# NavList:

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

**Re: Latitude by Lunar Distance**

**From:**George Huxtable

**Date:**2006 Oct 7, 11:23 +0100

Peter Fogg has put his finger on the principle involved, without being aware of it. Yes, if parallax and refraction are ignored, the separation from any point on Earth will be exactly the same, and should correspond to the predicted separation, calculated from the positions of the two bodies taken from the almanac, at a known GMT. It's on that parallax-and-refraction difference that the proposed method depends. That discrepancy, between that measured separation and the predicted separation, will depend on the parallax and the refraction (and the azimuth angle between the bodies). If a best-case situation is chosen (and there are plenty of stars to choose from) the azimuths of the two bodies will be closely the same, or nearly opposite. In that case, any arc-minute change in the Moon's parallax will be reflected in a discrepancy of a similar amount, between the observed distance and the almanac distance at that same moment. The Moon's parallax can vary, according to Cos (Moon altitude) x Moon HP. The Moon's HP (Horizontal Parallax, the parallax when the Moon is on the horizon, at zero altitude) is always somewhere near 1 degree. Let's take it to be exactly 1 degree. For example, take the situation when the Moon is high, say at 80 degrees, for which you would need to be near the tropics. That's when the proposed method would work best, because parallax is then changing most quickly with altitude when the Moon is highest. Then the parallax for 80 degrees alt is cos 80 x 1 degree =0 .1736 deg. or 10.41 minutes For 81 degrees it would be 0.1564, or 9.36 minutes, so just over 1 minute difference, for an altitude change of 1 degree. That means, if you can deduce the parallax to 1 arc-minute, then at those high Moon altitudes you can deduce the Moon altitude to the nearest degree, If you can do better with the parallax, by measuring lunar distance more precisely than 1 minute, you can determine Moon altitude correspondingly better. And this without being able to see a horizon. At lower altitudes, the method would become more insensitive. Having done that, starting from an estimated position, you can draw a position line on the chart, knowing the Moon's predicted GHA and dec, from the almanac, for that known GMT, and the deduced Moon altitude, just as if you had observed that Moon altitude, though a lot less precisely. As Alex has pointed out, even at its best, it would be twice as inaccurate as determining a lunar longitude by lunar distance without GMT, and that's difficult enough. I can see, then, how Frank's proposal could give one position line, placed at right angles to the azimuth of the Moon, on the chart, because the Moon's altitude has been deduced from its parallax. But it's only the Moon's altitude that can be deduced by this method, because it has such a large parallax. No information is provided on the altitudes of stars. I haven't yet understood Frank's explanation of how a second position line can be deduced by measuring the distance between the Moon and a second star, and, I too, ask for a more detailed explanation with a numerical example. George. contact George Huxtable at george@huxtable.u-net.com or at +44 1865 820222 (from UK, 01865 820222) or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK. ----- Original Message ----- From: "Peter Fogg"To: Sent: Saturday, October 07, 2006 1:39 AM Subject: [NavList 1373] Re: Latitude by Lunar Distance | On 10/5/06, Frank Reed wrote: | > | > | > Here's an example of fixing your position using angles measured from | > the Moon to stars only, no altitudes whatsoever. This approach assumes | > that GMT is a known quantity ... | | | If the effects of parallax and refraction are ignored, then the angular | distance between two celestial bodies is going to be identical from any | point on earth from which it can be observed at that moment? In other words; | this angular distance is invariant with respect to position? | | For this method to work the angular distance would need to change according | to the observer's position? | | Could Frank perhaps explain this process in more detail, including a fully | worked example? | | | | --~--~---------~--~----~------------~-------~--~----~ To post to this group, send email to NavList@fer3.com To unsubscribe, send email to NavList-unsubscribe@fer3.com -~----------~----~----~----~------~----~------~--~---