Drag Calculation of Round Hull

Discussion in 'Boat Design' started by myaney, Nov 20, 2021.

  1. myaney
    Joined: Nov 20, 2021
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    myaney New Member

    I am calculating the drag for a round hull - I know this form sucks, but it is only a transition state while the vessel gets up on foil.

    D = 1/2 × r x Cd x A x v^2

    My question is what to use for Reference Area? Planform Area, Wetted Area, or Frontal Area? The shape is an upside down spherical cap (bottom of a ball). High Reynold's number regime: 15-20 knots foiling speed with 12 meter LWL on the displacement hull initially, progressively decreasing as the hull lifts out of the water. Second question is what approximate Cd to use? Does 0.2 for a semi-streamlined half-body make sense as a rough approximation?
     
  2. kerosene
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    kerosene Senior Member

    I don't think your formula works for surface vessels. looks like a basic drag formula that does not account for wave making.
     
  3. BlueBell
    Joined: May 2017
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    BlueBell "Whatever..."

    Depending what your "take-off" speed is, a round cross section in your hull may be the best form to use along with a length-to-beam ratio of about 18:1.
    Unless you're looking to plane before lift-off.
     
  4. jehardiman
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    jehardiman Senior Member

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  5. Heimfried
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    Heimfried Senior Member

    A and Cd are interdependent. While in a common textbook A is the area of the obstacle in question projected to a plane perpendicular to the velocity vector of the streaming fluid, sometimes the planform area is used instead (e. g. Abbott, von Doenhoff: "Theory of Wing Sections"). Next question is what units do you use? And does "r" mean fluid density?
     
  6. fredrosse
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    fredrosse USACE Steam

    The proposed drag formula is for a continuum of constant density fluid, not involving free surface flow. For a floating spherical shape, with the static draft a small fraction of the sphere's radius, I would think that beyond V(knots) = 1.35 * SQRT(Waterlne Lengh, Feet) would be the practical limit at which point the spherical hull will start to dig itself into the water. This is the classical limit for displacement hulls. The spherical shape will be ideal to produce this problem, and its undesirable effect of resisting further velocity increase, with resistance forces behaving somewhat near a fifth power function of velocity. This is far worse than the classical second power function originally proposed. Planing speeds would generally require a transom with sharp separation, and a sphere here id very much not the proper shape to enter into the planing realm.
     
  7. Mr Efficiency
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    Mr Efficiency Senior Member

    Why do you want to start with something you say "sucks" ?
     
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  8. bajansailor
    Joined: Oct 2007
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    bajansailor Marine Surveyor

    +1 - what type of hydrofoil boat will this hull form be used for, and how big will it be?
    Why do you want it to be bubble shaped, when there are better hull shapes available?
     
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  9. kerosene
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    kerosene Senior Member

    As myaney is having connection issues on Mars I will guess that this is for a concept study.

    why cap of a sphere for a boat?
    Because tetrahedron was taken… B8F8FB85-B8F3-4F1E-B48A-DB7E68862B07.jpeg 0B1C72A1-8A40-4315-8E1A-01DF68D46AFF.jpeg

    It is nice that foils can make any random shape super feasible and efficient, And FAST! Check this one out:
    EAA43CDB-7AC8-4348-A432-761DEB59CEF9.jpeg
     
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  10. Mr Efficiency
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    Mr Efficiency Senior Member

    The compressed air drive sounds novel, but not likely effective.
     
  11. clmanges
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    clmanges Senior Member

    Yeah, the trouble with compressed air as propellant is that ... it's compressible.
     

  12. Ad Hoc
    Joined: Oct 2008
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    Ad Hoc Naval Architect

    The total Drag, or correctly, Resistance is more complex than that.
    You have the residuary resistance, viscous pressure resistance and frictional resistance.

    That one liner equations is just "one" part of the total resistance - drag.
     
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