Sorry to come a little bit late to this; I was busy with other projects.
This is a perfect example of a problem you encounter with 22.5-degree designs, something you can indeed solve with some mathematics. It is actually the only type of design that is discussed in this thread, even though it is not explicitly named. I know the theory behind this was explained in several posts many years ago, but they are difficult to follow as the pictures have since disappeared. There is also a new tool that has appeared that can be very useful. So, I would like to give here a comprehensive explanation, step by step with fresh illustrations. It is a procedure adapted from an article by Satoshi Kamiya in Tanteidan Magazine (
Folding Out 22.5 Degree and Relevant Angles, Tanteidan Magazine No. 115, p. 11 (2009)). It's going to be a bit lengthy, but I think it is worth it.
You start by assigning a unit length to some characteristic feature of your design from which it is easy to draw your pattern. Here, even though you say you started with a 10x10 grid and designed the crossed-out square in the lower part of the CP, it is actually easier to start with the three smaller squares at the top. So, let’s assign a unit length to these squares:
Next, we work out the size of the other features, step by step as we draw them. For this, we use three properties of triangles we encounter in 22.5-degree designs:
I use a personal notation to make it easier to read. “s” is actually the square root of 2, which is 1.4142… So, s times s, or s squared, or s^2, is 2. Using the above rules, we can deduce that the length of the diagonal of a small square is s and the height defined by the bisector (kite base) in the lowest of the three squares is s-1 (see rule #1 rotated 90 degrees):
Next, we look at the tiny triangle that bridges the gap between the small square and the (slightly) bigger square. It too has the same shape as rule #1 (not rotated) but has a height of (s-1) instead of 1. So, all its dimensions are multiplied by (s-1) and we end up with (s-1) times (s-1) or (s-1)^2 for its base. To get the size of the bigger square, we simply add the size of the smaller square and we end up with 1+(s-1)^2:
Here, we need a little bit of math in order to simplify things. You can skip to the following drawing if you are not interested in the details:
(s-1)^2 = (s-1)(s-1) = s^2-2s+1 = 2-2s+1 = 3-2s
At the third equal sign, I used the property s^2 = 2. With this simplification, the size of the bigger square is now 4-2s, so the original grid is now assigned a size of 2-s:
Summing up (literally), the three smaller squares at the top have a total height of 3 and the bottom part has five of the original grid size, or 5(2-s). The total height of the paper is 3+5(2-s) = 13-5s:
Now, we don’t like to have a negative sign in our paper size, so we have to use some more math to simplify this. This time, it is quite involved, so if you are not mathematically inclined, you either have to trust me or ask for some help. Here we go: we divide all dimensions by (2-s) and multiply them by s. So, we have the following transformations:
Did you follow me? In the first and last lines, I used the “multiply by the conjugate” technique to get rid of the denominator. See for example
here for an easy explanation. If you are serious about all this, you will have to get used to this technique. Sorry, there is no free lunch here!
But here is the reward: we now have a cleaner picture with numbers that are easier to work with:
Most importantly, these are numbers that
Satoshi Kamiya’s calculator can use. It is a calculator that he has developed for solving this kind of pattern. It is more useful in this case than Robert Lang’s Reference Finder. It is in Japanese, but it is easy to use if you have done the work above. You simply enter the total size of your paper in the two boxes at the top of the page and hit the button underneath. In this case, you enter “3” and “8” (because he wants the size to be entered as [] + []*s) and you get 19 possible solutions (with other sizes, you can get less or more or sometimes 0). He only shows the first one but there are “Previous” and “Next” buttons underneath to browse through all solutions.
Let’s look at the first solution. There are two drawings showing two specific folds we have to make (I have rotated them by 90 degrees):
Their intersection gives us a new reference specific to the dimension we gave. The calculator tells us this reference point lies at distance 3s from the left edge of the paper:
So, with a trisection, we get the vertical strip at distance s from the edge and we can start folding the CP:
If you don’t like the trisection, you can select the next solution in Satoshi Kamiya’s calculator, which gives you 4s instead of 3s. Unfortunately, there is no solution that gives you directly 5s, so we always have to do some more folding to get to the three upper squares (which require a trisection anyway).
There you have it: not only a way to fold your design on any size paper but a procedure you can apply to any other design you make in the 22.5-degree system. Good luck!