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If we take the convex hull of the 3D coordinates of the stars in Ursa Major and Ursa Minor: which hull has larger volume?

This is meant semi-humorous but we might also learn some interesting points from good answers, e.g. which metric to use when mass distorts space or how to measure with respect to frames of reference.

Update: to those asking about which stars to consider, just choose e.g. the canonical 7 stars of each. But this question is less about one arbitrary number and more about which things to consider, e.g. to show beginners that a bright star is not always close (and the other way around), and that size can be different in 3d than our arbitrary 2d-perspective.

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You'll have to decide which stars to consider. Each constellation covers a defined region of the sky and extends all the way to the edge of the observable universe. Assuming you define a reasonable set of stars for each constellation (the ones forming the outline of the bear, or the ones that have been assigned names of the form Greek letter Ursa (Majoris|Minoris)), I seriously doubt that relativistic distortion of space is going to be significant. –  Keith Thompson Jun 9 '14 at 20:08

3 Answers 3

up vote 6 down vote accepted

To calculate that you will have to build a 3D model based upon the distance of each of their stars and then calculate the resulting 3D object's volume. I know this may not answer the question, but I give the way to answering it. It's not an easy job.

Making a quick guess based upon the distances of the stars of both constellations in my opinion I guess the Small Dipper has a bigger volume since it's starts have bigger distance differences ranging from < 100 ly up to > 500 ly. In the case of the Big Dipper all distance are very similar since all stars come from the same open cluster. Their distance range between 78 and 124 ly.


Useful resources:

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thanks, this is a good start –  Bernd Elkemann Jun 9 '14 at 7:43

This obviously depends on which stars you choose, but in general Ursa Major is likely to win by a factor of about five(ish), since it occupies a much larger chunk of the sky (1280 square degrees) than Ursa Minor (256 square degrees). At a fixed magnitude, and therefore at a fixed visibility radius for each intrinsic luminosity, the volume will be (very roughly) proportional to the area of each constellation on the celestial sphere. (There are subtleties, such as these, and this only rigorously holds when there are many stars, but it is a good rule of thumb.)

The volumes themselves can quite easily be calculated in Mathematica using the curated data. There is one easy way, which selects what Wolfram Research thinks are the 'bright stars' in each constellation. If you do that, you get

$$\mathrm{Vol}(\mathrm{UMa})=9.20133\times 10^{20}\mathrm{AU}^3, \ \mathrm{Vol}(\mathrm{UMi})=1.7105\times 10^{20}\mathrm{AU}^3$$

which is the volume of the convex hull of the stars shown here:

Mathematica graphics

If you have Mathematica (v10+ only, I think) and you want to tinker with the code, here it goes:

     ConstellationData[Entity["Constellation", #], 
      EntityProperty["Constellation", "BrightStars"]], 
] & /@ {"UrsaMajor", "UrsaMinor"}

What one should really do is set a threshold magnitude and only count stars brighter than this, and then see the dependence of the volumes on the magnitude threshold. If you do this, you get pretty similar results:

Mathematica graphics

That is, Ursa Major has a consistently higher volume at all naked-eye magnitude thresholds.

For the curious, here's the code.

nakedEyeStarProperties = 
 StarData[#, {"Name", "Constellation", "ApparentMagnitude", 
     "HelioCoordinates", "RightAscension", "Declination"}
] & /@ StarData[EntityClass["Star", "NakedEyeStar"]]
constellationVolume[constellation_, magnitude_] := 
  selectedStars = 
    And[#[[2, 2]] == constellation, #[[3]] < magnitude] &];
  RegionMeasure[ConvexHullMesh[selectedStars[[All, 4]]]]
   {{m, constellationVolume["UrsaMajor", m]}, {m, 
     constellationVolume["UrsaMinor", m]}}
   , {m, 3, 6.5, 0.1}]\[Transpose]
 , AxesLabel -> {"magnitude threshold", 
   "constellation volume/\!\(\*SuperscriptBox[\(AU\), \(3\)]\)"}
 , PlotLegends -> {"Ursa Major", "Ursa Minor"}
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good answer thanks. will wait for some more people's feedback and then consider changing accepted-answer. –  Bernd Elkemann May 16 at 6:37

It's hard to say without some research, but I don't think the answer really matters much. Both constellations are chance alignments that have little or nothing to do with how close the stars really are to each other. It's not hard to envision a constellation composed of a few nearby stars and a few galaxies millions of light years away.

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Does not answer the question, only repeats it. –  Bernd Elkemann Jun 9 '14 at 5:58
@eznme "I don't think the answer matters much", followed by why the answer doesn't matter much. It's a silly question, with no real relationship to the constellation (which is an artificial construct anyway), and the answer has no general application to any other constellations, and no connection to anything beyond the angular separation of these unrelated objects. –  Marc Jun 9 '14 at 15:04
Then by all means downvote the question, or post a comment, or flag it if you think it doesn't belong here, or just ignore it. Answers are supposed to be answers. –  Keith Thompson Jun 9 '14 at 20:10
"This question has no meaningful answer" is an answer to a question that has no meaningful answer. –  Marc Jun 10 '14 at 0:04
Any answer to the question "how large is that constellation" is meaningless. Any such volume isn't defined. Would you mean the smallest sphere that would contain every star or galaxy that makes up the constellation or the polyhedron the stars and galaxies as vertices? When you've defined what you mean by the size, it tells you nothing. So one constellation has a larger volume than another because one of it's constituent "stars" is really a galaxy half a billion light years away. Constellations are human constructs, with no more intrinsic reality than hobbits. –  Marc Sep 18 '14 at 1:59

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