rotate node in 3D

Question

I am trying to draw this 3D shape

I draw the 4 arrows and the cylinder, but can't rotate the arrows in 3D to put them on the top.

\documentclass{standalone}
\usepackage[usenames,dvipsnames]{xcolor}
\usepackage{tikz}
\usetikzlibrary{shapes.geometric}
\usetikzlibrary{shapes.arrows}

\begin{document}    
%,rotate around={30:(0,1,1)}

\begin{tikzpicture}
        \node[cylinder, draw=white,fill=blue, shape aspect=0.7,rotate=90,minimum height=0.7cm, minimum width=2cm] {hh};
        \node [fill=red,single arrow, draw=none, rotate=90,minimum height=1cm,minimum width=0.7cm]at(0,0) {};
        \node [fill=blue,single arrow, draw=none, rotate=-90,minimum height=1cm,minimum width=0.7cm] at (0,1.5) {};
        \node [fill=red,single arrow, draw=none, rotate=-180,minimum height=1cm,minimum width=0.7cm]at(-0.75,0.7) {};
        \node [fill=blue,single arrow, draw=none, rotate=-360,minimum height=1cm,minimum width=0.7cm] at (0.75,0.7) {};
    \end{tikzpicture}
\end{document}

score 6 · Answer 1 · edited Apr 13 '17 at 12:35

6

Here's another version with uses the 3D library and paints a custom arrow.

The code between \makeatletter and \makeatother is used to correct an error in the implementation of canvas is xy plane at z=. As far as I know it was first fixed by Jake in TikZ: How to draw an isometric drawing.

Code

\documentclass[tikz, border=2mm]{standalone}
\usetikzlibrary{calc, 3d}

\makeatletter
    \tikzoption{canvas is xy plane at z}[]%
    {   \def\tikz@plane@origin{\pgfpointxyz{0}{0}{#1}}%
        \def\tikz@plane@x{\pgfpointxyz{1}{0}{#1}}%
        \def\tikz@plane@y{\pgfpointxyz{0}{1}{#1}}%
        \tikz@canvas@is@plane
    }
\makeatother  

\newcommand{\Arrow}[4]%
% start, direction, length, width
{   \fill[white] (#1) -- ++ (#2+90:#4/2) -- ++ (#2:#3-#4) -- ++ (#2+90:#4/2) -- ++ (#2-45:1.414214*#4) -- ++ (#2-135:1.414214*#4) -- ++ (#2+90:#4/2) -- ++ (#2-180:#3-#4) -- cycle;
}

\begin{document}    
\begin{tikzpicture}
[   xy/.style={canvas is xy plane at z=#1},
    x={(-20:1cm)},
    y={(20:1cm)},
    z={(90:1cm)},
]
    \fill[cyan!50!gray, draw=white] (-135:4) arc (-135:45:4) -- ++ (0,0,2) arc(45:-135:4) -- cycle;
    \fill[cyan!50!gray, draw=white] ($(225:4)+(0,0,2)$) arc (-135:225:4)
        coordinate[pos=0.20] (n1)
        coordinate[pos=0.55] (n2)
        coordinate[pos=0.70] (n3)
        coordinate[pos=0.95] (n4)
     -- cycle;
    \begin{scope}[xy=2]
        \foreach \A in {1,...,4}
        { \Arrow{{\A*90-60-180*mod(\A,2)}:{3.6-3.2*mod(\A,2)}}{\A*90-60+180}{3.2}{0.5}
        }
    \end{scope}
\end{tikzpicture}

\end{document}

Output

edited Apr 13 '17 at 12:35

Community

1

answered Dec 13 '15 at 21:41

Tom Bombadil

40,123

What is \tikzoption? Don't recall seeing that. Seems to be a noop here as removing it does not seem to change the image. Also it does not appear to be documented in the TikZ/PGF manual. – Peter Grill Dec 13 '15 at 23:45
That's a fix for the not properly working canvas is xy plane at z=. If you use it unpatched, it draws everything in planar coordinates instead of xyz if I recall correctly. – Tom Bombadil Dec 13 '15 at 23:54
I can't see any difference in the output with TexLive2013, 2014 and 2015, so perhaps that was patched a while ago. But, furthermore, I don't see any documentation on \tikzoption. I assume it just saves settings that are applied to the start of each tikzpicture. – Peter Grill Dec 14 '15 at 01:55
Hmm, probably it works here for some reason, but for instance \begin{tikzpicture} [ x={(-10:1cm)}, y={(200:1cm)}, z={(90:1cm)}, ] \draw[canvas is xy plane at z=0] (0,0) rectangle ++ (2,2); \draw[canvas is xz plane at y=0] (0,0) rectangle ++ (2,2); \draw[canvas is yz plane at x=0] (0,0) rectangle ++ (2,2); \end{tikzpicture} fails miserably. – Tom Bombadil Dec 14 '15 at 08:18
1

Wow, will have to remember this question if I run into that. BTW, if you have a reference to a question here that provides this fix you should post it in your answer. The best reference I can find is TikZ: How to draw an isometric drawing in tikz. – Peter Grill Dec 14 '15 at 11:41
Hmm, good idea! I just always remember that there was a catch and add it from an old answer of mine. – Tom Bombadil Dec 14 '15 at 11:45

score 4 · Accepted Answer · edited Dec 13 '15 at 21:51

A solution with arrows.meta library

\documentclass[border =3mm]{standalone}
\usepackage[usenames,dvipsnames]{xcolor}
\usepackage{tikz}
\usetikzlibrary{shapes.geometric}
\usetikzlibrary{arrows.meta,calc}

\begin{document}    
%,rotate around={30:(0,1,1)}
\tikzset{Ultra thick/.style={line width=2.5pt}}
\begin{tikzpicture}
        \node[cylinder, draw=white,fill=blue, shape aspect=0.7,rotate=90,minimum height=0.7cm, minimum width=2cm,aspect=1.5](c) {hh};
   \begin{scope}[white,-{Triangle[angle=45:1pt 2]},inner sep=0.5pt]
        \node(topt) at (c.-5){};
        \node(topl) at (c.50){};
        \node(topc) at ($(c.before top)!0.5!(c.after top)$){};
        \node(topb) at ($(topc)!-1!(topt)$){};
        \node(topr) at ($(topc)!-1!(topl)$){};
        % draw arrows      
        \draw[Ultra thick](topc)--(topt); 
        \draw[Ultra thick](topc)--(topb);
        \draw[Ultra thick](topr)--(topc);
        \draw[Ultra thick](topl)--(topc);
  \end{scope}
\end{tikzpicture}

\end{document}

Output [added by cfr]:

I hope that you do not mind my adding an image to show the output. What is the point of the hh? (I know it is in the original code, but it is not in the target picture and it looks rather odd to me. But perhaps I'm missing the significance.) — cfr, Dec 13 '15 at 21:52

rotate node in 3D

2 Answers2

Code

Output