# Most irregular

It’s a remarkable fact that there are only five regular convex polyhedra, that is, solid shapes whose flat faces are all identical regular polygons. The most familiar example is the cube, with 6 identical square faces. There are other, less regular but still orderly polyhedra, such as the truncated icosahedron, seen in a bloated form in some footballs.

But what, I wondered, about truly irregular polyhedra, where every face is a different irregular polygon? What would they look like? Last January I set out to make some and find out.

My irregular polyhedra are all based on spheres. Imagine placing a number of dots on a sphere. At each dot, let there be a plane just touching the sphere. These planes will all intersect each other, and if we remove the parts of each plane cut off by the neighbouring planes, we’re left with a polyhedron, with one face for each dot that we placed on the sphere. The nature of this polyhedron depends upon how the points are distributed over the sphere. In the polyhedra shown here, the placement of dots was random but subject to certain constraints.

I wrote some software that allowed me to choose how many faces I wanted, and to regulate how evenly spread over the sphere the dots were. I could preview the resulting polyhedron, and when I saw one that I liked, the program produced a set of images of the faces of the polyhedron, with numbered tabs on them. Then all I had to do was print them, cut them out, fold the tabs, and glue them all together. This is not a task for the impatient: the 83-hedron at the top took about 3 days.

The polyhedra shown here differ principally in how evenly the dots were spread over the imaginary internal sphere. After randomly placing the dots, the program simulated repulsion between them (as if they were electric charges). The longer this repulsion process went on, the more evenly distributed the dots became. For the 29-hedron, no repulsion process was done, and for the 43-hedron, the process was allowed to continue until the dots stopped moving; presumably this is now (in some undefined sense) as regular as a 43-hedron can get.

For the 29-hedron (right), I took advantage of the four-colour map theorem, which tells me that if I want to colour the faces such that no two neighbouring faces have the same shade, 4 shades of card are all I need. I leave it to you to convince yourself that if this theorem is true for flat maps, it must be true for maps on balls too. (If you don’t see where maps come into it, think of each face of the polyhedron as a country on a political map.)

None of this would have happened had it not been for a visit to Jenny Dockett to talk about her Illuminating Geometry project. It was while talking to Jenny that the idea for making irregular polyhedra came to me.

In 2015 I exhibited these polyhedra, and others, at Mini Maker Faire and Porty Art Walk in Edinburgh.

I wrote the software in Python. The internally-illuminated shapes are made out of layout paper (very thin paper), and the 29-hedron is made of thin card. I used UHU Office Pen glue (not to be confused with UHU Pen glue). This glue doesn’t wrinkle the thin paper, and dries at about the right speed, but has proved to be somewhat unreliable in humid weather.

## 4 thoughts on “Most irregular”

1. Hey Ben!

I’ve been trying to design an icosahedron with random flat sides, and came across your 83 sided icosahedron, which is really cool, and basically what I’m envisioning to do. I’m a woodworker, and I’m trying to create a sculpture where the base of the project starts with an icosahedron like yours, and I then continue to build on from there. I was wondering how you can generate software to do that, as I’m not very experienced with digital design. I can use google sketchup, but my knowledge is fairly limited. What I’d like to do is make something like yours, but only using four sides per shape. Most icosahedron images I find, use either all triangles or polygons, but I’d like to make irregular squares. Would your software be able to do that? I’d like to create a sphere about roughly 10-12″ in diameter (304mm) where the shapes vary in size from like 1-3 inches, and print it all out on paper, and then recreate it in wood. If you have any advice or suggestions, please let me know. Thanks!

1. bencraven says:

I’ve had a good think about your question, and although I haven’t been able to reach a cast-iron conclusion, I believe that what you hope to do isn’t possible. Here’s the main line of reasoning (it’s a bit complicated I’m afraid):

Any polyhedron has faces (the flat bits), edges (the ridges where 2 faces meet) and vertices (the peaks where the corners of 3 or more faces meet). There’s a formula that is true for all polyhedra:

F + V – E = 2

where F, V and E are the numbers of faces, vertices, and edges in the polyhedron. For example, a cube has 6 faces, 8 vertices, and 12 edges:
6 + 8 – 12 = 2 as predicted.

Now consider a polyhedron where all the faces are quadrilaterals (4-sided shapes). Each face touches 4 edges, but each edge is shared with the neighbouring face, so the number of edges will only be twice the number of faces. So E = 2 times F.

Let’s suppose, just for the sake of argument, that the polyhedron has 40 four-sided faces. From the above, that means there’ll be 80 edges. Putting these number into the formula gives us

40 + V – 80 = 2

which means that V must 42, ie this polyhedron must have 42 vertices. (In general, for a polyhedron with 4-sided faces, the formula tells us that the number of vertices must be 2 greater than the number of faces).

Just for now, let’s ignore the extra 2 vertices and suppose that we’ve got 40 faces and 40 vertices. Now each face has 4 corners, and so must take part in 4 vertices. But because there are as many vertices as faces, this means that there must be 4 faces meeting at each vertex. Now in a flat tiling like this there’s no problem having 4 squares meeting at a point everywhere.

[Imagine a bathroom wall tiled in the usual way with square tiles. I didn’t realise I couldn’t get an image in the comment]

We could also imagine a distorted version where the tiles are all different non-square quadrilaterals. But there aren’t any peaks (vertices) – it’s flat. Now imagine trying to distort the tiles to make a peak in one place. The angles of the tiles meeting at this peak will have to be made less than 90 degrees. That’s not a problem in itself, but we want every tile to have a vertex at *each* of its corners, which means that the shapes of the faces need to be 4-sided and with angles less than 90 degrees at *every*corner. Maybe this is too strict – perhaps there can be a bit of give-and-take, with each face having some faces bigger than 90 degrees compensated by some being much smaller, such that the average size of the angles is less than 90 degrees. But it turns out that the internal angles of any quadrilateral add up to exactly 360 degrees. So the average angle at each corner must be 90 degrees. It can’t be less. So it’s impossible to have a polyhedron (or even part of one) made of 4-sided faces where each vertex is the meeting of 4 faces.

Now in fact we have 2 more vertices than have faces. This means that 2 of the vertices must have 3 faces meeting at them, not 4. So it lets us off the hook at 2 vertices but doesn’t solve the problem elsewhere.

For this reason, I think what you’re hoping to do isn’t possible. We can’t make the angles at the corners of each face small enough.

I hope that makes some sort of sense. It’s a rather elaborate argument!

There is another reason why I don’t think it’s possible: I think I would have heard of it. The polyhedron would surely have a name of its own, and I don’t know of one.

Having said all that, if in the future you send me a photo of a wooden polyhedron, the faces of which all have 4 sides, I will be delighted to have been proved wrong and will write a special blog post on the subject!