Visual simulation of crack and bend generation in deteriorated films coated on metal objects: Combination of static fracture and position-based deformation

Abstract

Weathering, an expression of degradation caused by rain and wind, is essential for photorealistic computer graphics. One of the commonest targets of weathering is metal, which is omnipresent in reality. However, for the realistic reproduction of scenes, many of which display degradation, the application of rust-proof paint to metal surfaces cannot be ignored. In our study, we propose a weathering method for coated films on metal objects, which are modeled using a three-dimensional (3D) triangular polygon mesh and deformed by combining two kinds of simulations: static simulation, for determining fractures based on the balance of the internal forces, and position-based bend simulation for moving vertices according to geometric constraints. Our method can digitally reproduce the deterioration of coated films using complex 3D deformation, which is difficult to achieve by material manipulation only.

Appendix

We apply the triangle bending constraint in [14] to describe an angle constraint as a combination of simple length constraints to achieve high-quality simulations with fast convergence. As shown in Fig. A1a, a triangle bending constraint controls the angle formed by two connected edges by setting a length constraint on the distance between the center of the triangle, which has the two edges as sides, and the junction point of the edges.

Figure A1b shows a tetrahedron bending constraint, which controls the angle formed by two adjacent faces. A tetrahedron bending constraint sets a length constraint on the distance between the center of a tetrahedron, which has the two faces as surfaces, and the edge shared by the faces. Notice we assume that each polygon has a shape similar to an equilateral triangle and the distance is equal to the distance between the center and midpoint of the edge. If the ends of the edges are $\varvec{p}_1, \varvec{p}_2$ and the other vertices are $\varvec{p}_3, \varvec{p}_4$, the middle point of the edge $\varvec{v}$ and the center of the tetrahedron $\varvec{c}$ are given as:

$$\begin{aligned} \begin{aligned} \varvec{v} =\frac{1}{2}(\varvec{p}_1+\varvec{p}_2),\quad \varvec{c} =\frac{1}{4}(\varvec{p}_1+\varvec{p}_2+\varvec{p}_3+\varvec{p}_4). \end{aligned} \end{aligned}$$

Fig. 20

Comparison of two bending constraints. While the target of a triangle bending constraint a is the angle formed by the edges, the target of a tetrahedron bending constraint b is the angle formed by the faces

Fig. 21

Division of a tetrahedron bending constraint

Therefore, if the target value of the distance between $\varvec{v}$ and $\varvec{c}$ is $d_0$, the tetrahedron bending constraint $C_{\textrm{tetrahedron}}$ is described as:

$$\begin{aligned}{} & {} C_{\textrm{tetrahedron}} (\varvec{p}_1,\varvec{p}_2,\varvec{p}_3,\varvec{p}_4)=\Vert \varvec{v} - \varvec{c}\Vert -d_0\\{} & {} \qquad =\frac{1}{4}\Vert \varvec{p}_1+\varvec{p}_2-\varvec{p}_3 -\varvec{p}_4\Vert -d_0. \end{aligned}$$

However, similar to the angle constraints described in Sect. 4.3, tetrahedron bending constraints must also be divided to distinguish between mountain and valley folds (Fig. A2). Given $l^1=\Vert \varvec{v}-\varvec{p}_3\Vert , l^2=\Vert \varvec{v}-\varvec{p}_4\Vert$, the target value of the angle ${\phi }_0\; (0<\phi _0<2\pi )$, and the positions on the neutral axis $\varvec{v}^1=\varvec{v}+l^1\varvec{n}_{\textrm{neutral}},\varvec{v}^2=\varvec{v}+l^2\varvec{n}_{\textrm{neutral}}$, two tetrahedron bending constraints $C_{\textrm{tetrahedron}}^1$ and $C_{\textrm{tetrahedron}}^2$ are defined as:

$$\begin{aligned}{} & {} C_{\textrm{tetrahedron}}^1 (\varvec{p}_1,\varvec{p}_2,\varvec{p}_3,\varvec{v}_1) =\frac{1}{4}\Vert \varvec{p}_1+\varvec{p}_2-\varvec{p}_3 -\varvec{v}^1\Vert -d^1, \\{} & {} C_{\textrm{tetrahedron}}^2 (\varvec{p}_1,\varvec{p}_2,\varvec{p}_4,\varvec{v}_2) =\frac{1}{4}\Vert \varvec{p}_1+\varvec{p}_2-\varvec{p}_4 -\varvec{v}^2\Vert -d^2, \end{aligned}$$

where the constraint offsets $d^1$ and $d^2$ are given as:

$$\begin{aligned} d^1=\frac{l^1}{4}\sqrt{2\left( 1+\cos {\frac{\phi _0}{2}}\right) },\quad d^2=\frac{l^2}{4}\sqrt{2\left( 1+\cos {\frac{\phi _0}{2}}\right) }. \end{aligned}$$

Because $\varvec{v}_1, \varvec{v}_2$ are virtual points, their inverse masses are set to 0. Let $\varDelta \varvec{p}_1^1, \varDelta \varvec{p}_2^1$, and $\varDelta \varvec{p}_3^1$ be the corrections for $\varvec{p}_1, \varvec{p}_2$, and $\varvec{p}_3$ according to $C_{\textrm{tetrahedron}}^1$, and let $\varDelta \varvec{p}_1^2, \varDelta \varvec{p}_2^2$, and $\varDelta \varvec{p}_4^2$ be the corrections for $\varvec{p}_1, \varvec{p}_2$, and $\varvec{p}_4$ according to $C_{\textrm{tetrahedron}}^2$, respectively. If the inverse mass of $\varvec{p}_i$ is denoted as $w_i$ and the sums of the inverse masses $W=w_1+w_2+w_3+w_4$, $W^1=W-w_4$, and $W^2=W-w_3$ are defined, then Eq. (3) yields corrections for the positions, given as:

$$\begin{aligned} \begin{aligned}&\varDelta \varvec{p}_1^1=-\frac{2u^1 w_1}{W^1}(\varvec{v}-\varvec{v}^1), \quad \varDelta \varvec{p}_1^2=-\frac{2u^2 w_1}{W^2}(\varvec{v}-\varvec{v}^2),\\&\varDelta \varvec{p}_2^1=-\frac{2u^1 w_2}{W^1}(\varvec{v}-\varvec{v}^1), \quad \varDelta \varvec{p}_2^2=-\frac{2u^2 w_2}{W^2}(\varvec{v}-\varvec{v}^2),\\&\varDelta \varvec{p}_3^1=+\frac{2u^1 w_3}{W^1}(\varvec{v}-\varvec{v}^1), \quad \varDelta \varvec{p}_4^2=+\frac{2u^2 w_4}{W^2}(\varvec{v}-\varvec{v}^2),\\ \end{aligned} \end{aligned}$$

where strains $u^1$ and $u^2$ are given as:

$$\begin{aligned} u^1=1-\frac{d^1}{\Vert \varvec{v}-\varvec{v}^1\Vert },\quad u^2=1-\frac{d^2}{\Vert \varvec{v}-\varvec{v}^2\Vert }. \end{aligned}$$

Considering that the ratio of corrections is equal to the ratio of the inverse mass if the constraint is not divided, we represent the corrections $\varDelta \varvec{p}_1, \varDelta \varvec{p}_2, \varDelta \varvec{p}_3$, and $\varDelta \varvec{p}_4$ as follows:

$$\begin{aligned} \begin{aligned}&\varDelta \varvec{p}_1=(\varDelta \varvec{p}_1^1+\varDelta \varvec{p}_1^2)\times \frac{W^1 W^2}{W(W^1+W^2)},\\&\varDelta \varvec{p}_2=(\varDelta \varvec{p}_2^1+\varDelta \varvec{p}_2^2)\times \frac{W^1 W^2}{W(W^1+W^2)},\\&\varDelta \varvec{p}_3=\varDelta \varvec{p}_3^1\times \frac{W^1}{W},\\&\varDelta \varvec{p}_4=\varDelta \varvec{p}_4^2\times \frac{W^2}{W}. \end{aligned} \end{aligned}$$