Paleta de colores con su valor hex?

[tunait]

Bueno pues el script queda así

Código:

<script type="text/javascript">
<!-- Original:  D10n ([email protected]) -->
<!-- Web Site:  http://www.iinet.net.au/~biab -->

<!-- This script and many more are available free online at -->
<!-- The JavaScript Source!! http://javascript.internet.com -->

<!-- Begin
addary = new Array();           //red
addary[0] = new Array(0,1,0);   //red green
addary[1] = new Array(-1,0,0);  //green
addary[2] = new Array(0,0,1);   //green blue
addary[3] = new Array(0,-1,0);  //blue
addary[4] = new Array(1,0,0);   //red blue
addary[5] = new Array(0,0,-1);  //red
addary[6] = new Array(255,1,1);
clrary = new Array(360);
for(i = 0; i < 6; i++)
for(j = 0; j < 60; j++) {
clrary[60 * i + j] = new Array(3);
for(k = 0; k < 3; k++) {
clrary[60 * i + j][k] = addary[6][k];
addary[6][k] += (addary[i][k] * 4);
   }
}
var tunIex=navigator.appName=="Microsoft Internet Explorer"?true:false;
if(tunIex && navigator.userAgent.indexOf('Opera')>=0){tunIex = false}
function capture() {
var layobj = document.getElementById("wheel");
if(!tunIex) {
layobj.addEventListener("mousemove", moved, false);
}
else {
layobj.onmousemove = moved;
   }
}
function moved(e) {
y = 4 * ((!tunIex)?e.layerX:event.offsetX);
x = 4 * ((!tunIex)?e.layerY:event.offsetY);
sx = x - 512;
sy = y - 512;
qx = (sx < 0)?0:1;
qy = (sy < 0)?0:1;
q = 2 * qy + qx;
quad = new Array(-180,360,180,0);
xa = Math.abs(sx);
ya = Math.abs(sy);
d = ya * 45 / xa;
if(ya > xa) d = 90 - (xa * 45 / ya);
deg = Math.floor(Math.abs(quad[q] - d));
n = 0;
sx = Math.abs(x - 512);
sy = Math.abs(y - 512);
r = Math.sqrt((sx * sx) + (sy * sy));
if(x == 512 & y == 512) {
c = "000000";
}
else {
for(i = 0; i < 3; i++) {
r2 = clrary[deg][i] * r / 256;
if(r > 256) r2 += Math.floor(r - 256);
if(r2 > 255) r2 = 255;
n = 256 * n + Math.floor(r2);
}
c = n.toString(16);
while(c.length < 6) c = "0" + c;
}
document.f.t.value = "#" + c;
document.getElementById("wheel").style.backgroundColor = "#" + c;
return false;
}
//  End -->
</script>

Si lo quieren probar a ver si me dejé algo pendiente ...

Un saludo