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Today's sunrise/sunset calculation for 39.040759 °N, 77.04876 °W.Note: I am not using the appropriate number of significant figures to calculate sunrise, sunset, and solar noon. While it may appear that my calculations are accurate to one second, they actually have a precision of closer to 3 minutes (mostly due to approximation). Also, this calculation does not take into account the effect of air temperature, altitude, etc. Together, these may affect the time by 5 minutes or more. Find today's Julian date (days since Jan 1, 2000 + 2451545): Julian date: 2456066Now, calculate Jtransit at longitude 77.04876, start with n:
n* = (Jdate - 2451545 - 0.0009) - (lw/360)
n = round(n*) n* = (2456066 - 2451545 - 0.0009) - (77.04876/360) = 4520.7850756667 n = round(4520.7850756667) = 4521 Now J*:
J* = 2451545 + 0.0009 + (lw/360) + n
J* = 2451545 + 0.0009 + (77.04876/360) + 4521 = 2456066.2149243 Using J*, calculate M (mean anomaly) and then use that to calculate C and λ:
M = [357.5291 + 0.98560028 * (J* - 2451545)] mod 360
M = [357.5291 + 0.98560028 * (2456066.2149243 - 2451545)] mod 360 = 4813.6397953629 mod 360 = 133.63979536293 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(133.63979536293) + 0.0200 * sin(2 * 133.63979536293) + 0.0003 * sin(3 * 133.63979536293) = 1.3659458290383 We need λ which is the ecliptical longitude of the sun:
λ = (M + 102.9372 + C + 180) mod 360
λ = (133.63979536293 + 102.9372 + 1.3659458290383 + 180) mod 360 = 417.94294119197 mod 360 = 57.942941191971 Finally, calculate Jtransit:
Jtransit = J* + (0.0053 * sin(M)) - (0.0069 * sin(2 * λ))
Jtransit = 2456066.2149243 + (0.0053 * sin(133.63979536293)) - (0.0069 * sin(2 * 57.942941191971)) = 2456066.2125522 Now, to get an even more accurate number, recursively recalculate M using Jtransit until it stops changing. Notice how close the approximation was.
I1: M = 133.63745740121, C = 1.3659997994549, λ = 57.940657200669, Jtransit = 2456066.2125521
I2: M = 133.63745731126, C = 1.3659998015314, λ = 57.94065711279, Jtransit = 2456066.2125521 I3: M = 133.63745731126, C = 1.3659998015314, λ = 57.94065711279, Jtransit = 2456066.2125521 Ok, translate this into something we understand. i.e. When is Solar Noon?
Jtransit =
2456066.2125521 = 05/18/2012 at 13:06:04 -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(57.94065711279) * sin(23.45)) = 19.710085944241 Now we can go about calculating H (Hour angle):
H = arccos( [sin(-0.83) - sin(ln) * sin(δ)] / [cos(ln) * cos(δ)] )
H = arccos((sin(-0.83) - sin(39.040759) * sin(19.710085944241))/(cos(39.040759) * cos(19.710085944241))) = 108.07965011677 Just as above, calculate J*, but this time using hour-angle:
J** = 2451545 + 0.0009 + ((H + lw)/360) + n
J** = 2451545 + 0.0009 + ((108.07965011677 + 77.04876)/360) + 4521 = 2456066.5151456 We can use M from above because it really doesn't change that much over the course of a day, calculate Jset in the same way:
Jset = J** + (0.0053 * sin(M)) - (0.0069 * sin(2 * λ))
Jset = 2456066.5151456 + (0.0053 * sin(133.63745731126)) - (0.0069 * sin(2 * 57.94065711279)) = 2456066.5127734 Now I'm going to cheat and calculate Jrise:
Jrise = Jtransit - (Jset - Jtransit)
Jrise = 2456066.2125521 - (2456066.5127734 - 2456066.2125521) = 2456065.9123309 Using the same idea, figure out when sunrise and sunset are:
Jrise =
2456065.9123309 = 05/18/2012 at 05:53:45 -0500
Jset = 2456066.5127734 = 05/18/2012 at 20:18:23 -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. |