# NavList:

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

**Re: Direct methods**

**From:**George Huxtable

**Date:**2007 Nov 4, 11:55 -0000

Responding to my comment in 3760- | > Equations 7.5a to 7.5d take you to the latitude | > of the crossing; indeed, the two latitudes, that you have to choose between, | > for the two crossings. No analysis will choose for you which of those two | > you are at Ronald van Riet replied- | For starters, if you don't have a rough idea where you are, you are a | very poor navigator, so taking your dead reckoning position wil be | sufficient for eliminating ambiguity. Yes, that's exactly why I added the words (which Ronald omitted) "..., and it's hardly ever a problem to do so." | If you want a formal way of doing so, shoot a third star. | | IMHO there is no mathematical way to take away the ambiguity with just | the two star observations, just as it is impossible to define a plane | by just two points. Yes, of course, that's the case. Two position circles, if they intersect at all, intersect twice. There is no analysis of the numbers that will decide between the two solutions. Which was exactly what I wrote, above. That ambiguity is unavoidable. It can only be resolved by the navigator using commonsense, knowing roughly where he is. I doubt if Ronald and I disagree at all about any of that. The point I was trying to make, and which I suspect Ronald might have missed, is this. The equations (7.5e, 7.5f) that have been provided in the book to determine longitude from cos LHA1, add an EXTRA level of ambiguity to the unavoidable one that Ronald and I have discussed above. For EACH of the two unavoidable ambiguous latitudes those equations produce TWO possible longitudes. And it isn't always easy to determine which of those longitudes to choose, by commonsense. It gets particularly difficult if the azimuth of a star is anywhere near North or South of the observer. But THAT ambiguity IS possible to resolve, and without needing to refer to any DR position. The procedure described in my earlier attachment on intersecting circles is one way to do it, but it's not very elegant, and I would be interested to learn of a better one. How it works is this- First take one of the two possible values for latitude that have been determined, say for the northernmost of the two intersections. Draw (in your mind) that line of latitude. We know that it intersects each position circle twice, and we can calculate the four longitudes of those intersection points, using an equation such as 7.5f, applied to EACH circle. Remember that when 7.5f calculates LHA1 from its cosine, that has two solutions, equal positive and negative angles. But one of those longitudes, from circle 1, will be exactly equal to one of those longitudes, from circle 2. That must be the longitude of the wanted common intersection, between the two circles and the line of latitude. The other longitudes are discarded. Next, take the other possible answer for latitude, and calculate the longitude that corresponds to that, in just the same way. I hope that Ronald will re-read my posting and perhaps think again. George. contact George Huxtable at george---.u-net.com or at +44 1865 820222 (from UK, 01865 820222) or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK. --~--~---------~--~----~------------~-------~--~----~ To post to this group, send email to NavList@fer3.com To unsubscribe, send email to NavList-unsubscribe@fer3.com -~----------~----~----~----~------~----~------~--~---