Center of Effort Help

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

1. Joined: Oct 2010
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Location: Bainbridge Island WA

LTDboatdesignJunior 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. Joined: May 2004
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daiquiriEngineering 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. Joined: Oct 2010
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LTDboatdesignJunior 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

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Perm StressSenior 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. Joined: May 2004
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daiquiriEngineering and Design

Yes. How did you find the CE position of each sail?

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LTDboatdesignJunior Member

I found the center by drawing lines from the corners to the opposite centerline on the triangle. Pretty standard way i think.

7. Joined: May 2004
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Location: Italy (Garda Lake) and Croatia (Istria)

daiquiriEngineering 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 ):

"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

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latestarterSenior 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

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HakimKlunkerAndreas 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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