Introduction and stability calculations (MTc 1cm)

Discussion in 'Stability' started by david1123, Jul 28, 2014.

  1. david1123
    Joined: Jul 2014
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    Location: cardiff

    david1123 Junior Member

    Evening all first post,
    David here, just a quick intro. Currently studying a long distance course in marine surveying which is frustrating at times! I have a GH 31 no 72, gib-sea 80 and a 3/4 finished nice and beamy 18ft ply clinker open design intended for bass fishing which has been moth balled for the time being.

    This forum is an absolute gold mine! I was just wondering if any one would advise me on MTc 1cm calculations, here she goes!

    I understand that MTc 1cm = WxGML
    100L
    I also understand that GML equates to KB + BML - KG which I have the data for.

    However I have been confused by a worked example (pictured below) that I found
    Which states WxGML
    100L

    But the actual equation is 5000 x 9.81 x 104
    100 x 98

    There is no reference to how the figure of 9.81 is achieved unless this is a constant, which I doubt so I must being missing something.

    My actual assignment which I am not sure if I will get into trouble for quoting is: A vessel has an ā€˜Lā€™ of 100 metres and a displacement of 6300 tonnes. The mean draught is 6 metres, and the following parameters apply:
    KB= 3 metres
    BML Longitudinal= 104.7 metres
    KG 5.4 metres

    The centre of flotation (F) is amidships. A weight of 60 tonnes is moved from forward to aft over a distance of 50 metres. Draw a simple diagram or diagrams illustrating this situation and calculate the MCT 1 cm. List the new draughts forward and aft.
    I'm not looking for the answer just an explanation of the equation, the assignment data is just a reference to what I have. That 9.81 in the worked example has just stumped me!
    Kind regards.
    David.

    [​IMG]
     
  2. TANSL
    Joined: Sep 2011
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    TANSL Senior Member

    9.81 is the acceleration of gravity.
    The statement says that the boat has a mass of 5000 tons. In formula must work with weight, not mass, so you need put the value of the mass times the acceleration of gravity (you know: force = mass x acceleration)
    I think this is the explanation but it seems a little absurd to work with units of mass.
     
  3. david1123
    Joined: Jul 2014
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    Location: cardiff

    david1123 Junior Member

    Excellent, thank you,
    The ''worked example'' was just one I found on-line.
    I fully understand what you are explaining, 9.81 m/s is a constant. This is what has me stumped. For the data I have which specifies displacement in tonnes should I just proceed with the equation, calculating GML first
    GML = (KB + BML - KG) disregarding the 9.81 m/s gravity thing and proceeding with equation:

    MTc 1cm = WxGML
    100L

    OR: W x (KB + BML - KG)
    100L

    Which is essentially the same equation.

    Thank you for your reply, appreciate it.
    David.
     
  4. Rastapop
    Joined: Mar 2014
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    Location: Australia

    Rastapop Naval Architect

    No, don't ignore it.

    W = mg = 5000 * 9.81
     

  5. gonzo
    Joined: Aug 2002
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    gonzo Senior Member

    I agree with TANSL that it is a bit absurd to work with units of mass. As all the vessels we survey are on Earth gravity is always about the same, the cargo gets weighted and not massed. You would have to work backwards from the weight to mass and then from mass and gravity back to weight.
     
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