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: 2457837
Now, calculate J
n
_{*} = (J_{date} - 2451545 - 0.0009) - (l_{w}/360)n = round(n _{*})n _{*} = (2457837 - 2451545 - 0.0009) - (77.04876/360) = 6291.7850756667n = round(6291.7850756667) = 6292
Now J
J
_{*} = 2451545 + 0.0009 + (l_{w}/360) + nJ _{*} = 2451545 + 0.0009 + (77.04876/360) + 6292 = 2457837.2149243
Using J
M = [357.5291 + 0.98560028 * (J
_{*} - 2451545)] mod 360M = [357.5291 + 0.98560028 * (2457837.2149243 - 2451545)] mod 360 = 6559.1378912429 mod 360 = 79.137891242933 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(79.137891242933) + 0.0200 * sin(2 * 79.137891242933) + 0.0003 * sin(3 * 79.137891242933) = 1.8876435997446 We need λ which is the ecliptical longitude of the sun:
λ = (M + 102.9372 + C + 180) mod 360
λ = (79.137891242933 + 102.9372 + 1.8876435997446 + 180) mod 360 = 363.96273484268 mod 360 = 3.9627348426774
Finally, calculate J
J
_{transit} = J_{*} + (0.0053 * sin(M)) - (0.0069 * sin(2 * λ))J _{transit} = 2457837.2149243 + (0.0053 * sin(79.137891242933)) - (0.0069 * sin(2 * 3.9627348426774)) = 2457837.219178
Now, to get an even more accurate number, recursively recalculate M using J
I1: M = 79.142083628796, C = 1.8876672429464, λ = 3.9669508717426, J
_{transit} = 2457837.219177I2: M = 79.142082709966, C = 1.8876672377658, λ = 3.9669499477322, J _{transit} = 2457837.219177I3: M = 79.142082709966, C = 1.8876672377658, λ = 3.9669499477322, J _{transit} = 2457837.219177Ok, translate this into something we understand. i.e. When is Solar Noon?
J
_{transit} =
2457837.219177 = 03/24/2017 at 13:15:36 -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(3.9669499477322) * sin(23.45)) = 1.5775806809259 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(1.5775806809259))/(cos(39.040759) * cos(1.5775806809259))) = 92.349334681128
Just as above, calculate J
J
_{**} = 2451545 + 0.0009 + ((H + l_{w})/360) + nJ _{**} = 2451545 + 0.0009 + ((92.349334681128 + 77.04876)/360) + 6292 = 2457837.4714503
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} = 2457837.4714503 + (0.0053 * sin(79.142082709966)) - (0.0069 * sin(2 * 3.9669499477322)) = 2457837.475703
Now I'm going to cheat and calculate J
J
_{rise} = J_{transit} - (J_{set} - J_{transit})J _{rise} = 2457837.219177 - (2457837.475703 - 2457837.219177) = 2457836.9626511Using the same idea, figure out when sunrise and sunset are:
J
_{rise} =
2457836.9626511 = 03/24/2017 at 07:06:13 -0500
J _{set} =
2457837.475703 = 03/24/2017 at 19:25:00 -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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