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    Re: Time Sight Computations
    From: George Huxtable
    Date: 2004 Sep 29, 12:03 +0100

    Yesterday, I wrote-
    
    =============
    
    I think I can see where Chuck Taylor's discrepancy may have come from.
    
    As Henry Halboth has pointed out, the different methods of calculation
    should give indistinguishable results. He put his finger near it when he
    expressed suspicion about the conversion of the Sun dec into decimal
    degrees.
    
    What seems to have happened is this. The Sun dec. at the time of
    observation, which Chuck stated to be 1deg 17.6' S, should actually be 1deg
    16.7' S. At least, that's what my pocket-calculator program makes it to be;
    I haven't checked it against an almanac. Chuck seems to have made a
    transcription error.
    
    However, in Chuck's "Method 1" he has written-
    
    Dec = d = -1.27833
    
    This is the decimal equivalent of 1deg 16.7' S, so there's no error here.
    
    Then, in "Method 2" (and presumably, therefore, method 3) he has written-
    
    First compute polar distance:
    p = 90d + 1d 17.6' = 91d 17.6'
    
    So in this case he has used his incorrectly-transcribed value for Sun dec.
    
    I suggest (though I haven't taken the calculation further) that if Chuck
    corrects his transcription error and uses 1deg 16.7' S for Sun dec in each
    example, he will get good agreement between them.
    
    ======================
    
    However, I was quite wrong in assuming that was the origin of Chuck's
    discrepancy.
    
    In method 2, although Chuck wrote
    
    First compute polar distance:
    p = 90d + 1d 17.6' = 91d 17.6'
    
    The value of p he actually used, in his method 2 calculation, was not that
    stated value at all, but the correct value of 91deg 16.7'. Something else
    has gone wrong. But what?
    
    I copy Chuck's method 2 working, below.
    
    Method 2:  Using log tables
    
    First compute polar distance:
    p = 90d + 1d 17.6' = 91d 17.6'
    
    
    h    24d 54.3'
    L    48d 30.1'    log sec  9.12462
    p    91d 16.7'    log csc  0.00011
      2)164d 41.1'
    s    82d 20.6'    log cos  9.12462
    s-h  57d 26.3'    log sin  9.92573
                               -------
    t                 log hav  9.22991
    t    48d 40.1'
    
    If we add the four values in his table of logs to the right, we get not
    Chuck's stated value of 9.22991, but a wildly different number, 8.17508!
    
    It's suspicious that two entries in this sum have identical values, of
    9.12462, and it looks as if there may be a transcription error when Chuck
    has typed his numbers up for us.
    
    The suspicion falls on the tirst term in the sum, representing log sec 48d
    30.1', or log sec 48.5017. If I calculate this term, I make it to be
    0.17875, not 9.12462 as Chuck has it.
    
    So if we substitute 0.17875 as the first term in the sum, and add the other
    three terms to it, the result is 19.22921 or (neglecting tens) 9.22921.
    This is close to (but slightly different from) the figure Chuck arrives at,
    of 9.22991.
    
    And if we calculate or look up the angle for which log hav is 9.22991, we
    find the result to be 48d 37.6; exactly tha same as the result of method 1,
    just as it should be!
    
    So, something has gone wrong with Chuck's assessment. Either he has arrived
    at a value for log sec 48d 30.1' that's somehow .0007 too high, or he has
    worked out all the logs and their sum correctly and then mistranscribed
    9.22991 instead of 9.22921, as it should have been. And then the error has
    been further obscured by writing what is quite the wrong quantity in place
    of the first of those logs to be summed.
    
    ===========
    
    I ended my posting yesterday by adding (and it's now reinforced)-
    
    Such errors are easy to make; I expect that navigators in pre-calculator
    days were more fluent in numbers than we are today, and got things right a
    bit more often. I keep finding silly errors in my own numberings,
    particularly when working in sexagesimals.
    
    George.
    
    
    ================================================================
    contact George Huxtable by email at george---.u-net.com, by phone at
    01865 820222 (from outside UK, +44 1865 820222), or by mail at 1 Sandy
    Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
    ================================================================
    
    
    

       
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