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: 2458532
Now, calculate J
n
_{*} = (J_{date} - 2451545 - 0.0009) - (l_{w}/360)n = round(n _{*})n _{*} = (2458532 - 2451545 - 0.0009) - (77.04876/360) = 6986.7850756667n = round(6986.7850756667) = 6987
Now J
J
_{*} = 2451545 + 0.0009 + (l_{w}/360) + nJ _{*} = 2451545 + 0.0009 + (77.04876/360) + 6987 = 2458532.2149243
Using J
M = [357.5291 + 0.98560028 * (J
_{*} - 2451545)] mod 360M = [357.5291 + 0.98560028 * (2458532.2149243 - 2451545)] mod 360 = 7244.1300858429 mod 360 = 44.130085842932 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(44.130085842932) + 0.0200 * sin(2 * 44.130085842932) + 0.0003 * sin(3 * 44.130085842932) = 1.3534680343305 We need λ which is the ecliptical longitude of the sun:
λ = (M + 102.9372 + C + 180) mod 360
λ = (44.130085842932 + 102.9372 + 1.3534680343305 + 180) mod 360 = 328.42075387726 mod 360 = 328.42075387726
Finally, calculate J
J
_{transit} = J_{*} + (0.0053 * sin(M)) - (0.0069 * sin(2 * λ))J _{transit} = 2458532.2149243 + (0.0053 * sin(44.130085842932)) - (0.0069 * sin(2 * 328.42075387726)) = 2458532.2247713
Now, to get an even more accurate number, recursively recalculate M using J
I1: M = 44.139790972482, C = 1.3537009163543, λ = 328.43069188884, J
_{transit} = 2458532.2247708I2: M = 44.13979054244, C = 1.353700906036, λ = 328.43069144848, J _{transit} = 2458532.2247708I3: M = 44.13979054244, C = 1.353700906036, λ = 328.43069144848, J _{transit} = 2458532.2247708Ok, translate this into something we understand. i.e. When is Solar Noon?
J
_{transit} =
2458532.2247708 = 02/17/2019 at 12:23:40 -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(328.43069144848) * sin(23.45)) = -12.024965727305 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(-12.024965727305))/(cos(39.040759) * cos(-12.024965727305))) = 81.159985850139
Just as above, calculate J
J
_{**} = 2451545 + 0.0009 + ((H + l_{w})/360) + nJ _{**} = 2451545 + 0.0009 + ((81.159985850139 + 77.04876)/360) + 6987 = 2458532.4403687
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} = 2458532.4403687 + (0.0053 * sin(44.13979054244)) - (0.0069 * sin(2 * 328.43069144848)) = 2458532.4502152
Now I'm going to cheat and calculate J
J
_{rise} = J_{transit} - (J_{set} - J_{transit})J _{rise} = 2458532.2247708 - (2458532.4502152 - 2458532.2247708) = 2458531.9993264Using the same idea, figure out when sunrise and sunset are:
J
_{rise} =
2458531.9993264 = 02/17/2019 at 06:59:01 -0500
J _{set} =
2458532.4502152 = 02/17/2019 at 17:48:18 -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