Join Date: Jan 2007
Thanked 2 Times in 2 Posts
Rep Power: 11
Here's my wet finger guess:-
Take the mass of the sail and boom and multiply by the relative wind speed to get the maximum momentum. That's the worst impulse that has to be stopped. (It assumes that the sail and boom accelerate up to relative wind speed before being stopped, so its probably the worst case.)
So if your boom and sail together weigh 40 kg and the wind is about 10 m/s then your main sheet has to take a 400 kgm/s impulse. I guess that means, if the sheet or the shroud stops it in 10 cm, its going to have to apply 40000 N equivalent force. (Ouch!)
It looks a lot better if you can slow it down over say 4 metres, then the impulse a) doesn't build up and b) would a lot less than 1000 N, so effectively you only have to deal with the wind force.
The force of 20 knots (ca. 10 m/s) of wind on a 400 sqft (=53 sqm) is about 0.5x1.22x1x53x100 = 3233 N but maybe 2/3 of that will be carried by the mast, so the sheet tension will be about 1000 N, divided by the mechanical advantage of a 4 part block system will be about 250 N (about a 55lbf pull).
Sounds roughly right?