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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}
+
+\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{1cm}
+\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)\\}
+
+
+
+
+\begin{document}
+    %\header{BlattNr}{Tutor}{Abgabedatum}{Vorlesungsname}{Namen}{Semester}{Anzahl Aufgaben}
+    \header{3}{}{2015-12-05}{Mobile Robots}{
+    	\textit{Jan-Peter Hohloch}\\ \textit{Maximus Mutschler}
+    }{SS 15}{3}
+    \vspace{1cm}
+
+	\Aufgabe{}{8}
+	\begin{enumerate}[(a)]
+
+	\item
+	\begin{tikzpicture}
+	\draw[->] (1,0) -- (-3,0) node[anchor=north]{$x$ in m};
+	\draw[->] (0,0) -- (0,2) node[anchor=east] {$y$ in m};
+	
+	\draw	(0,0) node[anchor=north] {0}
+	
+	(-1,0) node[anchor=north] {-1}
+	(-2,0) node[anchor=north] {-2}
+	;
+	
+		\draw	(0,1) node[anchor=east] {1};
+	
+	\coordinate (A) at (0,0);
+	\coordinate (B) at (0,1);
+	\coordinate (C) at (-1,1);
+	\coordinate (D) at (-2,1);
+		
+	\draw [fill=blue] (A) circle (2pt) node [left] {x1};
+	\draw [fill=blue] (B) circle (2pt) node [right] {x2};
+	\draw [fill=blue] (C) circle (2pt) node [left] {x3};
+	\draw [fill=blue] (D) circle (2pt) node [left] {x4};	
+
+	\draw[red,->] (A) -- (0,1) ;
+	 \draw[blue,->] (A) -- (0,0.5) ;
+	\draw[blue,->] (B) -- (0,1.5) ;
+	\draw[blue,->] (C) -- (-1,0.5) ;
+	\draw[blue,->] (D) -- (-2,1.5) ;
+	
+	\draw[red,->] (B) arc (0:180:0.5) ;
+	\draw[red,->] (C) arc (0:180:0.5) ;
+
+	%TODO Trajectories
+	\end{tikzpicture}
+	\item $u_t= \begin{pmatrix}
+	\frac{2v_t-l\omega_t}{2}\\
+	\frac{2v_t+l\omega_t}{2}\\
+	\end{pmatrix}$
+	\\ $l=0.2m$ \\
+		\begin{itemize}
+		 \item$ v_0=1\\
+		 \omega_0=0 
+		 \\u_0= \begin{pmatrix}
+			\frac{2*1}{2}\\
+			\frac{2*1}{2}\\
+			\end{pmatrix}=
+			\begin{pmatrix}
+				1\\
+				1
+			\end{pmatrix} \frac{m}{s}$
+		\item$ \Delta s=r * \Delta \varTheta\\
+		r=0.5,\Delta \varTheta= \pi \\
+		 v_1= \frac{\Delta s}{\Delta t_1} 
+		= 0.5\pi\\
+		\omega_1=\frac{\Delta \varTheta}{\Delta t_1} 
+		 = \pi
+		\\
+		u_1= \begin{pmatrix}
+		\frac{2*0.5*\pi-0.2*\pi}{2}\\
+		\frac{2*0.5*\pi+0.2*\pi}{2}\\
+		\end{pmatrix} = \begin{pmatrix} 1.2567\\ 1.885 \end{pmatrix} \frac{m}{s}
+		$	
+			\item$ \Delta s=r * \Delta \varTheta\\
+			r=0.5,\Delta \varTheta= -\pi \\
+			v_2= \frac{\Delta s}{\Delta t_2} 
+			= -0.5\pi\\
+			\omega_2=\frac{\Delta \varTheta}{\Delta t_2} 
+			= -\pi
+			\\
+			u_2= \begin{pmatrix}
+			\frac{-2*0.5*\pi+0.2*\pi}{2}\\
+			\frac{-2*0.5*\pi-0.2*\pi}{2}\\
+			\end{pmatrix} = \begin{pmatrix} -1.2567\\ -1.885 \end{pmatrix} \frac{m}{s}
+			$	
+		\end{itemize}
+			
+	\item $l_\omega = 0.3m$ irrelevant\\ $l_a = 0.5m \\
+ \varPsi	=  tan^{-1}(\frac{l_a}{r_t})\\
+ v_t= \frac{\Delta s}{\Delta t}
+	$
+	\begin{itemize}
+		\item 	$u_1=\begin{pmatrix}
+			0^\circ\\
+			1 \frac{m}{s}
+		\end{pmatrix} $
+		\item 
+			$ \varPsi_1	=  tan^{-1}(\frac{0.5}{0.5}) = 90^\circ\\
+			v_1= 0.5*\pi= 90^\circ\\
+			u_1=\begin{pmatrix}
+			90^\circ\\
+			1.5708 \frac{m}{s}
+		\end{pmatrix} $
+		\item 	
+			$u_2=-u_1=\begin{pmatrix}
+			-90^\circ\\
+			-1.5708 \frac{m}{s}
+			\end{pmatrix} $
+		
+	\end{itemize}	
+	90 Grad Einschlagwinkel ist nicht möglich...
+\end{enumerate}
+	
+	
+	\Aufgabe{}{4}
+	
+	\Aufgabe{}{8}
+
+    \Aufgabe{}{6}
+        \begin{enumerate}[(a)]
+            \item Forward kinematic velocity: calculation of the robots' current pose with knowledge about the velocity and position of the last pose  \\
+            inverse kinematic velocity: given a pose the velocity and position needed to reach this pose are calulated
+            \item $^R\mathbf{v}=\begin{pmatrix}
+                    0.2\frac{m}{s} \\ 1 \frac{1}{s}
+                \end{pmatrix}$, $l=0.2m$\\
+                \begin{align}
+                    v_l &= \frac{2\cdot 0.2-0.2}{2}&=0.1\\
+                    v_r &= \frac{2\cdot 0.2+0.2}{2}&=0.3
+                \end{align}
+                $\Rightarrow \mathbf{u}_t=\left(0.1\frac{m}{s}, 0.3\frac{m}{s}\right)^T$
+            \item \begin{align}
+                    v &= \frac{0.3+0.5}{2}\frac{m}{s} &=0.4\frac{m}{s}\\
+                    \omega &= \frac{0.5-0.3}{0.2}\frac{1}{s}&=1\frac{1}{s}
+                \end{align}
+                    $\Rightarrow \mathbf{v}_t=\left(0.4\frac{m}{s},1\frac{1}{s}\right)$
+            \item \begin{align}
+                    \frac{1}{2}v_r+\frac{1}{2}v_l &=v\\
+                    \frac{1}{l}v_r -\frac{1}{l}v_l &=\omega
+                \end{align}$\Leftrightarrow$\begin{align}
+                    v_r+v_l &= 2v\\
+                    v_r-v_l &=l\omega\\
+                \end{align}
+                $\Rightarrow 2v_r =2v+l\omega\Leftrightarrow v_r=\frac{2v+l\omega}{2}$\\
+                $\Rightarrow \frac{2v+l\omega}{2}+v_l =2v \Leftrightarrow v_l=2v-\frac{2v+l\omega}{2}=\frac{2v-l\omega}{2}$\qed
+        \end{enumerate}
+     
+    
+
+\end{document}
+