Engine prototype #13
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These coordinates are of form $I=(c, r, x, y, z)$ where we think of $c$ as the co-radius, $r$ as the radius, and $x, y, z$ as the "Euclidean" part, which we abbreviate $E_I$. There is an underlying basic quadratic form $Q(I_1,I_2) = (c_1r_2+c_2r_1)/2 - x_1x_2 -y_1y_2-z_1z_2$ which aids in calculation/verification of coordinates in this representation. We have:
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These coordinates are of form $I=(c, r, x, y, z)$ where we think of $c$ as the co-radius, $r$ as the radius, and $x, y, z$ as the "Euclidean" part, which we abbreviate $E_I$. There is an underlying basic quadratic form $Q(I_1,I_2) = (c_1r_2+c_2r_1)/2 - x_1x_2 -y_1y_2-z_1z_2$ which aids in calculation/verification of coordinates in this representation. We have:
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| Entity or Relationship | Representation | Comments/questions |
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| Entity or Relationship | Representation | Comments/questions |
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| Sphere s with radius r>0 centered on P = (x,y,z) | $I_s = (1/c, 1/r, x/r, y/r, z/r)$ satisfying $Q(I_s,I_s) = -1$, i.e., $c = r/(\|P\|^2 - r^2)$. | Can also write $I_s = (\|P\|^2/r - r, 1/r, x/r. y/r, z/r)$—so there is no trouble if $\|E_{I_s}\| = r$, just get first coordinate to be 0. |
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| Sphere s with radius r>0 centered on P = (x,y,z) | $I_s = (1/c, 1/r, x/r, y/r, z/r)$ satisfying $Q(I_s,I_s) = -1$, i.e., $c = r/(\|P\|^2 - r^2)$. | Can also write $I_s = (\|P\|^2/r - r, 1/r, x/r, y/r, z/r)$ -- so there is no trouble if $\|P\| = r$; we just get first coordinate to be 0. |
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| Plane p with unit normal (x,y,z) through the point s(x,y,z) | $I_p = (-2s, 0, -x, -y, -z)$ | Note $Q(I_p, I_p)$ is still −1. |
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| Plane p with unit normal $(x,y,z)$ through the (Euclidean) point $(sx,sy,sz)$ | $I_p = (-2s, 0, -x, -y, -z)$ | Note $Q(I_p, I_p)$ is still −1. This plane is at distance $s$ from the origin. |
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| 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 $(x/\||P\||, y/\||P\||, z/\||P\||)$ . |
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| 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 $(\| |
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| ∞, 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. |
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| ∞, 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. |
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| P lies on sphere or plane given by I | $Q(I_P, I) = 0$ | |
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| P lies on sphere or plane given by I | $Q(I_P, I) = 0$ | |
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| 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. |
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| 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. |
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