Loads on an Asymmetrical Spinnaker Pole

[mww195]

I am in the process of designing a 36ft fast cruiser, and I would appreciate some advice on how to design the pole (bowsprit) for the asymmetrical spinnaker (approx 50sq m/550 sq ft). I have assumed the worst load scenario to be a 30 knot gust whilst reaching, which will result in a side force of around 1 ton in the sail. I have assumed the side force to be split between the sheet, the pole and the mast head, and using moments about the geometric centre, this would give around 350kg of side force on the pole. Bearing in mind the pole is 1.4m long, this will result in a considerable bending moment about point of fixture at the stem. I don't really want to have wire stays to support the pole, but with these loads I can't see another option. However looking at other similar designs, they are able to sail with an unsupported pole. My questions are:
1. Have I assumed the correct load case for the pole?
2. Have I assumed the correct distribution of side force on the sail acting on the pole end?
3. Does anyone have experience with a similar sized rig or have any other suggestions about pole arrangment?

Mike.
Bristol, UK

[zerogara]

Just some comments and thoughts.
A gust is usually in the area of 120% of the wind from about the same direction. So the boat is already moving in the "right" direction with a 25 knot wind, which is different than an imaginary situation of a boat standing still in calm weather and BRUFF a 30 knot gust hit it, as if a bomb exploded 45' off its stern.
At 25 knots the boat is also heeled therefore its affected sail area has decreased somewhat. Also at the point of sail where an asymmetriic is used the main and jib will have absorbed some of the impact before or about the same time, further heeling the boat if not knocking it down.
How much force can the spinnaker hold before it tears up? Should it not tear up before your pole and mountings break?
The spinaker size doesn't sound too big for the boat and 1.4m is not that large of a pole to require any stays.
What are you doing flying the spinaker in 25 knot wind, anyway?

[Tim B]

Mike,

I think you're on the right track. Dependant on what the spinnaker pole is made of, it's failure mode may be critical. For instance, steel or aluminium will bend, and as the material bends it will absorb an awful lot of energy. Carbon doesn't bend, it fails catastrophically, with is not what you want to happen for obvious reasons.

I think you might be over-predicting with a gust of 30 knots, 25kts might be closer, that will drop the loads a little. The force transmitted to the pole will not be like a point load on a cantilever. The force vector will point aft of a plane normal to the neutral axis. (ie. there will be some compression).

Could you describe how you got the sail side load? I just tried it and got a rather higher number. I haven't done a full force resolution though.

If you send me an e-mail ( Tim@MarineDesign.tk ) I'll happily cross-check your working, then send it to a freind of mine who is a composite stress engineer.

Tim Brocklehurst

[zerogara]

on thread 10109
Calculating Sail Power

Originally Posted by messabout says

:

C.A. Marchaj, an authority of considerable stature, gives us a little bit of information about this subject. He is mainly interested in the overturnig force of the sail, a matter of considerable importance. His equation is pretty simple really. It does not tell you what the driving force is but it will give you some clues.

F= 0.0034*A*V^2 Where F=force, A= sail area in square feet, V= wind velocity in knots. It is easy to see that the big factor is wind velocity squared.

That formula gives you about 1700lb, 25% than your calculation is, but it is simplistic and if that works for a tight flat main it doesn't for a spinaker.

I believe Tim B is correct on his comment that part of the force may be in compression, or tension as well, depending on the trim of the sail. The more of a compressive component you have the least effective your trim is. It is acting against the motion of the boat and that happens on an overly trimmed sail and by trying to use it to high up the wind. Most of this force is actually added to the sheet load and the mast/halyard.
A good way to be sure about this is to actually go out and try it in less extreme conditions and figure out a way to measure deflection of the pole. By measuring deflection on a pole of a given material you can take it back to the drawing board and figure out what would happen with double the frce etc. But then it would be too late to reengineer your deck feetings if they were under/over designed.
It is also standard practice for retracting poles to be tappered. The reason is that if a perfectly cylindrical sleeve is used for the cylindrical pole to retract to, any eventual bend in the future will make it stick.
Asymmetrics only have a place on planing hulls, a tactical advantage for them to run at speeds much higher than the true wind against a boat with a symmetric running off wind at less than true wind speed.

[dougfrolich]

I would base the force generated by the Spinn. on a worst case righting moment instead of wind speed. Say Rm35degrees. Then the Force generated from the Spinn would be RM35/SpinnCE above LWL. Figure the load distribution as Tack .3 and Head .4 and Clew .3 of the total. Draw the vectors assuming a beam reach, and calculate the tranverse bending, upward bending, and comp. Other loads to consider would be bending induced by waves.