Approximation of the Geodesic Curvature and Applications for Spherical Geometric Subdivision Schemes

Abstract

Many applications of geometry modelling and computer graphics necessite accurate curvature estimations of curves on the plane or on manifolds. In this paper, we define the notion of the discrete geodesic curvature of a geodesic polygon on a smooth surface. We show that, when a geodesic polygon P is closely inscribed on a $C^2$-regular curve, the discrete geodesic curvature of P estimates the geodesic curvature of C. This result allows us to evaluate the geodesic curvature of discrete curves on surfaces. In particular, we apply such result to planar and spherical 4-point angle-based subdivision schemes. We show that such schemes cannot generate in general $G^2$-continuous curves. We also give a novel example of $G^2$-continuous subdivision scheme on the unit sphere using only points and discrete geodesic curvature called curvature-based 6-point spherical scheme.

A. Appendix

Let $P^j:=\{p_i^j \, \in {\mathbb {R}}^2 \}$ and

$$\begin{aligned} \alpha _{i}^{j}:=\sphericalangle (\overrightarrow{p_{i-1}^{j}p_{i+1}^{j}},\overrightarrow{p_{i-1}^{j}p_{i}^{j}}), \qquad \beta _{i}^{j}:=\sphericalangle (\overrightarrow{p_{i}^{j}p_{i+1}^{j}},\overrightarrow{p_{i-1}^{j}p_{i+1}^{j}}), \qquad \delta _{i}^{j}:=\sphericalangle (\overrightarrow{p_{i}^{j}p_{i+1}^{j}},\overrightarrow{p_{i-1}^{j}p_{i}^{j}}). \end{aligned}$$

The planar angle-based 4-point scheme [6] is defined by:

$$\begin{aligned} \left\{ \begin{aligned}&p_{2i}^{j+1}=p_{i}^{j}, \\&p_{2i+1}^{j+1}=\displaystyle p_i^j+\frac{1}{2cos(\alpha _{2i+1}^{j+1})}\, R(\alpha _{2i+1}^{j+1})\, (p_{i+1}^j-p_i^j),\\ \end{aligned} \right. \end{aligned}$$

(24)

where

$$\begin{aligned} \alpha _{2i+1}^{j+1}=\beta _{2i+1}^{j+1}=\displaystyle \frac{\delta _{i}^{j}+\delta _{i+1}^{j}}{8}, \end{aligned}$$

and $R(\alpha _{2i+1}^{j+1})$ is the rotation matrix by angle $\alpha _{2i+1}^{j+1}$. We have:

Proposition A.1

Let $P^0=\{p_{-2},p_{-1},p,p_1,p_2\}$ be a planar polygon with equal edge lengths. If $\delta _{-1}^0=\delta _1^0$ and $\delta _{0}^0 \ne \delta _{1}^0$ (see Fig. 13), then $\displaystyle \lim _{j \rightarrow \infty }\, \frac{\delta _0^{j}}{e_0^{j}}=\pm \infty$.

Fig. 13

A first iteration of the polygon $P^0$

Proof

In the planar case, Eq. (22) become

$$\begin{aligned} \left\{ \begin{aligned}&\delta _{2i-1}^{j+1}=\displaystyle \frac{\delta _{i-1}^j+\delta _{i}^j}{4},\\&\delta _{2i}^{j+1}=\frac{-\delta _{i-1}^j+6\delta _i^j-\delta _{i+1}^j}{8},\\&\delta _{2i+1}^{j+1}=\displaystyle \frac{\delta _i^j+\delta _{i+1}^j}{4}. \end{aligned} \right. \end{aligned}$$

(25)

If we set $A=\begin{pmatrix} 2&{}2&{}0\\ -1&{}6&{}-1\\ 0&{}2&{}2\\ \end{pmatrix}$, then the above system becomes

$$\begin{aligned} \displaystyle \begin{pmatrix} \delta _{2i-1}^{j+1}\\ \delta _{2i}^{j+1}\\ \delta _{2i+1}^{j+1}\\ \end{pmatrix}=\frac{1}{8} A \begin{pmatrix} \delta _{i-1}^{j}\\ \delta _{i}^{j}\\ \delta _{i+1}^j\\ \end{pmatrix}. \end{aligned}$$

Without losing generality, we can suppose that $i=0$. We get

$$\begin{aligned} \displaystyle \begin{pmatrix} \delta _{-1}^{j+1}\\ \delta _{0}^{j+1}\\ \delta _{1}^{j+1}\\ \end{pmatrix}=\frac{1}{8} A \begin{pmatrix} \delta _{-1}^{j}\\ \delta _{0}^{j}\\ \delta _{1}^j\\ \end{pmatrix}. \end{aligned}$$

Hence

$$\begin{aligned} \displaystyle \begin{pmatrix} \delta _{-1}^{j}\\ \delta _{0}^{j}\\ \delta _{1}^{j}\\ \end{pmatrix}=\frac{1}{2^{3j}} A^j \begin{pmatrix} \delta _{-1}^{0}\\ \delta _{0}^{0}\\ \delta _{1}^0\\ \end{pmatrix}. \end{aligned}$$

Finally we obtain

$$\begin{aligned} \left\{ \begin{aligned} \displaystyle&\delta _{-1}^j=\Big ( \frac{-j}{2^{j+2}}+\frac{1}{2^{j+1}}+\frac{1}{2^{2j+1}}\Big ) \delta _{-1}^0+\frac{j}{2^{j+1}}\delta _{0}^0+ \Big ( \frac{-j}{2^{j+2}}+\frac{1}{2^{j+1}}- \frac{1}{2^{2j+1}} \Big )\delta _{1}^1,\\&\delta _0^j=\frac{-j}{2^{j+2}}\delta _{-1}^0+\frac{6}{2^{j+2}}\delta _{0}^0-\frac{-j}{2^{j+2}}\delta _{1}^0,\\&\delta _1^j=\Big ( \frac{-j}{2^{j+2}}+\frac{1}{2^{j+1}}-\frac{1}{2^{2j+1}}\Big ) \delta _{-1}^0+\frac{j}{2^{j+1}}\delta _{0}^0+ \Big ( \frac{-j}{2^{j+2}}+\frac{1}{2^{j+1}}+ \frac{1}{2^{2j+1}} \Big )\delta _{1}^1.\\ \end{aligned} \right. \end{aligned}$$

(26)

Since $\delta _{-1}^0=\delta _1^0$, Eq. (26) become

$$\begin{aligned} \left\{ \begin{aligned}&\delta _0^j=\displaystyle \frac{2+j}{2^{j+1}} \delta _0^0-\frac{j}{2^{j+1}}\delta _1^0,\\&\delta _1^j= \frac{j}{2^{j+1}} \delta _0^0+\frac{2-j}{2^{j+1}}\delta _1^0.\\ \end{aligned} \right. \end{aligned}$$

On the other hand, we have $\alpha _1^j=\alpha _{-1}^j=\displaystyle \frac{\delta _0^{j-1}+\delta _1^{j-1}}{8}$, then

$$\begin{aligned} \alpha _1^j=\displaystyle \frac{j(\delta _0^0-\delta _1^0)}{2^{j+2}}+\frac{2}{2^{j+2}}\delta _1^0. \end{aligned}$$

(27)

Let $e_0^j:=\Vert p_1^j-p\Vert =\Vert p_{-1}^j-p\Vert$. Since $e_0^{j+1}=\displaystyle \frac{e_0^j}{2\, cos(\alpha _1^{j+1})}$, this yields

$$\begin{aligned} e_0^j=\displaystyle \frac{e_0^0}{2^j} \prod _{k=1}^{j} cos(\alpha _1^k)^{-1}. \end{aligned}$$

Finally

$$\begin{aligned} \displaystyle \frac{ \delta _0^j}{ e_0^j}&=\frac{\displaystyle \frac{2+j}{2^{j+1}} \delta _0^0-\frac{j}{2^{j+1}}\delta _1^0 }{\displaystyle \frac{e_0^0}{2^j} \prod _{k=1}^{j} cos(\alpha _1^k)^{-1}}\end{aligned}$$

(28)

$$\begin{aligned}&=\displaystyle \frac{ j(\delta _0^0-\delta _1^0) +2\delta _0^0 }{2e_0^0\, \displaystyle \prod _{k=1}^{j} cos(\alpha _1^k)^{-1}}. \end{aligned}$$

(29)

By (27), the series $\displaystyle \sum _{j}\, |\alpha _1^j|$ is convergent, then $\displaystyle \sum _{j}\, (\alpha _1^j)^2$ is also convergent. By equivalence, the product $\displaystyle \prod _{k=1}^{j} \Bigg ( 1-\frac{(\alpha _1^k)^2}{8}\Bigg )^{-1}$ is also convergent and so is $\displaystyle \prod _{k=1}^{j} cos(\alpha _1^k)^{-1}$. Finally, $\displaystyle \lim _{j \rightarrow \infty }\, \frac{\delta _0^{j}}{e_0^{j}}=\pm \infty$ once $\delta _{0}^0 \ne \delta _{1}^0$ (depends on the sign of $(\delta _0^0-\delta _1^0)$). $\square$