## 2017: "Stargazing 2"

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### Re: 2017: "Stargazing 2"

Good thing we weren't the winning bidder on eBay for those Stinger missiles.

### Re: 2017: "Stargazing 2"

If you use 5000 for the number of naked-eye stars, and the triangle formula given here:

https://math.stackexchange.com/question ... n-a-circle

They say: n = N(N-1)(N-2) / 6

then the number of visible triangles is like 20,820,835,000. But that's the whole sky ... not what she'd see just walking out at any specific time.

And if you limit the triangles to just the really bright stars, say magnitude <+2.5, (maybe 46 at a time), then you get 15,180 triangles. So triangles are easy. She should be looking for squares or pentagons.

https://math.stackexchange.com/question ... n-a-circle

They say: n = N(N-1)(N-2) / 6

then the number of visible triangles is like 20,820,835,000. But that's the whole sky ... not what she'd see just walking out at any specific time.

And if you limit the triangles to just the really bright stars, say magnitude <+2.5, (maybe 46 at a time), then you get 15,180 triangles. So triangles are easy. She should be looking for squares or pentagons.

### Re: 2017: "Stargazing 2"

Heimhenge wrote:And if you limit the triangles to just the really bright stars, say magnitude <+2.5, (maybe 46 at a time), then you get 15,180 triangles. So triangles are easy. She should be looking for squares or pentagons.

Problem with that is: any 3 noncollinear points make a triangle. To make a square or higher-sided figure, they all gotta be in one plane. That makes things a lot more difficult.

Maybe just make triangle-based pyramids (4 stars needed)

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Former OTTer

Vote cellocgw for President 2020. #ScienceintheWhiteHouse http://cellocgw.wordpress.com

"The Planck length is 3.81779e-33 picas." -- keithl

" Earth weighs almost exactly π milliJupiters" -- what-if #146, note 7

### Re: 2017: "Stargazing 2"

cellocgw wrote:Heimhenge wrote:And if you limit the triangles to just the really bright stars, say magnitude <+2.5, (maybe 46 at a time), then you get 15,180 triangles. So triangles are easy. She should be looking for squares or pentagons.

Problem with that is: any 3 noncollinear points make a triangle. To make a square or higher-sided figure, they all gotta be in one plane. That makes things a lot more difficult.

Maybe just make triangle-based pyramids (4 stars needed)

real distance to stars never played a role in constellations.

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### Re: 2017: "Stargazing 2"

1) Take the (next) two nearest stars in the desired field of view/whole sphere, by angular distance

2) If both are noted as at one end of Sides whose other ends are upon the same third star, complete the Triangle ((and forget any other stray Sides associated to these points)), then skip back to (1) for the next go.

3) Skip back to (1) for next option if either(/both) already is in another Triangle.

4) Skip back to (1) for next option if either(/both) is already the end of a Side with a further Side (if covered by step (2), this is moot)

5) (MAYBE) Skip back to (1) for next option if either(/both) are

(( If not using (5), revise step (4) to permit mutli-branching searches and possibly long-chains but on the understanding that longer chains remain hypothetical. ))

6) (OPTIONAL) Skip back to (1) if any existing Side intersects the line between the current two stars (does not include those terminating at either!).

(( Depending on the geometries and other ruleset tuning, some cases of crossing may exist, like slightly irregular tetrahedral foursomes, but all examples that currently come to mind are contrived, e.g. one a super-low star count spread widely over a spherical field. Needs testing, maybe "Warning:" for, during development.))

7) Assign a speculative Side between those two points, then repeat for next option in (1).

With 3N stars, some form of mutually exclusive 'minima' of triangles that use up each and every stsr should be possible (but not necessarily non-overlapping). This may not be the most minimal set of triangles by particular counts, e.g. combined perimiter lengths or (angular) area contained within all triangles (summation) or any triangle (union), which is a difference especially useful obvious where an earlier (tightly packed) triangle lies within a later (wider) triangle. The summed area (large+small) is greater than the union area (the large alone). And both are larger in coverage than forming two non-intersecting/overlaying triangles from two of the outer vertices and one inner (less than ⅓rd the original Large area) and then again from the remaining two inner ones and one outer (an extremely pointy triangle with virtually no area).

Not that area (or minimalism thereof) might be the best measure. Looking for a metric of closest isoscelean perfection (regardless of size) and/or any pattern which deliberately avoids all inter-triangle edge intersections might be alternate criteria. And one could conceivably reweight the orderedness of potential side-pairings by combined (or similarity of!) magnitude to prioritise "obvious" triangles over two bright stars and a seeming afterthought of an edge-of-vision third partner.

That's assuming you don't actually want a geodesic sky. Connect, in order, all next closest pairs whose connection does not cross any prior connection for a full 'stardome' effect. Each star should connect to no less than three, and probably no more than~~six~~ five¹ neighbouring stars, with every possible >3-shape eventually bisected down to the best 3-shape available given the penultimate 4-shape. Although I can't guarantee that this 4-shape is necessarily the best through-route from shaving the 5-shape down, and so on.

(@cellogcw, to address your ninjaed point, I'm of course thinking of surface-warped shapes, as necessary above, rather than planar-limited ones. And, to address the tetrahedrality, the process can be extrapolated into higher dimensions if you're working with that, but I'd have to run through some exceptions to the planar assumptions regarding how edges and faces form surfaces (form hypersurfaces (form hyper**surfaces …)).)

¹ There are contrived exceptions (a single star centering a whole hemisphere otherwise empty, and an arbitrarily large number of stars in the antipode hemisphere to produce a 'rim' to spoke to across the other side of the gnomonic horizon), which I think ought to mostly be avoided in a sufficiently distributed set of randomised points; and I downgraded from the original six when I realised I was on a non-Euclidean plane with spherical aberration. But until I try it, I might yet be missing some of the finer intricracies.

2) If both are noted as at one end of Sides whose other ends are upon the same third star, complete the Triangle ((and forget any other stray Sides associated to these points)), then skip back to (1) for the next go.

3) Skip back to (1) for next option if either(/both) already is in another Triangle.

4) Skip back to (1) for next option if either(/both) is already the end of a Side with a further Side (if covered by step (2), this is moot)

5) (MAYBE) Skip back to (1) for next option if either(/both) are

(( If not using (5), revise step (4) to permit mutli-branching searches and possibly long-chains but on the understanding that longer chains remain hypothetical. ))

6) (OPTIONAL) Skip back to (1) if any existing Side intersects the line between the current two stars (does not include those terminating at either!).

(( Depending on the geometries and other ruleset tuning, some cases of crossing may exist, like slightly irregular tetrahedral foursomes, but all examples that currently come to mind are contrived, e.g. one a super-low star count spread widely over a spherical field. Needs testing, maybe "Warning:" for, during development.))

7) Assign a speculative Side between those two points, then repeat for next option in (1).

With 3N stars, some form of mutually exclusive 'minima' of triangles that use up each and every stsr should be possible (but not necessarily non-overlapping). This may not be the most minimal set of triangles by particular counts, e.g. combined perimiter lengths or (angular) area contained within all triangles (summation) or any triangle (union), which is a difference especially useful obvious where an earlier (tightly packed) triangle lies within a later (wider) triangle. The summed area (large+small) is greater than the union area (the large alone). And both are larger in coverage than forming two non-intersecting/overlaying triangles from two of the outer vertices and one inner (less than ⅓rd the original Large area) and then again from the remaining two inner ones and one outer (an extremely pointy triangle with virtually no area).

Not that area (or minimalism thereof) might be the best measure. Looking for a metric of closest isoscelean perfection (regardless of size) and/or any pattern which deliberately avoids all inter-triangle edge intersections might be alternate criteria. And one could conceivably reweight the orderedness of potential side-pairings by combined (or similarity of!) magnitude to prioritise "obvious" triangles over two bright stars and a seeming afterthought of an edge-of-vision third partner.

That's assuming you don't actually want a geodesic sky. Connect, in order, all next closest pairs whose connection does not cross any prior connection for a full 'stardome' effect. Each star should connect to no less than three, and probably no more than

(@cellogcw, to address your ninjaed point, I'm of course thinking of surface-warped shapes, as necessary above, rather than planar-limited ones. And, to address the tetrahedrality, the process can be extrapolated into higher dimensions if you're working with that, but I'd have to run through some exceptions to the planar assumptions regarding how edges and faces form surfaces (form hypersurfaces (form hyper**surfaces …)).)

¹ There are contrived exceptions (a single star centering a whole hemisphere otherwise empty, and an arbitrarily large number of stars in the antipode hemisphere to produce a 'rim' to spoke to across the other side of the gnomonic horizon), which I think ought to mostly be avoided in a sufficiently distributed set of randomised points; and I downgraded from the original six when I realised I was on a non-Euclidean plane with spherical aberration. But until I try it, I might yet be missing some of the finer intricracies.