Using font-relative distances in tikzpictures












6















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}
newdimenmydimAnewdimenmydimB
makeatletter
newcommand{showat}[1]{%
pgfextractymydimA{pgfpointanchor{A}{center}}
pgfextractymydimB{pgfpointanchor{B}{center}}
node[red, font=tiny, align=left] at(#1) {Before themydimA \ After themydimB \
Scale y thepgf@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}


enter image description here










share|improve this question























  • If you want to scale the whole picture you could use resizebox -- which scales everything (line width, text spacing)

    – Aubrey Blumsohn
    Feb 21 at 12:44











  • @AubreyBlumsohn, no, it's not a matter of rescaling the whole picture. I want to be able to say that a label should be at 1ex distance from an anchor, no matter the scale of the figure... it's for a much more complex case, really (see github.com/circuitikz/circuitikz/issues/91) .

    – Rmano
    Feb 21 at 12:47













  • I didn't understand your question.

    – AndréC
    Feb 21 at 14:07











  • Coordinates are transformed by scaling while node text isn't. I'm not sure why one would expect something else. Internally, TeX does not care if it is cm or ex, it's all sp.

    – TeXnician
    Feb 21 at 14:37











  • @AndréC Basically the question is how to get all these three tikzpictures to produce the same results while using coordinate transformations based on font dimensions like ex (relative to the node font).

    – TeXnician
    Feb 21 at 14:44


















6















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}
newdimenmydimAnewdimenmydimB
makeatletter
newcommand{showat}[1]{%
pgfextractymydimA{pgfpointanchor{A}{center}}
pgfextractymydimB{pgfpointanchor{B}{center}}
node[red, font=tiny, align=left] at(#1) {Before themydimA \ After themydimB \
Scale y thepgf@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}


enter image description here










share|improve this question























  • If you want to scale the whole picture you could use resizebox -- which scales everything (line width, text spacing)

    – Aubrey Blumsohn
    Feb 21 at 12:44











  • @AubreyBlumsohn, no, it's not a matter of rescaling the whole picture. I want to be able to say that a label should be at 1ex distance from an anchor, no matter the scale of the figure... it's for a much more complex case, really (see github.com/circuitikz/circuitikz/issues/91) .

    – Rmano
    Feb 21 at 12:47













  • I didn't understand your question.

    – AndréC
    Feb 21 at 14:07











  • Coordinates are transformed by scaling while node text isn't. I'm not sure why one would expect something else. Internally, TeX does not care if it is cm or ex, it's all sp.

    – TeXnician
    Feb 21 at 14:37











  • @AndréC Basically the question is how to get all these three tikzpictures to produce the same results while using coordinate transformations based on font dimensions like ex (relative to the node font).

    – TeXnician
    Feb 21 at 14:44
















6












6








6








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}
newdimenmydimAnewdimenmydimB
makeatletter
newcommand{showat}[1]{%
pgfextractymydimA{pgfpointanchor{A}{center}}
pgfextractymydimB{pgfpointanchor{B}{center}}
node[red, font=tiny, align=left] at(#1) {Before themydimA \ After themydimB \
Scale y thepgf@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}


enter image description here










share|improve this 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}
newdimenmydimAnewdimenmydimB
makeatletter
newcommand{showat}[1]{%
pgfextractymydimA{pgfpointanchor{A}{center}}
pgfextractymydimB{pgfpointanchor{B}{center}}
node[red, font=tiny, align=left] at(#1) {Before themydimA \ After themydimB \
Scale y thepgf@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}


enter image description here







tikz-pgf






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Feb 21 at 11:06









RmanoRmano

8,18121647




8,18121647













  • If you want to scale the whole picture you could use resizebox -- which scales everything (line width, text spacing)

    – Aubrey Blumsohn
    Feb 21 at 12:44











  • @AubreyBlumsohn, no, it's not a matter of rescaling the whole picture. I want to be able to say that a label should be at 1ex distance from an anchor, no matter the scale of the figure... it's for a much more complex case, really (see github.com/circuitikz/circuitikz/issues/91) .

    – Rmano
    Feb 21 at 12:47













  • I didn't understand your question.

    – AndréC
    Feb 21 at 14:07











  • Coordinates are transformed by scaling while node text isn't. I'm not sure why one would expect something else. Internally, TeX does not care if it is cm or ex, it's all sp.

    – TeXnician
    Feb 21 at 14:37











  • @AndréC Basically the question is how to get all these three tikzpictures to produce the same results while using coordinate transformations based on font dimensions like ex (relative to the node font).

    – TeXnician
    Feb 21 at 14:44





















  • If you want to scale the whole picture you could use resizebox -- which scales everything (line width, text spacing)

    – Aubrey Blumsohn
    Feb 21 at 12:44











  • @AubreyBlumsohn, no, it's not a matter of rescaling the whole picture. I want to be able to say that a label should be at 1ex distance from an anchor, no matter the scale of the figure... it's for a much more complex case, really (see github.com/circuitikz/circuitikz/issues/91) .

    – Rmano
    Feb 21 at 12:47













  • I didn't understand your question.

    – AndréC
    Feb 21 at 14:07











  • Coordinates are transformed by scaling while node text isn't. I'm not sure why one would expect something else. Internally, TeX does not care if it is cm or ex, it's all sp.

    – TeXnician
    Feb 21 at 14:37











  • @AndréC Basically the question is how to get all these three tikzpictures to produce the same results while using coordinate transformations based on font dimensions like ex (relative to the node font).

    – TeXnician
    Feb 21 at 14:44



















If you want to scale the whole picture you could use resizebox -- which scales everything (line width, text spacing)

– Aubrey Blumsohn
Feb 21 at 12:44





If you want to scale the whole picture you could use resizebox -- which scales everything (line width, text spacing)

– Aubrey Blumsohn
Feb 21 at 12:44













@AubreyBlumsohn, no, it's not a matter of rescaling the whole picture. I want to be able to say that a label should be at 1ex distance from an anchor, no matter the scale of the figure... it's for a much more complex case, really (see github.com/circuitikz/circuitikz/issues/91) .

– Rmano
Feb 21 at 12:47







@AubreyBlumsohn, no, it's not a matter of rescaling the whole picture. I want to be able to say that a label should be at 1ex distance from an anchor, no matter the scale of the figure... it's for a much more complex case, really (see github.com/circuitikz/circuitikz/issues/91) .

– Rmano
Feb 21 at 12:47















I didn't understand your question.

– AndréC
Feb 21 at 14:07





I didn't understand your question.

– AndréC
Feb 21 at 14:07













Coordinates are transformed by scaling while node text isn't. I'm not sure why one would expect something else. Internally, TeX does not care if it is cm or ex, it's all sp.

– TeXnician
Feb 21 at 14:37





Coordinates are transformed by scaling while node text isn't. I'm not sure why one would expect something else. Internally, TeX does not care if it is cm or ex, it's all sp.

– TeXnician
Feb 21 at 14:37













@AndréC Basically the question is how to get all these three tikzpictures to produce the same results while using coordinate transformations based on font dimensions like ex (relative to the node font).

– TeXnician
Feb 21 at 14:44







@AndréC Basically the question is how to get all these three tikzpictures to produce the same results while using coordinate transformations based on font dimensions like ex (relative to the node font).

– TeXnician
Feb 21 at 14:44












4 Answers
4






active

oldest

votes


















5














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}





share|improve this answer
























  • I used this and it works as supposed... thanks a lot!

    – Rmano
    Feb 21 at 16:12



















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:



output



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






share|improve this answer































    2














    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}
    newdimenmydimAnewdimenmydimB
    makeatletter
    newcommand{showat}[1]{%
    pgfextractymydimA{pgfpointanchor{A}{center}}
    pgfextractymydimB{pgfpointanchor{B}{center}}
    node[red, font=tiny, align=left] at(#1) {Before themydimA \ After themydimB \
    Scale y thepgf@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}


    screenshot






    share|improve this answer

































      2














      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}
      newdimenmydimAnewdimenmydimB
      makeatletter
      newcommand{showat}[1]{%
      pgfextractymydimA{pgfpointanchor{A}{center}}
      pgfextractymydimB{pgfpointanchor{B}{center}}
      node[red, font=tiny, align=left] at(#1) {Before themydimA \ After themydimB \
      Scale y thepgf@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}


      enter image description here






      share|improve this answer



















      • 1





        Thanks @marmot, +1! I already accepted Eric's answer (and implemented it), but I think both are valid.

        – Rmano
        Feb 21 at 16:13











      • @Rmano Yes, Eric was the first to point out what the problem is and provided an elegant way to fix it. You should definitely accept his nice answer.

        – marmot
        Feb 21 at 17:20











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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5














      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}





      share|improve this answer
























      • I used this and it works as supposed... thanks a lot!

        – Rmano
        Feb 21 at 16:12
















      5














      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}





      share|improve this answer
























      • I used this and it works as supposed... thanks a lot!

        – Rmano
        Feb 21 at 16:12














      5












      5








      5







      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}





      share|improve this answer













      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}






      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered Feb 21 at 14:45









      Eric MarsdenEric Marsden

      75637




      75637













      • I used this and it works as supposed... thanks a lot!

        – Rmano
        Feb 21 at 16:12



















      • I used this and it works as supposed... thanks a lot!

        – Rmano
        Feb 21 at 16:12

















      I used this and it works as supposed... thanks a lot!

      – Rmano
      Feb 21 at 16:12





      I used this and it works as supposed... thanks a lot!

      – Rmano
      Feb 21 at 16:12











      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:



      output



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






      share|improve this 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:



        output



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






        share|improve this answer


























          2












          2








          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:



          output



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






          share|improve this answer













          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:



          output



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







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Feb 21 at 14:59









          Pippip19Pippip19

          1,6538




          1,6538























              2














              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}
              newdimenmydimAnewdimenmydimB
              makeatletter
              newcommand{showat}[1]{%
              pgfextractymydimA{pgfpointanchor{A}{center}}
              pgfextractymydimB{pgfpointanchor{B}{center}}
              node[red, font=tiny, align=left] at(#1) {Before themydimA \ After themydimB \
              Scale y thepgf@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}


              screenshot






              share|improve this answer






























                2














                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}
                newdimenmydimAnewdimenmydimB
                makeatletter
                newcommand{showat}[1]{%
                pgfextractymydimA{pgfpointanchor{A}{center}}
                pgfextractymydimB{pgfpointanchor{B}{center}}
                node[red, font=tiny, align=left] at(#1) {Before themydimA \ After themydimB \
                Scale y thepgf@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}


                screenshot






                share|improve this answer




























                  2












                  2








                  2







                  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}
                  newdimenmydimAnewdimenmydimB
                  makeatletter
                  newcommand{showat}[1]{%
                  pgfextractymydimA{pgfpointanchor{A}{center}}
                  pgfextractymydimB{pgfpointanchor{B}{center}}
                  node[red, font=tiny, align=left] at(#1) {Before themydimA \ After themydimB \
                  Scale y thepgf@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}


                  screenshot






                  share|improve this answer















                  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}
                  newdimenmydimAnewdimenmydimB
                  makeatletter
                  newcommand{showat}[1]{%
                  pgfextractymydimA{pgfpointanchor{A}{center}}
                  pgfextractymydimB{pgfpointanchor{B}{center}}
                  node[red, font=tiny, align=left] at(#1) {Before themydimA \ After themydimB \
                  Scale y thepgf@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}


                  screenshot







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Feb 21 at 15:22

























                  answered Feb 21 at 15:08









                  AndréCAndréC

                  10.5k11548




                  10.5k11548























                      2














                      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}
                      newdimenmydimAnewdimenmydimB
                      makeatletter
                      newcommand{showat}[1]{%
                      pgfextractymydimA{pgfpointanchor{A}{center}}
                      pgfextractymydimB{pgfpointanchor{B}{center}}
                      node[red, font=tiny, align=left] at(#1) {Before themydimA \ After themydimB \
                      Scale y thepgf@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}


                      enter image description here






                      share|improve this answer



















                      • 1





                        Thanks @marmot, +1! I already accepted Eric's answer (and implemented it), but I think both are valid.

                        – Rmano
                        Feb 21 at 16:13











                      • @Rmano Yes, Eric was the first to point out what the problem is and provided an elegant way to fix it. You should definitely accept his nice answer.

                        – marmot
                        Feb 21 at 17:20
















                      2














                      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}
                      newdimenmydimAnewdimenmydimB
                      makeatletter
                      newcommand{showat}[1]{%
                      pgfextractymydimA{pgfpointanchor{A}{center}}
                      pgfextractymydimB{pgfpointanchor{B}{center}}
                      node[red, font=tiny, align=left] at(#1) {Before themydimA \ After themydimB \
                      Scale y thepgf@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}


                      enter image description here






                      share|improve this answer



















                      • 1





                        Thanks @marmot, +1! I already accepted Eric's answer (and implemented it), but I think both are valid.

                        – Rmano
                        Feb 21 at 16:13











                      • @Rmano Yes, Eric was the first to point out what the problem is and provided an elegant way to fix it. You should definitely accept his nice answer.

                        – marmot
                        Feb 21 at 17:20














                      2












                      2








                      2







                      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}
                      newdimenmydimAnewdimenmydimB
                      makeatletter
                      newcommand{showat}[1]{%
                      pgfextractymydimA{pgfpointanchor{A}{center}}
                      pgfextractymydimB{pgfpointanchor{B}{center}}
                      node[red, font=tiny, align=left] at(#1) {Before themydimA \ After themydimB \
                      Scale y thepgf@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}


                      enter image description here






                      share|improve this answer













                      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}
                      newdimenmydimAnewdimenmydimB
                      makeatletter
                      newcommand{showat}[1]{%
                      pgfextractymydimA{pgfpointanchor{A}{center}}
                      pgfextractymydimB{pgfpointanchor{B}{center}}
                      node[red, font=tiny, align=left] at(#1) {Before themydimA \ After themydimB \
                      Scale y thepgf@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}


                      enter image description here







                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered Feb 21 at 15:57









                      marmotmarmot

                      112k5140264




                      112k5140264








                      • 1





                        Thanks @marmot, +1! I already accepted Eric's answer (and implemented it), but I think both are valid.

                        – Rmano
                        Feb 21 at 16:13











                      • @Rmano Yes, Eric was the first to point out what the problem is and provided an elegant way to fix it. You should definitely accept his nice answer.

                        – marmot
                        Feb 21 at 17:20














                      • 1





                        Thanks @marmot, +1! I already accepted Eric's answer (and implemented it), but I think both are valid.

                        – Rmano
                        Feb 21 at 16:13











                      • @Rmano Yes, Eric was the first to point out what the problem is and provided an elegant way to fix it. You should definitely accept his nice answer.

                        – marmot
                        Feb 21 at 17:20








                      1




                      1





                      Thanks @marmot, +1! I already accepted Eric's answer (and implemented it), but I think both are valid.

                      – Rmano
                      Feb 21 at 16:13





                      Thanks @marmot, +1! I already accepted Eric's answer (and implemented it), but I think both are valid.

                      – Rmano
                      Feb 21 at 16:13













                      @Rmano Yes, Eric was the first to point out what the problem is and provided an elegant way to fix it. You should definitely accept his nice answer.

                      – marmot
                      Feb 21 at 17:20





                      @Rmano Yes, Eric was the first to point out what the problem is and provided an elegant way to fix it. You should definitely accept his nice answer.

                      – marmot
                      Feb 21 at 17:20


















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