NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Lat/Lon by "Noon Sun" & The Noon Fix PROVE IT
From: George Huxtable
Date: 2009 Apr 14, 14:21 +0100
From: George Huxtable
Date: 2009 Apr 14, 14:21 +0100
Brad proposed making a simulation test of the "latitude around noon" method, and I am happy to go along with the notion. But first, we need to fix a misconception. Brad wrote- "George - Please post ONE example where this method doesn't work. You have claimed that Lat/Lon by Noon Sun Method doesn't work at high latitudes (if I read your postings correctly)." No. I haven't ever claimed it "doesn't work". I have pointed out that its accuracy is degraded under certain conditions, calling for observation over a longer time-span, so shifting it further from noon. It will be interesting to see if that degredation brings it outside the claim of 5-mile accuracy, or not. Brad continued- Fine. Suppose your DR is around the Antarctic Circle. Make the declination as north as you would like it, within reason, as nobody expects this to work if you cannot see the sun. Provide the NavList and therefore Frank with the 13 altitudes and supposed times. You may even skew the data by injecting a course and a speed, up to 20 knots, your choice. I think we should avoid anything so extreme. I've suggested, in the past taking as an example a vessel approaching the Clyde in Winter, a regular route used by thousands of vessels, and which was particularly relevant during Word-war 2, when navigation aids were unavailable and there was no alternative to celestial nav. I will take as an example a voyage in a standard steamer of that era, doing 10 knots Northwards. In midwinter, we can take the decination to be -23.4�, and ignore any changes during the observation. Near midwinter, equation of time is also small, and for simplicity we can take that to be zero, and again ignore any change during the hour of an observation. Is that reasonable? The nominal position would be, I suggest, 56�N, 10�W, and we could take the vessel as being somewhere within a box centred on that position, the box being 1� in latitude and 2� in longitude. I will simulate an hour's worth of 13 observations, at 5-min intervals, to be centred around 12:40 GMT, which would correspond to the moment of noon when seen from a vessel at the centre of the box. I can promise that its actual position, at 12:40 will be somewhere between N55� 30' and N56� 30', and W9� to 10�. That actual position, at that moment, is what's being sought. Northward component of the nominal speed is taken to be 10 knots Northwards, but with an added Gaussian scatter in the actual northward component of +/- 1 knot (1 standard deviation) to account for uncertainties in the ship's log, course, and tidal stream. That actual speed will hold constant over the 1 hour observation period. Altitude corrections for index, dip, refraction, semidiameter are ignored, presuming that any such corrections have already been made precisely, and without error. The altitudes, then, are taken to be true altitudes of the centre of the Sun. Each of the 13 altitudes is calculated according to the formula- sin alt = sin lat sin dec + cos alt cos dec cos hour-angle An observational Gaussian scatter of +/- 1 arc-minute (1 standard deviation) is then applied to each calculated altitude. Is that accepted as fair? For each "data set" will be provided a set number, followed by a string of 13 altitudes, given as decimal degrees, to 3 decimal places. For each such numbered set, I will record the initial latitude and longitude chosen (and the actual speed), but not disclose them until later. The challenge, for anyone who wishes to accept it, is to use any method you choose to determine that latitude and longitude, as well as possible. It should be done for a number of runs, in order to compare, later, with the values that until then have been withheld. The scatter in the difference can then be compared with Frank Reed's expectation, with seems to be +/- 2 miles (= 2') in latitude, +/-5 miles (= 8.9') in longitude. I can provide an endless stream of such sets if requested; please suggest a convenient format (an Excel worksheet?). Gaussian simulation will be done using the Excel function- NORMSINV(RAND()) . I have convinced myself that it supplies a string of numbers averaging zero with a Gaussian distribution having a standard deviation of 1 unit. None of this is set in stone and if anyone has any objections or suggestions, say so now. I don't regard myself as particularly adept with Excel; if anyone would like to check over the spreadsheet, it will be provided. I have no idea whether Frank's criteria will be met; if they are, that will be to the credit of the method, because this is a demanding test. George. contact George Huxtable, at george@hux.me.uk or at +44 1865 820222 (from UK, 01865 820222) or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK. --~--~---------~--~----~------------~-------~--~----~ Navigation List archive: www.fer3.com/arc To post, email NavList@fer3.com To , email NavList-@fer3.com -~----------~----~----~----~------~----~------~--~---