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Today's sunrise/sunset calculation for 34.8444 °N, 82.3856 °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: 2460439Now, calculate Jtransit at longitude 82.3856, start with n:
n* = (Jdate - 2451545 - 0.0009) - (lw/360)
n = round(n*) n* = (2460439 - 2451545 - 0.0009) - (82.3856/360) = 8893.7702511111 n = round(8893.7702511111) = 8894 Now J*:
J* = 2451545 + 0.0009 + (lw/360) + n
J* = 2451545 + 0.0009 + (82.3856/360) + 8894 = 2460439.2297489 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 * (2460439.2297489 - 2451545)] mod 360 = 9123.6844308892 mod 360 = 123.6844308892 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(123.6844308892) + 0.0200 * sin(2 * 123.6844308892) + 0.0003 * sin(3 * 123.6844308892) = 1.5749118789416 We need λ which is the ecliptical longitude of the sun:
λ = (M + 102.9372 + C + 180) mod 360
λ = (123.6844308892 + 102.9372 + 1.5749118789416 + 180) mod 360 = 408.19654276814 mod 360 = 48.196542768143 Finally, calculate Jtransit:
Jtransit = J* + (0.0053 * sin(M)) - (0.0069 * sin(2 * λ))
Jtransit = 2460439.2297489 + (0.0053 * sin(123.6844308892)) - (0.0069 * sin(2 * 48.196542768143)) = 2460439.227302 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 = 123.68201918842, C = 1.574957189415, λ = 48.194176377833, Jtransit = 2460439.227302
I2: M = 123.68201924808, C = 1.5749571882941, λ = 48.194176436375, Jtransit = 2460439.227302 I3: M = 123.68201924808, C = 1.5749571882941, λ = 48.194176436375, Jtransit = 2460439.227302 Ok, translate this into something we understand. i.e. When is Solar Noon?
Jtransit =
2460439.227302 = 05/08/2024 at 13:27:18 -0500
Alrighty, now calculate how long the sun is in the sky at latitude 34.8444: Now we need to calculate δ which is the declination of the sun:
δ = arcsin( sin(λ) * sin(23.45) )
δ = arcsin(sin(48.194176436375) * sin(23.45)) = 17.255556612632 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(34.8444) * sin(17.255556612632))/(cos(34.8444) * cos(17.255556612632))) = 103.57524203466 Just as above, calculate J*, but this time using hour-angle:
J** = 2451545 + 0.0009 + ((H + lw)/360) + n
J** = 2451545 + 0.0009 + ((103.57524203466 + 82.3856)/360) + 8894 = 2460439.5174579 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 = 2460439.5174579 + (0.0053 * sin(123.68201924808)) - (0.0069 * sin(2 * 48.194176436375)) = 2460439.515011 Now I'm going to cheat and calculate Jrise:
Jrise = Jtransit - (Jset - Jtransit)
Jrise = 2460439.227302 - (2460439.515011 - 2460439.227302) = 2460438.939593 Using the same idea, figure out when sunrise and sunset are:
Jrise =
2460438.939593 = 05/08/2024 at 06:33:00 -0500
Jset = 2460439.515011 = 05/08/2024 at 20:21:36 -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. |