Drawing chain with TikZ

Question

I was answering a set of problems and found this picture. Since I am just starting to learn about TikZ, I have not been able to draw it.

enter image description here

You need to create a ring shape (node) that so that you can fill it. An arc version could handle the overlap. — John Kormylo, Oct 26 '13 at 03:25
I think you should not decide the accepted answer too early to avoid causing you to get confused and making you switch the accepted answer many times in a short period of time. Letting the time elapse about 12 hours or more seems to be good. :-) — kiss my armpit, Oct 26 '13 at 10:42
Up voting the answers that you think they are useful, good, etc is of course a good behavior. — kiss my armpit, Oct 26 '13 at 11:03

score 17 · Answer 1 · edited Apr 13 '17 at 12:36

Clipping, and it needs my paths.ortho library.

Improvements/Problems

Line widths are not accounted for.
Keys instead of arguments.
Keys also would allow to calculate a few things.
Recursion.
Adding missing upper and lower ring.

Code

\documentclass[tikz]{standalone}
\usetikzlibrary{paths.ortho}
\newcommand*\chainy[8]{% #1 = upper center       | #2 = lower center
                       % #3 = upper outer radius | #5 = lower outer radius
                       % #4 = upper inner radius | #6 = lower inner radius
                       % #7 = upper options      | #8 = lower options
\scope
  \clip (#1) -- ++(right:{#4}) arc[radius={#4}, start angle=0, delta angle=-90] --cycle
    ([shift=(right:{#3})] #1) coordinate (@)
    arc[radius={#3}, start angle=0, delta angle=-90] -- (#2) -- ++(right:{#5}) rl (@);
  \path[even odd rule, #8] (#2) circle [radius={#5}] (#2) circle [radius={#6}];
\endscope
\scope
  \clip (#2) -- ++(left:{#6}) arc[radius={#6}, start angle=180, delta angle=-90] --cycle
    ([shift=(left:{#5})] #2) coordinate (@)
    arc[radius={#5}, start angle=180, delta angle=-90] -- (#1) -- ++(left:{#3}) lr (@);
  \path[even odd rule, #7] (#1) circle [radius={#3}] (#1) circle [radius={#4}];
\endscope
\scope
  \clip (#1) rectangle ++ ({#3},{-#3});
  \path[even odd rule, #7] (#1) circle [radius={#3}] (#1) circle [radius={#4}];
\endscope
\scope
  \clip (#2) rectangle ++ ({-#5},{#5});
  \path[even odd rule, #8] (#2) circle [radius={#5}] (#2) circle [radius={#6}];
\endscope
}
\begin{document}
\begin{tikzpicture}[
  udlr/rl distance=+0pt, udlr/lr distance=+0pt,
  y=5mm, x=5mm, a/.style={fill=blue}, b/.style={fill=red}]
  \chainy{0,0}  {0,-9} {6}{5}{5}{4}{a}{b}
  \chainy{0,-9} {0,-16}{5}{4}{4}{3}{b}{a}
  \chainy{0,-16}{0,-21}{4}{3}{3}{2}{a}{b}
  \chainy{0,-21}{0,-24}{3}{2}{2}{1}{b}{a}
\end{tikzpicture}
\end{document}

Output

enter image description here

@kalakay I’m going to need more information to be able to help you. Error message? Unexpected output? UFOs instead of rings? You need to download the two files of the paths.ortho library I linked to. — Qrrbrbirlbel, Oct 26 '13 at 05:30
I did, but here is https://www.writelatex.com/read/jhjsfzfnwmqt — kalakay, Oct 26 '13 at 08:36
@kalakay: sorry, but you linked a document which exploit the PSTricks solution by Marienplatz. Moreover, as said by Qrrbrbirlbel, you need his library paths.ortho; I don't know how to include external libraries in writeLaTeX, but you definitely have to find a way. — Claudio Fiandrino, Oct 26 '13 at 09:24
I was upload the two files: tikzlibrarypaths.ortho.code.tex and tikzlibrarypaths.ortho.tex as Qrrbrbirlbel said, but in writeLaTeX the files changed to tikzlibrarypaths_ortho_code.tex and tikzlibrarypaths_ortho.tex, and don't work. — kalakay, Oct 26 '13 at 09:32

score 16 · Answer 2 · 2013-10-27T20:39:04.363

needs some time with xelatex. You can reduce the number of calculated polygons by modifying ngrid. r1 is the outer radius of the ring and r0 the radius of the ring itself:

\documentclass[border=5mm,pstricks,dvipsnames]{standalone}
\usepackage{pst-solides3d}    
\begin{document}

\begin{pspicture}[solidmemory](-3,-7)(3,10.2)
\psset{lightsrc=viewpoint,viewpoint=40 -10 0 rtp2xyz,Decran=100,ngrid=18 30,
       object=tore,r0=0.2,action=none}
\psSolid[r1=1,  RotY=90,        fillcolor=blue,  name=R1](0,0,3)
\psSolid[r1=0.9,RotX=90,RotZ=30,fillcolor=Brown, name=R2](0,0,1.5)
\psSolid[r1=0.8,RotY=90,        fillcolor=red,   name=R3](0,0,0.1)
\psSolid[r1=0.7,RotX=90,RotZ=30,fillcolor=yellow,name=R4](0,0,-1)
\psSolid[r1=0.6, RotY=90,        fillcolor=green,name=R5](0,0,-2)
\psSolid[object=fusion,base=R1 R2 R3 R4 R5, action=draw**]
\end{pspicture}

\end{document}

enter image description here

It's awesome, because it's real 3D! – Henri Menke Oct 27 '13 at 20:29 — Henri Menke, Oct 27 '13 at 20:29
I guess it is r0 and r1 instead of r1 and r2. – Svend Tveskæg Oct 27 '13 at 20:34 — Svend Tveskæg, Oct 27 '13 at 20:34

kiss my armpit · Answer 3 · 2013-10-26T17:10:11.947

14

A recommended solution with PSTricks. The fewer key strokes the code use, the more readable the code is and the easier the code maintenance is.

\documentclass[pstricks,border=20pt]{standalone}
\SpecialCoor
\psset{linewidth=.4,linecap=1}
\def\Atom#1#2#3#4#5{%
    \rput(0,#5){%
        \psarc[linecolor=#1](0,#3){!#3 .2 add}{180}{270}%
        \psarc[linecolor=#2](0,-#4){!#4 .2 add}{0}{180}%
        \psarc[linecolor=#1](0,#3){!#3 .2 add}{270}{360}%
    }%
}
\begin{document}
\begin{pspicture}(-4,-12.5)(4,4)
    \Atom{red}{blue}{4}{3}{0}
    \Atom{blue}{red}{3}{2}{-6}
    \Atom{red}{blue}{2}{1}{-10}
    \Atom{blue}{red}{1}{.5}{-12}
\end{pspicture}
\end{document}

enter image description here

edited Oct 26 '13 at 17:10

answered Oct 26 '13 at 09:12

kiss my armpit

36,086

My code above should be refactored to get a more friendly interface. I will do it later. – kiss my armpit Oct 26 '13 at 09:14
@kalakay: I am so sorry, I don't use TikZ. – kiss my armpit Oct 26 '13 at 09:23
Oh, it's OK, never mind. – kalakay Oct 26 '13 at 09:24
I can see some sign where the two half arch meet... not very pleasant. – Bakuriu Oct 26 '13 at 16:56
1

@Bakuriu: The unpleasant parts have been removed. – kiss my armpit Oct 26 '13 at 17:10

score 11 · Answer 4 · answered Oct 26 '13 at 10:17

enter image description here

A somewhat lazy approach without clipping, using a slightly overlapping sectors, implemented in the Asymptote module chainofrings.asy. The structure chainOfRings uses two functions, real Rscaled(int); and real rscaled(int); to define the major and minor radii of i-th ring. Figure 1c demonstrates how to use a sequence of explicitly defined radii.

% chain.tex :
%
\begin{filecontents*}{chainofrings.asy}
struct chainOfRings{
  int n;             // number of rings
  real w;
  pair origin;
  pen[] clrA={deepgreen,deepblue};
  pen[] clrB={white,lightyellow,palered};
  guide qring;
  real Rscaled(int);
  real rscaled(int);
  real eps;

  void drawHalf(int i,real Rt,real rt, pair p,real phi){
    qring=rotate(phi)*(arc((0,0),Rt,-eps,90+eps)--reverse(arc((0,0),rt,-eps,90+eps))--cycle);
    radialshade(shift(p)*qring
    ,clrA[i%clrA.length], p, (Rt+rt)*0.382
    ,clrB[i%clrB.length], p, (Rt+rt)*0.618
    );
    radialshade(shift(p)*rotate(180)*qring
    ,clrA[i%clrA.length], p, (Rt+rt)*0.382
    ,clrB[i%clrB.length], p, (Rt+rt)*0.618
    );
  }

  void drawChain(){
    pair p;
    real Rt,rt,dh;
    p=origin; Rt=Rscaled(0); rt=rscaled(0); 
    for(int i=0;i<n;++i){
      dh=rt;
      drawHalf(i,Rt, rt, p,0);
      Rt=Rscaled(i+1); rt=rscaled(i+1);
      w=Rt-rt;
      assert(Rt>rt && rt>0 && w>0);
      p+=(0,-(dh+Rt-w));
    }
    p=origin; Rt=Rscaled(0); rt=rscaled(0); 
    for(int i=0;i<n;++i){
      dh=rt;
      drawHalf(i,Rt, rt, p,90);
      Rt=Rscaled(i+1); rt=rscaled(i+1);
      w=Rt-rt;
      assert(Rt>rt && rt>0 && w>0);
      p+=(0,-(dh+Rt-w));
    }
  }

  void operator init(pair origin=(0,0),int n=3,real Rscaled(int),real rscaled(int)
    ,pen[] clrA={gray(0.5),gray(0.7)}
    ,pen[] clrB={white,black}
    ,real eps=0.1
  ){
    this.origin=origin; this.n=n;  
    this.Rscaled=Rscaled;
    this.rscaled=rscaled;
    this.clrA=copy(clrA);
    this.clrB=copy(clrB);
    this.eps=eps;
  }
}
\end{filecontents*}
%
\documentclass{article}
\usepackage{lmodern}
\usepackage{subcaption}
\usepackage[inline]{asymptote}
\usepackage{lmodern}
\begin{document}
\begin{figure}
\captionsetup[subfigure]{justification=centering}
    \centering
      \begin{subfigure}{0.3\textwidth}
\begin{asy}
settings.outformat="pdf";
size(8cm);
import chainofrings;
real Rscaled(int k){return 10-4/3*k;};
real rscaled(int k){return 7-k;};

chainOfRings(origin=(0,0),n=4,Rscaled,rscaled).drawChain();    
\end{asy}
%
\caption{Using default colors}
\label{fig:1a}
\end{subfigure}
%
\begin{subfigure}{0.3\textwidth}
\begin{asy}
settings.outformat="pdf";
size(8cm);
import chainofrings;
real Rscaled(int k){return 10-4/3*k;};
real rscaled(int k){return 7-k;};

chainOfRings cr=chainOfRings(origin=(30,0),n=6,Rscaled,rscaled
  ,clrA=new pen[]{deepred,deepblue}
  ,clrB=new pen[]{white,lightyellow,palered}
);
cr.drawChain();
\end{asy}
%
\caption{Using custom colors}
\label{fig:1b}
\end{subfigure}
%
\begin{subfigure}{0.3\textwidth}
\begin{asy}
settings.outformat="pdf";
size(8cm);
import chainofrings;
real Rscaled(int k){
  real[] R={20,10,5,10};
  return R[k%R.length];
};
real rscaled(int k){
  real[] r={18,6,4.5,7};
  return r[k%r.length];
};

chainOfRings cr=chainOfRings(origin=(30,0),n=8,Rscaled,rscaled
  ,clrA=new pen[]{deepred,deepblue}
  ,clrB=new pen[]{white,lightyellow,palered}
);
cr.drawChain();
\end{asy}
%
\caption{Using a set of radii}
\label{fig:1c}
\end{subfigure}
\caption{Chains of rings} \label{fig:1}
\end{figure}
\end{document}
%
% Process:
%
% pdflatex chain.tex
% asy chain-*.asy
% pdflatex chain.tex

Mark Wibrow · Answer 5 · 2013-10-27T16:24:46.067

The original drawing seems out of proportion, but anyway, this requires the latest CVS version of PGF for the math library.

Two versions are shown below. In the first version, the right half of each ring is drawn first 'top to bottom' and the the left halves are drawn 'bottom to top'.

\documentclass[border=0.125cm]{standalone}

\usepackage{tikz}
\usetikzlibrary{math}

\begin{document}

\begin{tikzpicture}[x=2.5pt, y=2.5pt, line cap=round, thick, font=\sf, >=stealth]
\tikzmath{
    %
    % Thickness
    \t = 1; 
    % Outer Diameters
    let \D = {{20, 16, 12, 8}};
    %
    \y = 0;
    \s = 0.5;
    %
    int \i;
    for \i in {0,...,3}{
        \r =  \D[\i]/2;
        if (\i > 0) then {
            \y = \y - \r - \D[\i-1]/2 + 2*\t + \s;
        };
        \p = int(mod(\i,2)*100);
        {
            \fill [orange!\p!red] (0,\y) ++(90:\r) arc (90:-90:\r)  -- ++(0,\t) arc (-90:90:\r-\t) -- cycle; 
        };
    };
    for \i in {3,...,0}{
        \r =  \D[\i]/2;
        if (\i < 3) then {
            \y = \y + \r + \D[\i+1]/2 - 2*\t - \s;
        };
        \p = int(mod(\i,2)*100);
        {
            % Overlap the arcs so no white lines in PDF viewers
            \fill [orange!\p!red] (0,\y) ++(85:\r) arc (85:275:\r)  -- ++(0,\t) arc (275:85:\r-\t) -- cycle; 
        };
    };
    int \M;
    \M1 = \D[0];
    \M2 = \D[0] - 2*\t;     
}

\draw [thick] (0, \M1/2)  -- ++(+20,0) ++(-5, 0) coordinate (a1);
\draw [thick] (0, -\M1/2) -- ++(+20,0) ++(-5, 0) coordinate (a2);

\draw [thick] (0, \M2/2)  -- ++(-20,0) ++(5, 0) coordinate (b1);
\draw [thick] (0, -\M2/2) -- ++(-20,0) ++(5, 0) coordinate (b2);

\draw [<->] (a1) -- (a2) node [midway, right] {\M1};
\draw [<->] (b1) -- (b2) node [midway, left] {\M2};

\end{tikzpicture}

\end{document}

enter image description here

But the 'two pass' system used above is a bit inefficient. Here is a version using layers, so the rings can be drawn 'in one go':

\documentclass[border=0.125cm]{standalone}

\usepackage{tikz}
\usetikzlibrary{math}
\pgfdeclarelayer{background}
\pgfsetlayers{background,main}
\begin{document}

\begin{tikzpicture}[x=2.5pt, y=2.5pt, line cap=round, thick, font=\sf, >=stealth]
\tikzmath{
    %
    % Thickness
    \t = 1; 
    % Outer Diameters
    let \D = {{20, 16, 12, 8}};
    %
    \y = 0;
    \s = 0.5;
    %
    int \i;
    for \i in {0,...,3}{
        \r =  \D[\i]/2;
        if (\i > 0) then {
            \y = \y - \r - \D[\i-1]/2 + 2*\t + \s;
        };
        \p = int(mod(\i,2)*100);
        {
            \fill [orange!\p!red] (0,\y) ++(90:\r) arc (90:360:\r)  -- ++(-\t, 0) arc (360:90:\r-\t) -- cycle; 
            \begin{pgfonlayer}{background}
                \fill [orange!\p!red] (0,\y) ++(95:\r) arc (95:-5:\r)  -- ++(-5:-\t) arc (-5:95:\r-\t) -- cycle;
            \end{pgfonlayer}
        };
    };
    int \M;
    \M1 = \D[0];
    \M2 = \D[0] - 2*\t;     
}

\draw [thick] (0, \M1/2)  -- ++(+20,0) ++(-5, 0) coordinate (a1);
\draw [thick] (0, -\M1/2) -- ++(+20,0) ++(-5, 0) coordinate (a2);

\draw [thick] (0, \M2/2)  -- ++(-20,0) ++(5, 0) coordinate (b1);
\draw [thick] (0, -\M2/2) -- ++(-20,0) ++(5, 0) coordinate (b2);

\draw [<->] (a1) -- (a2) node [midway, right] {\M1};
\draw [<->] (b1) -- (b2) node [midway, left] {\M2};

\end{tikzpicture}

\end{document}

The result is the same as before.

Or...

\documentclass[border=0.125cm]{standalone}

\usepackage{tikz}
\usetikzlibrary{math}
\pgfdeclarelayer{background}
\pgfsetlayers{background,main}

\newbox\ringbox
\def\ring#1#2#3#4{%
    \def\ringColor{#1}%
    \def\radius{#2}%
    \def\ringThickness{#3}%
    \def\highlightColor{\ringColor!25!white}
    \def\lowlightColor{\ringColor!35!black}
    \setbox\ringbox=\hbox{%
        \tikzmath{%
            \xf = #4;
            {
                \fill [even odd rule, \ringColor] 
                    circle [x radius=\xf*\radius, y radius=\radius] 
                    circle [x radius=\xf*\radius-\ringThickness, y radius=\radius-\ringThickness];
            };
            for \l in {0,0.5,...,5}{
                \t = \l*\ringThickness*3;
                \o = 0.05;
                \angleA = 45+\l*5;
                \angleB = 225-\l*5;
                \ry1 = \radius-\ringThickness/10*3;
                \ry2 = \radius-\ringThickness/10*7;
                \rx1 = \ry1 * \xf;
                \rx2 = \ry2 * \xf;
                {
                    \draw [\highlightColor, opacity=\o,line width=\t, line cap=round]
                        (\angleA:\rx1 and \ry1) arc (\angleA:\angleB:\rx1 and \ry1)
                        [rotate=180]
                        (\angleA:\rx2 and \ry2) arc (\angleA:\angleB:\rx2 and \ry2); 
                    \draw [\lowlightColor, opacity=\o,line width=\t, line cap=round]
                    (\angleA:\rx2 and \ry2) arc (\angleA:\angleB:\rx2 and \ry2)
                    [rotate=180]
                        (\angleA:\rx1 and \ry1) arc (\angleA:\angleB:\rx1 and \ry1);
                };
            };
        }%
    }%
    \begin{scope}
        \clip (0,0) -- (90:\radius) arc (90:365:\radius) -- cycle;
        \copy\ringbox
    \end{scope}
    \begin{pgfonlayer}{background}
        \begin{scope}
            \clip (0,0) -- (91:\radius) arc (91:0:\radius) -- cycle;
            \copy\ringbox
        \end{scope}
    \end{pgfonlayer}
}

\begin{document}

\begin{tikzpicture}

\ring{red}{10}{2}{1}
\tikzset{shift=(270:10+8-4)}
\ring{orange}{8}{2}{0.875}
\tikzset{shift=(270:8+6-4)}
\ring{red}{6}{2}{1}
\tikzset{shift=(270:6+4-4)}
\ring{orange}{4}{2}{0.875}

\end{tikzpicture}
\end{document}

enter image description here

They are magnetized chains as there is air gap between two adjacent chains. — kiss my armpit, Oct 26 '13 at 14:49
@Marienplatz so where the code says \s = 0.5; use \s = 0;. — Mark Wibrow, Oct 26 '13 at 19:49

score 10 · Answer 6 · answered Oct 26 '13 at 09:46

My answer is based on the solution of Qrrbrbirlbel but with a simpler TikZ approach.

There is two macros (\chainyr and \chainyl) to change the orientation of superpositions.

enter image description here

\documentclass{standalone}
\usepackage{tikz}
\newcommand*\chainyr[8]{% #1 = upper center       | #2 = lower center
                        % #3 = upper outer radius | #5 = lower outer radius
                        % #4 = upper inner radius | #6 = lower inner radius
                        % #7 = upper options      | #8 = lower options

  \begin{scope}[shift={(#2)}]
    \path[#8] (0:#6) -- (0:#5) arc (0:90:#5) -- (90:#6) arc (90:0:#6) -- cycle;
  \end{scope}
  \begin{scope}[shift={(#1)}]
    \path[#7] (0:#4) -- (0:#3) arc (0:-180:#3) -- (-180:#4) arc (-180:0:#4) -- cycle;
  \end{scope}
  \begin{scope}[shift={(#2)}]
    \path[#8] (90:#6) -- (90:#5) arc (90:180:#5) -- (180:#6) arc (180:90:#6) -- cycle;
  \end{scope}
}
\newcommand*\chainyl[8]{% #1 = upper center       | #2 = lower center
                        % #3 = upper outer radius | #5 = lower outer radius
                        % #4 = upper inner radius | #6 = lower inner radius
                        % #7 = upper options      | #8 = lower options

  \begin{scope}[shift={(#2)}]
    \path[#8] (90:#6) -- (90:#5) arc (90:180:#5) -- (180:#6) arc (180:90:#6) -- cycle;
  \end{scope}
  \begin{scope}[shift={(#1)}]
    \path[#7] (0:#4) -- (0:#3) arc (0:-180:#3) -- (-180:#4) arc (-180:0:#4) -- cycle;
  \end{scope}
  \begin{scope}[shift={(#2)}]
    \path[#8] (0:#6) -- (0:#5) arc (0:90:#5) -- (90:#6) arc (90:0:#6) -- cycle;
  \end{scope}
}
\begin{document}
\begin{tikzpicture}
  [line width=.1pt,
  a/.style={fill=orange,draw=orange},
  b/.style={fill=violet,draw=violet}]
  \chainyr{0,0}  {0,-9} {6}{5}{5}{4}{a}{b}
  \chainyl{0,-9} {0,-16}{5}{4}{4}{3}{b}{a}
  \chainyr{0,-16}{0,-21}{4}{3}{3}{2}{a}{b}
  \chainyl{0,-21}{0,-24}{3}{2}{2}{1}{b}{a}
\end{tikzpicture}
\end{document}

Thank you, Paul! This one seems to be more easily understood to me. — kalakay, Oct 26 '13 at 09:55

score 6 · Answer 7 · edited Oct 27 '13 at 18:14

Here is an alternative without paths.ortho.

enter image description here

Code:

Define a ring (called chain) via clip with radii and color parameters available. #1=outter circle, #2=color, #3=inner circle:

\documentclass[tikz,border=1cm]{standalone}

\usetikzlibrary{matrix, shapes, arrows, positioning}

\newcommand\chain[3]{%  
\begin{tikzpicture}
\clip (0,0) circle (#1 mm);
\fill[#2] (0,0) circle (#1 mm);
\clip (0,0) circle (#3 mm);
\fill[white] (0,0) circle (#3 mm);
\end{tikzpicture}
}

Put rings into a chain by node with coordinates (user defined), then fill the overlay again to show which is 'front' and which is 'behind'.

\begin{document}
\begin{tikzpicture}
\node at (0,0)          {\chain{10}{black}{9}};

\node at (0,-15mm) {\chain{8}{red}{7}};
\fill [black] (-60:9 mm) -- (-60:10 mm) arc (-60:-113:10 mm) -- (-109:9 mm) arc (-109: -60:9 mm);

\node at (0,-26mm) {\chain{6}{blue}{5}};
\fill [red,yshift=-15mm] (-50:7 mm) -- (-50:8 mm) arc (-50:-115:8mm) -- (-111:7 mm) arc (-111:-50:7 mm);

\node at (0,-33mm) {\chain{4}{yellow}{3}};
\fill [blue,yshift=-26mm] (-50:5 mm) -- (-50:6 mm) arc (-50:-115:6mm) -- (-111:5 mm) arc (-111: -50:5 mm);
\end{tikzpicture}
\end{document}

Thank you so much, it's work and answed, just put the measures. — kalakay, Oct 26 '13 at 09:43

Kpym · Answer 8 · 2018-04-11T19:24:37.190

3

This looks like a code golfing exercise ... here is a ~~264~~ 244 characters tikz solution ;)

\documentclass[tikz]{standalone}
\def~#1.{pic[#1]{p}[cm={-.84,0,0,.84,(a)}](a)}
\begin{document}
\tikz\path[transform shape,p/.pic={\fill(0,0)arc(90:-90:1)node(a){}++(0,-.4)arc(-90:90:1.4);}]
{~.~red.~.~red.}[xscale=-1]~.~red.~.~red.;
\end{document}

edited Apr 11 '18 at 19:24

answered Apr 11 '18 at 17:26

Kpym

23,002

Drawing chain with TikZ

8 Answers8

Improvements/Problems

Code

Output

Linked