Astronomy Principles and Practice Fourth Edition.pdf

410 Practical projects
Figure 24.5. Determination of the Sun’s azimuth
Now the situation is depicted in figure 24.5 showing that the apparent azimuth of a celestial object
is not altered by the effect of refraction.
By using the sine formula, we have
sin(90 − δ ⊙ )
sin(360 − A) = sin z
sin H
giving
Also, we have
sin z = − cos δ ⊙ sin H
.
sin A
sin z cos(360 − A) = cos(90 − δ ⊙ ) sin(90 − φ) − sin(90 − δ ⊙ ) cos(90 − φ)cos H
giving
sin z = sin δ ⊙ cos φ − cos δ ⊙ sin φ cos H
.
cos A
Hence, elimination of sin z gives
tan A =
sin H
cos H sin φ − tan δ ⊙ cos φ .
Example. On May 2nd 2000 the Sun’s centre was at the cross-wire of a theodolite at 10 h 42 m 10 s UT.
The reading on the horizontal circle was 222 ◦ 12 ′ ; the reading corresponding to the reference object was
108 ◦ 27 ′ . The coordinates of the observing site are known to be λ = 4 ◦ 18·′9Wandφ = 55 ◦ 54·′3N.
Sun’s declination
According to The Astronomical Almanac (AA) for May 2nd 2000 at 00 h (TDT),
By noting the declination for May 3rd 2000,
δ ⊙ =+15 ◦ 24 ′ 10·′′ 1
=+15 ◦ 24·′17.
The daily increase in δ ⊙ =+17·′75.