What is the reduction in metacentric height due to free surface when a tank 60 ft. wide and 60 ft. long is partially filled with saltwater? (The vessel's displacement is 10,000 tons.)
A. 3.00 feet B. 3.09 feet C. 3.15 feet D. 3.20 feet
A. 3.00 feet B. 3.09 feet C. 3.15 feet D. 3.20 feet
(l * b3)/(420 * Δ)
ReplyDelete(60 x 60 cubed)/(420 x 10000)=3.0857
ANS: Charley
the tank is filled with seawater so I didn't add the ratio.
ReplyDelete(r*l*b3)/(420*Δ)
I agree. "C."is the best answer.
ReplyDeleteFYI,, 420 is a constant = 12 x 35 cuft/LT. It is a nice short cut if you need to multiply 12 x 35 x displacement in LT like in the FS formula.
The result I have is 3.0857 and it is closer to answer "B" 3.09.
ReplyDelete