New hull speed equation

Discussion in 'Sailboats' started by JakeWood, Aug 7, 2023.

  1. baeckmo
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    baeckmo Hydrodynamics

    ....and if D/L represents "Displacement in long tons", what "ratio" are you referring to then?
     
  2. Robert Biegler
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    Robert Biegler Senior Member

    It is the D that refers to Displacement in long tons. I used the underlined font in an attempt to write a fraction. That is why the ratio got its own two lines. Read it as a fraction, then tell me what you think. If you prefer it written on a single line, here it is:

    Imperial measurements: D/L ratio = (Displacement in long tons)/(0.01*length in feet)^3

    Metric, made consistent with the imperial measurement version: D/L ratio =(Displacement in m^3)/(0.0328084*length in m)^3

    If you want to be really precise, you can correct for a long ton of 2240 lbs corresponding to 1016 kg, and ask about the temperature and salinity of the water that is being displaced, but that would be rather going off the topic of what the D/L ratio is that is usually cited in connection with boat speeds. Could we agree that the term "D/L ratio" is misleading when taken literally because it should be written "D/L^3 ratio" to accommodate dimensional analysis?
     
    Last edited: Aug 9, 2023
  3. DCockey
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    DCockey Participant

    The multiplier “1.34” is really a function of D/L ratio.
    Wow!
    I know Gerr tends towards the emperical but I thought he was more knowledgable than that.
     
  4. baeckmo
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    baeckmo Hydrodynamics

    Well, now you are referring to the "D/L ratio" as the slenderness ratio. Then we are back to the original statement. If we express it in writing, with your definition, it says that the "multiplier" is a function of slenderness ratio. This is, as explained by AdHoc a couple of notes back, fundamentally wrong; it has nothing to do with slenderness ratio.
     
  5. Robert Biegler
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    Robert Biegler Senior Member

    I think I have seen it called that as well, so fine by me.

    No, we most definitely are not. I was not and I am not commenting on Dave Gerr's empirical formula, or Carlos' interpretation of that formula, or Ad Hoc's comments. You told Carlos he got the dimensions wrong. Judging by the calculation he did for Beowulf, he did not. If you agree that this particular point of criticism was invalid, you should say so in a manner as public and as direct as your original statement. If you think your criticism is valid, you can explain why, and if I agree, I will say so. Otherwise, I am done with this.
     
  6. Stephen Ditmore
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    Stephen Ditmore Senior Member


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