center of lateral resistance

Discussion in 'Sailboats' started by Guest, Jul 23, 2003.

  1. tspeer
    Joined: Feb 2002
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    tspeer Senior Member

    Check your math:
    5 deg = 0.087 rad.

    Cla * a = 0.54

    As a practical matter, the lift curve slope is closer to 0.1 due to boundary layer displacement.

    And don't confuse the two-dimensional lift curve slope with the angle of attack required for a three-dimensional planform. Considerably more angle of attack is needed to reach the same level of lift.

    The apparent lift curve slope of planform is closer to

    a = a0 / (1 + a0 * 57.3 / (pi * AR * e))

    a = 3D lift curve slope, per deg
    a0 = 2D lift curve slope, per deg
    AR = aspect ratio
    e = Oswald efficiency factor
     
  2. RHough
    Joined: Nov 2005
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    RHough Retro Dude

    Thanks Tom,

    What is 57.3?

    I've been using .7-.8 for value of e, is that "close enough" for a sailboat?

    -Randy
     
  3. jam007
    Joined: Sep 2005
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    jam007 Junior Member

    57.3 = 180 / pi
    Conversion from degrees to radians.

    Anders M
     
  4. Raggi_Thor
    Joined: Jan 2004
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    Raggi_Thor Nav.arch/Designer/Builder

    I was just too quick to copy paste some formula...
     
  5. chandler
    Joined: Mar 2004
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    chandler Senior Member

    So is that where all you rocket scientists get your formulas, copying and pasting? For a while there I thought you all Knew and understood these formulas? Silly me.
     
  6. jam007
    Joined: Sep 2005
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    jam007 Junior Member

    No no Chandler donĀ“t confuse egineers with scientists. :)
     

  7. Raggi_Thor
    Joined: Jan 2004
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    Location: Trondheim, NORWAY

    Raggi_Thor Nav.arch/Designer/Builder

    Sorry again, Chandler :)
    One thing is copy/paste or look up in a reference. The most embarrassing is not to check the answer, is it reasonable. When dynamic pressure is 0.5xROxV^2, it is not reasonable to have a lift coefficient much larger than 0.5...
     
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