Inspired by the Paper.js "tadpoles" example. Built with D3.js. Feeling inspired? Fork this example and incorporate the Boids flocking algorithm.
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Spermatozoa
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<!DOCTYPE html> | |
<html> | |
<head> | |
<script type="text/javascript" src="http://mbostock.github.com/d3/d3.js?1.29.1"></script> | |
<style type="text/css"> | |
body { | |
background: #000; | |
} | |
ellipse { | |
fill: #fff; | |
} | |
path { | |
fill: none; | |
stroke: #fff; | |
stroke-linecap: round; | |
} | |
.mid { | |
stroke-width: 4px; | |
} | |
.tail { | |
stroke-width: 2px; | |
} | |
</style> | |
</head> | |
<body> | |
<script type="text/javascript"> | |
var w = 960, | |
h = 500, | |
n = 100, | |
m = 12, | |
degrees = 180 / Math.PI; | |
var spermatozoa = d3.range(n).map(function() { | |
var x = Math.random() * w, y = Math.random() * h; | |
return { | |
vx: Math.random() * 2 - 1, | |
vy: Math.random() * 2 - 1, | |
path: d3.range(m).map(function() { return [x, y]; }), | |
count: 0 | |
}; | |
}); | |
var svg = d3.select("body").append("svg:svg") | |
.attr("width", w) | |
.attr("height", h); | |
var g = svg.selectAll("g") | |
.data(spermatozoa) | |
.enter().append("svg:g"); | |
var head = g.append("svg:ellipse") | |
.attr("rx", 6.5) | |
.attr("ry", 4); | |
g.append("svg:path") | |
.map(function(d) { return d.path.slice(0, 3); }) | |
.attr("class", "mid"); | |
g.append("svg:path") | |
.map(function(d) { return d.path; }) | |
.attr("class", "tail"); | |
var tail = g.selectAll("path"); | |
d3.timer(function() { | |
for (var i = -1; ++i < n;) { | |
var spermatozoon = spermatozoa[i], | |
path = spermatozoon.path, | |
dx = spermatozoon.vx, | |
dy = spermatozoon.vy, | |
x = path[0][0] += dx, | |
y = path[0][1] += dy, | |
speed = Math.sqrt(dx * dx + dy * dy), | |
count = speed * 10, | |
k1 = -5 - speed / 3; | |
// Bounce off the walls. | |
if (x < 0 || x > w) spermatozoon.vx *= -1; | |
if (y < 0 || y > h) spermatozoon.vy *= -1; | |
// Swim! | |
for (var j = 0; ++j < m;) { | |
var vx = x - path[j][0], | |
vy = y - path[j][1], | |
k2 = Math.sin(((spermatozoon.count += count) + j * 3) / 300) / speed; | |
path[j][0] = (x += dx / speed * k1) - dy * k2; | |
path[j][1] = (y += dy / speed * k1) + dx * k2; | |
speed = Math.sqrt((dx = vx) * dx + (dy = vy) * dy); | |
} | |
} | |
head.attr("transform", function(d) { | |
return "translate(" + d.path[0] + ")rotate(" + Math.atan2(d.vy, d.vx) * degrees + ")"; | |
}); | |
tail.attr("d", function(d) { | |
return "M" + d.join("L"); | |
}); | |
}); | |
</script> | |
</body> | |
</html> |
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