I have a lot of points on the surface of the sphere. How can I calculate the area/spot of the sphere that has the largest point density? I need this to be done very fast. If
This is really just an inverse of this answer of mine
just invert the equations of equidistant sphere surface vertexes to surface cell index. Don't even try to visualize the cell different then circle or you go mad. But if someone actually do it then please post the result here (and let me now)
Now just create 2D cell map and do the density computation in O(N) (like histograms are done) similar to what Darren Engwirda propose in his answer
This is how the code looks like in C++
//---------------------------------------------------------------------------
const int na=16; // sphere slices
int nb[na]; // cells per slice
const int na2=na<<1;
int map[na][na2]; // surface cells
const double da=M_PI/double(na-1); // latitude angle step
double db[na]; // longitude angle step per slice
// sherical -> orthonormal
void abr2xyz(double &x,double &y,double &z,double a,double b,double R)
{
double r;
r=R*cos(a);
z=R*sin(a);
y=r*sin(b);
x=r*cos(b);
}
// sherical -> surface cell
void ab2ij(int &i,int &j,double a,double b)
{
i=double(((a+(0.5*M_PI))/da)+0.5);
if (i>=na) i=na-1;
if (i< 0) i=0;
j=double(( b /db[i])+0.5);
while (j< 0) j+=nb[i];
while (j>=nb[i]) j-=nb[i];
}
// sherical <- surface cell
void ij2ab(double &a,double &b,int i,int j)
{
if (i>=na) i=na-1;
if (i< 0) i=0;
a=-(0.5*M_PI)+(double(i)*da);
b= double(j)*db[i];
}
// init variables and clear map
void ij_init()
{
int i,j;
double a;
for (a=-0.5*M_PI,i=0;i1.0) c=1.0;
glColor3f(0.2,0.2,0.2); glCircle3D(x,y,z,x,y,z,0.45*da,0); // outline
glColor3f(0.1,0.1,c ); glCircle3D(x,y,z,x,y,z,0.45*da,1); // filled by bluish color the more dense the cell the more bright it is
}
}
//---------------------------------------------------------------------------
The result looks like this:

so now just see what is in the map[][] array you can find the global/local min/max of density or whatever you need... Just do not forget that the size is map[na][nb[i]] where i is the first index in array. The grid size is controlled by na constant and cm is just density to color scale ...
[edit1] got the Quad grid which is far more accurate representation of used mapping

this is with na=16 the worst rounding errors are on poles. If you want to be precise then you can weight density by cell surface size. For all non pole cells it is simple quad. For poles its triangle fan (regular polygon)
This is the grid draw code:
// draw cell quad grid (color is function of density)
int i,j,ii,jj;
double x,y,z,a,b,c,cm=1.0/10.0,mm=0.49,r=1.0;
double dx=mm*da,dy;
for (i=1;i1.0) c=1.0;
glColor3f(0.2,0.2,0.2);
glBegin(GL_LINE_LOOP);
abr2xyz(x,y,z,a-dx,b-dy,r); glVertex3d(x,y,z);
abr2xyz(x,y,z,a-dx,b+dy,r); glVertex3d(x,y,z);
abr2xyz(x,y,z,a+dx,b+dy,r); glVertex3d(x,y,z);
abr2xyz(x,y,z,a+dx,b-dy,r); glVertex3d(x,y,z);
glEnd();
glColor3f(0.1,0.1,c );
glBegin(GL_QUADS);
abr2xyz(x,y,z,a-dx,b-dy,r); glVertex3d(x,y,z);
abr2xyz(x,y,z,a-dx,b+dy,r); glVertex3d(x,y,z);
abr2xyz(x,y,z,a+dx,b+dy,r); glVertex3d(x,y,z);
abr2xyz(x,y,z,a+dx,b-dy,r); glVertex3d(x,y,z);
glEnd();
}
i=0; j=0; ii=i+1; dy=mm*db[ii];
ij2ab(a,b,i,j); c=map[i][j]; c=0.1+(c*cm); if (c>1.0) c=1.0;
glColor3f(0.2,0.2,0.2);
glBegin(GL_LINE_LOOP);
for (j=0;j1.0) c=1.0;
glColor3f(0.2,0.2,0.2);
glBegin(GL_LINE_LOOP);
for (j=0;j
the mm is the grid cell size mm=0.5 is full cell size , less creates a space between cells