4

I want to draw a batman logo. As it is symmetric about the vertical line, it should be more convenient to just define the right part (or the left part) first and then reflect it to get the other part. The important note you must consider is that the logo must be constructed by a closed curve that can be used as a clipping path.

According to pst-news13 on page 4, we have \reversepath that can be used to construct a curve in reversed coordinates.

Unfortunately, I failed to use \reversepath as shown in the following code.

\documentclass[pstricks,border=6pt]{standalone}
\SpecialCoor

\pstVerb
{
    /theta 72 def
    /Major 6.0 def
    /Minor 3.3 def
    % b a t p2c ---> x y
    % where b (semi-minor), a (semi-major), t (theta)
    /p2c {dup 3 1 roll cos mul 3 1 roll sin mul} bind def
}


\def\RightPart
{
    \psline(0,2.7)(0.5,2.7)(1,3.25)
    \psbezier(1.2,1.3)(1.3,1.0)(2.0,1.0)
    \psbezier(3.0,1.0)(3.0,2.2)(!Minor Major theta p2c)
    \psellipticarcn(0,0)(!Major Minor){(!Minor Major theta p2c)}{(!Minor Major theta neg p2c)}
    \psbezier(4,-2)(4,0)(2.2,-1.8)
    \psbezier(1.5,-1)(1,-1)(0,-3.2)
}

\def\LeftPart
{
    \scale{-1 1}
    \reversepath
    \psbezier(1.5,-1)(1,-1)(0,-3.2)
    \psbezier(4,-2)(4,0)(2.2,-1.8)
    \psellipticarcn(0,0)(!Major Minor){(!Minor Major theta p2c)}{(!Minor Major theta neg p2c)}
    \psbezier(3.0,1.0)(3.0,2.2)(!Minor Major theta p2c)
    \psbezier(1.2,1.3)(1.3,1.0)(2.0,1.0)
    \psline(0,2.7)(0.5,2.7)(1,3.25)
    \closepath
}


\begin{document}
\begin{pspicture}[showgrid=false](-7.85,-4.85)(7.85,4.85)
    \pscustom[dimen=middle]{\RightPart}
\end{pspicture}
\begin{pspicture}[showgrid=false](-7.85,-4.85)(7.85,4.85)
    \pscustom[dimen=middle]{\LeftPart}
\end{pspicture}
\begin{pspicture}[showgrid=false](-7.85,-4.85)(7.85,4.85)
    \pscustom[dimen=middle,fillstyle=solid,fillcolor=red]{\RightPart\LeftPart}
\end{pspicture}
\end{document}

enter image description here

So my question is how to use \reversepath in this case?

0

1 Answer 1

5

Having wasted much time for figuring out how \reversepath works, its behavior is now understood.

\reversepath affects the previous paths rather than the following paths (that most people probably assume). Its behavior is inconsistent with other macros (such as \scale, \translate, etc) defined for pscustom. It is one of many other inconsistency in PSTricks. But it is still fun!

\documentclass[pstricks,border=12pt]{standalone}


% b a t p2c ---> x y
% where b (semi-minor), a (semi-major), t (theta)
\pstVerb{/p2c {dup 3 1 roll cos mul 3 1 roll sin mul} bind def}


\def\RightPart
{
    \psline(0.5,2.7)(1,3.25)
    \psbezier(1.2,1.3)(1.3,1.0)(2.0,1.0)
    \psbezier(3.0,1.0)(3.0,2.2)(!3.3 6.0 72 p2c)
    \psellipticarcn(6.0,3.3){(!3.3 6.0 72 p2c)}{(!3.3 6.0 72 neg p2c)}
    \psbezier(4,-2)(4,0)(2.2,-1.8)
    \psbezier(1.5,-1)(1,-1)(0,-3.2)
}


\begin{document}

\begin{pspicture}[showgrid](-7,-4)(7,4)
\pscustom[dimen=middle,fillstyle=solid,fillcolor=orange,linewidth=6pt]
{
    \RightPart
    \reversepath
    \scale{-1 1}
    \RightPart
    \closepath
}
\end{pspicture}
\end{document}

enter image description here

3
  • And now Stiff Jokes will obtain a bounty from Stiff Jokes. ;-) Dec 30, 2013 at 10:16
  • @PrzemysławScherwentke: It is impossible for the current implementation of the StackExchange system. Stiff Jokes can only obtain a bounty from Steve Jobs or vice verse. Dec 30, 2013 at 12:21
  • No, this is not an inconsistency in PSTricks, but behaves like defined in the Postscript Language, see the redbook.
    – Christoph
    May 16, 2014 at 19:54

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