Manual steps to convert equatorial coordinates (RA, Dec)
to horizontal coordinates (Altitude,Azimuth)
First we define:
α= RA of the observed object
δ = Declination of the object, +ve for north of celestial equator, ve for south.
L = Longitude of the observer, +ve for west of Greenwich, ve if east.
φ = Latitude of the observer, +ve for northern hemisphere, ve for southern hemisphere.
h = Altitude, +ve above horizon, ve below horizon
H = Local Hour Angle, measuredwestwards from the South.
(H is the angle between the observer’s meridian and
the RA circle passing through the observed object.)
A = Azimuth, also measured westwards from the South.
(Hence, an observedobject which is exactly on the southern meridian has A = H = 00.)
Step 1.
From any almanac (e.g. Hong Kong Almanac), find out the GST (Greenwich Sidereal Time) at 0h UT. If you do not have almanac, additional calculations are required, see the Remarksection.
Step2. Calculate: GST at any given time, θ = 1.0027379 xGiven time in UT + GST at 0h UT
Step 3. Calculate: Local Sidereal Time, LST = θ Longitude (in hr, min, sec.)
Step 4. Calculate: Local Hour Angle, H = LST – RA of the observed object
Step 5. Convert: RA/Dec to Alt/AZ with thesetwo formulae
sin h = sin φ sin δ + cos φ cos δ cos H
tan A = sin H / (cos H sin φ – tan δ cos φ)
The above steps have not included the effect of atmospheric refraction which may affect the calculated Altitude by a small fraction of a degree.
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Example
Given: RA of observed object (Star Rigel in Orion)= 5h 14m 41s.4
Dec of observed object, δ= 80 11’ 54.9” = 8.1985830
Longitude of observer (Taipo site), L= 1140 13.2’ east = 7h 36m 52s.8
Latitude of observer,φ= 220 28.3’ north= 22.4716670
Date & Time of observation = 2003 Jan 1, 9:00 pm HKT
= 2003 Jan 1, 13h 0m 0s UT
(so the given time is 13h 0m 0s UT)
1 hour in longitude = 15 degrees
From Step 1
The “HK Almanac 2003” page 3 gives GST at 0h UT = 6h 40m 56s.0
From Step 2
GST at the given time, θ = 1.002 737 9 (13h 0m 0s) + 6h 40m56s.0
= 13h2m 8s.1 + 6h 40m 56s.0
= 19h 43m 4s.1
From Step 3
Local Sidereal Time, LST = 19h 43m 4s.1 (7h 36m 52s.8) = 27h19m 56s.9 = 3h19m 56s.9
From Step 4
Local Hour Angle, H = 3h19m 56s.9 5h 14m 41s.4 = 22h5m 15s.5 = 22h.087639 = 331.3145830
From Step 5, we can calculate Rigel’s altitude h
sin h = sin φ sin δ + cos φ cos δ cos H
= sin (22.4716670) sin (8.1985830) + cos (22.4716670) cos (8.1985830) cos (331.3145830)
= (0.382226523) (0.142604455) + (0.924068658) (0.989779758) (0.877268362)
= 0.054507205 + 0.802371095
= 0.74786389
h48.410 (final answer)
We can also calculate Rigel’s Azimuth A
tan A = sin H / (cos H sin φ – tan δ cos φ)
= sin (331.3145830) / [cos (331.3145830) sin (22.4716670)
tan (8.1985830) cos (22.4716670) ]
= 0.480000229 / [(0.877268362) (0.382226523) (0.144076956) (0.924068658)]
= 0.480000229 / 0.468452234
= 1.024651382
A =45.6980
=45.700eastwards from southor134.300 eastwards from north (final answer)
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Remark
If you do not have an almanac, try the following calculations which can replace the data in almanac within an accuracy of 1 second.
Calculate Julian Day (JD) at 0h UT. For year 1901 to 2099, it follows this equation
JD = 1721103.5 + INT(365.25y) + INT(30.6 m + 0.5) + D,
where INT means to take the Integer of the value inside the brackets. Our Gregorian calendar date is divided into year (Y), month (M) and days (D). To use the JDequation, we define:
For M < or = 2, y = Y 1, and m = M + 9
For M > 2, y = Y , and m = M 3
Example: On 2003 Jan 1, 0h UT, ybecomes 2002, mbecomes 10 and D is 1. Hence,
JD = 1721103.5 + INT (365.25 x 2002) + INT (30.6 x 10 + 0.5) + 1
= 1721103.5 + 731230 + 306 + 1 = 2452 640.5
Next, calculate GST at 0h UT with this formula
GST at 0h UT 6h41m 50s.548 + 8640 184s.813 T + 0s.0931 T2
where T = (JD – 2451 545) / 36525 (Note: This formula is valid only at0h UT.)
Using above example, we have T = (2452 640.5 – 2451 545) / 36525 = 0.029993155
GST at 0h UT of 2003 Jan 1
6h41m 50s.548 + 8640 184s.813 (0.029 993155) + 0s.0931 (0.029993155)2
= 6h41m 50s.548 + 259146s.359 + 0s.000
= 24110s.548 + 259146s.402
= 283256s.950
= 3.278436921 days
= 0.278436921 days
= 6h.68248610
= 6h 40m 56s.9(very close to the GST data given in HK Almanac 2003, page 3)
In fact this 6h 40m 56s.9 is the mean sidereal time at Greenwich. The apparent sidereal time at Greenwich can be obtained by adding a “nutation correction factor” which accounts for the periodic oscillation of the Earth’s pole from its mean position. But since top accuracy is not required, we can take GST at 0h UT = 6h 40m 56s.9 without clarifying it is mean or apparent sidereal time.
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ABASIC language program for converting RA / Dec to Alt / Az isAltAz.bas, downloadable from S&T software archives. You input RA (5,14,41.4), Dec (-8,11,54.9), Longitude (114.22), Latitude (22.471667) and GST of the given time (19,43,4.1), the programthen reports Altitude and Azimuth which shall have same answers as the manual calculations.The AltAz.bas is also downloadable from also need to download BASICA.exe(explained in running AltAz.bas.
It is important to note that in AltAz.bas, both latitude and longitude must be input in degrees with decimals; andlongitude is +ve if east of Greenwich, –ve if west. In the manual steps, however, longitude is +ve if west, ve if east. If you get awkward results of calculation, check that you have used correct sign convention and correct units in the formulae.
RADec_AltAz_conversion.doc Alan Chu 1 of 3 2010 July