Center of Effort Help

Discussion in 'Boat Design' started by LTDboatdesign, Oct 29, 2010.

  1. LTDboatdesign
    Joined: Oct 2010
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    Location: Bainbridge Island WA

    LTDboatdesign Junior Member

    I have looked through other posts on this issue and still can't figure it out. The books i have and from what i have seen in other posts say that you are supposed to measure back from a point (forward Lwl in my case) to the center of effort of each sail, then multiply that number by the area of the sail, repeat for all sails then divide that number by the total sail area.
    I have designed a cutter rig and when I do the math I am getting a huge number. has anyone else had these problems. I am using feet and inches for the length back and square feet for the area. I don't know what i'm doing wrong.

    Thanks Cody
     
  2. daiquiri
    Joined: May 2004
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    Location: Italy (Garda Lake) and Croatia (Istria)

    daiquiri Engineering and Design

    Remember that you have to do it in both X and Y direction. ;)
    For example (main + jib):
    Xce = (Xce,main*Amain + Xce,jib*Ajib) / (Amain + Ajib)
    Yce = (Yce,main*Amain + Yce,jib*Ajib) / (Amain + Ajib)

    My guess is that you have messed something up with the unit system. Can you show some numbers?
     
  3. LTDboatdesign
    Joined: Oct 2010
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    Location: Bainbridge Island WA

    LTDboatdesign Junior Member

    my numbers

    main = 179 sq ft
    Stay Sail = 118 sq ft
    Jib = 127 sq ft

    total area = 424

    measurements back from LWL

    Main 15.5 ft
    Stay 6.5 ft
    jib 2.75 ft

    so (15.5*179)+(6.5*118)+(2.75*127)/424

    that should be it, right

    I'm more concerned now with the horizontal ce i will find the vertical later.

    thanks for the help
     
  4. Perm Stress
    Joined: Sep 2009
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    Perm Stress Senior Member

    Horizontal and vertical CE's are interrelated: the higher the rig, -the more forward CE of sails should be... .
    Could you post some sketch?
     
  5. daiquiri
    Joined: May 2004
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    Location: Italy (Garda Lake) and Croatia (Istria)

    daiquiri Engineering and Design

    Yes. How did you find the CE position of each sail?
     
  6. LTDboatdesign
    Joined: Oct 2010
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    LTDboatdesign Junior Member

    I found the center by drawing lines from the corners to the opposite centerline on the triangle. Pretty standard way i think.
     
  7. daiquiri
    Joined: May 2004
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    Location: Italy (Garda Lake) and Croatia (Istria)

    daiquiri Engineering and Design

    It is the traditional geometrical method for estimation of the CE position, but it has not been confirmed at all by CFD and wind tunnel testing. In reality, CE of sails moves with angle of attack relative to the wind.
    Let me quote myself from another thread (don't want to re-type everyhing :D ):

    "the center of effort (C.E.) of the sails move aft with the increase of the A.A. To give an idea of this, increasing the A.A. from 10° to 20° can shift the C.E. from the initial 30% to about 40% of the mean sail chord.
    It means that for a, say, 42 yacht with a sail having some 14 feet mean chord, the C.E. can travel by more than 1 ft for 10° of change in A.A."


    Try searching this research paper in internet: "Computational and experimental study on performance of sails of a yacht" by Jaehoon Yoo and Hyoung Tae Kim.
    At page 1341 it contains a numerical example showing the huge difference between traditionally calculated CE (the one you did) and the CE obtained from experimental and CFD work.

    Cheers
     
  8. latestarter
    Joined: Jul 2010
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    latestarter Senior Member

    not quite, it should be

    [(15.5*179)+(6.5*118)+(2.75*127)]/424


    the answer to the above formula gives 9.175
     

  9. HakimKlunker
    Joined: Aug 2009
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    HakimKlunker Andreas der Juengere

    That makes 9.17 ft. That looks correct and I do not have the feeling that the number is huge.
     
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