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Here you can Sed posuere consectetur est at lobortis. Donec ullamcorper nulla non metus auctor fringilla. Maecenas sed diam eget risus varius blandit sit amet non magna. Donec id elit non mi porta gravida at eget metus. Praesent commodo cursus magna, vel scelerisque nisl consectetur et.

` ````
tt = 0
Z animate, ->
bg "#ccc";
# animated paper
fill "white"; noStroke();
rx = 0.3; ry = 0.3; tt += 0.1
push()
translate 0.5, 0.5; scale 1, 0.5; rotate QUART_PI
shadow color: '#000', alpha: 0.1, blur: 50, x: 0.001, y: 0.001
para 0, 0,
range: TWO_PI
f: (t) ->
s = 0.2 * cos(t * 6)
# http://math.stackexchange.com/questions/69099/equation-of-a-rectangle
sx = rx * (abs(cos t) * cos(t) + abs(sin t) * sin(t))
sy = ry * (abs(cos t) * cos(t) - abs(sin t) * sin(t))
cx = rx * cos(t * 3 + tt)
cy = ry * sin(t * 3 + tt)
x: cx * (0.2 * s) + sx
y: cy * (0.2 * s) + sy
pop()
# embuke
noFill(); stroke "#000"; strokeWidth 2
a = 0.1; a2 = a * 2
x = 0.15; y = 0.13
para x, y,
range: TWO_PI * 2
f: (t)->
x: a * cos(t) ** 3
y: a * sin(t) ** 3
l = (f) ->
para x, y,
range: [ -a, a2 ], step: a2, f: f, close: no
l (t) -> x: 0, y: t
l (t) -> x: t, y: 0
# lu
r = 0.15;
para 0.15, 0.45,
range: TWO_PI
f: (t)->
x: r * cos(t * 3) * (0.1)
y: r * sin t
close: yes
# at
strokeWidth 1
x = 0.15; y = 0.75
group ->
translate x, y;
rotate tt / 16
range 0, TWO_PI, TWO_PI / 50, (i)->
push()
rotate i;
para 0, 0, f: (t)-> x: t * 0.1, y: 0.01 * sin t * PI
pop()
circle 0, 0, 0.099
#pa
spiked = (r) ->
polar
step: 0.05
radius: ()->
0.2 + r * rand()
push()
strokeWidth 4
translate 0.64, 0.625
scale 0.31
circle(0.5, 0.5, 0.48);
spiked(0.25);
spiked(0.1);
fill('#000', 0.8);
circle(0.5, 0.5, 0.17);
fill('#ffffff', 0.08);
fill('#ffffff', 0.2);
circle(0.62, 0.36, .2);
circle(0.8, 0.5, .05);
pop();
```

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