Dualization method for ten-fold rotational symmetry

We now use the dualization method with grids made of several sets of parallel lines. It is important to take the same grids as earlier for the projection method, see “projection method for ten-fold rotational symmetry” and “Projection method for the Stampfli and the Socolar tiling“. Then we can see how these methods go together.

As an example look at the Stampfli tiling. In the projection method its points arise wherever two hexagons form a complete twelve-pointed star. These stars give in the dualization method a star of twelve rhombs surrounded by triangles. Similarly, looking at ten-fold rotational symmetry we get for each point of the projection method a star of ten rhombs from the dualization method.

Thus, we need the same setup as detailed in “Projection method – geometry and maths“. There are n sets of parallel lines that make up a 2n-fold rotational symmetry. The distance between parallel lines is equal to one. They are perpendicular to a vector (cos(i*PI/n), sin(i*PI/n)) , where i is between 0 and (n-1). One of these lines has a distance of  s_i=0.5+xTrans*cos(i*PI/n)+yTrans*sin(i*PI/n) from the origin, where xTrans and yTrans determine the position of the point of perfect 2n-fold rotational symmetry (see “How to shift the design“).

This is easy to program, see the next post. First sets of vertical lines at entire x-values are shifted by s_i in x-direction. Then they are rotated by an angle of PI*i/n and we put these sets together to make a grid. A typical result looks like this:

tenFoldDualizationThe center of ten-fold symmetry lies a bit at the right and above the center. With different colors for the rhombs we can better see large-scale structures and recognize the stars of ten rhombs. We see that they have indeed the same distribution as the points in “projection method for ten-fold rotational symmetry“. However, there is no match with the tiling of rhombs as shown above. Thus, this tiling is not self-similar. By the way, we get the self-similar Ammann-Beenker tiling for n=4.

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