Spherical projections SLJ 11/24

The idea is to have some data or curve or whatnot in 3d, do a flat projection
of that data onto a surface, and then project from 3d into 2d.  For example:
map a fractal onto a sphere, and then display it on the computer.

There is a basic problem with "holes", though -- when you map individual
(discrete) points to a sphere, some points on the sphere won't be covered.
Moreover, when you map from the sphere down to 2d, chances are that some
points will get skipped on the screen.  And so on.  One method is to use
a very fine grid in 3d, but this is obviously very inefficient.

The ideal situation would be to project only those points which need
projecting.  This leads naturally to the idea of starting with the
_projected point_ and working backwards to figure out where it lies on
the surface, and then figuring out which point in the original data set
corresponds to that point on the surface.

That is: instead of starting with the data, projecting it onto the surface,
and then projecting that surface from R3 to R2 to get a pixel on the
computer, we do exactly the opposite: start with the pixel on the computer,
trace it backwards onto the surface, and trace that backwards onto the 
original data set.  This way, we only use the points that we need.

Projecting from 3D into 2D amounts to drawing a line from the original
point (x0,y0,z0) through the origin (0,0,0) and figuring out where it
intersects a plane located at z=d (the plane is the computer screen):

	t*(x,y,z) = (x0,y0,d)

this tells us that t=d/z, which gives the classic projection equations:

		x0 = d * x/z,	y0 = d * y/z

i.e. divide by the z-coordinate and multiply by a magnification factor.

Now the idea is to do exactly the opposite: starting from a point (x0,y0,d),
we start drawing a line through the origin

		(x,y,z) = t*(x0,y0,d)			(line)

and figure out where that line intersects the surface.  The equation of
a general 3D surface is 

		f(x,y,z) = 0

For example, a sphere has the equation 

		f(x,y,z) = x^2 + y^2 + z^2 - r^2	(sphere)

The solution to the problem amounts to figuring out a value of t which
satisfies both equations.  Substituting from equation (line) above:

		f(x,y,z) = (t*x0)^2 + (t*y0)^2 + (d*t)^2 - r^2

We need f(x,y,z)=0 -- factoring out a t^2, the above simplifies to

		t^2 * (x0^2 + y0^2 + d^2) = r^2

		=> t = r/sqrt(x0^2 + y0^2 + d^2)

This is the value of t which gives the intersection point of the line
with the surface.  The x,y,z coordinates are given by equation (line) as:

		x = t*x0,  y = t*y0,  z = t*d

In summary: to project a fractal to a sphere (of radius r) to the screen, 
start with the screen coordinate (x0,y0,d) (d=mag factor), calcuate

		t = r/sqrt(x0^2 + y0^2 + d^2),
		x = t*x0,
		y = t*y0

and then do the fractal iteration on (x,y) (or they might be the coordinates
of some picture data, etc.).

So, problems to be solved:

	- Computing t is a real bitch!  The following might work:

	  o Write it as t = r/(d*sqrt(1 + (x0/d)^2 + (y0/d)^2))
	  o Construct a table of squares: f(x) = c*(x/d)^2 (one table
	    for each value of d needed)
	     * c = constant = scale factor to retain accuracy!
	  o Construct a table of g(x) = r/(d*sqrt(1+x/c))
	     * Probably need both integer and remainder

	  That way you could
		
		LDX X0
		LDY Y0
		CLC
		LDA TABLEF,X	;f(x0)
		ADC TABLEF,Y	;f(y0)
		TAX
		LDA TABLEG,X	;This is t

	  and then do a standard fast multiply on t and x0, etc.  Hopefully
	  the above method will be accurate enough to work.  Note that
	  d typically takes on values around 200, so x/d and y/d are usually
	  quite small numbers -- hence the need for the constant c above.

	- The sphere equation above is centered at the origin.  There are
	  actually two values of t, the + and - roots.  Using the other
	  root of t will change the sign of both x and y.

	- Moreover, a sphere centered off of the origin, or an arbitrary
	  surface, is much more difficult to handle.  I think by being
	  sneaky though that it can be handled without too much difficulty,
	  although it would involve a second set of projections.  Dunno.