# NavList:

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

**Re: Updated Transcript of Worsley's Log**

**From:**Lars Bergman

**Date:**2017 Jan 27, 07:12 -0800

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^{h}22^{m}6^{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