I would like to use PSTricks to draw a line. However, I only have one point, but I know at which angle I want to draw the line, and how long I want it.
Is there some way to draw a line from a point (x,y) at a given angle, for a given length?
Are there any macro packages that would help me out here?
PSTricks has polar coordinates: see page 13 in the manual.
\SpecialCoor
\psline[linewidth=2pt](4;50)(0,0)(4;10)
\psarc[arcsepB=2pt]{->}{3}{10}{50}
produces 
which will require that you move the coordinate system so that the origin is at the point you have.
Otherwise, you could use the more elaborate options of \SpecialCoor, listed at pp. 72. specifically the ([par]node) notation which describes
The position relative to node determined using the
angle,nodesepandoffsetparameters. E.g.,([angle=45]A),
but requires that you define the current point you have as a named node (not so painful).
Thus:
\documentclass{article}
\usepackage{pstricks}
\usepackage{pst-node}
\begin{document}
\begin{pspicture}(4,5)
\pnode(1,1){Y}
\SpecialCoor
\psline(Y)([angle=30,nodesep=2]Y)
\end{pspicture}
\end{document}
creates:
it is possible to draw a vectorline which always starts at the last current point. It needs only one coordinate the other one is the current point. All points are internally saved as nodes, which can be used later.
\documentclass{article}
\usepackage{pstricks-add,multido}
\begin{document}
\begin{pspicture}[showgrid,linewidth=1pt](10,10.4)
\psStartPoint[A](1,1)% nodes have the base name A
\psVector(3;30)\psVector(4;60)\psVector[linecolor=red](3;10)
\psVector[linestyle=dashed](4;110)
\psline{->}(A0)(A4)
\psStartPoint[B](1,1)\psset{markAngle}% nodes have the base name B
\psVector[linestyle=dashed](4;110)
\psVector[linecolor=red](3;10)
\psVector(4;60)\psVector(3;30)
\multido{\iA=0+1}{5}{\uput[0](A\iA){A\iA}\uput[180](B\iA){B\iA}}
\end{pspicture}
\end{document}
\psVectors enclosed by \pscustom still work? In my attempt, all the tails become at the same point rather than what they are supposed to be.
– kiss my armpit
Jul 18 '12 at 11:22
\psVectors inside the \pscustom for my answer here. But they do not work in \pscustom. As a replacement, I used \rlineto and it works.
– kiss my armpit
Jul 18 '12 at 14:34
you need to know a little bit of postscript language and reverse polish notation:
\documentclass{article}
\usepackage{pstricks-add}
\usepackage[top=3cm,bottom=3cm,left=3cm,right=3cm]{geometry}
\begin{document}
\begin{figure}[h]
\def\xA{1}
\def\yA{2}
\def\Length{4}
\def\Angle{45} % angles are in degrees
\begin{pspicture}(0,0)(6,6)
\psgrid
% define point A
\cnode*(\xA,\yA){0pt}{A}%
% define point B
\cnode*(! \Length\space \Angle\space cos mul \xA\space add \Length\space \Angle\space sin mul \yA\space add){0pt}{B}%
% draw line
\psline[linewidth=1pt](A)(B)%
\end{pspicture}
\end{figure}
\end{document}
\tikz \draw (0,0) -- (1,1) -- ++(canvas polar cs:angle=90,radius=1cm) -- +(1,0); or, concisely, \tikz \draw (0,0) -- (1,1) -- ++(90:1cm) -- +(1,0);. Is there not a similar way to use polar coordinates with PStricks that doesn't involve manually invoking cos and sin?
– Sharpie
Oct 07 '10 at 05:26
I think the simplest method is to use \SpecialCoor, polar coordinate form, and origin option as illustrated in the following example.
\documentclass{article}
\usepackage{pstricks}
\begin{document}
\begin{pspicture}[showgrid=true](5,5)
\SpecialCoor
% draw a line from a point (1,2) at angle 45 degrees and of length 3 unit.
\psline[origin={1,2}](3;45)
\end{pspicture}
\end{document}
To make sure the line length is 3 cm, I draw a red circle of radius 3 cm.
