Why is the measured distance uncertainty for M33 (835 ± 105 kiloparsec) six times bigger than the measured distance uncertainty for M31 (778 ± 17 kiloparsec)?

They are approximately at the same distance, are both spiral galaxies and in the same region of the sky (within 15 degrees).

  • $\begingroup$ perhaps it has to do with the position relative to milky way. For instance, sideways versus "on top". $\endgroup$
    – Jus12
    Aug 30, 2011 at 11:05
  • $\begingroup$ @Jus12: they are in the same region of the sky, on the order of 10 degrees apart. $\endgroup$ Aug 30, 2011 at 15:40
  • $\begingroup$ I think that has to do with size. andromeda galaxy measures 100,000 light years while the triangle only 60,000 $\endgroup$ Aug 31, 2011 at 13:04

1 Answer 1


I haven't trawled through the papers in detail, but I think the short answer is that in the Andromeda measurements, a variety of methods have converged, whereas in Triangulum, they don't agree as well. For example, if you look at standard methods like observing Cepheid variables or using the tip of the RGB, the errors are both quoted as 40 kpc or 25 kpc, respectively. But in Andromeda, the corresponding distances are 770 and 780 kpc, whereas in Triangulum they're 850 and 790 kpc.

Note that the Wikipedia page uses the eclipsing binary observation too. That gives the upper limit (940 kpc) so who knows if there's something about the stars' light that messes one about. Also, the Wiki page quotes a range instead of combining the averages properly. I'm sure a proper statistical combination of the measurements would give about 860±46 kpc. Still not as precise as for Andromeda, but better than the 105 kpc implied by the range.

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    $\begingroup$ To expand on (or maybe just distill) your answer, I think the question misinterprets and then correspondingly misrepresents the data presented in the Triangulum Wikipedia article. There, it's quoted as a range, not as a central value +- a precision, so it should not be interpreted as representing a valid experimental uncertainty, rather just a discrepancy between different measurements. $\endgroup$
    – Andrew
    Aug 30, 2011 at 11:38

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