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

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



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




\begin{document}
    %\header{BlattNr}{Tutor}{Abgabedatum}{Vorlesungsname}{Namen}{Semester}{Anzahl Aufgaben}
    \header{5}{}{2015-06-02}{Mobile Robots}{
    	\textit{Jan-Peter Hohloch}\\ \textit{Maximus Mutschler}
    }{SS 15}{2}
    \vspace{1cm}

	\Aufgabe{MATLAB}{10}
	\begin{enumerate}[(a)]
        \item Marlin F-033 specs from \url{https://www.alliedvision.com/cn/products/cameras/detail/f-033-1.html}:
            \begin{itemize}
                \item Resolution: 656x494
                \item Framerate: 73fps
                \item $73\frac{1}{s}\cdot 656\cdot 494\cdot 1$ Byte$=23656672\frac{\text{Byte}}{s}$
            \end{itemize}
        \item We'd expect the principle point to be at (0,0) if this correspondents to the middle of the picture. So it should be in Pixel 162032.
        \item after optimization:
        \begin{itemize}
            \item focal length: 780.37666, 776.84481
            \item Principle point: 304.55702, 336.54883
            \item distortion: -0.19833, 0.30243, 0.01975, -0.00276, 0.00000
            \item Pixel error: 0.95118, 0.87892
            \item The pixel error ist the standard deviation of the reprojection error in x- and y-Direction
        \end{itemize}
        %TODO
	\end{enumerate}
    \Aufgabe{}{10}
        \begin{enumerate}[(a)]
            \item $n=\begin{pmatrix}
                \frac{u-u_0}{k_u}\\
                \frac{v-v_0}{k_v}\\
                1
            \end{pmatrix}$
            \item $n_c\approx\begin{pmatrix}
                0.236\\0.255\\1
            \end{pmatrix},\ n_b\approx\begin{pmatrix}
                0.245\\0.258\\1
            \end{pmatrix}$
            \item $\alpha =\tan^{-1}\left(d(n_c,n_b)\right)\approx\tan^{-1}\left(\sqrt{\left(0.236-0.245\right)^2+\left(0.255-0.258\right)^2}\right)\approx 0.00980$
            \item $Z_C=\frac{20mm }{d(n_c,n_b)}=\frac{20mm}{ \sqrt{\left(0.236-0.245\right)^2+\left(0.255-0.258\right)^2}}\approx \frac{20mm}{0.00949}\approx 2107mm$
            \item $Z_C\cdot n=N\Rightarrow P_C=\begin{pmatrix}
                Z_C\cdot 0.236\\Z_C\cdot 0.255\\ Z_C
            \end{pmatrix}\approx \begin{pmatrix}
                497.2\\537.3\\ 2107
            \end{pmatrix}mm$
            \item Depending on how many distortion parameters we want to know, the system of linear equations may be of high dimension. In addition we can't sepeate by coordinate because the solutions for $x$ and $y$ depend on each other. Even for only computing radial distortion only up to the second degree we have to\\
            solve $\begin{pmatrix}
                x_d\\y_d
            \end{pmatrix}=\begin{pmatrix}
                x\\y
            \end{pmatrix}+k_1\left|\left|\begin{pmatrix}
                x\\y
            \end{pmatrix}\right|\right|_2^2+k_2\left|\left|\begin{pmatrix}
                x\\y
            \end{pmatrix}\right|\right|_2^4$ for $x$ and $y$.
        \end{enumerate}
\end{document}