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Polar zonohedra have an n-fold
symmetry axis and rhombic faces. There are n faces meeting at the top and
bottom vertex, and n-1 girdles of faces. The total number of faces is thus
n(n-1).
It can be shown that the vertices all lie on equally spaced planes normal to the polar axis, and that the longitudinal cross section of the zonohedron is approximately a sine curve. |

There are only n distinct edge directions in a polar zonohedron. We can
remove sets of parallel edges and *diminish* the zonohedron (left) or add a
new set of parallel edges distinct from the original edges and *augment*
the zonohedron. For the most part the results are of no great interest.

However, there is one exception. If n is odd, we can cut the zigzag series of edges that bisects the zonohedron, pull the two halves apart parallel to the axis, and add a set of edges joining the two halves. If the new edges are the same length as the original edges, the solid still is equilateral. We add a girdle of 2n faces around the equator with its zone axis parallel to the polar axis. |

For a simple polar zonohedron with n-fold symmetry there are two independent dimensions: the radius R and the polar axis length L. We define angle A = 360/n. We can find angle Q, the angle between one of the top or bottom edges and the polar axis. We can then find Vi, the face angle of the i-th girdle of rhombuses, and Ri, the radius of the i-th ring of vertices. The formulas are defined for edge length = 1.

- Length L of polar axis = nh = ncosQ
- Maximum radius R = sinQ/sin(A/2)
- SinVi/2 = sinQsin(iA/2)
- Ri = Rsin(iA/2)

For the extended equilateral zonohedron, we have

- Length L* of polar axis = nh+1 = ncosQ+1. Solving for Q, we get cosQ=(L*-1)/n
- Maximum radius R = sinQ/sin(A/2)
- SinVi/2 = sinQsin(iA/2)
- Ri = Rsin(iA/2)

Now, what are the face angles for the new girdle? We can plow through some
geometry or we can think a bit (oww - make the pain go *away*!). The new
girdle rhombuses consist of one of the original edges plus a new edge parallel
to the polar axis. All the original edges make an angle Q with the polar axis.
So the face angles of the equatorial girdle must be Q and 180-Q. That was a *lot*
easier than calculating.

The most interesting extended zonohedron is the n=5 case. If we solve for the case where the faces are congruent, we get, not the rhombic triacontahedron, but a flattened solid with 20 faces |

For n=5, we have: A=72, A/2=36, Sin(A/2)=0.5877....

For simplicity, let sin36=v and sin72=w

- For the isohedral case, solve for V1=180-V2
- SinV1/2=Rsin(A/2)sin(A/2)=Rv
^{2} - SinV2/2=Rsin(A/2)sin(2A/2)=Rvw=(w/v)SinV1/2
- Sin(V1/2)=sin((180-V2)/2)=sin(90-V2/2)=cos(V2/2)
- SinV2/2=(w/v)sin(V1/2), so sin
^{2}(v2/2)=(w/v)^{2}sin^{2}(v1/2) - cos
^{2}(v2/2)=1-(w/v)^{2}sin^{2}(v1/2)=sin^{2}(v1/2) - Thus sin
^{2}(v1/2)=1/(1+(w/v)^{2}), and sin(v1/2)=0.5257 - V1/2=31.72, thus V1=63.43, the angles for a rhombic triacontahedron.

Now let's calculate R and L* for the extended case.

- Sin(V1/2) = sinQsin(A/2), so sinQ=Sin(V1/2)sin(A/2)=.8944
- Q = 63.43 degrees also. Thus the equatorial girdle faces are congruent with all the other faces.
- Cos Q = 0.4772
- L* = 5cosQ+1 = 3.236 = sqrt(5)+1
- R = sinQ/sin(A/2) = 0.8944/sin(36) = 1.5217

It turns out that we get the rhombic triacontahedron by extending the 20-sided simple zonohedron. |

This sort of thing only works for n odd. What happens if we try it with a rhombohedron (n=3)?

The rhombohedron has six faces. We pull the solid apart and add a girdle of six more around the middle for a total of n=12. We can get a solid with congruent faces, and it turns out to be the rhombic dodecahedron viewed along one of its threefold symmetry axes.

So the rhombic dodecahedron is both a simple 4-zonohedron and an extended 3-zonohedron.

For n=3, we have: A=120, A/2=60, Sin(A/2)=sqrt(3)/2. We only have one girdle of faces, so we have to find the case V1=Q.

- V1=Q
- Sin(V1/2)=Rsin(A/2)sin(A/2)=3R/4
- sinQ = Rsin(A/2)
- But we can also write cosQ=cos(V1)=cos2(V1/2)
- cosQ=cos2(V1/2)=1-2sin
^{2}(V1/2)=1-2(9R^{2}/16)=1-9R^{2}/8 - Also, since sinQ = Rsin(A/2), cos
^{2}Q=1-3R^{2}/4 - Thus (3/2)cos
^{2}Q=1/2+cosQ or 3cos^{2}Q-2cosQ-1=0 - This factors to (3cosQ+1)(cosQ-1)
- Solving, we get cosQ=1 or 1/3. -CosQ=1 means Q=0, cosQ=1/3 means Q=109.471, the obtuse face angle of a rhombic dodecahedron. We get the obtuse angle because that's the one at the polar vertex.

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*Created 21 January 2001, Last Update
14 December 2009
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