# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Sight Reduction Formula Question**

**From:**Geoffrey Kolbe

**Date:**2005 Jan 8, 09:25 +0000

George Huxtable wrote: ========================== >Now consider the other formula, deriving Z from its sine. If, in this case, >that formula has given us sin Z = +0.98, what do we do then? Well, one >solution is that Z = 88.5 deg, which might well be the right answer. On the >other hand, sin 101.5 deg is also +0.98, so that's an equally valid answer. >It could be either, then. How do we discover which is the one we want? In >this case, there's no simple way, that I know of, to resolve the ambiguity >by using inspection and logic to determine whether the wanted result is >just North of East or just South of East. =========================== Geoffrey Kolbe comments: The attraction of creating two right angled triangles from the navigational triangle and using sines and cosines to solve for Hc and Z is that only one table is required. In Ageton's method, for example, a perpendicular is dropped from the geographical position of the celestial body to the observer's meridian to create the two right angled triangles. Ageton's tables are just 36 pages short. For the sight reduction part of my own "Long Term Almanac 2000 - 2050" I use what is essentially the Ageton method, (but keep the sines and cosines instead of inverting them to create secants and cosecants as Ageton did), and the tables occupy just 18 pages. George correctly identifies the choice of azimuth quadrant as being a potential problem where the celestial body lies near the observer's Prime Vertical. The ambiguity should only arise where the declination of the celestial body is less than the observer's estimated latitude, both are in the same North-South Hemisphere (have the 'same name') and where the LHA is less than 90 degrees or more than 270 degrees. Otherwise, as George says, the choice of azimuth quadrant is easily resolved by inspection. Where the above conditions apply and there is an ambiguity, it turns out that there is a simple test to resolve this problem. One of the angles it is necessary to calculate to derive Hc is the co-latitude for the intersection of the perpendicular dropped to the observer's meridian. If the latitude of this intersection is greater than the observer's estimated latitude, the azimuth quadrant will be a Northerly one where the positions are in the North, and vice versa. Geoffrey Kolbe.