Using font-relative distances in tikzpictures

Question

When I use scale in tikzpictures (without transform shape), the text size in nodes is not changed (which is good and well). Suppose now that I want to put some node in a scaled graph using a distance proportional to the node font; I naively supposed that I could use ex coords, but see below...

I also printed the values of the coordinates, and \pgf@yy which should give the unit vector, but there is something I miss here...

Is there a way to express a distance which is proportional to the (default) font size, independently from the scale?

\documentclass[border=10pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\newdimen\mydimA\newdimen\mydimB
\makeatletter
\newcommand{\showat}[1]{%
    \pgfextracty\mydimA{\pgfpointanchor{A}{center}}
    \pgfextracty\mydimB{\pgfpointanchor{B}{center}}
    \node[red, font=\tiny,  align=left] at(#1) {Before \the\mydimA \\ After \the\mydimB \\
    Scale y \the\pgf@yy};
}
\makeatother
\begin{document}
\begin{tikzpicture}[baseline]
    \draw (0,0)  --(1,0) coordinate(A) -- (2,0);
    \path (A) ++(0,1ex) coordinate(B);
    \node [anchor=base] at (B) {$R_g$};
    \showat{1,-1}
\end{tikzpicture}
\begin{tikzpicture}[baseline, scale=2]
    \draw (0,0)  --(0.5,0) coordinate(A) -- (1,0);
    \path (A) ++(0,1ex) coordinate(B);
    \node [anchor=base] at (B) {$R_g$};
    \showat{0.5,-0.5}
\end{tikzpicture}
\begin{tikzpicture}[baseline, scale=0.01]
    \draw (0,0)  --(100,0) coordinate(A) -- (200,0);
    \path (A) ++(0,1ex) coordinate(B);
    \node [anchor=base] at (B) {$R_g$};
    \showat{100,-100}
\end{tikzpicture}
\end{document}

Answer 1

Your 1em is being converted at read time into the coordinate system, which is then affected by the picture scale. You can use reset cm to locally reset the coordinate transformation matrix, locally cancelling the picture scale effect.

\documentclass[border=10pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}[baseline]
    \draw (0,0)  --(1,0) coordinate(A) -- (2,0);
    \path (A) ++(0,1em) coordinate(B);
    \node [anchor=base] at (B) {$R_g$};
\end{tikzpicture}
\begin{tikzpicture}[baseline, scale=2]
  \draw (0,0)  --(0.5,0) coordinate(A) -- (1,0);
  \path[reset cm] (A) ++(0,1em) coordinate(B);
  \node [anchor=base] at (B) {$R_g$};
\end{tikzpicture}
\begin{tikzpicture}[baseline, scale=0.01]
  \draw (0,0)  --(100,0) coordinate(A) -- (200,0);
  \path[reset cm] (A) ++(0,1em) coordinate(B);
  \node [anchor=base] at (B) {$R_g$};
\end{tikzpicture}
\end{document}

Answer 2

Resetting the scale locally is one option (see Marsden's answer), but there is a better way to go about things.

With scale=factor, all coordinates are scaled by factor, regardless of their dimension (while the unit vectors are unchanged, as you discovered). Instead, redefine the length of the unit vectors according to the desired scaling. For example, with y={(0cm,2cm)}, while (0,0) -- (1,0) will be twice the usual distance, coordinates with dimensions, such as (0,1ex), will be unchanged:

\begin{document}
\begin{tikzpicture}[baseline]
    \draw (0,0)  --(1,0) coordinate(A) -- (2,0);
    \path (A) ++(0,1ex) coordinate(B);
    \node [anchor=base] at (B) {$R_g$};
    \showat{1,-1cm}
\end{tikzpicture}
\begin{tikzpicture}[baseline, x={(2cm,0cm)},y={(0cm,2cm)}]
    % Identical code to previous picture:
    \draw (0,0)  --(1,0) coordinate(A) -- (2,0);
    \path[scale=1] (A) ++(0,1ex) coordinate(B);
    \node [anchor=base] at (B) {$R_g$};
    \showat{1,-1cm}
\end{tikzpicture}
\end{document}

Output:

Note how the distance between A and B is unchanged, while \the\pgf@yy has doubled, as expected.

Answer 3

Why don't you place B as itself a node (with relative placement)?

That would solve all your problems.

\documentclass[border=10pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\usetikzlibrary{positioning}
\newdimen\mydimA\newdimen\mydimB
\makeatletter
\newcommand{\showat}[1]{%
    \pgfextracty\mydimA{\pgfpointanchor{A}{center}}
    \pgfextracty\mydimB{\pgfpointanchor{B}{center}}
    \node[red, font=\tiny,  align=left] at(#1) {Before \the\mydimA \\ After \the\mydimB \\
    Scale y \the\pgf@yy};
}
\makeatother
\begin{document}
\tikzset{node distance=1ex}
\begin{tikzpicture}[baseline]
    \draw (0,0)  --(1,0) coordinate(A) -- (2,0);
    \node[above=of A,anchor=base](B){$R_g$};
    \showat{1,-1}
\end{tikzpicture}
\begin{tikzpicture}[baseline, scale=2]
    \draw (0,0)  --(0.5,0) coordinate(A) -- (1,0);
    %\path (A) ++(0,1ex) coordinate(B)node{y};
    \node[above=of A,anchor=base] (B) {$R_g$};
    \showat{0.5,-0.5}
\end{tikzpicture}
\begin{tikzpicture}[baseline, scale=0.01]
    \draw (0,0)  --(100,0) coordinate(A) -- (200,0);
    %\path (A) ++(0,1ex) coordinate(B)node{y};
    \node[above=of A,anchor=base] (B) {$R_g$};
    \showat{100,-100}
\end{tikzpicture}
\end{document}

Answer 4

Eric Marsden's answer explains nicely what's going on. However, I would just read out the relevant entry of the transformation matrix (it is the 22 entry) and "invert" it. This is a trick that I also used (without inversion, of course) to scale the line width with a shape because by default it doesn't scale.

\documentclass[border=10pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\newdimen\mydimA\newdimen\mydimB
\makeatletter
\newcommand{\showat}[1]{%
    \pgfextracty\mydimA{\pgfpointanchor{A}{center}}
    \pgfextracty\mydimB{\pgfpointanchor{B}{center}}
    \node[red, font=\tiny,  align=left] at(#1) {Before \the\mydimA \\ After \the\mydimB \\
    Scale y \the\pgf@yy};
}
\makeatother
\begin{document}
\begin{tikzpicture}[baseline]
    \draw (0,0)  --(1,0) coordinate(A) -- (2,0);
    \pgfgettransformentries{\tmp}{\tmp}{\tmp}{\myscale}{\tmp}{\tmp}
    \path  (A) ++(0,1ex/\myscale) coordinate(B);
    \node [anchor=base] at (B) {$R_g$};
    \showat{1,-1}
\end{tikzpicture}
\begin{tikzpicture}[baseline, scale=2]
    \draw (0,0)  --(0.5,0) coordinate(A) -- (1,0);
    \pgfgettransformentries{\tmp}{\tmp}{\tmp}{\myscale}{\tmp}{\tmp}
    \path  (A) ++(0,1ex/\myscale) coordinate(B);
    \node [anchor=base] at (B) {$R_g$};
    \showat{0.5,-0.5}
\end{tikzpicture}
\begin{tikzpicture}[baseline, scale=0.01]
    \draw (0,0)  --(100,0) coordinate(A) -- (200,0);
    \pgfgettransformentries{\tmp}{\tmp}{\tmp}{\myscale}{\tmp}{\tmp}
    \path  (A) ++(0,1ex/\myscale) coordinate(B);
    \node [anchor=base] at (B) {$R_g$};
    \showat{100,-100}
\end{tikzpicture}
\end{document}