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Re: Lunar trouble, need help
From: Kent Nordstr�m
Date: 2008 Jun 29, 21:47 +0200
From: Kent Nordstr�m
Date: 2008 Jun 29, 21:47 +0200
George Huxtable wrote [5582]: I didn't understand all that (and still don't) and asked for some relevant numbers to help me out. On the other hand, perhaps it's better if I explain how I would do that job myself, with numbers, to show how simple it can be. I have taken another look on the LD case given by Jeremy. I used the data given by George and put them into my own model to see if any difference between George�s and mine could be seen. Accordingly I have added the Moon�s SD and subtracted the suns SD when reducing th distance. So these are the data used as input: Height a.s.l. 32,29m Air pressure: 757,5 mm Hg Air temp: 36,55 degr. Celsius Moon observed UL: 61d 05m 16,44s Sun observed LL: 33d 07m 23,16s Distance observed: 86d 10m 18,2s My model gave the following result: Moon true altitude: 61d 05m 48,1s (George 61d 06m 51,5s) Sun true altitude: 33d 11s 42,8s (George 33d 11m 45s) Suns decl: 23d 02m 39,7s (George 23d 02,7m) d: 86d 09m 57,9s (George 86d 10m 11,3s) D: 85d 44m 49,2s (George 85d 44m 18,2s) First GMT: 06-25-54,2 Correction for 2nd difference: -2,78s GMT incl. correction for 2nd diff: 06-25-51,4 GHA: 276d 35m 44s (George 276d 20,9m) LHA: 59d 56m 50,4s (George 59d 56m 47,8s) MT (with the complicated way via Aries, RA): 16-03-12 MT-GMT=9-37-21 corresponding to E 144d 20,3m (George 143d 41,9m) (note that the ship steered 270d and we do not know the reckoned position at the LD obs. � only the fix before) As can seen from this my deviation from the reckoned position is 16m to east (George 22m to west). With this I dare say that there is nothing bad with my model even if there are different opinions how to model various conditions. Any model tries to reproduce reality but does not do it exactly. The very question is: What is right? Should be nice to know if Jeremy could provide a better opinion about his position at 06-24 on the 10th June. George claims: Instead of using the Sun's altitude, we could just as well have used the Moon's, which was measured for the same moment, and a similar calculation for the Moon tells me that Moon LHA = 26.9243�, and resulting WGL = 216.3176�. that also the moon could be used for finding the longitude. In theory this is of course possible. The way to calculate is the same. The result from my model using the moon is: LHA: 26d 56m 29,3s (George 26d 55m 27,5s) MT: 16-01-31,5 This means that 16-01-31,5 � 06-25-51 = 9-35-40 or equal to E 143d 55m (George 143d 40m 56,5s). Being a professional navigator myself (long time ago) I would say that the moon should be avoided for finding the MT and longitude due to its un-regular movements. Only if no other means are available the moon shall be used, then preferably in or close to the prime meridian. Kent N ----- Original Message ----- From: "George Huxtable"To: Sent: Wednesday, June 25, 2008 3:57 PM Subject: [NavList 5582] Re: Lunar trouble, need help Let's recap, a bit. This thread started with a message about a lunar distance from Jeremy [5413], the important bits being summarised below- "Here is my data around 0620 on 10 June 2008. 1) Sun LL: Hs 33deg 58.5� @ 06h 19m 13s UTC 2) Moon UL: 60deg 36.8� @ 06h 20m 44s 3) LD1: 86deg 10.3� @ 06h 21m 40s 4) LD2: 86deg 10.0� @06h 22m 20s 5) LD3: 86deg 10.2� @06h 22m 52s 6) LD4: 86deg 10.6� @06h 23m 24s 7) LD5: 86deg 10.6� @06h 24m 39s 8) Sun LL: 32deg 35.8� @ 06h 25m 19s 9) Moon UL: 61deg 49.2� @ 06h 26m 28s DR (didn�t take a fix) Latitude was 15deg 14.0�N and Longitude was 144-04�E. Ship was on course 270 at 12.0 knots. IE was 0.0 T/P was 98deg F and 1010MB. Height of eye is 106 feet. I averaged the two sun lines to get 33deg 17.2� at 06h 22m 15s, then the two moon altitudes to get 61deg 13.0� at 06h 23m 37s. Finally the 5 LD�s averaged to 86deg 10.34� at 06h 22m 59s." ============================= Frank suggested that that overlapping, rather than side-by-side, images of Sun and Moon had been measured, and I recalculated on that basis, arriving [5530] at a deduced GMT, from lunar distance, of 6h 24m 53s, which differs little from Jeremy's corresponding chronometer time of 6h 22m 59s GMT. Kent Nordstrom has questioned some aspects of that calculation, but I hope that these have now all been resolved, except that Kent seems unwilling to accept the reality of those overlapping images. Unless he does so (or something similar), then he will be faced with an unbridgeable error of 1 Sun diameter in the lunar distance, about 1 hour in the resulting GMT, and about 15� in the deduced longitude. But that's a matter for him. What remains in question is Kent's method of finding longitude from the altitude of an observed body (or two bodies), once the GMT had been established. He wrote, in [5577] - "My approach for calculating the MT is as I understand the very same as used in the old days. This is of course for finding the difference between the GMT and the MT, i.e. the longitude. It is a separate calculation. So let me outline what I do: 1.. Find the LHA at the predicted GMT (we know (roughly) the latitide, from the generation of true distances at say time 3, time 6 etc. we can predict the declination at the GMT. True altitude is from the altitude reduction. Aries GHA to the GMT can be predicted as well). 2.. Find the Aries GHA for upper meridian passage in Greenwich, (normally one day before). 3.. Convert this GHA to time. 4.. Correct for assumed longitude and then you get the time for Aries UMP on your location. 5.. Find the difference between Aries GHA and the body�s GHA at GMT. This is the RA. 6.. Find the sum of the LHA and the RA, which is the sideral time. 7.. Subtract the value for Aries UMP at the location. Now you have the difference in sideral time. 8.. Convert this difference in sideral time and you get the MT at observation. And now you can easliy find the difference between the GMT and the MT. I hope this explaination clarifies. I should also add that one must be very observant when doing this calculation, it is easy to do wrong." ==================================== I didn't understand all that (and still don't) and asked for some relevant numbers to help me out. On the other hand, perhaps it's better if I explain how I would do that job myself, with numbers, to show how simple it can be. Since the mid-1800's mariners have increasingly accepted that sea positions can be determined from intersecting position lines from observations of several bodies, drawn on a chart, rather than calling for a calculated value of lat and long, separately, as was previously the case. This is at the heart of what became called " the new navigation", as it's learned and accepted today. And it doesn't need me to explain in detail how that's done, once GMT is known from a lunar or a chronometer. You start with a presumed position, and the Nautical Almanac tells you dec and GHA of each body according to the GMT. Then Altitude-azimuth tables, or some computer, provides calculated altitudes, which you compare with measured altitudes, work out an intercept and azimuth for each body from the presumed position, draw in position lines, see where they cross, and the job's done, providing both lat and long. That could be done for Jeremy's Sun and Moon altitudes, using for GMT either his chronometer GMT, or the GMT derived from his lunar. (Being nearly 2 minutes different in time, these will provide longitudes that differ by about 30 arc-minutes.) Those position lines will both lie nearly North-South, so they will cut at a close angle, and latitude won't be nearly so well defined as is longitude. But there's an earlier way to do the job, which doesn't call on position lines or the New Navigation. It relies on the notion that, however difficult it may be for a navigator to determine longitude (without a chronometer), he always knows his latitude. That may or may not be the case in practice, depending on how clear his skies have been recently. Anyway, that's the assumption, and in this case we can take Jeremy's latitude as being his assumed lat. of N15� 14' (= +15.2333), and try to determine his longitude, on that basis. Closing our minds to the chronometer reading itself, we can say that [see 5130] according to his lunar, at a GMT of 6h 24m 53s, the measured corrected Sun altitude was 33� 13.1' or 33.2189� and for that GMT the 2008 Nautical Almanac predicted on 10 June Sun dec = N 23� 02.7' (= +23.0450�), and Sun GHA = 276� 20.9' (=276.2483�). Next we need to calculate the Local Hour Angle of the Sun at that moment, using the simple spherical triangle formula cos (HA) = (sin alt - sin lat sin dec) / (cos lat cos dec), which tells us that the Local Hour Angle of the Sun at that moment was + / - 59.9466�. All this, as far as I can tell, is what is covered in step 1 of Kent's suggested procedure. All the additional steps in that procedure can be removed, and replaced by the following- The Sun had by then long passed the meridian and was sinking in the sky so we therefore have to subtract Local Hour Angle of the Sun from its Greenwich Hour Angle, to get the observer's Westerly Greenwich Longitude (WGL), which ends up at +216.3017� or +216�18.1. Those who prefer a different convention can subtract that from 360� to get longitude = E 143�41.9. That's it! Job done! You can see that by using the GHA provided by a modern almanac, there's been no need to invoke sidereal time, Right Ascension, or Aries. ========================== Instead of using the Sun's altitude, we could just as well have used the Moon's, which was measured for the same moment, and a similar calculation for the Moon tells me that Moon LHA = 26.9243�, and resulting WGL = 216.3176�. 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. --~--~---------~--~----~------------~-------~--~----~ Navigation List archive: www.fer3.com/arc To post, email NavList@fer3.com To , email NavList-@fer3.com -~----------~----~----~----~------~----~------~--~---