tilable polyhedron with golden ratio

Higher-dimensional geometry (previously "Polyshapes").

tilable polyhedron with golden ratio

Postby ae » Thu Aug 17, 2006 7:37 pm

I am wondering if there is any class/group of polyhedra that involve golden ratio and
they tile the space.

In particular I have a 3D lattice generated by integer linear combinations columns of the matrix:
[ phi 0 1 ]
[ 1 phi 0 ]
[ 0 1 phi ]

I'm wondering about its Voronoi cell, also the polyhedron that is formed by vertices around a lattice point whose Voronoi cells share a face with the Voronoi cell of that lattice point.
ae
Mononian
 
Posts: 5
Joined: Fri Aug 11, 2006 8:37 pm

Postby wendy » Fri Aug 18, 2006 5:43 am

I will wait for the english translation of this to come out.
The dream you dream alone is only a dream
the dream we dream together is reality.

\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger
User avatar
wendy
Pentonian
 
Posts: 2014
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

Postby Keiji » Fri Aug 18, 2006 8:04 am

I believe ae was talking in English. :P
User avatar
Keiji
Administrator
 
Posts: 1984
Joined: Mon Nov 10, 2003 6:33 pm
Location: Torquay, England

Postby pat » Fri Aug 18, 2006 4:26 pm

Certainly, the Voronoi cell of any lattice tiles the span of the lattice. But, the other polyhedron you mentioned, it would be shocking to me if that also tiles for an arbitrary lattice.

Wendy, I'm not sure what language you were having trouble with in the original post. Maybe it's just Voronoi Cell.

A lattice is a vector space with integer coefficients. A lattice is said to be generated by a set of vectors S if every element of the lattice is a linear combination (with integer coeffiecients) of members of S. Put another way, a lattice is said to be generated by a set of vectors S if every element of the lattice is a sum of members of the closure (under additive inverses) of S.

For the matrix given, the following points (amongst infinitely many others) are also in the lattice: (phi,1,0), (-phi,-1,0), (phi,1+phi,1), (phi,1-phi,-1), (35*phi-3,35+2*phi,2-3*phi), etc.

A lattice generated by a set of vectors S is invariant under translation by any member of S.

For interesting lattices, there is a minimum distance between points in the lattice. A lattice could be degenerate. For example, the lattice formed by the vectors [ 1 ] and [ sqrt(2) ] is degenerate. There is no minimum distance between lattice points. We'll ignore degenerate lattices from here on.

Let's consider the lattice as a subset of R<sup>n</sup> for large enough n. The Voronoi cell of the origin is the set of points in R<sup>n</sup> which are as close or closer to the origin than any other point in the lattice. Because our lattice is not degenerate, this Voronoi cell is more than just the origin. It is, in fact, easy to show that it's a polytope. (For each point in the lattice other than the origin, make a line segment between the point and the origin. The perpendicular bisector of that segment is a hyperplane. Hang onto the (bounded) halfspace containing the origin. Now, take the intersection of all such halfspaces.)

Because of the invariance under translation by members of S, the Voronoi cell of every point is identical to the one for the origin. Also, by the way the Voronoi cells are constructed, there can be no gaps between adjacent Voronoi cells. So... they must tile.

Now, if you draw the Voronoi cell for every lattice point and highlight the lattice points, the final question is what does the polyhedron look like if you take for vertices all of the lattice points whose Voronoi cells share faces with the Voronoi cell of the origin. I'd have to think a bit more, but it seems to me that for an arbitrary lattice, this set of points might not be convex.
pat
Tetronian
 
Posts: 563
Joined: Tue Dec 02, 2003 5:30 pm
Location: Minneapolis, MN

Postby ae » Fri Aug 18, 2006 7:01 pm

Thanks Pat for clearing up my question. I think Wendy was complaining about the way I asked it.
[ phi 0 1 ]
M = [ 1 phi 0 ]
[ 0 1 phi ]


So, for this particular lattice that I mentioned, do you happen to know what its Voronoi cell is? I am also interested in the Voronoi cell of the lattice generated by inverse transpose of M.

As you mentioned the Voronoi cells would tile the space, and I am wondering if there is a known group of tileable polyhedra that involve the golden ratio (in a nontrivial way, i.e. scaling) in the coordinates of their vertices.
ae
Mononian
 
Posts: 5
Joined: Fri Aug 11, 2006 8:37 pm

Postby pat » Fri Aug 18, 2006 7:42 pm

Yep, I hadn't gotten that far... to try to determine the Voronoi cells of that lattice. I'll probably poke at it again later.
pat
Tetronian
 
Posts: 563
Joined: Tue Dec 02, 2003 5:30 pm
Location: Minneapolis, MN

Postby wendy » Sat Aug 19, 2006 7:24 am

I do know what the voronii cell is, since this is an alternate presentation of a lattice. It corresponds roughly to the mirror-margin face of the tiling.

Still, the voronii cell of a tiling over a span of three integer vectors, such as the cyclic permutation of (0,1,f), is typically a parallelopied of this shape. One could rearrange such a shape into something different, but it is usually better to eat this by some other means.

Of course, there are indeed rather interesting things that one can do with phi and 3d (and 4d) tilings. These are the various sections of the symmetries 4w (eg o5o5/2oAo) and 5w (o5o3o3o5/2o). This is a much more fruitful endeavour than trying to use simple cubic lattices.

One notes that pentagonal angles (such as icosahedra and dodecahedra) do occur in nature. There is some rather interesting work at, zB Kabai Sandor's site at http://www.kabai.hu/ . Sandor is a graphic artist, who among other things, looks at tilings of various rhombohedra derived from disecting the tricontahedron o3m5o.

W
The dream you dream alone is only a dream
the dream we dream together is reality.

\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger
User avatar
wendy
Pentonian
 
Posts: 2014
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

Postby ae » Sat Aug 19, 2006 10:26 pm

I doubt the Voronoi cell be a parallelopiped shape. only the orthogonal Cartesian lattice has a Voronoi cell which is a cube.
ae
Mononian
 
Posts: 5
Joined: Fri Aug 11, 2006 8:37 pm

Postby wendy » Sun Aug 20, 2006 7:41 am

I suppose. Anyway, i played around with the voronoi cell. The general cell is modularly equivalent to a hombohedron, but some bits can be moved about.

The shape is probably some kind of distorted rhombic dodecahedron, prehaps made out of golden rhombs.
The dream you dream alone is only a dream
the dream we dream together is reality.

\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger
User avatar
wendy
Pentonian
 
Posts: 2014
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

Postby pat » Sun Aug 27, 2006 5:01 pm

Here's what my raytracer says...

Image

Which, looks lots like "some kind of distored rhombic dodecahedron" to me, too.

This is the intersection of all of the halfspaces (which include the origin) with coordinates (and normals) < a * phi + c, b * phi + a, c * phi + b > where a, b, and c are integers with -2 <= a,b,c <= 2 (and not all simultaneously zero). I'm pretty sure this is the Voronoi polytope... since I also tried it with a,b,c running from -3 to +3 and got the same shape.

When I say a half-space has coordinate and normal < x, y, z >, I mean that the point < x, y, z > is on the boundary of the halfspace and that the normal to the halfspace is parallel to the vector < x, y, z >.
pat
Tetronian
 
Posts: 563
Joined: Tue Dec 02, 2003 5:30 pm
Location: Minneapolis, MN


Return to Other Geometry

Who is online

Users browsing this forum: No registered users and 8 guests

cron