The Dodecagon Pyramid Stack of Cylinders

I’ve been working on the problem of crystallizing rotating cylinders into Waterman polyhedra for ages. To some extent I’ve been doing it for over six years, ever since I got the initial idea of the crystal structure of Lignin spheres. However, for some reason I never got round to properly illustrating the shape as a 3D model until two months ago. One of the big problems I had was that I had done a rough illustration of the crystallization model three years ago, which I included in my first manuscript that was rejected. Then, I used the same rough model in the newer paper submitted a year ago,  which was also rejected, but which I at least managed to publish as a preprint. But the problem with this model was that it wasn’t exactly a crystallization model of a solid sphere, but only a model of how to crystallize a truncated octahedral crust with nothing inside and nothing outside. I had explained that the model depicted a shape within a Waterman polyhedron, which might have made sense to the reviewer, if they had took the time to learn what a Waterman polyhedron is. But I think they didn’t. So, to really illustrate the mechanism of crystallization, I need to illustrate each step in the process. Otherwise, a reviewer will again state that I’m taking steps that are a bit too unspecific.

The reason I wasn’t specific enough in step 2 three years ago when I first illustrated the model is the issue of memory overload. Or more specifically whenever I tried to make a full model that’s more complete, my computer froze. But there is a solution. I can learn the basic mathematical rules and then illustrate these rules with simpler models. This way I can either a) never need to make the larger model showing all the features at once, or b) I can build the model without errors with these rules. I know that option a) is easier but being able to do b) would be great. However, option b) only works if I don’t need to do any error corrections. Whenever I’ve made a model with thousands of objects, its manipulation becomes sluggish and any error correction becomes near-impossible, as any changes take forever to realize and when you need to make hundreds of small (or large) changes, the model always freezes.

Today I’m presenting the simplest possiblepyramidal stacking model with three slightly different repeating planes, which makes the final pyramid jagged. To start with a bang, I’ll first show what the pyramid looks like and only after that I’ll show how the shape is constructed.  So, here’s the shape:

https://video.wixstatic.com/video/ed0781_d76057b97e044f639efee0092d4ac127/1080p/mp4/file.mp4

And you might ask yourself "Well... how did I get here?"

Sorry about the gratuitous Talking heads reference. The basic shape above is the same as in the models I’ve been showing these last weeks or months, except in most of these models the hexagonal pyramidal stacking structure was mostly obscured by the rest of the model. The post where I came closest was “The Tricolor Spherical Crystal”, but even there I only showed a 2d projection of the crystal boundaries, where each of the cylinders were depicted with spheres. So, even though I’ve seen plenty of versions of the above shape, I haven’t included these in my posts. And that’s because the prior versions were part of larger structures and I had never reduced them into basic pyramids of cylinders.

If you look at the model above very carefully, you could just about equate three of the six jagged faces of the pyramid with the windmill blade shapes of the Tricolor Spherical Crystal. But if I hadn’t pointed this out, it wouldn’t have been at all clear. And probably still isn’t too clear. That’s why I’m cutting the shape, comprised of 9 layers dedecagons (= 12 sided polygons) into three sets of planes: planes (1,4,7), (2,5,8) and (3,6,9).

Let’s begin by look at planes (1,4,7). First, if one looks directly along the x-y projection (leftmost picture below), it’s impossible to separate planes 1 and 4 from plane 7. I.e. plane 1 is perfectly overlain on top of plane 4, which is overlain on top of plane 7. Only when one tilts the model (center below), will one see the shape of the smaller planes. The first plane isn’t actually a dodecagon, but a triangle, as there are only three cylinders in the shape. Only the fourth plane shows the dodecagon shape. However, even there the shape is so small that one can be skeptical why the crystallization of cylinder should produce specifically a dodecagon and not any other shape. On layer 7 it starts to become clear that the main crystallite is indeed a hexagon, but at the interface of two hexagonal crystallite there is no way for one cylinder to be part of both crystallites. For a while I considered that the cylinder would randomly become part of either triangular crystallite segments, I realized that I have electron microscopy evidence that the triangular crystallites fuse, which means that the arrangement of the cylinder on the interface must be halfway between the two larger segments. Basically, this means that even though the crystallite appear to be a hexagon, the soft edge between the segments makes it really a dodecagon. This might sound like a trivial issue to a layperson, but for an illustration base of physical entities, this is extremely important. And finally for clarity, if one tilts the model even more (right below), It’s possible to see each separate plane and that there indeed is sufficient space to fit two additional planes between the neighboring planes in this shape.

Next, we’ll look at planes (2,5,8). It can be difficult to see why the cylinders are offset between the neighboring planes, but when one remembers that the model of rotating objects assumes close packing of equal spheres, the offset becomes more obvious. This means that while the quasi-hexagons in planes (1,4,7) is not equilateral but rather a semi-regular or isogonal hexagon, with alternating lengths of the sides, plane (2,5,8) is. While in both plane groups the side lengths of the upper left, lower left and the right segments are 4 cylinders in both plane 7 and plane 8, the side lengths of the upper right, lower right and the left segments are increased from three to four cylinders from plane 7 to 8.

Finally, the planes (3,6,9) are also isogonal, but this time the side lengths of the upper right, lower right and the left segments are increased from four to five cylinders from plane 8 to 9.

This also means that planes 7,8,9 are a set, with lengths of the upper left, lower left and the right segments are 4 cylinders in each of the three planes. Having learned this rule, one can see that planes 4,5,6 are another set, where the lengths of the upper left, lower left and the right segments are 2 cylinders in each of the three planes, where the remaining three segments have lengths of 1, 2 and 3, respectively.

And to follow the logic, planes 1,2 and 3 are a set where the upper left, lower left and the right segments are 0 cylinders each, where the remaining three segments have lengths of 1, 0 and -1, respectively. The length of -1 might sound weird, but it just means that the hexagon with a side length of 0 is converted into a triangle when the side length drops to -1.

Now that I’ve made the quasi-hexagonal pyramidal stacks, this means I need to also make quasi-square pyramidal stacks. By the same logic, these stack will actually be octagonal. Again, I’m expecting this task to be relatively simple, but possibly still laborious. The only place where there might be some surprises are at the interfaces of crystal segments. I’m noticing that whenever I’m forced to abandon simplifications, new ideas emerge. And I think the square-hexagon interface is the only simplification left in the uncollapsed model. Of course, if I wish to convert the uncollapsed model to a collapsed on, I’m sure I’ll learn something new and interesting.