Mathematics desk
< April 6	<< Mar \| April \| May >>	April 8 >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

April 7

Crowding the sphere

How many points can we put on some sphere so that the distance between any two of these points exceeds the sphere radius? Alternatively I'll take distance along the sphere arc ("as the crow flies") if that's easier to calculate. I'm guessing around 8, assuming a cube shape distance between points is 1 while the distance to the center is half the space diagonal or ${\sqrt {\frac {3}{4}}}$ (by measuring spatial distance). Is it possible to do better than that? 93.136.165.205 (talk) 15:09, 7 April 2019 (UTC)[reply]

I think this is an instance of Tammes problem. (The article is a stub but it has links.) Unfortunately the sites I found on-line but not behind a paywall are rather cryptic, so I'm having trouble finding a specific answer. The general problem is an area of current research, but apparently it has been solved completely when the number of spheres is small. --RDBury (talk) 17:56, 7 April 2019 (UTC)[reply]

I had to use Wayback Machine, but [1] has a of table best-known (at the time) values for various numbers of spheres n and dimensions d. The d you want is 3, and the angle you want is 60°. From the table the best value of n is then 12, in which I assume the points are arranged like the vertices of an icosohedron. See also Kissing number problem for dimension 3. --RDBury (talk) 19:00, 7 April 2019 (UTC)[reply]

Hmm that's quite a bit more than I expected. ~~What about arranging them so that the mean of distances between each point and its respective nearest neighbor exceeds the radius? 93.136.95.179 (talk) 20:32, 7 April 2019 (UTC)~~ Scratch that I thought of some obvious trivial solutions.[reply]

I was interested in this because I've noticed that there's quite a lot of stars within 10 ly from Earth, yet the nearest star system to us is over 4 ly away and the next one is almost 6 ly. I wondered if that's statistically unusual, but probably not if you can fit twelve of them in this way. 93.136.95.179 (talk) 20:37, 7 April 2019 (UTC)[reply]

We have List of nearest stars and brown dwarfs if you want hard data. I'm not a statistician or an astronomer but I don't think there's much doubt that the distribution of stars is 'clumpier' than you'd expect in a random distribution. For example Alpha Centauri, the closest system to the sun, is actually three stars, and it seems pretty unlikely that they would be so close to each other by pure chance. I'd be surprised if astronomers haven't studied this in more detail so you might want to try the Science desk for more info. --RDBury (talk) 02:24, 8 April 2019 (UTC)[reply]

I'm wrong again. @RDBury: As far as I understand the problem statement, OP wants the distance not less than the radius being measured over the ball's surface ('along the sphere arc'), not over the chord. If I get it right, the minimum angular separation required would equal 1 radian ≈ 57.3°, for which 13 points are possible with 57.1367031 deg separation (according to the page you found). --CiaPan (talk) 08:14, 8 April 2019 (UTC)[reply]

Not an icosahedron, a cuboctahedron (or a triangular orthobicupola). This is the face-centred cubic packing. — I keep a collection of relevant links. —Tamfang (talk) 05:58, 10 April 2019 (UTC)[reply]

According to Regular Polytopes the central angle of an icosahedron edge is arctan(2) > π/3. —Tamfang (talk) 16:51, 10 April 2019 (UTC)[reply]