Sunrise equation

Suppose ${\displaystyle \phi _{N}}$ is a given latitude in Northern Hemisphere, and ${\displaystyle \omega _{\circ N}}$ is the corresponding sunrise hour angle that has a negative value, and similarly, ${\displaystyle \phi _{S}}$ is the same latitude but in Southern Hemisphere, which means ${\displaystyle \phi _{S}=-\phi _{N}}$, and ${\displaystyle \omega _{\circ S}}$ is the corresponding sunrise hour angle, then it is apparent that

${\displaystyle \cos \omega _{\circ S}=-\cos \omega _{\circ N}=\cos(-180^{\circ }-\omega _{\circ N})}$,

which means

${\displaystyle \omega _{\circ N}+\omega _{\circ S}=-180^{\circ }}$.

The above relation implies that on the same day, the lengths of daytime from sunrise to sunset at ${\displaystyle \phi _{N}}$ and ${\displaystyle \phi _{S}}$ sum to 24 hours if ${\displaystyle \phi _{S}=-\phi _{N}}$, and this also applies to regions where polar days and polar nights occur. This further suggests that the global average of length of daytime on any given day is 12 hours without considering the effect of atmospheric refraction.

${\displaystyle n=\lceil J_{\text{date}}-2451545.0+0.0008\rceil }$

where:

${\displaystyle n}$ is the number of days since Jan 1st, 2000 12:00.

${\displaystyle J_{\text{date}}}$ is the Julian date;

2451545.0 is the equivalent Julian year of Julian days for Jan-01-2000, 12:00:00.

0.0008 is the fractional Julian Day for leap seconds and terrestrial time (TT).

TT was set to 32.184 sec lagging TAI on 1 January 1958. By 1972, when the leap second was introduced, 10 sec were added. By 1 January 2017, 27 more seconds were added coming to a total of 69.184 sec. 0.0008=69.184 / 86400 without DUT1.

The ${\displaystyle \lceil \cdot \rceil }$ operation rounds up to the next integer day number n.

${\displaystyle J^{\star }=n-{\dfrac {l_{w}}{360^{\circ }}}}$

where:

${\displaystyle J^{\star }}$ is an approximation of mean solar time at integer ${\displaystyle n}$ expressed as a Julian day with the day fraction.

${\displaystyle l_{\omega }}$ is the longitude (west is negative, east is positive) of the observer on the Earth;

${\displaystyle M=(357.5291+0.98560028\times J^{\star }){\bmod {3}}60}$

where:

M is the solar mean anomaly used in the next three equations.

${\displaystyle C=1.9148\sin(M)+0.0200\sin(2M)+0.0003\sin(3M)}$

where:

C is the Equation of the center value needed to calculate lambda (see next equation).

1.9148 is the coefficient of the Equation of the Center for the planet the observer is on (in this case, Earth)

${\displaystyle \lambda =(M+C+180+102.9372){\bmod {3}}60}$

where:

λ is the ecliptic longitude.

102.9372 is a value for the argument of perihelion.

${\displaystyle J_{transit}=2451545.0+J^{\star }+0.0053\sin M-0.0069\sin \left(2\lambda \right)}$

where:

J_transit is the Julian date for the local true solar transit (or solar noon).

2451545.0 is noon of the equivalent Julian year reference.

${\displaystyle 0.0053\sin M-0.0069\sin \left(2\lambda \right)}$ is a simplified version of the equation of time. The coefficients are fractional days.

${\displaystyle \sin \delta =\sin \lambda \times \sin 23.4397^{\circ }}$

where:

${\displaystyle \delta }$ is the declination of the sun.

23.4397° is Earth's maximum axial tilt toward the sun [3]

This is the equation from above with corrections for atmospherical refraction and solar disc diameter.

${\displaystyle \cos \omega _{\circ }={\dfrac {\sin(-0.833^{\circ })-\sin \phi \times \sin \delta }{\cos \phi \times \cos \delta }}}$

where:

ω_o is the hour angle from the observer's meridian;

${\displaystyle \phi }$ is the north latitude of the observer (north is positive, south is negative) on the Earth.

For observations on a sea horizon needing an elevation-of-observer correction, add ${\displaystyle -1.15^{\circ }{\sqrt {\text{elevation in feet}}}/60}$, or ${\displaystyle -2.076^{\circ }{\sqrt {\text{elevation in metres}}}/60}$ to the −0.833° in the numerator's sine term. This corrects for both apparent dip and terrestrial refraction. For example, for an observer at 10,000 feet, add (−115°/60) or about −1.92° to −0.833°.[4]

${\displaystyle J_{rise}=J_{transit}-{\dfrac {\omega _{\circ }}{360^{\circ }}}}$

${\displaystyle J_{set}=J_{transit}+{\dfrac {\omega _{\circ }}{360^{\circ }}}}$

where:

J_rise is the actual Julian date of sunrise;

J_set is the actual Julian date of sunset.