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# Emulation and Simulation of Microelectronic Fabrication Processes

### A Calculating Filling Fractions from Level Set Values

Consider a single cell in a grid. In its centre, there is a point of a level set grid, $$\vec {O} = (0.5, 0.5, 0.5)$$. In order to approximate a surface corresponding to this level set point, a plane is constructed using $$\hat {n}$$, the normalised normal vector of the level set at $$\vec {O}$$. A point on the plane, $$\vec {P}$$ is then found by shifting $$\vec {O}$$ in the direction of $$\hat {n}$$ by the level set value at $$\Phi (\vec {O})$$:

$$\vec {P} = \vec {O} - \hat {n} \frac {\Phi (\vec {O})}{\mid \nabla \Phi (\vec {O}) \mid }$$

The plane approximating the surface inside the cell, $$p(\vec {x})$$ is therefore defined as

$$\hat {n} \cdot (\vec {x} - \vec {P}) = 0$$

Due to the symmetry of the cubic cell, the normal vector can always be mapped into one half/quarter of the first quadrant/octant, bounding the polar angles by $$0 < \theta \le \frac {\pi }{4}$$ and $$0 < \phi \le \frac {\pi }{4}$$.

#### A.1 2D Problem

In two dimensions the filling fraction is the area below the line describing the surface within the cell shown in Fig. A.1. This line is given by

$$p = \frac {\hat {n} \cdot \vec {P}}{n_y} - \frac {n_x}{n_y} x = q - \frac {n_x}{n_y} x$$

Therefore, the integral is bound to $$(0, 0) \le \vec {x} \le (1, 1)$$. However, the line might intersect the x-aligned edge of the cell at other points, for $$y = 0$$ and $$y = 1$$. These intersections, $$a$$ and $$b$$, are defined by:

$$a = p(y = 0) = \frac {n_y q}{n_x} = \frac {\hat {n} \cdot \vec {P}}{n_x} \quad ,$$

$$b = p(y = 1) = \frac {n_y}{n_x}q - \frac {n_y}{n_x} = \frac {\hat {n} \cdot \vec {P} - n_y}{n_x} \quad ,$$

where both a and b are bound to the interval [0, 1]. The area A is then given by

$$A = \int _{0}^{b} dx + \int _{b}^{a} q - \frac {n_x}{n_y} x dx$$

Solving this integral gives an expression for the area under curve within the cell, i.e. the filling fraction:

$$A = b + \frac {\hat {n} \cdot \vec {P}}{n_y}(a-b) - \frac {n_x}{2n_y}(a^2 - b^2)$$