calculating traveller/sheet loads for square sails

[DeAnander]

I have the math for wind pressure on a rectangular sail of known area... the tricky part is that the boomkin load is the rotational moment of the sail panel and the boomkin lead is downward, not horizontal. my sails are balanced (about 20 pct fwd of the mast) which complicates things a bit, but it's not a super tricky calculation if only the sheet were led horizontally to the boom end -- I sat down with a mech engineer from the workplace and we came up with something like 3/7 * F * L = T (where F is wind force on the sail area and L is the boom length to the sheet attachment, T is the tension or load on the sheet)... but when the sheet leads at an angle downward and sideways there's this 3D vector problem which frankly has me stumped.

my boomkin has no bobstay so if I only knew the force being applied to the after end of it, it would be a simple cantilevered beam problem -- and that I know how to solve.

it seems a simple question: how strong does a boomkin (or a mainsheet deck fitting for that matter) have to be? but it turns out to be kind of ugly in practise.

after staring at the problem for a while, neither I nor the mech engineer could think of any advantage to making the boomkin extend any further than the boom. and yet, iirc, most boomkins were traditionally longer than the boom... I could be remembering wrong, but visual memory of paintings and drawings of traditional boats suggests the boomkin longer than the mizzen boom by 2 or 3 feet.