# Arcs on a plane in 3-D drawing

I'd like to do the following figure using TikZ:

I would know how to to this using \clip, \draw, arc and \draw[->] in 2D. But how can I draw this on a plane that lies in the x-z plane?

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May be the tikz-3dplot package could be useful. –  Manuel Dec 23 '12 at 22:35
Could you please post a minimal example to get people started, using the commands you mentioned? It's no fun to start from scratch, but with the framework in place, it will be quite easy to get this into the x-z plane. –  Jake Dec 23 '12 at 23:06
@Jake: True, but I learned a lot going through the documentation myself. I posted my solution as an answer. –  queueoverflow Dec 24 '12 at 0:14

The tikz-3dplot package totally rocks. I just drew my picture in the x-y-plane and rotated it then. The only caveat is that rectangle does not work any more, one has to supply all four corners by hand.

\begin{tikzpicture}[scale=2]
\tdplotsetmaincoords{45}{200}

\begin{scope}[tdplot_main_coords]
\draw (0, 0) -- (5, 0) -- (5, 5) -- (0, 5) -- cycle;
\node[below right] at (0, 5) {$\R^3$};

\begin{scope}
\clip (0, 0) -- (5, 0) -- (5, 5) -- (0, 5) -- cycle;
\fill[opacity=20, color=black!20] ($(1, 1)+(45:3)+(0.5, 0)$) arc (0:360:0.5);
\begin{scope}[->, thick]
\draw (1, 1) -- ++(60:3) node[midway, below] {$t$};
\draw (1, 1) -- ++(100:2.5) node[midway, right] {$t_0$};
\draw (1, 1) -- ++(30:3.5) node[midway, below] {$t_1$};
\draw (1, 1) -- ++(0:1.5) node[midway, above] {$t$};
\end{scope}
\draw (4, 1) arc (0:360:3);
\draw (3.5, 1) arc (0:360:2.5);
\draw (4.5, 1) arc (0:360:3.5);
\draw (2.5, 1) arc (0:360:1.5);
\end{scope}

\draw[thick, ->] (5, 5, 0) -- ++(0, 0, 2) node[sloped, midway, above] {Zeitachse};
\end{scope}
\end{tikzpicture}

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