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    Re: Transcript Of Worsleys Navigational Log Book
    From: Brad Morris
    Date: 2017 Jan 29, 15:57 -0500

    I added a few columns to your spreadsheet. 

    They Are:
    W - Worsley's chronometer is running faster than he wants.  Seconds
    X - convert to minutes of arc
    Y - nautical miles of longitude at the LATITUDE.  Y27 (yellow) shows an accumulation of 19.4 miles.  I am sure you could make this 20
    Z - from your longitude, subtract out the minutes of arc (col x.)  IN MINUTES OF ARC
    AA - convert to whole degrees
    AB - the remaining minutes

    One debatable point.  Do we update DR positions with the chronometer?  I think NO!

    Please add a second course to the kml for comparison



    On Jan 29, 2017 12:35 PM, "Brad Morris" <NoReply_Morris@fer3.com> wrote:

    I will examine the areas you have requested for correction.  

    I had an thought about the chronometer error.
    1) Worsley gets the location of Point Wild correct to 1' in longitude, as plotted on Chart 29104.  How he does this is not important.  It's essentially correct.
    2) That means his chronometer correction makes the reading of his chronometer align with the longitude.
    3) On 7 May, Worsley writes that he should be 20 [nautical] miles astern, but that his latitude is essentially correct.
    4) 20 nautical miles at S55° is 34.8', call it 35' of arc.
    5) At 4 seconds of time per minute of arc, that's 140 seconds of time, or 2 minutes 20 seconds.
    6) the open boat journey takes 15 days total (24 April -》7 May)

    The one thing that Worsley never directly tells us is his RATE.  Suppose for a moment, that his rate is 10 seconds per day faster than he expected.  Then the rising sun's altitude will be lower in the sky than expected, leading him to believe that he was further east than he truly was.  This would barely affect his latitude around noon, as the sun's altitude would be only marginally wrong.

    That is the theory I am working with now.  It is simple, and requires no machinations about charts, Ethiopics or other external evidence.  

    Why would his rate change so dramatically?  Perhaps, after months on the ice, the "warmer" climate of the open ocean at Elephant Island and beyond allowed the mechanism to run more freely.  Other theoretical speculation can apply here as well.  To satisfy all of the listed points, his chronometer just has to run slightly faster than before.

    Your thoughts??


    On Jan 29, 2017 11:04 AM, "Lars Bergman" <NoReply_Bergman@fer3.com> wrote:

    Thank you, Brad.

    On the 28th,  "1m4' more slow" shall probably be "1m4s more slow", with or without underlined s.

    On the 26th,

       26   7
       30  10

    this shall probably read

    2.23.29      or     2.23.29
    2 26   7                  26   7
    2 30 10                  30 10
    ----------              ----------
    7 19.46                  79 46

    being the sum of three chronometer readings. Divide by three, and you get the mean value 2h26m35s shown below. The mean of the altitudes is perhaps calculated among the x:es top middle or down left?

    Apart from the question of CE determination, I now see only two (there may be more) remaining peculiarities:

    a) April 26th

    16 37
    73.20 x
    13 32 x
    59 47.44

    My impression is that this is the noon observation for latitude. I know that you insist that the last row is correct, but why is the observed position then given as 59°46' above? The next last row should be the declination, 13°32'34" within one or two seconds of arc. Could you please give it another look? As it is now it makes no sense.

    b) 4th May

    N36E 52 41

    Looks like course dlat dep, but the values don't correspond to each other. Please give it a look again!


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    File: 138045.worsleycalculations-rate.xlsx
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