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## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

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Re: Transcript Of Worsleys Navigational Log Book
Date: 2017 Jan 29, 12:09 -0500
Lars

I will examine the areas you have requested for correction.

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.

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

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

On the 26th,

2.23.29
26   7
30  10
----------
19.46

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°28'
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!

Lars

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