How can we display fireworks?

Question

Soon we can celebrate the New Year. How can we use TikZ for drawing fireworks? For example like this, photographed by Kurume-Shimin, found on Wikipedia:

Click on Image search for getting further impressions. On the Wikipedia page you can see further effects, just click the image. I would like to post a nice drawing on our blog for celebrating New Year, and as addition to the TikZ example gallery on this occasion.

Answer 1

Here is my take:

Update: More "bling bling", more compilation time... (and trying @Stefan's brilliant idea)

Update 2 Thanks to Stefan, I have seen that the previous version was only working in the QtikZ environment but nowhere(I guess) else. For some reason beyond me, the fadings are still not displayed correctly in Adobe Reader but at least it works (almost) properly in SumatraPDF on Windows. Sorry for the inconvenience and Happy New Year everyone!

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc,decorations.pathmorphing}
\pgfdeclareradialshading{someshade}{\pgfpointorigin}{color(0mm)=(pgftransparent!40);color(4mm)=(pgftransparent!50);color(8mm)=(pgftransparent!70);color(2cm)=(pgftransparent!100)}
\pgfdeclareradialshading{somenodeshade}{\pgfpointorigin}{color(0mm)=(pgftransparent!0);color(2mm)=(pgftransparent!0);color(5mm)=(pgftransparent!99);color(20mm)=(pgftransparent!100)}
\pgfdeclareradialshading{invertshade}{\pgfpointorigin}{color(0mm)=(pgftransparent!100);color(10mm)=(pgftransparent!95);color(16mm)=(pgftransparent!60);color(2cm)=(pgftransparent!0)}

\pgfdeclarefading{fadeit}{\pgfuseshading{someshade}}
\pgfdeclarefading{fadein}{\pgfuseshading{invertshade}}

\begin{document}
\begin{tikzpicture}[projectile/.style={decorate,decoration={random steps,segment length=2pt,amplitude=0.5pt}}]
\fill[black] (-4,-3) rectangle (6,4);

\begin{scope}[xshift=0cm,yshift=-0.4cm,transparency group]
\pgfsetfading{fadein}{\pgftransformshift{\pgfpointorigin}}
    \foreach \x in {0,10,..., 360}{\draw[blue!80!white,projectile,line width=1.1pt] (0,0) to [in=90] (10*rand+\x:rand*1mm+2cm);};
\end{scope}

\begin{scope}[xshift=2cm,yshift=1cm]
   \foreach \x in {0,12,..., 360}{\draw [yellow!5,thick,projectile] (0.7,0) to  (3*rand+\x :1mm*rand+2.2cm)  node%
[circle,inner sep=1mm,shade,shading=somenodeshade,opacity=1,outer sep=0] {};}
{\pgfsetfading{fadeit}{\pgftransformshift{\pgfpoint{2.5cm}{1cm}}}};\fill[white] (-3,-3) rectangle (3,3);
\end{scope}

\begin{scope}[xshift=3cm,yshift=-1cm]
\foreach \x in {0,10,..., 360}{\def\r1{rand}\draw [yellow]  ($(0,0)!abs{\r1}!(\x :5mm)$) to [in=90] ($(0,0)!abs{\r1}+0.2!(\x :8mm)$);}
{\pgfsetfading{fadeit}{\pgftransformshift{\pgfpoint{3cm}{-1cm}}}};\fill[yellow,opacity=0.6] (-3,-3) rectangle (3,3);
\end{scope}

\begin{scope}[xshift=-1cm,yshift=1.5cm]
   \foreach \x in {0,12,..., 360}{\def\r2{rand}\draw [red,line width=0.5pt]  ($(0,0)!abs{\r2}!(\x :3mm)$) -- ($(0,0)!abs{\r2}+0.1!(\x :7mm)$);}
{\pgfsetfading{fadeit}{\pgftransformshift{\pgfpoint{-1cm}{1.5cm}}}};\fill[red,opacity=0.6] (-3,-3) rectangle (3,3);
\end{scope}
\end{tikzpicture}
\end{document}

and the updated code gives (see the history for the previous code)

Answer 2

Update 4 : it's better with more rays and I add a decoration (Stephan's idea) but the result is aleatory. I change some arguments. With some shading in the center (idea from percusse and bloodworks):

\documentclass{minimal} 
\usepackage[dvipsnames,svgnames]{xcolor} 
\usepackage{tikz}
 \usetikzlibrary{calc} 

 \usetikzlibrary{decorations,decorations.pathmorphing} 

\begin{document}

\begin{tikzpicture}[decoration={bent,aspect=.3}]
  \draw[fill=black] (-8,-3) rectangle (8,5); 
  \foreach \colA/\colB/\sx/\sy in {red/yellow/-4/2,blue/green/-2/1,
                                   yellow/red/0/2,green/blue/4/2,yellow/white/3/1.5} 
{% 
   \draw[shading=radial, inner color=white, outer color=black, opacity=.5]  (\sx,\sy) circle (1cm);   
  \pgfmathparse{rnd}   
  \let\ri\pgfmathresult  
  \begin{scope}[xshift=\sx]
      \foreach \i in {0,...,150}{%  
         \pgfmathparse{rnd}
         \global\let\x\pgfmathresult 
         \pgfmathparse{rnd}  
         \global\let\y\pgfmathresult 
         \pgfmathparse{rnd} 
         \let\c\pgfmathresult 
         \pgfmathsetmacro{\col}{50*\c+50} 
           \draw[decorate,\colA!\col!\colB] %
           ($(360*\x:2*\ri*\y+1) +(\sx,\sy)$) to    ($(360*\x:1*\y) + (\sx,\sy)$);} 
  \end{scope}
} 
\end{tikzpicture} 
\end{document} 

Answer 3

Not quite as smart as the two above but well thats what i have. I (almost) used no commands starting with pgf.

\documentclass[border=5mm]{standalone}

\usepackage{tikz}
\usetikzlibrary{decorations}
\usetikzlibrary{decorations.pathmorphing} 

\begin{document}

\begin{tikzpicture}
 \draw[fill=black] (-5,-15) rectangle (25,15);
 \foreach \mya in {1,10,20}{
  \pgfmathrandominteger{\myb}{1}{10}
  \draw[shading=radial, inner color=white, outer color=black, opacity=.5](\mya , \myb) circle (5cm);
  \foreach \x in {-3.1,-2,-1.1,0.1,1.2,2.3,3.1}{%
  \foreach \y in {-3.2,-2.1,-1.2,0,1.1,2,3.2}{%
  \draw[red, thick, decorate, decoration={random steps,segment length=3pt,amplitude=1pt}, rotate around={10:(\mya , \myb)}, opacity=0.7](\mya , \myb)--(\mya+\x, \myb+\y)node[red, yshift=-.1mm]{*};%
  \draw[blue,decorate, decoration={random steps,segment length=3pt,amplitude=1pt}, rotate around={20:(\mya , \myb)}](\mya , \myb)--(\mya+\x, \myb+\y)node[red, yshift=-.1mm]{*};%
  \draw[red!50!yellow, decorate, decoration={random steps,segment length=3pt,amplitude=1pt}, rotate around={50:(\mya , \myb)}](\mya , \myb)--(\mya+\x,  \myb+\y);
 }
 }
 \draw[decorate, decoration=snake, red, thick] (0 , -13)--(\mya , \myb);
}


\end{tikzpicture}
\end{document}

It takes a while to compile!