L/B ratios the same for short and long hulls?

Discussion in 'Multihulls' started by BMP, Jun 15, 2022.

  1. BMP
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    BMP Junior Member

    I have read that wave drag becomes much less of an issue on hulls with a L/B of greater than ~8:1 and improves with higher ratios till wetted surface drag becomes the main event. Hypothetically if I have two hulls, one that is 8 (meters or feet take your pick) long with a beam of 1 and one that is 16 long with a beam of 2 they have the same L/B will but different LWLs all else being equal (L/D etc.) will they have similar performance?
    Will the performance scale with size if the ratios are the same? 2x long 2x fast?

    Sorry for the beginner questions and thanks for the help.
    Godspeed,
    Ben
     
  2. TANSL
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    TANSL Senior Member

    L/B is rather a ratio that could affect, more, the stability of the hulls. For the resistance to the formation of waves or the resistance due to friction, you must also take into account other ratios and coefficients in which the draft of the boat always intervenes.
     
  3. jehardiman
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    jehardiman Senior Member

    "Similar" being the operative word here. The total hydrodynamic resistance of vessel is comprised of two parts; the "skin friction" and the "form" or "residual" resistance. Skin friction is simply a function of the wetted surface and the speed through the water. The form resistance is more complicated; indeed the reason it is sometimes call the residual resistance is because it is what is left when the skin friction is subtracted from the total resistance. The easiest way to explain form resistance is to say it is the pressure variation over the wetted surface caused by moving the mass of the water out of the way and pulling it back in behind you. These pressure variations not only retard the vessel, they also cause the wake waves to be formed due to energy being put into the fluid. As the speed of the vessel changes, the form resistance changes based upon the interactions of pressure changes and energy already imparted to the fluid in the form of wake waves. Because wake waves propagate through the fluid proportional to the gravitational acceleration (g), Froude early on came up with a way to express this speed (V), length (L), and gravity interaction as the Froude Number (Fn): i.e. Fn =V/sqrt(gL). Generally; the terms Total resistance, skin friction, and form resistance are non-dimensionalized into coefficients. (see the referenced ITTC procedure)

    To make a long story short, the only way that two geometrically similar hulls have "similar performance" is that the coefficient of form resistance will be the same when both are operating at the same Froude number.

    https://ittc.info/media/1217/75-02-02-01.pdf
     
  4. Ad Hoc
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    Ad Hoc Naval Architect

    No, when you use a geosim (scaling) of one hull to another, just like from model to real vessel, the powering scales to the power of ^3.5, pro-rata.
     
  5. BMP
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    BMP Junior Member

    Great thanks.
     
  6. fallguy
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    fallguy Senior Member

    Lurking here and trying to understand.

    So, Wood's 20' Skoota ran quite nicely with a 20hp, then the Skoota 28 ran nicely with two of them.

    So, same hulls ftmp..8' longer waterline, double the horsepower..speeds probably close

    20=x
    28=1.4x
    1.4^3.5 = 3.25

    20hp to 40hp = 2.0

    Have I done it right? Or is the imperial throwing it all off? Or my poor interpretation, etc?
     
  7. Ad Hoc
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    Ad Hoc Naval Architect

    What does ftmp mean?

    Hard to tell, as i don't understand your nomenclature.
     
  8. fallguy
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    fallguy Senior Member

    ftmp is for the most part
    20 and 28 are waterline lengths

    I'm just trying to understand how to put the numbers you cited into an equation.

    it is very interesting, but quite sure I've misinterpreted
     
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  9. Ad Hoc
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    Ad Hoc Naval Architect

    Ok..so the two hulls are not geometrically the same then?...i.e. just at different scales.
    In this case, it does not apply.
     
  10. fallguy
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    fallguy Senior Member

    No, the are the same geometry.
     
  11. Ad Hoc
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    Ad Hoc Naval Architect

    Ok..but im getting confused by the words used to describe them...as this is contrary to your statement of 'ftmp' etc..

    So, just to be 100% clear. The two hulls are identical in every way, the L, B T etc all scale as the same and same shape etc, the only difference is that one hull is at a different scale to the other.
    Is that correct?
     
  12. fallguy
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    fallguy Senior Member

    Yes
    The geometry is the same, but the L and the B would be scaled.

    so 28' long scales to 20' as 1.4
    and the beam of the 20 foot is say 1.8' and the beam of the 28' is say 1.4•1.8 or 2.5'

    But my example doesn't matter all that much, I am just trying to wrap my head around what you mean by the powering scales to ^3.5...

    Like are you saying a vessel 140% larger, same geometry requires power of the original vessel ^3.5?

    so, power of vessel A is 20 hp
    20 hp^3.5=35771??

    I am just tryin to understand how the calc works because it is very interesting
     
  13. Ad Hoc
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    Ad Hoc Naval Architect

    In this sense then, no, they are not identical geometry, since their parameter of ratios are different.

    Since using the scale - f - Ok, we used to do this sum endlessly before computers came along and so always when time didn’t permit, were always looking for short circuits for the salesman waiting for the answer at the end of a fax line!
    That tells you how long ago I sued to do these quick 'sums'.

    If, for example, you have a hull which is geosimed and the scale = f.
    And this means the L, B and T everything is identical. Just like a scaled model in a test tank is an exact facsimile of the real vessel, just scaled smaller.

    So LN/LO = f

    Where LN = new length and LO = original length.

    So If we have the EHPO (Original naked resistance) of the original what would it be for a geosim (exact scale)? …well, if we ignore many of the basic inputs we can work out that it is simply EHPN = EHPO x f^3 ½ , that is f to the power 3 ½ or 3.5 whichever you prefer.

    How??...well, the EHP = resistance x speed.

    Resistance = frictional + residuary.

    Frictional is a function of:
    WSA = f^2
    Speed = f^1/2
    Rn = speed x length = f^1 ½

    And residuary is a function of f^3.

    Thus the resistance (total) = f^3 + (f^2 x f^(1/2 + 1/2)/f^(1 ½ + 1 ½ ) = f^3 the frictional “scales” cancel out)

    And since speed = f^1/2 the EHPN = EHPO x change of scale = EHPO x f^3 x f^1/2 = f^3.5

    Make sense?

    Since using the scale - f - as the base, everything can be calculated based upon the scale factor.
    Assuming the 2 hulls are an exact scale of each other.
     
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  14. fallguy
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    fallguy Senior Member

    Thank you. Here, Father's Day and I am cooking steaks. Yes, cooking, but per my request. I will look it over tomorrow. I suppose units may matter somewhat?

    And, Richard surely did things right. It is me that is guessing!!
     
    Last edited: Jun 19, 2022
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