It’s been quite a few days since my last post. The reason for this silence was natural: I was doing a lot of math to figure out how to describe the folding of the double helix. And as always, I was constantly going two steps forward and one step backwards. And in some cases, I was hitting a wall, until I realized where I had made mistakes.
The most important tool in all of this was visualization: with Blender, which allowed me to draw 3D structures, Powerpoint, which allowed me to make abstractions, which helped me figure out the general rules and finally paper and scissor, which allowed me to make the shape in ‘real life’.
After several false leads I was finally able to draw the ribbon below. It’s not pretty yet, but now that I know its general shape, I can finally make it pretty.
To begin with, I first thought of just using vectors for illustration, but soon realized that equilateral triangles depict the interactions better. The twisted ribbon shape clearly shows that there are two dots ‘behind’ each dot, one dot besides the dot and one dot in front of it. And both the dots behind and the dot next to the dot affect determine its location.
We can begin by randomly choosing the first dot. It doesn’t really matter what the x,y,z coordinate values of the first dot is. The second dot can be chosen also almost at random, the only requirement is that the distance between the first and the second dot is one, if we wish to normalize the shape (but even normalization is optional). The third dot already requires a bit more care, as its distance from both first and second dots should be one (when normalized). But still, the location can be in a circle drawn around a vector drawn from point one and point two. Only when we try to figure out the fourth point, should we begin to do some serious math.
If the dots were arranged to an infinitely long rod, the fourth dot would be a the fourth corner of a regular tetrahedron drawn from the equilateral triangle defined by dots one to three. However, if we assume that the shape isn’t completely linear the fourth dot is defined by a second equilateral triangle, with dots four, two and three. And not to bore you with details, the angle between the two triangles is arctan√8, or about 70.53°, if the rod is pretty much linear. In real life, if there is some curvature for the rod, the angle is higher than arctan√8, but it probably still rounds to 70.53°.
For a long time, I just couldn't figure out the mathematics, but I always knew the answer was about vectors. For a while I was confident, I would need matrices, but surprisingly, just regular vectors are enough. After a long time hitting my head against the wall, I began to see clearly. This was aided by Mathcad. Excel is pretty dismal with vectors, but if you know what you are looking for, you can just about make vectors in Excel. Anyhow, once I had a rough idea of what I was looking for, I could define the vectors with Mathcad and use it to derive cross products of the neighboring vectors. This allowed me to check my equations quite quickly and helped me simplify them considerably. I had started with such convoluted logic, that I was positively surprised by how simple the mathematics was.
As I mentioned, the structure can be defined by a two-dimensional ribbon, bent in two directions into a three-dimensional rod. The bending can be seen in the equation by the alteration of the sign of a single term: plus sign for the dots in one helix and a minus sign in the other helix. Here is the equation for the fourth dot:
Here, d4 and d2 are the x,y,z coordinates of the fourth and second dots. All of the v’s with arrows are vectors drown from one dot to the next. x marks the cross product and θ marks the angle between the two triangles.
The fifth dot is defined almost identically. However, as the angle between the neighboring triangles is 180° - arctan√8, we can make a ‘trick’ and just keep the equation above but reverse just one sign. And we get:
And the magical thing is that all of the dots with even numbers define one helix and the dots with odd numbers define another helix. But these helices aren’t independent. Rather the movement of the dots in the entangled helices determine the movement of the dots of each other.
Once I had the equations confirmed with Mathcad, I could laboriously add the equations to Excel. I might have gotten a better graph with Mathcad, but as I don’t know how to use it yet, for now, I’ll settle for Excel.
This is how the double helix of dots looks like with Excel:
Here I use a trick in Excel: while each dot is a point with no volume, I can introduce volume by making the marker for the dot t be a circle with just the right radius that the neighboring dots touch each other.
And here we see just one helix:
However, it needs to be remembered that the equations do not produce a single helix. As the dots in each helix are interconnected, you only get the next dot in one helix by defining the previous dot of the neighboring helix.
So, now that we have the equations for the helix, can we see the twisting and bending? The annoying answer seems to be:no. You see, the equations apply only in the case where the triangles form a uniform ribbon of uninterrupted triangles.
For a while I tried to apply other angles to the equations to see whether I could see bending, but to no avail! Then it hit me: the arcing of the ribbon causes it to ‘tear’. When this takes place, the triangles become connected to each other by only one vertex. In this case, the bending of the origami can be described by drawing circles connected to ‘rods’ whose length is the radius of curving. And around each smaller circle, a bigger circle with a radius of √3 ≈ 1.73 of the smaller circle to determine the location of the neighboring circle connected to a rod. The exact logic for all of this is a bit complicated to explain, especially as I’ve only realized it today and haven’t come up with a easily-understandable sentence. But below you can see the structure. Again, with the curving heavily exaggerated.
So, can I apply the equations above to this? Not straight away, at least. And this model still has a tiny problem. If you look closely, the structure isn’t a double-helical toroid: it still lacks the spinning of the triangles.
So, how will this be converted into a double-helical toroid? You need to introduce a second tear. After the second tear, there are no longer these triangular planes left. What you will be left with are vectors. But luckily, I think now I know exactly what to do, even if I don’t have the mathematics sorted out just yet.
I might be wrong, but I’ll probably be able to introduce the correct mathematical expression for the curving of the vectors already in my next post. I’ll keep my fingers crossed!
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