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Re: Updated Transcript of Worsley's Log
From: Brad Morris
Date: 2017 Jan 27, 12:58 -0500

Lars

Reply in parts

++++++

Part one

You wrote

16°43' Noon altitude of lower limb as observed with the sextant (in your paper 6°43')

16 52 30 Corrected altitude

73 7 30 Zenith distance

14 29 45 Declination

58 37 45 Resulting noon latitude

The 16°43' is close to the inner edge of the page. So much so that there is a tear in the page from the stitching holding the log together. What I see is

'6°43'

But I wholeheartedly agree. The corrected altitude of 16 52 30 can only come from an altitude in proximity, not from an altitude 10° away. Further (90°-16 52 30)=73 7 30. That makes perfect sense!

Mystery solved Lars! Well done!

This will be updated in the next version of the transcript, with a special note about the ripped page and the numerical value reconstruction.

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Part 2

You wrote (in extracted phrases)

As his traverse table doesn't reach as far as 660', he uses half that number [editors note: 330]

Linear interpolation then gives the increment as 21/60 · 4.8' = 1.7'

And you then wrote these numbers

274 660=364.8

184.5

179.7

4.8

2.1

182.4

Discussion:

Just under the 274, there is an extremely faint 1.7 Just under the 660, there is an extremely faint 330, half of 660. I picked these up with a negative (reverse black and white) exposure of the image I have of the log.

The 2.1 is very faint and malformed. The 2.1 could easily be a 1.6.

As to the resultant 182.4, upon closer inspection, the 2 could be a 3, with the 4 possibly a 9. So the potential readings are 182.4, 182.9, 183.4 and 183.9. But I note that 184.5 - 1.6 = 182.9, with the 184.5 being a confirmed value.

Again, under the 182.9, I observe 365.8, which is precisely 2*182.9. That appears to confirm the earlier parts.

364.8 is just not there. It's a reasonably clear 365.8. Worsley's 5's look like 3's, with an additional pen stroke. Note that the 365.8 here matches precisely to the 365.8 under the 182.9

After reviewing the data, here is what I would propose

274 660=365.8

1.7 330

184.5

179.7

4.8

1.6

182.9

365.8

If this seems right to you, as it does to me, we can update this section of the transcript.

I'd like to apologize for the terrible state of the transcript here. It's very faint. Worsley's penmanship is awful in this spot. He was performing a quick calculation he knew by heart, so he is just jotting down a few figures. I have difficulty reading anything at all here.

Brad

On Jan 27, 2017 10:40 AM, "Lars Bergman" <NoReply_Bergman@fer3.com> wrote:

Brad, you asked "Does 73.7.30 make more sense?" regarding April 29th. Yes, indeed! I would like to have it like this:
16°43' Noon altitude of lower limb as observed with the sextant (in your paper 6°43')
16 52 30 Corrected altitude
73   7 30 Zenith distance
14 29 45 Declination
58 37 45 Resulting noon latitude
This latitude is then moved southwards to the time of the am sight, by 9.8' of dlat (from " 12 9.8 6.9 = 13 "), making the latitude of the am obs 58°48'.
The longitude of the am obs is 3^h22^m6^s, which equals 50°31'30" (in your document given as 50°31'34"). This am longitude should now be brought forward by the 13' of dlong sailed between am sight and noon, resulting in a noon longitude of say 50°19'. Thus there is a 19' diff to the longitude stated by Worsley, 50°0'. This last value is used in the distance calculation to Wallis, and also brought forward to the next day, so the transcription seems to be correct. One possible explanation could be that Worsley by mistake swapped 13 to 31 and substracted the latter value from the am sight result.
Regarding the calculation of course and distance to "27^m W of Wallis" we can conclude the following:
The difference of latitude is 274' and dlong is 660', as correctly calculated by Worsley. Now he has to convert dlong to departure. As his traverse table doesn't reach as far as 660', he uses half that number and look up the values for 56° and 57° latitude, finding that 330' of dlong equals 184.5' of departure at 56° lat, and 179.7' at 57° lat. The difference between these values is 4.8'. These values are thus correct in your document.
The mean latitude is (54°4' + 58°38')/2 = 56°21'. Linear interpolation then gives the increment as 21/60 · 4.8' = 1.7' to be subtracted from 184.5', giving 182.8'. Alternatively you could say that 21' is very nearly 1/3 degree and one third of 4.8' is 1.6', giving a result of 182.9'.
If we assume that your 183.4' actually is 182.4', then Worsley has subtracted 2.1' from the 56°-value. This corresponds to a mean latitude of 56°26'. The mean lat is most easily found by adding half of dlat to the smallest latitude. If Worsley by mistake read dlat=284' instead of dlat=274' the result would have been 56°26'.
If this suggestion seems plausible, then the following would make sense:
274 660=364.8
184.5
179.7
   4.8
   2.1
182.4
These values fit the resulting " N53E 458m ". What is your opinion, Brad?
Lars

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