Adjustable shroud setup

[SteveC]

Chris,

I went through a rough calculation last night with my friend Maurice - does this make sense?

Fully loaded rig worse case gets hit with a 20mph (10m/sec) lump of air. Worse case force on rig F = 1/2mv2 where m = area x mass of air = 10m2 x 1000kg/870 balanced by weight on end of plank.

Force = 1/2 x 11.49kg x 100 = 574kg. Assume load shared between forestay, windward shroud and mainsheet but 60% taken by windward shroud = 344kg.

At 30mph of course this flies up to 775kg just about within the limit of the Harken wire fitting.

In practice this is static pressure with the boat stationary and sail at 90 degress to the wind. With an apparent wind angle of say 30 degrees to the sail this will reduce sideways pressure to 60% or 465Kg in a 30mph wind.

The above explains why most people use the wire blocks as they should never break but also why my solution didn't albeit with not much of a safety margin.

I think I'll change the blocks!

[Chris Maas]

Hmmm.... I am fairly sure the load on the shroud is limited by your righting moment. Once you are hiking off the end of the seat static loads are at their max. Sail area and wind strength have nothing to do with what that load is except that they affect when you get out to the end of the seat. So there is no more load on the shroud in 35 knots than there is in 15.

The pertinent numbers are your distance to weather of the center of buoyancy - pretty much the center line of the hull - and the distance of the chainplate from the center line of the boat.

You are right that the forestay shares some of the load but I don't think it's much and since I don't know how to calculate that I ignore it. Maybe it's around 10%?

I'm hoping someone real knowledge of engineering will step in and give us the real story.

[SteveC]

Yes for fully loaded rig you are correct as the balance of forces would dictate that the righting moment has to equal the force on the rig and the shrouds then share the overall load but this assumes you are level, fully powered up and with leverage force at the gunwale which will be about 5 times the weight of the helm.

My analysis I think is still valid for shock loading from gusts where the righting moment may not be equal to the rig load for a short period of time but the rig will have to absorb the power until the wind speed drops or the sail blades off. The windward shroud would probably take most of this load before the mast has time to bend or heel the boat.

Bit complicated, suppose the easiest way is to put a strain gauge on the cascade and then measure it.

[Jon Rawstron]

I'm hoping someone real knowledge of engineering will step in and give us the real story.

I will try.

Worst case scenario is hiking out off the end of the seat and the seat is all the way aft, just in front of the tiller arc. The leeward shroud is loose. Sailor weight is 200 pounds. The lateral component of the torque generated is matched by the lateral component of the shroud torque. But the shroud is not aligned with the torque vector so the headstay has to take some axial load. I did the vector mechanics on a spreadsheet (I would share the spreadsheet if asked). The forestay sees 118 lb, the shroud 1207 lb and the 3:1 line 402 lb. A Harken H313 block is rated for 1500 lb safe working load and is capable of handling the 1207 lb load. Harken builds an appropriate safety factor into the safe working load so no additional safety factor needs to be applied. I did a quick look at the H313. The limiting factor is likely to be the sheave. When used with the becket, the sheave only sees 2/3 of the load (806 lb), so there may be additional margin for our application.

I dismiss shock loading as much of an adder. If it becomes considerable your boat will flip instantly. The steady state torques involved here are many times what your car engine puts out.

[chrishampe]

has anyone thought about the effects of spreader deflection?