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    Re: Averaging
    From: Herbert Prinz
    Date: 2004 Oct 21, 16:17 -0400

    Bill has brought to my attention that the table that I posted earlier may be misleading
    in so far as it does not directly answer the question he asked, namely, "what is the
    maximum error resulting from linear averaging of a given data set of altitude
    I had tabulated the difference between the arithmetic mean of two altitudes and the
    actual altitude at the middle of the interval. In other words, the maximum error due to
    linear interpolation. My purpose was to demonstrate the non linearity of the curve. To
    find the answer to Bill's question, you have to apply the table in the data to your
    actual sample. For instance, if you take three sights, each 2 minutes apart, you have 2
    sights that each contribute an error of 14" and one that contributes 0". So the total
    error is 66% of what you find in column 4 m.
    For this purpose, the table was incomplete as it lacked the column for 2 minutes
    separation; so I add it here (and remove the one for 5).
       Alt             2m             4m             6m
    60 deg         0'03"          0'14"          0'31"
    75 deg         0'07"          0'30"          1'06"
    80 deg         0'11"          0'45"          1'40"
    85 deg         0'22"          1'30"          3'21"
    Now, if you take five sights in 1 minute intervals, you add the entries in the
    appropriate line from columns 2m and 4m  and multiply with 0.4 to find the error. For
    seven sights you add the entries from all three columns and multiply with 2/7 or 0.3.
    But remember, this is the MAXIMUM error. How bad it really is, say 30 deg of the
    meridian, I addressed in my previous message.
    Herbert Prinz

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