Last month I wrote about how to tackle pesky “evenly across” instructions. As promised here’s the follow-up for those times when the numbers aren’t so tidy.
For example: “Increase 6 stitches evenly across the 52 stitches of next row.”
52 ÷ 6 = 8.666667
You’ve got 6 groups of 8 stitches each and a few left over since 52 = 6 x 8 + a remainder of 4. Here’s how to configure the increases:
IN THE ROUND
Those numbers give you the following: (K8, m1) 6 times, k4.
That’s not very even, since you end up with 12 stitches between the first increase worked and the last one, split across the start of the round.
In situations like this, I like to write out the full round, to visualize what’s happening:
(K8, m1), (k8, m1), (k8, m1), (k8, m1), (k8, m1), (k8, m1), k4.
A good solution is to just pepper those extra 4 stitches along the round, changing four of the k8 to k9.
(K9, m1), (k9, m1), (k8, m1), (k9, m1), (k9, m1), (k8, m1).
If I’m working flat, in rows, it’s a little different. In the previous column, I said to put the increase in the middle of each repeat, for tidiness.
(K4, m1, k4), (k4, m1, k4), (k4, m1, k4), (k4, m1, k4), (k4, m1, k4), (k4, m1, k4), k4.
But this isn’t great when there’s a remainder number of stitches because, as in this specific case, it puts 4 stitches at the start and 8 at the end. Lopsided! (See the top line in the illustration below.)
To distribute remainders more evenly, you can just split them between the start and end of the row. (See those stitches on the move in the middle line of the illustration below.)
In other words having a remainder gives you a different way to handle it … put the increase at the end of the repeat (like you do for working in the round), and the remainder at the end of the row (as illustrated in the bottom line of the illustration below):
(K6, m1), (k8, m1), (k8, m1), (k8, m1), (k8, m1), (k8, m1), k6.
There’s another possibility: You could take stitches from some of the repeats and stick them at the end. If I steal a stitch from 3 of the repeats, and add them to the end, the spacing alternates between 7 and 8 all the way along the row:
(K7, m1), (k8, m1), (k7, m1), (k8, m1), (k7, m1), (k8, m1), k7.
They’re different, yes, but both solutions work toward more evenly spacing the increases. Remember, because the pattern wasn’t specific, you can do it however you want, however it makes sense to you!
The same tricks work when decreasing, of course. Just remember that the decrease uses up 2 of the stitches in each group.
WHAT IF IT’S A GIANT REMAINDER?
What happens if there are lots of stitches or lots of repeats to fuss with? You really get to dig into the math with this one.
Look at this one, taken from a real pattern I once knit:
Increase 41 stitches evenly distributed through 197 of the next row.
197 ÷ 41 = 4.804878.
Which is to say, 197 = 41 x 4 + a remainder of 33 stitches.
Which gives us
(K4, m1) 41 times, k33.
Because there are many more stitches in the remainder than in the individual repeats, distribute the remainders throughout the row. (Splitting the 33 between the start and end of the row wouldn’t really improve things.)
If I want to keep 4 at the start and the end, I’d subtract 29 from the end of the row, and pepper those stitches through the row:
(K4, m1) 12 times, (k5, m1) 29 times, k4.
If you enjoy number puzzles, you could start to play with it a bit more, distributing the twelve (k4, m1) repeats within the twenty-nine repeats of (k5, m1)
How about this?
[(K4, m1), (k5, m1) twice] 12 times, (k5, m1) 5 times, k4.
Or even this?
(K5, m1) 3 times, [(K4, m1), (k5, m1) twice] 12 times, (k5, m1) 2 times, k4.
I love these sorts of instructions because there’s so many ways to come up with an answer. And they’re all right, they all work.
Writing it out might seem like a bit of a pain, but I find it very worthwhile: it helps me count up and make sure that I’m working the right number of increases and decreases, and it helps me keep track as I work the instruction. (Also, I’ve got good notes if I need to repeat the step on the second sleeve …)