Is there any easy way to draw a ruled surface like a hyperbolic paraboloid in TikZ?

Question

At the risk of asking YET ANOTHER TIKZ question on a general TeX forum :), I was wondering if there was any easy way to draw a ruled surface like a hyperbolic paraboloid in TikZ? I'm not particular about the surface per se: I just need some eye candy for a nontrivial looking surface, and the hyperbolic paraboloid is a good example because of the negative curvature. I checked texample.net and google, and while I found a PSTricks package that can do this, it doesn't fit my workflow with pdflatex and Beamer.

Answer 1

PGFplots can do a reasonably good job with not too complex 3D plots. Here's an example of a hyperbolic paraboloid:

\documentclass{article}
\usepackage{pgfplots}

\begin{document}
\begin{tikzpicture}
\begin{axis}
\addplot3 [surf,shader=flat,draw=black] {x^2-y^2};
\end{axis}
\end{tikzpicture}
\end{document}

Answer 2

Using class beamer and running it with xelatex to get directly a pdf. It shows a 2D view of 3D curves in the x-y plane.

\documentclass{beamer}
\usetheme{Warsaw}
\usepackage{pst-solides3d}

\begin{document}
\begin{frame}{Title}
\psset{arrowlength=3,arrowinset=0,unit=0.6,
       viewpoint=50 30 25 rtp2xyz,Decran=50,lightsrc=viewpoint}
\begin{center}
\begin{pspicture}(-2,-4)(7,8)
\axesIIID[linecolor=gray](0,0,0)(7,7,7)
\psSolid[ngrid=.3 .3,object=grille,base=1.5 6.5 1.5 6.5,
         linewidth=0.4pt,linecolor=gray!50,action=draw]%
\psPoint(4,4,4 5 sub 2 exp 4 5 sub 2 exp sub 6 div 5 add){P}
\psPoint(4,4,0){Po}
\pcline[linecolor=red,linestyle=dashed](P)(Po)\Aput{$z=f(x,y)$}
\psSurface[ngrid=.3 .3,fillcolor=green!30,incolor=gray!30,intersectiontype=0,
  intersectionplan={
    [0 0 1 -6.5]
    [0 0 1 -6.1]
    [0 0 1 -5.7]
    [0 0 1 -5.3]
    [0 0 1 -4.9]},intersectioncolor=(bleu),intersectionlinewidth=1,
  linewidth=0.4pt,algebraic](1.5,1.5)(6.5,6.5){ ((y-5)^2-(x-5)^2)/6+5 }
\psdot[linecolor=red,dotscale=0.7](P)
\psdot[dotscale=0.7](Po)
\psPoint(1.5,6.5,0){D}\uput[90](D){$D$}
\psPoint(2,6.5,6.5 5 sub 2 exp 2 5 sub 2 exp sub 6 div 5 add){S}\uput[0](S){$S$}
%% Contouring on xy plane for z=6.5 6.1 5.7 5.3 4.9
%% Explicit representation: z=((y-5)2-(x-5)2)/6+5
%% Parametric representation of z=f(x,y)
%% x=x(x)=x
%% y=y(x)=sqrt((x-5)^2+6*(z-5))+5
%% z=z(x)=0
\psset{object=courbe,r=0,linecolor=blue,resolution=360,function=Fxy,algebraic}
\defFunction{Fxy}(x){x}{-sqrt((x-5)^2+6*(6.5-5))+5}{0}
\psSolid[range=3.155 6.5]
\defFunction{Fxy}(x){x}{-sqrt((x-5)^2+6*(6.1-5))+5}{0}
\psSolid[range=2.6 6.5]
\defFunction{Fxy}(x){x}{-sqrt((x-5)^2+6*(5.7-5))+5}{0}
\psSolid[range=2.15 6.5]
\defFunction{Fxy}(x){x}{-sqrt((x-5)^2+6*(5.3-5))+5}{0}
\psSolid[range=1.75 6.5]
\defFunction{Fxy}(x){x}{sqrt((x-5)^2+6*(5.3-5))+5}{0}
\psSolid[range=4.35 5.7]
\defFunction{Fxy}(x){-sqrt((x-5)^2+6*(5-4.9))+5}{x}{0}
\psSolid[range=1.6 6.5]
\defFunction{Fxy}(x){sqrt((x-5)^2+6*(5-4.9))+5}{x}{0}
\psSolid[range=3.7 6.3]
\end{pspicture}
\end{center}
\end{frame}
\end{document}

Answer 3

Just because it is fun (I don't claim that this is a good way to draw this), using the fact that the surface is ruled):

\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{calc}

\begin{document}

\definecolor{startcolor}{named}{red}
\definecolor{endcolor}{named}{blue}

\newcommand\steps{250}
\newcommand\lineAstart{0,0,0}
\newcommand\lineAend{1,0,1}
\newcommand\lineBstart{0,1,1}
\newcommand\lineBend{1,1,0}

\begin{tikzpicture}[x={(-3.5cm,-2cm)},y={(10cm,-1cm)},z={(0,7cm)}]
    \draw[->] (0,0) -- (1,0,0);
    \draw[->] (0,0) -- (0,1,0);
    \draw[->] (0,0) -- (0,0,1);
    \foreach \n in {0,1,...,\steps} {
        \pgfmathparse{\n/\steps*100}
        \let\i\pgfmathresult
        \draw[ultra thick,color={startcolor!\i!endcolor}]
            ($(\lineAstart)!{\n/\steps}!(\lineAend)$) --
            ($(\lineBstart)!{\n/\steps}!(\lineBend)$);
    }
\end{tikzpicture}

\end{document}

---

More interestingly, one can draw the two families of lines:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc,intersections}

\begin{document}

\tikzset{f1/.style={red}}
\tikzset{f2/.style={blue}}
\newcommand\steps{40}

\begin{tikzpicture}[x={(-3.5cm,-2cm)},y={(10cm,-1cm)},z={(0,7cm)}]
    \coordinate (lineAstart) at (1,0,1);
    \coordinate (lineAend) at (0,0,0);
    \coordinate (lineBstart) at (1,1,0);
    \coordinate (lineBend) at (0,1,1);

    \draw[->] (0,0) -- (1,0,0);
    \draw[->] (0,0) -- (0,1,0);
    \draw[->] (0,0) -- (0,0,1);

    %
    % Draw the first family of lines
    %

    \draw[name path global={f1l0},name path global={f1l0-short},f1] (lineAstart) -- (lineBstart);

    \foreach \n in {1,...,\steps} {
        \pgfmathparse{\n/\steps*100}
        \let\i\pgfmathresult
        \pgfmathtruncatemacro\p{\n-1}

        % Create the path of the lines
        \edef\optname{name path global={f1l\n}}
        \expandafter\path\expandafter[\optname]
            ($(lineAstart)!{\n/\steps}!(lineAend)$) --
            ($(lineBstart)!{\n/\steps}!(lineBend)$);
        % Draw the correct bits of the lines
        \edef\optname{name intersections={of={f1l\n} and f1l\p},name path global={f1l\n-short}}
        \expandafter\draw\expandafter[\optname,f1]
            (intersection-1) -- ($(lineBstart)!{\n/\steps}!(lineBend)$);
        \edef\optname{name intersections={of={f1l\n} and f1l0}}
        \expandafter\draw\expandafter[\optname,f1]
            ($(lineAstart)!{\n/\steps}!(lineAend)$) -- (intersection-1);
    }

    %
    % Draw the second family of lines
    %

    \foreach \n in {\steps,...,0} {
        \pgfmathparse{\n/\steps*100}
        \let\i\pgfmathresult
        \pgfmathtruncatemacro\p{\n+1}

        % Create the paths
        \edef\optname{name path global={f2l\n}}
        \expandafter\path\expandafter[\optname,shorten <=4pt]
            ($(lineAstart)!{\n/\steps}!(lineBstart)$) --
            ($(lineAend)!{\n/\steps}!(lineBend)$);

        % Draw the correct bits
        \ifnum\n=\steps % handle the first line separately
            \draw[f2] (lineBstart) -- (lineBend);
        \else
            % Note: one should actually find the intersection with the correct line of the first family.
            % However, this is rather complicated. The following gives a good approximation when the
            % number of lines is high enough.
            \edef\optname{name intersections={of={f2l\n} and f2l\p,total=\noexpand\total}}
            \expandafter\draw\expandafter[\optname,f2]
                \ifnum\total=1
                   ($(lineAstart)!{\n/\steps}!(lineBstart)$) -- (intersection-1);
                \else
                    ($(lineAstart)!{\n/\steps}!(lineBstart)$) -- ($(lineAend)!{\n/\steps}!(lineBend)$);
                \fi;
        \fi

    }
\end{tikzpicture}

Or combine the the pictures: