Emulation and Simulation of
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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 \(\Phi (\vec {O})\):
\( \seteqsection {A} \)
\begin{equation} \vec {P} = \vec {O} - \hat {n} \frac {\Phi (\vec {O})}{\mid \nabla \Phi (\vec {O}) \mid } \end{equation}
The plane approximating the surface inside the cell, \(p(\vec {x})\) is therefore defined as
\( \seteqsection {A} \) \( \seteqnumber {2} \)
\begin{equation} \hat {n} \cdot (\vec {x} - \vec {P}) = 0 \end{equation}
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
\( \seteqsection {A} \) \( \seteqnumber {3} \)
\begin{equation} p = \frac {\hat {n} \cdot \vec {P}}{n_y} - \frac {n_x}{n_y} x = q - \frac {n_x}{n_y} x \end{equation}
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:
\( \seteqsection {A} \) \( \seteqnumber {4} \)
\begin{equation} a = p(y = 0) = \frac {n_y q}{n_x} = \frac {\hat {n} \cdot \vec {P}}{n_x} \quad , \end{equation}
\( \seteqsection {A} \) \( \seteqnumber {5} \)
\begin{equation} 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 , \end{equation}
where both a and b are bound to the interval [0, 1]. The area A is then given by
\( \seteqsection {A} \) \( \seteqnumber {6} \)
\begin{equation} A = \int _{0}^{b} dx + \int _{b}^{a} q - \frac {n_x}{n_y} x dx \end{equation}
Solving this integral gives an expression for the area under curve within the cell, i.e. the filling fraction:
\( \seteqsection {A} \) \( \seteqnumber {7} \)
\begin{equation} A = b + \frac {\hat {n} \cdot \vec {P}}{n_y}(a-b) - \frac {n_x}{2n_y}(a^2 - b^2) \end{equation}
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