I would like to be able to shade a torus in TikZ to make it look like a surface in 3D, similarly to how the sphere in this code is shaded to give it a 3D feel but the opacity is such that it does not look like a solid ball:

\draw (0,0) circle (1cm);
\draw (-1,0) arc (180:360:1cm and 0.5cm);
\draw[dashed] (1,0) arc (0:180:1cm and 0.5cm);
\shade[ball color=blue!10!white,opacity=0.50] (0,0) circle (1cm);

Here is my attempt with the torus:

\draw (0,0) ellipse (1.6 and .9);
\shade[ball color = blue!10!white, opacity=0.5] (0,0) ellipse (1.6 and .9);
\clip (0,1.3) circle (1.55);
\fill[white] (0,-1.27) circle (1.55);
\path[rounded corners=24pt] (-.9,0)--(0,.6)--(.9,0) (-.9,0)--(0,-.56)--(.9,0);
\draw[rounded corners=28pt] (-1.1,.1)--(0,-.6)--(1.1,.1);
\draw[rounded corners=24pt] (-.9,0)--(0,.6)--(.9,0);
\draw[dashed] (0,-.9) arc (270:90:.2 and .365);
\draw (0,-.9) arc (-90:90:.2 and .365);

This is naive, since I'm using the same shading as for a ball, and I'm using clipping hacks to remove the shading from the hole. My attempt looks bad because it does not accurately reflect how light would hit the surface of a torus. Any tips for producing a realistically shaded torus would be greatly appreciated.

  • 2
    Render it using a proper 3D programme - TikZ is not designed for this. Also, please post compilable code rather than mere fragments. – cfr Apr 25 '18 at 2:04
  • 1
  • ... and here are some more, just drop the helix ... – user121799 Apr 25 '18 at 3:23
  • I have seen both of these posts, but they are not helpful for me - either because I don't like how they look or because I don't understand how sketch works. I'm not something that looks truly 3D, I would only like to add an extra light spot onto the torus to make it look better than what I currently have. – rosterherik Apr 25 '18 at 3:40

That's not a too serious answer, but just to tell you that if you have an idea how the shading should look like, you can reverse engineer it with pgfplots. Here is an example.

enter image description here

       view={30}{60},axis lines=none
       samples=60, point meta=x+3*z*z-0.25*y,
       domain=0:2*pi,y domain=0:2*pi,
       z buffer=sort]

The trick is to play with point meta. Is this shading realistic? Certainly not (unless you have some crazy light sources). Could you make it realistic? Yes, if you know what you're doing or study the asymptote manual long enough. So if you want something realistic, use asymptote. If you want a cartoon and like to play with pgfplots, you may potentially find this useful.

  • Thank you for being so helpful marmot. This is closer to what I want than what I have, so it is useful. Is there a way to make this picture translucent? – rosterherik Apr 25 '18 at 3:43
  • @rosterherik I am not sure I understand "translucent". Do you mean transparent? – user121799 Apr 25 '18 at 3:57
  • @rosterherik To best way to thank on TeX.SX is to up-vote and award green ticks to what you think is the best answer :) – user30471 Apr 25 '18 at 4:27
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    @rosterherik I agree that it is good to wait before accepting an answer but there is no reason why you cannot up-vote marmot's nice answer now: currently only one person has voted for the TeX.SX rodent and it is not you! – user30471 Apr 25 '18 at 10:52
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    @rosterherik :) – user30471 Apr 26 '18 at 7:52

I know that the question was asked two years ago, but maybe another point of view could still be useful.

I try to mimic a ray-tracer to obtain the shading of the torus. The main elements in the code below are the two unitary vectors pointing towards the observer, I'll call it w, and towards the light source, s. (The sun position suggested by the image is not quite correct due to a certain reflection angle for the highlights, but this is not so important.)

The torus ("centered" at the origin of the coordinate system) is constructed using a quadrilateral mesh; the points of the mesh are given by a classical parametrization. The inner product of w with a point (more precisely, with its position vector) of the mesh decides if the point is visible. For a visible quadrilateral, the inner product of s with one of its vertices controls the color of the quadrilateral which eventually yields the shading.

Note that we need the three components of 3D points and vectors for various computations and we cannot recuperate them from a TikZ \path coordinate command.

enter image description here

\usetikzlibrary{cd, arrows, matrix, intersections, math, calc}
\xdefinecolor{O}{RGB}{255, 102, 17}
\xdefinecolor{B}{RGB}{17, 87, 221}


  real \slongit, \slatit, \sunx, \suny, \sunz;  % towards the light source 
  real \ry, \rz, \longit, \latit, \tox, \toy, \toz;
  real \newxx, \newxy, \newyx, \newyy, \newzx, \newzy;  
  integer \Ny, \Nz, \prevj, \prevk, \aj, \ak;
  \slongit = -60; \slatit = 45;
  \sunx = sin(\slongit)*cos(\slatit);
  \suny = sin(\slatit);
  \sunz = cos(\slongit)*cos(\slatit);
  % j moves around Oy and k moves around Oz.
  % They describe full circles of radii \ry and \rz respectively.
  \ry = 4;
  \rz = 1.5;
  \longit = 25;
  \latit = 35;
  \tox = sin(\longit)*cos(\latit);
  \toy = sin(\latit);
  \toz = cos(\longit)*cos(\latit);
  \newxx = cos(\longit); \newxy = -sin(\longit)*sin(\latit);
  \newyy = cos(\latit);
  \newzx = -sin(\longit); \newzy = -cos(\longit)*sin(\latit);
  \Nz = 36;  % 60;
  \Ny = 84;  % 120;
  \ktmp = \Nz-1; 
  \jtmp = \Ny-1;
  \aj = 10;
  \ak = 0;
  function isSeen(\j, \k) {
    let \px = cos(360*(\k/\Nz))*cos(360*(\j/\Ny));
    let \py = -sin(360*(\k/\Nz));
    let \pz = cos(360*(\k/\Nz))*sin(360*(\j/\Ny));
    let \res = \px*\tox + \py*\toy + \pz*\toz;
    if \res>0 then {return 1;} else {return 0;};
  function inLight(\j, \k) {%
    let \px = cos(360*(\k/\Nz))*cos(360*(\j/\Ny));
    let \py = -sin(360*(\k/\Nz));
    let \pz = cos(360*(\k/\Nz))*sin(360*(\j/\Ny));
    return {\px*\sunx + \py*\suny + \pz*\sunz};
\begin{tikzpicture}[every node/.style={scale=.8},
  z={(\newzx cm, \newzy cm)},
  x={(\newxx cm, \newxy cm)},
  y={(0 cm, \newyy cm)},
    int \j, \k;
    real \tmp;
    for \j in {0, 1, ..., \Ny}{%
      for \k in {0, 1, ..., \Nz}{%
        \test{\j,\k} = isSeen(\j, \k);
        if \test{\j,\k}>0 then {%
          \tmp{\j,\k} = int(100*inLight(\j,\k)));
          if \tmp{\j,\k}>0 then {%
          else {%
        } else {};
  % coordinate system $Oxyz$; first layer
  (0, 0, 0) -- (\ry, 0, 0)
  (0, 0, 0) -- (0, 0, \ry);

  % points (P-\j-\k)
  \foreach \j in {0, ..., \Ny}{%
    \foreach \k in {0, ..., \Nz}{%
      ( {( \ry+\rz*cos(360*(\k/\Nz)) )*cos(360*(\j/\Ny))},
      {( \ry+\rz*cos(360*(\k/\Nz)) )*sin(360*(\j/\Ny))} )
      coordinate (P-\j-\k);

  % "squares"---the mesh
  \foreach \k [remember=\k as \prevk (initially 0)] in {1, ..., \Nz}{%
    \foreach \j [remember=\j as \prevj (initially 0)] in {1, ..., \Ny}{%
          (P-\j-\prevk) -- (P-\prevj-\prevk)
          -- (P-\prevj-\k) --(P-\j-\k) -- cycle;
          (P-\j-\prevk) -- (P-\prevj-\prevk)
          -- (P-\prevj-\k) --(P-\j-\k) -- cycle;

  % longitude cycle
  \foreach \k [remember=\k as \prevk (initially 0)] in {1, ..., \Nz}{%
      \draw[red, thick] (P-\aj-\k) -- (P-\aj-\prevk);
      \draw[red, very thin, opacity=.4] (P-\aj-\k) -- (P-\aj-\prevk);

  % latitude cycle
  \foreach \j [remember=\j as \prevj (initially 0)] in {1, ..., \Ny}{%
      \draw[red, thick] (P-\j-\ak) -- (P-\prevj-\ak);
      \draw[red, very thin, opacity=.3] (P-\j-\ak) -- (P-\prevj-\ak);
  % coordinate system $Oxyz$; second layer
  \draw[green!50!black, -{Latex[length=5pt, width=5pt]}]
  (\ry+\rz, 0, 0) -- (8, 0, 0) node[right] {$x$};
  \draw[green!50!black, -{Latex[length=5pt, width=5pt]}]
  (0, 0, 0) -- (0, 6, 0) node[above] {$y$};
  \draw[green!50!black, -{Latex[length=5pt, width=5pt]}]
  (0, 0, \ry+\rz) -- (0, 0, 8) node[below left] {$z$};


Some explanations about the code.

  1. The components of w are \tox, \toy, and \toz, where
    \tox = x_w = sin\longit cos\latit
    \toy = y_w = sin\latit
    \toz = z_w = cos\longit cos\latit
    The angles \longit and \latit represent the longitude and the latitude, respectively. See also my answer at Drawing a wedge of a torus in Asymptote for more details .

  2. The components of s are \sunx, \suny, and \sunz. They are computed similarly using \slongit and \slatit.

Note that for these two vectors, zero longitude corresponds to the plane x=0 in the coordinate system Oxyz.

  1. The mesh on the torus is defined by the points (P-\j-\k) constructed from seeing the torus as the surface of revolution around the Oy axis of a circle of radius \rz in the plane Oxy. The distance from the center of the circle to Oy is \ry>\rz.

  2. isSeen and inLight are the functions based on inner products I was referring to above.

  3. The hidden part of the cycles is controlled manually based on isSeen output.

Of course, if the number of points of the mesh is increased, the result is smoother. But the compilation asks for a large amount of patience. The image below is obtained setting \Nz = 60 and \Ny = 120. I also set \slongit=-110.

enter image description here

  • 1
    Where is your code ? – projetmbc Aug 12 '20 at 12:52
  • Sorry. As projetmbc noticed, I had forgotten the code in the first version of my answer. Now it is there. – Daniel N Aug 12 '20 at 14:17

Not an answer just an explanation of a setting or two available in @marmot 's excellent answer that may answer your question of translucency to an acceptable level (marmots that is) so if we take the above answer and tweak the one line (I found 50 to be a safer memory value than 60) the Opacity looks better around 7.5 where the far walls are ghosted through the surface materials, personally think this is more understandable than covering with broken lines see Drawing Torus with semi-dashed line on it

   samples=50, point meta=x+3*z*z-0.25*y,

enter image description here

And the main collection of toroidal answers are at How to draw a torus

  • 1
    Looks great!!!! P.S. the normal at a given point is given by n={cos(\u)*cos(\v), cos(\v)*sin(\u),sin(\v)}. Playing with this one may obtain a realistic shading (point meta = vector of light source . normal). – user121799 Mar 10 '19 at 3:51

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