Figure
1: Spiral configuration
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 Nth hexagon after the first revolution indicated by S0.
The angle
is
the rotation angle around the center rotating along the top edge of
the hexagon from P0 to P1.(red spiral). The
rotation angle around the center rotating along the bottom edge from
P4 to P3 (blue spiral) has the same size
.
The angle
is
the rotation angle around the center rotating along at the middle of
the hexagon from P5 to P2 (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 P0
over the angle
brings
us to S0 . If there were only 1 spiral (M=1) then S1
would coincide with P5: 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 S0
to Y0 . After the 2 revolutions there are still M
rotations over angle
needed
for having the same direction to the center, in this case from Y0
to Z0 . 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 P0
and P4 , and on the other hand the circular curve along P0
over an angle of
;
the latter curve approximates the spiral curve from P0 to
N0 . This choice of the ratio has a clear human
interpretation.
The points P0, Q0, R0, and S0 lie on a logarithmic spiral by construction. The factor F between subsequent points is based on the quotient between P0 and S0 :
(4) 
S0 has
been obtained from P0 by rotating over
.
Hence:
(5) 
Bear in mind that
the 4 mentioned points traverse in opposite direction compared to the
spiral path from P0 to S0 . 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 P0 and P1
overlap, and also the points P3 and P4 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].
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Special case: triangle
If angle
equals zero, the points P2 and P5 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.
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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
