Using TikZ, how to draw an elliptical arc starting from point A to point B with the origin as its center where both radii are given?

Question

I am new to TikZ and reading its manual now. For the sake of exercises, I want to draw Batman's logo as follows:

enter image description here

Using TikZ, how to draw an elliptical arc starting from point A to point B with the origin as its center?

\documentclass{article}
\usepackage{tikz}


\usepackage[active,tightpage]{preview}
\PreviewEnvironment{tikzpicture}
\PreviewBorder=12pt


\begin{document}
\begin{tikzpicture}

    \filldraw[fill opacity=0.5,fill=yellow] (0,0) ellipse (7.0 and 4.3);
    \draw[line width=5mm] (0,0) ellipse (6.5 and 3.8);
    \draw[line width=2pt,dashed,white] (0,0) ellipse (6.0 and 3.3);

    \draw[line width=2pt]   
                            (0,2.7) -- (0.5,2.7) -- (1,3.25) 
                            .. controls (1.2,1.3) and (1.3,1.0) .. 
                            (2.0,1.0) 
                            .. controls (3.0,1.0) and (3.0,2.2) .. 
                            (2,3.1);

    %How to draw an elliptical arc from (2,3.1) to (3.2,-2.8)?
    %\draw ()   arc ();

    \draw[line width=2pt] 
            (3.2,-2.8) 
            .. controls (4,-2) and (4,0) ..
            (2.2,-1.8)
            .. controls (1.5,-1) and (1,-1) ..
            (0,-3.2);
\end{tikzpicture}
\end{document}

Note: Horizontal and vertical radii are given.

Bonus question: Instead of drawing the left part by using the same method as I did for the right part, how to use reflection technique to get a complete Batman's logo?

@AndrewStacey: Is there a simpler way for beginners like me? — kiss my armpit, Mar 9, 2012 at 12:39
percusse has answered in the meantime, but I was thinking that Will Hunting's answer would be easy to adapt to this situation since in that answer the centre is explicitly calculated and then the arc drawn between two of the points. You already have the centre, so don't need to compute it. The only bit missing is whether it is possible to deform it to an ellipse rather than a circle. — Andrew Stacey, Mar 9, 2012 at 12:44

percusse · Accepted Answer · 2012-03-09 14:21:31Z

I would use a slightly lower level PGF (Portable Graphics Format) command \pgfpatharcto for this ~~and I have guessed the 11cm by eyeballing~~. For the bonus I created a scope where I defined the x unit vector to be (-1,0) so whatever is drawn inside, the horizontal coordinate will be reversed. Hence, I can literally copy the right part code into left part except the arc command which accepts absolute coordinates (unless there is any coordinate transformation is going on).

\documentclass{article}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
    \filldraw[fill opacity=0.5,fill=yellow] (0,0) ellipse (7.0 and 4.3);
    \draw[line width=5mm] (0,0) ellipse (6.5 and 3.8);
    \draw[line width=2pt,dashed,white] (0,0) ellipse (6.0 and 3.3);

    \draw[line width=2pt]   (0,2.7) -- (0.5,2.7) -- (1,3.25) 
                            .. controls (1.2,1.3) and (1.3,1.0) .. 
                            (2.0,1.0) .. controls (3.0,1.0) and (3.0,2.2) .. (2,3.1);
    \draw[line width=2pt] 
            (3.2,-2.8) .. controls (4,-2) and (4,0) .. (2.2,-1.8)
            .. controls (1.5,-1) and (1,-1) ..(0,-3.2);

    \pgfsetlinewidth{2pt}
    \pgfmoveto{\pgfpoint{2cm}{3.1cm}}
    \pgfpatharcto{6cm}{3.3cm}{0}{0}{0}{\pgfpoint{3.2cm}{-2.8cm}}\pgfusepath{stroke};

\begin{scope}[x=-1cm]
    \draw[line width=2pt]   (0,2.7) -- (0.5,2.7) -- (1,3.25) 
                            .. controls (1.2,1.3) and (1.3,1.0) .. 
                            (2.0,1.0) .. controls (3.0,1.0) and (3.0,2.2) .. (2,3.1);
    \draw[line width=2pt] 
            (3.2,-2.8) .. controls (4,-2) and (4,0) .. (2.2,-1.8)
            .. controls (1.5,-1) and (1,-1) ..(0,-3.2);
\end{scope}
    \pgfsetlinewidth{2pt}
    \pgfmoveto{\pgfpoint{-2cm}{3.1cm}}
    \pgfpatharcto{6cm}{3.3cm}{0}{0}{1}{\pgfpoint{-3.2cm}{-2.8cm}}\pgfusepath{stroke};
\end{tikzpicture}
\end{document}

enter image description here

EDIT2 : Filling the curve was much difficult with the code above so making a continuous path seemed easier since Damien already provided the necessary coordinates. There is simply no complication other than renaming the commands. The \pgfpathcurveto is equivalent to \draw (coord1) ..controls (support1) and (support2) .. (coord2) except that you have to move to (coord1) first. After the coordinate transformation, it is again a matter of copy-paste.

\documentclass{article}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}

    \filldraw[fill opacity=0.5,fill=yellow] (0,0) ellipse (7.0 and 4.3);
    \draw[line width=5mm] (0,0) ellipse (6.5 and 3.8);
    \draw[line width=2pt,dashed,white] (0,0) ellipse (6.0 and 3.3);

\pgfsetlinewidth{2pt}
\pgfpathmoveto{\pgfpoint{0}{2.7cm}}
\pgfpathlineto{\pgfpoint{0.5cm}{2.7cm}}
\pgfpathlineto{\pgfpoint{1cm}{3.25cm}}
\pgfpathcurveto{\pgfpoint{1.2cm}{1.3cm}}{\pgfpoint{1.3cm}{1cm}}{\pgfpoint{2cm}{1cm}}
\pgfpathcurveto{\pgfpoint{3cm}{1cm}}{\pgfpoint{3cm}{2.2cm}}{\pgfpoint{2cm}{3.1cm}}
\pgfpatharcto{6cm}{3.3cm}{0}{0}{0}{\pgfpoint{3.2cm}{-2.8cm}}
\pgfpathcurveto{\pgfpoint{4cm}{-2cm}}{\pgfpoint{4cm}{0}}{\pgfpoint{2.2cm}{-1.8cm}}
\pgfpathcurveto{\pgfpoint{1.5cm}{-1cm}}{\pgfpoint{1cm}{-1cm}}{\pgfpoint{0cm}{-3.2cm}}
\pgftransformcm{-1}{0}{0}{1}{\pgfpointorigin} % This is the coordinate change from x to -x
\pgfpathcurveto{\pgfpoint{1cm}{-1cm}}{\pgfpoint{1.5cm}{-1cm}}{\pgfpoint{2.2cm}{-1.8cm}}
\pgfpathcurveto{\pgfpoint{4cm}{0cm}}{\pgfpoint{4cm}{-2cm}}{\pgfpoint{3.2cm}{-2.8cm}}
\pgfpatharcto{6cm}{3.3cm}{0}{0}{1}{\pgfpoint{2cm}{3.1cm}}
\pgfpathcurveto{\pgfpoint{3cm}{2.2cm}}{\pgfpoint{3cm}{1cm}}{\pgfpoint{2cm}{1cm}}
\pgfpathcurveto{\pgfpoint{1.3cm}{1cm}}{\pgfpoint{1.2cm}{1.3cm}}{\pgfpoint{1cm}{3.25cm}}
\pgfpathlineto{\pgfpoint{0.5cm}{2.7cm}}
\pgfpathclose
\pgfusepath{fill,stroke}   


\end{tikzpicture}
\end{document}

EDIT3 Alfred, join this curve nicely will ya? So we draw one half and carry on backwards for a proper line join at the tail and fill.

enter image description here

Admit it: you had this already coded and were just waiting for someone to post a question. Right? — Andrew Stacey, Mar 9, 2012 at 12:45
@AndrewStacey Hahaha, actually I was trying to fix this answer while this was posted. — percusse, Mar 9, 2012 at 12:49
It does seem fit for the shape library to create a pure shape with prober anchors etc. Who don't need a \node[batman] {Bat me up};... — nickpapior, Mar 9, 2012 at 13:51

5 revs · Accepted Answer · 2013-02-19 02:14:54Z

enter image description here

\documentclass[pstricks,border={-8.5mm -5.5mm -8.5mm -5.0mm}]{standalone}
\usepackage{pst-eucl}
\usepackage{pst-grad}
\usepackage{graphicx}

\newpsstyle{A}
{
    fillstyle=gradient,
    gradbegin=red,
    gradend=yellow,
    gradangle=30,
    gradmidpoint=0.5,
    linestyle=none
}

\newpsstyle{B}{
    fillstyle=gradient,
    gradbegin=darkgray,
    gradend=black,
    gradangle=45,
    gradmidpoint=1,
    linestyle=none  
}


\psset
{
    runit=\psunit,
    fillstyle=solid,
    PointName=none,
    PointSymbol=none,
}

\pstVerb
{
    /theta 72 def
    /Major 6.0 def
    /Minor 3.3 def
}

\def\DeclareNodes
{
    \pstGeonode
        (0,-8){Bottom}
        (0,8){Top}
        (0,0){center}
        (0,2.7){A}
        (0.5,2.7){B}
        (1,3.25){C}
        (1.2,1.3){D}
        (1.3,1.0){E}
        (2.0,1.0){F}
        (3.0,1.0){G}
        (3.0,2.2){H}
        (!Minor Major theta PtoC){I}
        (!Minor Major theta neg PtoC){J}
        (4,-2){K}
        (4,0){L}
        (2.2,-1.8){M}
        (1.5,-1){N}
        (1,-1){O}
        (0,-3.2){P}
    \pstOrtSym{Bottom}{Top}{B,C,D,E,F,G,H,I,J,K,L,M,N,O}
}

\def\RightPart
{
    \psline(A)(B)(C)
    \psbezier(D)(E)(F)
    \psbezier(G)(H)(I)
    \psellipticarcn[dimen=middle](center)(!Major Minor){(I)}{(J)}
    \psbezier(K)(L)(M)
    \psbezier(N)(O)(P)
}


\def\LeftPart
{
    \psbezier(O')(N')(M')
    \psbezier(L')(K')(J')
    \psellipticarcn[dimen=middle](center)(!Major Minor){(J')}{(I')}
    \psbezier(H')(G')(F')
    \psbezier(E')(D')(C')
    \psline(B')
}


\begin{document}


\begin{pspicture}[showgrid=false](-7.85,-4.85)(7.85,4.80)
    \psellipse[style=A](0,0)(!Major 1 add Minor 1 add)
    \psellipse[style=B](0,0)(!Major 0.75 add Minor 0.75 add)
    \psellipse[style=A](0,0)(!Major 0.25 add Minor 0.25 add)
    \DeclareNodes
    \pscustom*{\RightPart\LeftPart\closepath}
\end{pspicture}


\begin{pspicture}[showgrid=false](-7.85,-4.85)(7.85,4.80)%(-7,-4)(7,4)
    \psellipse[style=A](0,0)(!Major 1 add Minor 1 add)
    \psellipse[style=B](0,0)(!Major 0.75 add Minor 0.75 add)
    \psellipse[style=A](0,0)(!Major 0.25 add Minor 0.25 add)
    \begin{psclip}{\DeclareNodes\pscustom[linewidth=6pt]{\RightPart\LeftPart\closepath}}
        \rput(0,0){\includegraphics[width=2\dimexpr7.85\psunit\relax]{example-grid-100x100pt}}
    \end{psclip}
\end{pspicture}


\end{document}

I get "\psunit - unrecognized command", same for psclip
– InLaw
Mar 2 at 18:54 — InLaw, Mar 2 at 18:54

Alain Matthes · Accepted Answer · 2012-03-09 18:17:14Z

Not really an answer, but only to get an homogeneous version that I can transform for use with pgfornament. I used pgfqpoint but in this case I need to add the unity (cm) for each numbers.

\begin{tikzpicture}
  \pgfsetlinewidth{0.4 pt}
  \pgfpathellipse{\pgfqpoint{0 cm}{0 cm}}
                 {\pgfqpoint{7 cm}{0 cm}}
                 {\pgfqpoint{0 cm}{4.3cm}} 
  \pgfsetfillopacity{.5}  
  \pgfsetfillcolor{yellow}
  \pgfusepath{fill,stroke}   
  \pgfsetlinewidth{30\pgflinewidth}
  \pgfpathellipse{\pgfqpoint{0 cm}{0 cm}}
                 {\pgfqpoint{6.5 cm}{0 cm}}
                 {\pgfqpoint{0 cm}{3.8 cm}}
  \pgfusepath{stroke}
  \pgfsetlinewidth{0.4 pt}       
  \pgfpathmoveto{\pgfqpoint{0 cm}{2.7 cm}}
  \pgfpathlineto{\pgfqpoint{0.5 cm}{2.7 cm}}
  \pgfpathlineto{\pgfqpoint{1 cm}{3.25 cm}}
  \pgfpathcurveto{\pgfqpoint{1.2 cm}{1.3 cm}}
                 {\pgfqpoint{1.3 cm}{1 cm}}
                 {\pgfqpoint{2 cm}{1 cm}}
  \pgfpathcurveto{\pgfqpoint{3 cm}{1 cm}}
                 {\pgfqpoint{3 cm}{2.2 cm}}
                 {\pgfqpoint{2 cm}{3.1 cm}}
  \pgfpatharcto{6 cm}{3.3 cm}{0}{0}{0}
               {\pgfqpoint{3.2 cm}{-2.8 cm}}
  \pgfpathcurveto{\pgfqpoint{4 cm}{-2 cm}}
                 {\pgfqpoint{4 cm}{0 cm}}
                 {\pgfqpoint{2.2 cm}{-1.8 cm}}
  \pgfpathcurveto{\pgfqpoint{1.5 cm}{-1 cm}}
                 {\pgfqpoint{1 cm}{-1 cm}}
                 {\pgfqpoint{0 cm}{-3.2 cm}}
  \pgftransformcm{-1}{0}{0}{1}{\pgfpointorigin}
  \pgfpathcurveto{\pgfqpoint{1 cm}{-1 cm}}
                 {\pgfqpoint{1.5 cm}{-1 cm}}
                 {\pgfqpoint{2.2cm}{-1.8cm}}
  \pgfpathcurveto{\pgfqpoint{4 cm}{0 cm}}
               {\pgfqpoint{4 cm}{-2 cm}}
               {\pgfqpoint{3.2 cm}{-2.8 cm}}
  \pgfpatharcto{6 cm}{3.3 cm}{0}{0}{1}
               {\pgfqpoint{2 cm}{3.1 cm}}
  \pgfpathcurveto{\pgfqpoint{3 cm}{2.2 cm}}
               {\pgfqpoint{3 cm}{1 cm}}
               {\pgfqpoint{2 cm}{1 cm}}
   \pgfpathcurveto{\pgfqpoint{1.3cm}{1 cm}}
               {\pgfqpoint{1.2cm}{1.3 cm}}
               {\pgfqpoint{1 cm}{3.25 cm}}
   \pgfpathlineto{\pgfqpoint{0.5 cm}{2.7 cm}}
   \pgfsetfillopacity{1}  
   \pgfsetfillcolor{black} 
   \pgfusepath{fill,stroke}
\end{tikzpicture}

The shape : (but I have some problems)

enter image description here

Stack Exchange Network

current community

your communities

more stack exchange communities

Using TikZ, how to draw an elliptical arc starting from point A to point B with the origin as its center where both radii are given?

3 Answers 3

Your Answer

Not the answer you're looking for? Browse other questions tagged tikz-pgf or ask your own question.

Linked

Hot Network Questions

Using TikZ, how to draw an elliptical arc starting from point A to point B with the origin as its center where both radii are given?

3 Answers 3

Your Answer

Sign up or log in

Post as a guest

Not the answer you're looking for? Browse other questions tagged tikz-pgf or ask your own question.

Linked

Related

Hot Network Questions