diff --git a/mr/UB3/CalcPhi.png b/mr/UB3/CalcPhi.png
new file mode 100644
index 0000000..6ae9458
--- /dev/null
+++ b/mr/UB3/CalcPhi.png
Binary files differ
diff --git a/mr/UB3/mr3.pdf b/mr/UB3/mr3.pdf
index 7c29377..b70f701 100644
--- a/mr/UB3/mr3.pdf
+++ b/mr/UB3/mr3.pdf
Binary files differ
diff --git a/mr/UB3/mr3.tex b/mr/UB3/mr3.tex
index 7988b0e..36172e5 100644
--- a/mr/UB3/mr3.tex
+++ b/mr/UB3/mr3.tex
@@ -170,54 +170,54 @@
 			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\\
+			$ \varPsi_1	=  tan^{-1}(\frac{0.5}{0.5}) = 45^\circ\\
+			v_1= 0.5*\pi= 1.5708 \frac{m}{s}\\
 			u_1=\begin{pmatrix}
-			90^\circ\\
+			45^\circ\\
 			1.5708 \frac{m}{s}
 		\end{pmatrix} $
 		\item 	
 			$u_2=-u_1=\begin{pmatrix}
-			-90^\circ\\
+			-45^\circ\\
 			-1.5708 \frac{m}{s}
 			\end{pmatrix} $
 		
 	\end{itemize}	
-	90 Grad Einschlagwinkel ist nicht möglich...
+
 \end{enumerate}
 	
 	
 	\Aufgabe{}{4}
-	
+	$ r=1m \\
+	l_\omega = 0.3m\\
+	l_a=0.5m
+	$
+	\begin{enumerate}[(a)]
+		\item 
+		Assumed left rotation:\\
+		$\varPsi_l = tan^{-1}(\frac{l_a}{r-0,5l_\omega}) = tan^{-1}(\frac{0.5}{1-0.5*0.3})= 30.47^\circ\\
+		\varPsi_r = tan^{-1}(\frac{l_a}{r+0,5l_\omega}) = tan^{-1}(\frac{0.5}{1+0.5*0.3})=23.5^\circ
+		$
+		
+		
+			\begin{figure}[!htb]
+			
+				\centering
+				\includegraphics[width=0.6\textwidth]{CalcPhi.png}
+				\caption{\label{fig:Psi01} Determination of $\varPsi_r$ and $\varPsi_l$}
+				
+			\end{figure}
+		\item
+		$sl = 2*PI* (1-0.5*0.3) = 5.341m	\\
+		sr = 2*PI* (1+0.5*0.3)  =7.225m\\
+		sl -sr = 1.885m
+		$
+			
+	\end{enumerate}
+
 	\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}
+