How can I calculate the tonnage

Discussion in 'Boat Design' started by peter radclyffe, Aug 18, 2013.

  1. peter radclyffe
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    peter radclyffe Senior Member

    does 900 tons sound close
     
  2. Mike Graham
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    Mike Graham Junior Member

    I can see if I can post something clearer later. For what purpose are you computing this displacement? What is your background?

    What kind of tons? 900 tons sounds high.
     
  3. peter radclyffe
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    peter radclyffe Senior Member

    900 metric tonnes, i need an idea of the weight, my background is boatbuilding
     
  4. peter radclyffe
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    peter radclyffe Senior Member

    at first i thought 700 tons
     
  5. peter radclyffe
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    peter radclyffe Senior Member

  6. gwboats
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    gwboats Naval Architect

    Displacement Estimate

    I would say you are close with your first weight, Peter. My estimate for that hull using a reasonable sailing waterline shape would be 760 tonnes.

    Cheers,
    Graham Westbrook
    (I can do the modern shapes as well as the traditional ones ;))
     
  7. Mike Graham
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    Mike Graham Junior Member

    700 tonnes sounds a lot more reasonable than 900.
     
  8. Mike Graham
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    Mike Graham Junior Member

    You are going to want to compute one cross-sectional area or many cross-sectional areas to compute the displacement.

    The better method is to calculate cross-sectional areas for a bunch of evenly-spaced stations. There are tons of guides for this on the internet for doing this by hand (google "calculating displacement simpson's rule"). There is also lots of software that can do this. A free option is Freeship. (I can't remember for sure that Freeship can take the table of offsets directly--if not, Michlet can turn the table of offsets into a Freeship file, I believe.)

    The worse method is to assume a prismatic coefficient and just have to calculate one cross-sectional area, the greatest. It's a lot less accurate, but it's enough to progress your number from "wild guess" to "rough estimate".

    -------

    The prismatic coefficient is the volume displaced compared to the volume displaced by a vessel with the same maximum submerged area of a cross-section, but with the same cross-section for the whole ship.

    This graphic someone has posted here before explains it really well:
    [​IMG]

    (It mentions the prismatic coefficients for kayaks ranging from 0.45-0.65. Yours will probably be higher than a typical kayak. I threw the number 0.6 out there, but it could vary a lot. I wouldn't want to just assume one myself, but would go with my method one.)

    In this case, you would come up with the equation

    Displacement = Prismatic coefficient * Maximum cross-sectional submerged area * waterline length * density of water

    -------

    How do you get the many/one area you need? You can get it using a computer, use graph paper and count roughly, break out your trusty planimeter, or take some measurements and integrate numerically using Simpson's rule or similar.

    If your midship section is your largest submerged area, then then it's about 70% full. Assuming a prismatic coefficient of 0.6, your displaced volume is

    0.6 * 0.7 * 54m * 11.4m * 2.65m = 685 m^3

    or 700 metric tons of seawater.

    (You might tack on the keel if it's very volumous.)
     
  9. peter radclyffe
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    peter radclyffe Senior Member

    54 x 11.4 x 2.6 x .44 equals 704
     
  10. nzboy
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    nzboy Senior Member

    rough guess tonnage

    Good guess Mike except I would tend to use a figure of 9.2m as a waterline beam so calc would be about 552m2 which I think I would be fairly safe with .Where Peter did you find 900 ton
     
  11. peter radclyffe
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    peter radclyffe Senior Member

    thank you Mike thats very helpful, i still dont understand simpsons ,im not good at maths. i understand block co effifient, i think, but it appears to be a guess as to what fraction it is, depending on hull type, up to about .66.
     
  12. peter radclyffe
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    peter radclyffe Senior Member

    nz boy how can you use 9.2 when the wl beam is 11.4,or is that an average, what am i missing
     
  13. peter radclyffe
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    peter radclyffe Senior Member

    what is 552m2, the waterline surface ? or something else
     
  14. peter radclyffe
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    peter radclyffe Senior Member

    54 x 11.4 x 2.6 x .47,,,,,,,,,,,752
     

  15. nzboy
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    nzboy Senior Member

    I apologise your right I mistook 11.4 as beam
     
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