[Tex/LaTex] TikZ: Smooth curve through three points without changing direction

tikz-pgf

In another question (see here) I was asking about how to put the control points of a curve on the tangent of a specific curve point. While the given answers where helpful for that particular arrangement of points, I would like to have a solution for a more general approach to that problem. It is important to me, that the curve does not change its direction at any given point. Consider the following example:

\documentclass[border=5pt,tikz]{standalone}

\usepackage{tikz}
\usetikzlibrary{calc,intersections,through}
\tikzset{point/.style={circle,inner sep=0pt,minimum size=3pt,fill=red}}

\begin{document}
\begin{tikzpicture}

\coordinate [label=left:$A$] (A) at (0,0);
\coordinate [label=left:$B$] (B) at (-.2,3);
\coordinate [label=left:$C$] (C) at (-.7,7);

\draw[thin,blue] plot [smooth] coordinates { (A) (B) (C) };

\node[point] at (A) {};
\node[point] at (B) {};
\node[point] at (C) {};

\end{tikzpicture}
\end{document}

This produces a smooth curve through points A, B and C, as you can see here:

However, I want the curve to leave A at an angle of 90 degrees and arrive at C at an angle of -45 degrees, running smoothly through B without changing its direction at any given point. So in this example the curve should always continue to turn to the left. Changing the above \draw statement to this

\draw[thin,green] (A) .. controls +(90:.3)  and +(-85:.2) .. (B) .. controls +(95:.2) and +(-45:.3) .. (C);

produces a curve that is close to what I need.

However, I have to guess about the angle of both control points surrounding B. From the blue plot above I figured that those angles are close to -85 and 95, respectively. But I would rather like to calculate these angles instead of eyeballing them.

So is there a way to determine the input and output angles of a bezier curve through B so that the curve does not change its direction anywhere?

Best Answer

Here's an answer to the question using the hobby package from Curve through a sequence of points with Metapost and TikZ.

\documentclass{article}
%\url{https://tex.stackexchange.com/q/309388/86}

\usepackage{tikz}
\usetikzlibrary{calc,intersections,through,hobby}
\tikzset{point/.style={circle,inner sep=0pt,minimum size=3pt,fill=red}}

\begin{document}
\begin{tikzpicture}[use Hobby shortcut]

\coordinate [label=left:$A$] (A) at (0,0);
\coordinate [label=left:$B$] (B) at (-.2,3);
\coordinate [label=left:$C$] (C) at (-.7,7);

\draw[thin,blue] plot [smooth] coordinates { (A) (B) (C) };

\node[point] at (A) {};
\node[point] at (B) {};
\node[point] at (C) {};

\draw[thin,green,out angle=90,in angle=-45] (A) .. (B) .. (C);
\draw[thin,red] (A) .. (B) .. (C);

\end{tikzpicture}
\end{document}

This produces:

I wasn't aware of the issue with coordinates on the curve (as mentioned in the comments). I'll have to have a look into that to see what's going on with that.

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] TikZ: Drawing an arc from an intersection to an intersection

A solution which allows to draw intersection segments of any two intersections is available as tikz library fillbetween.

This library works as general purpose tikz library, but it is shipped with pgfplots and you need to load pgfplots in order to make it work:

\documentclass{standalone}

\usepackage{tikz}
\usepackage{pgfplots}
\usetikzlibrary{fillbetween}

\begin{document}

\begin{tikzpicture}

\draw [name path=red,red] (120:1.06) circle (1.9);
%\draw [name path=yellow,yellow] (0:1.06) circle (2.12);
\draw [name path=green,green!50!black] (0:0.77) circle (2.41);
\draw [name path=blue,blue] (0:0) circle (1.06);

% substitute this temp path by `\path` to make it invisible:
\draw[name path=temp1, intersection segments={of=red and blue,sequence=L1}];
\draw[red,-stealth,ultra thick, intersection segments={of=temp1 and green,sequence=L3}];

\end{tikzpicture}

\end{document}

enter image description here

The key intersection segments is described in all detail in the pgfplots reference manual section "5.6.6 Intersection Segment Recombination"; the key idea in this case is to

create a temporary path temp1 which is the first intersection segment of red and blue, more precisely, it is the first intersection segment in the Left argument in red and blue : red. This path is drawn as thin black path. Substitute its \draw statement by \path to make it invisible.
Compute the desired intersection segment by intersecting temp1 and green and use the correct intersection segment. By trial and error I figured that it is the third segment of path temp1 which is written as L3 (L = left argument in temp1 and green and 3 means third segment of that path).

The argument involves some trial and error because fillbetween is unaware of the fact that end and startpoint are connected -- and we as end users do not see start and end point.

Note that you can connect these path segments with other paths. If such an intersection segment should be the continuation of another path, use -- as before the first argument in sequence. This allows to fill paths segments:

\documentclass{standalone}

\usepackage{tikz}
\usepackage{pgfplots}
\usetikzlibrary{fillbetween}

\begin{document}

\begin{tikzpicture}

\draw [name path=red,red] (120:1.06) circle (1.9);
%\draw [name path=yellow,yellow] (0:1.06) circle (2.12);
\draw [name path=green,green!50!black] (0:0.77) circle (2.41);
\draw [name path=blue,blue] (0:0) circle (1.06);

% substitute this temp path by `\path` to make it invisible:
\draw[name path=temp1, intersection segments={of=red and blue,sequence=L1}];
\draw[red,fill=blue,-stealth,ultra thick, intersection segments={of=temp1 and green,sequence=L3}]  
    [intersection segments={of=temp1 and green, sequence={--R2}}]
;

\end{tikzpicture}

\end{document}

enter image description here

Best Answer

Related Solutions

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

[Tex/LaTex] TikZ: Drawing an arc from an intersection to an intersection

Related Question