[Tex/LaTex] tikz: draw a piece of a path between given coordinates

tikz-pgf

I have a path in my application that I want to split up in smaller paths. These, I would like to draw with a label attached to them. Take the following image as an example:

What I am trying to achieve

The upper arc is split up into three segments by the horizontal lines and by the two lines coming out of the center of the circle. I draw all three parts of the path and attach a label to them.

Now, the way I do this now is a bit tedious. For each segment I have to reconstruct the path by hand: I have to specify how the path is continued at each intersection. In this case the intersection is simple and the continuation is clear, but if the path is complicated this becomes more difficult.

What I would like to do is the following:

Define a full path, in this case the upper arc from the the left all the way to the right. This path could be arbitrarily complicated.
Find coordinates in the path by intersecting with other paths.
Have a command to say: draw the path from the first intersection coordinate to the second and assign a label to this segment. Draw the path from the second intersection to the third, etc.

How can I achieve this?

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{calc,shapes.multipart,fit,shapes,calc,backgrounds,decorations.pathreplacing,decorations.markings,intersections}

\begin{document}

\begin{tikzpicture}[scale=2]
\path[name path=t] (1,0) to[out=90,in=180] (2,1) to[out=0,in=90] (3,0); % top arc
\path [name path=ml] (2,0) -- (1,1);
\path [name intersections={of=ml and t}];
\coordinate (A) at (intersection-1);
\path [name path=l] (2,0) -- (3,1);
\path [name intersections={of=l and t}];
\coordinate (B) at (intersection-1);

\draw[thick,draw=blue,] (3,0) to[out=90,in=315] node[midway,above right] {$c$} (B) to[out=135,in=45] node[midway,above] {$b$} (A) to[out=225,in=90] node[midway,above left] {$a$} (1,0);
\draw[thick,draw=blue,] (1,0) -- (0,0) node[midway,above] {};

\draw[thick,draw=blue,] (1,0) to[out=-90,in=180] node[midway,below left] {} (2,-1);
\draw[thick,draw=blue,] (2,-1) to[out=0,in=-90] node[midway,below right] {} (3,0);
\draw[thick,draw=blue,] (4,0) -- (3,0) node[midway,above] {};

\draw[thick,draw=blue,] (2,0) -- node[midway,right] {} (2,-1);
\draw[thick,draw=blue,] (A) -- node[midway,below left] {} (2,0);
\draw[thick,draw=blue,] (2,0) -- node[midway,below right] {} (B);
\end{tikzpicture}
\end{document}

Best Answer

This is a very late answer and the tools used in it were, if I am not mistaken, not available at the time you asked the question. I found this question with this query, and the purpose of this answer is just to mention that now rather convenient means to do what you suggest. The new tool is the pgfplots library fillbetween that allows you to split paths in segments, which you can redraw, fill, or use in other ways.

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{pgfplots}
\usetikzlibrary{calc,shapes.multipart,fit,shapes,calc,backgrounds,decorations.pathreplacing,decorations.markings,intersections}
\usepgfplotslibrary{fillbetween}
\begin{document}

\begin{tikzpicture}[scale=2]
\path[name path=t] (1,0) to[out=90,in=180] (2,1) to[out=0,in=90] (3,0); % top arc
\path [name path=ml] (2,0) -- (1,1);
\path [name intersections={of=ml and t}];
\coordinate (A) at (intersection-1);
\path [name path=l] (2,0) -- (3,1);
\path [name intersections={of=l and t}];
\coordinate (B) at (intersection-1);


\draw[thick,draw=blue,] (3,0) to[out=90,in=315] node[midway,above right] {$c$} (B) to[out=135,in=45] node[midway,above] {$b$} (A) to[out=225,in=90] node[midway,above left] {$a$} (1,0);
\draw[thick,draw=blue,] (1,0) -- (0,0) node[midway,above] {};

\draw[thick,draw=blue,] (1,0) to[out=-90,in=180] node[midway,below left] {} (2,-1);
\draw[thick,draw=blue,] (2,-1) to[out=0,in=-90] node[midway,below right] {} (3,0);
\draw[thick,draw=blue,] (4,0) -- (3,0) node[midway,above] {};

\draw[thick,draw=blue,] (2,0) -- node[midway,right] {} (2,-1);
\draw[thick,draw=blue,] (A) -- node[midway,below left] {} (2,0);
\draw[thick,draw=blue,] (2,0) -- node[midway,below right] {} (B);

% draw the left segment
\draw[thick,red,
        intersection segments={of=t and ml,sequence={L1}}];
% construct an auxiliary path
\path[name path=aux,%draw=green,thick,
        intersection segments={of=t and ml,sequence={L0}}];
% draw the middle segment
\draw[thick,green!60!black,
        intersection segments={of=aux and l,sequence={L1}}];
% draw the right segment
\draw[thick,purple,
        intersection segments={of=aux and l,sequence={L0}}];
\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] How to draw an elliptical arc in TikZ given only opposite extreme points (vertices) of the ellipse

I think that this problem is quite different that what the title implies. If you are are not doing these kind of figure often, you can just apply a scale=0.7 to yield the desired result:

enter image description here

Code:

\documentclass[11pt]{article}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath,amssymb,amsthm,amsfonts,graphicx}
\usepackage{tikz}
\usetikzlibrary{calc,decorations.markings,positioning}

\begin{document}
\begin{center}
\begin{tikzpicture}

\draw[postaction={decorate,decoration={markings,
      mark=at position 0.7 with {\node (a) {};}}}]
      (0,0) ellipse (0.6 and 0.4);

\draw[postaction={decorate,decoration={markings,
      mark=at position 0.7 with {\node (b) {};}}}]
      (0,0) ellipse (1.2 and 1);

\node (c) at ($(a)!0.5!(b)$) {};

\begin{scope}[red, shift={(c)},x={(a)}, scale=0.7]
\draw (1,0) arc (0:180:1 and 0.3);
\draw[dashed] (-1,0) arc (180:360:1 and 0.3);
\end{scope}
\end{tikzpicture}
\end{center}
\end{document}

Best Answer

Related Solutions

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

[Tex/LaTex] How to draw an elliptical arc in TikZ given only opposite extreme points (vertices) of the ellipse

Code:

Related Question