Approximate Position of the Sun (Altitude and Azimuth) from any Location at any Time (for low accuracy calculation)
based on Yallop, Nautical Almanac Office,
NAO Technical Note No. 46 (1978)
(Dave Laney, SAAO, (021) 447-0025 for comments)
examples-
(a) Cape Town Feb 15 10:30 1995
(b) Bloemfontein May 20 13:35 1996
(c) Johannesburg Sept 25 16:45 1997
(1) find Y, the year minus 1900:
(a) Y = 95
(b) 96
(c) 97
(2) find Z(J) from this table:
Jan J= 1 Z(J)=-0.5* Jul J= 7 Z(J)=180.5
Feb 2 30.5* Aug 8 211.5
Mar 3 58.5 Sep 9 242.5
Apr 4 89.5 Oct 10 272.5
May 5 119.5 Nov 11 303.5
Jun 6 150.5 Dec 12 333.5
(* reduce by one for a leap year)
(a) Z(J) = 30.5
(b) 119.5
(c) 242.5
(3) find D the number of days from this formula:
D = integer(365.25 x Y) + Z(J) + K + UT/24
where K is the day of the month and UT is the universal time
(a) D = int(365.25 x 95) + 30.5 + 15 + 8.500/24 = 34743.854
(b) int(365.25 x 96) + 119.5 + 20 + 11.583/24 = 35203.983
(c) int(365.25 x 97) + 242.5 + 25 + 14.750/24 = 35697.115
(4) find T the fraction of a julian century from this formula:
T = D/36525
(a) T = 0.9512349
(b) 0.9638325
(c) 0.9773337
(5) find L the mean longitude of the sun from this formula:
L = 279.697 + 36000.769 x T
(a) L = 34524.885 => 324.885 (removing multiples of 360 degrees)
(b) 34978.408 => 58.408
(c) 35464.462 => 184.462
(6) find M the mean anomaly of the sun from this formula:
M = 358.476 + 35999.050 x T
(a) M = 34602.029 => 42.029 (removing multiples of 360 degrees)
(b) 35055.530 => 135.530
(c) 35541.561 => 261.561
(7) find epsilon the obliquity from this formula:
epsilon = 23.452 - 0.013 x T
(a) epsilon = 23.4396
(b) 23.4395
(c) 23.4393
(8) find lambda the ecliptic longitude of the sun from this formula:
lambda = L + (1.919 - 0.005 x T) x sin(M) + 0.020 x sin(2M)
(a) lambda = 324.885 + 1.9142 x 0.6695 + 0.020 x 0.9946 = 326.186
(b) 58.408 + 1.9142 x 0.7005 + 0.020 x -0.9998 = 59.729
(c) 184.462 + 1.9141 x -0.9892 + 0.020 x 0.2903 = 182.574
(9) find alpha the right ascension of the sun from this formula:
alpha = arctan (tan(lambda) x cos(epsilon)) in same quadrant as
lambda
(a) alpha = 328.428
(b) 57.537
(c) 182.362
(10) find delta the declination of the sun from this formula:
delta = arcsin (sin(lambda) x sin(epsilon))
(a) delta = -12.789
(b) 20.093
(c) -1.024
(11) to proceed you need to know LONG the east-longitude of your location:
east-longitude latitude
Windhoek 17.10 -22.57
Cape Town 18.37 -33.92
P.E. 25.67 -33.97
Bloemfontein 26.12 -29.20
Johannesburg 28.00 -26.25
Durban 30.93 -29.92
(12) find HA the hour angle of the sun from this formula:
HA = L - alpha + 180 + 15 x UT + LONG
(a) HA = 324.885 - 328.428 + 180 + 15 x 8.500 + 18.37 = -37.673
(b) 58.408 - 57.537 + 180 + 15 x 11.583 + 26.12 = 20.736
(c) 184.462 - 182.362 + 180 + 15 x 14.750 + 28.00 = 71.350
(13) find the altitude of the center of the sun ALT from this formula:
ALT [degrees] =
ARCSIN [ SIN(LAT) x SIN(DEC) + COS(LAT) x COS(DEC) x COS(HA) ]
(a) ALT = ARCSIN ( -.5580 x -.2214 + .8298 x .9752 x .7915 ) = 49.822
(b) ARCSIN ( -.4879 x .3435 + .8729 x .9391 x .9352 ) = 36.800
(c) ARCSIN ( -.4423 x -.0182 + .8969 x .9998 x .3198 ) = 17.147
(14) find the azimuth of the sun AZ from this formula:
AZ [degrees] = ARCTAN [ SIN(HA) /
(COS(HA) x SIN(LAT) - TAN(DEC) x COS(LAT) ]
(a) AZ = ARCTAN [ -.6112/ ( .7915 x -.5580 - -.2270 x .8298 ) ]
= ARCTAN ( 2.4130 ) = 67.49 {i.e. east of true north}
(b) ARCTAN [ .3541/ ( .9352 x -.4879 - .3658 x .8729 ) ]
ARCTAN ( -0.45656 ) = -24.54 = 335.46 {i.e. west of true north}
(c) ARCTAN [ .9475/ ( .3198 x -.4423 - -.01787 x .8969 ) ]
ARCTAN ( -7.5546 ) = -82.46 = 277.54 {i.e. west of true north}
COMPARISON WITH COMPUTER ALMANAC PROGRAM
rough calculation here computer almanac calculation difference
(a) alt = 49.8, az = 67.5 ALT = 49.8, AZ = 67.5 none
(b) alt = 36.8, az = 335.5 ALT = 36.8, AZ = 335.5 none
(c) alt = 17.1, az = 277.5 ALT = 17.1, AZ = 277.5 none
Last updated: 23 Nov 2009
