[math-fun] "Kepler's Law" for Somsky Planets
"Kepler's Law" for Somsky Planets =========Warren D Smith===Aug 2015===== Probably everybody knew this, but it is worth saying explicitly. If you have a sun of radius S, and antisun (outer ring) of radius R, and their two centers are separated by some fixed distance D, then the locations of all the planet centers lie on an ELLIPSE with foci being the sun and antisun centers. PROOF: For a planet of radius P, the distance from a planet center to sun center is S+P, the distance from planet center to ring center is R-P, hence this distance sum is S+P+R-P=S+R=constant independent of P. That law describes the ellipse. QED -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
<< antisun (outer ring) >> --- Kuiper belt, perhaps? WFL On 8/6/15, Warren D Smith <warren.wds@gmail.com> wrote:
"Kepler's Law" for Somsky Planets =========Warren D Smith===Aug 2015=====
Probably everybody knew this, but it is worth saying explicitly. If you have a sun of radius S, and antisun (outer ring) of radius R, and their two centers are separated by some fixed distance D, then the locations of all the planet centers lie on an ELLIPSE with foci being the sun and antisun centers.
PROOF: For a planet of radius P, the distance from a planet center to sun center is S+P, the distance from planet center to ring center is R-P, hence this distance sum is S+P+R-P=S+R=constant independent of P. That law describes the ellipse. QED
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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It seems also possible to likewise explore gear thingies on the sphere S^2 and the hyperbolic plane H^2, their being the other unique simply connected surfaces of constant curvature besides the plane (up to scaling).* Via some stereographic projection** or its inverse, any finite configuration of nonoverlapping but possibly-tangent circular disks on the plane R^2 or sphere S^2 corresponds to nonoverlapping disks on the other one, having the same tangencies. The radii will almost always change, as well as their ratios. QUESTION: If the disks' radii in a planar configuration are all rationally related, can the projection point be chosen so the same is true of the corresponding gears on the sphere? (I'm fairly certain the answer is Yes: Choose a rotation matrix having only rational entries to rotate the chosen projection point to the North pole, then use the usual formula for stereographic projection: (x,y,z) on S^2 go to (X,Y) = (x/(1-z), y/(1-z)) —Dan __________________________________________________________________________________________________ * In a small enough region of either of S^2 or H^2, the surface approximates the plane well enough that any realistic bunch of gears will behave as close as desired — but that won't be exact, so let's ignore this case. ** Stereographic projection and its inverse take circles to circles. The exact projection depends on a choice of the projection point among uncountably many possibilities. If this point is inside one of the gears on the sphere, then the corresponding gear in the plane will have its teeth pointed inward — and vice versa. But the projection point should not be chosen *on* any circle = disk boundary in S^2, or else the corresponding "circle" in the plane will be a straight line.
On Aug 6, 2015, at 9:55 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
<< antisun (outer ring) >> --- Kuiper belt, perhaps? WFL
Mesh conditions aren't going to be preserved under projection, however, and that's the hard part. On 08/06/15 10:58, Dan Asimov wrote:
It seems also possible to likewise explore gear thingies on the sphere S^2 and the hyperbolic plane H^2, their being the other unique simply connected surfaces of constant curvature besides the plane (up to scaling).*
Via some stereographic projection** or its inverse, any finite configuration of nonoverlapping but possibly-tangent circular disks on the plane R^2 or sphere S^2 corresponds to nonoverlapping disks on the other one, having the same tangencies.
The radii will almost always change, as well as their ratios.
QUESTION: If the disks' radii in a planar configuration are all rationally related, can the projection point be chosen so the same is true of the corresponding gears on the sphere?
(I'm fairly certain the answer is Yes: Choose a rotation matrix having only rational entries to rotate the chosen projection point to the North pole, then use the usual formula for stereographic projection:
(x,y,z) on S^2 go to (X,Y) = (x/(1-z), y/(1-z))
—Dan __________________________________________________________________________________________________ * In a small enough region of either of S^2 or H^2, the surface approximates the plane well enough that any realistic bunch of gears will behave as close as desired — but that won't be exact, so let's ignore this case.
** Stereographic projection and its inverse take circles to circles. The exact projection depends on a choice of the projection point among uncountably many possibilities. If this point is inside one of the gears on the sphere, then the corresponding gear in the plane will have its teeth pointed inward — and vice versa.
But the projection point should not be chosen *on* any circle = disk boundary in S^2, or else the corresponding "circle" in the plane will be a straight line.
On Aug 6, 2015, at 9:55 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
<< antisun (outer ring) >> --- Kuiper belt, perhaps? WFL
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No, but judicious choice of projection point should be able to result in rationally related radii. Isn't the condition of meshability basically the question of RRR ? —Dan
On Aug 6, 2015, at 11:31 AM, William R Somsky <wrsomsky@gmail.com> wrote:
Mesh conditions aren't going to be preserved under projection, however, and that's the hard part.
On 08/06/15 10:58, Dan Asimov wrote:
It seems also possible to likewise explore gear thingies on the sphere S^2 and the hyperbolic plane H^2, their being the other unique simply connected surfaces of constant curvature besides the plane (up to scaling).*
Via some stereographic projection** or its inverse, any finite configuration of nonoverlapping but possibly-tangent circular disks on the plane R^2 or sphere S^2 corresponds to nonoverlapping disks on the other one, having the same tangencies.
The radii will almost always change, as well as their ratios.
QUESTION: If the disks' radii in a planar configuration are all rationally related, can the projection point be chosen so the same is true of the corresponding gears on the sphere?
(I'm fairly certain the answer is Yes: Choose a rotation matrix having only rational entries to rotate the chosen projection point to the North pole, then use the usual formula for stereographic projection:
(x,y,z) on S^2 go to (X,Y) = (x/(1-z), y/(1-z))
—Dan __________________________________________________________________________________________________ * In a small enough region of either of S^2 or H^2, the surface approximates the plane well enough that any realistic bunch of gears will behave as close as desired — but that won't be exact, so let's ignore this case.
** Stereographic projection and its inverse take circles to circles. The exact projection depends on a choice of the projection point among uncountably many possibilities. If this point is inside one of the gears on the sphere, then the corresponding gear in the plane will have its teeth pointed inward — and vice versa.
But the projection point should not be chosen *on* any circle = disk boundary in S^2, or else the corresponding "circle" in the plane will be a straight line.
On Aug 6, 2015, at 9:55 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
<< antisun (outer ring) >> --- Kuiper belt, perhaps? WFL
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participants (5)
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Dan Asimov -
Dan Asimov -
Fred Lunnon -
Warren D Smith -
William R Somsky