Sheet tension. Is this correct?

> 10 years ago
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sirgallivant
sirgallivant
NSW
1531 posts
NSW, 1531 posts
10 Sep 2015 6:12pm
Multiply sail area in sq. feet by wind speed in knots squared, dividing the result by 232.
Ex: 150 sq feet x 15 knot2 (225) : 232=145 pounds

Yara
Yara
NSW
1322 posts
NSW, 1322 posts
10 Sep 2015 9:39pm
It is really a lot more complicated than that. You could tie yourself in theoretical knots, but why not get practical. Hook a simple spring balance onto the end of the sheet and measure the tension. Note the wind speed. The force on the sheet will vary with the square of the wind speed. EG double the wind speed and the force will increase by 4.
The force is proportional to the area, so double the sail area and you will have double force.

All approximations, of course, but better than trying to do pure theory.
rumblefish
rumblefish
TAS
824 posts
TAS, 824 posts
11 Sep 2015 8:17am
I have always used this formula
Genoa
www.harken.com.au/content.aspx?id=9096
Main
www.harken.com.au/content.aspx?id=9094

Traveller load is usually 0.2 of mainsheet load
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