Fringed Loomwork: the Mathematical Approach

By Jane Tyson

(Jane is an Australian. She has not misspelled things and is not wrong using "maths" where Americans would use "math.")

They say you learn something new every day. Well, perhaps it's old, but new to me/you anyway. I must have been away from school on the day they taught arithmetic progression.1 Lucky for me, my daughter wasn't!

I was struggling with trying to guesstimate how much thread I'd need for the weft of a fringed choker, given that I wanted to make it in one piece.

Adding in thread is a pet peeve of mine, so I much prefer to brave knots and tangles than add to an existing piece of thread. Rachel (my daughter) said, "that's easy," and whipped out her maths exercise book with its formulae for Arithmetic Progression.

["Adding in thread": you prefer to brave knots and tangles? Wouldn't you get knots if you had to add thread? Do you mean you would prefer tangles to knots?]

It almost worked like a charm. It certainly was much closer than I would have guessed by using a measuring tape to measure the thread I needed. I felt that "near enough" would be "good enough," but I was about a 30cm (a foot) and several curses short. I think I have identified the error and corrected it.

The choker is made up of two 'arms' and a V-shaped fringe.

If you work out the amount of weft thread needed for one "arm" and half the fringe (see below), all you need to do is double the total, then double the total again to account for the weft thread on the return pass.

O.K. here goes:

1. The "arm" of the choker is a rectangle, so all you need to do is measure how wide it is (in this case, 4/10 of an inches or 1 cm). Multiply that by the number of rows of beads (50), giving you 20 inches or 50 cm. This number is quadrupled, as I just explained. Now we have 80 inches (6' 8") or 2 metres.

2. That is pretty straightforward. It is the fringe that is tricky. Here we use arithmetic progression. The formula for it is: s = n ¸ 2(2a + (n-1)d)

Where S = the sum of the length of the fringes. (This is the answer we want).

n = the number of fringes (38 in this case)

a = the length of the shortest fringe (1 inch or 2.5 cm)

d = the difference in length between the first and second fringe (1/8 inch or 3 mm).

Using metrics this would be:

S = 38 ¸ 2 (2 x 25mm + [38 ¸ 1] x 3)
S = 19 x (50 +111)
S = 19 x 161
S = 3.059 m

Or in the English system:

S = 38 ¸ 2 (2 x 1 + [38 ¸ 1] X 1/8)
S = 19 x (2 + 4 5/8)
S = 19 x 6 5/8
S = 125 7/8 inches
S = 3 yards, 1 1/2 feet

Now we multiply this answer by 2 because we only measured half the fringe.

S = 6.118 m or 7 yards.

Add the "arms" we calculated above:

+ 2 m or 80 inches (6' 8")

S = 8.118 m or 9 yards, 8 inches

This gives you an exact measurement of the distance travelled, but it still needs some refining. You will need to add some thread for the start thread, the end tail and 'turns' between rows. You should allow 3mm (1/8 inch) for each turn = 174 x 3mm = 0.522mm (1' 9 3/4") and 10cm (4") each for the start thread and end tail.

The grand total comes to 8.84 metres or 9 yards, 2 feet and 5 3/4 inches).*

After all of this, you probably think that it is easier to make a rough guess and be done with it!

* In reality, this is slightly long because the longest, center fringe should not be doubled, but it is better to have too much thread than too little and the difference is rather small.

P.S. I (pfjr) did the conversions between English and metric. Guess which one was much easier to work with.

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