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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)}
+\end{align*}
+\end{block}}
+\end{column}
+\begin{column}{0.48\textwidth}
+\uncover<10->{%
+\begin{block}{Animation}
+\centering
+\bild{11}{i00}
+\bild{12}{i01}
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+\end{block}}
+\end{column}
+\end{columns}
+\end{frame}
+\egroup