Calculate Inclination of pontoon

Discussion in 'Stability' started by Robert Wilkinson, Feb 11, 2018.

  1. Robert Wilkinson
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    Robert Wilkinson Junior Member

    I am trying to workout how to arrive at the answer for the following question.

    A rectangular pontoon is 12m long, 4.2m broad and 2.1m deep and it floats in fresh water. The pontoon has a displacement of 65 metric tons unloaded and the centre of gravity is 0.6m up from the bottom and 20 mm aside from the vertical centreline.
    Determine inclination of centreline when pontoon floats freely?
    If pontoon becomes partially floaded, find inclination of centreline when depth of water inside pontoon at centre is 200 mm?

    Answer from book is given as 58 minutes and 4 degrees, 40 minutes.

    For the first part I tried calculating it using GG' = inclination * (I/V) where inclination is in radians, I = second moment of inertia and V = volume displaced.
    I used GG'= 20mm, I = (12 * 4.3^3)/12 and V = 65m^3 and I got inclination = 60.32 minutes or just over a degree.
    I was just thinking that 20mm is probably the perpendicular distance from centreline, so when pontoon is inclined you will have to workout horizontal GG' somehow.

    I'd just like to hear some thoughts on this from anyone!
     
  2. upchurchmr
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    upchurchmr Senior Member

    Trying to get your homework done?
     
  3. TANSL
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    TANSL Senior Member

    Forgive my way of expressing myself and, please, do not be offended. I want to help. Are you an amateur trying to learn more about naval architecture or are you a student of some branch related to naval architecture that tries to get someone to solve his problems? In the first case, I will be happy to help you. In the second case, the best advice I can give you is to study a bit.
     
  4. Robert Wilkinson
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    Robert Wilkinson Junior Member

    No it's not homework, so this is not to assist me with a qualification.
    I am interested in naval architecture and have a hobby on the side,where I am trying to answer questions from a book.
    May be you can show me how to arrive at the answers of 58 minutes for when pontoon floats freely and 4 degrees, 40 minutes when depth of water inside pontoon at centre is 200 mm.
     
    Last edited: Feb 19, 2018
  5. Robert Wilkinson
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    Robert Wilkinson Junior Member

    My answer of 60.32 minutes for the first part of the question when pontoon floats freely relates to the horizontal distance from the vertical centreline, but with the way that the questions is worded "centre of gravity is 0.6m up from the bottom and 20 mm aside from the vertical centreline" that 20mm aside will be on an angle, I 'm not sure how to calculate horizontal distance from this .
    The answer is 58 minutes, not too far away from mine of 60.32 minutes, so I am nearly on the right track.
    What are your thought?
     
  6. TANSL
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    TANSL Senior Member

    To do it one way quite correct, you would have to calculate the curve of the righting arms of the ship (curve GZ) as if the weight of the tank were in the center line, then calculate the heeling arm due to the transferred weight, for each heel degree , which is a cosinosoidal curve (see picture), and calculate the first point of intersection of the two curves. This method supposes to forget the effect of the free surfaces (if the volume of the tank made it necessary), but I do not think that the exercise has tried to complicate so much the thing.
    Snap2.jpg
     
    Last edited: Feb 19, 2018
  7. Robert Wilkinson
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    Robert Wilkinson Junior Member

    I'm still confused how to deal with the fact that the centre of gravity is 20mm aside from the vertical centreline.

    From your transfer of weights image
    Me = P * (h * cos * theta + v * sin * theta)

    theta = Heeling angle, I can't use that same symbol you have in image
    so from this would h * cos * theta = 20mm aside from vertical centreline and v * sin * theta = 0.6m up from the bottom be correct

    But still some uncertainties like what would P be when barge is unloaded, would it just be mass of pontoon = 65 Tonnes

    And then you had Ha = Me / Ship displacement (or in this case Pontoon displacement)

    From my understanding that I have, in order for the barge to float freely, the overturning moment = right moment
    But what is the overturning moment when pontoon is unloaded, this is another uncertainty of mine for this question.

    When the barge becomes partially flooded with 200mm of water
    then assuming walls of pontoon are thin, that mass of water = 12 * 4.2 * 0.2 = 10.08 m^3 or 10.08 Tonnes
    It's centre of gravity will be 0.1 m up from the bottom on vertical centre line
    so from that we can take moments to find combined centre of gravity, the only thing is how do you handle original 20mm aside from vertical centreline for this.

    Still an unsolved question for me, may I please have some more help.

    Just curious as to where you got the Transfer of Weights image from.
     

  8. TANSL
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    TANSL Senior Member

    It is necessary to clarify that initially you must calculate as if the weight were in the center line. This defines the curve of the righting arms. Then the weight is shifted and the curve of the heeling arms is traced by changing the list (cosinusoidal curve). The point of intersection of both curves will mark the equilibrium heel.
    Although I suppose I could deduce the equation of the heeling arm, at this moment I am not in a position to do so. That is the formula that the Spanish Administration requests to be applied for the ships under its jurisdiction. At the moment I do not know anything else.
     
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