just say that the cube is 10% less dense than saltwater, since 10% of it is sticking out of the water (assuming that the cube is equal in density throughout)? Then you could take the equivalent weight of the cube volume of saltwater and multiply it by 0.90 to get the line strength. You couldn't do this w. non-cube shapes or variable density objects though (the latter because any part of something that is in water that sticks out of the water is 100% negative buoyancy.)

 

[Diver0001]

oh snap.

R..

 

[nimoh]

You need to make some assumptions to answer the second question, so you may as well make assumptions that simplify the question such as you are not wearing anything such as a wetsuit that will change buoyancy at depth, and assuming that 1ft3 of air at 1ata weighs the same as 1ft3 of air at 1.3ata since the difference is insignificant, then the answer is quite simple:

if it takes 1/3ft3 of air to make you neutral at the surface, then it will take you 1/3ft3 of air at 10' remain neutral.

 

[boulderjohn]

You need to make some assumptions to answer the second question, so you may as well make assumptions that simplify the question such as you are not wearing anything such as a wetsuit that will change buoyancy at depth, and assuming that 1ft3 of air at 1ata weighs the same as 1ft3 of air at 1.3ata since the difference is insignificant, then the answer is quite simple:

if it takes 1/3ft3 of air to make you neutral at the surface, then it will take you 1/3ft3 of air at 10' remain neutral.

I really didn't understand the question the way it was worded, and I still don't. Your answer makes sense with one interpretation of what the question means--in fact, it was how I interpreted it at first. Then I wasn't sure and decided that they were asking what would happen to that 1/3 cubic foot as you ascend. That is a basic Boyle's law question very similar to one I ask DM candidates, except that I include altitude issues in mine. Until you actually sit down and so the math, you will not believe how much difference it can make.

 

[nimoh]

I really didn't understand the question the way it was worded, and I still don't. Your answer makes sense with one interpretation of what the question means--in fact, it was how I interpreted it at first. Then I wasn't sure and decided that they were asking what would happen to that 1/3 cubic foot as you ascend. That is a basic Boyle's law question very similar to one I ask DM candidates, except that I include altitude issues in mine. Until you actually sit down and so the math, you will not believe how much difference it can make.

I interpreted it as what happens when starting at the surface and then descending to 10'. I was willing to ignore altitude since it wasn't mentioned. However, does altitude really have a significant impact on this calculation as you suggest?

I mean, if you are at an elevation where the atmospheric pressure is .9ata rather than 1 ata, and you go to a depth of 33', are you now at 2 ata, or 1.9? In other words, I think descending 33' would cause a 1ata change no matter what the starting pressure.

 

[boulderjohn]

I interpreted it as what happens when starting at the surface and then descending to 10'. I was willing to ignore altitude since it wasn't mentioned. However, does altitude really have a significant impact on this calculation as you suggest?

I mean, if you are at an elevation where the atmospheric pressure is .9ata rather than 1 ata, and you go to a depth of 33', are you now at 2 ata, or 1.9? In other words, I think descending 33' would cause a 1ata change no matter what the starting pressure.

First of all, if you are at altitude, you will be in fresh water, so 1 ATA = 34 feet.

Your ATA at depth is equal to the pressure of the water plus the pressure of the atmosphere, so in your example, the pressure at 34 feet would be 1.9, not 2.0. But an atmospheric pressure of 0.9 is not all that high. I dive locally at 0.83. I own a Shearwater Predator, which adjusts for altitude automatically. When I am diving air (.21 oxygen), it gives me my partial pressure in red numbers because it considers the mix--the air I breathe every day--to be dangerously hypoxic.

The closer you are to the surface, the bigger the difference, because the air pressure is a bigger part of the total pressure. The deeper you are, the more the total ATA is comprised of water pressure. The pressure at what would normally be 4 ATA at sea level will be 3.83 where I live--not that much difference. It makes a huge difference, though, as you ascend from that depth. Calculate the volume of a bubble in your body or the air in a BCD that starts at a volume of 1 whatever and goes to the surface. It will be 4 whatevers at sea level. It will be 4.62 whatevers in Denver.

I took my first DPV (scooter) instruction at Jefferson Lake in Colorado, at a altitude of over 11,000 feet. That made the atmospheric pressure 0.67. The instructor laid a line on the bottom and had us practice go from one end to the other, turn around, and head back, over and over and over again. One end was at roughly 34 feet deep--or 1.67 ATA. The other end was at 10 feet--or .96 ATA. That means that the volume of air in our wings very nearly doubled in that short trip from the deep end to the shallow end, and then the volume changed the other way on the return trip. We were supposed to be working on controlling the scooters, a new experience for us, but we were instead working primarily on high speed buoyancy changes.

 

[nimoh]

First of all, if you are at altitude, you will be in fresh water, so 1 ATA = 34 feet.

Your ATA at depth is equal to the pressure of the water plus the pressure of the atmosphere, so in your example, the pressure at 34 feet would be 1.9, not 2.0. But an atmospheric pressure of 0.9 is not all that high. I dive locally at 0.83. I own a Shearwater Predator, which adjusts for altitude automatically. When I am diving air (.21 oxygen), it gives me my partial pressure in red numbers because it considers the mix--the air I breathe every day--to be dangerously hypoxic.

The closer you are to the surface, the bigger the difference, because the air pressure is a bigger part of the total pressure. The deeper you are, the more the total ATA is comprised of water pressure. The pressure at what would normally be 4 ATA at sea level will be 3.83 where I live--not that much difference. It makes a huge difference, though, as you ascend from that depth. Calculate the volume of a bubble in your body or the air in a BCD that starts at a volume of 1 whatever and goes to the surface. It will be 4 whatevers at sea level. It will be 4.62 whatevers in Denver.

I took my first DPV (scooter) instruction at Jefferson Lake in Colorado, at a depth of over 11,000 feet. That made the atmospheric pressure 0.67. The instructor laid a line on the bottom and had us practice go from one end to the other, turn around, and head back, over and over and over again. One end was at roughly 34 feet deep--or 1.67 ATA. The other end was at 10 feet--or .96 ATA. That means that the volume of air in our wings very nearly doubled in that short trip from the deep end to the shallow end, and then the volume changed the other way on the return trip. We were supposed to be working on controlling the scooters, a new experience for us, but we were instead working primarily on high speed buoyancy changes.

Well, altitude does not mean fresh water, but yes, I meant to put 34'

Also, I get the impression that you would like to use this thread to demonstrate how much you know about altitude diving even though the OP didn't mention it. It is mostly good information, just not sure how relevant using a Shearwater Predator to prevent hypoxia on a 10 foot dive is or a scooter ride at a depth of 11,000 feet is either.

 

[boulderjohn]

Well, altitude does not mean fresh water, but yes, I meant to put 34'

Also, I get the impression that you would like to use this thread to demonstrate how much you know about altitude diving even though the OP didn't mention it. It is mostly good information, just not sure how relevant using a Shearwater Predator to prevent hypoxia on a 10 foot dive is or a scooter ride at a depth of 11,000 feet is either.

I was simply responding to the argument--which I have herd often--that altitude does not need to be a consideration in diving.