# Extended Polar Zonohedra

Steven Dutch, Natural and Applied Sciences, University of Wisconsin - Green Bay
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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)=Rv2
• 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 sin2(v2/2)=(w/v)2sin2(v1/2)
• cos2(v2/2)=1-(w/v)2sin2(v1/2)=sin2(v1/2)
• Thus sin2(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)=1-2sin2(V1/2)=1-2(9R2/16)=1-9R2/8
• Also, since sinQ = Rsin(A/2), cos2Q=1-3R2/4
• Thus (3/2)cos2Q=1/2+cosQ or 3cos2Q-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.

Created 21 January 2001, Last Update 14 December 2009

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