3

With nodes (expected output)

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-node}
\begin{document}
\begin{pspicture}[showgrid=t,dimen=m](-2,0)(2,4)
    \pnode(0,3){A}
    \pnode(!1 210 PtoC 3 add){B}
    \pnode(!1 -30 PtoC 3 add){C}
    \qdisk(A){3pt}
    \qdisk(B){3pt}
    \qdisk(C){3pt}
    \pscircle(A){1}
    \psarc[linecolor=red,origin={A}](A){1}{(B)}{(C)}
\end{pspicture} 
\end{document}

enter image description here

With literals

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-node}
\begin{document}
\begin{pspicture}[showgrid=t,dimen=m](-2,0)(2,7)
    \qdisk(0,3){3pt}
    \qdisk(!1 210 PtoC 3 add){3pt}
    \qdisk(!1 -30 PtoC 3 add){3pt}
    \pscircle(0,3){1}
    \psarc[linecolor=red,origin={0,3}](0,3){1}{(!1 210 PtoC 3 add)}{(!1 -30 PtoC 3 add)}
\end{pspicture}
\end{document}

enter image description here

Question

How to fix the second one such that its output is similar to the first one?

1

In your second example the coordinates seems to be calculated relative to (0,3), so you don't have to add 3. Nodes you set before are fixed and not relative.

\documentclass[pstricks,border=12pt]{standalone}
\usepackage{pst-node}
\begin{document}
\makeatletter
\begin{pspicture}[showgrid=t,dimen=m](-2,0)(2,7)
    \qdisk(0,3){3pt}
    \qdisk(!1 210 PtoC 3 add){3pt}
    \qdisk(!1 -30 PtoC 3 add){3pt}
    \pscircle(0,3){1}
    \psarc[linecolor=red](0,3){1}{(!1 210 PtoC)}{(!1 -30 PtoC)}
\end{pspicture}
\makeatother
\end{document}
  • Nodes are fixed in the current node dictionary – user2478 Dec 3 '18 at 11:48

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