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    Re: Sun semidiameter
    From: Bill B
    Date: 2007 Apr 14, 23:08 -0400

    Alex
    
    I see your point, that the N.A. lives up to it's accuracy claim.
    
    Past that it gets fuzzy for me.  For the 23, 24, 25, August, 2007 pages N.A.
    SD is 15!8.  For the 26, 27, 28 August, 2007 pages N.A. SD is 15!9.  So it
    is safe to assume that mean SD will be 15!85 somewhere between the 25th and
    28th.
    
    Yet we see SD's from other sources of:
    
    23 Aug, 2007   15.82
    24 Aug, 2007   15.83
    25 Aug, 2007   15.83
    26 Aug, 2007   15.83
    27 Aug, 2007   15.84
    28 Aug, 2007   15.84
    
    If we take the average of the 23,24.25 page, 15.83
    If we take the average of the 26, 27, 28 page, 15.84
    
    No way either rounds up to 18.9.
    
    --My $10 pocket calculator claims to display 10 digits but works to 11
    digits and rounds the last up at 5.
    
    --My version of Excel claims to work to 13 figures and drops the last
    digit with no attempt to round up or down.
    
    --Even if, as you surmise, the N.A. may drop the last digit the result is
    still 15.8, not 15.9 based on Frank's numbers, and surveying numbers down to
    tenths of a second.
    
    I suspect something else is at work here, perhaps the N.A. algorithm?
    
    Bill
    
    
    
    Bill wrote
    
    > for 14 May 2007 I find agreement in the semidiameter of the sun; 15!83
    > (Frank) and  15' 49.6" = 15!83 (Cadastral).  Yet the nautical almanac lists
    > SD as 15.9.  I see this happening on other dates as well.  I would expect it
    > could happen near a rounding, point, say 15!9 one page and 16!0 the next,
    > but that thought does check out.
    >
    > Why the difference in SD?
    
    Alex Replied:
    
    The date May 14 is actually close to the date
    May 15/16 where the almanac data changes to 15.8
    The explanation in the Almanac (p. 254) says
    that the almanac data are accurate to 0.1'
    which means that the difference between the tabulated
    value and the true value is less than 0.1'.
    In your example, asuming that the value given by Frank
    is exact, the difference is 0.07' which is
    consistent with the Almanac statement of accuracy.
    But this statement does not imply that the last digit
    is actually correct. So there is no contradiction.
    
    The reason of these errors is probably the rounding
    subroutine. To get the last digit really true,
    yoiu have to do all computations with one more digit,
    and then apply an exact rounding algorithm, while
    in most cases they just cut off the last digit
    instead. Because nobody really cares about the exactness
    of the
    decimal minute digit.
    
    
    
    
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