calculating traveller/sheet loads for square sails

[DeAnander]

I have been trying to figure out the force on the mizzen boomkin for my three masted traditional (square sailed) vessel, to spec a replacement spar.

Googling around I found Brion's article at the Harken site, which included a "traveller car load calculator" for Marconi rigs. that formula, if I've transcribed it correctly, is

(E**2 * P**2 * .00431 * V**2) / (sqrt(P**2 + E**2) * (E - X) )

where E is boom length, P is luff length, V is wind speed in knots, and X is the distance from the AFT end of the boom to the mainsheet attachment point.

anyway, I figured this was written for a Marconi triangular sail, so for my rectangular sails I should multiply the luff length by 2 (doubling the area so it approximates my square sail of the same boom length).

I ran this model for my 125 sf mizzen with a 12 foot mizzen boomkin, the object of the exercise being to figure out the force on the boomkin mount point (your basic cantilever beam problem).

I was surprised by the magnitude of the numbers I ended up with -- I was getting more force on the mizzen boomkin than on the mast at the partners, which doesn't seem right. surely the load on the boomkin can only be a fraction of the load on the mast...

I'm not good enough at the physics to figure out where the .00431 constant is coming from or how to tweak this formula for rectangular sails, so I appeal to wiser (or at least more mathematically literate) heads -- how would you rewrite this formula for squaresails?

Could you share the url for that article by Brion? I'm not sure I understand the formula as you've written it.

[DeAnander]

http://www.harken.com/traveler/TravelerTuning.php

I'm not at all sure on what this formula is based upon; it appears to be something specific to Harken though it is reprinted in Brion Toss's 'Apprentice.

The constant used here is likewise not explained, but it is modified upward for metric measures and so is likely related to surface area. My guess is that it approximates developed force at sea level/high humidity of air.

::doodles with numbers for Cape Dory 25D::

ML = (105.0625 * 742.5625 * 0.00431 * 100) / (29.1140000687 * 8.75)
ML = 131.99214373252

This seems rather excessive for 10 knots of wind, for one thing.

I don't think this is a reasonable formula to determine the kind of load might develop on your mizzen. A simpler method to get an estimate would be to multiply maximum sail area carried by an estimated dynamic pressure (q) for the maximum wind that much sail would be carried in, times the height of the center of effort of the sail above the step.

To figure out the dynamic pressure you can work out the formula:

q = (p / 2) * V^2

where q is the dynamic pressure, p = 0.002378 slugs per ft^3 (standard sea level conditions estimate), and V = wind speed in ft per second.

Or you could check this quick ugly table:

[code]
V in Beaufort q
knots number lbs per ft^2
----- ------- -----------
10 3 0.33
20 5 1.3
30 7 3
40 8 5
50 10 8
60 11 12
70 12 17
80 13 22
90 15 28
100 16 34[/code]
(Earl R. Hinz "The Complete Book of Anchoring and Mooring" 2d edition)

[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.

There might be a slight advantage in sail shape at narrow sheeting angles, but it's a mizzen so this isn't really relevant. Maybe when setting a riding sail dead flat. On the other hand, in traditional working craft the boomkin (bumkin, or outrigger earlier still) might be used for other purposes than just sheeting the jigger.

It still comes back to the mizzen transverse load, though, when determining what might end up coming down the mizzen sheet. And that can be estimated by RM * .5 * .5beam at mizzen chainplates, * the safety factor. (Skeene's 8th ed.) Cunliffe and Leather talk about unstayed bowsprits (no bobstays), basically pointing out they were excessively over-engineered for the compression loads and then thoroughly abused, and very rarely carried away. Rendering outriggers were literally lashed aside the sternpost when and if they were needed.

The traditional method of figuring out appropriate scantlings may be required here. Build something by guess, then start removing material until it breaks. Then fix it just stronger than it was before it broke.