[Tex/LaTex] How to put a mark on an arc (tikz,decoration)

decorationstikz-pgf

I have a problem with this code :

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{decorations.markings}
\begin{document}

\begin{tikzpicture}[decoration={markings, mark = at position .5 with
 {\draw (-2pt,-2pt) -- (2pt,2pt)  (2pt,-2pt) -- (-2pt,2pt);}}]    

% wrong 
%\draw [postaction={decorate}] (0,0) -- ++(146:1) arc (146:157:1)--(0,0);
%fine
\draw [postaction={decorate}] (0,0) -- ++(146:1.2) arc (146:157:1.2)--(0,0);
\end{tikzpicture}

\end{document}

The line after % wrong gives an error dimension too large. I think there is a math problem because the arc or the radius is too small or perhaps both.
A solution is (another one is to use a scale >1 but for the fine line don't use a scale like .8 :

\documentclass{article}
\usepackage{tikz,fp}
\usetikzlibrary{decorations.markings,fixedpointarithmetic}

\begin{document}

\begin{tikzpicture}[fixed point arithmetic,
                    decoration={markings, mark = at position .5 with
       {\draw (-2pt,-2pt) -- (2pt,2pt)  (2pt,-2pt) -- (-2pt,2pt);}}]     

\draw [postaction={decorate}] (0,0) -- ++(146:1) arc (146:157:1)--(0,0);
\end{tikzpicture}

\end{document}

But I can't use it because the time for the compilation is too long.

Is it possible to speed up the last code or perhaps someone knows another way to get the mark without an error ? Perhaps I use decoration in a wrong way!

Best Answer

The problem is in the implementation of the mathematical function veclen. In short, it doesn't just do (x^2 + y^2)^{1/2} (How could it? Taking square roots isn't even possible theoretically let alone the issues of precision.) but does something more complicated that at some stage involves dividing by one of the components. You can test this by trying a simple:

\pgfmathparse{veclen(0.00006,0.00006)}

(the number was chosen as that happens to be the number that TeX barfs on when trying Altermundus' example. Multiply by 10 and it all works again.)

Looking at the implementation of veclen I noticed that it did some initial scaling to take into account large numbers. So I added some lines to test also for small numbers as well. And that appears to solve the problem.

(Unfortunately, I couldn't get \pgfmathredeclarefunction to work properly. Rather than debug that, I just \let the old function to \relax and overwrote it.)

Code:

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{decorations.markings}
\makeatletter
\let\pgfmath@function@veclen\relax
\pgfmathdeclarefunction{veclen}{2}{%
  \begingroup%
  \pgfmath@x#1pt\relax%
  \pgfmath@y#2pt\relax%
  \ifdim\pgfmath@x<0pt\relax%
  \pgfmath@x-\pgfmath@x%
  \fi%
  \ifdim\pgfmath@y<0pt\relax%
  \pgfmath@y-\pgfmath@y%
  \fi%
  \ifdim\pgfmath@x=0pt\relax%
  \pgfmath@x\pgfmath@y%
  \else%
  \ifdim\pgfmath@y=0pt\relax%
  \else%
  \ifdim\pgfmath@x>\pgfmath@y%
  \pgfmath@xa\pgfmath@x%
  \pgfmath@x\pgfmath@y%
  \pgfmath@y\pgfmath@xa%
  \fi%
  % We use a scaling factor to reduce errors.
  % First, see if we should scale down
  \let\pgfmath@tmp@scale=\divide
  \let\pgfmath@tmp@restore=\multiply
  \ifdim\pgfmath@y>10000pt\relax%
  \c@pgfmath@counta1500\relax%
  \else%
  \ifdim\pgfmath@y>1000pt\relax%
  \c@pgfmath@counta150\relax%
  \else%
  \ifdim\pgfmath@y>100pt\relax%
  \c@pgfmath@counta50\relax%
  \else%
  % Not scaling down, should we scale up?
  \let\pgfmath@tmp@scale=\multiply
  \let\pgfmath@tmp@restore=\divide
  \ifdim\pgfmath@y<0.00001pt\relax%
  \c@pgfmath@counta1500\relax%
  \else%
  \ifdim\pgfmath@y<0.0001pt\relax%
  \c@pgfmath@counta150\relax%
  \else%
  \ifdim\pgfmath@y<0.001pt\relax%
  \c@pgfmath@counta50\relax%
  \else
  \c@pgfmath@counta1\relax%
  \fi%
  \fi%
  \fi%
  \fi%
  \fi%
  \fi%
  \pgfmath@tmp@scale\pgfmath@x\c@pgfmath@counta\relax%
  \pgfmath@tmp@scale\pgfmath@y\c@pgfmath@counta\relax%
  \pgfmathreciprocal@{\pgfmath@tonumber{\pgfmath@y}}%
  \pgfmath@x\pgfmathresult\pgfmath@x%
  \pgfmath@xa\pgfmath@tonumber{\pgfmath@x}\pgfmath@x%
  \edef\pgfmath@temp{\pgfmath@tonumber{\pgfmath@xa}}%
  %
  % Use A+x^2*(B+x^2*(C+x^2*(D+E*x^2))) 
  % where
  % A = +1.000012594
  % B = +0.4993615349 
  % C = -0.1195159052
  % D = +0.04453994279
  % E = -0.01019210944
  %
  \pgfmath@x-0.01019210944\pgfmath@xa%
  \advance\pgfmath@x0.04453994279pt\relax%
  \pgfmath@x\pgfmath@temp\pgfmath@x%
  \advance\pgfmath@x-0.1195159052pt\relax%
  \pgfmath@x\pgfmath@temp\pgfmath@x%
  \advance\pgfmath@x0.4993615349pt\relax%
  \pgfmath@x\pgfmath@temp\pgfmath@x%
  \advance\pgfmath@x1.000012594pt\relax%
  \ifdim\pgfmath@y<0pt\relax%
  \pgfmath@y-\pgfmath@y%
  \fi%
  \pgfmath@x\pgfmath@tonumber{\pgfmath@y}\pgfmath@x%
  % Invert the scaling factor.
  \pgfmath@tmp@restore\pgfmath@x\c@pgfmath@counta\relax%
  \fi%
  \fi%
  \pgfmath@returnone\pgfmath@x%
  \endgroup%
}

\makeatother

\begin{document}

\pgfmathparse{veclen(0.00006,0.00005)}
Vector length is: \pgfmathresult

\begin{tikzpicture}[decoration={markings, mark = at position .5 with
 {\draw (-2pt,-2pt) -- (2pt,2pt)  (2pt,-2pt) -- (-2pt,2pt);}}]    

% wrong 
\draw [postaction={decorate}] (0,0) -- ++(146:1) arc (146:157:1) -- (0,0);
%fine
\draw [postaction={decorate}] (2,0) -- ++(146:1.2) arc (146:157:1.2) -- (2,0);
\end{tikzpicture}

\end{document}

Result:

decorated curve

As can be seen, the accuracy is not great! However, at that level of precision then it's probably not all that important. Perhaps a better implementation would be to test if the components of the vector are less than some small number and then simply return the maximum of the two: at that level, the difference between the sup norm and the l^2 norm is not a lot! (A slightly more sophisticated version would have a switch that returned the sup norm or the l^1 norm depending on whether it was acceptable to underestimate or overestimate.)

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