Semi displacement powercat

Discussion in 'Boat Design' started by Sailcy, Mar 23, 2016.

  1. fallguy
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    fallguy Senior Member

    I would like to know how you calc the LD ratio Ad Hoc. Thank you. I can’t get to your number of 6.1. ...Dan

    Units maybe?

    For a cat is it per hull displacement?

    Are you using meters per English ton? Seems the only way I can get to it..

    I want to calculate it for my boat is all.
     
    Last edited: Jan 12, 2018
  2. Ad Hoc
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    Ad Hoc Naval Architect

    Perhaps you missed it:

     
  3. fallguy
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    fallguy Senior Member

    If we take 8=28 feet divide the cube root of displacement; I need the cube root of 43, which isn’t logical.

    If we take 8=8.5m divide the cube root of displacement; I need the cube root of 1.2. And that might be thousand kg? which is 2688#.

    When I did the calc for my boat; I get 7.61.

    Is it m/metric ton? I think last night I used the number 6.1 that you originally referenced. And using the same units; I arrive at 6048# per hull?

    We can take this offline if you prefer; I genuinely just want to understand my boat.

    In a hurry now. Genuine thanks.
     
  4. JSL
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    JSL Senior Member

    for US/Imperial units the formula I have used is
    Displacement Length Ratio- as referred to an all the Naval Architecture books/papers etc I have seen
    Disp (tons)/ (0.01 LWL)^3
    ex: 10 long tons on 35'
    10/ (35 x .01)^3 =
    10/0.0428 = 233
    for a mono-hull sailboat, fairly moderate
     
    Last edited: Jan 14, 2018
  5. Heimfried
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    Heimfried Senior Member

    I understand Ad Hoc's answer this way:
    Ratios are usually just a number without a dimension or unit (at least in engineering).
    Therefore displacement is to understand as displacement volume, because displacementVOL^(1/3) = cubic root of displacementVOL is a length, same dimension as the waterline lenght.

    In this case the LD ratio = lenght/displacementVOL^(1/3) is independent of the used units. The lenght in meters divided by cubic root of (DISPVOL in m³) is the same number as the length in feet divided by cubic root of (DISPVOL in cubic feet).
     
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  6. fallguy
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    fallguy Senior Member

    Thank you

    Okay, I like the math now.

    But how did he ascertain the displacement volume for the Wood’s cat then? He only knew the length that I could tell.

    How can I determine the displacement volume if I have displacement weight. Is it a straight conversion? Thanks for any help-just trying to learn a bit. Back to my spreadsheets.
     
  7. TANSL
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    TANSL Senior Member

    Divide the weight by the density of the water and you will obtain the volume. Use the correct units in each case.
     
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  8. fallguy
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    fallguy Senior Member

    Thanks-will do.

    Thanks to all. It should be doable now.

    Still not sure how ad hoc got the Skoota 28 numbers-maybe he looked at a study plan.
     
    Last edited: Jan 12, 2018
  9. Ad Hoc
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    Ad Hoc Naval Architect


  10. fallguy
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    fallguy Senior Member

    I ran all the Wood's 28’ plus powercats save the 36’ and I got freshwater numbers of....
    6.1 for the Skoota 28
    7.5 for the Skoota 32 fixed
    7.6 for the Skoota 32 DM (the one I am building)

    Probably the -28- got a bit heavy because it is ply; the other two are foam sandwich. Not sure I will come in as light, time will tell.

    Thanks again.
     
    Last edited: Jan 12, 2018
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