A Fractured Octahedron

Sometimes, when using Stella 4d (available here) to make various polyhedra, I lose track of how I got from wherever I started to the final step. That happened with this fractured version of an octahedron.

Octahedra and Truncated Cubes Can Fill Space Without Leaving Any Gaps

Augmented Trunc Cube

I created this using Stella 4d, which you can try for free right here. It’s much like a tessellation, but in three dimensions instead of two.

Octahedra and Cuboctahedra Can Fill Space Without Leaving Any Gaps

Augmented Cubocta

I created this image using Stella 4d, which you can try for free right here. It’s much like a tessellation, but in three dimensions instead of two.

Some Ten-Part Polyhedral Compounds

While examining different facetings of the dodecahedron, I stumbled across one which is also a compound of ten elongated octahedra.

Faceted Dodeca and compound of ten elongated octahedra.gif

Here’s what this compound looks like with the edges and vertices hidden:

Faceted Dodeca and compound of ten elongated octahedra without edges and vertices.gif

Next, I’ll put the edges and vertices back, but hide nine of the ten components of the compound. This makes it easier to see the single elongated octahedron which is still shown.

Faceted Dodeca one part of ten with edges and vertices.gif

Here’s what this elongated octahedron looks like with all those vertices and edges hidden from view.

Faceted Dodeca one part of ten.gif

I made all these polyhedral transformations using Stella 4d, a program you can try for yourself at this website. Stella includes a “measurement mode,” and, using that, I was able to determine that the short edge to long edge ratio in these elongated octahedra is 1:sqrt(2).

The next thing I wanted to try was to make the octahedra regular. Stella has a function for that, too, and here’s the result: a compound of ten regular octahedra.

compound of ten regular octahedra.gif

My last step in this polyhedral exploration was to form the dual of this solid. Since the octahedron’s dual is the cube, this dual is a compound of ten cubes.

compound of ten cubes.gif

Honeycomb Made of Cuboctahedra and Octahedra

Augmented Cubocta.gif

This is the three-dimensional version of what is called a tessellation in two dimensions. It fills space, and can be continued in all directions.

Software used: Stella 4d, available here.

Augmenting the Octahedron with Octahedra, Repeatedly.

This is an octahedron.

Octa

If you augment each face of an octahedron with more octahedra, you end up with this.

Augmented Octa

One can then augment each triangular face of this with yet more octahedra.

Aug Augmented Octa

Here’s the next iteration:

Aug Aug Augmented Octa

This could, of course, go on forever, but one more step in the series is all you will see here. I don’t want to get caught in an infinite loop.

next aug Octa

Performing various manipulations of polyhedra is easy with Stella 4d: Polyhedron Navigator, which I used to make all five of these rotating images. If you’d like to try this program for yourself, just check out http://www.software3d.com/Stella.php.

The Compounds of Five Octahedra and Five Cubes, and Related Polyhedra

This is the compound of five octahedra, each a different color.

Cubes 5

Since the cube is dual to the octahedron, the compound of five cubes, below, is dual to the compound above.

cubes-5a

Here are five cubes and five octahedra, compounded together, and shown with the same five colors used above.

Cubes 5 and octahedra 5

This is the same compound, except with all squares/cubes having one color, and triangles/octahedra having another, made by changing the color-scheme used by Stella 4d (the program I use to make these images; it’s available here).

Cubes 5 and octahedra 5 colored by face-type

After seeing the two-color version of this ten-part compound, I decided to start stellating it, looking for stellations with an interesting appearance. Here is the 23rd stellation of the ten-part compound, colored by face-type.

Compound of 5 Cubes and dual 23rd stellation

Next, the 27th stellation, which is chiral, unlike the stellation showed above.

Compound of 5 Cubes and dual 27th stellation

The 33rd stellation also has an interesting appearance (using, I admit, completely subjective criteria for “interesting”), while still having easily-noticable differences to the stellations shown above.

Compound of 5 Cubes and dual 33rd stellation

At the 35th stellation, another interesting chiral polyhedron is found. Unexpectedly, its direction of “twist” appears opposite that seen in the 27th stellation. (It could well be that this “twist-reversal” is a common phenomenon in stellation-series — simply one I have never noticed before.)

Compound of 5 Cubes and dual 35th stellation

Next, the ten-part compound’s 39th stellation.

Compound of 5 Cubes and dual 39th stellation

After the 39th stellation, I entered a sort of “desert,” with many stellations in a row which did not strike me as interesting, often with only tiny differences between one and the next. The 194th stellation, though, I liked.

Compound of 5 Cubes and dual 194th stellation

Although I liked the 194th stellation, I didn’t want to risk trudging through another “desert” like the one which preceded it, so I jumped ahead to the final valid stellation, after which the series “wraps around” to its beginning.

Compound of 5 Cubes and dual final valid stellation

Next, I made another rotating image of this final valid stellation, this time with the color-scheme set to “rainbow color mode.”

Compound of 5 Cubes and dual final valid stellation rbc

I couldn’t resist taking this one stellation further, to see the beginning of the stellation-series, since I knew I might have entered it somewhere in the middle, rather than at the beginning.

Compound of 5 Cubes and dual final valid stellation next one rbc

What I found, I immediately recognized as the rhombic triacontahedron. In some ways, this was surprising, and in other ways, it was not. The compound of five cubes is, itself, a stellation of the rhombic triacontahedron — but what I started stellating also included the compound of five octahedra, which, so far as I know, is not part of the rhombic triacontahedron’s (very) long stellation-series. Also, I know what the rhombic triacontahedron’s final stellation looks like, and it isn’t the final stellation shown above, but is, instead, this:

final-stellation-of-the-rhombic-triaconta

To try to better-understand just what was going on here, I went back, and deliberately left out the five-cube part of the ten-part compound (which is a stellation of the rhombic triacontahedron), which left me just with the compound of five octahedra — and then I had Stella produce this compound’s final stellation.

Octahedra 5 final stellation

This was another polyhedron I recognized: the final stellation of the icosahedron. To verify that my memory was correct, I stellated it one more time. Sure enough, this is what I got:

Octahedra 5 final stellation one more

This reminded me that the compound of five octahedra is the second stellation of the icosahedron, helping to explain some of this. I also noticed that the five-octahedron compound can be seen as a faceting of the icosidodecahedron. (The icosidodecahedron is dual to the rhombic triacontahedron, and faceting is the reciprocal function of stellation.) However, I have no idea why the final stellation of the ten-part compound above appears as it does.

It is my opinion that a productive polyhedral investigation usually does more than answer questions; it also raises new ones. At least in my mind, that’s exactly what has happened. Therefore, I think this was a perfectly good way to begin the new year.