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Daily Question for March 4, 2018

A cargo of 30 tons is to be loaded on deck 30 feet from the ship's centerline. The ship's displacement
including the 30 tons cargo will be 9,000 tons and the GM will be 5 feet. The list of the vessel after loading this cargo will be __________.

A.  1.14°  B.  2.05°   C.  2.31°   D. 3.40°

Comments

  1. JAYAVIGNESHMarch 4, 2018 at 5:48 AM

    1.14°

    ReplyDelete
  2. Friends of STABILITY AND TRIM FOR THE SHIP'S OFFICERMarch 4, 2018 at 2:42 PM

    angle of list = ArcTan ( Weight x Dist from Centerline)/(GM x Displacement)
    angle of list = ArcTan (30 tons x 30 feet)/(5 feet x 9000 tons) = 0.020
    angle of list = ArcTan (0.020) = 1.14576 degrees, A is the best answer.

    In Stability and Trim for the Ship's Officer, 4th Ed. check out page 115, Derivation of the Inclining Experiment Formula.

    ReplyDelete
  3. Friends of STABILITY AND TRIM FOR THE SHIP'S OFFICERMarch 4, 2018 at 2:46 PM

    Here is a video explaination of how to solve this type of question. In the video they mention THEORY OF MOMENTS I prefer WEIGHTED AVERAGE.

    https://www.youtube.com/watch?v=vIY_RFxnNXw

    ReplyDelete
  4. UnknownMarch 5, 2018 at 8:59 AM

    tanθ=(w*d)/(GM*Δ)
    30*30/5*9000=0.02
    arc_tan=1.145762838 or 1.15°

    ANS: ALPHA is closest

    ReplyDelete

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