theorie magic

exercise3/Theorie.tex

@@ -25,10 +25,24 @@ The more obtuse the triangle, the more acute the angle of the corner of the voro
 \subsection*{(b)}
 The formula for calculating the surface of a triangle is: $\frac{1}{2}*h*b$
 As for the red triangle: $\frac{1}{2}*h*b$
-with: $h=\frac{||p_i - p_j||}{2}$ and $b=b_1 + b_2$ where $b_1$ is left part of the baseline facing $\alpha_{ij}$ and $b_2$ is the right one facing $\beta_{ij}$.\\
-
-\subsection*{(c)}
-
+We define: $h=\frac{||\underline{p}_i - \underline{p}_j||}{2}$ and $b=b_1 + b_2$ where $b_1$ is left part of the baseline facing $\alpha_{ij}$ and $b_2$ is the right one facing $\beta_{ij}$. In addition we define the vertex of the traingle facing $\alpha_{ij}$ to be $A$ and the vertex of the triangle facing $\beta_{ij}$ to be $B$. The distance between $p_i$ and $A$ (R) is equal to $\frac{||\underline{p}_i-\underline{p}_j||}{2 sin(\alpha_{ij}}$. It can be assumed that the angle $\underline{p}_S \underline{A} \underline{p}_i=\alpha_{ij}$ with $p_S$ as the vertex where the red triangle crosses the edge from $\underline{p}_i$ to $\underline{p}_j$. As a result it can be assumed that $cos(\alpha_{ij})= \frac{b_1}{R}$ and also $b_1 = cos(\alpha_{ij}) \times R$\\
+As a result the surface of the left part of the red triangle can be calculated:
+\begin{align*}
+A_{\alpha}(i,j)&=\frac{1}{2}\times b_1 \times \frac{||\underline{p}_i- \underline{p}_j||}{2}\\
+A_{\alpha}(i,j)&=\frac{1}{4}\times cos(\alpha_{ij}) \times R\times ||\underline{p}_i- \underline{p}_j||\\
+A_{\alpha}(i,j)&=\frac{1}{4}\times \frac{cos(\alpha_{ij})}{2\times sin(\alpha_{ij})} \times R \times {||\underline{p}_i- \underline{p}_j||}^2\\
+A_{\alpha}(i,j)&=\frac{1}{8}\times cot(\alpha_{ij}) \times R \times {||\underline{p}_i- \underline{p}_j||}^2\\
+\end{align*}
+The same goes for the right side and as a result if we combine both:
+\begin{align*}
+A(i,j)&=\frac{1}{8}\times cot(\alpha_{ij}) \times R \times {||\underline{p}_i- \underline{p}_j||}^2 + \frac{1}{8}\times cot(\beta_{ij}) \times R \times {||\underline{p}_i- \underline{p}_j||}^2\\
+A(i,j)&=\frac{1}{8}\times (cot(\alpha_{ij})+cot(\beta_{ij})) \times R \times {||\underline{p}_i- \underline{p}_j||}^2\\
+\end{align*}
+If we sum up all the triangles T with the index $j$ of the voronoi-area this is the result:
+\begin{align*}
+A_{voronoi-area}(i)&=\sum_{j\in T(i)} \frac{1}{8}\times (cot(\alpha_{ij})+cot(\beta_{ij})) \times R \times {||\underline{p}_i- \underline{p}_j||}^2\\
+A_{voronoi-area}(i)&=\frac{1}{8}\times \sum_{j\in T(i)} (cot(\alpha_{ij})+cot(\beta_{ij})) \times R \times {||\underline{p}_i- \underline{p}_j||}^2\\
+\end{align*}
 \subsection*{3.1.2}
 \subsection*{(a)}
 We know every vertex($v$) has three edges($e$) and 2 faces($f$). In addition every edge($e$) can separated into two half edges ($e_h$), this means $e_h = 2*e = 6*v$.\\