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    Re: Averaging
    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.
    
    
    

       
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