## Saturday, April 14, 2007

### More Mathematical Musings

Issue 4, Revision 1:
I spotted a minor flaw in yesterday's musings on Issue 4 (how many blocks will be in the longest row of this blanket) because I assumed the afghan was square (and you know what happens when you assume. . .).

I can write two simultaneous equations about this blanket. First, the ratio of length to width is 90 inches to 75 inches and the number of rows in the length (L) versus the number of rows in the width (W) will have the same ratio!

L/W = 90/75
= 6/5

Multliplying both sides by W, I get
L = 6W/5

The second equation has to do with the number of blocks in total, which is the number of rows in the width (W) times the number of rows in the length (L).

Total number of blocks = 19,360 blocks
LW = 19,360

I can substitute L = 6W/5 into the formula and get,
(6W/5)W = 19,360
6W*W/5 = 19,360
6W*W = (19,360)5 = 96,800
W*W = 96,800/6 = 16,133 1/3
W ~ 127

Then I can solve for L:
L = 6W/5
= 6(127)/5
= 762/5
~ 152

Issue 5:
I started to wonder on what skein (Y) will I turn the corner for the width (N = 127). Each skein has 605 blocks, so the number of blocks used will be 605Y. The total number of blocks can be calculated from Issue 2 from yesterday's musings for the 127th row.

Total number of blocks = N(N+1)/2
= 127(128)/2
= 127(64)
= 8,128

605Y = 8,128
Y = 8,128/605
~ 13

I should be in middle of the 14th skein when I make the first turn.

Issue 6:
One way to check how accurate my calculations are I wanted to forecast on what row I expect to switch skeins. If I miss the mark by a wide margin, it could either be due to poor quality control on the length of yarn or faulty assumptions on my part. For example, I expect to switch to the third skein on row 48. Solving this problem involves the quadratic equation formula!

1 skein = 605 blocks
2 skeins = 1210 blocks

Total number of blocks = N(N + 1)/2
1210 = (N*N + N)/2
(1210)2 = N*N + N
2420 = N*N + N
0 = N*N + N + (-2420)

Compare that to the quadratic equation:

0 = a(N*N) + b(N) + c

For purposes of the quadratic equation formula,
a = 1
b = 1
c = -2420

So, I have two possible solutions:

N = 0.5a(-b - SQRT(b*b - 4ac)
OR
N = 0.5a(-b + SQRT(b*b - 4ac)

The first will not work because I get a negative answer, and the number of rows must be positive:
N = 0.5*1*(-1 - SQRT(1*1 - 4*1*(-2420))

Here is the second possibility:

N = 0.5*1*(-1 + SQRT(1*1 - 4*1*(-2420))
= 0.5(-1 + SQRT(1 + 9680))
= 0.5(-1 + SQRT(9681))
= 0.5(-1 + 98.39)
= 0.5(97.39)
~ 48

If I miss my predictions by a wide margin, then I will need to retrench and rethink my plans before it is too late and I am forced to do a massive yank-fest.

Out of curiosity, I wanted to see if this way of calculation would predict turning the first corner on the 14th skein, and it did!

Okay, okay, I promise . . . no more links between algebra and crocheting for awhile!

P.S. By tomorrow, I should be on my third skein. That will be "only" thirty to go!