Stable Length Estimates of Tube-Like Shapes

Abstract

Motivated by applications in biology, we present an algorithm for estimating the length of tube-like shapes in 3-dimensional Euclidean space. In a first step, we combine the tube formula of Weyl with integral geometric methods to obtain an integral representation of the length, which we approximate using a variant of the Koksma–Hlawka Theorem. In a second step, we use tools from computational topology to decrease the dependence on small perturbations of the shape. We present computational experiments that shed light on the stability and the convergence rate of our algorithm.

Appendix: Bumps

In this appendix, we study the unit bump, which we define by rotating two quarter circles of unit radius, as illustrated in Fig. 6. It consists of a hemi-sphere, for 0≤x≤1, and a quarter torus, for −1≤x≤0. Our main result is that the total mean curvature of the unit bump is

$$\begin{aligned} \operatorname{Mean}_{\rm bump} =& (4-\pi) \pi. \end{aligned}$$

(49)

Note that scaling the bump by a factor of ε changes its area by a factor of ε ² and its total mean curvature by a factor of ε.

Fig. 6

The unit bump obtained by rotating two quarter circles about the vertical axis. The two dotted circles are the cross-sections of the torus delimiting the bump. We compute the two principal curvatures at the solid point by projecting the blue circle to the plane with angle φ (Color figure online)

Two Elementary Results

It is easy to see that the total mean curvature over the hemi-sphere is $\operatorname{Mean}_{S} = 2 \pi$. The computation of the total mean curvature of the quarter torus requires some preparations. First, we recall the substitution method for integration. Specifically, we set x=sinα, dx=cosα dα, and notice that ${1}/{\sqrt{1-x^{2}}} = 1/\cos\alpha$. Hence

$$\begin{aligned} \int_0^1 \frac{1}{\sqrt{1-x^2}} \,\mathrm{d}x =& \int_0^{\pi/2} \frac{\cos\alpha}{\cos\alpha} \,\mathrm{d} \alpha, \end{aligned}$$

(50)

which gives $\frac{\pi}{2}$. Second, we compute the curvature of an ellipse at the endpoints of its short axis. Assuming the half-length of the long axis is 1 and that of the short axis is cosα, this curvature is cosα. To see this, recall that the unit circle is the best approximating circle of the parabola with formula $y = \frac{1}{2} x^{2} - 1$. Imagine the drawing in three dimensions and rotate the plane of the drawing about the horizontal axis until the projection of the circle back to the original plane is the ellipse with desired axes. The projection of the paraboloid satisfies $y = \frac{\cos\alpha}{2} x^{2} - \cos\alpha$. The best approximating circle thus has radius 1/cosα, as required.

Quarter Torus

We now compute the area and the total curvature of the quarter torus, obtained by rotating the circle that is the graph of $f(x) = 2 - \sqrt{1-x^{2}}$. As before, we have $\sqrt{1 + f'(x)^{2}} = 1 / \sqrt{1-x^{2}}$. There area is therefore

$$\begin{aligned} \operatorname{Area}_{T} =& 2 \pi\int_{-1}^{0} f(x) \sqrt {1+f'(x)^2} \,\mathrm{d}x \\ =& 2 \pi\int_{-1}^{0} \biggl( \frac{2}{\sqrt{1-x^2}} - 1 \biggr) \,\mathrm{d}x, \end{aligned}$$

(51)

which evaluates to 2π(π−1). To compute the total mean curvature, we first get the two principal curvatures. Along the rotating quarter circle, it is −1. In the other direction, we get it by projecting the horizontal circle with radius f(x); see Fig. 6. The angle of the projection is φ=arcsinx. Hence, x=sinφ and $\cos\varphi= \sqrt{1-x^{2}}$. The projection of the circle is an ellipse with axes of half-lengths f(x) and f(x)cosφ. It follows that the second principal curvature is cosφ/f(x). We get the total mean curvature by integrating the contributions of the two principal curvatures separately:

$$\begin{aligned} \operatorname{Mean}_{T,1} =& - \frac{1}{2} \operatorname{Area}_{T} , \end{aligned}$$

(52)

$$\begin{aligned} \operatorname{Mean}_{T,2} =& 2 \pi\int_{-1}^{0} f(x) \sqrt{1+f'(x)^2} \frac{\cos\varphi}{2 f(x)} \,\mathrm{d}x \\ =& 2 \pi\int_{-1}^{0} \frac{1}{2} \, \mathrm{d}x , \end{aligned}$$

(53)

which gives $\operatorname{Mean}_{T,1} = - \pi(\pi-1)$ and $\operatorname{Mean}_{T,2} = \pi$. We get

$$\begin{aligned} \operatorname{Mean}_{\rm bump} =& \operatorname{Mean}_{S} + \operatorname{Mean}_{T,1} + \operatorname{Mean}_{T,2} \\ =& 2 \pi- \pi(\pi- 1) + \pi, \end{aligned}$$

(54)

which evaluates to (4−π)π, as claimed.