I'm sure some of you can shorten this c… by cantelope

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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cantelope

function u(t) {

F=(v)=>("0"+((.5+v/2)*255|0).toString(16)).slice(-2)
for(i=1e3;i--;x.fillStyle="#"+F(S(v=i/70-t*4))+F(C(v))+F(-S(v)))x.fillRect(5*i,0,5,2e3)

Jun 27 2017 11:55 PM UTC

}// 140/140

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I'm sure some of you can shorten this considerably.

u/alien

good sir what does F=(v)=> are you doing some functional programming here?
u/joeytwiddle

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u/cantelope

@alien the F=(v)=> is indeed a function, using the short ES6 arrow syntax. Its purpose is to return a hex value between 0 and 255, (with a preceding 0 when value is less than 16). the hex code is then combined with 2 other hex codes to form a complete, valid color code in the form of #000000. There is a shorter way to do this but it results in exponentially fewer colors. sorry for delayed response.
u/New_Core

Using rgb colors instead of hex you dont need to struggel with converting. a=130;for(i=1e3;i--;x.fillStyle=R(a*S(v=i/70-t*4)+a,a*C(v)+a,-a*S(v)+a))x.fillRect(5*i,0,5,2e3) a here uses 130 of the 255 colors to match the same colors.
u/alien

thank you for the response @catelope


      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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