On Jun 14, 2006, at 1:50 AM, didi bruckmayr wrote:

> > has anyone taken a look at the spherical harmonics at: > http://astronomy.swin.edu.au/~pbourke/surfaces/sphericalh/ > has anyone written an expression for this "simple formula"?

XYZ Eval(double theta,double phi, int *m) { double r = 0; XYZ p;

r += pow(sin(m[0]*phi),(double)m[1]); r += pow(cos(m[2]*phi),(double)m[3]); r += pow(sin(m[4]*theta),(double)m[5]); r += pow(cos(m[6]*theta),(double)m[7]);

p.x = r * sin(phi) * cos(theta); p.y = r * cos(phi); p.z = r * sin(phi) * sin(theta);

return(p); }

is probably easiest to accomplish with two cascaded expressions. One to compute radius, r, and then passed to a second one to convert to xyz based on r, theta, and phi.

In the above, replace m[N] with in[N] (to create adjustable parameters); replace theta with norm[0]*TWOPI, and phi with norm[1] *PI (to convert our normalized X and Y axis values to normalized radian based coordinates across the surface of the sphere), and that should be it.

So the expression for r would be:

"pow(in[0]*PI*norm[1],in[1]) + pow(cos(in[2]*PI*norm[1]),in[3]) + pow(sin(in[4]*TWOPI*norm[0]),in[5]) + pow(cos(in[6]*TWOPI*norm[0]), [7])"

then your output expressions for xyz would be the following (where theta and phi are replaced by the same, and r is replaced by in[0]). You could do with a single instance of jit.expr by replacing r with the first expression above, but it will be more expensive to compute for each x, y, and z rather than once for all, as I've suggested.

"in[0]*sin(PI*norm[1])*cos(TWOPI*norm[0])" "in[0]*cos(PI*norm[1])" "in[0]*sin(PI*norm[1])*sin(TWOPI*norm[0])"

Quick hack example below. There's some stupid problem with updating the float values and jit.expr, which requires banging the input again, which I'll look into (might be some bug for jit.expr float input), but this should give a decent example of how to convert such stuff into jit.expr expressions.

-Joshua