how to find metacenter, center of buyoncy, center of gravity?

Discussion in 'Boat Design' started by andikarahma, Dec 16, 2013.

  1. andikarahma
    Joined: Mar 2013
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    andikarahma New Member

    im designing a boat with dimension in cm
    the lenght of the small boat is 150cm, calculated weight is 15kg.
    how to find metacenter (M), center of buyoncy (B), center of gravity (G) without depending on software?

    im newby here, need help....
  2. Ad Hoc
    Joined: Oct 2008
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    Ad Hoc Naval Architect


    Well, firstly you must ask yourself the question how do you calculate the metacentre? By that I mean, what methods/formulas are used?

    To calculate the KM, you can get this from knowing that BM = I/V

    And since you hull is prismatic, B is easy to calculate as is the KB, and also the I, since it is a simple rectangle shape.

    So once you calculate the KB, the I and then the BM, you can find the KM.

    I = bd^3/12
    V = volume (buoyancy)

    KB = centroid of the simple prismatic shape, at the given draft.

    So KM = KB + BM
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  3. Mr Efficiency
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    Mr Efficiency Senior Member

    Is that just one cross section of the boat illustrated ? If this is a theoretical exercise, you will need a proper drawing of it, (unless it actually is a prism), if it is a practical exercise to help create a boat for actual use, the position of the COG etc will be somewhat academic given people moving around such a small boat will alter them drastically.
  4. NavalSArtichoke
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    NavalSArtichoke Senior Member

    The location of the CG is the tricky part.

  5. philSweet
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    philSweet Senior Member

    Just to add a bit to Ad Hoc's explanation...

    There are two M's, a longitudinal one and a transverse one. For the transverse M, Mt, you use Ix, for the longitudinal M, Ml, use Iy. I is the second moment of area of the waterplane.

    I for Mt is waterline beam cubed times waterline length divided by 12 for a rectangular waterplane. You have to first solve for the waterline beam at 15 kg. Use the entire beam, not the 1/2 beam, in the above equation.
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