# [pstricks] behavior of \parametricplotThreeD

Nitecki, Zbigniew H. Zbigniew.Nitecki at tufts.edu
Fri Dec 30 23:19:59 CET 2011

Thanks---I did catch the missing mul.  But even with that, the curve seems to lie not in the yz-plane, but in the plane x=y (which would explain why with Alpha at its default value of 45 one is looking into the curve).

I didn't know about the use of vertical bars to separate components.  Is this a recent development?

Zbigniew Nitecki
Department of Mathematics
Tufts University
Medford, MA 02155

telephones:
Office    (617)627-3843
Dept.    (617)627-3234
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http://www.tufts.edu/~znitecki/

On Dec 30, 2011, at 15:29, Herbert Voss wrote:

Am 30.12.2011 16:45, schrieb Nitecki, Zbigniew H.:
I am trying to draw a hyperbola in the xy-plane, and another, similar one in the yz-plane.
When I use \parametricplotThreeD, I get the first one, but the second ends up as a line segment on the z-axis.

More precisely, the code below (trying to draw one branch of each) leads to the attached picture.  Is the problem somehow related to the fact that the first entry in the second (blue) plot is zero?

\begin{pspicture}(-3,-3.5)(3,3.5)

I just realized that your function is not correct:

\listfiles
\documentclass{minimal}
\usepackage{pst-3dplot}
\pagestyle{empty}

\begin{document}

\begin{pspicture}(-8,-3.5)(3,3.5)
%\psset{Alpha=30,Beta=10}
\pstThreeDCoor
\parametricplotThreeD[algebraic](-2,2){0|t|sqrt(4*t^3+36)/3}
\parametricplotThreeD(-2,2){
0
t
t 3 exp 4 mul 36 add sqrt 3 div }

%t dup t mul 4 mul 36 add sqrt 3 div %% Your function
\end{pspicture}

\end{document}

your code needs one more mul. However, use the algebraic
mode also seen in my example, which makes live easier.

Herbert
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