For some reason my phone does not requi… by Qwitter

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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function u(t) {

a=530;m=t%28;c.width=2e3;x.beginPath();for(i=1;i<m+1;i++){p=3*t+i*6.28/m;x.lineTo(a*(2+C(p)),a*(1+S(p)))}with(x)closePath(),fill(stroke())

Jan 21 2018 5:00 PM UTC

}// 138/140

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For some reason my phone does not require the x.beginPath(); command. (?)

u/sigveseb

afaik you don't need beginPath at all, since setting the width of a canvas resets it completely, including the current path
u/Qwitter

Thank you for your answer. It does work the same w/o beginPath on the phone's Safari browser, and i was able to get a little color with x.fillStyle=R(p);. However, it just makes a way to many verticied jumbled circular shape w/o it on the edge explorer (its a bad habit) on my computer. Another major difference I have seen between the phone andthe computer is in the results of using the 'HSVA' color coordinates, sometimes they look vastly different, but im not sure under what conditions that happens. Probably not worth investigating, but I'll note them here to spread some awareness.
u/Qwitter

*HSLA
u/Xen

part way between square and pentagon it gives a cube optical illusion ;)


      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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