On Sat, 12 Mar 2005 03:15:42 GMT, "
wrote:
But that formula for z does not appear to be correct!

Hi Art,

As I said, you are not working in Cartesians in the first place with
"polygons." It is a formula for computer generation of the surface.

If you absolutely need to understand the formula, read through the
code that is used to construct the distribution curve:
#define sz 2000000
#define bc 19
static double fr() {return rand()/(double)~(131);}
static double sq(double x){return x*x;}
int hi[bc]; int out=0;
int main(){
{int j=bc; while(j--) hi[j]=0;}
{int j=sz; while(j--) {
double x = fr()*1.4-.7, y=fr()*1.4-.7, z=fr();
if(sq(x*x+y*y+z*z) z) {
++hi[(int)(z/sqrt(x*x+y*y+z*z)*bc)];}
else ++out;}}
{int j; for(j=0; jbc; ++j) printf("%d %d %8.2f\n",
j, hi[j], (sz-out)*(j+.5)/(bc*bc/2));}
printf ("%d out of box.\n", out);
}

If you still don't understand, the polar coordinate formulas are just
as useful, simpler, and take very little work to construct a "polygon"
that obtains complete closure. It is, after all, a construction much
like any of a number of classic curves. You need only conform to the
requirements of a Lambertian surface or emitter to obtain the curve
you describe.

73's
Richard Clark, KB7QHC