Lift bag calculations

[n03]

Does anyone know how to work out this problem?

An object with a volume of one cu. ft., weighing 150 lbs. is embedded in the sediment at a depth of 99 ft. in the ocean harbor. This object is best lifted to the surface by:
1) a line tied to a boat at the surface
2) Two 75-lb. lift bags weighing five lbs. each
3) Three 50-lb. lift bags weighing four lbs. each
4) Three 1,000-lb. lift bags and two 50-lb. lift bags

Thanks!

 

[victorzamora]

Piece of cake, kinda.

You should know what one cubic foot of water weighs, right? If not, it's 62.4 lbs/ft3. That means that your item has a total negative buoyancy of.............84.6 lbs (150-62.4). All you need is 84.6 pounds of lift.

Option 2 gives you 140lbs of lift.....(75-5)*2.
Option 3 gives you 138lbs of lift.....(50-3)*3.
Option 4 gives you 3100lbs of lift....obviously.

So now you've narrowed it down to a few options (all of them, still). I don't know what your book is asking for, because the object being embedded in the sediment makes me think it'd be best to use a line tied to a boat at the surface, but loose sediment could allow for two 75# lift bags to be the best option. Get why I said "kinda"?

 

[hyrax]

Does anyone know how to work out this problem?

An object with a volume of one cu. ft., weighing 150 lbs. is embedded in the sediment at a depth of 99 ft. in the ocean harbor. This object is best lifted to the surface by:
1) a line tied to a boat at the surface
2) Two 75-lb. lift bags weighing five lbs. each
3) Three 50-lb. lift bags weighing four lbs. each
4) Three 1,000-lb. lift bags and two 50-lb. lift bags

Thanks!

Where did you get this questions? Based on simple lift bag calculations none of the answers are correct, but the embedded in sediment is the part that I'm confused about. For basic lift bag think of it this way. One cubic foot displaces about 64 pounds. So you can say your weight in the water is 86 pounds. That means you need 86 pounds of lift to lift the weight. As far as the embedded in sediment goes you need to break the suction between the sediment and the object and I wouldn't do that with a lift bag, I would do it by removing sediment.

---------- Post added August 14th, 2014 at 03:17 PM ----------

Piece of cake, kinda.

You should know what one cubic foot of water weighs, right? If not, it's 62.4 lbs/ft3. That means that your item has a total negative buoyancy of.............84.6 lbs (150-62.4). All you need is 84.6 pounds of lift.

Option 2 gives you 140lbs of lift.....(75-5)*2.
Option 3 gives you 138lbs of lift.....(50-3)*3.
Option 4 gives you 3100lbs of lift....obviously.

So now you've narrowed it down to a few options (all of them, still). I don't know what your book is asking for, because the object being embedded in the sediment makes me think it'd be best to use a line tied to a boat at the surface, but loose sediment could allow for two 75# lift bags to be the best option. Get why I said "kinda"?

62.4....64.... I just went by memory, whats 1.6 pounds between friends.

 

[Hickdive]

If it's embedded then a line to the surface and single 1000lb lift bag tied to the line so that, fully inflated, it's just at the surface. Do this at low tide.

Inflating the bag will take less compressed air as it's at the surface (not 99') and can probably be done with a length of garden hose to avoid having divers in the water to do the inflating.

When the tide comes in the big bag will break the object clear of the bottom suction (the bag can't go further than the surface so no uncontrolled lift will result) then tow the whole a few yards so the object clear of its own scour. Deflate the bag to lower the object back to the sea bed, remove it from the line, then haul the object to the surface.

One dive to tie the line to the object and tie the lift bag to the line at the surface. No divers trying to fill bags at 99' down. No divers in the water when lifting. No calculations required.

 

[n03]

Where did you get this questions? Based on simple lift bag calculations none of the answers are correct, but the embedded in sediment is the part that I'm confused about. For basic lift bag think of it this way. One cubic foot displaces about 64 pounds. So you can say your weight in the water is 86 pounds. That means you need 86 pounds of lift to lift the weight. As far as the embedded in sediment goes you need to break the suction between the sediment and the object and I wouldn't do that with a lift bag, I would do it by removing sediment.

---------- Post added August 14th, 2014 at 03:17 PM ----------



62.4....64.... I just went by memory, whats 1.6 pounds between friends.

Thank you all for the quick replies! Yes, I did get to the point that I have negative 84 lb buoyancy, and in theory that's how much lift power I need to counteract it, except for the buried in sediment part... who knows how much suction the sediment has! And how should I apply the depth information (99 ft)?

From analyzing the answer options #2 and #3 look very similar, so it seems like one of these two should be correct. They offer the same lift power, except #3 is 2 lb heavier than #2, but does that matter? Does it matter if you use 2 bags vs. 3 bags? Is there some rule that using the fewest number of bags is optimal?

So in what scenario would one use a line tied to a boat on the surface? Is there any chance that's the correct answer?

 

[rhwestfall]

If it's embedded then a line to the surface and single 1000lb lift bag tied to the line so that, fully inflated, it's just at the surface. Do this at low tide.

Inflating the bag will take less compressed air as it's at the surface (not 99') and can probably be done with a length of garden hose to avoid having divers in the water to do the inflating.

When the tide comes in the big bag will break the object clear of the bottom suction (the bag can't go further than the surface so no uncontrolled lift will result) then tow the whole a few yards so the object clear of its own scour. Deflate the bag to lower the object back to the sea bed, remove it from the line, then haul the object to the surface.

One dive to tie the line to the object and tie the lift bag to the line at the surface. No divers trying to fill bags at 99' down. No divers in the water when lifting. No calculations required.

man would I like to share a few pints with you!

 

[Rich Keller]

An object with a volume of one cu. ft., weighing 150 lbs. is embedded in the sediment at a depth of 99 ft. in the ocean harbor. This object is best lifted to the surface by:
1) a line tied to a boat at the surface
2) Two 75-lb. lift bags weighing five lbs. each
3) Three 50-lb. lift bags weighing four lbs. each
4) Three 1,000-lb. lift bags and two 50-lb. lift bags

I would ignore the internal volume part of this question all together. If it is embedded in sediment there is probably no air in it to begin with and you can not put air into it until you break it free from the bottom. Also the weight of the object will shift in its rigging unless that void is 100% full of air and sealed but that seems unlikely with an object that has been on the bottom long enough to become embedded.

We also do not know what the object is made of so there is no way to calculate how much it weighs after you factor in the amount of water it displaces.

The weight of the lift bags is meaningless as well, the amount the bag is rated to lift has nothing to do with how much the bag itself weighs.

You last option is over kill.

I use a 1" polypropylene line to a boat on the surface that I run through a continues feed come-along to pick up objects this size. This is attached to a davit on the boat so it will not only get the object to the surface it can take it out of the water and place it on the deck as well. This method only requires the amount of air I need to breath while I am rigging the object.

 

[Rich Keller]

If it's embedded then a line to the surface and single 1000lb lift bag tied to the line so that, fully inflated, it's just at the surface. Do this at low tide.

Inflating the bag will take less compressed air as it's at the surface (not 99') and can probably be done with a length of garden hose to avoid having divers in the water to do the inflating.

When the tide comes in the big bag will break the object clear of the bottom suction (the bag can't go further than the surface so no uncontrolled lift will result) then tow the whole a few yards so the object clear of its own scour. Deflate the bag to lower the object back to the sea bed, remove it from the line, then haul the object to the surface.

One dive to tie the line to the object and tie the lift bag to the line at the surface. No divers trying to fill bags at 99' down. No divers in the water when lifting. No calculations required.

This is more commonly done by tying the line to the boat and waiting for the tide to come up instead of using a lift bag but is usually not used for an object this small. This is usually done with two boats on the surface attached to a boat on the bottom that needs to be lifted. Rig it tight at low tide, wait for the tide to come up, tow it to shore until it is grounded again, when the tide is out rig it tight again, keep repeating until the boat is above water on an outgoing tide near the shore, then pump it out and re-float it when the tide is up again.

You idea of a 1000lb lift bag would work better if the bag were tied 5' below the surface then filled, it will break the suction with out running away on you and having to wait for the tide. You could then go back down and add 150lbs worth of lift directly to the 150lb object. This will also negate the depth that you are working in as the bags will be full to capacity so as the object rises the excess air created as the water pressure decreasing will just vent out the bottom of the bag so it will rise at a steady rate.

If all you have is 150lbs worth of lift to work with then you could tie a line to the boat, let the incoming tide break the suction, then attach your bags to the object.

 

[Akimbo]

…An object with a volume of one cu. ft., weighing 150 lbs. is embedded in the sediment at a depth of 99 ft. in the ocean harbor….

Depth is irrelevant and you don’t have enough data. There is a major factor called “stiction” that can easily exceed the gross weight, let alone submerged weight, of an object by many times. That is the very principal that mushroom, Bruce/claw, and Danforth anchors work on.

You have enough information to calculate the submerged weight of the object but not what it will take to break if free from the bottom. In any case, you better get out of the way when that baby lets loose because it will breach the surface like a Trident Missile… OK maybe a Sperm whale.

---------- Post added August 14th, 2014 at 05:37 PM ----------

… We also do not know what the object is made of so there is no way to calculate how much it weighs after you factor in the amount of water it displaces…

What am I missing?

… An object with a volume of one cu. ft., weighing 150 lbs….

150 Lbs – 64 Lbs displaced weight of the sea water (rounded average) = 86 Lbs. — sounds like an estimate for a block of concrete. All the answers are bogus in the real world.

n03, I hope you are not actually paying these guys to teach you salvage! BTW, the average weight of seawater world-wide is 64.1 Lbs/Ft³ (rounded) and pure fresh water is 62.3 Lbs/Ft³ (rounded).

Seawater - Wikipedia, the free encyclopedia

 

[Rich Keller]

What am I missing?

The question is poorly asked so I am making an assumption here. I am assuming that the object weighs 150lbs on the surface. If it is 150lbs of steel it will not displace anywhere near as much water as a 150lbs of concert will. Whatever the amount of water the object displaces is subtracted from the objects weight on the surface thereby reducing the amount of lift needed as long as the object remains in the water.