I prefer this with a slower cycle by ha… by bligglenuber

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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bligglenuber

function u(t) {

c.width=198;a=6+C(t/9)
for(i=-99;i<99;i+=a)for(j=-56;j<56;j+=a)k=i*j,f=k*k*(1+C(t)),x.fillStyle=R(f,f/9),x.fillRect(99+i,56+j,a*S(k),a*C(k))

Feb 28 2017 2:00 PM UTC

}// 140/140

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I prefer this with a slower cycle by having t/99 instead of 9, but I couldn't find another place to drop a character

u/iverjo

You could move f= inside R, like this: c.width=198;a=6+C(t/9) for(i=-99;i<99;i+=a)for(j=-56;j<56;j+=a)k=i*j,x.fillStyle=R(f=k*k*(1+C(t)),f/9),x.fillRect(99+i,56+j,a*S(k),a*C(k))
u/bligglenuber

Ah, good shout, thanks


      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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