NavList:

David Pike proposed a problem to find latitude from two star altitudes only: Pollux, declination 27°57.8’, SHA 243°16.0’, Ho 62°54.4’, and Capella, Dec 46°1.6’, SHA 280°20.6’ and Ho 57°12.0’. As several have pointed out, this problem has two solutions, as the circles of equal altitude for each star on the earth intersect in two locations. The Longitudes of the intersections depend upon the time - not specified - so the Longitudes are not determined.

To find those solutions for Latitude we have to do a little math. We start with the formula for the Navigational Triangle:
  sin(Hc) = sin(Lat)sin(Dec) + cos(Lat)cos(Dec)cos(LHA). 
Substituting in the values for sin(Hc), sin(Dec) and cos(Dec) for 
Pollux and after a little rearranging we get:
  cos(LHA) = (0.89026 - 0.46890sin(Lat))/0.88325cos(Lat)
This is an equation with two unknowns, LHA of Pollux (LHAP) and Latitude.

We can write a similar equation for Capella. In addition, there is a relationship between the LHA’s of Pollux and Capella: If the Longitude of the intersection of the circles is between the GHA’s of the stars, then the LHAP + LHAC = SHA(Capella) - SHA(Pollux) = 37.0766 degrees. If the Longitude of the intersection (solution) is outside that range, then the absolute value of the difference of LHA’s is equal to 37.0766 degrees.

This is a set of simultaneous transcendental equations and it has to be solved numerically. I developed a simple program for my calculator to generate a little table of Latitudes with their corresponding LHAP, LHAC, LHAP + LAHC, and LHAC - LHAP. The solutions are found by inspection of the table. You can expand the table around the Latitudes where the LHAC +LHAP or the LHAC - LHAP is near 37.0766 to determine the Latitude of the solution to any required accuracy.

The solutions I found were:
  Latitude 14.707 °, LHAP 25.4800°, LHAC 11.5956° and
  Latitude 53.1655°, LHAP 13.4613°, LHAC 50.5314°
You know which solution you are at by whether you are wearing a bathing suit or a down parka.

The approach described here seems pretty straightforward, but I had a few false starts. In particular, the rough sketch I made of the relative star positions and the circles of equal altitude indicated that the intersections were both between the GHA’s of the stars. Turns out, it is hard to sketch less than great circles on a sphere on a flat sheet of yellow paper. When I couldn’t find the northern solution, I had to re-think the possible relationships between the LHA’s and realize that LHAC - LHAP = 37.0766 would also give a solution.

Wendel Brunner