https://en.wikipedia.org/wiki/Esprit_Jouffret

Years of studying about four dimensions caused me to see this as 8 pyramidial like polyhedra thing

Is this shape part of the hypercube derived polychora?

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https://en.wikipedia.org/wiki/Esprit_Jouffret

Years of studying about four dimensions caused me to see this as 8 pyramidial like polyhedra thing

Is this shape part of the hypercube derived polychora?

Years of studying about four dimensions caused me to see this as 8 pyramidial like polyhedra thing

Is this shape part of the hypercube derived polychora?

- Secret
- Trionian
**Posts:**162**Joined:**Tue Jul 06, 2010 12:03 pm

I'd say it is a 24-cell: the numbers 2,4,6,8 seem to be vertices where several octahedral meet, those octahedra then are connected to some other vertices, for a total of 24 (you can see the biggest number is 24).

When this actually is a 24-cell, it can indeed be constructed from a hypercube. In dynkin-notation, the hypercube is given an x4o3o3o, and the 24-cell is given as o4o3x3o. Thus when one takes the dual of a hypercube (the 16-cell o4o3o3x), and one moves the edges radially outward, one gets o4o3x3x. This process is quite similar to standard truncation. The tetrahedra of the 16-cell are changed into truncated tetrahedra, and the vertices change into octahedra. When one then moves these octahedra radially inward, the truncated tetrahedra are changed into octahedra and one obtains the 24-cell o4o3x3o.

More directly, it can also be constructed from a hypercube by moving the squares radially outwards, one then obtains a x4o3x3o, with rhombicuboctahedra x4o3x, octahedra o3x3o and triangular prisms x x3o. One then again moves the octahedra radially inwards, which changes the rhombicuboctahedra into octahedra, and the triangular prisms are collapsed into triangles. One then again obtains a 24-cell.

When this actually is a 24-cell, it can indeed be constructed from a hypercube. In dynkin-notation, the hypercube is given an x4o3o3o, and the 24-cell is given as o4o3x3o. Thus when one takes the dual of a hypercube (the 16-cell o4o3o3x), and one moves the edges radially outward, one gets o4o3x3x. This process is quite similar to standard truncation. The tetrahedra of the 16-cell are changed into truncated tetrahedra, and the vertices change into octahedra. When one then moves these octahedra radially inward, the truncated tetrahedra are changed into octahedra and one obtains the 24-cell o4o3x3o.

More directly, it can also be constructed from a hypercube by moving the squares radially outwards, one then obtains a x4o3x3o, with rhombicuboctahedra x4o3x, octahedra o3x3o and triangular prisms x x3o. One then again moves the octahedra radially inwards, which changes the rhombicuboctahedra into octahedra, and the triangular prisms are collapsed into triangles. One then again obtains a 24-cell.

- student91
- Tetronian
**Posts:**310**Joined:**Tue Dec 10, 2013 3:41 pm

Still more directly:

take the tesseract x4o3o3o and therefrom then use the centers of the squares.

What you then get as hull of those 24 points, is the 24-cell.

Or alternatively, you could use the tesseract and its dual, the 16-cell.

If the 16-cell would be scaled such that its edges are sqrt(2) times those of the cube, then the hull of that compound also would result in the 24-cell.

--- rk

take the tesseract x4o3o3o and therefrom then use the centers of the squares.

What you then get as hull of those 24 points, is the 24-cell.

Or alternatively, you could use the tesseract and its dual, the 16-cell.

If the 16-cell would be scaled such that its edges are sqrt(2) times those of the cube, then the hull of that compound also would result in the 24-cell.

--- rk

- Klitzing
- Pentonian
**Posts:**1347**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

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