5

I am trying to draw a red border of a Moebius strip. I have tried several things, playing with shader=faceted and mesh but I could not achieve this.

enter image description here

The code : 

\documentclass[english]{article}
\usepackage{pgf,pgfplots}
\begin{document}
    \begin{tikzpicture}
    \begin{axis}[
    hide axis,
    view = {40}{40}
    ]
    \addplot3 [
    surf,
    colormap={blackwhite}{gray(0cm)=(1); gray(1cm)=(0.5)},
    shader     = faceted interp,opacity = 0.7,
    %shader = interp,
    point meta = x,
    samples    = 40,
    samples y  = 4,
    z buffer   = sort,
    domain     = 0:360,
    y domain   =-0.5:0.5
    ] (
    {(1+0.5*y*cos(x/2)))*cos(x)},
    {(1+0.5*y*cos(x/2)))*sin(x)},
    {0.5*y*sin(x/2)}
    );
    \end{axis}
    \end{tikzpicture}
\end{document}

Do you have an idea?

naphaneal
  • 2,614
BlueTiger
  • 53

1 Answers1

4

Welcome to TeX-SE! You could just draw the contours separately.

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{pgfplots}
\pgfplotsset{compat=1.16}
\begin{document}
 \begin{tikzpicture}
    \begin{axis}[
    hide axis,
    view = {40}{40}
    ]
    \addplot3 [color=blue,thick,    
    domain     = 360:720,samples y=0,
    ] (
    {(1+0.5*0.5*cos(x/2)))*cos(x)},
    {(1+0.5*0.5*cos(x/2)))*sin(x)},
    {0.5*0.5*sin(x/2)}
    );
    \addplot3 [
    surf,
    colormap={blackwhite}{gray(0cm)=(1); gray(1cm)=(0.5)},
    shader     = faceted interp,opacity = 0.7,
    %shader = interp,
    point meta = x,
    samples    = 40,
    samples y  = 4,
    z buffer   = sort,
    domain     = 0:360,
    y domain   =-0.5:0.5
    ] (
    {(1+0.5*y*cos(x/2)))*cos(x)},
    {(1+0.5*y*cos(x/2)))*sin(x)},
    {0.5*y*sin(x/2)}
    );
    \addplot3 [color=blue,thick,    
    domain     = -140:497.5,samples y=0,samples=(640/360)*24+1,
    ] (
    {(1+0.5*0.5*cos(x/2)))*cos(x)},
    {(1+0.5*0.5*cos(x/2)))*sin(x)},
    {0.5*0.5*sin(x/2)}
    );
    \end{axis}
 \end{tikzpicture}
\end{document}

enter image description here