[math-fun] Tetrahedral hypercube projection
Let u, v, w, and x be the vectors from the center of a tetrahedron to the four vertices. Does the zonohedron formed by the fifteen vectors u, v, w, x, u+v, u+w, u+x, v+w, v+x, w+x, u+v+w, u+v+x, u+w+x, v+w+x, and u+v+w+x=0 have a name? Better, are there images of it on the web? Better still, are there gifs or videos that show it rotating in a way that brings its three-dimensionality to the fore? I ask because this seems like a nice way of looking at a hypercube (leaving aside the defect that two of the vertices of the hypercube coincide). Jim Propp
I believe you are talking about the rhombic dodecahedron. --Jeannine On Sat, May 12, 2018, 7:09 PM James Propp <jamespropp@gmail.com> wrote:
Let u, v, w, and x be the vectors from the center of a tetrahedron to the four vertices. Does the zonohedron formed by the fifteen vectors u, v, w, x, u+v, u+w, u+x, v+w, v+x, w+x, u+v+w, u+v+x, u+w+x, v+w+x, and u+v+w+x=0 have a name? Better, are there images of it on the web? Better still, are there gifs or videos that show it rotating in a way that brings its three-dimensionality to the fore?
I ask because this seems like a nice way of looking at a hypercube (leaving aside the defect that two of the vertices of the hypercube coincide).
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
But the rhombic dodecahedron has 14 vertices, and I thought mine had 15. Did I miscount? ... Oh wait, 0 is interior to the tetrahedron with vertices u, v, w, and x, so it can't be a vertex! Can someone explain to me why Jeannine is right (now that I believe her)? Jim On Sat, May 12, 2018 at 7:22 PM, Jeannine Mosely <j9mosely@gmail.com> wrote:
I believe you are talking about the rhombic dodecahedron.
--Jeannine
On Sat, May 12, 2018, 7:09 PM James Propp <jamespropp@gmail.com> wrote:
Let u, v, w, and x be the vectors from the center of a tetrahedron to the four vertices. Does the zonohedron formed by the fifteen vectors u, v, w, x, u+v, u+w, u+x, v+w, v+x, w+x, u+v+w, u+v+x, u+w+x, v+w+x, and u+v+w+x=0 have a name? Better, are there images of it on the web? Better still, are there gifs or videos that show it rotating in a way that brings its three-dimensionality to the fore?
I ask because this seems like a nice way of looking at a hypercube (leaving aside the defect that two of the vertices of the hypercube coincide).
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Coxeter's "Regular Polytopes" has a lot about this and similar zonohedral projections of orthotopes. WFL On 5/13/18, James Propp <jamespropp@gmail.com> wrote:
But the rhombic dodecahedron has 14 vertices, and I thought mine had 15. Did I miscount?
... Oh wait, 0 is interior to the tetrahedron with vertices u, v, w, and x, so it can't be a vertex!
Can someone explain to me why Jeannine is right (now that I believe her)?
Jim
On Sat, May 12, 2018 at 7:22 PM, Jeannine Mosely <j9mosely@gmail.com> wrote:
I believe you are talking about the rhombic dodecahedron.
--Jeannine
On Sat, May 12, 2018, 7:09 PM James Propp <jamespropp@gmail.com> wrote:
Let u, v, w, and x be the vectors from the center of a tetrahedron to the four vertices. Does the zonohedron formed by the fifteen vectors u, v, w, x, u+v, u+w, u+x, v+w, v+x, w+x, u+v+w, u+v+x, u+w+x, v+w+x, and u+v+w+x=0 have a name? Better, are there images of it on the web? Better still, are there gifs or videos that show it rotating in a way that brings its three-dimensionality to the fore?
I ask because this seems like a nice way of looking at a hypercube (leaving aside the defect that two of the vertices of the hypercube coincide).
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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I can answer my own question: Let the vectors u, v, w, and x be (1,1,1), (-1,-1,1), (-1,1,-1), and (1,-1,-1). Then u+v = (0,0,2), u+w = (0,2,0), u+x = (2,0,0), v+w = (-2,0,0), v+x = (0,-2,0), w+x = (0,0,-2), u+v+w = (-1,1,1), u+v+x = (1,-1,1), u+w+x = (1,1,-1), and v+w+x = (-1,-1,-1), which exactly correspond to the coordinates given in https://en.wikipedia.org/wiki/Rhombic_dodecahedron#Cartesian_coordinates . Jim On Sat, May 12, 2018 at 10:40 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Coxeter's "Regular Polytopes" has a lot about this and similar zonohedral projections of orthotopes. WFL
On 5/13/18, James Propp <jamespropp@gmail.com> wrote:
But the rhombic dodecahedron has 14 vertices, and I thought mine had 15. Did I miscount?
... Oh wait, 0 is interior to the tetrahedron with vertices u, v, w, and x, so it can't be a vertex!
Can someone explain to me why Jeannine is right (now that I believe her)?
Jim
On Sat, May 12, 2018 at 7:22 PM, Jeannine Mosely <j9mosely@gmail.com> wrote:
I believe you are talking about the rhombic dodecahedron.
--Jeannine
On Sat, May 12, 2018, 7:09 PM James Propp <jamespropp@gmail.com> wrote:
Let u, v, w, and x be the vectors from the center of a tetrahedron to the four vertices. Does the zonohedron formed by the fifteen vectors u, v, w, x, u+v, u+w, u+x, v+w, v+x, w+x, u+v+w, u+v+x, u+w+x, v+w+x, and u+v+w+x=0 have a name? Better, are there images of it on the web? Better still, are there gifs or videos that show it rotating in a way that brings its three-dimensionality to the fore?
I ask because this seems like a nice way of looking at a hypercube (leaving aside the defect that two of the vertices of the hypercube coincide).
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Yes, this is a wonderful (if degenerate) way of looking at the hypercube. Degenerate because the nearest and furthest vertices appear to coincide at (0,0,0), when in fact they are far apart in 4-space. This is exactly analogous to the isometric projection of the 3-cube into 2-space (or any isometric projection of an N-cube into (N-1)-space). The standard optical illusions that use 3d to 2d isometric projection work just fine up a dimension using the rhombic icosahedral forms, namely: the reversible staircase (concave convex), the impossible triangle (which becomes the impossible skew quadrilateral), and the endlessly rising staircase (which works poorly in 3d to 2d — the loop is a kite when really it wants to be a digon — but works beautifully in 4d to 3d — the loop is a triangle). As you probably know, the fact that you were able to use such nice integer coordinates for the vertices of the tetrahedron in 3-space is a consequence of the dimension of our space being one less than a power of 2. This does not happen in general…for instance the isometric projection of the cube into a plane results in a regular hexagon, which does NOT align nicely with a square grid. The 2-d analog of your idea is to map the four basis vectors u, v, w, and x onto four segments radiating from a point at 45° angles, which leads to one of the most standard (and pretty) pictures of the hypercube. When I first started exploring Hypercubes as a kid, I realized you could of course generalize this for any dimension, and draw yet higher dimensional cubes. The six cube drawing is particularly pretty; I drew up to 9 by hand. Pushing your idea upward, there is no nice way to project 5 basis vectors into 3-space, but there is a completely symmetrical way to project 6 basis vectors into 3 space, namely vectors radiating out from the origin to half the vertices of an icosahedron (the other six vectors point in the opposite directions). And this leads to the rhombic triacontahedron, which I'm sure Jeanine and Fred are familiar with, with 30 golden mean rhombic faces. This projection is the one that leads to the structure of quasicrystals. I haven't tried pushing this higher…I wonder what zonohedron one would get starting from 15 vectors that point out to half the vertices of a dodecahedron? That seems like the next thing to try. Does anyone know? — Scott On Sat, May 12, 2018 at 8:07 PM, James Propp <jamespropp@gmail.com> wrote:
I can answer my own question:
Let the vectors u, v, w, and x be (1,1,1), (-1,-1,1), (-1,1,-1), and (1,-1,-1). Then u+v = (0,0,2), u+w = (0,2,0), u+x = (2,0,0), v+w = (-2,0,0), v+x = (0,-2,0), w+x = (0,0,-2), u+v+w = (-1,1,1), u+v+x = (1,-1,1), u+w+x = (1,1,-1), and v+w+x = (-1,-1,-1), which exactly correspond to the coordinates given in https://en.wikipedia.org/wiki/Rhombic_dodecahedron#Cartesian_coordinates .
Jim
On Sat, May 12, 2018 at 10:40 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Coxeter's "Regular Polytopes" has a lot about this and similar zonohedral projections of orthotopes. WFL
On 5/13/18, James Propp <jamespropp@gmail.com> wrote:
But the rhombic dodecahedron has 14 vertices, and I thought mine had
Did I miscount?
... Oh wait, 0 is interior to the tetrahedron with vertices u, v, w, and x, so it can't be a vertex!
Can someone explain to me why Jeannine is right (now that I believe her)?
Jim
On Sat, May 12, 2018 at 7:22 PM, Jeannine Mosely <j9mosely@gmail.com> wrote:
I believe you are talking about the rhombic dodecahedron.
--Jeannine
On Sat, May 12, 2018, 7:09 PM James Propp <jamespropp@gmail.com> wrote:
Let u, v, w, and x be the vectors from the center of a tetrahedron to the four vertices. Does the zonohedron formed by the fifteen vectors u, v, w, x, u+v, u+w, u+x, v+w, v+x, w+x, u+v+w, u+v+x, u+w+x, v+w+x, and u+v+w+x=0 have a name? Better, are there images of it on the web? Better still, are there gifs or videos that show it rotating in a way that brings its three-dimensionality to the fore?
I ask because this seems like a nice way of looking at a hypercube (leaving aside the defect that two of the vertices of the hypercube coincide).
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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On May 13, 2018, at 12:43 PM, Scott Kim <scott@scottkim.com> wrote:
I haven't tried pushing this higher…I wonder what zonohedron one would get starting from 15 vectors that point out to half the vertices of a dodecahedron? That seems like the next thing to try. Does anyone know?
The dodecahedron has 20 vertices. Are you thinking about the 15 two-fold axes of the icosahedron/dodecahedron, i.e. vectors that point to the edge mid-points? If so, then the answer is: https://en.wikipedia.org/wiki/Truncated_icosidodecahedron -Veit
The zonohedron based on the ten directions of the dodecahedron's long diagonals is the rhombic enneacontahedron. Note that Zometool is the ideal material for making models of these structures because the ball design is constrained to maintain parallelism in space. Use just yellow for the rhombic dodecahedron or the rhombic enneacontahedron, just blue for the truncated icosidodecahedron, and just red for the rhombic triacontahedron. George http://georgehart.com On 5/14/2018 7:44 AM, Veit Elser wrote:
On May 13, 2018, at 12:43 PM, Scott Kim <scott@scottkim.com> wrote:
I haven't tried pushing this higher…I wonder what zonohedron one would get starting from 15 vectors that point out to half the vertices of a dodecahedron? That seems like the next thing to try. Does anyone know?
The dodecahedron has 20 vertices. Are you thinking about the 15 two-fold axes of the icosahedron/dodecahedron, i.e. vectors that point to the edge mid-points? If so, then the answer is:
https://en.wikipedia.org/wiki/Truncated_icosidodecahedron
-Veit
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Is there a Zome-system kit for building large structures strong enough to climb on? Some of the large structures at G4G13 seemed Zome-ish (but not strong enough to stand up to kids climbing on them). Jim Propp On Monday, May 14, 2018, George Hart <george@georgehart.com> wrote:
The zonohedron based on the ten directions of the dodecahedron's long diagonals is the rhombic enneacontahedron.
Note that Zometool is the ideal material for making models of these structures because the ball design is constrained to maintain parallelism in space. Use just yellow for the rhombic dodecahedron or the rhombic enneacontahedron, just blue for the truncated icosidodecahedron, and just red for the rhombic triacontahedron.
George http://georgehart.com
On 5/14/2018 7:44 AM, Veit Elser wrote:
On May 13, 2018, at 12:43 PM, Scott Kim <scott@scottkim.com> wrote:
I haven't tried pushing this higher…I wonder what zonohedron one would get starting from 15 vectors that point out to half the vertices of a dodecahedron? That seems like the next thing to try. Does anyone know?
The dodecahedron has 20 vertices. Are you thinking about the 15 two-fold axes of the icosahedron/dodecahedron, i.e. vectors that point to the edge mid-points? If so, then the answer is:
https://en.wikipedia.org/wiki/Truncated_icosidodecahedron
-Veit
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The "mega-zome" materials you saw at G4G-13 (white metal struts with machined aluminum connectors) is strong enough to climb on, but is not commercially available. It was produced with only the "blue" directions of Zometool (i.e., the fifteen 2-fold axis directions of the icosahedral symmetry group). George http://georgehart.com On 5/14/2018 9:30 AM, James Propp wrote:
Is there a Zome-system kit for building large structures strong enough to climb on?
Some of the large structures at G4G13 seemed Zome-ish (but not strong enough to stand up to kids climbing on them).
Jim Propp
On Monday, May 14, 2018, George Hart <george@georgehart.com> wrote:
The zonohedron based on the ten directions of the dodecahedron's long diagonals is the rhombic enneacontahedron.
Note that Zometool is the ideal material for making models of these structures because the ball design is constrained to maintain parallelism in space. Use just yellow for the rhombic dodecahedron or the rhombic enneacontahedron, just blue for the truncated icosidodecahedron, and just red for the rhombic triacontahedron.
George http://georgehart.com
On 5/14/2018 7:44 AM, Veit Elser wrote:
On May 13, 2018, at 12:43 PM, Scott Kim <scott@scottkim.com> wrote:
I haven't tried pushing this higher…I wonder what zonohedron one would get starting from 15 vectors that point out to half the vertices of a dodecahedron? That seems like the next thing to try. Does anyone know?
The dodecahedron has 20 vertices. Are you thinking about the 15 two-fold axes of the icosahedron/dodecahedron, i.e. vectors that point to the edge mid-points? If so, then the answer is:
https://en.wikipedia.org/wiki/Truncated_icosidodecahedron
-Veit
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Thanks George. Didn't know about that shape. Yeah, got my 15s and 10s mixed up, but nice to know there are polyhedra for both 10 and 15. On Mon, May 14, 2018 at 5:51 AM, George Hart <george@georgehart.com> wrote:
The zonohedron based on the ten directions of the dodecahedron's long diagonals is the rhombic enneacontahedron.
Note that Zometool is the ideal material for making models of these structures because the ball design is constrained to maintain parallelism in space. Use just yellow for the rhombic dodecahedron or the rhombic enneacontahedron, just blue for the truncated icosidodecahedron, and just red for the rhombic triacontahedron.
George http://georgehart.com
On 5/14/2018 7:44 AM, Veit Elser wrote:
On May 13, 2018, at 12:43 PM, Scott Kim <scott@scottkim.com> wrote:
I haven't tried pushing this higher…I wonder what zonohedron one would get starting from 15 vectors that point out to half the vertices of a dodecahedron? That seems like the next thing to try. Does anyone know?
The dodecahedron has 20 vertices. Are you thinking about the 15 two-fold axes of the icosahedron/dodecahedron, i.e. vectors that point to the edge mid-points? If so, then the answer is:
https://en.wikipedia.org/wiki/Truncated_icosidodecahedron
-Veit
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participants (6)
-
Fred Lunnon -
George Hart -
James Propp -
Jeannine Mosely -
Scott Kim -
Veit Elser