Lift bag calculations

[Akimbo]

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.

True on all counts, but I think you missed that it has a fixed volume.

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

 

[Rich Keller]

True on all counts, but I think you missed that it has a fixed volume.

That was another assumption I was making. The question says nothing about what is inside that space or if the space is sealed or open. The way they are asking these questions they could be referring to the size of the object. I assume in post 7 that being down there long enough to be embedded even if it was a sealed space full of air that the air was no longer inside the object. Given the choices he has for answers I see nothing that relates to 86lbs being needed to lift the object anyway.

 

[Mark Michaud SELAUSAR]

[video=youtube_share;pYPtsu6FOBg]http://youtu.be/pYPtsu6FOBg[/video]

Hickdive has a good solution. All the recreational lift formulas make something out of nothing. It sounds good in theory but it ain't. Throw that crap away. Deeper than about 30' use open bottom bags. Above 30' use open or closed bags. If an object weighs 100 lbs you need 100 lbs of lift. If an object weighs 1000 lbs you need 1000 lbs of lift. More lift is ok but too little is unsafe. More lift can help with suction and debris being piled around it. Sometimes you have to jet an area, before you can lift, because bags won't work well when the object is large. If you can cradle an object it is better and more safe. But like I said, all this weight vs displacement vs salt water vs fresh water vs how many beers I have had........Keep it simple. BTW, I have lifted hundreds of objects from trawlers, to sailboats, to flatboats, to vehicles, to logs, to concrete gargoyles.

 

[eponym]

For working purposes assume that it will take ten times as much lift as the object actually weighs in water to free it from the suction of embedment. Only a sufficiently-large boat and line and tide (or gradual tensioning with swells) will do the job.

-Bryan

 

[n03]

OK, but for the purpose of solving this textbook problem - what formulas do I need, and what assumptions should I make? I think we all agreed that the object has negative 86 lb buoyancy. This makes me think answer choices #2 and #3 might be correct. Is #2 better than #3 because it involves fewer lift bags? Or is it better to use more lift bags?

 

[Rich Keller]

If you have any interest in seeing a search and recovery forum added to Scuba Board let the moderators know.

http://www.scubaboard.com/forums/suggestions/422428-search-recovery-forum.html

 

[Bob DBF]

OK, but for the purpose of solving this textbook problem - what formulas do I need, and what assumptions should I make? I think we all agreed that the object has negative 86 lb buoyancy. This makes me think answer choices #2 and #3 might be correct. Is #2 better than #3 because it involves fewer lift bags? Or is it better to use more lift bags?

1) a line tied to a boat at the surface

Screwing around with lift bags if you don't have to, and without knowing about how much lift you might actually need is a disaster in the making. Imagine your arm under the second 1000# lift bag when the cube breaks loose, depending what happens next you may be attached to the lift and headed for the surface at a great rate of speed.

I'd go with Rich Keller's method, safe and effective.



Bob
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I think that advocating unsafe and dangerous practices is both stupid and foolish. That is why I don't tell people to do what I do. Dsix36

 

[Rich Keller]

OK, but for the purpose of solving this textbook problem - what formulas do I need, and what assumptions should I make? I think we all agreed that the object has negative 86 lb buoyancy. This makes me think answer choices #2 and #3 might be correct. Is #2 better than #3 because it involves fewer lift bags? Or is it better to use more lift bags?

If you take into consideration the assumptions the people asking the question are making, even though some of those assumptions are wrong, I would go with both options 1&3. Option 1 will pull the object out of the bottom so the suction will no longer be a factor in the lift. There is no way to calculate the suction because you do not know how deeply the object is buried, the shape of the object or what the bottom is made of. Option 3 will get you closest to the amount of lift needed if you only use 2 of the 3 lift bags. They seem to think that the weight of the lift bag should be figured into this so they think the 50lb lift bag that weighs 4lbs will only lift 46lbs and that the object weighs 86lbs in the water. Using 2 fully inflated 50lb bags in their opinion will only lift a total of 92lbs there by giving you a lift that is at the maximum only 6lbs positive regardless of the depth.

OR

Options 1&2 using only 1 of the lift bags. Option 1 still breaks the suction but you would need to leave the line attached. Option 2 using 1 bag makes the weight of the object at the end of the line 16lbs in the water. You could then pull the object surface and partially inflate the remaining bag.

---------- Post added August 18th, 2014 at 12:51 PM ----------

PS: If you can only pick one option I would choose #1. Every way I look at this problem starts with the line connected to the boat pulling the object out of the bottom. On a calm day you would have to wait for the tide to come in but if there are waves that would probably be enough to break suction in a few minutes.