[Tex/LaTex] How to decorate a path with arrows at certain coordinates

decorationsintersectionstikz-pgf

Next example (taken from TiKZ manual) shows how to find intersection points between two paths.

Once you know intersection points it's easy to decorate them with any marking. But if I want to use an arrow following the path I need to convert coordinate to position (percentage of total length) over the path.
Do you know how to do it?

To draw the example I've used a trial and error method to place arrows. I would like to use (i-\s) coordinates.

\documentclass[border=2mm,tikz]{standalone}
\usetikzlibrary{intersections,decorations.markings}

\begin{document}
\begin{tikzpicture}[]
\clip (-2,-2) rectangle (2,2);
\draw [name path=curve 1] (-2,-1) .. controls (8,-1) and (-8,1) .. (2,1);
\draw [name path=curve 2] (-1,-2) .. controls (-1,8) and (1,-8) .. (1,2);
\fill [name intersections={of=curve 1 and curve 2, name=i, total=\t}]
[red, opacity=0.5, every node/.style={above left, black, opacity=1}]
\foreach \s in {1,...,\t}{(i-\s) circle (1pt) node {\footnotesize\s}};
\draw[decorate,decoration={ 
     markings,% 
     mark=at position .72  with {\arrow[red,line width=1pt]{>}},%
     mark=at position .595  with {\arrow[blue,line width=1pt]{>}},%
     mark=at position .225  with {\arrow[green,line width=1pt]{>}}}]% 
     (-2,-1) .. controls (8,-1) and (-8,1) .. (2,1);
\end{tikzpicture}
\end{document}

enter image description here

EDIT: I will try to explain it better.

In my example I've used mark=at position .72 where .72 was decided after several drawings. I would like to use something like mark=at position corresponding to (i-4).
where (i-4) is one of the intersection point or any other specific point over the path but with an unknown position. This way would be possible to decorate a path with an arrow from, for example, (i-4) to (i-9) or would be possible to provide an exact answer to
Decorate path with arrows at nodes. Provided answers also position the arrow with some approximation and not with the precision that TiKZ shows us all around.

Best Answer

The following excerpts of text can be found on pages 243 and 244 of the manual:

One final word of warning: Decorations can be pretty slow to typeset and they can be inaccurate.

and

Due to the limits on the precision in TEX, some inaccuracies in positioning when crossing input segment boundaries may occasionally be found.

That is, the accuracy of the feature decoration is limited and we will see it now. The picture below was built by my code and it is at the end of this answer.

(in)accurate of decorations

The yellow curve is the representation of the respective control curve with nodes. Within these we have identified the points of intersection and colored red. The blue points are the representation of these same intersection points, but using the decorations feature (strange, huh?). The green arrows are a way I found to solve your problem.

This is a slow solution since I did the intersection of the path with itself to turn it into a set of nodes. I believe there must be a more elegant way of doing this. =)

etoolbox package is used to facilitate the management of variables and because of the comparisons.

Furthermore, it was necessary to add a margin of error at the time of comparison. Depending on what you want to do, this margin may need to be increased or decreased.

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

\usepackage{calc}
\usepackage{etoolbox}

\usetikzlibrary{intersections, calc, shapes.arrows, decorations.markings}

\newcounter{cnt}
\newcommand\setpoint[2]{\csxdef{point#1}{#2}}
\newcommand\getpoint[1]{\csuse{point#1}}

\newcounter{curveindex}

\begin{document}
    \begin{tikzpicture} 

        \clip (-2,-2) rectangle (2,2);

        \draw [name path=curve 1] (-2,-1) .. controls (8,-1) and (-8,1) .. (2,1);
        \draw [name path=curve 2] (-1,-2) .. controls (-1,8) and (1,-8) .. (1,2);

        \path [name intersections={of=curve 1 and curve 1, name=i, total=\t, sort by = curve 1}] node {\xdef\totalone{\t}};
        \path [name intersections={of=curve 1 and curve 2, name=j, total=\t, sort by = curve 1}] node {\xdef\totaltwo{\t}};

        \pgfmathsetmacro{\myunit}{1/(\totalone)}

        \newdimen\xone
        \newdimen\yone
        \newdimen\xtwo
        \newdimen\ytwo

        \foreach \q in {1, ..., \totaltwo}
        {
            \pgfextractx{\xtwo}{\pgfpointanchor{j-\q}{center}}
            \pgfextracty{\ytwo}{\pgfpointanchor{j-\q}{center}}

            \foreach \p in {1, ..., \totalone}
            {
                \pgfextractx{\xone}{\pgfpointanchor{i-\p}{center}}
                \pgfextracty{\yone}{\pgfpointanchor{i-\p}{center}}

                \ifboolexpr{
                    test {\ifdimless{\xtwo - 0.5pt}{\xone}} and test {\ifdimless{\xone}{\xtwo + 0.5pt}}
                    and
                    test {\ifdimless{\ytwo - 0.5pt}{\yone}} and test {\ifdimless{\yone}{\ytwo + 0.5pt}}
                }{
                    \setpoint{\q}{\p}
                }{
                }
            }
        }

        \edef\mypath{}
        \pgfmathsetmacro{\lastbutone}{\totalone - 1}
        \foreach \k in {1, ..., \lastbutone}
        {
            \xdef\mypath{\mypath (i-\k) -- }
        }
        \edef\mypath{\mypath (i-\totalone)}

        \draw [yellow, line width = 1mm, opacity = 0.5] \mypath;

        \foreach \k in {1, ..., \totaltwo}
        {
            \pgfmathsetmacro{\positioning}{\getpoint{\k} * \myunit}
            \draw [decorate, decoration = {markings, mark = at position \positioning with {
                \node [
                    circle
                    , fill = blue
                    , inner sep = 1pt
                ]{};
            }}] \mypath;
        }

        \foreach \k in {1, ..., \totaltwo}
        {
           \node [
                circle
                , fill = red
                , inner sep = 1pt
                , opacity = 0.7
            ] at (i-\getpoint{\k}) {};
        }

        \foreach \k in {1, ..., \totaltwo}
        {
            \pgfmathsetmacro{\lastpoint}{\getpoint{\k} - 1}
            \draw [decorate, decoration = {markings, mark = at position -0.1pt with {
                \node [
                    single arrow
                    , fill = green
                    , anchor = east
                    , minimum size = 2mm
                    , inner sep = 1pt
                    , single arrow head extend = 2pt
                    , transform shape
                    , opacity = 0.8
                ]{};
            }}] (i-\lastpoint) -- (i-\getpoint{\k});
        }

    \end{tikzpicture}
\end{document}

Related Solutions

[Tex/LaTex] Rotate a node but not its content: the case of the ellipse decoration

The source of the difficulty is that ellipses are constructed in a particular way in TikZ. They are paths that start from the x-axis and proceed counter-clockwise around their centre. The vast majority of the time, the exact parametrisation doesn't matter. You appear to have found the one situation where it does!

In the actual question, you only want to be able to mirror the ellipse, and so draw it starting from the negative x-axis (the title of the question suggests a more flexible approach). That's actually not too hard since we can exploit the symmetry of the ellipse. The key is to provide it with a negative x-radius, since then it will start from the negative x-axis (and proceed clockwise, but we could correct for that by negating the y-radius as well). To do this, we interrupt the call from the node shape to the drawing command and flip the sign of the x-radius. The simplest way to do this is to redefine the \pgfpathellipse macro to do the negation and then call the original macro. The following code does this.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{decorations,shapes,decorations.markings}
\makeatletter
\let\origpgfpathellipse=\pgfpathellipse
\def\revpgfpathellipse#1#2#3{%
  #2%
  \pgf@xa=-\pgf@x
  \origpgfpathellipse{#1}{\pgfqpoint{\pgf@xa}{0pt}}{#3}}
\makeatother

\tikzset{
  reversed ellipse/.style={
    ellipse,
    reverse the ellipse%
  },
  reverse the ellipse/.code={
    \let\pgfpathellipse=\revpgfpathellipse
  }
}
\begin{document}
\begin{tikzpicture}
\node[ellipse,
  draw,
  postaction={
    decorate,
    decoration={
      markings,
      mark=at position 1 with {
        \arrow[line width=5pt,blue]{>}
      }
    }
  }
] at (0,0) {hello world};

\node[reversed ellipse,
  draw,
  postaction={
    decorate,
    decoration={
      markings,
      mark=at position 1 with {
        \arrow[line width=5pt,blue]{>}
      }
    }
  }
] at (0,-2) {hello world};
\end{tikzpicture}

\end{document}

Here's the result:

rotated ellipse

(the arrow got clipped, but you can see where it lies)

[Tex/LaTex] drawing arrows on a circle path and a tangent arrow in tikz

Yes, it is ;-) Try something and when you are stuck, ask for specific help. Good people from the community will help.

enter image description here

\documentclass[tikz]{standalone}
\usetikzlibrary{shapes}
\begin{document}

\xdef\Rad{2}
\newcommand{\ARW}[2][]{%
    \foreach \ang in {#2}{%
        \draw[#1] (\ang:\Rad)--(\ang+1:\Rad) ;
    }
}

\begin{tikzpicture}[>=stealth]

\draw[thick] (0,0) circle (\Rad) ;

\node[draw,circle,fill=white] (N1) at (30:\Rad) {A node} ;
\node[draw,fill=white] (N2) at (190:\Rad) {Another node} ;
\node[draw,fill=white,rounded corners=2mm] (N3) at (225:\Rad) {A different node} ;
\node[draw,fill=white] (N4) at (340:\Rad) {A node} ;


\begin{scope}[shift={(110:\Rad)},rotate=20]
\node[draw,fill=red!20,star,star points=4] (N5) at (3,0) {Here} ;
\node[draw,fill=blue!20,regular polygon,regular polygon sides=5]
    (N6) at (-3,0) {There} ;
\draw[->,thick] (N6)--(0,0)--(N5) ;
\end{scope}

\ARW[thick,red,<-]{50.5, 197, 236, 347};

\end{tikzpicture}
\end{document}

Best Answer

Related Solutions

[Tex/LaTex] Rotate a node but not its content: the case of the ellipse decoration

[Tex/LaTex] drawing arrows on a circle path and a tangent arrow in tikz

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