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: 2458102
Now, calculate J
n
_{*} = (J_{date} - 2451545 - 0.0009) - (l_{w}/360)n = round(n _{*})n _{*} = (2458102 - 2451545 - 0.0009) - (77.04876/360) = 6556.7850756667n = round(6556.7850756667) = 6557
Now J
J
_{*} = 2451545 + 0.0009 + (l_{w}/360) + nJ _{*} = 2451545 + 0.0009 + (77.04876/360) + 6557 = 2458102.2149243
Using J
M = [357.5291 + 0.98560028 * (J
_{*} - 2451545)] mod 360M = [357.5291 + 0.98560028 * (2458102.2149243 - 2451545)] mod 360 = 6820.3219654429 mod 360 = 340.32196544293 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(340.32196544293) + 0.0200 * sin(2 * 340.32196544293) + 0.0003 * sin(3 * 340.32196544293) = -0.65771884187013 We need λ which is the ecliptical longitude of the sun:
λ = (M + 102.9372 + C + 180) mod 360
λ = (340.32196544293 + 102.9372 + -0.65771884187013 + 180) mod 360 = 622.60144660106 mod 360 = 262.60144660106
Finally, calculate J
J
_{transit} = J_{*} + (0.0053 * sin(M)) - (0.0069 * sin(2 * λ))J _{transit} = 2458102.2149243 + (0.0053 * sin(340.32196544293)) - (0.0069 * sin(2 * 262.60144660106)) = 2458102.2113774
Now, to get an even more accurate number, recursively recalculate M using J
I1: M = 340.31846958735, C = -0.65783076298761, λ = 262.59783882437, J
_{transit} = 2458102.2113763I2: M = 340.31846845924, C = -0.65783079910417, λ = 262.59783766014, J _{transit} = 2458102.2113763I3: M = 340.31846845832, C = -0.65783079913358, λ = 262.59783765919, J _{transit} = 2458102.2113763I4: M = 340.31846845832, C = -0.65783079913358, λ = 262.59783765919, J _{transit} = 2458102.2113763Ok, translate this into something we understand. i.e. When is Solar Noon?
J
_{transit} =
2458102.2113763 = 12/14/2017 at 12:04:22 -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(262.59783765919) * sin(23.45)) = -23.243040235959 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(-23.243040235959))/(cos(39.040759) * cos(-23.243040235959))) = 70.852416571588
Just as above, calculate J
J
_{**} = 2451545 + 0.0009 + ((H + l_{w})/360) + nJ _{**} = 2451545 + 0.0009 + ((70.852416571588 + 77.04876)/360) + 6557 = 2458102.4117366
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} = 2458102.4117366 + (0.0053 * sin(340.31846845832)) - (0.0069 * sin(2 * 262.59783765919)) = 2458102.4081885
Now I'm going to cheat and calculate J
J
_{rise} = J_{transit} - (J_{set} - J_{transit})J _{rise} = 2458102.2113763 - (2458102.4081885 - 2458102.2113763) = 2458102.014564Using the same idea, figure out when sunrise and sunset are:
J
_{rise} =
2458102.014564 = 12/14/2017 at 07:20:58 -0500
J _{set} =
2458102.4081885 = 12/14/2017 at 16:47:47 -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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