Expert: Philip Stahl - 8/3/2007

Question
Is it possible to find a lat/long on earth on a specific historical date given the Az and El of the sun on that day? If so, could you guide me in the formulation? Specifically, where would i be standing in The Bull run area of Virginia on 10 May 1863 at 1515 if the Az=80.24512 & El= 16.42940?

Answer
Hello,


The problem is that what you seek to find (longitude and latitude on a given date, for a given time and locale) requires working from THREE different coordinate systems (horizon, hour angle and Right Ascension).

Let's start out with your "givens": A, and a (or Az and El in your parlance).

First you need to change a (altitude) to the zenith distance z:

z =  (90 - a)

Then you would need to obtain (from the appropriate) astronomical spherical traingle, the hour angle (h) using:

tan (h) = - sin (z) sin (A)/  {cos (z)*cos (lat) - sin(z)* cos(A)* sin(lat)

You can see the basic problem immediately, to find h you need to have the cosine and sine of the latitude - but that is what you are also seeking to find. (You need the hour angle to get the Right Ascension, then the sidereral time and the longitude )

The only way to break the "loop" is if you can find the Sun's declination for that day and time. (There may be declination tables available that do that. You may inquire at the U.S. Naval Observatory. In fact they may have an entire program to do this sort of thing)

Anyway, if the Sun's declination can be obtained, one can make use of the spherical traingle formula:

sin (Decl.)  = sin (lat) cos(z) + cos (lat) sin(z) cos (A)

Then with the known values of z, A, and Decl. substituted in one can easily solve for (lat). Once this is obtained, one can obtain tan(h) above, and hence h, the hour angle.

With the hour angle obtained, the apparent sidereal time (AST) is found from :

AST =  RA + h

That is, the Sun's Right Ascension on the date plus the hour angle. How to get the Sun's right ascension? Probably the same source where you obtained its declination (to break out of the "too many unknowns" loop noted earlier!)

Assuming you can do that - you can then obtain the longitude on the date from:

L =  AST - GAST

That is, your apparent sidereal time at the location minus the Greenwich apparent sidereal time on the same day.

This is the general way the formulation would proceed, and as I noted its sucess depends on breaking out of the too few knowns knot.

A much simpler approach is just to get a suitable computer program and let it do the work for you. I believe one that did this was called, 'Skyglobe' if I am not mistaken. You may also want to check the program 'Cybersky'. Google these options and see what you come up with. Failing that, I'd contact the Naval Observatory as I am sure they have a program or two to cover your sort of scenario.

Hope this helps!