# Arrows, nomenclature and shaded area

I want to make a diagram similar to this one However, I have only been able to do this: With the code

\begin{tikzpicture}

\draw ([shift={(1,0)}]19.8028:7.69) arc[radius=7.69, start angle=19.8028, end angle= 24.2458] -- ([shift={(1,0)}]24.2458:8.31) arc[radius=8.31, start angle=24.2458, end angle= 19.8028] -- cycle;

\draw ([shift={(1,0)}]155.7542:7.69) arc[radius=7.69, start angle=155.7542, end angle= 160.1972] -- ([shift={(1,0)}]160.1972:8.31) arc[radius=8.31, start angle=160.1972, end angle= 155.7542] -- cycle;

\draw ([shift={(1,0)}]267.0380:5.69) arc[radius=5.69, start angle=267.0380, end angle= 272.962] -- ([shift={(1,0)}]272.962:6.31) arc[radius=6.31, start angle=272.962, end angle= 267.0380] -- cycle;

\end{tikzpicture}


Certainly, the location of the holes are different, what I need in my code is to add: 1.- The shaded area 2.- The nomenclature (symbols), the dimensioning. 3.- The arrows and the nodes in the center of the holes.

Any help is appreciated.

One possible approach (knowing how you calculate the angles of the wholes, it may be possible to more easily draw the arrows of the angles):

\documentclass[tikz]{standalone}

\begin{document}

\begin{tikzpicture}

\begin{scope}[even odd rule]
\end{scope}

\draw[fill=white] (19.8028:7.69) node (a1) {} arc[radius=7.69, start angle=19.8028, end angle= 24.2458] --  (24.2458:8.31) node (a2) {} arc[radius=8.31, start angle=24.2458, end angle= 19.8028] -- cycle;
\path (a1) -- (a2) node[inner sep=2pt,circle,fill,pos=.5] (x1) {};

\draw[fill=white] (155.7542:7.69) node (b1) {} arc[radius=7.69, start angle=155.7542, end angle= 160.1972] -- (160.1972:8.31) node (b2) {} arc[radius=8.31, start angle=160.1972, end angle= 155.7542] -- cycle;
\path (b1) -- (b2) node[inner sep=2pt,circle,fill,pos=.5] (x2) {};

\draw[fill=white] (267.0380:5.69) node (c1) {} arc[radius=5.69, start angle=267.0380, end angle= 272.962] -- (272.962:6.31) node (c2) {} arc[radius=6.31, start angle=272.962, end angle= 267.0380] -- cycle;
\path (c1) -- (c2) node[inner sep=2pt,circle,fill,pos=.5] (x3) {};

\draw (0,0) -- (x1) node[pos=.5,fill=white,circle] {$r_1$};
\draw (0,0) -- (x2) node[pos=.5,fill=white,circle] {$r_2$};
\draw (0,0) -- (x3) node[pos=.5,fill=white,circle] {$r_3$};

\draw (0,0) -- ++(3,0);
\draw[->] (0:2cm) arc (0:22.0243:2cm) node[pos=.5,label={0:$\phi_1$}] {};
\draw[->] (0:1.5cm) arc (0:157.9757:1.5cm) node[pos=.5,fill=white,circle] {$\phi_2$};
\draw[->] (0:1cm) arc (0:270:1cm) node[pos=.5,fill=white,circle] {$\phi_3$};

\end{tikzpicture}

\end{document}


Result: • Use \varphi instead of \phi for the different phi symbol (φ). – Jasper Habicht Nov 2 '18 at 12:55
• Thank you so much!. I got the figure I wanted. – Alfredo Nov 2 '18 at 13:32

Just for fun: you can get the wedge-shaped nodes also by using nonlinear transformations, particularly the so-called \polartransformation from the pgfmanual.

\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{calc}
\usepgfmodule{nonlineartransformations}
\makeatletter
% from https://tex.stackexchange.com/a/434247/121799
\tikzdeclarecoordinatesystem{polar}{
\tikz@scan@one@point\relax(#1)
\polartransformation
}
% from the pgfmanual
\def\polartransformation{% from the pgfmanual section 103.4.2
\pgfmathsincos@{\pgf@sys@tonumber\pgf@x}%
\pgf@x=\pgfmathresultx\pgf@y%
\pgf@y=\pgfmathresulty\pgf@y%
} % note: the following should work with arbitrary (nonlinear) transformations
\makeatother
\begin{document}

\begin{tikzpicture}
\draw (0,0)circle(3) (0,0)circle(7.5);
\filldraw[even odd rule,fill=gray!50] (0,0)circle(3.5) (0,0)circle(7);
\begin{scope}[transform shape nonlinear=true]
\pgftransformnonlinear{\polartransformation}
\node[draw,inner xsep=4mm/4,inner ysep=2mm,fill=white] (box1) at (2,4) {};
\node[draw,inner xsep=4mm/5,inner ysep=2mm,fill=white] (box2) at (3.5,5) {};
\node[draw,inner xsep=4mm/6,inner ysep=2mm,fill=white] (box3) at (9.5,6) {};
\end{scope}
\draw(0,0)-- (2.5,0);
\foreach \X in {1,2,3}
{\fill (polar cs:box\X.center) coordinate (p\X) circle (2pt);
\draw[-] (0,0) -- (polar cs:box\X.south) node[midway,fill=white]{$r_\X$};
\draw[-latex] let \p1=(p\X), \n1={veclen(\y1,\x1)},
\n2={ifthenelse(atan2(\y1,\x1)<0,360+atan2(\y1,\x1),atan2(\y1,\x1))}
in (0:{\n1/(1.6*\X+ifthenelse(\X==3,0.5,0))}) arc(0:\n2:{\n1/(1.6*\X+ifthenelse(\X==3,0.5,0))})
node[pos={1/(\X+1)},fill=white]{$\varphi_\X$};}
\end{tikzpicture}
\end{document} This is a minimal working example of your desired output.

\documentclass[12pt]{article}
\usepackage[utf8]{inputenc}

\usepackage{tikz}

\begin{document}

\scalebox{0.7}{
\begin{tikzpicture}

\fill[gray!20,even odd rule] (1,0) circle (9.2) (1,0) circle (4.2);

\draw ([shift={(1,0)}]19.8028:7.69) arc[radius=7.69, start angle=19.8028, end angle= 24.2458] -- ([shift={(1,0)}]24.2458:8.31) arc[radius=8.31, start angle=24.2458, end angle= 19.8028] -- cycle;
\draw ([shift={(1,0)}]155.7542:7.69) arc[radius=7.69, start angle=155.7542, end angle= 160.1972] -- ([shift={(1,0)}]160.1972:8.31) arc[radius=8.31, start angle=160.1972, end angle= 155.7542] -- cycle;
\draw ([shift={(1,0)}]267.0380:5.69) arc[radius=5.69, start angle=267.0380, end angle= 272.962] -- ([shift={(1,0)}]272.962:6.31) arc[radius=6.31, start angle=272.962, end angle= 267.0380] -- cycle;

\draw [thick] (1,0) -- (8.4,3) node [midway,fill=white] {$r_1$};
\draw [thick] (1,0) -- (-6.4,3) node [midway,fill=white] {$r_2$};
\draw [thick] (1,0) -- (1,-6) node [midway,fill=white] {$r_3$};
\draw [thick] (1,0) -- (3.2,0);

% angles
% \draw (starting point coordinates) arc (starting angle:ending angle:radius)
\draw[->] (2, 0) arc (20:265:1) node [midway,fill=white,font=\tiny] {$\phi_{3}$};
\draw[->] (2.5, 0) arc (20:130:1.5) node [midway,fill=white,font=\tiny] {$\phi_{2}$};
\draw[->] (3.2, 0) arc (20:40:2) node [midway,fill=white,font=\tiny] {$\phi_{1}$};

\end{tikzpicture}
}

\end{document}


All you have to do is to change the angles and the position of the nodes.

This is the output: • Thank you!, the only one observation is that the holes are not in white. – Alfredo Nov 2 '18 at 13:34