Tag Archives: Snowball blocks

Wind River Beauty, Math Part 1

In 2017 Jim and I drove thousands of miles in a number of different trips. When you’re in the car together that much, literally a few inches apart, it helps to have entertainment. Fortunately, we like to talk with one another, so the types of things one might muse about silently instead become topics for conversation. For example, after noticing a stop sign and considering how it looks like a snowball block, I asked, “If you start with a square and want to make a regular octagon from it, how do you calculate the length so each of the 8 sides is the same?” Huh?

Okay, look at the two illustrations below. The one on the left is a stop sign. It’s a regular octagon, meaning that all of the angles are equal and all of the lengths are equal. The diagonal segments of the octagon are the same lengths as the horizontal and vertical segments. I noted the dimensions as a for the vertical and horizontal segments, and c for the diagonal segments. As you can see, a = c. (Click the image to open the gallery and see larger.) The segment lengths are all the same. The dotted segment noted as b is not part of the octagon. If you extend the vertical and horizontal lines to create a square, b is the extension.

On the right is an illustration of a square, red & white snowball block. (This specific snowball block is designed to pair with something like a 9-patch block.) For the octagon (white, 8-sided shape,) the angles are all the same. However, the lengths of the octagon line segments are not the same. The diagonal segments of the octagon c are longer than the horizontal and vertical segments a. Why? For this particular block, each side of the square is cut in thirds; a = b. Going down the left side of the square, the top red segment b is equal in length to the center white segment a, which is equal to the lower red segment. The equal lengths make it easy to pair this block with a 9-patch. But the equal lengths of a and b mean the diagonals c are longer by a factor of 1.414. The general idea is the same for all snowball blocks, with the length of c the diagonal dependent on the length of the two triangle legs. See the primer on the Pythagorean theorem at the bottom of the post if you want to know more. 

So how do you take a square and make a regular octagon from it? I’m not bad at math but will be the first to admit I didn’t learn my geometry. Jim worked it out for me. What he found is

b = .707a
2b = 1.414a
==> side of the square = 2b + a = 2.414a

Why does it matter? At the time it was just curiosity, but I quickly found a project to apply it. In spring of 2018 I took a workshop with Toby Lischko on making New York Beauty blocks. She taught a simple way to use curved rulers and paper piecing to create these lovely, complex blocks. This was mine.

After I made it, I thought about how to use it to center a quilt. My design idea would work best if the center was a regular octagon.

With Jim’s formula in hand, knowing the size of the square, I solved for a and b, which let me know how big to cut the stitch-and-flip squares to make the corners.

I wanted a center block finishing at 17″. (Why 17″? That comes later, more math!) That means
side of the square = 2b + a = 2.414a
17 = 2.414a
a = 17/2.414 (just dividing both sides by the same number to solve for a)
a = 7.04, or just barely over 7″.

Since the finished side of the square is 17″ and a (the center segment) is 7″, the other two segments b are 5″ each. I cut my stitch and flip corners 5.5″ each. This is the result. The finished length of the diagonals (along the purple/orange seam) is the same as the finished length of the orange segment along the horizontal and vertical sides of the square. 

Pythagorean Theorem

Here’s the concept. The picture below shows a triangle that is 1″ on the vertical and horizontal sides. The diagonal measures 1.414″, which is the square root of 2″. (Check with your calculator if you don’t believe me.)

Sq_rt_of_2
For a right triangle, the square of the length of the diagonal (hypotenuse) is equal to the sum of the squares of the other two sides. We often see this expressed as a² + b² = c². To find c, take the square root of c².

In the case to the left, a = 1; a² = 1; b = 1; b² = 1; a² + b² = c² = 2; c is the square root of 2, or 1.414.

In fact, the diagonal of every square is 1.414 times the length of the side. So
length x 1.414 = diagonal.

1″ x 1.414 = 1.414″
2″ x 1.414 = 2.828″ or close to 2 7/8″
3″ x 1.414 = 4.242″ or close to 4 1/4″
4″ x 1.414 = 5.656″ or close to 5 5/8″
and so on.

That also means that if I know the diagonal of a square, I can find the length using
diagonal/1.414 = length. For example
6″/1.414 = 4.243″, or very close to 4 1/4″.
This is also useful in the next step of the Wind River Beauty.

Agreed, you gotta be something of a math nerd to work through all this. I’m glad all my quilts don’t require this process, but it’s a wonderful tool to use for a few.

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