... or how a little knowledge of Geometry can be a Good Thing, especially for Stash-Busters ...
Recently, I was doing a spot of stash-busting, and I decided to knit my sister a little shawl from some ribbon-y yarn in stash. It's a straightforward matter to knit a triangular shawl, especially in garter stitch, and we probably all know how it's done ... cast on 2 stitches, and knit them, then on each following row, increase one stitch at the beginning of the row, usually by knit 1, knit 1 tbl, then knit to the end of the row. When the resulting triangle is big enough for what we want, we cast off. So far, not very difficult ... but how do we know if we have enough yarn for what we want to do?
And that's when Geometry can be our Best Friend ... I had four balls of Veritas 'Colorado' (bought in Veritas in Mons, Belgium), that's this stuff here
no yardage on the ball band, though I would guess it was around about the 60 yard mark, it's a type of slightly glittery 'railroad' ribbon, with little tufts along the edges, and when knitted up, it looks rather like this
I had originally intended to knit a scarf with it, but decided to be a bit cleverer ... hence the shawl, but I wasn't really sure just how far the 240 yards would go, and I didn't want to end up with something too small to be of any use. Enter Pythagoras, and one of the only two formulas that most people know (the other one is e=mc^2)! Knitting the shawl in the way I've described gives you an isosceles triangle (two sides the same length, two angles the same), but more importantly, the angle between the two short sides is a right angle, like this
and once I knew how big a triangle one ball would make (12 inches on each of the short sides and about 17 inches on the long side), it was a relatively simple matter to work out that 4 balls would give me a triangle twice as long on each side (ie 24 inches for each short side, and about 34 inches on the long side), like this
each right-angled triangle in the diagram represents one ball of yarn. Interestingly, to make that triangle three times as long on every side as the original, you need 9 balls
... I will leave it to you to work out how many balls you need to knit a triangle 4 or 5 times as long on each side, but here's the sequence - 1, 4, 9, ..., ... - see if you can fill in the missing numbers - no, there is no prize for this (just a sense of satisfaction).
That's all well and good, but what if I'd had only 2, 3 or 5 balls (or some other number of balls of yarn not in that sequence?) Thanks to knowing a bit about right-angled triangles and their properties, I can work that out too, based on the dimensions of a triangle made from one ball of yarn, with the aid of a calculator that can do square roots, or using MS-Excel.
And just so that you can do it too - here's how: Use your calculator to find out the square root of the number of balls you have, eg. the square root of two is a nasty number that goes on for ever, but the first few digits are 1.414213562 (in case you're interested, this type of number is called an irrational number), but you can round it to something more reasonable, say 1.41 (to 2 dp). Now multiply the sides of the triangle you get from 1 ball of yarn by 1.41, ie. if the sides from 1 ball are 10 inches, 10 inches, and 14.14 inches, then 2 balls will make a triangle with sides 14.14 inches, 14.14 inches and 20 inches.
If you want to do the same thing in Excel, type this formula into the first cell (cell A1) in the top left corner of the spreadsheet, '=sqrt(x)' (leave out the quote marks, and replace x with the number of balls of yarn). That gives you the multiplier you need. Now type into another cell '=A1*y' where y is the length of one of the two short sides of the original triangle, and hit enter. Do the same sort of thing in another cell for the long side of the triangle, and now you know how big your finished triangle will be.
(ETA - It occurs to me that some people don't have access to either the right kind of calculator, or MS-Excel, so just for you, here are the square roots for the numbers 1 - 10 (where these are irrational, I've simplified them to 2 decimal place which should be quite sufficient for most purposes)
1, 1.41, 1.73, 2, 2.24, 2.45, 2.65, 2.83, 3, 3.16)