breaking the 10-fold rotational symmetry

We modify the “Projection method for 10-fold rotational symmetry“. For details see “Projection method – geometry and maths“. To destroy the symmetry we shift the sets of lines alternatingly away or towards the origin. Until now we set the position of the lines with s_i=0.5 for all sets with indices i. This gave us a regular decagon at the origin. Here we alternate and use s_i=0.5+a for all even i and s_i=0.5-a for all odd i. Thus we get a pentagram around the origin and a five-fold rotational symmetry.

We won’t find any ten-pointed stars for most values of the constant a. However, around a=0.4 we get some interesting results that depend strongly on the exact value of a. You can do your own experiments with the processing 2 code published in this blog. Here is an example for a=0.405:

breaking10

As before, the center of perfect 5-fold rotational symmetry lies near the upper right corner. Here, we seem to get a tiling with a large number of different tiles. It is most certainly not self-similar and quite different to the Penrose tiling.

This entry was posted in Quasiperiodic design, Tilings and tagged , , , , , . Bookmark the permalink.

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