# Draw picture in tikz

I would like to draw a picture as follows:

but I don't know how to start. It consists of two connected semi-rings, $\Omega_1$ and $\Omega_2$. Anyone can help me?

• There are examples in the pgfmanual that show circles/arcs, angles, etc. So that should get you started. – Skillmon May 23 '19 at 13:04

You may combine this with Skillmon's answer.

\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{arrows.meta,bending}
\begin{document}
\begin{tikzpicture}
\draw (0:4) arc(0:180:4) node[midway,above]{WALL} -- (180:0.5) node[midway,above]{$\Omega_{13}$}
arc (180:0:0.5)
node[midway,above]{$\Omega_{11}$}-- cycle node[midway,above]{$\Omega_{34}$};
\draw (0,0) -- (35:4);
\draw[-{Latex[bend]}] (1.5,0) arc(0:35:1.5) node[near end,right]{$\theta$};
\draw[-Latex] (0,0) -- (70:3) node[near end,right]{$\Omega$};
\begin{scope}[yshift=-2pt,red]
\draw (0:4) arc(0:-180:4) node[midway,below]{WATER}
-- (-180:0.5) node[midway,below]{$\Omega_{23}$}
arc(-180:0:0.5)
node[midway,below]{$\Omega_{21}$} -- cycle node[midway,below]{$\Omega_{24}$};
\draw (0,0) -- (-35:4);
\draw[-{Latex[bend]}] (3,0) arc(0:-35:3) node[midway,right]{$\varphi$};
\foreach \X in {1.75,2.25,2.75}
{\draw[-Latex] (1,-\X) -- (-1,-\X);}
\end{scope}
\end{tikzpicture}
\end{document}


To get you started, I've drawn the PokeBall:

EDIT: Fixing some issues, some calculations were wrong (at least based on the names of the variables).

\documentclass[tikz,border=3.14]{standalone}

\pgfkeys
{%
/tikz/kim/.cd
,gap/.initial=.02cm
,inner/.initial=.2cm
}
\newcommand*\kimvalue[1]{\pgfkeysvalueof{/tikz/kim/#1}}

\begin{document}
\begin{tikzpicture}
\pgfmathsetmacro\kimAngleI{asin(\kimvalue{gap}/\kimvalue{inner})}
\draw[red]
arc
[%
start angle={-180+\kimAngleO},
end angle={-\kimAngleO},
]
-- ({-\kimAngleI}:\kimvalue{inner})
arc
[%
start angle={-\kimAngleI},
end angle={-180+\kimAngleI},
]
-- cycle
;
\draw
arc
[%
start angle=180-\kimAngleO,
end angle=\kimAngleO,
]
-- ({\kimAngleI}:\kimvalue{inner})
arc
[%
start angle=\kimAngleI,
end angle=180-\kimAngleI,