[Math] Cyclic-Harmonic Curve on the Golden Ratio
This page describes the formula of cyclic-harmonic curve applied the golden ratio. This page also publishes the SageMath script to create the figures of multiple cyclic-harmonic curves.
(There is the Japanese(日本語) page.)
(Last updated date: February 21, 2021)
Preface
While studying Rose curve (or Rhodonea curve) on the Web, I came to know about “cyclic-harmonic curve” 1 2 3.
In the figures drawing this curve, there are figures composed of large and small petals. The figure shown above is one of them. This time, I applied the Golden ratio to the lengths of the petals in these figures. The formula is explained below.
Moreover, I wrote the SageMath script to create the figures of multiple cyclic-harmonic curves applied the golden ratio. The script is published in the last half of this page.
Cyclic-harmonic curve on the golden ratio
Cyclic-harmonic curve formula
The cyclic-harmonic curve is represented by the following polar equation.
\[r = a \cos \left( \frac{n}{d} \, \theta \right) + b\]
- \(n\) and \(d\) are positive integers.
- \(a\) and \(b\) are positive real numbers.
- \(\theta\) changes from \(0\) to \(2d \pi\).
By the way, if \(b\) in this expression is set to \(0\), it corresponds to the Rose curve.
Under the following conditions, the figure that draws the cyclic-harmonic curve will look like it is composed of large and small petals.
- \(n \geq d\)
- \(a>b\)
At this time, the number of large and small petals is the same, and is calculated by the following formula.
\[\frac{n}{\mathrm{gcd}(n,d)}\]
Also, the lengths of large and small petals can be calculated by the following formulas.
- The major axis length of the large petal ( \(l\) in the figure below) is \(a+b\).
- The major axis length of the small petal ( \(s\) in the figure below) is \(a-b\).
Applying the golden ratio
This time, the ratio of the lengths of large and small petals is set to the Golden ratio as follows.
\[s : l = (l-s) : s = 1 : \frac{1 + \sqrt{5}}{2}\]
Substituting the expressions for the petal’s length into this expression yields the following.
\[(a-b) : (a+b) = 1 : \frac{1 + \sqrt{5}}{2}\]
The variable \(a\) in the cyclic-harmonic curve equation corresponds to the amplification of the figure. Therefore, to simplify the formula, set \(a=1\). Then the above expression becomes:
\[(1-b) : (1+b) = 1 : \frac{1 + \sqrt{5}}{2}\]
\[\therefore \; (1-b) \times \frac{1 + \sqrt{5}}{2} = 1+b\]
Solving this equation for \(b\) 4 gives:
\[b = \sqrt{5} -2 \approx 0.236 >0\]
Since \(a\) and \(b\) have been decided, calculating the lengths of large and small petals are as follows, respectively. 5
\[\begin{align*} l & = 1 + \left( \sqrt{5} -2 \right) \approx 1.236 \\ s & = 1 - \left( \sqrt{5} -2 \right) \approx 0.764 \end{align*}\]
Finally, substituting the determined values of \(a\) and \(b\) into the cyclic-harmonic curve equation, we get:
\[r = \cos \left( \frac{n}{d} \, \theta \right) + \sqrt{5} -2\]
After that, by substituting various numerical values for \(n\) and \(d\) in this equation, various figures composed of golden ratio can be drawn.
Examples are shown below.
Example 1 : \(n=1\) , \(d=1\)
For \(n\) and \(d\), the combination of \(n=1\) and \(d=1\) is the simplest example.
The formula used and the results of running SageMath version 9.0 are shown below.
\[r = \cos \theta + \sqrt{5} -2\]
sage: var('θ') # Create a symbolic variable θ. You only have to do it once.
sage: polar_plot(cos(θ)+sqrt(5)-2, (θ, 0, 2*pi), plot_points=200)
In the figure above, you can see that the lengths of the large and small petals are equal to the calculated values.
Example 2 : \(n=3\) , \(d=1\)
\[r = \cos 3 \theta + \sqrt{5} -2\]
sage: polar_plot(cos(3*θ)+sqrt(5)-2, (θ, 0, 2*pi), plot_points=200)
Example 3 : \(n=5\) , \(d=3\)
\[r = \cos \left( \frac{5}{3} \, \theta \right) + \sqrt{5} -2\]
sage: polar_plot(cos((5/3)*θ)+sqrt(5)-2, (θ, 0, 2*3*pi), plot_points=200*3)
Example 4 : \(n=12\) , \(d=7\)
In the three examples above, you have seen that the petal lengths are equal to the calculated values, so, in the following examples, I will not display the axes so that you can enjoy the beauty of the figure.
\[r = \cos \left( \frac{12}{7} \, \theta \right) + \sqrt{5} -2\]
sage: polar_plot(cos((12/7)*θ)+sqrt(5)-2, (θ, 0, 2*7*pi), plot_points=200*7, axes=False)
Example 5 : \(n=25\) , \(d=6\)
\[r = \cos \left( \frac{25}{6} \, \theta \right) + \sqrt{5} -2\]
sage: polar_plot(cos((25/6)*θ)+sqrt(5)-2, (θ, 0, 2*6*pi), plot_points=200*6, axes=False)
Example 6 : \(n=49\) , \(d=6\)
\[r = \cos \left( \frac{49}{6} \, \theta \right) + \sqrt{5} -2\]
sage: polar_plot(cos((49/6)*θ)+sqrt(5)-2, (θ, 0, 2*6*pi), plot_points=200*6, axes=False)
Script to create figures of multiple cyclic-harmonic curves
The following is a SageMath script for creating multiple figures at a time from the above cyclic-harmonic curve equation, and an example of creating it.
stop_int = 50 # >= 2
file_ext = "svg"
#file_ext = "png"
var('θ') # Create a symbolic variable 'θ'
for numer in range(1, stop_int): # Numerator
for denom in range(1, (numer + 1)): # Denominator
if gcd(numer, denom) != 1:
continue
c_h_curve = polar_plot( cos((numer/denom)*θ)+sqrt(5)-2, \
(θ, 0, 2*denom*pi), plot_points=200*denom)
file_name = "c_h_curve_" + str(numer) + "_" + str(denom) + \
"_gold." + file_ext
c_h_curve.save(file_name, axes=False)
print(file_name + " saved.")
To run this script, save a copy of the script in a file (for example make_figures.sage
) and run the following command in your terminal:
$ sage make_figures.sage
This script creates diagrams of every combination of \(n\) and \(d\) values in the range of \(1\) to \((\mathrm{stop\_int} -1)\). However, some combinations of \(n\) and \(d\) values that result in a diagram of the same shape are not output.
For example, the following three cyclic-harmonic curve equations are the same shape.
\[\begin{align*} r & = \cos \left( \frac{3}{2} \, \theta \right) + \sqrt{5} -2 \\[0.5em] r & = \cos \left( \frac{6}{4} \, \theta \right) + \sqrt{5} -2 \\[0.5em] r & = \cos \left( \frac{9}{6} \, \theta \right) + \sqrt{5} -2 \end{align*}\]
This script outputs only the first figure and does not output other figures of the same shape after that. As a result, all the figures output by this script will be different.
You can change the number of diagrams created by changing the value of the \(\mathrm{stop\_int}\) variable in the script. Examples are shown below.
\(\mathrm{stop\_int}\) | Number of diagrams |
---|---|
30 | 270 |
50 | 754 |
100 | 3004 |
When I checked the created graphs when \(\mathrm{stop\_int} =100\), the cyclic-harmonic curves above \(n=50\) were all similar. So when you try this script, I think \(\mathrm{stop\_int} =50\) is enough.
In addition, although the figures are output in SVG format in this script, the figures can be output in another format by changing the extension of the output file name. For example, when outputting figures in PNG format, change the commented out part in the script.
If you are interested, please try it.
Postscript
Some examples of colored cyclic-harmonic curve are shown in another page.
Robert E. Moritz, The General Theory of Cyclic-Harmonic Curves. Annals of Mathematics Second Series, Vol. 23, No. 1 (Sep., 1921), pp. 29-39 (11 pages).
DOI: 10.2307/1967779
Documentation on JSTOR↩︎Sonja Gorjanc, Ema Jurkin, Generalized Rose Surfaces and their Visualizations. 2013.
Documentation on arXiv↩︎The calculation procedure is shown below.
\[\begin{align*} (1-b) \times \frac{1 + \sqrt{5}}{2} & = 1+b \\[0.5em] (1-b) \left(1+ \sqrt{5} \right) & = 2 \, (1+b) \\[0.5em] \left(1+ \sqrt{5}\right) - \left(1+ \sqrt{5}\right)b & = 2+2b \\[0.5em] -\left(1+ \sqrt{5}\right)b -2b & = 2 - \left(1+ \sqrt{5}\right) \\[0.5em] \left(1+ \sqrt{5}\right)b +2b & = -2 + \left(1+ \sqrt{5}\right) \\[0.5em] \left(3+ \sqrt{5}\right)b & = -1 + \sqrt{5} \\[0.5em] b & = \frac{-1 + \sqrt{5}}{3+ \sqrt{5}} \\[0.5em] & = \frac{\left(-1 + \sqrt{5}\right) \left(3- \sqrt{5}\right)}{\left(3+ \sqrt{5}\right) \left(3- \sqrt{5}\right)} \\[0.5em] & = \frac{-3 + \sqrt{5} + 3 \sqrt{5} -5}{9-5} \\[0.5em] & = \frac{-8 + 4 \sqrt{5}}{4} \\[0.5em] & = -2 + \sqrt{5} \\[0.5em] & = \sqrt{5} -2 \end{align*}\]↩︎
The verification procedure is shown below.
\[\begin{align*} \frac{l}{s} & = \frac{1 + \left( \sqrt{5} -2 \right)}{1 - \left( \sqrt{5} -2 \right)} \\[0.5em] & = \frac{-1 + \sqrt{5}}{3 - \sqrt{5}} \\[0.5em] & = \frac{\left(-1 + \sqrt{5} \right) \left(3 + \sqrt{5} \right)}{\left(3 - \sqrt{5} \right) \left(3 + \sqrt{5} \right)} \\[0.5em] & = \frac{-3 - \sqrt{5} + 3 \sqrt{5} +5}{9-5} \\[0.5em] & = \frac{2 + 2 \sqrt{5}}{4} \\[0.5em] & = \frac{1 + \sqrt{5}}{2} \\[0.5em] & = \varphi \end{align*}\]↩︎
Comments
Post a Comment