by
Juegen Giesen

The observer is located at the center of his "celestial
sphere" with zenith Z above
his head and the horizon N-E-S-W. The
Sun, Moon or any other celestial body can be identified by the two
coordinates altitude h
and azimuth alpha (horizontal
coordinates). Altitude is the angular distance above the horizon
(0 < h < 90°), and azimuth the angular distance, measured
along the horizon, westwards from the south point S (in astronomy)
or eastwards from the north point N in nautics (0 < alpha <360°).
The daily movement of an object - resulting from the rotation of
the Earth on its axis - starts when it rises
at (1). At (2) it passes across the observer's meridian
NZS, reaching its maximum altitude above horizon (transit,
culmination), and it sets at (3).
Please watch the Java applet "Apparent
Movement of a Star" (with kind permission of Walter
Fendt).
The horizontal coordinates of an object depend on the location
of the observer on the Earth (and on time). In astronomy equatorial
coordinates are commonly used when giving the position of
an object on the celestial sphere. The equatorial system is based
on the celestial equator, which is the great circle obtained by
projecting the Earth's equator on to the celestial sphere, the equatorial
plane being perpendicular to the Earth's axis of rotation.

The first equatorial coordinate is declination
delta, measured in degrees north and south of the celestial equator
(N: 0° < delta < 90°, S: 0° > delta >
- 90°. The second coordinate, may be the hour
angle tau, measured along the equator from the meridian
S-NP-N ob the observer to the hour circle
SP-St-NP of the star St. The hour angle corresponds to the length
of sidereal time elapsed since the body St last made a transit of
the meridian.
A screen shot of Walter Fendts applet
shows the relationship of the two systems:
To convert equatorial coordinates hour angle and declination (tau,
delta) to horizontal coordinates azimith and altitude (az, h), the
"nautical triangle" NP-Ze-St is used (NP-Ze = 90° - beta, NP-St
= 90° - delta, Ze-St = 90° - h). From spherical trigonometry
we get:
sin h = sin beta sin delta + cos beta cos delta
cos tau
tan az = (- sin tau) / (cos beta tan delta -
sin beta cos tau)
Example:
Observer O at geogr. lat. beta=50° N and long 10° E, on
1991/05/19 at 13:00 UT
will see a star of right ascension RA=55.8° and declination
delta=19.7°
at azimuth az=43.6° and altitude h=53.4°
(Sidereal time is 81.7°, hour angle is 25.9°)
The second equatorial coordinate may also be right
ascension RA, measured in hours, minutes and seconds of time,
taking into account the rotation of the celestial sphere once in
24 hours of sidereal time. The zero point for right ascension is
taken as the northern vernal equinox.
This is one of the two points at which the celestial equator intersects
the ecliptic (the plane of the Earth's orbit around the Sun).
Right ascension RA, hour angle tau and sidereal time theta are
related by:
tau = theta - RA
Example: compute the postion
of the Sun on 1991/05/19 at 15:00 CEST
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1. Conversion of date and time
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local time to universal time UT
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15 h CEST = 13 h UT
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convert to ephemeris time
Julian Day of 1991/ 5/19 at 13
UT
number of Julian centuries since 2000/01/01 at 12 UT (JD
= 2451545.0)
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JD = 2448396.0416666665
T = (JD - 2451545.0 ) / 36525
= - 0.08621378051563282
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2. Astronomical algorithms:
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compute ecliptic longitude
of the Sun
mean longitude (degree)
mean anomaly (degree)
true longitude (degree)
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L0 = - 2823.296016°
M = - 2746.085122°
L = - 2821.936289° = 58.063711°
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convert ecliptic longitude
to right ascension RA and declination delta
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RA = 3 h 43 min 14 s = 55.80978°
delta = 19° 44' N
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compute sideral time (degree)
at Greenwich
local sidereal time at lattitude 10° E
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theta0 = -1136448.3019° = 71.698°
theta = theta0 + 10° = 81.698°
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compute local hour angle (degree)
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tau = teta - RA = 81.698° - 55.810°
= 25.9° from S
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3. Final results
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convert (tau, delta) to
horizon coordinates (h, az) of the observer (50° N, 10°
E)
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altit. h = 53.4 °
azim. az = 223.6° from
N
azim. az = 223.6° - 180° = 43.6°
from S
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