Marbling-based creative modelling

Abstract

Most mathematical marbling simulations generate patterns for texture mapping and surface decoration. We explore the application of three-dimensional deformations inspired by mathematical marbling as a suite of tools to enable creative shape design. Our tools are expressed as analytical functions of space and are volume-preserving vector fields, meaning that the modelling process preserves volumes and avoids self-intersections. Complicated deformations are easily combined to create complex objects from simple ones. To achieve smooth and high-quality shapes, we also present a mesh refinement and simplification algorithm adapted to our deformations. We show a number of examples of shapes created with our technique in order to demonstrate its power and expressiveness.

Appendices

Appendix 1: Volume-preserving deformations

In this section, we argue that the tools we define lead to volume-preserving deformations of space. We adopt seven transform functions. Of these, the line, comb, shell, vortex, wave and welding patterns are homeomorphisms (i.e. continuous bijections) of $\mathbb {R}^3$, but the sphere insertion tool is not. For a continuously differentiable vector field $\varvec{u}$, the volume form is preserved under the flow of a solenoidal vector field $\bigtriangledown \cdot \varvec{u} = 0$ [1].

We can regard our tools as vector fields by considering the vector by which they displace every point in space. That is, we define a vector field $\varvec{u}(x,y,z)=f(\varvec{P})-\varvec{P}$. For each of the first five tools mentioned above, it is easy to prove that its divergence equals zero.

Fig. 10

Sphere insertion tool satisfies the volume-preserving property

The divergence of the welding function is not well defined as x, y or z goes to infinity. So we cannot use the divergence test for the welding function. But it is volume preserving for bounded x, y, z. Slice 3-space into infinitesimally thin slices perpendicular to the twist axis. Each slice is then simply rotated around the twist axis. Because the rotation does not change the areas in each slice, the volumes are preserved.

The sphere insertion function is not continuously differentiable around the injection point $\varvec{C}$, so its divergence is not well defined. However, we can show that this transform preserves the volume of all neighbourhoods not containing $\varvec{C}$. As shown in Fig. 10, consider the neighbourhood having solid angle $\varOmega$ of the spherical shell centred on $\varvec{C}$ having inner radius a and outer radius b. Its volume is $(4/3)\pi \varOmega (b^3-a^3)$. If a sphere of radius e is injected at $\varvec{C}$, the new spherical region centred on $\varvec{C}$ will have volume $(4/3)\pi e^3$ and increase the radius from $\varvec{C}$ of all other points. The radially symmetric expansion does not change the solid angle $\varOmega$. The radius of the outer shell increases from b to $\root 3 \of {b^3+e^3}$; the inner shell radius increases from a to $\root 3 \of {a^3+e^3}$. Because $(b^3+e^3)-(a^3+e^3)=b^3-a^3$, the volume of this neighbourhood remains $(4/3)\pi \varOmega (b^3-a^3)$. Neighbourhoods with other shapes can be assembled from these shell fragments, each preserving its volume under injection, so long as each fragment does not include $\varvec{C}$. Therefore, the sphere insertion function is volume preserving at all locations except the point of injection, $\varvec{C}$.

Appendix 2: Exponential decay mode

Consider an unbounded plane containing a two-dimensional incompressible laminar flow. Given x, y as the coordinates of the point $\varvec{P}$, associated with f(x, y) is a vector field $\varvec{H}(x,y)=\varvec{F}(x,y)-(x,y)$ returning the vector displacement at each coordinate. Along the y axis, we introduce a displacement $\alpha , \varvec{H}(0,y)=(0,\alpha )~~~[\varvec{F}(0,y)=(0,y+\alpha )]$. This displacement will not affect points far away from the x axis; so the limit of $\varvec{H}(x,y)$ tends to zero as the magnitude of x grows. Because the fluid is incompressible, the divergence of $\varvec{H}$ is zero everywhere. Because its flow is laminar, it is uniform in the direction of motion, y: $\frac{\partial \varvec{H}_y}{\partial y} = 0.$

Thus, $\varvec{H}(x,y)$ depends only on x. Furthermore, only the y component of $\varvec{H}(x,y)$ depends on x. Let $f(x)=\varvec{H}_y(x,0)$; then, $f(0)=\alpha$. f(x) is even; the displacements to either side of $x=0$ will be equal and in the same direction. So we will consider f(x) for $x\ge 0$ only. Because the flow is laminar, displacements induced by $\alpha$ travel along the x axis should be proportional to $\alpha$. Let $A=f(b)$. Then, f(2b) will be reduced from A by the same proportion as A was reduced from $\alpha$:$f(2b) = \frac{A^2}{\alpha }$. Thus, $\frac{f(2b)}{f(0)} = \frac{A^2}{\alpha ^2} = \frac{f(b)^2}{f(0)^2}$.

The only continuous real functions satisfying these constraints are $f(x)=\alpha \lambda ^{|x|}$ with independent parameter $0<\lambda <1$ related to the viscosity.