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  #1  
Old 04-21-2008
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Calculating forces on a boom during a Gybe...

I've had a pet project in my head for a while that I'd like to flesh out and implement. I'm trying to figure out how to calculate the load placed on the boom (or mainsheet/deck fitting) during an gybe.

You see there are many devices in the rock climbing, caving and technical rescue field that are designed to limit a shock load, distribute the load over time, and absorb a shock load. There are also devices designed to take a shock load and dissipate it by allowing a rope to slide through a plate at a controlled rate.

It seems to me that some of these devices, cleverly implemented, could reduce the risk of injury and damage from the accidental gybe that never, ever happens.

The only problem is that I can't figure out exactly where to start with calculating the forces involved. I know my mainsail is about 400sqft and my aluminum boom is of "normal" weight (which I could look up).

Anyone know how to figure forces for an oh %$&! gybe or even a regular 20knot downwind gybe?

Here are some of the toys I've been looking at:
KONG - Carabiners and Safety Equipment (.Doc.408)
Yates SCREAMERS

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Old 04-21-2008
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MedSailor-

Why re-invent the wheel. They already make several devices to do just that. The Dutchman Boombrake, the Scott BoomLock, and the Wichard Gyb'Easy are all designed to slow the boom in the case of an accidental gybe.
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Old 04-21-2008
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Quote:
Originally Posted by sailingdog View Post
MedSailor-

Why re-invent the wheel. They already make several devices to do just that. The Dutchman Boombrake, the Scott BoomLock, and the Wichard Gyb'Easy are all designed to slow the boom in the case of an accidental gybe.
- Sure, but I hope someone answers his question. I like to learn the theory. Sometimes knowing the applied principle allows us to develop better, cheaper, custom solutions- and maybe he's got a whole lot of climbing gear lying around.
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Old 04-21-2008
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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?
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Old 04-21-2008
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Actually, the bom and sail may be moving faster than windspeed, since they're accelerating around a pivot point and rotational acceleration can exceed linear acceleration.

Part of the problem with calculating the force exerted during a gybe is there are so many variables. Angle of the sails to the wind, speed of the wind, speed of the boat, changes in momentum as the boat heels, etc.
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Old 04-21-2008
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Quote:
Originally Posted by sailingdog View Post
Actually, the bom and sail may be moving faster than windspeed, since they're accelerating around a pivot point and rotational acceleration can exceed linear acceleration.
Get a patent on that quick SD - it could be worth a fortune to a windfarmer.
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Old 04-21-2008
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Idiens—

Same thing happens if you drop a baseball bat... the center of the mass of the bat may be moving at the gravitational acceleration, but if the bat is rotating, one end is most definitely moving faster than gravity could possibly accelerate it at.
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You know what the first rule of sailing is? ...Love. You can learn all the math in the 'verse, but you take
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Old 04-21-2008
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SD- Alternatively, when you drop your bat, the air resistance slows your handle more than your head. What would happen to your bat in a vacuum?
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How about some crude tests?

- a fish scale with a slide on it that saves the highest weight pulled? Lash it between the boom and a cleat and gybe away-

- tie bits of small cordage with a known breaking strength?
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Most fish scales don't go high enough to deal with the forces generated by a boom in a crash gybe in higher winds.
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You know what the first rule of sailing is? ...Love. You can learn all the math in the 'verse, but you take
a boat to the sea you don't love, she'll shake you off just as sure as the turning of the worlds. Love keeps
her going when she oughta fall down, tells you she's hurting 'fore she keens. Makes her a home.

—Cpt. Mal Reynolds, Serenity (edited)

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