Saturday, March 12, 2011

How big is the pi(e)? or The "joys" of designing

     If you ask my FIL, the bigger the pie, the better. Lucky this isn't the case with the Birthday Pi. Last Thursday I managed to knit two swatches using different yarns. Below you'll find calculations done using those samples.
     This is how they look before blocking:


                                                              Fingering weight yarn

     I used Valley Yarns Franklin natural cone.  My swatch had 30 sts and 30 rows, it turned out to be 20 cm x 15 cm, blocked very harshly.
     Here goes the math (the "joys" I referred to on the title of this entry).
I know that 30 rows measures 15 cm (sorry folks, this is complicated enough without me trying to do it using imperial measurements) and I also know the the full shawl has 200 rounds.  I multiplied the number of rounds by the height of my swatch and divided it by the number of rows I actually knit on the swatch, giving me an approximate measurement for the radius of the shawl = 100cm.

     There are a number of variables I didn't include in my calculations as I didn't want to make it more complicated than my brains could handle.
- The gauge when knitting back and forth and when knitting in the round tends to be different,
- the way tension is applied to a rectangular piece and to a circular piece is different - on a rectangular piece all stitches will have a similar size, but on a circular item the inner stitches tend to be longer the the outer ones,
- on this particular shawl, the density of the fabric (the amount of holes vs. the amount of actual stitches) varies from one chart to the other...

Td;dr version: the shawl, done in fingering weight on size 4.5mm needles will be approximately 200cm in diameter (78.74 inches) - not counting the edging. The shawl will use approximately 1097 m(1200 yards) of yarn.

     To find out how much yarn we'd need I calculated the area of the swatch, the area the finished shawl will (if my math is correct) have, frogged the swatch, measured the length of yarn I used.  Once I had those numbers I multiplied the area of the shawl (31416 cm2) by the length of yarn used in the swatch (8m) and divided it by the area of the swatch (300cm2).
     Dizzy yet?  The final number is an approximation of the total length we'd need = 835m (916 yards).  Due to some of the reasons I pointed above and to the fact that I didn't take the edging into the calculation, I'd feel more comfortable sticking with the 1200yards I recommended already - we can always find some use for extra yarn, but running short is no fun.

                                                                    Lace weight yarn
     For this swatch I used Jaggerspun Heather 2/20 and size 3.5mm needles.  My swatch was 30 stitches by 30 rows and the measurements, after blocking were 16 cm x 11 cm.
     Using the same calculations as the ones for the fingering yarn this is what I got: the shawl will be approximately 130 cm(51.18 inches) in diameter - without the edging and will probably require a similar amount of yarn.

     Final considerations:  If you want a smaller shawl and will be using fingering weight yarn, you can make the final chart smaller - I'll provide alternatives.  If you want a bigger shawl and will be using lace weight, you can repeat one of the charts after knitting all of the charts for the body.


  1. Thanks so much for head is reeling a bit and I'm making sure I have at at least 1200, but I am heartened to know you've figured it all out scientifically - thanks...

  2. Pseudo-scientifically would be more appropriate. The calculations above are far from accurate. Usually I knit a few samples and have test knitters knitting a pattern before it's released, so I have a better idea of how much yarn is required. With this one, I just didn't have the time to do anything...sigh.