The Sleeve Problem

(Originally posted 15 January 2021)

Usually when I’m writing for this blog, I’m trying to communicate both to engineers / nerds / machines-and-math people and to knitters / that other kind of nerd / fabric people. Usually. This post is going to end up rolling towards no-holds-barred nerddom. There’s math. There’s spreadsheets. I’m going to say the name of some classes I took in engineering school. We’re going places in this one, folks. As much as you’re normally not obligated to read these, you’re definitely not obligated to read this one if that’s not something that seems interesting to you. This is about me solving a math problem. The end result is nice to have, and now that it exists, I’ll share it so you don’t ever have to actually do the solving yourself if you don’t want to, and if you want to leave it at the end of this paragraph, that’s totally fine and dandy. Here comes the math.

Now that many of the knitters are gone (that’s bold, Rowan, assuming anyone was even reading this from the get-go), I guess I have to explain some knitting stuff, finally. It’s a fairly common problem in knitting (especially with what I’m doing and the way in which I’m doing it) that you start off with some number of stitches (horizontal dimension), and you have some number of rows (vertical dimension) over which to decrease to some new number of stitches. For a sleeve, I start off with a very wide piece where the shoulder is, and over the length of the sleeve, I’ll decrease down to something more narrow where the wrist is. The whole thing looks like a kind of lumpy trapezoid. Maybe I’ll come back and populate this with images, we’ll see.

For this type of machine, it’s easiest to decrease stitches on the outside (far left and far right) of the needle bed. This is because the further in you go, you have to pick up the stitches on the outside and move them in, which gets pretty annoying if you’re doing more than four or five. That leaves me with a maximum of 2 decreases per row, because I can decrease 1 on the left, and 1 on the right, every time. Man, I need to explain how this machine works in another post. The question becomes: How often do I need to decrease stitches in order for me to get from length A to length B fairly evenly, while having a symmetrical end product that is to-spec? Here, let’s define the space.

The Problem:

Decrease S stitches over R rows.

In one row, you can decrease either 0, 1, or 2 stitches.

When I was doing this, it’s about all I thought about for like a day and a half, and I managed to do it in two and a half different ways. The first two approaches really walked so my final iteration could run. I have so many pictures of my whiteboard from these couple of days, I’m sure it makes me look like I’ve lost my mind. I had a great time working on this.

Approach 1: Quick and dirty

Treat it like any line with a slope. You’re decreasing 2 stitches per row, so you should do a decreasing row every 2R/S rows. Rows are discrete, so do the ceiling or the floor of that number, whichever has the least error, and go for it. I could have stopped here.

The problem is that, depending on what your parameters are and how close they are to multiples of each other, you can end up with some pretty egregious error. For the second sweater, I was trying to decrease 75 stitches over 204 rows, which would have meant decreasing every 5th or 6th row. That left us decreasing 7 fewer or 5 more stitches than I need. For some things that’s ok, but 1. that’s an almost 10% error, and 2. the whole point of this is to have clothes that fit people, and I don’t think I can do a great job of that if my stuff can be an inch off in either direction. I did this in Excel, and ended up implementing not only a tool where you could input gauge and pick a tolerance in inches, but I set up a data table that would highlight any combination of S and R from 1-400 in blue if it was within your tolerance, and red if it wasn’t. It’s a neat looking table, I think. Here’s what it looked like for that example, with a tolerance of 1”: