DIVE PROBLEM... test your physics!

[KangarooSeatbelt]

Ok guys.. I'm taking a class and the instructor has given us a dive physics problem... see if you can solve it! I'd like to see how you got there, as well as the actual answer.... we had a lot of fun with it, so I thought I'd see what everyone here could do.

You find a 5000 pound anchor with a volume of 10ft3 (10 cubic feet) in 68 ft of sea water. You want to get it to the surface so you can use it for your decoration on the front lawn of your house. Fortunately, you happen to have a 6000 lb lift bag. You also have an Aluminum 80, a Steel 104, and a steel 72. Do you have enough air to get it to the surface?
Approximately what volume of air is required to lift the anchor?

See what you can do... remember boyle's law and to take into consideration the weight of air..

-Sam

 

[Charlie99]

See what you can do... remember boyle's law and to take into consideration the weight of air..

In real life you wouldn't know either the weight of the anchor nor it's volume close enough that the weight of the air is relevant.

5000 pounds/64pound/cuft seawater = 78 cu ft.
The anchor's volume is 10, so you need 68cu ft.
68 cu ft at 66' is 3x68=204 cu ft of air needed. (66' is close enough to 68' for my calcs.).

I'd get another lift bag and also have at least 300cf on hand, since you have to break the anchor out of the mud.

Don't forget that you need to breath something, too.

And the real question: HOW DO YOU HANDLE THE ANCHOR ONCE YOU GET IT TO THE SURFACE?

 

[Diver0001]

Hmmmm.....first let's convert it to useful measurements:

5000 lbs is 2268kg
10 ft3 is 283 litres
68 ft is 20.73 metres or 3.06 ata

In terms of the tanks you didn't say what pressure they were filled to but assuming they were more or less full:

an AL 80 holds about 2200 litres of air
and ST 104 holds about 3000 litres of air
and a ST 72 holds about 2000 litres of air
so in total you have 7200 litres of air available

given that the anchor weighs effectively 1985kg under water then you'd need about 3.06*1985 litres of air to lift it. That's 6074 litres. That much air weighs about 7 kg so you'd need another 7 odd litres bringing the total to about 6100 litres.

We'll have to assume the liftbag is neutral.

About 37% of the liftbag will breach the surface

And you'd have to make the dive on about 1/2 of a steel 72.

R..

 

[Wijbrandus]

It's problems like this that tell me I'll never be a salvage expert.

Why do you want to take the anchor? Just observe all the fish and life that has grown upon it. It's a little mini-reef.

 

[Diver0001]

You wouldn't salvage a 5000lb anchor like this. You would lower a cable from a winch and haul it up the old fashioned way. Some things are only interesting for the puzzle.

R..

 

[Diver0001]

BTW, you can also work out which kind of metal the anchor was made of based on the available information. Any takers?

R..

 

[hecker-the-wrecker]

You wouldn't salvage a 5000lb anchor like this. You would lower a cable from a winch and haul it up the old fashioned way. Some things are only interesting for the puzzle.

R..

Very true, I own a towing company, how about a 35ft freightliner with a 425hp cat engine, 100,000 lb deck winch with 750 ft of 5/8 cable??

 

[Lead_carrier]

Ok guys.. I'm taking a class and the instructor has given us a dive physics problem... see if you can solve it! I'd like to see how you got there, as well as the actual answer.... we had a lot of fun with it, so I thought I'd see what everyone here could do.

You find a 5000 pound anchor with a volume of 10ft3 (10 cubic feet) in 68 ft of sea water. You want to get it to the surface so you can use it for your decoration on the front lawn of your house. Fortunately, you happen to have a 6000 lb lift bag. You also have an Aluminum 80, a Steel 104, and a steel 72. Do you have enough air to get it to the surface?
Approximately what volume of air is required to lift the anchor?


-Sam

None. I call a metal salvage company, tell them that I have a 5000 # anchor and will sell it to them half price if they will haul it off. AFTER I cash the check I'll tell them where it is.

 

[lamont]

weird thread ressurection. there were two of these threads. this is the shorter one. longer one is here:

http://www.scubaboard.com/t73437-.html

 

[ScubaDadMiami]

You find a 5000 pound anchor with a volume of 10ft3 (10 cubic feet) in 68 ft of sea water. You want to get it to the surface. . . . Fortunately, you happen to have a 6000 lb lift bag. You also have an Aluminum 80, a Steel 104, and a steel 72. Do you have enough air to get it to the surface?
Approximately what volume of air is required to lift the anchor?

I'd explain the solution a slightly different way. The anchor weighs 5000 pounds as a given. Because it displaces (takes up the space of) ten cubic feet of seawater, that means that ten cubic feet's worth of pressure is already trying to lift the anchor from the bottom before we even attach our lift bag.

Sea water weighs 64 pounds per cubic foot. So, we have 64 pounds of sea water per cubic foot x 10 cubic feet displaced providing upward force: 64 x 10 = 640 pounds of upward force. Therefore, we take the 5,000 pounds that the anchor weighs and subtract the amount of force that is trying to lift it by the displaced sea water to find out how much lift we need to make it neutral: 5,000 - 640 = 4,360 pounds of lift needed. (If you want to, you can add a pound or something more than this since we have made it neutral at this point but not actually positively bouyant. However, for our purposes, we will leave the number at this, the 4,360.)

Okay, now we need to figure out how much air we need to do the job of creating this much lift using our lift bag. We take the 4,360 pounds of lift we need and translate that into how much sea water we will need to displace, which we already know weighs 64 pounds per square foot. So, we take the weight of lift we need and divide this by the weight of sea water: 4,360 / 64 = 68.125, the amount of these 64 pound cubes of sea water still needed to be displaced by air in our lift bag.

Now, all we need to do is to consider that we are at a depth of 68 feet and that we will need to bring enough gas with us to produce this displacement at this depth rather than at the surface. This is due to increased ambient pressure from the weight of the atmosphere over us and the water on top of us when we are at the bottom when compared to the surface, where only the weight of the atmosphere is surrounding us.

Since we are at 68 feet of sea water, we need to factor in how much absolute pressure surrounds us at that depth when compared to the one atmosphere we live in at the water's surface. 68 feet divided by 33 feet (the pressure of one atmosphere absolute) = 2.064. We then need to add the one atmosphere of the ocean of air on top of us to this, 2.064 + 1 = 3.064 atmospheres absolute pressure surrounding us when at 68 feet of sea water as compared to only 1 ata on the surface.

Because of this, at 68 feet of sea water, we will need 3.064 times the amount of gas we need at the surface to produce this lift. 3.064 * the 68.125 one foot cubes of seawater on the bottom needed to be displaced means that we need to carry 208.735 cubic feet of gas just to get neutral.

Of course, we did not include the weight of the lift bag and lines in this equation. We also did not produce any positive lift but just got the anchor neutral. But we have a great starting point from which to select tanks and gas volume from the cylinders we have to choose from in this example.