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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: 2458169

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

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

n* = (2458169 - 2451545 - 0.0009) - (77.04876/360) = 6623.7850756667
n = round(6623.7850756667) = 6624

Now J*:

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

J* = 2451545 + 0.0009 + (77.04876/360) + 6624 = 2458169.2149243

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 * (2458169.2149243 - 2451545)] mod 360 = 6886.3571842029 mod 360 = 46.357184202932

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(46.357184202932) + 0.0200 * sin(2 * 46.357184202932) + 0.0003 * sin(3 * 46.357184202932) = 1.4058312221553

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

λ = (M + 102.9372 + C + 180) mod 360

λ = (46.357184202932 + 102.9372 + 1.4058312221553 + 180) mod 360 = 330.70021542509 mod 360 = 330.70021542509

Finally, calculate Jtransit:

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

Jtransit = 2458169.2149243 + (0.0053 * sin(46.357184202932)) - (0.0069 * sin(2 * 330.70021542509)) = 2458169.2246492

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 = 46.366769019948, C = 1.4060518434819, λ = 330.71002086343, Jtransit = 2458169.2246486
I2: M = 46.366768409996, C = 1.4060518294435, λ = 330.71002023944, Jtransit = 2458169.2246486
I3: M = 46.366768409996, C = 1.4060518294435, λ = 330.71002023944, Jtransit = 2458169.2246486

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

Jtransit = 2458169.2246486 = 02/19/2018 at 12:23:29 -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(330.71002023944) * sin(23.45)) = -11.226520292448

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(-11.226520292448))/(cos(39.040759) * cos(-11.226520292448))) = 81.839253339693

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

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

J** = 2451545 + 0.0009 + ((81.839253339693 + 77.04876)/360) + 6624 = 2458169.4422556

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 = 2458169.4422556 + (0.0053 * sin(46.366768409996)) - (0.0069 * sin(2 * 330.71002023944)) = 2458169.4519798

Now I'm going to cheat and calculate Jrise:

Jrise = Jtransit - (Jset - Jtransit)

Jrise = 2458169.2246486 - (2458169.4519798 - 2458169.2246486) = 2458168.9973173

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

Jrise = 2458168.9973173 = 02/19/2018 at 06:56:08 -0500
Jset = 2458169.4519798 = 02/19/2018 at 17:50:51 -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