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Thanks to Lars Bergman, for contributing the following thoughts about 
calculating the altitudes, to go with a lunar distance observation. Sorrry 
for the delay in responding, but my old boat has taken some priority just 
now: but that has provided a bit of useful pondering-time.

Lars wrote-

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George wrote, in reply to Douglas Denny, in 14133:
“If the time was known, there would be no point in taking a lunar. That's 
the purpose of measuring a lunar-distance; to determine the time. So 
there's a bit of a paradox, in calculating altitudes with which to clear a 
lunar, for which time and position are both needed. To get round it, you 
have to assume a trial longitude first, and then use the lunar observation 
to refine that assumption.”
And in 14145:
“I don't know of any texts which have considered this question of a need 
for reiteration when altitudes have been calculated, and wonder whether 
travellers were even aware of its potential for causing error. Can anyone 
point to such a publication?”
---
In above context, the purpose of measuring a lunar-distance is to determine 
a standard time, i.e. the time at Greenwich. To calculate an altitude, 
knowledge of Greenwich time is not necessary, but the local time must be 
known. Latitude is assumed known and an approximate Greenwich time in order 
to find the declination from the almanac. Thus, an altitude observation to 
determine the local time should therefore be executed as near in time to 
the distance observation as possible. The error in time then depends only 
on the regularity of the watch and of the ship’s (or traveller’s) reckoning 
between the two observations.
To find the altitude corrections, some rework may be necessary. Normally, 
parallax and refraction are found from the apparent altitude, thus all 
tables are arranged with apparent altitude as argument. When altitudes have 
been calculated, some iterations are necessary to find the proper values. 
The aim is to end up with an apparent altitude that, reduced in the normal 
way, gives the same value of true altitude as that found by calculation.
In some Swedish manuals for navigation (e.g. Pettersson in 1876) the 
process of calculating altitudes are described in some detail, involving 
iteration not of the lunar calculation but of the altitude calculations.

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On reflection, I think that Lars has got the matter right. As long as an 
observation to determine local time has been made, at or near to the time 
of the lunar, that can provide sufficiently good data on which to base the 
necessary altitude calculations. In which case, the laborious iteration 
procedure that I had thought necessary could be circumvented. Thanks to 
Lars for putting me right.

However, sometimes a lunar is taken for a different purpose than to 
establish an immediate longitude. Sometimes, it might just be to set, or 
correct the error of the chronometer, and it might not be thought 
necessary, in mid-ocean, to make such a time-sight for local time, there 
and then. Can the altitudes be successfully calculated under those 
circumstances, I wonder?

And I still wonder about the use of Frank's lunar-distance calculator, at-
http://www.historicalatlas.com/lunars/lunars_v4.html , to solve the 
converse problem from that it was intended for.  I would be grateful for 
any step-by-step guidance on how to go about using that calculator to 
establish the GMT at the moment of a lunar distance observation, from an 
unknown longitude, without observing the altitudes. Can it be done? How 
much iteration is involved? With observed altitudes, there seems to be no 
such problem.

Here is the example I asked about, once again-

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"A navigator is somewhere on the Equator, at
lat = 0º, and knows it from his previous observations. We know (though he
doesn't) that at midnight, 00:00 hrs at the start of 26 March 2005, he is
exactly on the Greenwich meridian, at long = 0º, at which moment he takes a
precise lunar distance between the Moon's near limb (it's full Moon, so
either limb will do) and Regulus, of 36º 48.9'.
However, not knowing his exact longitude, he guesses it to be 01º 00' 
East."

===========

Can anyone explain to me the most efficient way to go about it, using that 
lunar calculator?

George.

contact George Huxtable, at george{at}hux.me.uk
or at +44 1865 820222 (from UK, 01865 820222)
or at 1 Sandy Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.

   

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