Re: formulas

[Dick Morrisson <rmorrisson@unidial.com>]

That was pretty close Ed, except the volume of a sphere is 4/3(pi)r^3 (cubed, not
squared).  By the way, thanks for your inputs on concrete hulls, I'm beginning to
get swayed towards them.

EdwMullin@aol.com wrote:

> Figure  (pi)r^2 x lenght of cylinder (each the sub hull and tower)
> then   4/3(pi)r^2 for the ends and em all up, should be ball park for the
> cubic foot displacement.
> where pi= ~3.1415    r=radius in ft,  and ^2 is squared.
>
> Some one correct me if I got them wrong, Its been a while since school...  :)