Question of the day
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03-21-2006, 08:28 PM,
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Question of the day
I have a sailboat that is 29 feet long, weights 2700 lbs.
It displaces 11 cu/ft. how many 500 lb and how many 250lb lift bags will it take to raise it? Next question is the sailboat is in 73 feet of water. the air supply is on the surface and you need 100 psi at the bottom to fill the bags, so what would you set the pressure regulator at the suface at? yes max a trick question, think out side the box grumpie |
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03-21-2006, 09:35 PM,
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Re: Question of the day
Is the boat right side up or upside down? Just mentally rigging it...Initially takes about 10 percent of the weight more than the object to "Break it Free" of the bottom. My calculations (best guess) 5-500lb, 4-250lb. Alternate - Stern of boat,500, 250, 500, 250, bow 500. If your comressed air source is at the surface and you send 100psi down the hose, you should still have 100psi at the bottom of two and a half atmospheres unless you have an incompetent compressor or restrictions in the air hose. It's Already Under Pressure!!!
My name is Lisa and I'm a SCUBAholic. It's been toooo long since my last dive!
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03-22-2006, 07:09 PM,
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Re: Question of the day
One cu/ft of fresh water weights 62.4 lbs. 11x62.4=704 lbs minus the boat weight 2700=1996 lbs. this is what you will have to lift.
inorder to get 100 lbs of pressure at 73 feet, you will have to muiltply 73 feet by .432 lbs=31.54 add this too 100lbs =132.54 lbs at the regulator. PS make sure that the hose is rated for a minium of 250 psi. good job you had the idea..............///// |
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03-22-2006, 07:13 PM,
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Re: Question of the day
73 divided 33=2.22+1=3.22 atm. depth divided by 33+1 gives you atms.
grumpie |
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03-23-2006, 10:25 AM,
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Re: Question of the day
For salt water. For fresh water its divided by 34. sagacity
That which does not kill you, makes you stronger.<br />Blood makes the grass grow, kill, kill, kill!
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