Circular shading

Question

I would like to draw a mirror and its thickness, using Tikz. I'll show you the final result I'd like to achieve

I managed to write this code (using tikz package)

    \shade[shading angle= 90] (75:3)
    -- (75:3.5) arc (75:-75:3.5)
    -- (-75:3) arc (-75:75:3);

which gives back this picture

I wish to have a different shading effect, like a ring one. Can someone help me please? Thank you in advance.

ADDENDUM: I'm gettin really mad. I tried to swap inner color with outer color to obtain this effect

but this is what actually latex gives me back

I'll leave you the code I used

\tikzfading[name=fade out,inner color=transparent!100,outer color=transparent!0]
\begin{scope}[xshift=6cm]
    \clip (75:-3)
    -- (75:-2.5) arc (75:-75:-2.5)
    -- (-75:-3) arc (-75:75:-3);
    \fill[path fading=fade out] (0,0) circle[radius=-3];
\end{scope}

If I keep outer and inner colors like the case before, I obtain this

I'm not understading why tikz is acting like this. Theoretically if I swap the colors, I should get what I am supposed to get, but I'm clearly wrong and I don't know why.

I added a shading halo to address the second part of your problem in an addendum to my answer. — , Dec 17 '19 at 22:02

score 22 · Accepted Answer · 2019-12-17T22:00:54.760

Welcome! These are two basic options from the pgfmanual:

circle with fuzzy edge 15 percent
fade out

Example:

\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{fadings}
\tikzfading[name=fade out,inner color=transparent!0,outer color=transparent!100]
\begin{document}
\begin{tikzpicture}
 \begin{scope}
 \clip (75:3)
    -- (75:3.5) arc (75:-75:3.5)
    -- (-75:3) arc (-75:75:3);
  \fill[path fading=circle with fuzzy edge 15 percent] (0,0) circle[radius=3.5cm];
 \end{scope}    
 \begin{scope}[xshift=5cm]
 \clip (75:3)
    -- (75:3.5) arc (75:-75:3.5)
    -- (-75:3) arc (-75:75:3);
  \fill[path fading=fade out] (0,0) circle[radius=3.5cm];
 \end{scope}    
\end{tikzpicture}
\end{document}

You can use this e.g. as follows:

\documentclass[tikz,border=3mm]{standalone}
\usepackage{contour}
\contourlength{0.05em}
\usetikzlibrary{fadings,decorations.markings,calc,angles,quotes}
\tikzset{->-/.style={decoration={% https://tex.stackexchange.com/a/39282/194703
  markings,
  mark=at position #1 with {\arrow{>}}},postaction={decorate}},
  ->-/.default=0.5}
\begin{document}
\begin{tikzpicture}[>=stealth,every label/.append style={black},
    bullet/.style={circle,fill=black,inner sep=0.1em}]
 \begin{scope}
 \clip (75:3)
    -- (75:3.5) arc (75:-75:3.5)
    -- (-75:3) arc (-75:75:3);
  \fill[path fading=circle with fuzzy edge 15 percent] (0,0) circle[radius=3.5cm];
 \end{scope}
 \draw (-3,0) coordinate (L) -- (4,0) coordinate (R) (3,0)
  node[bullet,label=below right:\contour{white}{$V$}](V){};
 \pgfmathsetmacro{\myangle}{atan2(3*sin(30),2+3*cos(30))}
 \draw[thick,orange,->-] (-2,0) node[bullet,label=below:$O$](O){} 
    -- (30:3) node[bullet,label={above right:\contour{white}{$P$}}] (P){};  
 \draw[dashed,blue] (0,0) node[bullet,label=below:$C$] (C){} -- (P);
 \path[overlay] (P) -- ++ (180+60-\myangle:1) coordinate (Q);
 \draw[thick,red,->-] (P) -- 
  (intersection cs:first line={(L)--(R)}, second line={(P)--(Q)}) 
  node[bullet,label=below:$I$] (I){};
 \path[blue]  
 pic [draw, "$\scriptstyle\alpha$",angle eccentricity=1.25,angle radius=1cm] {angle=R--O--P}
 pic [draw, "$\scriptstyle\beta$",angle eccentricity=0.7,angle radius=0.8cm] {angle=R--C--P}
 pic [draw, "$\scriptstyle\gamma$",angle eccentricity=0.7,angle radius=0.8cm] {angle=R--I--P};
\end{tikzpicture}
\end{document}

Or declare your own.

\documentclass[tikz,border=3mm]{standalone}
\usepackage{contour}
\contourlength{0.05em}
\usetikzlibrary{fadings,decorations.markings,calc,angles,quotes}
\pgfdeclareradialshading{invircle}{\pgfpointorigin}{
  color(0pt)=(pgftransparent!100); color(22.5bp)=(pgftransparent!80);
  color(25bp)=(pgftransparent!20); color(50bp)=(pgftransparent!00)}%
\pgfdeclarefading{inverted fading}{\pgfuseshading{invircle}}%

\tikzset{->-/.style={decoration={% https://tex.stackexchange.com/a/39282/194703
  markings,
  mark=at position #1 with {\arrow{>}}},postaction={decorate}},
  ->-/.default=0.5}
\begin{document}
\begin{tikzpicture}[>=stealth,every label/.append style={black},
    bullet/.style={circle,fill=black,inner sep=0.1em}]
 \begin{scope}
 \clip (75:3)
    -- (75:3.5) arc (75:-75:3.5)
    -- (-75:3) arc (-75:75:3);
  \fill[path fading=inverted fading] (0,0) circle[radius=3.5cm];
 \end{scope}
 \draw (-3,0) coordinate (L) -- (4,0) coordinate (R) (3,0)
  node[bullet,label=below right:\contour{white}{$V$}](V){};
 \pgfmathsetmacro{\myangle}{atan2(3*sin(30),2+3*cos(30))}
 \draw[thick,orange,->-] (-2,0) node[bullet,label=below:$O$](O){} 
    -- (30:3) node[bullet,label={above right:\contour{white}{$P$}}] (P){};  
 \draw[dashed,blue] (0,0) node[bullet,label=below:$C$] (C){} -- (P);
 \path[overlay] (P) -- ++ (180+60-\myangle:1) coordinate (Q);
 \draw[thick,red,->-] (P) -- 
  (intersection cs:first line={(L)--(R)}, second line={(P)--(Q)}) 
  node[bullet,label=below:$I$] (I){};
 \path[blue]  
 pic [draw, "$\scriptstyle\alpha$",angle eccentricity=1.25,angle radius=1cm] {angle=R--O--P}
 pic [draw, "$\scriptstyle\beta$",angle eccentricity=0.7,angle radius=0.8cm] {angle=R--C--P}
 pic [draw, "$\scriptstyle\gamma$",angle eccentricity=0.7,angle radius=0.8cm] {angle=R--I--P};
\end{tikzpicture}
\end{document}

ADDENDUM: As for the addendum to your question: here is a shading that produces something of that sort. It was obtained by looking up the definition of circle with fuzzy edge 10 percent and to "invert it", as you suggest.

\documentclass[tikz,border=10pt]{standalone}
\usetikzlibrary{decorations.markings,calc,angles,quotes}
\pgfdeclareradialshading{halo}{\pgfpointorigin}{
  color(0pt)=(pgftransparent!100); color(22.5bp)=(pgftransparent!100);
  color(25bp)=(pgftransparent!00); color(50bp)=(pgftransparent!00)}%
\pgfdeclarefading{halo}{\pgfuseshading{halo}}%
\tikzset{->-/.style={decoration={% https://tex.stackexchange.com/a/39282/194703
  markings,
  mark=at position #1 with {\arrow{>}}},postaction={decorate}},
  ->-/.default=0.5}
\begin{document}
\begin{tikzpicture}[>=stealth,every label/.append style={black},
    bullet/.style={circle,fill=black,inner sep=0.1em}]
\begin{scope}[xshift=6cm]
    \clip (75:-3)
    -- (75:-2.5) arc (75:-75:-2.5)
    -- (-75:-3) arc (-75:75:-3);
    \fill[path fading=halo] (0,0) circle[radius=3cm];
\end{scope}
\draw (-2,0) coordinate (L) -- (7,0) coordinate (R) 
  (3,0)  node[bullet,label=below left:$V$](V){}
  (6,0)  node[bullet,label=below:$C$](C){}
  (-1,0)  node[bullet,label=below:$O$](O){}
  ($(C)+(160:3)$) node[bullet,label=above:$P$](P){};
  \draw[orange,->-,thick] (O) -- (P);
  \draw[red,->-,thick] let \p1=($(P)-(C)$),\p2=($(P)-(O)$),
   \n1={180-atan2(\y1,\x1)},\n2={atan2(\y2,\x2)} in
  (P) -- ++({180-(2*\n1+\n2)}:2);
  \draw[dashed, shorten >=-2cm] (C) -- (P);
\end{tikzpicture}
\end{document}

@user203430 Remember to mark this answer as accepted if you think it adequately answered your question! — sintax, Dec 17 '19 at 19:16
is there any way to get the shading like the one obtained with the first method? Why if I swap inner with outer color, I got that result? — frad, Dec 18 '19 at 11:03
@user203430 Please note that fade out fades out over the full radius. However, we clip only a tiny shell of the circle, along which the color does not vary very much. If you remove the \clip ....; you will see what I mean. — , Dec 18 '19 at 15:25

Circular shading

1 Answers1

Linked