This discussion is taken from an article I posted to synergetics-l 
on August 7, 1999.  It describes a correspondence between quadrays
and 4-dimensional Cartesian coordinates.  I was not the first to
notice this; see the posting "Re: coordinate systems" by John Conway 
for a more general discussion of this kind of correspondence.  My
quadray formulas page gives program code derived using these ideas.


Consider the plane defined by the 3-D Cartesian coordinate equation
x+y+z=0.  This plane passes through the origin, and is parallel to
the plane passing through the points (1,0,0), (0,1,0), and (0,0,1).

Any point (x,y,z) in 3-space can be projected onto the plane x+y+z=0
by subtracting (x+y+z)/3 from all three coordinates.  E.g., the point 
(1,4,4)'s projection onto the plane is (-2,1,1).  (A point's
projection onto a plane is the point on the plane closest to the
original point.)

Now, recall the plane geometry analogue to quadrays:  a 3-ray system 
whose basis vectors are separated from one another by 120 degrees.
Any two points (a,b,c) and (a+k,b+k,c+k) are considered equivalent.
Although we think of this system as a form of plane geometry, 3-ray 
coordinates can be interpreted directly as 3-D Cartesian coordinates.  
The 3-ray system is then seen a geometry of the plane x+y+z=0, where 
all points in 3-space that project to the same point on that plane 
are considered equivalent.

If this isn't clear yet, visualize the plane x+y+z=0.  Then project
the three points (1,0,0), (0,1,0), and (0,0,1) onto that plane.  
The projected points are equidistant from the origin, and are 
separated by 120 degrees.  They are the endpoints of basis vectors 
for a 3-ray geometry on the plane x+y+z=0.  They aren't one unit 
away from the origin any more, so we're dealing with a scaled 3-ray 
system.  But it works the way we expect a 3-ray system to work.

What good is this observation?  For one, it lets us derive 3-ray 
formulas for most any geometric function we want (distance, dot 
product, etc.) by using established methods for manipulating points 
in Cartesian 3-D space.  But it also suggests how to do analogous 
manipulations on quadray coordinates, and that's why I brought it up.

Quadray coordinates can be interpreted as Cartesian hyperspace 
coordinates, where all points that project to the 3-space w+x+y+z=0 
are considered equivalent.  (I called the 4th axis in hyperspace 'w'; 
I don't know if there's a traditional letter for it.)  With this in 
mind, one can easily derive quadray formulas for various operations.

Kirby's normal form for a quadray is the one with no negative 
coordinates and at least one zero coordinate.  For now, consider a
different normal form, specifically the one where the four coordinates 
sum to zero.  Any point (a,b,c,d) can be converted to this normal form 
by subtracting (a+b+c+d)/4 from each coordinate.  This is equivalent
to projecting a point in Cartesian hyperspace onto the 3-space defined 
by w+x+y+z=0.

To find distance from the origin using quadray coordinates, normalize 
the point's coordinates and use the standard Cartesian distance formula 
sqrt(a^2+b^2+c^2+d^2).  To scale so that basis vectors have unit length, 
multiply the result by sqrt(4/3).  The table below gives the unscaled 
result for various points, distances with unit basis vectors, and Kirby's 
distance as described at http://www.grunch.net/synergetics/quadintro.html.

                                     Cartesian   dist. w/unit    Kirby's
point       normalized point         distance    basis vectors   distance

(1,0,0,0)   ( 3/4,-1/4,-1/4,-1/4)    sqrt(3)/2   1               sqrt(6)/2
(0,1,0,1)   (-1/2, 1/2,-1/2, 1/2)    1           sqrt(12)/3      sqrt(2)
(1,0,1,2)   (   0,  -1,   0,   1)    sqrt(2)     sqrt(24)/3      2

This technique also gives a simpler formula for dot product than the 
one I'd described a few days ago.  Normalize the points, take the dot 
product the usual (a1*a2+b1*b2+c1*c2+d1*d2) way, and multiply by 4/3.  
That scaling factor is the square of the one used for distances, 
which makes sense because the geometric interpretation of the dot 
product includes the product of two lengths.

This isomorphism also helps in seeing why the determinant expression I 
gave for quadray cross product works.  I'll explain if anyone wants to see.

To summarize:  3-ray plane geometry coordinates can be interpreted as 
the 3-D Cartesian coordinates of points on the plane x+y+z=0.  Similarly, 
quadray coordinates can be interpreted as 4-D Cartesian coordinates of 
points in the 3-space w+x+y+z=0.  I went through the 3-ray example first 
because it can be visualized.  I can't visualize hyperspace, but one
doesn't have to in order to do analogous manipulations with quadrays.