[pstricks] Current point

Hensh, Richard hensh at math.msu.edu
Tue Dec 4 19:49:26 CET 2007

Is there a way to move the current point without relying on the \pscustom environment (which doesn't seem to work)? For example, I have written the following postscript:

        /polar@ {dup cos 3 1 roll sin 3 1 roll dup 4 1 roll mul 3 1 roll mul} def % (r,theta) -> (x,y)

Now, I can specify the rectangular point (\sqrt{3},1) using:

\pnode(!2 30 polar@){A}

which is just fine whenever I want the point (0,0) to be the current point. Now suppose that I wish to measure a 30 degree angle from some other point. Then I find myself using the \pstTranslation macro, which can become very tedious.

For example, in the code at the end of this message, I wish to draw a line DE parallel to AC.

I would prefer something like:

\cp(2,0)        %%Now (2,0) is the current point.
\pnode(!6 60 polar@){C}
\cp(0,0)        %%The default

instead of

 \pnode(!6 60 polar@){C}

as shown below. It's not a big deal but it would be nice.



        \pnode(!6 60 polar@){C}
        \uput{6pt}[45](E){{ $E$}}


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