notes/inversive.md

@@ -41,3 +41,23 @@ I will have to work out formulas for the Euclidean distance between two entities
 In this vein, it seems as though if J1 and J2 are the reps of two points, then Q(J1,J2) = d^2/2. So then the sphere centered at J1 through J2 is (J1-(2Q(J1,J2),0,0,0,0))/sqrt(2Q(J1,J2)). Ugh has a sqrt in it. Similarly for sphere centered at J3 through J2, (J3-(2Q(J3,J2),0000))/sqrt(2Q(J3,J2)). J1,J2,J3 are collinear if these spheres are tangent, i.e. if those vectors have Q-inner-product 1, which is to say Q(J1,J3) - Q(J1,J2) - Q(J3,J2) = 2sqrt(Q(J1,J2)Q(J2,J3)). But maybe that's not the simplest way of putting it. After all, we can just say that the cross-product of the two differences is 0; that has no square roots in it.
 
 One conceivable way to canonicalize lines is to use the *perpendicular* plane that goes through the origin, that's uniquely defined, and anyway just amounts to I = (0,0,d) where d is the ordinary direction vector of the line; and a point J in that plane that the line goes through, which just amounts to J=(r^2,1,E) with Q(I,J) = 0, i.e. E\dot d = 0. It's also the point on the line closest to the origin. The reason that we don't usually use that point as the companion to the direction vector is that the resulting set of six coordinates is not homogeneous. But here that's not an issue, since we have our standard point coordinates and plane coordinates; and for a plane through the origin, only two of the direction coordinates are really free, and then we have the one dot-product relation, so only two of the point coordinates are really free, giving us the correct dimensionality of 4 for the set of lines. So in some sense this says that we could take naively as coordinates for a line the projection of the unit direction vector to the xy plane and the projection of the line's closest point to the origin to the xy plane. That doesn't seem to have any weird gimbal locks or discontinuities or anything. And with these coordinates, you can test if the point E=x,y,z is on the line (dx,dy,cx,cy) by extending (dx,dy) to d via dz = sqrt(1-dx^2 - dy^2), extending (cx,cy) to c by determining cz via d\dot c = 0, and then checking if d\cross(E-c) = 0. And you can see if two lines are parallel just by checking if they have the same direction vector, and if not, you can see if they are coplanar by projecting both of their closest points perpendicularly onto the line in the direction of the cross product of their directions, and if the projections match they are coplanar.
+
+#### Engine Conventions
+
+The coordinate conventions used in the engine are different from the ones used in these notes. Marking the engine vectors and coordinates with $'$, we have
+$$I' = (x', y', z', b', c'),$$
+where
+$$\begin{align*}
+x' & = x & b' & = b/2 \\
+y' & = y & c' & = c/2. \\
+z' & = z
+\end{align*}$$
+The engine uses the quadratic form $Q' = -Q$, which is expressed in engine coordinates as
+$$Q'(I'_1, I'_2) = x'_1 x'_2 + y'_1 y'_2 + z'_1 z'_2 - 2(b'_1c'_2 + c'_1 b'_2).$$
+In the `engine` module, the matrix of $Q'$ is encoded in the lazy static variable `Q`.
+
+In the engine's coordinate conventions, a sphere with radius $r > 0$ centered on $P = (P_x, P_y, P_z)$ is represented by the vector
+$$I'_s = \left(\frac{P_x}{r}, \frac{P_y}{r}, \frac{P_z}{r}, \frac1{2r}, \frac{\|P\|^2 - r^2}{2r}\right),$$
+which has the normalization $Q'(I'_s, I'_s) = 1$. The point $P$ is represented by the vector
+$$I'_P = \left(P_x, P_y, P_z, \frac{1}{2}, \frac{\|P\|^2}{2}\right).$$
+In the `engine` module, these formulas are encoded in the `sphere` and `point` functions.