I have access to "rise" and "set" times for astronomical objects, and want to determine the corresponding times of culmination. Is there a reliable and accurate way to do this? It isn't clear to me from the sources of the data I have whether the times I have are for the geometric centers of the objects, or take into account refraction, or the angular extents of the objects, so I'm hoping there is an approach that is insensitive to these factors. In particular, I'm interested in transit times for the Sun.


2 Answers 2


To a very good approximation, the transit time can be estimated as the arithmetic mean of the rise and set times. However, this will not work accurately for rapidly moving objects (e.g. Moon). I followed the link in your question not realizing it was in the Mathematica forum and I'm not familiar with how Mathematica computes these times. However, I am willing to bet that atmospheric refraction is accounted for in the traditional way, that is by adding 34 arc minutes to the object's altitude as rise/set. Refraction acts to increase an object's altitude, but the effect is zero at the zenith. Anyway, if refraction is accounted for, the rise/set times should be for either the center of the object's disk or the upper limb of the object's disk. There are two conventions used here. In the U.S. all sunrise/sunset/moonrise/moonset times are for upper limb of Sun's/Moon's disk. However, most European almanacs tabulate these times for the center of Sun's/Moon's disk. All other celestial objects are too small to show a naked eye disk, so the rise/set times refer to the disk's center.

  • $\begingroup$ How close an approximation is the mean though? For "fixed" stars it will be close, but for the Sun, the rate of change of azimuth varies with time (due to the eccentricity of the Earth's orbit, mostly; and to the fact that angle between the ecliptic and the horizontal system is changing), so the time it takes to traverse azimuths less than 180° will (almost) never be precisely the time it takes to traverse azimuths greater than 180°. (Right?) $\endgroup$
    – orome
    Oct 22, 2012 at 20:00
  • $\begingroup$ To give a sense of the accuracy I need: I want to use the solar transit times to calculate the difference in the RA of the sun between successive (actual) solar days. (Among the things I'll be doing is comparing those values the the corresponding values for successive mean solar days.) $\endgroup$
    – orome
    Oct 22, 2012 at 20:18
  • $\begingroup$ If you want to program the computation yourself, then I highly recommend a copy of Astronomical Algorithms by Jean Meeus (2nd edition, Willmann-Bell, 1998) for the algorithms. My own book, Fundamental Ephemeris Computations (Willmann-Bell, 1999) does too, and includes source code in C. Any modern desktop (or tablet or smartphone) planetarium app should give quite accurate results. Finally, solar-noon.com and esrl.noaa.gov/gmd/grad/solcalc and usno.navy.mil/USNO/astronomical-applications/data-services are all good, especially the third one. $\endgroup$
    – user11266
    Oct 22, 2012 at 21:53
  • $\begingroup$ Thanks; good sources. Mostly I'm trying (hoping) to use the tools I have at hand without writing any more code myself than is necessary, but that tool (Mathematica) seems, the more I learn, to be unreliable and inaccurate (not to mention cumbersome to use), and I may just toss it. $\endgroup$
    – orome
    Oct 22, 2012 at 21:59

To determine transit UTC transit time of a celestial object;

(360 plus your Longitude) minus star SHA= GHA Aries

Get GHA Aries from The Nautical Almanac here; www.TheNauticalAlmanac.com

To determine transit time of the Sun on your meridian (Longitude);

Sun GHA = your Longitude.

Let's say you're located at W 078 degrees 42 minutes. Date- November 18, 2015.

The Sun will be on your meridian at about UTC 17:00

  • $\begingroup$ Hello, and welcome to Physics Stack Exchange. This looks like the beginnings of a good answer, but it's a bit confusing because of the organization and formatting. Cleaning it up would make it more likely to be up-voted and accepted. $\endgroup$ Nov 19, 2015 at 1:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.