I'm trying to calculate the solar declination at the solstices, solar altitude at solar noon, and hour angles at sunrise and sunset to build a standing stone calendar at latitude 53.8 degrees South.

Problem is: I have an axial tilt of 14.92 degrees, and I can't find any resources for calculating solar position from the surface of bodies that have different obliquities.

Other than my axial tilt, my planet's orbital characteristics are similar to Mars in some ways and Earth in others. It has a 24.02 hour day and orbits at 1.55 AU.

Given axial tilt, local solar day, and latitude of the observer, is this enough to determine hour angle h0 at the very least?


<3 R


1 Answer 1


Upon re-reading the equations, I have discovered that the h0 is the angle from solar noon and that therefore my initial evaluation of the equation makes more sense than I thought. Add and subtract the result of the following equation from 90 degrees to get the h0 or hour angle of sunrise and sunset.

$\cos \omega _{\circ }={\dfrac {\sin(-0.83^{\circ })-\sin \phi \times \sin \delta }{\cos \phi \times \cos \delta }}$

where lowercase omega sub zero is the hour angle of sunrise and sunset;

lowercase phi is the northern latitude (southern latitudes are evaluated as negative);

lowercase delta is the declination of the sun.


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