Get partly filled circle symbol scale linearly with parameter

Question

I want to create a \rating command that draws a partly filled circle, according to a numeric parameter between 0 and 100:

Question: How can I modify the expression for determining the angles in arc, so that the filling rises more steadily over the whole value range? Right now, ratings from 0 to 20 look empty, 80 to 100 look completely filled and only in between there is a steady rise.

Clarification: Rather than a mathematically rigoros expression, I am looking for the simplest expression that looks right, i.e. that \rating{25}, \rating{50} and \rating{75} visually look like 1/4, 2/4 and 3/4.

\documentclass{minimal}
\usepackage{tikz}

\newcommand{\rating}[1]{
\begin{tikzpicture}
\fill[lightgray] ({270-1.8*#1}:1ex) arc ({270-1.8*#1}:{270+1.8*#1}:1ex) -- cycle;
\draw[black, thin, radius=1ex] (0,0) circle;
\end{tikzpicture}}

\begin{document}
This really should scale\dots
\rating{0} \rating{10} \rating{20} \rating{30} \rating{40} \rating{50}
\rating{50} \rating{60} \rating{70} \rating{80} \rating{90} \rating{100}
\dots linearly.
\end{document}

Answer 1

As an aside, please see Why should the minimal class be avoided?

Here is a solution using a clipped path in TikZ. Note that the clip path cannot have extra options added, so I've drawn the circle with your specifications separately, outside the scope of the \clip path.

\documentclass{article}
\usepackage{tikz}

\newcommand{\rating}[1]{%
  \begin{tikzpicture}[x=1ex,y=1ex]
    \begin{scope}
      \clip (0,1) circle (1);
      \fill[lightgray] (-1,0) rectangle (1,#1/50);
    \end{scope}
    \draw[black, thin, radius=1] (0,1) circle;
  \end{tikzpicture}%
}

\begin{document}
This really should scale\dots
\foreach \h in {0,5,...,100} {\rating{\h}}
\dots linearly.
\end{document}

Answer 2

All you need is a different parameterisation of the angle; see below for details.

Additionally,

1. since you seem to use \rating "inline", you may want to simply use \tikz instead of a tikzpicture environment;
2. avoid spurious spaces in the definition of \rating (instead, insert a space between calls to \rating, wherever one is needed);
3. instead of hardcoding 1ex as the radius, make it an optional argument, for more flexibility;
4. as already pointed out by Paul in his answer, avoid the minimal class when using tikz.

---

\documentclass{article}
\usepackage{tikz}
\usepackage{amsmath}

\newcommand{\rating}[2][1ex]{%
  \pgfmathsetmacro\th{asin(#2/50-1)}% (theta angle of polar coordinates)
    \tikz{%
      \fill[lightgray] (\th:#1) arc (\th:-180-\th:#1) -- cycle;
      \draw[black, thin, radius=#1] (0,0) circle;
    }%
}

\begin{document}
This really should scale\dots
\foreach \h in {0,10,...,100} {\rating{\h} }
\dots linearly.

\section*{Parameterisation explained}

Let~\(R\) be the disk radius. In polar coordinates,
%
\[
  y = R \sin \theta \,.
\]
%
Therefore,
%
\[
  \theta = \arcsin \frac{y}{R}  \,.
\]
%
Because the quantity~\(y/R\) varies over~\([-1,1]\)
but you want a parameter~\(u\) that varies over~\([0,100]\),
the right affine transformation will get you there:
%
\[
  \frac{y}{R} = \frac{u}{50} - 1 \,.
\]
%
\end{document}

Answer 3

Another way with TikZ but without explicit clipping:

\documentclass[border=5]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\tikzset{circle mark/.pic={
  \draw [path picture={%
    \fill (path picture bounding box.south west) |-  
      ($(path picture bounding box.south east)!#1/100!(path picture bounding box.north east)$) 
      |- cycle;}] circle [radius=1ex];
}}
\begin{document}
\foreach \i in {0,10,...,100}
  \tikz\path pic{circle mark=\i};
\end{document}

Answer 4

Just for fun with PSTricks.

\documentclass[preview,border=12pt,varwidth]{standalone}
\usepackage{pstricks}


\newcommand\rating[2][.25pt]{%
\psset{unit=#1,runit=#1}
\pspicture(100,100)
\psclip{\pscircle[linestyle=none,linewidth=0](50,50){50}}
    \psframe*[linecolor=gray](100,#2)
\endpsclip
\pscircle[dimen=m](50,50){50}
\endpspicture}

\begin{document}
Linearly filled circles:\\
\foreach \i in {0,10,...,100}{\rating{\i}}
\end{document}

Answer 5

Run with xelatex

\documentclass{article}
\usepackage{pstricks}

\def\rateColor{black!40}
\newcommand\rating[2][]{%
  \ifx\relax#1\relax\else\gdef\rateColor{#1}\fi
  \psset{unit=1ex}%
  \pspicture[dimen=inner](-1.2,-1.2)(1.2,1.2)
    \psclip{\pscircle{1}}
      \psframe*[linecolor=\rateColor](-1,-1)(!1 #2 100 div 2 mul 1 sub)
    \endpsclip\pscircle{1}
  \endpspicture}

\begin{document}
This really should scale \dots

\tiny
\multido{\iA=0+10}{21}{\ifnum\iA>100\expandafter\rating\expandafter{\the\numexpr200-\iA}\else\rating{\iA}\fi} \par
\normalsize
\rating[red]{0} 
\multido{\iA=10+10}{20}{\ifnum\iA>100\expandafter\rating\expandafter{\the\numexpr200-\iA}\else\rating{\iA}\fi} \par
\Large
\rating[blue]{0} 
\multido{\iA=10+10}{20}{\ifnum\iA>100\expandafter\rating\expandafter{\the\numexpr200-\iA}\else\rating{\iA}\fi} \par
\end{document}