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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: 2459767

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

n* = (2459767 - 2451545 - 0.0009) - (77.04876/360) = 8221.7850756667
n = round(8221.7850756667) = 8222

Now J*:

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

J* = 2451545 + 0.0009 + (77.04876/360) + 8222 = 2459767.2149243

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 * (2459767.2149243 - 2451545)] mod 360 = 8461.3464316429 mod 360 = 181.34643164293

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(181.34643164293) + 0.0200 * sin(2 * 181.34643164293) + 0.0003 * sin(3 * 181.34643164293) = -0.044074509372298

We need λ which is the ecliptical longitude of the sun:

λ = (M + 102.9372 + C + 180) mod 360

λ = (181.34643164293 + 102.9372 + -0.044074509372298 + 180) mod 360 = 464.23955713356 mod 360 = 104.23955713356

Finally, calculate Jtransit:

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

Jtransit = 2459767.2149243 + (0.0053 * sin(181.34643164293)) - (0.0069 * sin(2 * 104.23955713356)) = 2459767.21809

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 = 181.34955170634, C = -0.044176624759821, λ = 104.24257508158, Jtransit = 2459767.2180903
I2: M = 181.34955205194, C = -0.044176636070611, λ = 104.24257541587, Jtransit = 2459767.2180903
I3: M = 181.34955205194, C = -0.044176636070611, λ = 104.24257541587, Jtransit = 2459767.2180903

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

Jtransit = 2459767.2180903 = 07/06/2022 at 13:14:03 -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(104.24257541587) * sin(23.45)) = 22.688247553597

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(22.688247553597))/(cos(39.040759) * cos(22.688247553597))) = 111.0542795454

Just as above, calculate J*, but this time using hour-angle:

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

J** = 2451545 + 0.0009 + ((111.0542795454 + 77.04876)/360) + 8222 = 2459767.5234084

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 = 2459767.5234084 + (0.0053 * sin(181.34955205194)) - (0.0069 * sin(2 * 104.24257541587)) = 2459767.5265744

Now I'm going to cheat and calculate Jrise:

Jrise = Jtransit - (Jset - Jtransit)

Jrise = 2459767.2180903 - (2459767.5265744 - 2459767.2180903) = 2459766.9096062

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

Jrise = 2459766.9096062 = 07/06/2022 at 05:49:49 -0500
Jset = 2459767.5265744 = 07/06/2022 at 20:38:16 -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.