Theorie added

exercise2/Theorie.tex

@@ -60,5 +60,5 @@ Aus der Vorlesung: $dA = \sqrt{det I_{\tau}^f} \,du\,dv$\\
 \end{align*}
 \\
 $\Rightarrow A = \int_{0}^{2\pi} \int_{0}^{2\pi}  r(r(cos(u))+R) \,du\,dv = 2 \pi r \times  \int_{0}^{2\pi}  r(cos(u))+R \,dv =\\\\ 2 \pi r [Rv + r(sin(v)]_{0}^{2\pi} = 2 \pi r \times (2\pi R) = \underline{\underline{4\pi^2Rr}} $
-t\end{document}
+\end{document}
 

exercise3/Theorie.tex

@@ -0,0 +1,52 @@
+\documentclass[12pt, a4paper]{article}
+
+%packages
+\usepackage[ngerman]{babel}
+\usepackage[utf8x]{inputenc}
+%Formel packages
+\usepackage{amsmath}
+\usepackage{amsthm}
+\usepackage{amsbsy}
+\usepackage{amssymb}
+
+
+\usepackage{enumerate}
+\usepackage{geometry}
+\geometry{a4paper, top=25mm, left=18mm, right=15mm, bottom=30mm,
+headsep=10mm, footskip=12mm}
+
+
+\begin{document}
+
+\section*{Computer Graphics - Excercise 2}
+\subsection*{3.1.1}
+\subsection*{(a)}
+The more obtuse the triangle, the more acute the angle of the corner of the voronoi area gets. Because the both sides of this corner has to be in a 90° angle to the sides of the traingle. 
+\subsection*{(b)}
+The formula of 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 $b=$
+
+\subsection*{(c)}
+
+\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 $e_h = 2*e = 6*v$.\\
+If we add the memory for every part it results in a formula:\\
+\begin{center}
+memory $= v*16\text{bytes} + e*4\text{bytes} + e_h*16\text{bytes} + f*4\text{bytes}$\\
+$\rightarrow \text{through assumptions:}$\\
+memory $= v*(16+3*4+6*16+2*4)\text{bytes}=v*132\text{bytes}$
+\end{center}
+\subsection*{(b)}
+Because in a quad mash two triangles are combined into one quad, the resulting faces will be reduced by a half. The resulting ratio will be 1:3:1 (for v : e : f)
+\subsection*{(c)}
+We know every vertex($v$) has three edges($e$) and one face($f$). In addition every edge($e$) can separated into two half edges ($e_h$), this means $e_h = 2*e = 6*v$.\\
+If we add the memory for every part it results in a formula:\\
+\begin{center}
+memory $= v*16\text{bytes} + e*4\text{bytes} + e_h*16\text{bytes} + f*4\text{bytes}$\\
+$\rightarrow \text{through assumptions:}$\\
+memory $= v*(16+3*4+6*16+1*4)\text{bytes}=v*132\text{bytes}$
+\end{center}
+\end{document}
+