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I would like to plot the following surface (it coincides with the xy-plane "behind" the back slopes of the parabolas).

My problem is that I don't know how to plot the straight lines in my drawing well. Perhaps this is easily doable with a some loop? All I know how to do is plot the parabola curves. I have included the code doing that below.

enter image description here

\documentclass{scrartcl}
\usepackage{pgfplots}
\begin{document}
\begin{tikzpicture}
\begin{axis}[axis lines=center, view={210}{50}] \addplot3[variable=t,mesh,domain=-4:4] (t,{ (1/2)*t^2}, 0);
\addplot3[variable=t,mesh,domain=-2:2] (t,{ 2*t^2}, 0);
\addplot3[variable=t,mesh,domain=-2:2] (t,{ 2*t^2}, {abs(t)});
\end{axis}
\end{tikzpicture}
\end{document}
Phelype Oleinik
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1 Answers1

4

It would have been much easier if pgfpots did not delay its expansions.

\documentclass{scrartcl}
\usepackage{pgfplots}
\begin{document}
\begin{tikzpicture}
\begin{axis}[axis lines=center, view={210}{50}] 
\addplot3[variable=t,mesh,domain=-4:4] (t,{ (1/2)*t^2}, 0);
\addplot3[variable=t,mesh,domain=-2:2] (t,{ 2*t^2}, 0);
\addplot3[variable=t,mesh,domain=-2:2] (3*t,{ 2*t^2}, {abs(t)});
\addplot3[variable=t,mesh,domain=-2:2] (5*t,{ 2*t^2}, 0);
\foreach \Z in {1,...,8}
{\def\Y{\Z}\relax
\pgfmathsetmacro{\X}{sqrt(2*\Y)}\relax
\edef\temp{\noexpand\draw[-] (axis cs:\X,\Y,0) -- (axis cs:3*\X/2,\Y,\X/2);}
\temp
\edef\temp{\noexpand\draw[-] (axis cs:-\X,\Y,0) -- (axis cs:-3*\X/2,\Y,\X/2);}
\temp
\edef\temp{\noexpand\draw[-] (axis cs:-\X,\Y,0) -- (axis cs:\X,\Y,0);}
\temp
\edef\temp{\noexpand\draw[-] (axis cs:2.5*\X,\Y,0) -- (axis cs:3*\X/2,\Y,\X/2);}
\temp
\edef\temp{\noexpand\draw[-] (axis cs:-2.5*\X,\Y,0) -- (axis cs:-3*\X/2,\Y,\X/2);}
\temp
}
\end{axis}
\end{tikzpicture}
\end{document}

enter image description here