I'm sure I'm just doing something dumb, but:

  • Yildun has a declination of approximately 86.5. This puts it 3.5 degrees from the celestial north pole.

  • Per the snapshot below, Yildun's azimuth can be as high as 4.17 degrees.

  • Since the celestial north pole always has an azimuth of 0, wouldn't this put Yildun 4.17 degrees from the pole, contradicting the distance calculated from declination?

What am I missing?

enter image description here

EDIT: I thought I'd figured this out:

  • On Earth, one degree of latitude is not the same as one degree of longitude (except at the equator).

  • Analogously, in the sky, one declination degree is not the same as one right ascension degree.

I thought the situation for right ascension applied to azimuth, but it doesn't.

Both azimuth and declination are measured on "great circles" (declination is measured on a half circle, but same general idea), so degrees in azimuth should equal degrees in declination.

  • $\begingroup$ What is the latitude of the observer? $\endgroup$ – asawyer Dec 10 '13 at 19:21
  • $\begingroup$ Oh, this is stellarium's default for Albuquerque, NM, which is about 35N and -106.5W. However, I think this question is latitude independent. $\endgroup$ – user21 Dec 10 '13 at 20:26

It turns out I was right when I thought I was wrong.

Consider the azimuth circle for 85 degrees elevation. It's very small. For example, 85 degrees elevation and 0 azimuth is only 10 degrees away from 85 degrees elevation and 180 azimuth.

Therefore, the azimuth circle for a given elevation is smaller than a great circle. It's only cosine(elevation) the size of a great circle.

At 35 degrees, the azimuth circle is cosine(35 degrees) = 0.819152 the size of a great circle, so 4.17 degrees in azimuth = 4.17 degrees * 0.819152 = right around 3.5 degrees on a great circle.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy