Tag Archives: Pythagorean theorem

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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Math Is AWESOME!

Yes, awesome, inspiring feelings of awe; magnificent, amazing, stunning. Math rocks!

I truly enjoyed creating Untied, with its free-form, no math construction. Though I used a ruler for parts of it, it wasn’t because the outcome depended on the measurement. Instead, at those instances I was only interested in straight, and sometimes parallel, lines.

20160401_182836However, I love the challenge of a technically difficult quilt, too. For a change of pace, I began a brand new project. The inspiration for it is a piece of fabric I bought several years ago. It’s a fussy historical print that I’ve always loved. However, it’s fussy and historical, and the colors are just off, all of which have made it hard to use. If it is cut into small patches, the impact of the print disappears, but large pieces would require designing just for it. So I am.

I started by pulling from stash, which is how almost all my quilts begin. I pulled all the blues that could work with that print, which meant they had to have a tinge of green and a little grey. It’s the color I learned as French blue. As it turns out, I don’t have a lot of blue with that, and all of it is in scraps or small pieces, other than the inspiration print. We’ll see how far I get before needing to buy something.

Burgundy reds, creams, browns, and cheddar oranges also came out of stash, including from my scrap drawer. I’m trying to commit to using my scraps more effectively. They come in handy, as I’ll explain later.

My center block is 18″ finished. I knew I wanted to turn it on point twice. The method is exactly the same as used for the economy block setting. This is also called “square in a square.”

20160401_182819

And this is where the first round of harder quilt math comes in. (It’s not very hard, just a thing to learn, or store so you can look it up.) As noted in the linked post, when setting a square on point twice, the finished size of the resulting block is twice the dimensions of the initial finished square. For my 18″ block, I would end up with a 36″ center, once trimmed and finished. (See below for the calculation and reason why.) 

What a great way to quickly increase the size of a quilt!

20160401_182927

After trimming, I added a 1″ border all the way around, taking it to 38″ square. That’s an odd size.

Options for a 36″ Edge
Most quilt blocks are square, or are rectangular with length twice the width, like flying geese blocks. I typically divide a border length into a number of equal increments to find how many blocks could fit along the edge. For example, if I put a block border directly along the 36″ center, I would divide 36 in various ways to get possibilities. Let’s start with whole inches.
36/18 = 2. I could have 18 blocks measuring 2″.
36/12 = 3. I could have 12 blocks measuring 3″.
36/9 = 4. I could have 9 blocks measuring 4″.
36/6 = 6. I could have 6 blocks measuring 6″.
But remember we could also go to half-inches, such as
36/8 = 4.5. I could have 8 blocks measuring 4.5″.
36/24 = 1.5. I could have 24 blocks measuring 1.5″.

You can see there are actually infinite variations, though I like to stick to the ones that are easy to construct.

Options for a 38″ Edge
But I didn’t want to put a block border directly against the large center. I wanted the hard line of a narrow border before anything else, to contain the pale toile. That gave me 38″, which is harder to divide nicely than 36″.
38/19 = 2. I could have 19 blocks measuring 2″, but this was too narrow to work well with the proportions of the center. And 19 is a prime number, so I couldn’t subdivide it into whole numbers.
I could shift to fractionals, such as
38/8 = 4.75. Sure… This actually would work fine with something like HST or hourglasses.

My go-to blocks are half-square triangles and hourglasses. I don’t want this quilt to look like others I’ve made, so it’s good to try something different.

Then I had a thought, a math thought! What if I divide 38 by 1.414? (That actually was the first thought. Then I thought…) If I turn blocks on point rather than set them straight, I would need to know the number of blocks using the math for the diagonal.

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 Pythagorean Theorem
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; b = 1; 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″.
38″/1.414 = 26.875″. Hooray!!

Huh? Why is that good? 26.875 is very close to 27, which is a very easy number to use. For instance,
27/9 = 3.

I could use 9 blocks measuring 3″, turned on point. (There were other options, too, such as 6 blocks at 4.5″.)

3 x 1.414 = 4.242, the diagonal of a 3″ square, and the width of a 3″ block when turned on point.
4.242 x 9 = 38.178, or very close to 38″.

So if I use 9 3″ blocks turned on point, I’ll have the length needed for a 38″ border.

(And to review the concept in another way, let’s go back to the 18″ block. When I turn it on point twice, I’m doing this:
18″ x 1.414 = 25.452″
25.452″ x 1.414 = 36″.
This is the same as 18″ x 1.414² = 18″ x 2 = 36″.) 

I chose to use the historical print and other blue scraps as the 3″ finish squares. For the background fabric, a light background would give good value contrast for the blues so they stand out well. I picked a pink and tan print on pale cream. The pink works because the reds have a pink cast. Right now I’m still working completely from stash and scraps. I was able to cut most of the setting triangles from a couple of larger pieces, but for the last few I had to sew scraps together to cut patches.

20160401_184553

Can you find the seams where the setting fabric scraps had to be sewn together?

With the on-point border added, my current center is 46.5″ finished, another weird number. I plan to add an unpieced strip next, but I haven’t decided its width. I could add a strip to take it to 48″, 49″, 50″… So the next design decision, really, is the border after that, which will determine the width of the strip.

20160401_183045

I feel really fortunate to have the math skills as part of my craft toolbox. You can make beautiful quilts without knowing how to do any of this, but knowing increases the options open to me. Math is awesome!