Need some help with a math problem...

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Dec 6, 2003
295
Macgregor 26D Pollock Pines, Ca.
I'm trying to figure out how much bouancy a 4' diameter hollow sphere creates. If I've got it right, the volume of a sphere equals (4 pi x radius cubed divided by 3 ?) so for a 4' diameter that comes out to 33.5 cubic feet. A cubic foot of seawater weighs 64 lbs. so 33.5 times 64 equals 2144 lbs. Am I correct in assuming that bouancy equals weight of the water displaced? If so, would the bouancy of a 4' diameter sphere be 2144 lbs. minus the weight of the sphere? Any help or insight would be greatly appreciated!
 
Dec 2, 2003
4,245
- - Seabeck WA
Sounds right,

That's how you determine the actual weight of a concrete block used for a mooring, but not being a math wiz,,,,
 
T

T J Furstenau

Correct

Your math is correct. Of course, this assumes that the sphere is fully submerged, and that the diameter of the sphere is fixed. (i.e. As the sphere is submerged, atmospheric pressure does not result in the sphere shrinking)
 
Dec 6, 2003
295
Macgregor 26D Pollock Pines, Ca.
A truly awesome bulletin board

It just amazes me that you can post a question here and in less than an hour over 40 people have seen it and you get your answer. Thanks, T.J. and Fred. And hey Fred, are you saying that you take the weight of the dried concrete and then subtract its volume in water weight to know how much it will weigh in the water? I watched them drop one of those things from a barge one time and it was huge, shaped like a pier block, only about 8' on each side and must have weighed 10 tons!
 
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