If someone can shave 2 characters off, … by DirtyDan

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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DirtyDan

remix of d/7009 by u/DirtyDan

function u(t) {

with(x){lineWidth=b=44
fillStyle="#FFF"
for(i=0;i<b;i++)beginPath(),arc(960+S(a=t*i/b*4)*(d=i*t*S(t)),540+C(a)*d,i*3,0,b/7),stroke(),fill()}

Feb 20 2018 3:01 AM UTC

}// 140/140

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If someone can shave 2 characters off, we can invert the colors! Like in d/7011

u/akm

here, I shaved off 10 chars: with(x)for(fillStyle="#FFF",i=0;i++<44;lineWidth=44)beginPath(arc(960+S(a=t*i/11)*(d=i*t*S(t)),540+C(a)*d,i*3,0,7),fill(stroke()))


      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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