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: 2458439
Now, calculate J
n
_{*} = (J_{date} - 2451545 - 0.0009) - (l_{w}/360)n = round(n _{*})n _{*} = (2458439 - 2451545 - 0.0009) - (77.04876/360) = 6893.7850756667n = round(6893.7850756667) = 6894
Now J
J
_{*} = 2451545 + 0.0009 + (l_{w}/360) + nJ _{*} = 2451545 + 0.0009 + (77.04876/360) + 6894 = 2458439.2149243
Using J
M = [357.5291 + 0.98560028 * (J
_{*} - 2451545)] mod 360M = [357.5291 + 0.98560028 * (2458439.2149243 - 2451545)] mod 360 = 7152.4692598029 mod 360 = 312.46925980293 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(312.46925980293) + 0.0200 * sin(2 * 312.46925980293) + 0.0003 * sin(3 * 312.46925980293) = -1.4325367485165 We need λ which is the ecliptical longitude of the sun:
λ = (M + 102.9372 + C + 180) mod 360
λ = (312.46925980293 + 102.9372 + -1.4325367485165 + 180) mod 360 = 593.97392305442 mod 360 = 233.97392305442
Finally, calculate J
J
_{transit} = J_{*} + (0.0053 * sin(M)) - (0.0069 * sin(2 * λ))J _{transit} = 2458439.2149243 + (0.0053 * sin(312.46925980293)) - (0.0069 * sin(2 * 233.97392305442)) = 2458439.2044506
Now, to get an even more accurate number, recursively recalculate M using J
I1: M = 312.45893690269, C = -1.4327688928836, λ = 233.9633680098, J
_{transit} = 2458439.2044492I2: M = 312.45893549599, C = -1.4327689245145, λ = 233.96336657147, J _{transit} = 2458439.2044492I3: M = 312.45893549599, C = -1.4327689245145, λ = 233.96336657147, J _{transit} = 2458439.2044492Ok, translate this into something we understand. i.e. When is Solar Noon?
J
_{transit} =
2458439.2044492 = 11/16/2018 at 11:54:24 -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(233.96336657147) * sin(23.45)) = -18.771670089453 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(-18.771670089453))/(cos(39.040759) * cos(-18.771670089453))) = 75.171381924425
Just as above, calculate J
J
_{**} = 2451545 + 0.0009 + ((H + l_{w})/360) + nJ _{**} = 2451545 + 0.0009 + ((75.171381924425 + 77.04876)/360) + 6894 = 2458439.4237337
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} = 2458439.4237337 + (0.0053 * sin(312.45893549599)) - (0.0069 * sin(2 * 233.96336657147)) = 2458439.4132586
Now I'm going to cheat and calculate J
J
_{rise} = J_{transit} - (J_{set} - J_{transit})J _{rise} = 2458439.2044492 - (2458439.4132586 - 2458439.2044492) = 2458438.9956398Using the same idea, figure out when sunrise and sunset are:
J
_{rise} =
2458438.9956398 = 11/16/2018 at 06:53:43 -0500
J _{set} =
2458439.4132586 = 11/16/2018 at 16:55:05 -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. |

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