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Latitude Longitude
°N °W
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: 2458102

Now, calculate Jtransit at longitude 77.04876, start with n:

n* = (Jdate - 2451545 - 0.0009) - (lw/360)
n = round(n*)

n* = (2458102 - 2451545 - 0.0009) - (77.04876/360) = 6556.7850756667
n = round(6556.7850756667) = 6557

Now J*:

J* = 2451545 + 0.0009 + (lw/360) + n

J* = 2451545 + 0.0009 + (77.04876/360) + 6557 = 2458102.2149243

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 * (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 Jtransit:

Jtransit = J* + (0.0053 * sin(M)) - (0.0069 * sin(2 * λ))

Jtransit = 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 Jtransit until it stops changing. Notice how close the approximation was.

I1: M = 340.31846958735, C = -0.65783076298761, λ = 262.59783882437, Jtransit = 2458102.2113763
I2: M = 340.31846845924, C = -0.65783079910417, λ = 262.59783766014, Jtransit = 2458102.2113763
I3: M = 340.31846845832, C = -0.65783079913358, λ = 262.59783765919, Jtransit = 2458102.2113763
I4: M = 340.31846845832, C = -0.65783079913358, λ = 262.59783765919, Jtransit = 2458102.2113763

Ok, translate this into something we understand. i.e. When is Solar Noon?

Jtransit = 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(ln) * sin(δ)] / [cos(ln) * 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*, but this time using hour-angle:

J** = 2451545 + 0.0009 + ((H + lw)/360) + n

J** = 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 Jset in the same way:

Jset = J** + (0.0053 * sin(M)) - (0.0069 * sin(2 * λ))

Jset = 2458102.4117366 + (0.0053 * sin(340.31846845832)) - (0.0069 * sin(2 * 262.59783765919)) = 2458102.4081885

Now I'm going to cheat and calculate Jrise:

Jrise = Jtransit - (Jset - Jtransit)

Jrise = 2458102.2113763 - (2458102.4081885 - 2458102.2113763) = 2458102.014564

Using the same idea, figure out when sunrise and sunset are:

Jrise = 2458102.014564 = 12/14/2017 at 07:20:58 -0500
Jset = 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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