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as written in the title, I'd like to draw a circle in the YZ plane, at the moment i wrote static coordinates but if it's possible to let tikz calculate the perspektive it would be way better. Thank you in advance:)

Here is my (therefor maybe useless) code + png: -- I think, I have to work with \usepackage{tikz-3dplot}, but don't know how to do.. enter image description here

\documentclass{standalone}
\usepackage{tikz}
\begin{document}

\begin{tikzpicture}

%3D-Koordinatensystem:
\draw[->, >=latex] (-0.5,0,0) -- (7,0,0) node[below]{};
\draw[->, >=latex] (0,-0.5,0) -- (0,3,0) node[left]{}; 
\draw[->, >=latex] (0,0,-0.5) -- (0,0,5) node[left]{};

\coordinate (0) at (-1.5,1);

\coordinate (1) at (-2,-0.62);
\coordinate (2) at (-1,-0.62);
\coordinate (3) at (0,-0.62);
\coordinate (4) at (1,-0.62);
\coordinate (5) at (2,-0.62);
\coordinate (6) at (3,-0.62);
\coordinate (7) at (4,-0.62);
\coordinate (8) at (5,-0.62);
\coordinate (9) at (6,-0.62);
\coordinate (10) at (7,-0.62);
\coordinate (10) at (8,-0.62);
\coordinate (10) at (9,-0.62);


\draw[thick, dashed] (0) circle[x={(1cm,1.4cm)},y={(1cm,-1cm)},x radius=1, y radius=0.8] node [above=14mm, left=4mm] {$(t_o,x_o)$} ;
\draw[thick] (0) circle[x={(1cm,1.4cm)},y={(1cm,-1cm)},x radius=0.4, y radius=0.3]; 

\filldraw (0) circle (0.25mm) node [below=1.5mm, right=0.5mm] {$\delta$};

\draw (0) -- (-1.05,1) node [above=10mm, left=1mm] {$\varepsilon$};
\draw (0) -- (-1,2.62);

\filldraw (-2,-0.62)circle (0.2mm);




\draw [line width=0.8] (1) to [out=350,in=130] (2); 
\draw [line width=0.8] (2) to [out=320,in=140] (3);
\draw [line width=0.8] (3) to [out=320,in=140] (4);
\draw [line width=0.8] (4) to [out=320,in=140] (5);
\draw [line width=0.8] (5) to [out=320,in=140] (6);
\draw [line width=0.8] (6) to [out=320,in=140] (7);
\draw [line width=0.8] (7) to [out=320,in=140] (8);
\draw [line width=0.8] (8) to [out=320,in=140] (9);


\end{tikzpicture}
\end{document}
2
  • 2
    Btw, if you want people to answer your questions then you should consider accepting answers to some of your previous posts.
    – user30471
    Feb 14, 2018 at 9:24
  • Thank you for the hint, I'm new to this forum and it took me a bit to realize.
    – Don Ky
    Feb 14, 2018 at 12:16

1 Answer 1

2

Perhaps something like this:

enter image description here

This was done using a parametric plot of a circle using tikz's 3-D coordinates:

 \draw[smooth] plot (0,{2.2+0.6*cos(\x r)},{2.2+0.6*sin(\x r)});

Here is the full code:

\documentclass[tikz, border=2mm]{standalone}
\begin{document}

\begin{tikzpicture}

%3D-Koordinatensystem:
  \draw[->, >=latex] (-0.5,0,0) -- (7,0,0) node[below]{};
  \draw[->, >=latex] (0,-0.5,0) -- (0,3,0) node[left]{};
  \draw[->, >=latex] (0,0,-0.5) -- (0,0,5) node[left]{};

  \coordinate (0) at (-1.5,1);

  \coordinate (1) at (-2,-0.62);
  \coordinate (2) at (-1,-0.62);
  \coordinate (3) at (0,-0.62);
  \coordinate (4) at (1,-0.62);
  \coordinate (5) at (2,-0.62);
  \coordinate (6) at (3,-0.62);
  \coordinate (7) at (4,-0.62);
  \coordinate (8) at (5,-0.62);
  \coordinate (9) at (6,-0.62);
  \coordinate (10) at (7,-0.62);
  \coordinate (10) at (8,-0.62);
  \coordinate (10) at (9,-0.62);

  \draw[smooth] plot (0,{2.2+0.6*cos(\x r)},{2.2+0.6*sin(\x r)});
  \draw[smooth, dashed] plot (0,{2.2+1.4*cos(\x r)},{2.2+1.4*sin(\x r)});

  \draw(0,2.2,2.2)--++(0.55,0.75,0)node[above right]{$\varepsilon$};
  \draw(0,2.2,2.2)--node[below]{$\delta$} ++(0.5,0,0);

  \filldraw (0, 2.2, 2.2) circle (0.2mm);

  \draw [line width=0.8] (1) to [out=350,in=130] (2);
  \draw [line width=0.8] (2) to [out=320,in=140] (3);
  \draw [line width=0.8] (3) to [out=320,in=140] (4);
  \draw [line width=0.8] (4) to [out=320,in=140] (5);
  \draw [line width=0.8] (5) to [out=320,in=140] (6);
  \draw [line width=0.8] (6) to [out=320,in=140] (7);
  \draw [line width=0.8] (7) to [out=320,in=140] (8);
  \draw [line width=0.8] (8) to [out=320,in=140] (9);

\end{tikzpicture}
\end{document}

I have positioned the circles somewhat randomly and, in particular, have not lined them up with your "wave", but these details are easy enough to adjust.

2
  • Perfect, thank you very much. It solves completly my problem (thank you furthermore for the accepting hint)
    – Don Ky
    Feb 14, 2018 at 12:21
  • @DonKy Glad to help!
    – user30471
    Feb 14, 2018 at 13:57

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