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+\documentclass[a4paper,12pt]{scrartcl}
+\usepackage[ngerman]{babel}
+\usepackage{graphicx} %BIlder einbinden
+\usepackage{amsmath} %erweiterte Mathe-Zeichen
+\usepackage{amsfonts} %weitere fonts
+\usepackage[utf8]{inputenc} %Umlaute & Co
+\usepackage{hyperref} %Links
+\usepackage{ifthen} %ifthenelse
+\usepackage{enumerate}
+\usepackage{pdfpages}
+\usepackage{algpseudocode} %Pseudocode
+\usepackage{dsfont} % schöne Zahlenräumezeichen
+\usepackage{amssymb, amsthm} %noch stärker erweiterte Mathe-Zeichen
+\usepackage{tikz} %TikZ ist kein Zeichenprogramm
+\usetikzlibrary{trees,automata,arrows,shapes}
+\usepackage{qtree}
+\usepackage{listings}
+\lstset{language=Matlab}
+
+\pagestyle{empty}
+
+
+\topmargin-50pt
+
+\newcounter{aufgabe}
+\def\tand{&}
+
+\newcommand{\makeTableLine}[2][0]{%
+  \setcounter{aufgabe}{1}%
+  \whiledo{\value{aufgabe} < #1}%
+  {%
+    #2\tand\stepcounter{aufgabe}%
+  }
+}
+
+\newcommand{\aufgTable}[1]{
+  \def\spalten{\numexpr #1 + 1 \relax}
+  \begin{tabular}{|*{\spalten}{p{1cm}|}}
+    \makeTableLine[\spalten]{A\theaufgabe}$\Sigma$~~\\ \hline
+    \rule{0pt}{15pt}\makeTableLine[\spalten]{}\\
+  \end{tabular}
+}
+
+\def\header#1#2#3#4#5#6#7{\pagestyle{empty}
+\begin{minipage}[t]{0.47\textwidth}
+\begin{flushleft}
+{\bf #4}\\
+#5
+\end{flushleft}
+\end{minipage}
+\begin{minipage}[t]{0.5\textwidth}
+\begin{flushright}
+#6 \vspace{0.5cm}\\
+%                 Number of Columns    Definition of Columns      second empty line
+% \begin{tabular}{|*{5}{C{1cm}|}}\hline A1&A2&A3&A4&$\Sigma$\\\hline&&&&\\\hline\end{tabular}\\\vspace*{0.1cm}
+\aufgTable{#7}
+\end{flushright}
+\end{minipage}
+\vspace{0.5cm}
+\begin{center}
+{\Large\bf Assignment #1}
+
+{(Hand-in date #3)}
+\end{center}
+}
+
+
+
+%counts the exercisenumber
+\newcounter{n}
+
+%Kommando für Aufgaben
+%\Aufgabe{AufgTitel}{Punktezahl}
+\newcommand{\Aufgabe}[2]{\stepcounter{n}
+\textbf{Exercise \arabic{n}: #1} (#2 Punkte)}
+
+\newcommand{\Normal}[3]{\mathcal{N}\left(#1,#2,#3\right)}
+\newcommand{\Normalf}[3]{\frac{\left|#3\right|^{-\frac{1}{2}}}{\sqrt{2\pi}}e^{-\frac{\left(#1 - #2 \right)^2\cdot \left(#3\right)^{-1}}{2}}}
+
+\begin{document}
+    %\header{BlattNr}{Tutor}{Abgabedatum}{Vorlesungsname}{Namen}{Semester}{Anzahl Aufgaben}
+    \header{12}{}{2015-07-21}{Mobile Robots}{
+        \textit{Jan-Peter Hohloch}\\ \textit{Maximus Mutschler}
+    }{SS 15}{3}
+    \vspace{0.2cm}
+    \Aufgabe{Moment of Inertia}{6}
+        \begin{enumerate}[(a)]
+            \item $I=m_c\begin{pmatrix}
+                0&0&0\\0&0&0\\0&0&0
+            \end{pmatrix}+m_m\left(\begin{pmatrix}
+                0&0&0\\0&0.04&0\\0&0&0.04
+            \end{pmatrix}+\begin{pmatrix}
+                0.04&0&0\\0&0&0\\0&0&0.04
+            \end{pmatrix}\right.\\\left.+\begin{pmatrix}
+                0&0&0\\0&0.04&0\\0&0&0.04
+            \end{pmatrix}+\begin{pmatrix}
+                0.04&0&0\\0&0&0\\0&0&0.04
+            \end{pmatrix}\right)\\=0.2kg\cdot m^2\begin{pmatrix}
+                0.08&0&0\\0&0.08&0\\0&0&0.16
+            \end{pmatrix}$
+            \item $I=m_c\begin{pmatrix}
+                0&0&0\\0&0&0\\0&0&0
+            \end{pmatrix}+m_m\left(\begin{pmatrix}
+                0.02&0.02&0\\0.02&0.02&0\\0&0&0.04
+            \end{pmatrix}+\begin{pmatrix}
+                0.02&-0.02&0\\-0.02&0.02&0\\0&0&0.04
+            \end{pmatrix}\right.\\\left.+\begin{pmatrix}
+                0.02&0.02&0\\0.02&0.02&0\\0&0&0.04
+            \end{pmatrix}+\begin{pmatrix}
+                0.02&-0.02&0\\-0.02&0.02&0\\0&0&0.04
+            \end{pmatrix}\right)\\=0.2kg\cdot m^2\begin{pmatrix}
+                0.08&0&0\\0&0.08&0\\0&0&0.16
+            \end{pmatrix}$\\
+            The result is the same, that however isn't very suprising since the Moment of Inertia should be independent of the rotation around the z-axis. %TODO: really?
+        \end{enumerate}
+    \Aufgabe{Hexacopter}{8}
+        \begin{enumerate}[(a)]
+            \item $^Bp_1=\begin{pmatrix}
+                1\\0\\0
+            \end{pmatrix}l,\ ^Bp_2=\begin{pmatrix}
+                \frac{1}{2}\\\frac{\sqrt{3}}{2}\\0
+            \end{pmatrix}l,\ ^Bp_3=\begin{pmatrix}
+                -\frac{1}{2}\\\frac{\sqrt{3}}{2}\\0
+            \end{pmatrix}l$
+        \end{enumerate}
+\end{document}
+