Howdy y'all,
How would one go about calculating various pressures exerted by a sail in a given wind? IE: how much pressure is the sail putting on the halyard or outhaul in a given wind? My guess is you'd have to convert lbs of pressure to newtons, but it's been too long since I tool physics to go from there...any takers?
How to calculate pressures on a sail?
-
Figment
- Damned Because It's All Connected
- Posts: 2847
- Joined: Tue Apr 08, 2003 9:32 am
- Boat Name: Triton
- Boat Type: Grand Banks 42
- Location: L.I. Sound
Question:
How much pressure does a sail put on a halyard or outhaul?
Quick answer:
Nearly infinite. At least, it starts out that way, but we're dealing with elastic elements, and as they give, the angle of attack increases and the forces reach a realistic equilibrium.
Very Rough No-Calculus Required Answer:
It all starts with the boat. The sails can only exert as much force as the hullform will resist.
What is the righting moment of the boat? The Rigger's Apprentice (or was that Good Old Boat?) gives as good an explanation as any of the accurate-enough process of "inclining" the boat to establish Righting Moment.
Envision a cross-sectional diagram of the boat. Somewhere down below the waterline is the boat's Center Of Gravity (mass). The centerline of the boat at waterline level (not really, but close enough) is the Axis Of Rotation (heel). Somewhere up in the rig is the sailplan's Center Of Effort.
The COG imparts the Righting Moment about the AOR, counterbalanced by the sailplan's COE. Multiply RM by the disance from COG to AOR. Divide by the distance from AOR to COE. The product of this is the Force produced by the sailplan.
From the sail plan, establish how much of that Force is borne by each sail, determined by each sail's percentage of the whole. So now you have established the Total Force generated by the Mainsail, for example.
Focusing on the outhaul, assuming a loosefooted main....
The outhaul can be assumed to bear half of the Force of the main, the luff attachments bear the other half.
The force of the main is lateral, but is translated to a pure tension force in the outhaul itself. To find this force in the outhaul, an angle between the straight axis of the boom and the actual (slightly deviated to leeward) direction of the outhaul must be assumed. Academically, assume that it's 5 degrees. The outhaul Tension=(1/sin5)(1/2 sail Force).
Applying fictional numbers to the above, if Mainsail Force is 2,000lb, and the angle is 5 degrees, the outhaul Tension = 11,473lbs.
If we reduce that angle to 1 degree (closer to reality), the outhaul Tension = 57,298lbs
As the angle reduces, the Tension increases exponentially, eventually to infinity.
So go easy on that mainsheet! ;) And don't forget to multiply a safety factor of 3 to guard against shock loading. Just ask Team New Zealand.
But if you just want to know what size line you need, go to
http://www.yalecordage.com/html/pleasure.html and click on "Selection Guide". =P
How much pressure does a sail put on a halyard or outhaul?
Quick answer:
Nearly infinite. At least, it starts out that way, but we're dealing with elastic elements, and as they give, the angle of attack increases and the forces reach a realistic equilibrium.
Very Rough No-Calculus Required Answer:
It all starts with the boat. The sails can only exert as much force as the hullform will resist.
What is the righting moment of the boat? The Rigger's Apprentice (or was that Good Old Boat?) gives as good an explanation as any of the accurate-enough process of "inclining" the boat to establish Righting Moment.
Envision a cross-sectional diagram of the boat. Somewhere down below the waterline is the boat's Center Of Gravity (mass). The centerline of the boat at waterline level (not really, but close enough) is the Axis Of Rotation (heel). Somewhere up in the rig is the sailplan's Center Of Effort.
The COG imparts the Righting Moment about the AOR, counterbalanced by the sailplan's COE. Multiply RM by the disance from COG to AOR. Divide by the distance from AOR to COE. The product of this is the Force produced by the sailplan.
From the sail plan, establish how much of that Force is borne by each sail, determined by each sail's percentage of the whole. So now you have established the Total Force generated by the Mainsail, for example.
Focusing on the outhaul, assuming a loosefooted main....
The outhaul can be assumed to bear half of the Force of the main, the luff attachments bear the other half.
The force of the main is lateral, but is translated to a pure tension force in the outhaul itself. To find this force in the outhaul, an angle between the straight axis of the boom and the actual (slightly deviated to leeward) direction of the outhaul must be assumed. Academically, assume that it's 5 degrees. The outhaul Tension=(1/sin5)(1/2 sail Force).
Applying fictional numbers to the above, if Mainsail Force is 2,000lb, and the angle is 5 degrees, the outhaul Tension = 11,473lbs.
If we reduce that angle to 1 degree (closer to reality), the outhaul Tension = 57,298lbs
As the angle reduces, the Tension increases exponentially, eventually to infinity.
So go easy on that mainsheet! ;) And don't forget to multiply a safety factor of 3 to guard against shock loading. Just ask Team New Zealand.
But if you just want to know what size line you need, go to
http://www.yalecordage.com/html/pleasure.html and click on "Selection Guide". =P
