In your face sunset/sunrise calculator

New! Enter your latitude and longitude.

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: 2455265

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

n_* = (J_date - 2451545 - 0.0009) - (l_w/360)
n = round(n_*)

n_* = (2455265 - 2451545 - 0.0009) - (77.04876/360) = 3719.7850756667
n = round(3719.7850756667) = 3720

Now J_*:

J_* = 2451545 + 0.0009 + (l_w/360) + n

J_* = 2451545 + 0.0009 + (77.04876/360) + 3720 = 2455265.2149243

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 * (2455265.2149243 - 2451545)] mod 360 = 4024.1739710829 mod 360 = 64.173971082933

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(64.173971082933) + 0.0200 * sin(2 * 64.173971082933) + 0.0003 * sin(3 * 64.173971082933) = 1.7391717149522

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

λ = (M + 102.9372 + C + 180) mod 360

λ = (64.173971082933 + 102.9372 + 1.7391717149522 + 180) mod 360 = 348.85034279789 mod 360 = 348.85034279789

Finally, calculate J_transit:

J_transit = J_* + (0.0053 * sin(M)) - (0.0069 * sin(2 * λ))

J_transit = 2455265.2149243 + (0.0053 * sin(64.173971082933)) - (0.0069 * sin(2 * 348.85034279789)) = 2455265.2223131

Now, to get an even more accurate number, recursively recalculate M using J_transit until it stops changing. Notice how close the approximation was.

I1: M = 64.181253498592, C = 1.7392744584796, λ = 348.85772795707, J_transit = 2455265.2223118
I2: M = 64.181252165325, C = 1.7392744396719, λ = 348.857726605, J_transit = 2455265.2223118
I3: M = 64.181252165784, C = 1.7392744396784, λ = 348.85772660546, J_transit = 2455265.2223118
I4: M = 64.181252165784, C = 1.7392744396784, λ = 348.85772660546, J_transit = 2455265.2223118

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

J_transit = 2455265.2223118 = 03/09/2010 at 12:20:07 -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(348.85772660546) * sin(23.45)) = -4.4105117045377

Now we can go about calculating H (Hour angle):

H = arccos( [sin(-0.83) - sin(l_n) * sin(δ)] / [cos(l_n) * cos(δ)] )

H = arccos((sin(-0.83) - sin(39.040759) * sin(-4.4105117045377))/(cos(39.040759) * cos(-4.4105117045377))) = 87.487115710637

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

J_** = 2451545 + 0.0009 + ((H + l_w)/360) + n

J_** = 2451545 + 0.0009 + ((87.487115710637 + 77.04876)/360) + 3720 = 2455265.4579441

We can use M from above because it really doesn't change that much over the course of a day, calculate J_set in the same way:

J_set = J_** + (0.0053 * sin(M)) - (0.0069 * sin(2 * λ))

J_set = 2455265.4579441 + (0.0053 * sin(64.181252165784)) - (0.0069 * sin(2 * 348.85772660546)) = 2455265.4653316

Now I'm going to cheat and calculate J_rise:

J_rise = J_transit - (J_set - J_transit)

J_rise = 2455265.2223118 - (2455265.4653316 - 2455265.2223118) = 2455264.979292

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

J_rise = 2455264.979292 = 03/09/2010 at 06:30:10 -0500
J_set = 2455265.4653316 = 03/09/2010 at 18:10:04 -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.