Logarithmic spiral tiling of hexagons
Tis Veugen
tis.veugen@gmail.com
4 August 2021
Introduction
Robert Fathauer presents in [1] a logarithmic spiral tiling of quadrilaterals. In this paper an analoguous kind of tessellation is addressed using hexagons having all similar shape. We also give examples of degenerated hexagons yielding triangles and quadrilaterals.
Analysis
Figure 1 shows a logarithmic spiral configuration of hexagons. In this example the number of spiral arms are given by N=7 and M=3.
Figure 1: Spiral configuration
The vertices of all hexagons lie on the M spirals. In case of a single spiral (M=1) it is obvious that all vertices must lie on this spiral. In case of M=2 the 2 pairs of vertices that lie the fartest away from the spiral center and the closest to the center are on the same spiral, whereas the 2 middle vertices lie on the other spiral. For 3 or more spirals the vertices of a hexagon lie on 3 different spirals.
In the figure the “base” hexagon has a grey border. The spirals decrease in clockwise direction to the center. From the figure can be observed that there are N rays of hexagons towards the center, with N-1 rays in between those rays, and M-1 rays between the base hexagon and the N^{th} hexagon after the first revolution indicated by S_{0}.
The angle is the rotation angle around the center rotating along the top edge of the hexagon from P_{0} to P_{1}.(red spiral). The rotation angle around the center rotating along the bottom edge from P_{4} to P_{3} (blue spiral) has the same size . The angle is the rotation angle around the center rotating along at the middle of the hexagon from P_{5} to P_{2} (green spiral). If a hexagon rotates around the center along the 3 colored spirals over the sum angle it arrives at the next hexagon. Hence, the sum angle is written as:
(1)
Rotating P_{0} over the angle brings us to S_{0} . If there were only 1 spiral (M=1) then S_{1} would coincide with P_{5}: the hexagons would lie against each other. With M bigger than 1, there are M-1 hexagons in between.
The computation of the size of is easier when making another revolution, in this example from S_{0} to Y_{0} . After the 2 revolutions there are still M rotations over angle needed for having the same direction to the center, in this case from Y_{0} to Z_{0} . However, for an extra swing in the rays, we introduce an artificial extra skew parameter. So, satisfies:
(2)
There are no restrictions for and , apart from the fact that they must be non-negative.
The exponential growth parameter k will be derived from the ratio between horizontal and vertical characteristics of the hexagon.
(3)
In concreto, this ratio is the quotient of the distance between points P_{0} and P_{4} , and on the other hand the circular curve along P_{0} over an angle of ; the latter curve approximates the spiral curve from P_{0} to N_{0 }. This choice of the ratio has a clear human interpretation.
The points P_{0}, Q_{0}, R_{0}, and S_{0} lie on a logarithmic spiral by construction. The factor F between subsequent points is based on the quotient between P_{0} and S_{0} :
(4)
S_{0} has been obtained from P_{0} by rotating over . Hence:
(5)
Bear in mind that the 4 mentioned points traverse in opposite direction compared to the spiral path from P_{0} to S_{0} . For this reason the rotation angle is computed as a negative angle by subtracting an extra . Taking this into account and combining (4) and (5) gives:
(6)
or:
(7)
Using F, formulas for all vertices of the base hexagon can be derived:
(8a)
(8b)
(8c)
(8d)
(8e)
Substitution of (8d) in (3) leads to an expression for k :
(9)
For convenience, define the following variables:
(10)
(11)
With these we can convert (9) to:
(12)
For solving this quadratic equation we take the root with the negative sign, because that agrees with the simple case of zero skew:
(13)
So, the growth factor k becomes:
(14)
Because the value of is smaller than 1, the “growth” factor k has a negative value. This means that the base hexagon becomes smaller and smaller when rotating towards the center, as expected.
Special case: quadrilateral
When is chosen to be zero, the points P_{0} and P_{1} overlap, and also the points P_{3} and P_{4} overlap. As a consequence the hexagon becomes a quadrilateral. Figure 2 shows two examples of degenerated hexagons becoming a quadrilateral; in Figure 2a the spiral has the same number of spiral arms as above: N=7 and M=3, whereas in Figure 2b the spiral has N=3 and M=6 resembling the structure of Fathauer’s ducks spiral [2].
Figure 2a: Quadrilateral spiral: N=7, M=3 |
Figure 2b: Quadrilateral spiral: N=3, M=6 |
Special case: triangle
If angle equals zero, the points P_{2} and P_{5} overlap. The hexagon becomes then a pair of triangles like a diabolo. Figure 3a shows the resulting spiral with the same number of spiral arms as above: N=7 and M=3. Figure 3b shows a version with deformed edges.
Figure 3a: Triangle spiral: N=7, M=3 |
Figure 3b: Deformed triangle spiral: N=7, M=3 |
References
[1] Robert W. Fathauer, “Tessellations: Mathematics, Art, and Recreation”, AK Peters/CRC Press, 2021
[2] Robert W. Fathauer, “New duck tessellation based on a spiral tiling of quadrilaterals”, https://twitter.com/robfathauerart/status/1355884521120382984, 31 Januari 2021