# Drawing circular sector regions

I would like to round the corners of the gray filled areas. The problem seems to be in how I am building the borders as separate pieces as opposed to one contiguous command that can use the "rounded corners" optional argument. It seems that this piecewise work is necessary as I cannot figure out a satisfactory way to draw an arc in anything but a counterclockwise fashion.

Failing a solution to rounding the shaded region, how might we round the unshaded boxes?

\documentclass[border=10pt,multi,tikz]{standalone}

\begin{document}

\begin{tikzpicture}
\foreach \m in {0,...,2}{

\draw [white, fill=gray!50!white] (0:0) -- ({15+120*\m}:2.25) arc ({15+120*\m}:{129+120*\m}:2.25) -- (0:0);
\draw [white, fill=white] (0:0) -- ({15+120*\m}:1.75) arc ({15+120*\m}:{129+120*\m}:1.75) -- (0:0);

% draws border around shaded region
\draw ([shift=({15+120*\m}:1.75)]0:0) arc ({15+120*\m}:{129+120*\m}:1.75);
\draw ([shift=({15+120*\m}:2.25)]0:0) arc ({15+120*\m}:{129+120*\m}:2.25);
\draw ({15+120*\m}:1.75)--({15+120*\m}:2.25);
\draw ({129+120*\m}:1.75)--({129+120*\m}:2.25);
}

\foreach \m in {0,...,2}{
\draw ([shift=({63+120*\m}:1.65)]0:0) arc ({63+120*\m}:{177+120*\m}:1.65);
\draw ([shift=({63+120*\m}:2.35)]0:0) arc ({63+120*\m}:{177+120*\m}:2.35);
\draw ({63+120*\m}:1.65)--({63+120*\m}:2.35);
\draw ({177+120*\m}:1.65)--({177+120*\m}:2.35);
}

\foreach \n in {1,...,15}{\draw[fill=black] (\n*24:2) circle (0.05);};
\end{tikzpicture}

\end{document}


Could someone suggest a better way of constructing this diagram in general?

• Welcome! Could you please make your code compilable so we don't have to guess? – cfr Feb 4 '17 at 2:59
• I added what I think should be sufficient to make it comply now. – Laars Helenius Feb 4 '17 at 3:59

Like this?

\documentclass[border=10pt,multi,tikz]{standalone}

\begin{document}
\begin{tikzpicture}
\foreach \m in {0,1,2}{
\draw [fill=gray!50!white, rounded corners]
({ 15+120*\m}:2.25) arc ({15+120*\m}:{129+120*\m}:2.25) --
({129+120*\m}:1.75) arc ({129+120*\m}:{15+120*\m}:1.75) -- cycle;
}
\foreach \m in {0,1,2}{
\draw [rounded corners]
( 63+120*\m:1.65) arc (6 3+120*\m:177+120*\m:1.65) --
(177+120*\m:2.35) arc (177+120*\m: 63+120*\m:2.35) -- cycle;
}
\foreach \n in {1,...,15}{\draw[fill=black] (\n*24:2) circle (0.05);};
\end{tikzpicture}
\end{document}


Edit: Now I complete redrawing of image. In drawing arc I first determine starting point of arc and than draw closed line in which rounded corners work. You can determine all used angles separately, if you estimate that this will generalize and simplify proposed solution.

Addedndum (1): inspired by comment of Paul Gaborit the code can be with use of the option rotate around={120*\m:(0,0) further simplified:

\documentclass[tikz,border=10pt]{standalone}

\begin{document}
\begin{tikzpicture}
\foreach \m in {0,1,2}{
\draw [fill=gray!50!white, rounded corners, rotate around={120*\m:(0,0)}]
( 15:2.25) arc ( 15:129:2.25) --
(129:1.75) arc (129: 15:1.75) -- cycle;
}
\foreach \m in {0,1,2}{
\draw [rounded corners, rotate around={120*\m:(0,0)}]
( 63:1.65) arc ( 63:177:1.65) --
(177:2.35) arc (177: 63:2.35) -- cycle;
}
\foreach \n in {1,...,15}{\draw[fill=black] (\n*24:2) circle (0.05);};
\end{tikzpicture}
\end{document}


Addedndum (2): or by use of backgroudns library you can draw both arcs group in one loop:

\documentclass[tikz,border=10pt]{standalone}
\usetikzlibrary{backgrounds}

\begin{document}
\begin{tikzpicture}
\foreach \m in {0,1,2}
{
\scoped[on background layer]
\draw [fill=gray!50!white, rounded corners, rotate around={120*\m:(0,0)}]
( 15:2.25) arc ( 15:129:2.25) --
(129:1.75) arc (129: 15:1.75) -- cycle;
\draw [rounded corners, rotate around={120*\m:(0,0)}]
( 63:1.65) arc ( 63:177:1.65) --
(177:2.35) arc (177: 63:2.35) -- cycle;
}
\foreach \n in {1,...,15}{\draw[fill=black] (\n*24:2) circle (0.05);};
\end{tikzpicture}
\end{document}


In the both added solutions the obtained result is the same as in the first one.

• You can remove the +120*\m part of coordinates' computation using the rotate=120*\m option. – Paul Gaborit Feb 4 '17 at 9:07
• @PaulGaborit, thank you for point. See addendum in my answer! – Zarko Feb 4 '17 at 10:45
• (+1) For figuring out what was wanted, just for one thing :-). I thought the inner curves were deliberate and tried to preserve them ... – cfr Feb 4 '17 at 16:18
• @cfr, thank you. in this question i had rather rare "bright" moment (translated literally), that I figured out, what is OP demand. Usually I had problems with this. to be honest, your answer help mi in this. – Zarko Feb 4 '17 at 16:23
• I realise I've also suggested backgrounds in my new version. However, I made sure my solution was otherwise different, I think. – cfr Feb 4 '17 at 23:26

I would do it like this :

## The code

\documentclass[border=10pt,multi,tikz]{standalone}

\def\r{2}
\newcommand\myAngle[1]{24*#1}
\newcommand\myArc[3] %\myArc{startAngle}{endAngle}{halfLineWidth} with startAngle <
{
(#1:\r+#3)
arc [radius = #3, start angle=#1, end angle=#1-180]
arc [radius = \r-#3, start angle=#1, end angle=#2]
arc [radius = #3, start angle=#2+180, end angle=#2]
arc [radius = \r+#3, start angle=#2, end angle=#1]
}
\newcommand\smallSnake[2]
{
\draw [smallSnake] \myArc{\myAngle{#1}}{\myAngle{#2}}{.2};
}
\newcommand\largeSnake[2]
{
\draw [largeSnake] \myArc{\myAngle{#1}}{\myAngle{#2}}{.3};
}

\begin{document}

\begin{tikzpicture}
\tikzset
{
smallSnake/.style=
{
fill=blue,
fill opacity=.2,
},
largeSnake/.style=
{
fill=red,
fill opacity=.3,
},
}
\largeSnake{3}{8}
\largeSnake{9}{13}
\largeSnake{14}{17}
\smallSnake{1}{5}
\smallSnake{6}{11}
\smallSnake{12}{15}

\foreach \n in {1,...,15}
{
\node (dot-\n) [fill=black, circle, inner sep=1pt, draw] at (\myAngle{\n}:\r) {};
};

\end{tikzpicture}
\end{document}


# EDIT

Originally, I thought that the inner curves in the example were intentional, since the original question said that the filling worked just fine and only the bordering was problematic. This seemed plausible because preserving the inner curves made life much more difficult and I posted only an improved version, rather than a complete solution.

Since this was a PITA to do and somebody might find it useful sometime, I will leave it below. Meanwhile, other good answers have been posted to the real question, one which correctly read the user's mind and one which responded, I think, to the edited question clarifying that mind's contents.

I therefore propose a different solution which leverages some of TikZ built-in libraries.

This makes it possible to draw everything but the final black framing lines in one step:

• the black outline is drawn first, in a preaction;
• the grey filling is drawn next;
• the black dots are added last, in a postaction.

By doing this on a background layer, it is possible to draw the frame over the top in a second step, but as part of the same loop.

The filling in this solution is actually a line, a little thinner than the black line drawn behind it. Round arrow caps are used to end the fat lines, rather than rounded corners, and the circles are drawn as markings.

\documentclass[border=10pt,multi,tikz]{standalone}
\usetikzlibrary{arrows.meta,bending,decorations.markings,backgrounds}
\begin{document}
\begin{tikzpicture}
\foreach \i in {0,1,2}
{
\scoped [on background layer]{
\draw
[
preaction={draw=black, line width={5mm+.8pt}, shorten >=-.4pt, shorten <=-.4pt},
gray!50!white,
{Round Cap[length=4pt,bend]}-{Round Cap[length=4pt,bend]},
line width=5mm,
postaction={decorate},
decoration={markings, mark=between positions 0.1 and 0.92 step 0.2 with {\filldraw [black, line width=.4pt, -] circle (0.05);} }
] ({15+120*\i}:2) arc ({15+120*\i}:{129+120*\i}:2) ;
}
\draw ({63+120*\i}:2.35) arc ({63+120*\i}:{177+120*\i}:2.35) -- ({177+120*\i}:1.65) arc  ({177+120*\i}:{63+120*\i}:1.65) -- cycle;
}
\end{tikzpicture}
\end{document}


# Original solution to original version of question

This isn't perfect and it isn't elegant, but perhaps it is an improvement, at least. It uses the useful inverse clipping provided in Paul Gaborit's answer.

\documentclass[border=10pt,multi,tikz]{standalone}
% code for inverse clipping from Paul Gaborit's answer at https://tex.stackexchange.com/a/59168/

\tikzset{%
invclip/.style={%
clip,insert path={%
{%
[reset cm]
(-16383.99999pt,-16383.99999pt) rectangle (16383.99999pt,16383.99999pt)
}%
}%
},
}
\begin{document}
\begin{tikzpicture}
\foreach \m in {0,...,2}{
\begin{scope}
\begin{pgfinterruptboundingbox}
\path [invclip, rounded corners=4pt]  (0:0) -- ({15+120*\m}:1.75) arc ({15+120*\m}:{129+120*\m}:1.75) -- (0:0) circle (1.65);
\end{pgfinterruptboundingbox}
\draw [fill=gray!50!white, rounded corners=4pt] (0:0) -- ({15+120*\m}:2.25) arc ({15+120*\m}:{129+120*\m}:2.25) -- (0:0);
\end{scope}
}
\fill [white] (0,0) circle (1.65);
\foreach \m in {0,...,2}{
\draw [fill=white] ([shift=({63+120*\m}:1.65)]0:0) arc ({63+120*\m}:{177+120*\m}:1.65);
\draw  ([shift=({63+120*\m}:2.35)]0:0) arc ({63+120*\m}:{177+120*\m}:2.35);
\draw  ({63+120*\m}:1.65)--({63+120*\m}:2.35);
\draw  ({177+120*\m}:1.65)--({177+120*\m}:2.35);
}
\foreach \n in {1,...,15}{\draw[fill=black] (\n*24:2) circle (0.05);};
\end{tikzpicture}
\end{document}
`

• Thanks for the attempt. Your shaded areas are missing a black border on the inner radius and the corners along the inner radius aren't right either. I just noticed the same error in my work too. Given the relative simplicity of the object I am trying to draw, it is amazing to me how seemingly difficult it is to pull off correctly! – Laars Helenius Feb 4 '17 at 4:04
• @LaarsHelenius I understood that you wanted the inner corners like that, as you said that the filling was fine. Hence, I thought I had to preserve them, which made it much trickier. Had I understood what was wanted, it would have been a lot simpler! – cfr Feb 4 '17 at 16:20
• I didn't notice the error in my original work until I saw your example. I went back and edited my question as a result. The misunderstanding was all my fault. Thank you for your efforts despite my lack of clarity. – Laars Helenius Feb 4 '17 at 16:36
• @LaarsHelenius Please see edited answer above. (I know you have other solutions now, but here's another way of doing it.) – cfr Feb 4 '17 at 23:20
• @LaarsHelenius Note that your actual question is a lot easier than the one I tried to answer originally!! – cfr Feb 4 '17 at 23:27