Basic Hydrofoil lift/speed calculator

Discussion in 'Hydrodynamics and Aerodynamics' started by revintage, Sep 23, 2018.

  1. revintage
    Joined: Nov 2016
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    revintage Senior Member

    Did this basic calculator, based on Ray Vellinga´s metric approximation, L= 4.06 X V^2 X S X Cl.

    Wanted an easy tool for finding out needed foil area for a chosen take-off speed, for my planned wave piercing hydrofoil tri project, Windknife.

    The formula could eventually be adjusted to get better ballpark figures.

    Any suggestions?

    www.racerdirekt.com/takeoff.html

    Screenshot:

    takeoff.png
     
    Last edited: Sep 24, 2018
  2. BlueBell
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    BlueBell Ahhhhh...

    I guess you mean drag is not a variable in the equation.
    But it would be accounted for in the "fudge factor" 4.06, no?

    Not sure why you'd adjust the constant (4.06).
    Isn't it already adjusted?

    Suggestions on what?
    Maybe it's a language thing but I don't follow your post...
     
  3. revintage
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    revintage Senior Member

    4.06 "fudge factor"?

    Why not find out; search Vellinga and metric in this forum.

    Awaiting suggestions from someone of those who has the knowledge, to see if the figures are realistic.
     
  4. BlueBell
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    BlueBell Ahhhhh...

    Ohhhhh, yes, the figures are realistic.
     
  5. Doug Halsey
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    Doug Halsey Senior Member

    That formula tells you how much foil area you need to get the required lift at given values of CL & boatspeed, but that only answers part of the question of whether or not you can fly.

    As mentioned earlier, the drag also has to be considered. If taking off causes the drag to increase, you may slow down & drop back into the water. Or, given enough thrust, you might stay foilborne, but would probably be slower.

    Unfortunately, you need more than just one simple equation to determine whether or not that will happen.
     
  6. revintage
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    revintage Senior Member

    Doug, I am aware of the simplification. But for me, the power needed to get the expected lift is about drag. But maybe this to simplify things to much? If we assume that enough sailarea/engine is available to overcome the drag, would it then be an acceptable formula to use? Actually I tried the formula on the Broomstick data and found it realistic, CL=0.7 gave take-off at 10knots. Anywhere near your IRL figures?
     
  7. Doug Halsey
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    Doug Halsey Senior Member

    Sure, but it's just the same old formula (the definition of CL), with the constant adjusted to be consistent with one particular set of units.
     
  8. BlueBell
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    BlueBell Ahhhhh...

    Revintage,

    I am familiar with Ray's book, I have read it.
    It's very basic but helpful to a newbie.

    You don't need to adjust the formula.
    You need to overcome the drag of your vessel.
    If you can make 6-knots maximum without foils, then design for a take-off speed of 5-knots as you'll have the additional drag of the foils and struts to overcome.
     
  9. BlueBell
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    BlueBell Ahhhhh...

    Hi Doug,

    Drag has already been considered, otherwise you wouldn't be able to make take off speed.

    Drag will not increase after lift-off, it will decrease.
    It will then begin to build as speed increases on foil.

    In it's simplicity, the formula negates drag.
    If you can't make take-off speed, then you won't fly.
     
  10. Doug Halsey
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    Doug Halsey Senior Member

    It's entirely possible for the drag to increase after take-off, if you push for the lowest possible takeoff speed. Taking off in a surface-piercing foiler reduces the foil area (which can even cause a stall) and the span (which causes an increase in induced drag).

    Even in a Moth (with submerged foils), there are times when you can get the hull out of the water, but can't sustain it.
     
  11. BlueBell
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    BlueBell Ahhhhh...

    "Pushing" for the lowest take off speed will result in lower overall foiling speeds as your foils will be inherently slow foils.
    Pushing for the highest take off speed will make for smaller, higher speed foils.
     
  12. Doug Halsey
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    Doug Halsey Senior Member

    That's true, but I was referring to pushing for the lowest take-off speed for a given set of foils.
     
  13. BlueBell
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    BlueBell Ahhhhh...

    Why would you want to minimize take-off speed?
    I want the highest take-off speed possible, even if it means adding flaps.
     
  14. Doug Halsey
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    Doug Halsey Senior Member

    Presumably, you would want to be able to fly as soon as possible (i.e. - at the lowest speed). That's not always optimal though.
     

  15. BlueBell
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    BlueBell Ahhhhh...

    Ah, multi-purpose.
    I gotcha.

    Two different, interchangeable inverted T's then.
    Low speed and high speed.

    Or, a wing that resembles an airliners, that is... hugely deformable.

    Sorry for the thread drift there Revintage.

    Edit: There are two different threads I'm following and I'm getting CROSSTHREADED!
    (That would be a terrible name for a boat.)
    They both relate to small foils.
     
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