Trying to work out unstayed wooden mast scantlings

Discussion in 'Sailboats' started by Peter Vella, Aug 5, 2018.

  1. MikeJohns
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    MikeJohns Senior Member

    Weren't you just policing another thread on language regarding "Woofing up incorrect woody perennial plants" see what G translate makes of that :)

    Whenever I run through some established procedural method I check it wherever possible with fundamental formula. I've even found significant errors in class rules (BV) reviewing someone else's results doing that. If I'd just checked they put the right numbers into the formula it would have failed.

    AH's approach is laudable and perfectly sensible and he's gone to some effort in his post. It deserves better.
  2. TANSL
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    TANSL Senior Member

    OK, Mike Johns, AH deserves something better and from this post I'm going to shut up, but is not hes demonstration a bit obsolete? Are they really his own formulas? When I write 2 + 2 = 4, could I say that I am using my own formula? Yes, of course, I can but it is not a formula invented by me. I think calling it "my own formula" would be very pretentious and it would not be true. And now, I stop intervening in this thread commenting on the youthful exercises of AH. Google does not let me continue, although, as a good friend, it has encouraged me by saying that it is better to speak poorly a language that is not yours than do not speak it.
  3. tlouth7
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    tlouth7 Junior Member

    I appreciate that this thread is a little old, but would like to address the point about loading only at the top vs uniform loading:

    The moment at the base of a cantilever with load applied at the tip (OP's case) is double that with the load uniformly distributed along it (As per Skene's "Jib headed Cat"). The mast will therefore need to be twice as strong at the partners. Because the bending moment decreases linearly from partners to tip (rather than quadratically) you will want less taper for the same amount of mast bend.
  4. LP
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    LP Flying Boatman

    I feel like there may be a misapplication formulas here. The 'unstayed jib headed cat' leads me to believe that, while there is no stay, a jib is still being flown from the mast so that loading is accounted for in the strength of the mast and may contribute unnecessary strength (weigh) to the mast. The cubed root formula on the other hand is designed specifically for a freestanding mast with no "stays or shrouds" and might be a better starting point for your calculations.

    If you are wanting to determine ARM, I would start with the center of effort of your sails to determine bending moment at the pardners. To be clear though, you can't use this in the Skene freestanding mast formula as it's design to work with mast length and not moment arms. You'll have to dig into your own engineering here. Typically, the main mast is designed to withstand the full righting moment of the hull, but as a design choice, you can certainly choose to design it for anticipated max loads. From here, you can size the mast diameter at the pardners and apply standard taper ratios to the rest of the mast. I think that you'll find that only half the sail load is carried at the masthead(parrels). It could be more, but could also be less based on your sail plan.

    I wrote a paper a few years ago that I had planned to post here on, but never refined it enough to post it here. It's actually, a derivation of the formula found in Skene and may be of interest to members here. Here is a cut and paste:


    Designing a free standing mast or Me and Skene.

    Skene has a simple formula for determining the size of a round, solid, freestanding mast.

    His equation converted to computerese is D = (16(P)(L)(SF)/15700)^.333

    Where: (P) = Wind Pressure = Sail Area X 1.15 to 1.50, based on the boat.

    (L) = Mast length in feet

    (SF) = Safety Factor (1.5 to 3.5 based on the boat)

    15700 = Fiber Stress of Spruce times Pi.

    While the formula is simple, you are left with two adjustable variables that immediately adds a level of complexity to the solution. I wish I could definitively say what the right choice for P and SF will be, but this is all part of the fun (challenge) of designing your own free standing mast.

    I felt inclined to dissect this formula before blindly throwing a set of numbers and hoping to get some valuable information back out. The biggest concern with a Free Standing Mast (FSM) is the bending moment created at the deck by the force of the wind on the sails. Or perhaps more correctly, the force created by the righting moment of the hull as it resists the pressure of the sails. Regardless, there is a peak of internal stress in the mast at the deck that needs to be reckoned with. We’ll visit a couple of simple engineering formulae, combine them and see how it compares to Skene’s formula.

    Bending Stress (BS) in a beam can be determined the internal moment (M), the distance from it’s centroid to it’s edge (y) and it’s moment of inertia (I):


    Where (I), for a solid circle is:

    I=((pi) r^4)/4

    Since y = r in this instance, when we combine I into the BS equation, we get:

    BS=(4M)/((pi) r^3)

    Solve for r and we get”

    r = (4M/(pi)BS)^.333

    Since D= 2r we need to multiply each side by 2.

    2r = D = 2(4M/(pi)BS)^.333 = (32M/(pi)BS)^.333

    Let’s convert some terms and see if we can get this exercise finished up. Our Moment(M) in this case is the Mast Length (L) times the wind pressure (P), or M = (P)(L). Bending Stress is (BS) limited by fiber stress where earlier 15700=(pi)(fiber stress of Spruce) was defined. So, (pi)(BS) becomes 15700. Exchanging these terms, we get:

    D = (32M/(pi)BS)^.333 becomes

    D = (32(P)(L)/15700)^.333 per Skene though.........................,

    D = (16(P)(L)/15700)^.333

    Close, but no cigar. This had me perplexed for I while, but I think I can speculate, that since Skene is using the FULL length of the mast times the full pressure of the wind, he needs to account for vertical Center of Effort (COE) in his formula and does so by using the constant 16 instead of the 32 that I derived from my calculations (formulas). I made no account for COE and simply used L (mast length) in derivation of the formula.

    So what does this tell us? Skene is placing the CE of the sail half up the mast. Maybe a bit conservative with regard to most sailplans today. The CE on my triangular sail is 44% of the way up the mast. Not a big difference, but enough to reduce the size of my mast. I might even speculate that this formula leans towards masts for sails with four corners over three as the four cornered sail will carry it’s CE higher than a three corner of equivalent height or it could simply be account for mast bury.

    All in all, this was simply an exercise to help me wrap my head around the numbers that Skene presents in his section on mast design.
  5. Chuck Losness
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    Chuck Losness Senior Member

    Years ago Eric Sponberg posted a paper that he had written about the design of freestanding masts. In that paper he compared wind pressure on the sail to the righting moment at 30 degrees and found that numbers were very close to each other. Copy is attached.

    Attached Files:

  6. messabout
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    messabout Senior Member

    Please gentlemen, give it a rest you two.
  7. LP
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    LP Flying Boatman

    Eric was actually comparing the hull righting moment to the heeling forces of the anticipated sails loads within a specific set of parameters. The parameters being the maximum wind value in a close reach situation that the vessel would be expected to sustain prior to reefing. The fact that the righting moment and the heeling forces were in close agreement, he felt that he had a good match of sails to the vessel and a validation of the stresses that were going to be experienced in the masts, particularly at the partners where loads would be greatest.
  8. tlouth7
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    tlouth7 Junior Member

    I believe the term "jib headed cat" refers to a rig with no foresail (cat) where the mainsail is triangular rather than having a gaff (jib headed). This appears to be an archaic term for what we would typically call a "marconi" or "bermuda" cat rig.

    A lugsail will impose discrete loads at the yard and boom, rather than distributed along the entire luff (as per a "jib headed cat"). Additionally the sail area will be distributed differently (more biased towards the top of the sail), so CoE will be higher. A free standing mast can be considered to be a cantilever beam with its root at the partners*. Moments from multiple loads can be added together using the principle of linear superposition.

    e.g for a lugsail moment at the partners = Force at yard (e.g. 1/2 total force on sail) * height from yard to partners + force at boom (e.g 1/2 total) * height from boom to partners

    e.g for a triangular bermudan cat moment at partners = total force on sail * (height from partners to boom + 1/3 * height from boom to head)

    *for the purpose of finding moments & bending stresses. For deflections an additional rotation must be considered. Page 6 of the linked databook [1] gives formulae for calculating rotation and deflection at the root of beams under various load conditions.

  9. LP
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    LP Flying Boatman

    Ahhh.... I Gaggled when I should have Googled, Very nice. Thanks for the correction.

    Note to self: Jib-headed is referencing the sail head, not the mast head. That's what I get for trying to guess the etymology of a word. :oops:

    Good luck on your masts. Maybe, you will post your progress on your boat.

  10. sharpii2
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    sharpii2 Senior Member

    A few years ago I tried to design an un-stayed mast. Here is the proceedureI used:


    1.) I determined to the best of my ability the Maximum Righting Moment of my proposed boat then multiplied that by
    a safety factor of 1.5.

    2.) I drew a sail design which would have the desired SA to get the lengths of the Mast,Yard and Boom.

    3.) I treated the contact points of the Yard and Boom on the mast as point loads on the mast with the load of each
    assumed to be proportionate to each spar length. In my case it was easy. The Yard and Boom are the same length. I
    ignored the sheet loads as they are a small portion of the rig loads on a balanced lug (if it were a dipping lug
    rig, I would calculate it differently*).

    4.) I divided the Max Righting Moment x the safety factor by ((half the distance between the
    yard and Boom)+the distance from the Boom to the WL) Then divided this by two and assigned this product to each of the two load points on the

    5.) I multiplied each by their respective distances from the Mast Partners (in inches) then added these two products
    together. This gave me the theoretical bending moment imposed on the Mast.

    6.) I divided this sum by the allowed fiber stress of the wood to get the Sectional Modulus I would need for the
    mast (at the Partners).

    7.) I drew and calculated the Sectional Moduli in a number of attempts until I found one which was just slightly
    greater than what I needed, which could be built with the lumber store dimensions I would have access to.

    8.) I tapered each end of the mast to about two thirds its maximum thickness to encourage it to bend along its
    length rather than just at the partners. Doing this made it somewhat lighter too.

    *I would consider the sheet line and the tack line to be taking half the Max Righting load,divided by the heeling arm, times safety factor
    between them, leaving half of it for the point load where the Yard crosses the Mast. I would also use a bigger
    safety factor, perhaps 2.0. This would be the equivalent applying the Max Righting Moment, divided by the heeling arm**, to the one point load
    where the Yard crosses the mast.

    **Half the distance between the foot of the sail and the Yard parallel + the distance between the foot of the sail and the WL.
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