ured along the celestial equator; and the declination, which is the angular distance of the
body from the celestial equator as measured along an hour circle.
The relative coordinates of a body at a given instant are obtained by observation; the
absolute coordinates are obtained by consulting an almanac of astronomical data. The lat-
itude of the observer's site equals the angular distance of the observer's zenith from the
celestial equator as measured along the meridian. In the astronomical triangle PZS in Fig.
11, the arcs represent the indicated coordinates, and angle Z represents the azimuth of the
body as measured from the north.
Calculating the azimuth of the body yields
^21/;Z= SIn(^)SIn(S-.)
cos S cos (S-p)where L = latitude of site; h = altitude of star; p = polar distance = 90° - declination; S =
1
X 2 (L + h + /?); L = 41°20'; h = 46°48'; p = 90° - 7°58' = 82°02'; S = V 2 (L + h+p) =
85°05'; S-L = 43°45'; S-h = 38°17'; S-p = 30 OS'.
Then
Iogsin43°45' = 9.839800
Iogsin38°17' 9.792077
9.631877
Iogcos85°05' = 8.933015
Iogcos3°03' = 9.999384 8.932399
2 log tan Y 2 Z = 0.699478
log tan^1 X 2 Z = 0.349739(^1) X
2 Z = 65°55'03.5" Z= 131°50'07"
- Verify the solution by calculating Z in an alternative manner
To do this, introduce an auxiliary angle M 9 defined by
cosz?
cos^2 M= — (17)
sin h sin LThen
cos (180° - Z) = tan h tan L sin^2 M (18)
Then
Iogcos82°02' = 9.141754
Iogsin46°48' = 9.862709
Iogsin41°20' = 9.819832 9.682541
2 log cos M = 9.459213
log cos M = 9.729607
log sin M= 9.926276
2 log sin M = 9.852552
Iogtan46°48' = 0.027305
Iogtan41°20' = 9.944262
log cos (180° -Z) = 9.8241 19
Z= 131°50'07", as before