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The attachment Brad sent, showing Worsley's calculations when nearing (but 
still well out-of-sight-of) South Georgia, is intriguing. I've attached 
Brad's transcription again here.

Those were crucial observations for the James Caird. That was the only day 
on which Worsley was able to observe both a noon Sun altitude, for latitude, 
and an away-from-noon Sun altitude, which with that latitude, could then 
provide longitude, though both were somewhat uncertain. It was because of 
those uncertainties that Shackleton decided to head for the South coast of 
the island, which then called for that famous mountain-crossing to be made.

The attached page appears to deal only with the morning Sun sight, and I 
would guess that the reduction for the later noon observation has been 
detailed on another page. At noon, the Sun disc was unclear, so that Worsley 
had to do his best by observing the height of the centre of the fuzz, rather 
than a limb.

I am a bit hampered by not having a 1916 Nautical Almanac. Many such 
almanacs appear to have been digitised, but not, it seems, that one. I can 
look it up next visit to the Bodleian library in Oxford, but if anyone knows 
of a public source, or has a copy himself, that would be helpful; either the 
British or US version. All that's needed is, for May 7th (and for May 6th 
and 8th also) Sun declination and Equation of Time at Greenwich noon. Of 
course, such backwards predictions are available from modern sources, but it 
would be nice to see what the Almanac actually states. For comparison, I've 
just taken these quantities for the same day, May 7th, of a recent year, 
which will be within a few arc-minutes of the right value.

Brad was right to deduce that the navigation was done by calculating a 
longitude numerically, and not by a St Hilaire technique calling for Sun 
azimuth and providing a position line. It would be interesting to discover 
whether Worsley had been using those same old techniques, rather than the 
"new navigation",  the previous year, at times when Endurance had been a 
real moving ship, and he had a full chartroom at his disposal.

After all that it's been through, it's remarkable that the log can be read 
at all, and clearly Brad has made a good job of transcribing it, because 
much of the numbering seems to be consistent. However, there are several 
things about it that I don't understand (yet). With Brad's help, we might 
yet unravel the whole thing.

==============================

The following details are mainly for Brad, in the hope that he can use them 
to improve and annotate his transcription. It remains somewhat tentative, 
and calls out for corrections. Here goes-

Left hand column. This first figure, 5.10.14, is, I expect, related to 
chronometer reading, in hours, minutes, seconds. Whether it's actually GMT, 
or perhaps time-to-go until noon GMT, I havent deduced yet (and suspect the 
latter). I presume the 13-minute adjustment must be a correction for 
presumed chronometer error, but if so, it's surprising that it's given only 
in whole minutes, not minutes-and-seconds. The next number will, I presume, 
be the correction for Equation of Time at noon of that day, to end up with a 
corrected time of 5h 26m 47s.

For now, we will leave this column and move right to what I take to be the 
Sun lower-limb altitude reading of 9º49' The letters and numbers that follow 
it, (P.M. N68ºE14 5.2.13.0=23') make no sense to me.
Below that is a correction of 8' (the 54º, crossed out, was presumably a 
mistake, as Brad states). It's likely that 8' is  Worsley's standard 
correction, made each time for a Sun altitude, combining semidiameter, dip, 
and his sextant's index error.
Now things get interesting, as he computes the hour-angle of the Sun, from 
standard trig formulae. I will attach a page fron "Norie's Navigation" 
(1900) page 326 that explains the procedure, which is his Method 1..
Start with the corrected altitude, of 9º57'. Add to it the estimated 
latitude, of 54º 33' This differs a bit from his stated noon latitude of 54º 
38', perhaps because of the Northing estimated in the interval to noon. Also 
add the South Polar distance, of 106º 52', to arrive at a sum, for the 
three, of 171º 22'. (I need to explain South Polar distance, which is the 
angle between the Sun's declination at that moment which is presumably 16º 
52' N, and the South Pole, which is 90ºS, so therefore 90º +16º 52'. (To be 
accurate, what's required is thedeclination, not at Greenwich noon, but a 
few hours before then, when the observation was made. Without the right 
Almanac, I don't know whether or not that was made, but it probably was)).

Those three angles added up to 171º 22' (I hope you're checking Worsley's 
arithmetic along with me). That sum is to be halved, to give 85º 11', which 
we can call the "half-sum", and the sum itself will not be used further. And 
next, we take away, from that half-sum, the number we first thought of, the 
altitude of 9º 57', to arrive at 75º 44', which we can call the "remainder". 
Now we have obtained all the necessary angles for the spherical-trig 
calculation, to be done by 5-figure logs. All the numbers are to be treated 
as positive ones.

What Worsley had to do next is to add together the following

latitude           54º 33'  log    sec = .23658
polar dist.    106º 52'  log cosec = .01910
half-sum        85º 41'  log    cos = .87662
remainder     75º 44'   log     sin = .98640
   and adding all these together  =  .11870

Check it yourself. You will note that in this operation Worsley simply 
discards any whole numbers that may arise to the left of the decimal point 
He just doesn't care. What any such integer would do is to multiply or 
divide the end-result by a factor of 10, and Worsley knows perfectly well 
(within that factor of 10) the ball-park figure in which his end-result has 
to lie. This was a common attitude taken by navigators, who are repeating, 
day after day, similar calculations in which only the fine details vary, and 
they always knew, more or less, what the answer was going to be.

Of course, Worsley would have had (by that time, soggy with damp) tables 
giving directly logs of all the trig functions, which you may not own. But 
you can always get logs (to base 10) of angles (in degrees, not radians) in 
two successive steps of a scientific calculator, or a computer. Perhaps 
three steps, because you will probably need to get sec from 1 / cos, and 
cosec from 1 / sin.

If you compare those columns of numbers, above, with the table Brad has 
provided, you will see that Brad has transcribed every digit correctly, but 
there are some intervening dots and colons which could be removed, and the 
log numbers near the bottom have slipped upwards a notch; there should be a 
blank space to the right of 171.22.

I diverted when we had calculated that log of .11870, so let's get back to 
it now. This is actually the log haversine (log hav) of the hour angle. 
Worsley's log trig tables would have included such a table, for which you 
have to search for the value .11870, and find the hour angle, or the time 
before local apparent noon (at 1 hour = 15 º) that it corresponds to. For 
nearly the whole range of times that are practical with this method, the 
appropriate prefix integer was 9 (and Worsley would have known that well), 
so if we look up the time corresponding (or nearly so) tp 9.11870, we find 
that the nearest entry is for 9.118711, for 2hrs 50m 03sec, which was the 
result we needed  (or if you prefer, an equivalent angle of 42º 30' 45").

Here we need another diversion, for those that haven't come across 
haversines before. The haversine is another trig function, which has the 
advantage that it never goes negative, so is particularly suitable for log 
calculations. Hav of angle A is defined as (1 - cos A) /2, and is always in 
the range 0 to +1. [You may occasionaly come across the versine of an angle, 
which is (1 - cos A), in the range 0 to +2, and logically, a haversine is 
half of the versine.]  Occasionally a log hav table might also be named a 
"log sine square" table, because it also happens that hav A= (sin A/2) 
squared.

Anyway, let's get back to the hour angle Worsley deduced, of 2hrs 50min 
03sec. If we go back to the furthest left-hand column, we left off after 
writing down 5 26 27, which seems to be the time of that morning observation 
from Greenwich apparent noon. Now we see that below it, Worsley has written 
in his deduced time to local apparent noon, from the Sun altitude, 2 hrs 50 
min 03 sec. And then he subtracts, to provide the time difference between 
them, 2hr 36 min 44 sec, which is the time difference between local apparent 
noon and Greenwich apparent noon. And then he has converted that into an 
angle difference, at 15º per hour, to be 39º 11', which is, of course, the 
local Westerly longitude.

The next entry, adding another 25' to that longitude, is a bit of a puzzle 
to me. We see that the latitude he has underlined, in the top right corner, 
differs a bit from the value that was used in the previous calculation. Did 
this reflect a revision of the previous presumed latitude, which resulted 
from the noon observation? And then, did he somehow adjust the longitude to 
allow for that revision in latitude? That's one possibility, but there may 
be others. The revised value, marked "39º 34' Noon", can't reflect ground 
gained in the period up to noon, because it shows a 25-minute increase in 
Westerly longitude, and the Caird was travelling largely Eastward.

I take it that the underlined values at top right of the page indicate 
Worsley's best estimates of position at noon on May 7th, and fit in, as far 
as one can tell, with the position marked on his map.

Just below, the figures corresonding to Bird Island make little sense. Its 
modern position is 54º S, 38ºW, within a very few miles.

Below that is a mention of "Laith Harb.", which should be Leith, and a 
figure 51, which might show its distance from Bird Island, and the 121 may 
represent a total estimate of miles-to-go, to Leith, before the plan 
changed. There's a lot of unwarranted supposition in that.

I hope that the analysis I've provided will give Brad enough clues to tease 
out the final details, perhaps in conjunction with other pages from the log.. 
It's been an enjoyable bit of detective work, and may well have errors to be 
uncovered.

George.

contact George Huxtable, at  george@hux.me.uk
or at +44 1865 820222 (from UK, 01865 820222)
or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
----- Original Message ----- 
From: "Brad Morris" 
To: 
Sent: Thursday, February 26, 2009 4:07 PM
Subject: [NavList 7441] Re: How Worsley Navigated


I surrender George.  With the dispute in maps, I wanted to see the
navigational log to attempt to resolve the dispute.  I retract my
speculation.  Yes, I have read the books, I just haven't referenced them
recently.

I wish I knew the location of the Caird, so I could obtain better images
of the sextant.  If we accept the image of the sextant used, then I can
tell you that there almost no doubt in my mind that the sextant I have
is the twin of the one used.  Further, there are two components missing,
for which we can see spaces in the box, now vacant.  The first component
missing is a shade that attaches to the back of the telescope tubes.  It
goes into a slot on the right hand wall.  The greatest angle clamp and
one of the two eye shades are there.  The gap is visible right beyond it
for that second eye shade.  The other component missing from the sextant
used is an inverting scope.  It is held in the block of wood with the
large round hole, just over the index mirror.  I can see the other
components are present.  I can see nothing else missing.  Note that the
binoculars are standing in the case at the mounting location, but not
held in the case as designed.

Now that I am done "contemplating the angle of my knob" (hehehe), I can
easily give you one of the reductions, attached herein.

Best Regards
Brad







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