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Engine prototype #13

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glen merged 133 commits from engine-proto into main 2024-10-21 03:18:48 +00:00
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notes/inversive.md
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@ -11,7 +11,7 @@ These coordinates are of form $I=(c, r, x, y, z)$ where we think of $c$ as the c
| Point P with Euclidean coordinates (x,y,z) | $I_P = (\|P\|^2, 1, x, y, z)$ | Note $Q(I_P,I_P) = 0$.  Because of this we might choose  some other scaling of the inversive coordinates, say $(\||P\||,1/\||P\||,x/\||P\||,y/\||P\||,z/\||P\||)$ instead, but that fails at the origin, and likely won't have some of the other nice properties listed below.  Note that scaling just the co-radius by $s$ and the radius by $1/s$ (which still preserves $Q=0$) dilates by a factor of $s$ about the origin, so that $(\|P\|, \|P\|, x, y, z)$, which might look symmetric, would actually have to represent the Euclidean point $(x/\||P\||, y/\||P\||, z/\||P\||)$ . | | Point P with Euclidean coordinates (x,y,z) | $I_P = (\|P\|^2, 1, x, y, z)$ | Note $Q(I_P,I_P) = 0$.  Because of this we might choose  some other scaling of the inversive coordinates, say $(\||P\||,1/\||P\||,x/\||P\||,y/\||P\||,z/\||P\||)$ instead, but that fails at the origin, and likely won't have some of the other nice properties listed below.  Note that scaling just the co-radius by $s$ and the radius by $1/s$ (which still preserves $Q=0$) dilates by a factor of $s$ about the origin, so that $(\|P\|, \|P\|, x, y, z)$, which might look symmetric, would actually have to represent the Euclidean point $(x/\||P\||, y/\||P\||, z/\||P\||)$ . |
| ∞, the "point at infinity" | $I_\infty = (1,0,0,0,0)$ | The only solution to $Q(I,I) = 0$ not covered by the above case. | | ∞, the "point at infinity" | $I_\infty = (1,0,0,0,0)$ | The only solution to $Q(I,I) = 0$ not covered by the above case. |
| P lies on sphere or plane given by I | $Q(I_P, I) = 0$ | | | P lies on sphere or plane given by I | $Q(I_P, I) = 0$ | |
| Sphere/planes represented by I and J are tangent | $Q(I,J) = 1$ (??, see note at right) | Seems as though this must be $Q(I,J) = \pm1$  ? For example, the $xy$ plane represented by (0,0,0,0,1)  is tangent to the unit circle centered at (0,0,1) rep'd by (0,1,0,0,1), but their Q-product is -1. And in general you can reflect any sphere tangent to any plane through the plane and it should flip the sign of $Q(I,J)$, if I am not mistaken. | | Sphere/planes represented by I and J are tangent | If $I$ and $J$ have the same orientation where they touch, $Q(I,J) = -1$. If they have opposing orientations, $Q(I,J) = 1$. | For example, the $xy$ plane with normal $-e_z$, represented by $(0,0,0,0,1)$, is tangent with matching orientation to the unit sphere centered at $(0,0,1)$ with outward normals, represented by $(0,1,0,0,1)$. Accordingly, their $Q$-product is −1. |
| Sphere/planes represented by I and J intersect (respectively, don't intersect) | $\|Q(I,J)\| < (\text{resp. }>)\; 1$ | Follows from the angle formula, at least conceptually. | | Sphere/planes represented by I and J intersect (respectively, don't intersect) | $\|Q(I,J)\| < (\text{resp. }>)\; 1$ | Follows from the angle formula, at least conceptually. |
| P is center of sphere represented by I | Well, $Q(I_P, I)$ comes out to be $(\|P\|^2/r - r + \|P\|^2/r)/2 - \|P\|^2/r$ or just $-r/2$ . | Is it if and only if ?   No this probably doesn't work because center is not conformal quantity. | | P is center of sphere represented by I | Well, $Q(I_P, I)$ comes out to be $(\|P\|^2/r - r + \|P\|^2/r)/2 - \|P\|^2/r$ or just $-r/2$ . | Is it if and only if ?   No this probably doesn't work because center is not conformal quantity. |
| Distance between P and R is d | $Q(I_P, I_R) = d^2/2$ | | | Distance between P and R is d | $Q(I_P, I_R) = d^2/2$ | |