5

As we know that in TikZ, if unit is not mentioned in the coordinate in TikZ, it takes cm by default.

When I extract the coordinates, I was expecting the unit as cm. But TikZ shows the unit in the coordinate as pt.

How does TikZ determine the unit of measure in a coordinate if no unit is specified.

MWE:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{decorations.pathmorphing}
\usetikzlibrary{decorations.markings}
\usetikzlibrary{calc}
\usetikzlibrary{positioning}
\tikzset{zigzag/.style={decorate,decoration=zigzag}}
\begin{document}

\newdimen\XCoord
\newdimen\YCoord
\newcommand*{\ExtractCoordinate}[1]{\path (#1); \pgfgetlastxy{\XCoord}{\YCoord};}

\begin{tikzpicture}
  \coordinate (c) at (0,-2);
  \coordinate (d) at (4,-2);
  \coordinate (e) at (2,-4);
  \draw[thick,red,zigzag,postaction={
    decoration={
        markings,
        mark=at position 0.7 with { \coordinate (x); },
        mark=at position 0.5 with { \coordinate (singularity); },
    },
    decorate
  }] (-2,0) coordinate(a) -- (2,0) coordinate(b);

  \draw[thick,fill=blue!20] (c) -- (b) -- (d) -- (e) -- cycle;
  \draw[thick,postaction={
    decoration={
        markings,
        mark = at position 0.7 with \coordinate (y);
    },
    decorate
  }] (a) -- (c);
  \draw[thick,red,dashed] (x) -- (y);

  \node[above = 10ex of singularity,red] (sn) {singularity};
  \draw[red,->] (sn) -- ($(singularity)+(0,1)$);
  \ExtractCoordinate{x};
  \node[above] at (\XCoord,\YCoord) {(\XCoord,\YCoord)};
\end{tikzpicture}

\end{document} 

enter image description here

12
  • Not an answer, but the information may be useful: tex.stackexchange.com/a/20069/579 Feb 28 '19 at 3:33
  • Barbara I would like to know the default unit of measure when unit of measure is not mentioned Feb 28 '19 at 3:35
  • 1
    Internally TikZ works with pt. You have a coordinate system in which the unit vectors have length 1cm. Does that make sense? The IMHO clearest discussion on this can be found at tex.stackexchange.com/a/31606/121799 .
    – user121799
    Feb 28 '19 at 3:38
  • How did you measure the distance though? How did you make sure you are not magnifying the document when measuring?
    – zyy
    Feb 28 '19 at 3:39
  • @zyy As far as I can see, there is only the word singularity, not a real singularity, so you can use the Euclidean metric to a good approximation. ;-)
    – user121799
    Feb 28 '19 at 3:41
7

You can always convert everything from pt to cm or back by multiplying by the ratio 1pt/1cm or its inverse. (If that's not what you're after, I will be happy to remove the post.)

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{decorations.pathmorphing}
\usetikzlibrary{decorations.markings}
\usetikzlibrary{calc}
\usetikzlibrary{positioning}
\tikzset{zigzag/.style={decorate,decoration=zigzag}}
\begin{document}

\newdimen\XCoord
\newdimen\YCoord
\newcommand*{\ExtractCoordinate}[1]{\path (#1); \pgfgetlastxy{\XCoord}{\YCoord};}

\begin{tikzpicture}
  \coordinate (c) at (0,-2);
  \coordinate (d) at (4,-2);
  \coordinate (e) at (2,-4);
  \draw[thick,red,zigzag,postaction={
    decoration={
        markings,
        mark=at position 0.7 with { \coordinate (x); },
        mark=at position 0.5 with { \coordinate (singularity); },
    },
    decorate
  }] (-2,0) coordinate(a) -- (2,0) coordinate(b);

  \draw[thick,fill=blue!20] (c) -- (b) -- (d) -- (e) -- cycle;
  \draw[thick,postaction={
    decoration={
        markings,
        mark = at position 0.7 with \coordinate (y);
    },
    decorate
  }] (a) -- (c);
  \draw[thick,red,dashed] (x) -- (y);

  \node[above = 10ex of singularity,red] (sn) {singularity};
  \draw[red,->] (sn) -- ($(singularity)+(0,1)$);
  \ExtractCoordinate{x};
  \node[above] at (\XCoord,\YCoord) {%
  (\pgfmathparse{\XCoord*1pt/1cm}\pgfmathprintnumber{\pgfmathresult}\,cm,%
  \pgfmathparse{\YCoord*1pt/1cm}\pgfmathprintnumber[fixed,precision=2]{\pgfmathresult}\,cm)};
\end{tikzpicture}
\end{document} 

enter image description here

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