Here's how to use it for the year 1997.
Y = 1997;
y = 1997/1000 = 1.997 then
JD1 = 1721139.2855 + 365.2421376 (Y) + 0.0679190 y^2 - 0.0027879 y^3 = 1721139.2855 + 729388.549 + 0.27086 - 0.02220 = 2450528.083
After you have computed the above Julian Date, you can convert this data into a month, day and year using:
Add 0.5 to JD1 above = 2450528.583
Set I = integer part of JD = 2450528
Set F = fractional part of JD = 0.583
I is larger than 2299160 so calculate:
A = integer part of ( I - 1867216.25)/36524.25 = 15.9705
B = I + 1 + A - integer part of (A/4 = 3.99)
= 2450528 + 1 + 15.9705 - 3
= 2450541.971
Now calculate C = B + 1524 = 2452065.971
Calculate D = integer part of (C - 122.1)/365.25 = 6713
Calculate E = integer part of 365.25 x D = 2451923
Calculate G = integer part of ( C - E)/30.6001 = 4
Calculate d = C - E + F - integer part of (30.6001 x G)
= 2452065.971 - 2451923 + 0.583 - 122.40
= 21.154
This is the day of the month including its decimal fraction
ie the 21st day at 0.154x24 = 03:41
Calculate m = G - 1 if G is less than 13.5 = 4 - 1 = 3
This is the month number ie March
Calculate y = D - 4716 if m is more than 2.5 = 1997
This is the year ie 1997, which we already knew.
The next step of the calculation is to use this estimated JD to compute the apparent longitude of the Sun in the sky.
First calculate T:
T = (JD1 - 2415020.0)/36525 = (2450528.083 - 2415020.0)/36525
= 0.97216
Next, compute the Mean Solar Longitude from:
L = 279.69668 + 36000.76892 T + 0.0003025 T^2
= 35278.20448 degrees
= 358.20448
The Apparent Solar Longitude is given by:
Lapp =L + C - 0.00569 - 0.00479 sin(Omega)
where
C = (1.919460 - 0.004789T - 0.000014 T^2) sin(M) +
(0.0200094 - 0.000100 T) sin(2M) +
0.000293 sin(3M)
= 1.9147911 sin(M) + 0.0199122 sin(2M) + 0.000293 sin(3M)
and M = 358.47583 + 35999.04975 T - 0.000150 T^2 - 0.0000033 T^3
= 35355.31189 degrees
or = 75.31189 degrees
so that C = 1.8521847 + 0.009767 -0.00021 = 1.8617 degrees
Finally,
Omega = 259.18 - 1934.142 T
= -1621.115 degrees
= 178.885 degrees
Then
Lapp = 358.20448 + 1.8617 - 0.00569 - 0.00479 sin(178.885)
= 360.06039 degrees.
The correction to be applied to the previous JD is then
JD2 = JD1 + 58 sin(-Lapp)
On my calculator I got a predicted date of JD = 2450528.11 which was 0.6 days later than the forecast of March 20, 1997 at 13:53. If you now use this new JD2 as the input to the above calculation, the next iteration will probably give you a better match from JD3 = JD2 + 58sin(-Lapp) where Lapp is now the new apparent longitude based on JD2.
Copyright 1997 Dr. Sten Odenwald
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