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Re: sub weight
Steve,
Your hatch would be one half of an oblate spheroid, therefore it would be 2.42/2
= 1.21 cubic feet. Since you have two ends the sum of both halves would be one
"whole" oblate spheroid per the formula.
Dick
STEVE McColman wrote:
> >From: Dick Morrisson <rmorrisson@unidial.com>
> >Reply-To: personal_submersibles@psubs.org
> >To: personal_submersibles@psubs.org
> >Subject: Re: sub weight
> >Date: Mon, 31 May 1999 12:34:52 -0700
> >
> >Steve,
> >The formula for the volume of an oblate spheroid (an ellipse revolved about
> >its
> >minor axis) is 4/3 times pi times a times b squared, where "a" is the
> >radius of
> >the minor axis and "b" is the radius of the major axis and pi is ~3.14 .
> >For
> >your example "a" would equal 7" and "b" would equal 15". I believe this is
> >the
> >correct formula for the volume but it hasn't stood up to the "fork" test
> >yet so
> >double check it before going too far with your design.
> >Dick
>
> ----------------------------------------------------
> >
> Im still a little confused, I want to find out how many cubic feet of air
> is in my hatch. My hatch is 24" Dia. and 7"wide. 1.33 x 3.14 x .58 x 1 =
> 2.42 cu/ft Is that right???
>
> Ends are 30" Dia. and 7" wide.
> My ends---- 1.33 x 3.14 x .58 x 1.56 = 3.8 cu/ft in EACH end ????
>
> Steve McColman, Vancouver Island
>
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