# Polyhedrons V

My cover picture for Astronomical Calendar 2015

showed the stars as stellated polyhedrons, and there was much more I would have liked to say about these shapes.  I put some into posts here: Polyhdrons II, III, IV.

The simplest of all straight-edged solids is the tetrahedron.  A solid can’t have fewer faces, yet its mere four have endless aspects and connections.  One of the minor remarks I cut out was that “If the tetrahedron is flattened, its four faces become an equilateral triangle stellated.  Or, put the other way, you make a paper tetrahedron by drawing an equilateral triangle and stellating it.”

I mentioned that hidden in the tetrahedron is circularity: the cylinder that, pinched in two places, becomes a tetrahedral milk carton.  But didn’t have space to illustrate it –

– also that hidden in the tetrahedron is a squareness.  It’s all triangles, but if you cut it into two equal parts, the two new surfaces are squares.  I would have liked to show the paper model I once made of this, but must have thrown it away.  But this is what it looked like:

If you were sawing a wooden tetrahedron, you would start by making marks at the halfway points along four of the six edges.  Each of the two new faces made by the cut is a square.  And each of the two new solids has five faces: the square, two isosceles triangles, and two isosceles trapezoids.  Quite un-simple chunks!  The fun of making the model is that you toss these two chunks, with their total of ten faces, on the table and ask your friends to fit them together into something with only four faces.  Yes, you now easily see how to do this, but that’s only because you’ve just been reading about tetrahedrons.

Alastair McBeath told me he had been wondering about the possibility of paper polyhedral star globes, and asked whether I had seen any.  Yes: James Weightman in 2013 sent me this:

It is the all-sky map from my Astronomical Calendar, cut into a large number of polygons so that, folded and glued together by their tabs, they become the faces of a polyhedron, which is an approximation to the celestial sphere.

It would be a celestial sphere backwards, that is, with the constellations correctly mapped as seen from the outside but mirror images of themselves as seen from the center; as in ancient celestial spheres such as the one carried by the sculpture called the Farnese Atlas; and as distinct from modern transparent celestial globes into which you peer, to see the constellations the right way around on the far side.

The polyhedron James chose was the truncated icosahedron.  An icosahedron has 20 (Greek eikosi) triangular faces, 5 of which meet at each of 12 vertices.

Therefore, truncating – slicing off – a vertex produces a pentagonal face.  Doing the same to all the vertices produces 12 such pentagons, and by chopping off the corners of the 20 original triangles reduces them to hexagons.  So the result is 32 faces in all.  The more faces, the closer the approximation to a sphere, which could be considered a polyhedron with an infinite number of faces.

James used a “Photoshop add-on” called Flexify2.  It probably can make all the polyhedrons I laboriously figured out how to calculate, and more, but I’m glad I got as far as I did in understanding them.

A further challenge, and I don’t know whether it would be solved by trigonometry or sheer dextrous fingers, would be to make a celestial sphere by taping together 89 pieces, of very varied size and irregular shapes, for the 88 constellations (including the two separated parts of Serpens).  This would be like building a house, or a vault, out of – well, I can’t think of any object in nature that’s shaped like the constellation Draco.

## 13 thoughts on “Polyhedrons V”

1. chesscanoe says:

I’d love to get the *jpg’s as well as others I’m sure. Perhaps you might care to post a URL?

2. Martin Young says:

I believe Pythagoras himself didn’t ever “prove” his theorem. To quote from Arthur Koestler’s “The Sleepwalkers”…….”there is no obvious relationship between the lengths of the sides of a right angled triangle; but if we build a square over each side, the areas of the two smaller squares will exactly equal the area of the larger. ….If such wonderfully ordered laws, hitherto hidden from the human eye, could be discovered by the contemplation of number-shapes, ……..soon all the secrets of the universe would be revealed through them? “

3. chesscanoe says:

If your further challenge is to make a convex polygon of 89 faces, I suspect it would have to agree with Euler’s Characteristic of 2 from #Vertices – #Edges + #Faces =2. I leave it to the reader to solve. :-)

1. Yes, I went into that in the Ast.Cal. 2015 cover picture story. Euler’s formula is said to be true of all polyhedrons, including stellated ones – just about all shapes, therefore – except some ones with concave parts. For the 89-faced polyhedron, you could actually find the number of vertices (joining-points of constellation boundaries) and edges (segments of boundary between these points, ignoring twists) by counting them on the map of the sky. I don’t propose to do such counting. –Actually, it could probably be done from the catalog of constellation-boundary points, drawn up by Gould and then Delporte, of which I got a form from somewhere and manipulated into a file of my own, from which I plot the constellation boundaries. The file contains 782 lines, some being repeated at junctions of boundaries. So I’d have to write a program to count segments and junctions, which also I don’t propose to do just now.

4. Michael Dempster says:

Ha! Is Weightman’s polyhedral celestial sphere available anywhere? As PDF download? And thanks, Guy, for the ongoing flow of wonderful stuff!

1. I have suggested to James Weightman that he reply on this.

2. James Weightman tells me that celestial sphere templates like his are not in production, but “It is quite easy for people… to obtain and use the Flaming Pear software to produce their own examples.” Presumably you could produce the polyhedron templates that way, but they would be without the star maps; James used a flatbed scanner to scan, at 300 dots per inch, the star charts from my Astronomical Calendar, but how he applied those images to the templates I don’t know.

He has sent me images of the celestial sphere template in three forms – truncated icosahedron as I showed before, octahedron, and a 72-faced form much like a traditional sphere; these are jpg images and I believe I could send them to anyone who asks. I suppose you could then print them, on (large) paper, cut, fold, and tape them.

1. Michael Dempster says:

Well, if you are indeed willing and have time to forward the jpg images to me, I’d love it! Thanks!

1. Michael Dempster says:

Thanks, Guy.

5. John Goss says:

Very unusual and very intriguing!

1. Jack Gambino says:

Soccer, anyone? or futbol as it’s known in most other countries outside of the USofA. the soccer ball has a few pentagonal patches amongst the most hexagonals.
And I’ve been told there are many other ‘proofs’ for the Pythagorean theorum, one involving a circle around the triangle itself, but I don’t know that one. I just drop a line from the right angle to hypotenuse and so the similar triangle proof. It got me a 100% on the NYState Regents.

6. Jack Gambino says:

Geometry was one of most favorite subjects in high school,,,, De Witt Clinton…. I score 100% on the Regents.. I had to prove the Pythagorean theorom… I can still do it today,,, but don’t remember the rules for which I can do it.