(This was submitted as an article for publication in Knotting Matters, the quarterly publication of the International Guild of Knot Tyers, and also to The InterKnot, the newsletter of the North American Branch of the IGKT, on March 14, 1999.)

A Mathematical Discussion of the Single Line Turk's Head Knot.

A question that comes up from time to time in discussing the Turk's Head knot involves the length of the line that is required for a given size and style of knot. In my own knotting, which involves precious wire, it is pretty important to know how much material will be required to complete the piece, so I have developed a couple of mathematical models which help me to estimate how much to start with. These models work fine for regular line, too, of course, so it seems to me I should share this knowledge with the knottying public.

First off, one must understand the structure of the singleline Turk's Head (TH) knot. Any possible "singleline" TH has one path through the entire knot, and this path can be considered to consist of a number of segments, each of which is the length of a line connecting a bight on one side of the knot with a similar bight on the other side, along the path. I provide an illustration which shows that this segment is the hypoteneuse of a right triangle, the other two sides of which are, respectively, the width of the overall knot, and an easily computed fraction of the circumference of the knot. We are speaking, here, of an ideal line of zero thickness, with the bights replaced by sharp changes of direction, and the segments being rulerstraight. The length of a segment is easily found with the help of Pythagoras' Theorem.

For example:

A 5L11B knot consists of eleven segments joining each lower bight with the upper bight twoandahalf bights to the right, and another eleven segments similarly extending to the upper bight 2.5 segments to the left. (Note that that is half the number of leads. For a 4L knot it would be 2 segments to the right and 2 to the left, for a 7L knot you would use 3.5, as shown in the illustration.)

Observing that the total circumference is 11 bights, the length of the fraction of the circumference is found to be 2.5/11ths, or 5/22nds, of the total circumference of the knot.

Therefore, should you wish to tie an ideal (remember, infinitely thin line, no smooth curves) 5L11B Turk's Head with a width of 1 inch and a diameter of 5 inches, you would need:

{(5*pi*5/22)^2 + (1)^2}^(1/2)
(the squareroot of the sum of the diameter times pi times fivetwentyseconds, squared, and the width, squared)

*22 (the number of segments) =81.56 inches

This is very cumbersome unless you're used to it, but having worked through one of these it is clear that the same results can be obtained by multiplying the number of leads by the circumference (pi times the diameter) to get one leg of the right triangle, and multiplying the width by twice the number of bights, to obtain the other leg.

{(5*pi*5)^2 + (1*22)^2)}^(1/2)

=81.56 inches

Much easier, as I'm sure you will agree.

In the real world, there may be a considerable difference between this result and your actual usage of material, so this should be regarded as a conservative estimate! The curvature of the line, and its diameter, will both add to the overall length required, but as a rough estimate it is a good start.

Looking at a nicely tied Turk's Head, it occurred to me that the path described a cosine through a cylindrical space, and while I was sure that I could use threedimensional equations to describe it, I chickened out and decided to model it as the same curve over a linear distance in two dimensions, that distance being the product of the number of leads and the circumference of the knot. I will not weary the reader with the calculus required to determine the length of a curve in space, but having performed this calculation on the same starting dimensions and obtained a number somewhat less than 1% larger than the previous result, I would venture to say that there is little to choose between them for accuracy, and a great advantage in using the simpler one with a fudgefactor of two or three percent over the calculated result.



