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\begin{center}
{\Large\bf Assignment #1}

{(Hand-in date #3)}
\end{center}
}



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%Kommando für Aufgaben
%\Aufgabe{AufgTitel}{Punktezahl}
\newcommand{\Aufgabe}[2]{\stepcounter{n}
\textbf{Exercise \arabic{n}: #1} (#2 Punkte)}

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\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 $^Bp_1=\begin{pmatrix}
                1\\0\\0
            \end{pmatrix}l,\ ^Bp_2=\begin{pmatrix}
                \frac{1}{2}\\\frac{\sqrt{3}}{2}\\0
            \end{pmatrix}l,\ ^Bp_3=\begin{pmatrix}
                -\frac{1}{2}\\\frac{\sqrt{3}}{2}\\0
            \end{pmatrix}l\\
            ^Bp_4=\begin{pmatrix}
                -1\\0\\0
            \end{pmatrix}l,\ ^Bp_5=\begin{pmatrix}
                -\frac{1}{2}\\-\frac{\sqrt{3}}{2}\\0
            \end{pmatrix}l,\ ^Bp_6=\begin{pmatrix}
                \frac{1}{2}\\-\frac{\sqrt{3}}{2}\\0
            \end{pmatrix}l$
            \item $T_\Sigma=k_T\sum_{i=1}^6 \omega_i^2$
            \item $\tau_Q=k_Q \begin{pmatrix}
                0\\0\\-\omega_1^2+\omega_2^2-\omega_3^2+\omega_4^2-\omega_5^2+\omega_6^2
            \end{pmatrix}$
            \item  $l\begin{pmatrix}
            0 \\ -k_T\omega_1^2\\ 0
            \end{pmatrix},\ \tau_{T2} = l\begin{pmatrix}
            \frac{\sqrt{3}}{2}k_T\omega_2^2 \\ -\frac{1}{2}k_T\omega_2^2\\ 0
            \end{pmatrix},\ \tau_{T3} = l\begin{pmatrix}
            \frac{\sqrt{3}}{2}k_T\omega_3^2 \\ \frac{1}{2}k_T\omega_3^2\\ 0
            \end{pmatrix}, \\
            \tau_{T4} = l\begin{pmatrix}
             0\\ k_T\omega_4^2\\ 0
            \end{pmatrix}, \
            \tau_{T5} = l\begin{pmatrix}
            -\frac{\sqrt{3}}{2}k_T\omega_5^2 \\ \frac{1}{2}k_T\omega_5^2\\ 0
            \end{pmatrix}, \
            \tau_{T6} = l\begin{pmatrix}
            -\frac{\sqrt{3}}{2}k_T\omega_6^2 \\ -\frac{1}{2}k_T\omega_6^2\\ 0
            \end{pmatrix} \\
            \tau= \begin{pmatrix}
            l \frac{\sqrt{3}}{2}k_T(\omega_2^2+ \omega_3^2-\omega_5^2 -\omega_6^2)\\
            l\frac{1}{2}k_T(-2\omega_1^2-\omega_2^2+\omega_3^2+2\omega_4^2 +\omega_5^2 -\omega_6^2)\\
            k_Q (-\omega_1^2+\omega_2^2-\omega_3^2+\omega_4^2-\omega_5^2+\omega_6^2)
            \end{pmatrix}
            $
            \item $
	           \Gamma =  \begin{pmatrix}
	           k_T&k_T&k_T&k_T&k_T&k_T \\
		       0 & \frac{\sqrt{3}}{2}lk_T & \frac{\sqrt{3}}{2}lk_T & 0& -\frac{\sqrt{3}}{2}lk_T& -\frac{\sqrt{3}}{2}lk_T \\
		        -lk_T &\frac{1}{2}lk_T &-\frac{1}{2}lk_T & lk_T&\frac{1}{2}lk_T &-\frac{1}{2}lk_T  \\
		       -k_Q &k_Q & -k_Q &k_Q &-k_Q&k_Q \\
	           \end{pmatrix}
            $
        \end{enumerate}
        \Aufgabe{Newton-Euler Equations}{6}
        \begin{enumerate}
        	\item $T_{max} = k_T*w_{i,max}^2 =3.9115 N$
        	\item $a_{max} = \begin{pmatrix}
        	0 \\ 0\\ \frac{T_{\Sigma,max}}{m}
        	\end{pmatrix}- ^BR_w ^wg =\begin{pmatrix}
        	0\\0\\ \frac{15.646N}{1kg}
        	\end{pmatrix}-\begin{pmatrix}
        	0\\0\\ 9.87 \frac{m}{s^2}
        	\end{pmatrix} =\begin{pmatrix}
        	0\\0\\ 5.836 \frac{m}{s^2}
        	\end{pmatrix} $
            \item $\tau=\begin{pmatrix}
                lk_T\omega_2^2-lk_T\omega_4^2\\
                -lk_T\omega_1^2+lk_T\omega_3^2\\
                k_Q(-\omega_1^2+\omega_2^2-\omega_3^2+\omega_4^2)
            \end{pmatrix}$, $\alpha=I^{-1}\tau\approx \begin{pmatrix}
                142.857&0&0\\
                0&142.857&0\\
                0&0&83.333
            \end{pmatrix}\tau$\\
            maximal acceleration around x and y in both directions is the same:\\
            $\alpha_{max}=\begin{pmatrix}
                142.857(lk_T\omega_2^2-lk_T\omega_4^2)\\
                142.857(-lk_T\omega_1^2+lk_T\omega_3^2)\\
                83.333\cdot k_Q(-\omega_1^2+\omega_2^2-\omega_3^2+\omega_4^2)
            \end{pmatrix}$\\
            $\alpha_{x,max}\approx 142.857\frac{1}{kgm^2}(lk_T\omega_2^2)\approx 558.785l\frac{1}{s^2}$\\
            $\alpha_{z,max}\approx 83.333\frac{1}{kgm^2}\cdot k_Q(\omega_2^2+\omega_4^2)\approx 15.917\frac{m}{s^2}$
        	\item \begin{itemize}
                \item $\alpha_{x,max}\approx 558.785l\frac{1}{s^2}$\\
            where:
                $\omega_2=838\frac{rad}{s},\omega_4=0\frac{rad}{s}, \omega_1=\omega_3=\sqrt{\frac{\frac{9.81N}{k_T}-\omega_2^2}{2}}\approx 727.660\frac{rad}{s} < 838\frac{rad}{s}$\checkmark\\
                accordingly for $-x,y,-y$
                \item $\text{max: } \alpha_z  \text{ s.t. max: }\tau_{z}=k_Q(-\omega_1^2+\omega_2^2-\omega_3^2+\omega_4^2)$ s.t.\\
                    $\omega_2=\omega_4,\ \omega_1=\omega_3,\ \omega_1^2+\omega_2^2+\omega_3^2+\omega_4^2=\frac{9.81N}{k_T}$\\
                    $\Rightarrow \omega_2=\omega_4=838\frac{rad}{s}\\
                    \Rightarrow \omega_1^2+\omega_3^2=\frac{9.81N}{k_T}-2\cdot \left(838\frac{rad}{s}\right)^2\\
                    \Rightarrow \omega_1=\omega_3=\sqrt{\frac{9.81N}{2k_T}-838^2\frac{rad^2}{s^2}}\approx 422.334\frac{rad}{s}$\vspace{5mm}\\
                    $\Rightarrow\alpha_{z,max}=83.333\frac{1}{kgm^2}\cdot k_Q(-422.334^2+838^2-422.334^2+838^2)\frac{rad^2}{s^2}\approx 11.875\frac{1}{s^2}$
            \end{itemize}

        \end{enumerate}
\end{document}