Static stability(Academic problem)

Discussion in 'Stability' started by diegoperrera, Oct 1, 2014.

  1. Nick_Sinev
    Joined: Aug 2014
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    Nick_Sinev Junior Member

    sharpii2, how to calculate the moment of inertia I have explained in the 3d comment in this thread (just for the case somebody has forgotten the formula). Look at the beginning of the discussion.
    You can leave the length of the barge as letter "L". When dividing I / V the length will be cancelled.

    The answer is six:
    1) I'm pretty sure in my calculations.
    2) cmckesson got the same result.
     
  2. sharpii2
    Joined: May 2004
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    sharpii2 Senior Member

    Come to think of it, you're probably right.

    Just because the CG is higher than the Meta Center Radius does not necessarily mean a boat will capsize.

    It may float at a list.

    According to my calculations, there is about 0.167 m of free board left, once the sixth blue box is added.

    If the barge ship rolls ever so slightly, the Center of Buoyancy (CB) will shift slightly to the low side.

    The CG will shift in that direction too. But if the CB shifts further than the CG, some self righting moment will be created.
     
  3. Nick_Sinev
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    Nick_Sinev Junior Member

    There is nothing to think about.

    In case of 6 boxes the problem is equivalent to cuboid
    width = 3m
    height = 2.5m
    length = L
    uniform density = 1/3

    I = (L * 27*m^3 )/12
    V = 1/3 * 3m * 2.5m * L
    I / V = 0.9 m
    K_B = 1/2 * 1/3 * 2.5m
    K_M = K_B + I / V
    K_M ~ 1.32m

    K_G = 1/2*2.5m
    K_G = 1.25m

    1.32m > 1.25m
     
  4. philSweet
    Joined: May 2008
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    philSweet Senior Member

    One peculiarity of this problem is that the size and density of the boxes vis a vis the barge is such that the KG is a linear function of the number of boxes. That is not true in general, but it is true in this problem and that creates a simple way to get to the answer if you happen to notice it.

    Hopefully, nobody comes away thinking this is a general truth. It is just a quirk of this problem's construction.
     

  5. sharpii2
    Joined: May 2004
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    sharpii2 Senior Member

    I have to admit that I don't understand.

    The blue boxes have a density that is twice that of water, where the Barge ship has one that is one third that of water.

    It appears that though I may be off, that if I am, its by about 0.12 m.

    Now I see.

    You just imagined the blue boxes as simply increasing the section height of the barge ship by 0.25 m each. Each blue box is six times the density of the barge ship, but one sixth as wide.

    My method, though long winded, should have produced the same result. I must have made a small mathematical error somewhere.
     
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