Posted: 10/7/2013 9:43:31 AM EDT
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In a perfect world: if I have a 20000 pound brick of steel and I want it to float; it must be shaped to displace at least 20001 pounds of water; Yes? |
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Nope, just 20000 pounds, although that is neutral buoyancy. If some freeboard is needed (i.e., float with part of the steel out of the water), then it will need to be shaped to displace more weight.
Now for fun - Your brick of steel is a cube 3.41 feet on a side. The fresh water you'll need to displace is a cube 6.83 feet on a side. (0.29 lbf/in^3 steel, 62.4 lbf/ft^3 fresh water). Reshape the brick to displace 320.5 ft^3 of water. |
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Quoted: You are technically correct. However you would want the object to displace more than that for practical reasons. I guess that there will be a MAX payload of 2500 pounds including persons and gear/fuel/other assorted crap. @1/3 displacement remains. |
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Quoted:
Nope, just 20000 pounds, although that is neutral buoyancy. If some freeboard is needed (i.e., float with part of the steel out of the water), then it will need to be shaped to displace more weight. Now for fun - Your brick of steel is a cube 3.41 feet on a side. The fresh water you'll need to displace is a cube 6.83 feet on a side. (0.29 lbf/in^3 steel, 62.4 lbf/ft^3 fresh water). Reshape the brick to displace 320.5 ft^3 of water. Shape it into something that looks strikingly like a boat.  
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Quoted:
Shape it into something that looks strikingly like a boat. Quoted:
Quoted:
Nope, just 20000 pounds, although that is neutral buoyancy. If some freeboard is needed (i.e., float with part of the steel out of the water), then it will need to be shaped to displace more weight. Now for fun - Your brick of steel is a cube 3.41 feet on a side. The fresh water you'll need to displace is a cube 6.83 feet on a side. (0.29 lbf/in^3 steel, 62.4 lbf/ft^3 fresh water). Reshape the brick to displace 320.5 ft^3 of water. Shape it into something that looks strikingly like a boat. Not my job. |
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Quoted:
Nope, just 20000 pounds, although that is neutral buoyancy. If some freeboard is needed (i.e., float with part of the steel out of the water), then it will need to be shaped to displace more weight. Now for fun - Your brick of steel is a cube 3.41 feet on a side. The fresh water you'll need to displace is a cube 6.83 feet on a side. (0.29 lbf/in^3 steel, 62.4 lbf/ft^3 fresh water). Reshape the brick to displace 320.5 ft^3 of water. A hollow cube 6.83 feet on a side, of course  Notably, the cubic void is roughly equivalent to the original cube size of 3.4 feet per side.
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Quoted:
A hollow cube 6.83 feet on a side, of course Notably, the cubic void is roughly equivalent to the original cube size of 3.4 feet per side.Quoted:
Quoted:
Nope, just 20000 pounds, although that is neutral buoyancy. If some freeboard is needed (i.e., float with part of the steel out of the water), then it will need to be shaped to displace more weight. Now for fun - Your brick of steel is a cube 3.41 feet on a side. The fresh water you'll need to displace is a cube 6.83 feet on a side. (0.29 lbf/in^3 steel, 62.4 lbf/ft^3 fresh water). Reshape the brick to displace 320.5 ft^3 of water. A hollow cube 6.83 feet on a side, of course Notably, the cubic void is roughly equivalent to the original cube size of 3.4 feet per side.Shhhh. We're hunting wabbits. |
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Don't start that freaky shit here! Innocent thread and you have to go there. F-ing evil movie. Nice.
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