How many cuboctahedra in the 24 cell?

[Sphericality]

Can anyone clarify for me how many cuboctahedra there are in the 24 cell?
You can easilly see one of them inside this projection.


I believe I can trace 8 of them, but I am not sure if there are more.
I would appreciate it if you can also explain how to calculate this effectively.

This quote from wikipedia seems to imply to me that there might be 12 (or 24?) cuboctahedra -
"The cell-first parallel projection of the 24-cell into 3-dimensional space has a cuboctahedral envelope."
https://en.wikipedia.org/wiki/24-cell

[Klitzing]

This quest is quite easy.

Just consider, as can be seen from your pic too, that the icositetrachoron can be oriented octahedron first. Then it has 3 vertex layers, the nearest one describes the chosen octahedron, the equatorial is the searched for cuboctahedron, and the final one is the opposite octahedron. Now remember that "icositetra" is the greekish word for 24, that is, it has 24 octahedra all over. (In fact it is known also by the name 24-cell.) As any opposite pair of octahedra describes a cuboctahedron inbetween, thus you'll have 24/2 = 12 equatorial cuboctahedra for the various possible orientations.

--- rk

[Marek14]

Alternate calculation would work like this:

The verf of icositetrachoron is a cube. A cube can be sliced diagonally through two opposite edges to have a 1 by sqrt(2) rectangle on the cut, which is the verf of a cuboctahedron. There are six possible ways to do this (because cube has six pairs of opposite edges).

So each vertex of icositetrachoron belongs to six equatorial cuboctahedra (they must be equatorial because the cut passes through the center of verf). There is 24 vertices altogether and each cuboctahedron has 12 vertices, so their total number is 24 * 6 / 12 = 12.

[Sphericality]

Thank you both for the above answers, which help me to understand the icositetrachoron's elegant internal structure better!

Now that you have clarified things for me ...
In the version I use as my avatar image it is easy to see 4 cuboctahedra, one between each opposite pair of edges.
I believe the other 8 each include 3 consecutive 'outer' edges.

[Mercurial, the Spectre]

Easy. There are 12 pairs of opposite octahedra, each containing a cuboctahedron at the middle. Similar to how you picture the four pairs of opposite triangles in the cuboctahedron, which makes for 4 equatorial hexagons in total.