Implicit surfaces from polygon soup with compactly supported radial basis functions

Abstract

This paper presents a method for generating implicit surfaces from polygon soups based on compactly supported radial basis functions (CSRBFs). The surface is represented as the zero level set of an implicit function which interpolates the polygonal data with their outward normal constraints. By specifying two parameters, the support size and the shape parameter, users can flexibly control the accuracy of the reconstructed surfaces. For determining coefficients of RBFs, our method uses a quasi-interpolation framework to avoid solving a large linear system, which allows processing large meshes efficiently and robustly. Moreover, a relationship between the shape parameter and the support radius is provided for the quasi-solution validity, and an error bound of the reconstructed surfaces approximating the original models is deduced through the rigorous theoretical analysis.

Appendix: Integral over the triangles

Here, we discuss the following integrals:

$$\begin{aligned} \begin{aligned}&\underset{\varDelta ABC}{\int }\phi (||\mathbf{q }-\mathbf{p }||) \,{\hbox {d}}\mathbf{p }, \underset{\widehat{q_{1}Oq_{2}}}{\int }\phi (||\mathbf{q }-\mathbf{p }||) \,{\hbox {d}}\mathbf{p }. \end{aligned} \end{aligned}$$

For the triangle region $\varDelta \mathbf ABC$ in Fig. 2, we define

$$\begin{aligned} \begin{aligned} h&=R^{2}+\lambda ||\mathbf{Aq}||^{2},l_{1}=||\mathbf{AC}||, l_{2}=||\mathbf{AB}||,\\ a&=\mathbf{Aq}\cdot \frac{\mathbf{AC}}{l_{1}}, b=\mathbf{Aq}\cdot \frac{\mathbf{AB}}{l_{2}},c=\cos A. \end{aligned} \end{aligned}$$

For the sector region $\widehat{q_{1}Oq_{2}}$ in Fig. 2, we define

$$\begin{aligned} r=\sqrt{R^{2}-||\mathbf{Oq }||^{2}},\mathbf e _{1}=\frac{\mathbf{Oq }_{2}}{r}, \mathbf e _{2}=\mathbf{n }\times \mathbf e _{1}, \end{aligned}$$

(9)

where $\mathbf{n }$ is the unit normal vector of the triangle.

• 1.
  $$\begin{aligned} \begin{aligned}&\underset{\varDelta ABC}{\int }\phi (||\mathbf{q }-\mathbf{p }||) \,{\hbox {d}}\mathbf{p }=\delta _{1}^{1}+\delta _{2}^{1}+\delta _{3}^{1}. \delta _{1}^{1}\\&\quad =\frac{l_{1}l_{2}\sin A}{\lambda ^{2}R^{2}} (\frac{h}{2}-\frac{1}{3}\lambda al_{1}-\frac{1}{3}\lambda bl_{2}+ \frac{1}{12}\lambda l_{1}^{2}+\frac{1}{12}\lambda l_{2}^{2}\\&\qquad +\frac{1}{12}\lambda cl_{1}l_{2}). \delta _{2}^{2}\\&\quad =-\frac{l_{1}l_{2}}{\lambda }(1+\frac{1}{\lambda })\sin A. \delta _{3}^{1}\\&\quad =R^{2}(1+\frac{1}{\lambda })^{2}l_{1}l_{2}\sin A\delta _{4}^{1}. \delta _{4}^{1}\\&\quad =\int _{0}^{1}\,du\int _{0}^{1-u}\frac{1}{\delta (u,v)}\,{\hbox {d}}v. \delta (u,v)\\&\quad =h+\lambda l_{2}^{2}u^{2}+\lambda l_{1}^{2}v^{2}-2\lambda bl_{2}u-2\lambda al_{1}v+ 2\lambda cl_{1}l_{2}uv. \end{aligned} \end{aligned}$$
• 2.
  $$\begin{aligned} \begin{aligned}&\underset{\widehat{q_{1}Oq_{2}}}{\int }\phi (||\mathbf{q }-\mathbf{p }||) \,{\hbox {d}}\mathbf{p }=\delta _{1}^{2}+\delta _{2}^{2}+\delta _{3}^{2}. \delta _{1}^{2}\\&\quad =\frac{\alpha r^{2}}{4\lambda ^{2}R^{2}} ((2+\lambda )R^{2}+\lambda ||\mathbf{Oq }||^{2}). \delta _{2}^{2}\\ {}&\quad =-\frac{\alpha }{\lambda }(1+\frac{1}{\lambda })r^{2}. \delta _{3}^{2}\\&\quad =\frac{\alpha R^{2}}{2\lambda }(1+\frac{1}{\lambda })^{2} \ln \frac{R^{2}(1+\lambda )}{R^{2}+\lambda ||\mathbf{Oq }||^{2}}. \end{aligned} \end{aligned}$$

In I, we cannot get a closed-form solution of the two-dimensional integral $\delta _{4}^{1}$. Defining

$$\begin{aligned}&W=h+\lambda l_{2}^{2}(1-c^{2})u^{2}+2(ac-b)\lambda l_{2}u-\lambda a^{2},\\&\beta _{1}=cl_{2}u-a,\\&\beta _{2}=l_{1}(1-u)+cl_{2}u-a, \end{aligned}$$

then, we have

$$\begin{aligned} \int _{0}^{1}\,du\int _{0}^{1-u}\frac{1}{\delta (u,v)}\,{\hbox {d}}v= \int _{0}^{1}\,du\int _{\beta _{1}}^{\beta _{2}}\frac{1}{W+\lambda t^{2}}\,{\hbox {d}}t. \end{aligned}$$

For the integral $\tau (u) = \int _{\beta _{1}}^{\beta _{2}}\frac{1}{W+\lambda t^{2}}\,{\hbox {d}}t$, we can obtain the following results according to the value of W.

• 1. If $W>0$,then

  $$\begin{aligned} \tau (u) = \frac{1}{\sqrt{\lambda W}} \left( \arctan \left( \sqrt{\frac{\lambda }{W}}\beta _{2}\right) -\arctan \left( \sqrt{\frac{\lambda }{W}}\beta _{1}\right) \right) . \end{aligned}$$

• 2. If $W=0$,then

  $$\begin{aligned} \tau (u) = \frac{1}{\lambda } \left( \frac{1}{\beta _{1}}-\frac{1}{\beta _{2}}\right) . \end{aligned}$$

• 3. If $W<0$,then

  $$\begin{aligned} \tau (u) = \frac{1}{2\sqrt{-\lambda W}} \left( \ln \left|\frac{\sqrt{\frac{-\lambda }{W}}\beta _{2}-1}{\sqrt{\frac{-\lambda }{W}}\beta _{2}+1}\right|-\ln \left|\frac{\sqrt{\frac{-\lambda }{W}}\beta _{1}-1}{\sqrt{\frac{-\lambda }{W}}\beta _{1}+1}\right|\right) . \end{aligned}$$

After that, the integral $\int _{0}^{1}\tau (u)\,{\hbox {d}}u$ is computed with a numerical method (the composite Simpson rule [24]) and the final equations are not given out here.