research-article

Interlocking Spiral Drawings Inspired by M. C. Escher’s Print Whirlpools

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Kwok Wai ChungAuthors Info & Claims

ACM Transactions on Graphics, Volume 42, Issue 2

Article No.: 18, Pages 1 - 17

https://doi.org/10.1145/3560711

Published: 22 November 2022 Publication History

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Abstract

Whirlpools, by the Dutch graphic artist M. C. Escher, is a woodcut print in which fish interlock as a double spiral tessellation. Inspired by this print, in this article we extend the idea and present a general method to create Escher-like interlocking spiral drawings of N whirlpools. To this end, we first introduce an algorithm for constructing regular spiral tiling T. Then, we design a suitable spiral tiling T and use N copies of T to compose an interlocking spiral tiling K of N whirlpools. To create Escher-like drawings similar to the print, we next specify realization details of using wallpaper templates to decorate K. To enhance the aesthetic appeal, we propose several measures to minimize motif overlaps of the spiral drawings. Technologically, we develop algorithms for generating Escher-like drawings that can be implemented using shaders. The method established is thus able to generate a great variety of exotic Escher-like interlocking spiral drawings.

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Appendices

A Proof of Proposition 3.1

Differentiating \(z_{11}\) and \(z_{12}\) with respect to \(t\), we have

\begin{equation} {\left\lbrace \begin{array}{ll} \frac{\mathrm{d}z_{11}}{\mathrm{d}t} = \lambda _1(k + i)e^{(k + i) t}, \\ \frac{\mathrm{d}z_{21}}{\mathrm{d}t} = \lambda _1(k + i)e^{(k + i) t + \frac{2 \pi }{N} i}. \end{array}\right.} \end{equation}

(40)

Since \(k + i = e^{i\alpha } \sqrt {1 + k^2}\), where \(\tan \alpha = \frac{1}{k}\) and \(B_1 = z_{11} (\frac{\pi }{2} + \frac{\pi }{N})\), the first equation of Equation (40) at \(B_1\) becomes

\begin{equation} \frac{\mathrm{d} z_{11}}{\mathrm{d}t} \bigg |_{B_1} = \lambda _1 \sqrt {1 + k^2} \cdot e^{k (\frac{\pi }{2} + \frac{\pi }{N})} \cdot e^{i \left[ (\frac{\pi }{2} + \frac{\pi }{N}) + \alpha \right]}. \end{equation}

(41)

Let \(B_N\) be the intersection of \(z_{11}\) and \(z_{N1}\) on line segment \(\overline{A_1 A_N}\), as shown in Figure 4. Since \(\angle B_N A_1 C_1 = \frac{\pi }{2} + \frac{\pi }{N}\), \(B_N\) is given by

\begin{equation} B_N = z_{11} \left(-\frac{\pi }{2} - \frac{\pi }{N} \right). \end{equation}

(42)

Now,

\begin{equation} B_1 = e^{\frac{2 \pi }{N}} B_N = e^{\frac{2 \pi }{N}} z_{11} \left(-\frac{\pi }{2} - \frac{\pi }{N} \right) = z_{21} \left(-\frac{\pi }{2} - \frac{\pi }{N} \right). \end{equation}

(43)

That is, \(z_{21} = B_1\) when \(t = -\frac{\pi }{2} - \frac{\pi }{N}\). Then, the second equation of Equation (40) at \(B_1\) becomes

\begin{equation} \begin{split} \frac{\mathrm{d} z_{21}}{\mathrm{d}t} \bigg |_{B_1} & = \lambda _1(k + i) e^{(k + i) (- \frac{\pi }{2} - \frac{\pi }{N}) + \frac{2 \pi }{N} i} \\ &= \lambda _1 \sqrt {1 + k^2} \cdot e^{-k (\frac{\pi }{2} + \frac{\pi }{N})} \cdot e^{i \left[(-\frac{\pi }{2} + \frac{\pi }{N}) + \alpha \right]}. \end{split} \end{equation}

(44)

From Equations (41) and (44),

\begin{equation} \arg \left(\frac{\mathrm{d} z_{11}}{\mathrm{d}t} \bigg |_{B_1} \right) - \arg \left(\frac{\mathrm{d} z_{21}}{\mathrm{d}t} \bigg |_{B_1} \right) = \pi . \end{equation}

(45)

Therefore, the two tangential vectors are in opposite direction.

To prove part (b), we note that

\begin{equation} |\overline{A_1 B_1}| = \lambda _1 e^{k(\frac{\pi }{2} + \frac{\pi }{N})} \end{equation}

(46)

and

\begin{equation} |\overline{A_2 B_1}| = \lambda _1 e^{-k(\frac{\pi }{2} + \frac{\pi }{N})}. \end{equation}

(47)

Since \(|\overline{A_1 A_2}| = |\overline{A_1 B_1}| + |\overline{A_2 B_1}| = 2 \sin \frac{\pi }{N}\), we obtain

\begin{equation} \left[ \lambda _1 e^{k (\frac{\pi }{2} + \frac{\pi }{N})} + \lambda _1 e^{-k (\frac{\pi }{2} + \frac{\pi }{N})} \right] = 2 \sin \frac{\pi }{N}, \end{equation}

(48)

from which Equation (24) follows. This completes the proof.

B Proof of Proposition 3.2

We omit the proof of part (a) as it is similar to that of Proposition 3.1. As for part (b), we note that

\begin{equation} |\overline{A_1 E_1}| = \lambda _2 e^{k (\frac{\pi }{2} + \frac{\pi }{N})} \end{equation}

(49)

and

\begin{equation} |\overline{A_2 E_1}| = |\overline{A_1 E_N}| = \lambda _2 e^{k (\frac{3\pi }{2} - \frac{\pi }{N})}. \end{equation}

(50)

Since \(|\overline{A_1 A_2}| = |\overline{A_1 E_1}| + |\overline{A_2 E_1}| = 2 \sin \frac{\pi }{N}\), we obtain

\begin{equation} \left[ \lambda _2 e^{k (\frac{\pi }{2} + \frac{\pi }{N})} + \lambda _2 e^{k (\frac{3\pi }{2} - \frac{\pi }{N})} \right] = 2 \sin \frac{\pi }{N}, \end{equation}

(51)

from which Equation (30) follows. This completes the proof.

References

[1]

E. Akleman. 2009. Twirling sculptures. J. Math. Arts 3, 1 (2009), 1–10.

Abstract

Supplementary Material

A Proof of Proposition 3.1

B Proof of Proposition 3.2

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