#d3 #perspective #octahedron by f8f82804

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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f8f82804

remix of d/10658 by u/f8f82804

function u(t) {

for(c.width|=i=13;i--;x.lineTo(967+(y-1?0:1e3*S(p)/z),1e3*y/z))y='011112101211'[i],z=y-1?3:3+C(p=t+1.57*'012340302014'[i]);x.fill('evenodd')

Oct 25 2018 5:41 PM UTC

}// 140/140

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#d3 #perspective #octahedron

u/pavel

Here is 12 characters saved: c.width^=i=0;for(y of'111121012110')x.lineTo(960+y%2*S(p=t+1.6*'123403020140'[i++])*(z=1600/(3+y%2*C(p))),y*z);x.fill('evenodd')
u/f8f82804

using 1.6 to approximate Math.PI/2 is OK, but you should change all the 4s to 0s in the second data string, just to minimze the aggregate angular error. I used 4 for clarity, because 1.57 is accurate enough that the error is hard to notice. With 1.6 you can notice that one edge is longer than the others. c.width^=i=0;for(y of'111121012110')x.lineTo(960+y%2*S(p=t+1.6*'123003020100'[i++])*(z=1600/(3+y%2*C(p))),y*z);x.fill('evenodd')


      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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