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path: root/vorlesungen/slides/7/interpolation.tex
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%
% interpolation.tex -- slide template
%
% (c) 2021 Prof Dr Andreas Müller, OST Ostschweizer Fachhochschule
%
\bgroup
\def\bild#1#2{\only<#1|handout:0>{\includegraphics[width=\textwidth]{../slides/7/images/interpolation/#2.png}}}
\begin{frame}[t]
\setlength{\abovedisplayskip}{5pt}
\setlength{\belowdisplayskip}{5pt}
\frametitle{Interpolation}
\vspace{-20pt}
\begin{columns}[t,onlytextwidth]
\begin{column}{0.48\textwidth}
\begin{block}{Aufgabe}
Finde einen Weg $g(t)\in \operatorname{SO}(3)$ zwischen
$g_0\in\operatorname{SO}(3)$
und
$g_1\in\operatorname{SO}(3)$:
\[
g_0=g(0)
\quad\wedge\quad
g_1=g(1)
\]
\end{block}
\vspace{-10pt}
\uncover<2->{%
\begin{block}{Lösung}
$g_i=\exp(A_i) \uncover<3->{\Rightarrow A_i^t=-A_i}$
\begin{align*}
\uncover<4->{A(t) &= (1-t)A_0 + tA_1}\uncover<8->{ \in \operatorname{so}(3)}
\\
\uncover<5->{A(t)^t
&=(1-t)A_0^t + tA_1^t}
\\
&\uncover<6->{=
-(1-t)A_0 - t A_1}
\uncover<7->{=
-A(t)}
\\
\uncover<9->{\Rightarrow
g(t) &= \exp A(t) \in \operatorname{SO}(3)}
\\
&\uncover<10->{\ne 
\exp (\log(g_1g_0^{-1})t) g_0}
\end{align*}
\end{block}}
\end{column}
\begin{column}{0.48\textwidth}
\uncover<11->{%
\begin{block}{Animation}
\centering
\ifthenelse{\boolean{presentation}}{
\bild{12}{i00}
\bild{13}{i01}
\bild{14}{i02}
\bild{15}{i03}
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\bild{60}{i48}
\bild{61}{i49}
\bild{62}{i50}
}{
\includegraphics[width=\textwidth]{../slides/7/images/interpolation/i25.png}
}
\end{block}}
\end{column}
\end{columns}
\end{frame}
\egroup