Home Astronomy Sun, Moon & Earth Applet Ein vereinfachtes Verfahren zur Berechnung der Sonnenhöhe mit Tabellenkalkulation finden Sie hier  Basics of Positional Astronomy The observer is located at the centre 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 (transit), and it sets at (3). Only fixed stars (constant declination) reach the greatest altitude above horizon (culmination) on the meridian. Details about the difference between transit and culmination Please watch the Java applet Apparent Movement of a Star (on this server, 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 of 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 Apparent Movement of a Star shows the relationship of the two systems:   To convert equatorial coordinates hour angle (tau) and declination (delta) to horizontal coordinates azimuth (az) and altitude (h), the "nautical triangle" NP-Ze-St is used: NP-Ze = 90° - beta (geogr. latitude 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: An observer O at geogr. latitude beta=50° N and longitude 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 Animation: Sidereal Time and Solar Time

 Example compute the position of the Sun on 1991/05/19 at 15:00 CEST Berechnung des Sonnenstandes 2. Conversion of date and time: local time to universal time UT 15 h CEST = 13 h UT convert time: Julian Day of 1991/ 5/19 at 13 UT Julian day of 2000/01/01 at 12 UT number of Julian days since 2000/01/01 at 12 UT number of Julian centuries since 2000/01/01 at 12 UT used by the algorithm for L JD = 2448396.04167 JD = 2451545.0 -3148.95833 T = - 3148.95833/36525 = 0.086213780 2. Astronomical algorithms: apparent longitude latitude B is assumed to be zero L = 58.06° convert ecliptic longitude to right ascension RA and declination delta RA = 55.81° delta = 19.73 compute sidereal time (degree) at Greenwich local sidereal time at longitude 10° E local hour angle theta0 = 71.698° theta = theta0 + 10° = 81.698° tau = theta - RA = 81.698° - 55.81° = 25.89° 3. Final results: convert (tau, delta) to horizon coordinates (h, az) of the observer (50° N, 10° E) The function atan2(numerator,denominator) should be used to avoid ambiguity. altitude angle: h = 53.4 ° azimuth angle: az = 223.6° from N azimuth angle: az = 223.6° - 180° = 43.6° from S

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 Web Links How to compute the day/night terminator Books Montenbruck, Oliver / Pfleger, Thomas: Astronomie mit dem Personal Computer; mit CD-ROM, Springer Berlin, 4. Aufl. 2004, ISBN 3-540-21204-3 Montenbruck, Oliver / Pfleger, Thomas: Astronomy on the Personal Computer. with CD-ROM; Springer Berlin, 4th ed. 2004, ISBN 3-540-67221-4 Meeus, Jean: Astronomical Algorithms Willmann-Bell; Hardcover (1st ed. 1991), ISBN: 0943396352 Willmann-Bell; Hardcover (2nd ed. 1998), ISBN: 0943396611 Meeus, Jean: Astronomical Formulae for Calculators; Willmann-Bell; Softcover, ISBN: 0943396220