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Let's recap, a bit.

This thread started with a message about a lunar distance from Jeremy 
[5413], the important bits being summarised below-

"Here is my data around 0620 on 10 June 2008.

1) Sun LL: Hs 33deg 58.5� @ 06h 19m 13s UTC
2) Moon UL:  60deg 36.8� @ 06h 20m 44s
3) LD1: 86deg 10.3� @ 06h 21m 40s
4) LD2: 86deg 10.0� @06h 22m 20s
5) LD3: 86deg 10.2� @06h 22m 52s
6) LD4: 86deg 10.6� @06h 23m 24s
7) LD5: 86deg 10.6� @06h 24m 39s
8) Sun LL: 32deg 35.8� @ 06h 25m 19s
9) Moon UL: 61deg 49.2� @ 06h 26m 28s

DR (didn�t take a fix) Latitude was 15deg 14.0�N and Longitude was
144-04�E.  Ship was on course 270 at 12.0 knots.  IE was 0.0 T/P was
98deg F and 1010MB.  Height of eye is 106 feet.

I averaged the two sun lines to get 33deg 17.2� at 06h 22m 15s, then
the two moon altitudes to get 61deg 13.0� at 06h 23m 37s.  Finally the
5 LD�s averaged to 86deg 10.34� at 06h 22m 59s."

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

Frank suggested that that overlapping, rather than side-by-side, images of 
Sun and Moon had been measured, and I recalculated on that basis, arriving 
[5530] at a deduced GMT, from lunar distance, of 6h 24m 53s, which differs 
little from Jeremy's corresponding chronometer time of 6h 22m 59s GMT.

Kent Nordstrom has questioned some aspects of that calculation, but I hope 
that these have now all been resolved, except that Kent seems unwilling to 
accept the reality of those overlapping images. Unless he does so (or 
something similar), then he will be faced with an unbridgeable error of 1 
Sun diameter in the lunar distance, about 1 hour in the resulting GMT, and 
about 15� in the deduced longitude. But that's a matter for him.

What remains in question is Kent's method of finding longitude from the 
altitude of an observed body (or two bodies), once the GMT had been 
established. He wrote, in [5577]  -

"My approach for calculating the MT is as I understand the very same as used
in the old days.

This is of course for finding the difference between the GMT and the MT,
i.e. the longitude. It is a separate calculation. So let me outline what I
do:

  1.. Find the LHA at the predicted GMT (we know (roughly) the latitide,
from the generation of true distances at say time 3, time 6 etc. we can
predict the declination at the GMT. True altitude is from the altitude
reduction. Aries GHA to the GMT can be predicted as well).
  2.. Find the Aries GHA for upper meridian passage in Greenwich, (normally
one day before).
  3.. Convert this GHA to time.
  4.. Correct for assumed longitude and then you get the time for Aries UMP
on your location.
  5.. Find the difference between Aries GHA and the body�s GHA at GMT. This
is the RA.
  6.. Find the sum of the LHA and the RA, which is the sideral time.
  7.. Subtract the value for Aries UMP at the location. Now you have the
difference in sideral time.
  8.. Convert this difference in sideral time and you get the MT at
observation.
And now you can easliy find the difference between the GMT and the MT. I
hope this explaination clarifies. I should also add that one must be very
observant when doing this calculation, it is easy to do wrong."

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

I didn't understand all that (and still don't) and asked for some relevant 
numbers to help me out. On the other hand, perhaps it's better if I explain 
how I would do that job myself, with numbers, to show how simple it can be.

Since the mid-1800's mariners have increasingly accepted that sea positions 
can be determined from intersecting position lines from observations of 
several bodies, drawn on a chart, rather than calling for a calculated value 
of lat and long, separately, as was previously the case. This is at the 
heart of what became called " the new navigation", as it's learned and 
accepted today. And it doesn't need me to explain in detail how that's done, 
once GMT is known from a lunar or a chronometer. You start with a presumed 
position, and the Nautical Almanac tells you dec and GHA of each body 
according to the GMT. Then Altitude-azimuth tables, or some computer, 
provides calculated altitudes, which you compare with measured altitudes, 
work out an intercept and azimuth for each body from the presumed position, 
draw in position lines, see where they cross, and the job's done, providing 
both lat and long. That could be done for Jeremy's Sun and Moon altitudes, 
using for GMT either his chronometer GMT, or the GMT derived from his lunar. 
(Being nearly 2 minutes different in time, these will provide longitudes 
that differ by about 30 arc-minutes.) Those position lines will both lie 
nearly North-South, so they will cut at a close angle, and latitude won't be 
nearly so well defined as is longitude.

But there's an earlier way to do the job, which doesn't call on position 
lines or the New Navigation. It relies on the notion that, however difficult 
it may be for a navigator to determine longitude (without a chronometer), he 
always knows his latitude. That may or may not be the case in practice, 
depending on how clear his skies have been recently.

Anyway, that's the assumption, and in this case we can take Jeremy's 
latitude as being his assumed lat. of N15� 14' (= +15.2333), and try to 
determine his longitude, on that basis. Closing our minds to the chronometer 
reading itself, we can say that [see 5130] according to his lunar, at a GMT 
of 6h 24m 53s, the measured corrected Sun altitude was 33� 13.1' or 33.2189� 
and for that GMT the 2008 Nautical Almanac predicted on 10 June Sun dec = N 
23� 02.7' (= +23.0450�), and Sun GHA = 276� 20.9'  (=276.2483�).

Next we need to calculate the Local Hour Angle of the Sun at that moment, 
using the simple spherical triangle formula

cos (HA) = (sin alt - sin lat sin dec) / (cos lat cos dec),

which tells us that the Local Hour Angle of the Sun at that moment was + / - 
59.9466�.

All this, as far as I can tell, is what is covered in step 1 of Kent's 
suggested procedure. All the additional steps in that procedure can be 
removed, and replaced by the following-

The Sun had by then long passed the meridian and was sinking in the sky so 
we therefore have to subtract Local Hour Angle of the Sun  from its 
Greenwich Hour Angle, to get the observer's Westerly Greenwich Longitude 
(WGL), which ends up at +216.3017� or +216�18.1. Those who prefer a 
different convention can subtract that from 360� to get longitude = E 
143�41.9. That's it! Job done!

You can see that by using the GHA provided by a modern almanac, there's been 
no need to invoke sidereal time, Right Ascension, or Aries.

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

Instead of using the Sun's altitude, we could just as well have used the 
Moon's, which was measured for the same moment, and a similar calculation 
for the Moon tells me that Moon LHA = 26.9243�, and resulting WGL = 
216.3176�.

George.

contact George Huxtable at george@huxtable.u-net.com
or at +44 1865 820222 (from UK, 01865 820222)
or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK. 


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