I am trying to visualize what looks like a particular polyhedron(?), but I am having a bout of stupidity and cannot figure something simple about it.
I am describing the shape, section-wise:
Consider the unit circle on the xy-plane. Put vertexes at exp(2*k*Pi/8*i), k \in {0,1,...,7}
Rotate the unit circle around the x-axis by Pi/4.
Consider the new unit circle on the xy-plane. Put again vertexes on the new unit circle at exp(2*k*Pi/8*i), k \in {0,1,...,7} (vertex duplicity allowed).
Rotate the new unit circle around the x-axis by Pi/4 (this will make the first unit circle perpendicular to the xy-plane).
Repeat 7 times, until the first unit circle rotates back to its initial position.
What is the name of the resultant polyhedron? It has a total of 26 vertexes.
If anyone has any online refs of the resultant polyhedron or can cook up some Maple code to visualize this polyhedron (preferably with rotational animation), I'd appreciate it.
On 6 Oct, 11:53, "I.N. Galidakis" <morph...@olympus.mons> wrote:
> What is the name of the resultant polyhedron? It has a total of 26 vertexes.
I doubt it would have a special name.
> If anyone has any online refs of the resultant polyhedron or can cook up some > Maple code to visualize this polyhedron (preferably with rotational animation), > I'd appreciate it.
That's beyond my ability, but consider this description. It has the north and south poles as vertices. It has eight vertices, equally spaced, on the equator. These eight vertices define eight meridians. The remaining 16 vertices lie two on each of these meridians, at 45 degrees north and south. Thus the polyhedron has 56 edges and 32 faces. Sixteen of the faces are congruent isosceles triangles, with one vertex at a pole. The other sixteen are congruent isosceles trapezia.
"I.N. Galidakis" <morph...@olympus.mons> writes: > I am trying to visualize what looks like a particular polyhedron(?), but I > am > having a bout of stupidity and cannot figure something simple about it.
> I am describing the shape, section-wise:
> Consider the unit circle on the xy-plane. Put vertexes at exp(2*k*Pi/8*i), > k \in > {0,1,...,7}
> Rotate the unit circle around the x-axis by Pi/4.
> Consider the new unit circle on the xy-plane. Put again vertexes on the new > unit > circle at exp(2*k*Pi/8*i), k \in {0,1,...,7} (vertex duplicity allowed).
> Rotate the new unit circle around the x-axis by Pi/4 (this will make the > first > unit circle perpendicular to the xy-plane).
> Repeat 7 times, until the first unit circle rotates back to its initial > position.
> What is the name of the resultant polyhedron? It has a total of 26 > vertexes.
> If anyone has any online refs of the resultant polyhedron or can cook up > some > Maple code to visualize this polyhedron (preferably with rotational > animation), > I'd appreciate it.
-- Robert Israel isr...@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
Robert Israel wrote: > "I.N. Galidakis" <morph...@olympus.mons> writes:
>> I am trying to visualize what looks like a particular polyhedron(?), but I >> am >> having a bout of stupidity and cannot figure something simple about it.
>> I am describing the shape, section-wise:
>> Consider the unit circle on the xy-plane. Put vertexes at exp(2*k*Pi/8*i), >> k \in >> {0,1,...,7}
>> Rotate the unit circle around the x-axis by Pi/4.
>> Consider the new unit circle on the xy-plane. Put again vertexes on the new >> unit >> circle at exp(2*k*Pi/8*i), k \in {0,1,...,7} (vertex duplicity allowed).
>> Rotate the new unit circle around the x-axis by Pi/4 (this will make the >> first >> unit circle perpendicular to the xy-plane).
>> Repeat 7 times, until the first unit circle rotates back to its initial >> position.
>> What is the name of the resultant polyhedron? It has a total of 26 >> vertexes.
>> If anyone has any online refs of the resultant polyhedron or can cook up >> some >> Maple code to visualize this polyhedron (preferably with rotational >> animation), >> I'd appreciate it.
