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Re: Averaging
From: Bill B
Date: 2004 Oct 20, 14:13 -0500
From: Bill B
Date: 2004 Oct 20, 14:13 -0500
>> In the worst case scenario, what would be the difference between the >> averaged Ho and an Ho from from an observation taken at the average time >> used? > The following table shows the maximum difference near meridian transit between > the arithmetic mean of two sun altitudes separated by n minutes, and the actual altitude > at half time: > > Alt 4m 5m 6m > 60 deg 0'14" 0'21" 0'31" > 75 deg 0'30" 0'46" 1'06" > 80 deg 0'45" 1'10" 1'40" > 85 deg 1'30" 2'20" 3'21" Found Herbert's theoretical figures a wee-bit high (on averaging error due to treating a non-linear function as a linear function) to pass my common-sense test. But in the words my former Indiana Congressman Earl Landgrebe on Richard Nixon's problems, "My mind is made up. Don't confuse me with the facts :-) Forced to consider the facts by Herbert, computed Hc at LAN for the Sun at various latitudes on the dates of equinox and solstices, and took their 2-minute differences. NOTE: I did not go as far as picking the exact longitude for the GMT they occurred, but rather used 000 longitude. I still find his numbers too high. Following are the differences between Hc at 2 minutes before LAN and Hc at the exact meridian passage. N Lat 22/06/04 22/09/04 21/12/04 00 .3' 25.8' .3' 10 .5' .7' .3' 20 1.9' .3' .2' 23.5 23.2' .2' .2' 30 .9' .2' .2' 40 .3' .1' .1' 50 .2' .0' .1' 60 .1' .0' .1' 70 .1' .0' ---- 80 .0' .0' ---- 90 .0' .1' ---- Example average for N 20 with HC approx.86.5d, difference of 1.9': 22 June, 2004, N 20, Hc dif 1.9' 11:59:50 86d 31.7' 12:00:50 86d 33.1' 12:01:50 86d 33.6' 12:02:50 86d 33.1' 12:03:50 86d 31.7' Average: Averaged 12:01:50 Hc 86d 33.6' Averaged 12:01:50 Hc 86d 32.6' Shift 1.0' Not confining the exercise to meridian passage, we could expect altitude differences up to 30' in 2 minutes or 1d in 4 minutes. 22 September, 2004 Lat 00, Lon 000, 17:09-17:13 GMT Time Hc 17:09:00 010d 52.6' 17:10:00 010d 37.6' 17:11:00 010d 22.6' 17:12:00 010d 07.6' 17:13:00 009d 56.6' 4-mintute difference: 59.8' Actual 17:11:00 Hc 010d 22.6' Average 17:11:00 Hc 010d 23.4' Shift from averaging 000d 00.8' What did this beginner learn from the exercise? * Alex and Herbert are correct,the error is more pronounced for high altitude bodies, and they are unacceptable targets for averaging except for all but last-ditch efforts, especially at meridian passage; even if double-second-differences and problems with the operator holding the sextant vertical are discounted. * Shy away from averaging a body with a declination within 10d-20d of the observer's latitude. (If it would have trouble casting a significant shadow of the mast on the deck, leave it alone. This is cel nav SOP. * Herbert could have conjured up a more extreme LAN "worst-case scenario" had he chosen. O declination on the equator at an equinox, or N 23d 26.4'd at the summer solstice, but did not. His answer was, however, not within the confines of the stated hypothetical question--5 sights averaged--Ha within altitude 30d-70d. Having attempted to establish a benchmark, I leave it to the statisticians to elucidate me. Bill > > I mention the worst case scenario just for fun: If the sun goes through the > zenith.... > > The following table shows the maximum difference near meridian transit between > the arithmetic mean of two sun altitudes separated by n minutes, and the actual altitude > at half time: > > Alt 4m 5m 6m > 60 deg 0'14" 0'21" 0'31" > 75 deg 0'30" 0'46" 1'06" > 80 deg 0'45" 1'10" 1'40" > 85 deg 1'30" 2'20" 3'21" > > You see that in most 'normal' situations there is not much of an error, but it > becomes worse RAPIDLY with increasing altitude as well as with increasing time > span.