From 6dfa27d91207b39176a71728a9d1b0b385d01ffa Mon Sep 17 00:00:00 2001
From: Raph Levien <raph@google.com>
Date: Mon, 18 Nov 2024 11:29:32 -0800
Subject: [PATCH] Rough draft of hyperbezier blog post

It needs quite a bit of work, especially figures to illustrate the various curves, but I figured I'd push what I have.
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+---
+layout: post
+title:  "The mathematical beauty of hyperbezier curves"
+date:   2024-11-18 11:29:42 -0800
+categories: [curves]
+---
+I have been on a decades-long quest to find a curve family better suited for interactive design than the trusty cubic Bézier. My [PhD thesis] proposed Spiro, which is better in some ways (smoother curves) but is also less versatile and expressive. In particular, it does a poor job representing curves with high tension. I now propose a curve family which I believe is simply better than Béziers. I call it "hyperbezier," both because it is based on Béziers, and also because it is related to hypergeometric functions and can reasonably well approximate hyperbolas.
+
+The curve family is given as a [Cesàro equation], curvature as a function of arc length:
+
+$$\kappa(s) = \frac{p_1s+p_0}{(q_2s^2+q_1s+q_0)^{1.5}}$$
+
+The curve family contains within its parameter space two beautiful curves: Euler spiral and circle involute. It is also a versatile mimic
+
+## Related curves
+
+By considering various ranges of the parameter space, the hyperbezier either contains or closely approximates quite a large variety of familiar curves, including Euler spiral, circle involute, cubic Béziers, hyperbolas.
+
+### Euler spiral
+
+Setting $$q_1$$ and $$q_2$$ to zero, the hyperbezier reduces to a simple Euler spiral.
+
+### Circle involute
+
+Setting $$p_1$$ and $$q_1$$ to one, the others to zero, the hyperbezier reduces to a circle involute, given by the Cesàro equation $$\kappa(s)=\frac{1}{\sqrt{s}}$$. This curve is probably best known as the ideal profile for [gear teeth][Gears] without slipping, but has other fascinating properties. In particular, it is the only curve I know other than circular arc that is its own parallel curve.
+
+The circle involute has a cusp of order 3/2, and in this way is related to the semi-cubical parabola, which has the same cusp.
+
+## Whewell equation
+
+In general, when a curve is given as a Cesàro equation, a good way to evaluate it is to convert it into a [Whewell equation] (tangent angle as a function of arc length) by integrating the curvature, then doing numerical integration of the sine and cosine of that angle to give Cartesian coordinates.
+
+Somewhat remarkably, the Whewell form of the hyperbezier is extremely similar to its Cesàro form:
+
+$$\theta(s) = \frac{p_1's+p_0'}{\sqrt{q_2s^2+q_1s+q_0}}$$
+
+The quadratic polynomial in the denominator is the same, but the linear equation in the numerator is a straightforward transform of the Cesàro parameters (this is assuming that the coefficients have been normalized so that $$q_0 = 1$$):
+
+$$p_0' = \frac{2p_0q_1 - p_1q_1^2/q_2}{4q_2 - q_1^2} - \frac{p_1}{q_2}$$
+$$p_1' = \frac{4p_0q_2 - 2p_1q_1}{4q_2 - q_1^2}$$
+
+This transform follows fairly straightforwardly from the following integral, which has a pleasing symmetry:
+
+$$\int \frac{p_1s+p_0}{(as^2 + 1)^{1.5}} ds = \frac{p_0s-p_1/a}{\sqrt{as^2 + 1}} + C$$
+
+## The exponent
+
+The exponent in the denominator is 1.5, which may seem like an odd choice. I experimented also with both one and two. The former would be appealing; it would essentially write the curvature as a simple rational function, evoking [Padé approximation]. However, it is not numerically robust. To represent high-tension curves, the parameters 
+
+A choice of two is practical. The integrals don't come out quite as nicely, and we don't get the circle involute (instead, a logarithmic spiral).
+
+[PhD thesis]: TODO
+[Cesàro equation]: https://en.wikipedia.org/wiki/Ces%C3%A0ro_equation
+[Whewell equation]: https://en.wikipedia.org/wiki/Whewell_equation
+[Gears]: https://ciechanow.ski/gears/
+[Padé approximation]: https://en.wikipedia.org/wiki/Pad%C3%A9_approximant