The altitude of a star as it transits your meridian is found to be 45o along a vertical circle at azimuth 180o, the south point. Find the declination of the star.
Solution:
From the celestial pole geometry (e.g. in Fig. 3 of post from Aug. 12):
90o
= z + a max
Where a max is the altitude at meridian transit (hence a maximum)
But if: z = a max = 45o
Then: 90o - z = 90o - a max = f (latitude)
For which your latitude is inserted.
But d (decl.) = z + φ
For example, take Barbados' latitude of 13 deg north.
Then:
d (decl.) = z + f = (- 45o) + 13o = - 32o
Since the zenith distance z plus altitude (a max) must equal 90 degrees and we know CE (celestial equator) defines 0 degrees declination, then in this case (Barbados' location) the star’s altitude of a max = 45 deg shows it to be SOUTH of CE. How much? 90 deg - 45 deg = 45 deg. (The (-ve) sign implies direction below (south) of CE.)
The diagram below shows the geometry of the situation:
At the zenith the declination is + 13o on the basis of projection. Also, the north celestial pole (NCP) must be 13 degrees above the northern horizon. The star S is 45o above the observer's southern horizon. We know the celestial equator (CE) is 0 o declination. Then the position of the star S must be 32 degrees south of it, or as shown in the computation, d = - 32o, which would be its declination.
As we can infer: Z CE = 13o
CE S = 32 o
The altitude a max = 45o as shown
The zenith distance z = Z CE + CE S = 13o + 32 o = 45 o
Thus, 90o = z + a max
Since the zenith distance z plus altitude (a max) must equal 90 degrees and we know CE (celestial equator) defines 0 degrees declination, then in this case (Barbados' location) the star’s altitude of a max = 45 deg shows it to be SOUTH of CE. How much? 90 deg - 45 deg = 45 deg. (The (-ve) sign implies direction below (south) of CE.)
The diagram below shows the geometry of the situation:
At the zenith the declination is + 13o on the basis of projection. Also, the north celestial pole (NCP) must be 13 degrees above the northern horizon. The star S is 45o above the observer's southern horizon. We know the celestial equator (CE) is 0 o declination. Then the position of the star S must be 32 degrees south of it, or as shown in the computation, d = - 32o, which would be its declination.
As we can infer: Z CE = 13o
CE S = 32 o
The altitude a max = 45o as shown
The zenith distance z = Z CE + CE S = 13o + 32 o = 45 o
Thus, 90o = z + a max
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