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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: 2457933

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

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

n* = (2457933 - 2451545 - 0.0009) - (77.04876/360) = 6387.7850756667
n = round(6387.7850756667) = 6388

Now J*:

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

J* = 2451545 + 0.0009 + (77.04876/360) + 6388 = 2457933.2149243

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 * (2457933.2149243 - 2451545)] mod 360 = 6653.7555181229 mod 360 = 173.75551812293

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(173.75551812293) + 0.0200 * sin(2 * 173.75551812293) + 0.0003 * sin(3 * 173.75551812293) = 0.20404629244323

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

λ = (M + 102.9372 + C + 180) mod 360

λ = (173.75551812293 + 102.9372 + 0.20404629244323 + 180) mod 360 = 456.89676441538 mod 360 = 96.896764415376

Finally, calculate Jtransit:

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

Jtransit = 2457933.2149243 + (0.0053 * sin(173.75551812293)) - (0.0069 * sin(2 * 96.896764415376)) = 2457933.2171459

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 = 173.75770774322, C = 0.20397501021382, λ = 96.898882753434, Jtransit = 2457933.2171462
I2: M = 173.75770803328, C = 0.203975000771, λ = 96.898883034052, Jtransit = 2457933.2171462
I3: M = 173.75770803328, C = 0.203975000771, λ = 96.898883034052, Jtransit = 2457933.2171462

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

Jtransit = 2457933.2171462 = 06/28/2017 at 13:12:41 -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(96.898883034052) * sin(23.45)) = 23.270175034137

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(23.270175034137))/(cos(39.040759) * cos(23.270175034137))) = 111.65748716234

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

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

J** = 2451545 + 0.0009 + ((111.65748716234 + 77.04876)/360) + 6388 = 2457933.525084

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 = 2457933.525084 + (0.0053 * sin(173.75770803328)) - (0.0069 * sin(2 * 96.898883034052)) = 2457933.5273059

Now I'm going to cheat and calculate Jrise:

Jrise = Jtransit - (Jset - Jtransit)

Jrise = 2457933.2171462 - (2457933.5273059 - 2457933.2171462) = 2457932.9069866

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

Jrise = 2457932.9069866 = 06/28/2017 at 05:46:03 -0500
Jset = 2457933.5273059 = 06/28/2017 at 20:39:19 -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