"bgimeno" <bgimeno[at]persistencia[dot]org> wrote:
> // ------------------------------------------------------
> #while (Num_Tree<5000)
>   #local rnd_tree = seed (Num_Tree);
>   object {Tree
>           translate x*60       // code A
>           rotate <rand(rnd_tree),rand(rnd_tree),rand(rnd_tree)>*360
>           }
>   #local Num_Tree = Num_Tree +1 ;
> #end
> // ------------------------------------------------------
> #while (Num_Tree<5000)
>   #local rnd_tree = seed (Num_Tree);
>   object {Tree
>           translate y*60       // code B
>           rotate <rand(rnd_tree),rand(rnd_tree),rand(rnd_tree)>*360
>           }
>   #local Num_Tree = Num_Tree +1 ;
> #end
> // ------------------------------------------------------

That isn't really surprising.  Keep in mind that rotating by <rand,rand,rand>
doesn't necessarily do what you think it does.  It rotates by all three axes in
sequence, x, then y, then z, so if you start with a point ON the x-axis, at
least one rotation won't do anything.  And there's certainly no guarantee
they'll be uniformly distributed.  In other words, it doesn't mean the code
rotates around a random axis by a random angle, in fact it rotates around three
well-defined axes by random angles.

If you want random points on a sphere, I suggest one of two methods.

1) Use a random point in a box, call it p, and if vlength(p<1) it's inside a
sphere so you can use vnormalize(p) as your random point on the surface.  This
is inefficient, but looks something like:

#declare p = <rand(),rand(),rand()>*2-1;
#while( vlength(p)>1 )
     #declare p = <rand(),rand(),rand()>*2-1;
#end
#declare p = vnormalize(p);

2) The document I posted a link to has a better answer on p. 19:

r1 = rand()
r2 = rand()
x = 2 cos (2*pi*r1) * sqrt( r2*(1-r2) )
y = 2 sin (2*pi*r1) * sqrt( r2*(1-r2) )
z =  1-2*r2

This is just spherical coordinates, so I think you can rephrase it (according to
the document) as

vrotate( vrotate( y, 180/pi * acos( rand() ) * x ), 360*rand() * y )

I haven't tested this, but I think it would work.

 - Ricky

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