Instructions for the Compound of Five Octahedra
This is an explanation I posted on the Zome mailing list a while ago.
I just created the most extreme Zome creation I've ever worked on. It's a compound of five octahedra (shown at http://mathworld.wolfram.com/Octahedron5-Compound.html) and it stands well over 2 feet tall. It's geometrically difficult (I think it uses all of the greenline directions) and uses a LOT of parts. Do not attempt unless you have:
30 long blue
30 short blue
60 short half-green
120 long half-green <--- most people don't have this
60 long green
152 nodes (I think)
I initially didn't think this model was Zome constructable. Actually first I thought it was, and then I thought it wasn't, but eventually I found out it really is, and without too much technical difficulty (as long as you don't consider building an octahedron circumscribed about an icosahedron difficult.) The real difficulty is figuring out what parts to use, and where. I'll describe the construction first, then a few cool things I've noticed about it, and then I'll give details as to how I derived the oddball sizes in the construction.
MODEL PREP:
Before you build this model, you may want to build the a single octahedron - icosahedron compound, so you can see a bit more easily how five octahedra will work. To do this, build an icosahedron out of short blues (or mediums and scale the rest up by one as well.) Then build a little "pyramid" on a pair of the icosahedron faces as follows: on the edge the faces share, finish a triangle pointing directly away from the model center by using two short half-greens, and then connect this outer vertex to the two far icosahedron vertices (still on this pair of faces) using medium half-greens. You now have one green pyramid that covers 2 icosahedron faces. You'll build a total of 6 of these around the icosahedron such that each short half-green is paired with a medium half-green to make a straight line the length of a long half-green (but with an icosahedron vertex splitting it in half.) Now you have an icosahedron inscribed in an octahedron.
To imagine the compound of five octahedra, imagine you took a bunch more of these pyramids, and built them with five-fold symmetry around each icosahedron vertex. If you actually try this, you'll find that your medium half-greens intersect each other at very odd locations (that is, it's not at a short half-green length, or any other standard Zome length for that matter.) It turns out we have to scale the model up so that each short becomes a short plus a long in order to have the intersection work out to a normal length.
MODEL INSTRUCTIONS:
Build an icosahedron through whatever method you prefer (red starburst, freehand, etc), where each side is a long blue plus a short blue. This model stands about 2 feet tall. Now, off of each icosahedron edge, build an isosceles triangle, where the outer two edges are made out of one long and one short half-green each (I put the short half-greens toward the outer vertex of the triangle, to present a uniform appearance.) Each of these triangles should point directly away from the center of the icosahedron. What you have at this point should be a big blue ball with a bunch of green spikes sticking out at 90 degree angles.
Now, from each icosahedron vertex, you can build a line to the tip of the triangle on the opposite edge, using a long green and a long half-green together. Attach the long green to the icosahedron vertex in the right direction, and then attach the long half-green to that to finish the edge (here, the order is important, due to the next step.) You can also put the first long green and long half-green together first and treat them as a single strut, as long as you make sure they're in the right order (though once they start intersecting, you have to deal with them individually.) You'll do this a total of 5 times per vertex, or 60 times total. This means on each face, 3 of these lines will cross, right at the node where the long green meets the long half-green (but only if they're in the right order.) Now you have the compound of five octahedra circumscribed about an icosahedron. If you want just the five octahedra, you can pull out the internal bracing as you build the model or at the very end, or you can freehand the whole thing (which isn't a lot harder than freehanding an octahedron -- each edge is now made up of a short half-green, a long half-green, a long green, and a long half-green in that order, and at any vertex the two opposite edges should "match".)
COOL PROPERTIES:
So far, all I've really come up with is that this model has cool symmetries. In particular, it has 2-fold, 3-fold, and 5-fold axes, which I think are exactly the same as the buckyball. I'd be interested to see if someone actually classified this, though. It's also easy to build in paper -- you just need a bunch of triangles with side lengths T^2, 1+T^2, and 2T (T = tau, the golden ratio, about 1.618), though I wouldn't have known that if I hadn't first built it in Zome.
MODEL DESIGN PROCESS
(WARNING: this is very technical!)For advanced readers (mostly, those familiar with Zomod and/or the coordinate system described by Walter M. Venable) who are wondering why this model is built in the size it is (and how I came up with such bizarre edge lengths), here's the explanation of how I designed it. I suggest you build the model, or at least a portion of it, before trying to read this so you can see what I'm talking about.
A while back, I designed something called "Kepler's Obsession", which was meant to become a commercial Zome product but hasn't yet. It's all of the platonic solids inscribed in one another, and is shown in a scaled version on the new advanced math kit instructions. I have one on display on top of my bookshelf, next to my wife's paper models of platonic and archimedean solids. I was looking at that and wondering what would happen if I put the compound of five cubes inside it (adding four cubes to the one already there) and also if I did the same with the ten tetrahedra. I realized the dodecahedron would stay in the same place, and so would the icosahedron, but that the octahedron shared the same symmetries as the cube and so would also end up in a compound of five. I wanted to build it as soon as possible. I decided it'd be smarter to build the five octahedra from scratch as the base for the new model (adding the tetras, cubes, etc. later) than to try to build all of these inside an existing model, so I set about designing the Zome model of the compound of five octahedra.
I started out by building a short blue icosahedron and then inscribing it in an octahedron made of short and medium half-greens. I thought from here I could just add four more octahedra (in five-fold symmetry about one icosahedron vertex) to get the compound. I tried it, but quickly ran into difficulty -- the edges intersected. I tried to build the model scaled up by tau (the golden ratio; I now refer to it as T for ease of notation.) I couldn't quite get the edge intersection right, though, so I scaled it up again and tried with combinations of the smallest pieces I had. Nothing seemed to work, though -- I kept ending up having to bend the model to get the edges to intersect at a node. Finally I decided it wasn't possible to build, but being a math nerd, I had to prove it to be satisfied. I decided to mathematically locate where in space the octahedron edges intersected, and then find what length strut I'd need to reach that point.
I started out by setting up a coordinate system as described in "A coordinate system for the Zometool Modeling System" by Walter M. Venable (for those who know Zomod, it sets up the x axis as R10+, y as R3+, and z as R6+ and then derives coordinates for the rest of the directions based on this.) I then built an icosahedron face using R10+3, R4+3, and whatever connects the two. Underneath that I built a pyramid out of medium blues (R5+2 was one of the struts) and on top I built a pyramid out of long yellows (T9+3), and then connected the two pyramid vertices with T6-3. I knew that since the compound of five octahedra had threefold symmetry about the icosahedron face, the intersection of the octahedra edges had to be along that line. I next built a triangle off of R4+3 using long half-greens to find my octahedron vertex, and I knew the intersection I was looking for had to be on the line between that and R10+3.
So, I had to find the intersection of two lines in threespace. One line (from the tip of the green triangle to the vertex of the icosahedron) had endpoints (T+1,0,0) and (1/4, 3T/4+1/2, -T/4-1/4) and the other (between the pyramid vertices) had endpoints (T/2+1/2, 1/2, T/2) and (T/2+1/2, T/2+1/2, -T/2-1/2). I knew the x-coordinate of the intersection would be T/2+1/2 (because the second line has both endpoints at that x-coordinate) so I was really interested in finding the y and z coordinates. I did this by simply projecting everything into the y-z plane and renaming my axes x' and y', and then solving the linear system
y' = (3T/4+1/2)/(T/4+1/4) * x'
y'-1/2 = 1/(1+T) * (x'+T/2)
Some ugly algebra yields
y' = (3T+2)/(4+2T)
x' = -z = (T^2)/(4+2T)
So, my point of intersection was ( T/2+1/2, (3T+2)/(4+2T), -(T^2)/(4+2T) ), and the length of strut I was interested in would go from there to (1/4, 3T/4+1/2, -T/4-1/4). Subtracting those gives us a vector that simplifies to 1/2 * [ T+1/2, (3T+2)(-T/(4+2T)), -(T^2)(-T/(4+2T)) ] which has to be in the direction of a greenline -- the diagonal between R7- and R12+, which is 1/2 * [2T-1, -T, T-1]. Therefore, the first vector is a multiple of the second -- and just dividing the x coordinates gives us the scaling factor, (T+1/2)/(2T-1) where 1 would be the length of a medium green (the diagonal of a short blue square). The rest of the algebra does work out here (I currently have it written up on my whiteboard with "some algebra shows..." like a true mathematician), so the first vector really is a multiple of the second. This meant I needed a strut of length (T+1/2)/(2T-1), which I thought was impossible.
After a lot of algebraic manipulation, I converted (T+1/2)/(2T-1) to (T^4)/2(T^2+1), which I also thought was impossible to build in Zome -- I could build a strut of length (T^4)/2 all right, but that other stuff in the denominator just made things impossible, so I had to get rid of it. Suddenly it came to me -- if I just scaled the whole model by (T^2+1)/T^2, then the strut I'd need would be of length (T^2)/2, which is a long half-green. This meant my long-blue icosahedron face would instead be long blue plus short blue struts, my long half-green triangle edges would be long half-green plus short half-green struts, and the length of the edge I'd been working on would be a super-long half-green plus a medium half-green, or a long half-green plus two medium half-greens (same as a long green), of which a long half-green would be on one side of the intersection and the remaining length would be on the other side. This explains all of the strut lengths in the model described above.