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Daily Question for Jan 4, 2018

On a vessel of 10,000 tons displacement, compute the reduction in metacentric height due to free surface in a hold having free water on the tank top. The hold is 40 feet (12.19 m) long and 50 feet ( 15.24 m)wide. The reduction in metacentric height is __________.

A, 1.1 feet (0.335 m)

B. 1.2 feet (0.366 m)

C. 1.3 feet (0.393 m)

D. 1.5 feet (0.457 m)

Comments

  1. UnknownJanuary 4, 2018 at 3:08 PM

    B

    ReplyDelete
  2. Friends of STABILITY AND TRIM FOR THE SHIP'S OFFICERJanuary 4, 2018 at 9:47 PM

    The basic Free Surface Correction for a Rectangular Tank is

    (length of the tank)x(width of the tank)^3/12 x Displaced Volume

    Where:
    Length of the tank top = 40 feet
    Width of the tank top = 50 feet
    Displacement = 10,000 tons
    Displaced Volume = (35 cubic feet/ton)x 10,000 tons = 350,000 cubic feet

    Evaluating the formula

    (40 ft)(50ft)(50ft)(50ft)/12(350,000cubic feet) = 1.1904 ft ANS

    So the best answer is B. and I AGREE with Ken!

    Do you agree or disagree?

    ReplyDelete

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