Crafting by Concepts: Fiber Arts and Mathematics by Sarah-Marie Belcastro (ed.), Carolyn Yackel (ed.)

By Sarah-Marie Belcastro (ed.), Carolyn Yackel (ed.)

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Extra info for Crafting by Concepts: Fiber Arts and Mathematics

Example text

K10, sl m, (m1, k6) 4 times, sl m, k10 (28 sts between markers). Row 14: K10, sl m, (k9, m1) 3 times, k1, sl m, k10 (31 sts between markers). Row 16: K10, sl m, (m1, k10) 3 times, k1, sl m, k10 (34 sts between markers). Row 18: Change to CC. K10, sl m, k6, (m1, k12) 2 times, m1, k4, sl m, k10 (37 sts between markers). Row 20: K10, sl m, k18, m1, k19, m1, sl m, k10 (39 sts between markers). Row 22: K10, sl m, k10, m1, k19, m1, k10, sl m, k10 (41 sts between markers). Row 24: Change to MC. K10, sl m, k20, m1, k21, sl m, k10 (42 sts between markers).

However, it’s clear that we’ll need more than two strands: when we want to return to red we’ll have only knit 4k 5 stitches and the end of our red yarn strand is at stitch 2k 5 mod k. ) but we need to knit in yellow for another 2k 5 stitches to get to the end of the first strand of red. In total, we need six strands, three of each color. ) This, by the way, creates stripes that precess. That is, they travel upwards in the direction opposite that of the knitting, as in Figures 5 and 6. It turns out that stripes that progress can be created by knitting 2k 3 stitches in each color and using four strands, two of each color.

To answer this, we will start by making a list of mathematical constraints that are induced by the knitting practicalities of the problem. Then, we will solve the resulting equations. Suppose we have mk r stitches for each color and n colors. ) Without loss of generality, we may assume that gcd(m, r) = 1. Segments of color that are more than one round long produce horizontal stripes, so in order to retain helical striping m < r so that mk r < k. To obtain stripes of height h, the second occurrence of the first color must overlap the first occurrence of the first color by a fraction of the color seg1 th ment length, namely (h−1) h .

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