Page 206 - The Complete Rigger’s Apprentice

and aristocrat David Ryder-Turner came up with the Compression load? Does the old wire or hardware
sliding-gunter arrangement shown in Figure 6-11. show signs of excess strain? It might turn out that,
The first thing I want to do is check David’s though we would never stay with 8.58 degrees for a
work. Ahem. What I mean is that I’m interested in larger, more heavily-loaded boat, it might be okay
finding out what angles the shrouds and forestay for this small, light daysailer.
make relative to the mast. For most vessels this Another example: The right triangle for the fore-
angle should be 10 to 12 degrees, but in this case stay has a base of 2 feet 6 inches and a height of
the rig is intended more to check the motion of the 15 feet 6 inches. The tangent of the angle in question
mast than to turn it into a compression member, is given by Base ÷ Height = 2.5 ÷ 15.5 = 0.1612903,
and low stresses are involved. A somewhat lesser and the Arctangent = 9.162347. The forestay angle
staying angle ought to suffice. To find out what the relative to the mast is 9.2 degrees. Because staying
designed angle is, I’ll construct a right triangle for angles are inversely proportional to standing-rigging
the shrouds based on the sail and deck plans shown loads, the tangent and arctangent functions are of
in Figure 6-11. The base is 28 inches and the length particular value to the rigger.
is 15 feet 6 inches. In a right triangle, the ratio of
base/height is the “tangent” of the angle opposite Hypotenuse and Wire Length Of all the words of
the base. (Consult any high school geometry text rigger’s wit, the saddest are these: “It doesn’t fit!”
for a fuller explanation of trigonomic functions; Thimbles and soft eyes affect finished length, but
your local library would be a good place to start.) of course the major determinant of rig fit is length
In this case, the tangent is 0.1507936, and a tan- between terminals. Because the sail plan appears to
gent of 0.1507936 means that the angle at the top show wire length, many people scale their lengths
(represented by the symbol Ø) is 8.58 degrees. How from it. Their rigs are too short. “Ahh,” say others,
do I know? Because my calculator tells me so. It has “but that’s why we have the rigging plan” (Figure
a key on it marked “tan-1,” a function sometimes 6-12), and they scale their lengths from it. Their rigs
written out as “arc- tangent.” If 0.1507936 is the are too short. The sail plan has no depth, so does
tangent of the angle Ø, then Ø is the arctangent of not show length-consuming athwartships shroud
0.1507936. If you enter 0.1507936 on your calcu- travel, although it is accurate for fore-and-aft wires,
lator and then hit “tan-1,” you’ll get 8.58 degrees, which have no athwartships travel. The rigging plan
the angle of the shrouds relative to Katy’s mast. has no depth, so does not show length-consuming
A glance at the staying angle-stress curve (Figure fore-and-aft travel, although it is accurate for the
5-5) shows me that this is a very tight angle. This rare shroud that has no fore-and-aft travel. Each
is characteristic of catboat shrouds, as the mast plan shows a dimension that the other does not; our
is always well forward, where the boat narrows. I problem is how to get the vertical, athwartships,
could widen the angle by shortening the mast, but and fore-and-aft dimensions in one place. That’s
I don't want to lose sail area, but I could lower the why I used the sail and deck plan to construct the
upper attachment point of the shrouds instead. This right triangle for Katy’s shroud. Height was taken
would create a bit of a cantilever above the shrouds, off the sail plan, and fore-and-aft and athwartships
but only a short one. Bringing the attachment down travel off the deck plan. We could now measure wire
one foot would widen the angle to a bit over 9 length from our diagram, but for greater precision
degrees. I could also add "channels" for the chain- let’s return to trigonometry, which tells us that “the
plates (see below) to widen the staying base. With a square of the hypotenuse is equal to the sum of the
1
2 ⁄2"channel, we'd have almost a 10 degree staying squares of the other two sides.” For Katy’s forestay,
angle. Whether any of this would be worth doing the sum of the squares is 35,496 inches. The square
would be a matter of determining cost and benefit. root of this is 15 feet 8 inches, the length of the stay.
How much would the changes reduce shroud load? Some fine points:

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