Halved pentagonal duocomb

The  is a regular skew polyhedron in 4-dimensional Euclidean space. It can be constructed by halving the pentagonal duocomb, and the two are abstractly equivalent. Halving the gives the pentagrammic duocomb, and if you halve the  three times you return to the original pentagonal duocomb.

Vertex coordinates
Its vertex coordinates are the same as those of a pentagonal duoprism of edge length $$\sqrt{2}/2$$. For the halved pentagonal duocomb of edge length 1 these can be given as:
 * $$\left(0,\,\sqrt{\frac{5+\sqrt{5}}{20}},\,0,\,\sqrt{\frac{5+\sqrt{5}}{20}}\right)$$,
 * $$\left(0,\,\sqrt{\frac{5+\sqrt{5}}{20}},\,\pm\frac{\sqrt{3+\sqrt{5}}}{4},\,\sqrt{\frac{5-\sqrt5}{80}}\right)$$,
 * $$\left(0,\,\sqrt{\frac{5+\sqrt{5}}{20}},\,\pm\frac{\sqrt{2}}{4},\,-\sqrt{\frac{5+2\sqrt{5}}{40}}\right)$$,
 * $$\left(\pm\frac{\sqrt{3+\sqrt{5}}}{4},\,\sqrt{\frac{5-\sqrt{5}}{80}},\,0,\,\sqrt{\frac{5+\sqrt{5}}{20}}\right)$$,
 * $$\left(\pm\frac{\sqrt{3+\sqrt{5}}}{4},\,\sqrt{\frac{5-\sqrt{5}}{80}},\,\pm\frac{\sqrt{3+\sqrt{5}}}{4},\,\sqrt{\frac{5-\sqrt{5}}{80}}\right)$$,
 * $$\left(\pm\frac{\sqrt{3+\sqrt{5}}}{4},\,\sqrt{\frac{5-\sqrt{5}}{80}},\,\pm\frac{\sqrt{2}}{4},\,-\sqrt{\frac{5+2\sqrt{5}}{40}}\right)$$,
 * $$\left(\pm\frac{\sqrt{2}}{4},\,-\sqrt{\frac{5+2\sqrt{5}}{40}},\,0,\,\sqrt{\frac{5+\sqrt{5}}{20}}\right)$$,
 * $$\left(\pm\frac{\sqrt{2}}{4},\,-\sqrt{\frac{5+2\sqrt{5}}{40}},\,\pm\frac{\sqrt{3+\sqrt{5}}}{4},\,\sqrt{\frac{5-\sqrt{5}}{80}}\right)$$,
 * $$\left(\pm\frac{\sqrt{2}}{4},\,-\sqrt{\frac{5+2\sqrt{5}}{40}},\,\pm\frac{\sqrt{2}}{4},\,-\sqrt{\frac{5+2\sqrt{5}}{40}}\right)$$.