New Year Fireworks by keithclark

function u(t) {

c.width=1920 // clear the canvas
for(i=0;i<9;i++)
x.fillRect(400+i*100+S(t)*300,400,50,200) // draw 50x200 rects

} // 122/140

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    u(t) is called 60 times per second.
    t: Elapsed time in seconds.
    S: Shorthand for Math.sin.
    C: Shorthand for Math.cos.
    T: Shorthand for Math.tan.
    R: Function that generates rgba-strings, usage ex.: R(255, 255, 255, 0.5)
    c: A 1920x1080 canvas.
    x: A 2D context for that canvas.

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keithclark

function u(t) {

for(x.fillRect(x.globalAlpha=.2,q=t%2,2e3,2e3),i=125;i--;x.clearRect(S(t|1)*760+S(i)*v+960,(q*q/2-C(i))*v+380,l=6-q*3,l))v=q*200-S(i*.2)*i*q

Dec 31 2018 9:44 PM UTC

}// 140/140

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New Year Fireworks

Show more comments…
u/13thptr

elegant :)
u/keithclark

change S(i*.2)*i*q for S(i*t>>4)*i*q for varying firework effects (2 chars over the limit tho)
u/katkip

haha i like the variation
u/JWinslow23

If q will only ever be 0 or 1 (t%2), couldn't this just be for(x.fillRect(x.globalAlpha=.2,q=t%2,2e3,2e3),i=125;i--;x.clearRect(S(t|1)*760+S(i)*v+960,(q/2-C(i))*v+380,l=6-q*3,l))v=q*200-S(i*.2)*i*q? That way, the variation also works.
u/keithclark

I opted for what I had because it held the initial shape in the sky for longer before gravity takes hold. Your version works and if you're happy to trade off the initial shape, you could also drop the \2 and save another 2 chars
u/keithclark

... also q can be any value between 0 and 1
u/JWinslow23

Ah, my mistake, I forgot about floats
u/lionleaf

Nice one! #nye #fireworks
u/iverjo

Beautiful!
u/vain

3... 2... 1... ignition... beautiful firework


      u(t) is called 60 times per second.
      t: elapsed time in seconds.
      c: A 1920x1080 canvas.
      x: A 2D context for that canvas.
      S: Math.sin
      C: Math.cos
      T: Math.tan
      R: Generates rgba-strings, ex.: R(255, 255, 255, 0.5)

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