Graphics, Figures & Tables ⇒ pst-poly | Center of a Polygon (again)

[svend_tveskaeg]

Hi all.

First: On http://latex-community.org/forum/viewtopic.php?t=21601, Thorsten pointed me to an example on how to define the center of a \psPolygon. I am not sure if the following has anything to do with this.

Consider the following MWE:

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\documentclass{article}

\usepackage{auto-pst-pdf,pst-poly,pstricks-add}
\usepackage{expl3}

\ExplSyntaxOn
  \cs_new_eq:NN
    \calculate
  \fp_eval:n
\ExplSyntaxOff

\begin{document}

\begin{figure}
\def\AntalSider{8 }
\def\Sidelaengde{60 }
\def\RadiusB{78.39377789258259 }
%\def\RadiusB{\calculate{\Sidelaengde/(2*sin(180/\AntalSider))} }
\def\RelVinkel{\calculate{360/\AntalSider} }
\def\Rotation{\calculate{\RelVinkel/2} }
 \centering
  \begin{pspicture}[showgrid=false](156.8,156.8)
  \psset{unit=0.05cm}
   \pnode(!\RadiusB \RadiusB){A}
   \pscircle(A){\RadiusB}
   \PstPolygon[PolyNbSides=\AntalSider,PolyRotation=\Rotation,unit=\RadiusB]
  \end{pspicture}
\end{figure}

\end{document}

This compiles just fine but when

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\def\RadiusB{78.39377789258259 }

is commented out instead of

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\def\RadiusB{\calculate{\Sidelaengde/(2*sin(180/\AntalSider))} }

it will not compile. I simply do not understand this since

Code: Select all

60/(2*sin(180/8)) = 78.39377789258259

I would really like to have the latter of the two definitions of \RadiusB so that I do not have to re-calculate the radius of the circle (\RadiusB) manually every time I change the number of vertices of the polygon (\AntalSider) or the length of the sides in the polygon (\Sidelaengde).

Thank you in advance!