What am I looking for?

from any point Latitude, Longitude and date D

I want to know where the sun is in the sky at any given time anywhere on our planet. I even don't know what the outcome will be for this calculation. I Imagine an angle and a height.

Can someone send me on the right path for doing this?

I am not looking for a quick formula that let's me calculate this bur rather understand the components :)


1 Answer 1


There are many resources that will let you compute the position of the Sun in a coordinate system with its center at the center of the Earth. The answer will be an angle above or below the equator, an angle eastward from the vernal equinox (first point of Aries, and a distance. Books such as Astronomical Algorithms 2nd ed. by Jean Meeus have information on how to transform such a position to a horizon system centered at an observer on the surface of the Earth. The answer will be an angle from the north, an angle above the horizon, and the distance. Wikipedia has articles about software libraries that will do this kind of calculation, such as "SOFA (astronomy)" and "Naval Observatory Vector Astrometry Subroutines". These subroutines are quite complicated because they are meant to be suitable for the most critical work, with angles accurate to a few milliarcseconds. The Meeus formulas are much simpler.

  • 2
    $\begingroup$ I am not yet ready to accept your answer yet. I am looking at Naval Observatory Vector Astrometry Subroutines right now as I can read C like English and I need this to build a simulation of the most important celestial bodies (sun, moon, planets and the visible stars) for archaeological uses, I thank you for already pointing me in the right direction. $\endgroup$
    – Napivo
    Aug 31, 2015 at 21:53
  • $\begingroup$ Your question implies an observer who can see the star and write down an observation, with a date that can be converted into a calendar that we understand. In considering the calendar conversion, read up on Delta T. If you deal with prehistoric times, the uncertainty in the astronomy data increases. $\endgroup$ Sep 2, 2015 at 12:13

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