Adjustable shrouds

Discussion in 'Sailboats' started by disciple, Mar 4, 2007.

  1. disciple
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    disciple Junior Member

    Why not shortin the windward shrouds on each tack, all you would need is a hinge at the base of the mast and two leaverd catches one on each side. You could shorten the leeward one just befor each tack. I'm shoure thay would would be cheeper than any canting keel. P.S (Where is the spell check :eek: )
     
  2. Doug Lord

    Doug Lord Guest

    Mast Cant

    It can work-some ORMA 60 multies and one brand new multi-the 23' Exploder use mast cant to enhance power. But it can never be as effective on a mono as a 55° canting keel-you simply can't angle the mast far enough and when you do the hull is at a very large angle of heel. That said a combination of some mast cant and a canting keel has been used on mono's effectively.
    I think there was an English dinghy that used a system similar to what you described.
     
  3. PAR
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    PAR Yacht Designer/Builder

    I've debated using a continuous shroud, completely rounding the hull and going up the other side. Could offer some canting stick appeal, but it raises a number of issues, some not easily sorted and I missed on the weekly Lotto again, so my budget for hair brained ideas is as slim as it was last week.
     
  4. LP
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    LP Flying Boatman

    ideas

    I'm good at hairbrained ideas.

    Seriously though a monohull operates in a state of balanced equilibrium where if can't the mast up, you power up the sail, but you also increase the heeling moment. That force needs to be counteracted to keep the sail powered, otherwise the boat heels over more and you're at about the same point as before only closer to your heel angle of vanishing stability.

    I'm not trying rain on your parade. Finding ways to stiffer your vessel, make it stand up straighter, would acomplish the same goal.

    Going hairbrain though, you could put a hiking plank on your boat and have a mechanical interconnect between it and the shrouds so it would pull the mast to windwand with the plank.
     
  5. DanishBagger
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    DanishBagger Never Again

    What about the adjustable shrouds from www.precourt.ca

    That way you won't need bronze or steel highfield levers.

    (I am in no way affilitated with him, but have bought a couple of his stuff, and it's pretty neat, saving weight, no electrolysis etc.)
     
  6. DanishBagger
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    DanishBagger Never Again

    It just struck me that you obviously mean shortening it so much that you're canting the mast rather much.
     
  7. disciple
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    disciple Junior Member

    All i am saying is that, for say every 10 degrees of mast cant to windward the keel is effectively lifted to windward by
    10 degrees depending on where the mast is hinged cabin sole (good), coach roof (Not as effective).
    So anyone who races any one design mono-hull and can tilt his/her rig to windward on each tack will sail the boat stiffer ie 5 degrees and beat the competition.
    ps
    i grewup in nassau bahamas and my dad had a bahamian sailing sloop i know all about hiking planks and falling off them.
     
  8. tspeer
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    tspeer Senior Member

    It would only work if the amount you took in on one side was the same as the amount you let out on the other side. It's straightforward to show whether or not this is the case.

    Let theta be the angle you want to cant the mast (positive to starboard), and h be the height of the mast. The distance between the chainplates is b, and for simplicity I'll assume the chain plates are level with the mast step and in line with the mast. Let Lp be the length of the port shroud and Ls the length of the starboard shroud. With zero cant, the length of the two shrouds, Lp0 and Ls0, are the same:

    Lp0 = Ls0 = sqrt(h^2 + (b/2)^2)

    When you cant the mast, the shroud lengths become:

    Lp1 = sqrt((h*cos(theta))^2 + (b/2 + h*sin(theta))^2)
    Ls1 = sqrt((h*cos(theta))^2 + (b/2 - h*sin(theta))^2)

    When you add the combined length of the shrouds before and after canting you get:

    Lp0 + Ls0 = Lp1 + Ls1 + DeltaL

    The difference, DeltaL, has to be zero if the scheme is to work properly. All that remains is to substitute Lp0, Ls0, Lp1, and Ls1 into the equation above and solve for DeltaL. When you do, you get:

    DeltaL = sqrt(4*h^2 + b^2) - 1/2*sqrt(b^2 - 4*b*h*sin(theta) + h^2) - 1/2*sqrt(b^2 + 4*b*h*sin(theta) + h^2)

    This is definitely not zero. The shrouds will go slack as you cant the mast.

    You might be able to handle this with a clever cam under the deck that not only moved the center of the continous shroud to the side, but also moved it down (or fore-aft) so as to tighten up the shrouds. For modest amounts of cant (say, <30 deg) and a mast height several times the beam, the difference (DeltaL) is on the order of 1% of the shroud length, so this should be manageable.
     
  9. DanishBagger
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    DanishBagger Never Again


  10. PAR
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    PAR Yacht Designer/Builder

    Interesting Tom, I was thinking along those lines, though tinkered with a tackle arrangement rather then a cam. My approach was for experimenting on a reasonable small (22 - 26') craft to limit cost exposure. The calculations say it would work out, but the "clever" part has the other half wanting me to get through the show season first and a few other projects next, before I can go out an play in the sand box again. You see, my last romp didn't work out quite as well as I had envisioned (and a friend got 3 stitches in his head too), but damn we were doing well over 20 knots, under sail, when the "incident" occurred. It's amazing how much equipment you can break at those speeds.
     
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