New! Enter your latitude and longitude.Today's sunrise/sunset calculation for 39.040759 °N, 77.04876 °W.
Find today's Julian date (days since Jan 1, 2000 + 2451545): Julian date: 2459116
Now, calculate J
n
_{*} = (J_{date} - 2451545 - 0.0009) - (l_{w}/360)n = round(n _{*})n _{*} = (2459116 - 2451545 - 0.0009) - (77.04876/360) = 7570.7850756667n = round(7570.7850756667) = 7571
Now J
J
_{*} = 2451545 + 0.0009 + (l_{w}/360) + nJ _{*} = 2451545 + 0.0009 + (77.04876/360) + 7571 = 2459116.2149243
Using J
M = [357.5291 + 0.98560028 * (J
_{*} - 2451545)] mod 360M = [357.5291 + 0.98560028 * (2459116.2149243 - 2451545)] mod 360 = 7819.7206493629 mod 360 = 259.72064936293 We need to calculate the equation of center, C:
C = (1.9148 * sin(M)) + (0.0200 * sin(2 * M)) + (0.0003 * sin(3 * M))
C = 1.9148 * sin(259.72064936293) + 0.0200 * sin(2 * 259.72064936293) + 0.0003 * sin(3 * 259.72064936293) = -1.8767854155646 We need λ which is the ecliptical longitude of the sun:
λ = (M + 102.9372 + C + 180) mod 360
λ = (259.72064936293 + 102.9372 + -1.8767854155646 + 180) mod 360 = 540.78106394737 mod 360 = 180.78106394737
Finally, calculate J
J
_{transit} = J_{*} + (0.0053 * sin(M)) - (0.0069 * sin(2 * λ))J _{transit} = 2459116.2149243 + (0.0053 * sin(259.72064936293)) - (0.0069 * sin(2 * 180.78106394737)) = 2459116.2095213
Now, to get an even more accurate number, recursively recalculate M using J
I1: M = 259.71532413302, C = -1.8767502117342, λ = 180.77577392129, J
_{transit} = 2459116.2095227I2: M = 259.71532547501, C = -1.8767502206078, λ = 180.7757752544, J _{transit} = 2459116.2095227I3: M = 259.71532547455, C = -1.8767502206048, λ = 180.77577525394, J _{transit} = 2459116.2095227I4: M = 259.71532547455, C = -1.8767502206048, λ = 180.77577525394, J _{transit} = 2459116.2095227Ok, translate this into something we understand. i.e. When is Solar Noon?
J
_{transit} =
2459116.2095227 = 09/23/2020 at 13:01:42 -0500
Alrighty, now calculate how long the sun is in the sky at latitude 39.040759: Now we need to calculate δ which is the declination of the sun:
δ = arcsin( sin(λ) * sin(23.45) )
δ = arcsin(sin(180.77577525394) * sin(23.45)) = -0.30871076152915 Now we can go about calculating H (Hour angle):
H = arccos( [sin(-0.83) - sin(l
_{n}) * sin(δ)] / [cos(l_{n}) * cos(δ)] )H = arccos((sin(-0.83) - sin(39.040759) * sin(-0.30871076152915))/(cos(39.040759) * cos(-0.30871076152915))) = 90.818276961155
Just as above, calculate J
J
_{**} = 2451545 + 0.0009 + ((H + l_{w})/360) + nJ _{**} = 2451545 + 0.0009 + ((90.818276961155 + 77.04876)/360) + 7571 = 2459116.4671973
We can use M from above because it really doesn't change that much over the course of a day, calculate J
J
_{set} = J_{**} + (0.0053 * sin(M)) - (0.0069 * sin(2 * λ))J _{set} = 2459116.4671973 + (0.0053 * sin(259.71532547455)) - (0.0069 * sin(2 * 180.77577525394)) = 2459116.4617957
Now I'm going to cheat and calculate J
J
_{rise} = J_{transit} - (J_{set} - J_{transit})J _{rise} = 2459116.2095227 - (2459116.4617957 - 2459116.2095227) = 2459115.9572497Using the same idea, figure out when sunrise and sunset are:
J
_{rise} =
2459115.9572497 = 09/23/2020 at 06:58:26 -0500
J _{set} =
2459116.4617957 = 09/23/2020 at 19:04:59 -0500
A couple anomalies occur. At high latitudes, you will sometimes get H = 0. This means that either the sun does not rise (in the winter) or the sun does not set (in the summer) on that day. |

Copyright ©1997 - 2011, Bulent Yilmaz