Draft of a oil tanker

Discussion in 'Boat Design' started by gust, Sep 16, 2016.

  1. gust
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    gust Junior Member

    Allright. So my displacement is (approximately) = 240*32*12*1000*0.9 = 82944000kg.
    Isn't that a bit high?? (my DWT is 60000)

    I'm asking all of this because I need to calculate forces acting on a ship due to wind and current using the ocimf guidelines.

    The part about the ballast load is what I don't get. The say ballast = 40% of draft. Does that makes sense to you????
     
  2. TANSL
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    TANSL Senior Member

    specific weight for sea water use to be 1025 kg/m3.
    Ballast = 40% of draft does not mean anything.
    I guess the volume of ballast should be given as % of the total load (60% as I say) or as a % of full load displacement (40% as you say).
    In any case, again, the ballast should be the required to fully submerge the propeller, regardless of % of what we are talking about.
    Note that a very important force is exerted by the wind on that part of the ship that is above the waterline.
     
  3. JSL
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    JSL Senior Member

    gust: add the correction for specific gravity (assuming you are in sea water) and the Cb (assuming the ship has some form and is not simply a box)
    82944000 = 82944 tonnes x 1.025 (for sea water) = 85017 x 0.85 = 72,265 tonnes.
     
  4. sharpii2
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    sharpii2 Senior Member

    Not really. Maybe even a bit low. See, your first number is in kilograms; your second one is in tons (1000's of kg). So, if the ship was 60,000 tons deadweight (my guess the ship itself, with no cargo) and only 82,944 tons, loaded, that would mean the 60,000 ton ship could only carry 22,944 tons of cargo, a little over one third its weight.

    Perhaps someone better informed could enlighten us on this.

    My flying guess is the 60,000 ton number includes ballast, which the ship must carry when it has no cargo.

    The 40% of draft number may be how much deeper the ship must float, than absolutely empty, when it has no cargo. The ballast could be to get it to float that deep.
     
  5. TANSL
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    TANSL Senior Member

    I think we're doing something silly talking about figures when we do not know the shapes of the boat. For example, a Cb = 0.85 or 0.95 makes a big difference in displacement. Could you show us a skeg of the ship?.
     
  6. gust
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    gust Junior Member

    I really appreciate your help guys, sadly I don't know anything else about the ship :(, thats why I'm using estimates and guesses.

    I found a reference saying that for Oil tankers Cb=0.9

    The 40% draft thing must is a missreading from my part, in any case I don't know many info that the guides request so there's no point in guessing what it means :eek:.

    Re-calculated displacement = 85867776kg = 85867,776t, using Cb = 0.9 and wd = 1025kg/m3. TY!!

    My real real project is simulating a multybuoy moored tanker btw :D.
     
  7. TANSL
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    TANSL Senior Member

    You will need superstructure area to calculate wind's force on it.
     
  8. mydauphin
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    mydauphin Senior Member

    Varies greatly depending on cargo...
     
  9. TANSL
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    TANSL Senior Member

    Any calculation must be carried on a specific laod condition. Thus, it depends a lot on cargo or on ballast content.
     
  10. gust
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    gust Junior Member

    Yes, I'm using 60000dwt and 40% of that for ballast load.

    Does anyone know any reference where I can find oil tankers of 60000dwt and similar dimensions (240x32x20m)???

    Maybe using a real ship could work for me to have better stimations for the calculations I'm doing.
    Thank you guys.
     

  11. TANSL
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    TANSL Senior Member

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