I have no problem with those three great circles (of the same color in this wikipedia link - you may have to push the broken image). But mating five of the copies is my problem.
Of course I mate the origins, but what next?
I guess I meant polyhedron’s projection on a sphere. I managed to open the STL, but I still don’t know how to design it myself. Maybe some ratios would help me through this…
Here is the link to their help forum.
Many years ago, I assembled a truncated icosahedron in a very early precursor of Solidworks. Rather than projecting anything onto a sphere, I simply created the faces and (very accurately) manipulated them together. It took a while, but it worked.
It’s been a while since I used Solidworks. I’ve pretty much switched to Fusion 360.
You want to draw lines on a sphere that will end up as the shape. As you noted, you can accomplish this with 5 sets of 3 great circles, with the 3 each being orthogonal. And you seem to have no problem with placing the first of those sets, since that doesn’t require any special angles or anything.
What I would do is this: note a single quadrant of one of the great circles, for instance the black circle (from your pic) around the equator. There are only two points of interest beyond the axis-aligned ones. I’m not easily finding what those coords are right now, but if you decode DPRK’s STL file into ASCII form, you should be able to extract them. The first point has a black/red/blue intersection and the second has all 5 colors.
From there, you can create points along your existing great circles, and then create others relative to the new points. For instance, if you create 4 points surrounding the fore-most black point (they are all just permuted versions of the two points from above), you can create the first of the orange, blue, red, and green great circles, since you just need two points and a center.
An alternative might be to get the center coords of a dodecahedron. Or even construct one yourself via the dihedral angles. Then, from the center of each pentagon you can trace an axis that intersects the sphere. Then, from each of those points, draw great circles 36 degrees apart (you won’t need all 5 great circles on each face of course, since other intersections will get filled out as you go).
I opened the file (in the link which DPRK shared) in ASCII format. Too many six decimal cordinates. Are those numbers rounded or exact - well, I’m still looking for a simpler solution. Once I sit down at my desktop computer, I seem to have patience for about seven minutes…
Look for the points where z is zero and both x and y are positive. Two of those points should be (1, 0, 0) and (0, 1, 0) if it was aligned sensibly to start with. Then there should be two more points like (+x0, +y0, 0) and (+x1, +y1, 0).
The coordinates are going to be rounded. The exact values will be some irrational number (i.e., infinite number of decimal places).
Thanks for your immediate reply. My seven minutes of this week is already gone. Is there a ratio 2:1 between two of the edges in these tringles?
It seems like any of the great circles is intersected 4 times by the smaller edge and 8 times by the greater side.
It must have been a lot easier for mr Disdyakis himself without sw.
Looks close, but according to this page the answer is no:
http://dmccooey.com/polyhedra/DisdyakisTriacontahedron.html
In fact it doesn’t appear to even be a right triangle, which is mildly surprising.
That may be one of the least surprising things about it. If it were a right angle, that would make it lie flat where four of them abut.
The STL file has all of the coordinates as single-precision floats. Assuming that is not adequate, Solidworks has an “equations” feature, so you do not need to be concerned about loss of precision.
ETA @Dr.Strangelove 's page has a text file http://dmccooey.com/polyhedra/DisdyakisTriacontahedron.txt with all the coordinates, which you could use
Ahh, I missed that link. That’s very handy. Their coordinates are scaled so that the extremal points are at +/-C8, but aside from that, what I said above about the points should hold. In particular, you can create just a handful of points, mirror/rotate them, and then create the remaining via intersections. In particular, you can start with these points:
V2 = ( C8, 0.0, 0.0)
V4 = (0.0, C8, 0.0)
V14 = ( C1, C7, 0.0)
V22 = ( C6, C3, 0.0)
These are the “first quadrant” ones along the equator. The rest should be derivable from them.
I don’t know about Solidworks, but Fusion 360 has a neat feature to split across the silhouette. So you can create an axis perpendicular to each great circle, and split the silhouette from there. You’ll end up with quasi-triangular wedges corresponding to the faces.
That’s interesting! What hardware and software would you need to be able to print such stuff, including Crafter_Man’s Roman dodecahedron, in 3D?
If you download or create an STL file, then for a one-off order you don’t need to do any printing/machining yourself, eg Shapeways will print it for you in steel, gold, sandstone, aluminium, bronze, plastic— whatever you want.
To make your own computer models, obviously any CAD program can export the right kind of file, or else a 3d modelling program like Blender. As for polyhedra specifically, I recall that @Chronos was flogging 3D-printed dice, so he may have some specific advice on avoiding certain pitfalls.
Fusion 360 is an extremely powerful system for designing… almost anything. It’s free for non-commercial use.
For printing, you can certainly use a standard filament 3D printer. But for large objects, you can cut the faces with a laser cutter. Here’s a modest icosahedron I built recently:
Thank you both very much! There is something for me to do for some time here.
This page has a downloadable Roman dodecahedron model
If you have a stencil cutter like Dr.Strangelove, or are handy with shears:
lol @ “Mr Disdyakis”, by the way 
Anyway, this guy, mentioned in Chronos’s thread,
http://thedicelab.com/d120.html
appears to be selling d120’s