# center of effort gaff sail

Discussion in 'Sailboats' started by susho, Mar 9, 2009.

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### sushoComposite builder

hi,

In the dutch waterscouts and seacadet corps we use a small sailing boat called a lelievlet. It is a steel multi chined boat , 5.6 metres in length.
It has a fault in it's design: the center of effort in the sails , and the lateral point are out of balance. I already have got the lateral point, But I need to have the center of effort.
I've got a book here wich explains how to determine it graphically, but when I've got the two triangles , I need to combine the two centrepoints in to one. and the book isn't very clear in that point.

Can someone help me?

Thank you,
Johannes

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### matoiJunior Member

You can connect the individual effort centers with a line. Then divide this line into two - in proportions which relate to each other in the same way as areas of the two triangles relate to each other. The combined CoE should be in place where your line is divided. Of course, this point of division should be closer to the CoE of the triangle which has bigger area.

For example if your first triangle is 3 sq m, and the other 2 sq m, then you should divide the line so that x1:x2=3:2

If you make any sketches, please post them - it would be interesting to see what you've come up with. You can take a look at my amateurish atempts with a Wayfarer dinghy rig:

Good luck.

m

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### sushoComposite builder

Thank you, makes sense, and I already tried it before, but I couldn't believe it was this simple

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### TcubedBoat Designer

The geometrical method is to connect the centers of the two triangles with a line. At the end of this line, at the center of triangle B, draw a new line with a length that represents the area of triangle A at right angles to the line connecting the centers of the triangles. Now repeat with a line proportional to surface B starting at center of A, normal to the line joining the centers but going the other way. Finally, draw a line connecting the ends of these two new lines. Where this line intersects the first line (that connects the centers) is the center of the total area.

Example; A has 150 M^2 , B 250 M^2.
1-find center of A and center of B
2-Join centers with a line. Call this AB
3-Draw a line 90 deg from AB starting at point A 25 cM long. Call this AB'
4-Draw a line 90 deg from AB (but 180 deg from AB' ) starting at point B 15 cM long. Call this BA'
5-Join A' to B' with a line and mark where A'B' intersects AB

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### sushoComposite builder

that was the method used in the book, But I couldn't find how to determine the length of those parallels. I've got all the info I need now, thank you.
It turns out that the center of effort is almost directly above the lateral point.

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### PARYacht Designer/Builder

There's a much easy way of finding a center of the sail.

Draw a line from the center of the foot (along it's length) to the center of the gaff and another from the center of the luff to the center of the leach. The resulting "X" will place the CE in a more realistic location for balancing your sail plan over the CLP. And you don't have to use a calculator or any geometry.

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### alan whiteSenior Member

Don't be surprised if the sail has exactly the correct lead according to the recommended percentage. There is a lot of witchcraft to balancing a sailboat involving many subtle factors (hull shape, rudder area, etc.)... and the sails must be in good condition and set properly for the wind conditions.
It would be unusual for an established class to be so far off the mark, as early refinement should have done away with the worst of design flaws.
Are all of the boats in the class the same?

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### sushoComposite builder

The class is developed right after WW2. Then it had 16m2 sail area and the mast at the front bulkhead. That changed to 12.15, and the mast placed on the daggerboardcase. We don't know if they changed more. There where 3 producers of them, now 2. And a newer design wich sails slightly better. (with a slightly bigger skeg)