Calling all lunarians

jdc

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Now that us astro-nav aficionados have been woken up by GHA's interesting post, here's something I've been meaning to seek opinions on for a while.

A proposed method for deriving time and longitude from lunar observations

There are some problems with using lunar distances to derive time and longitude:

(i) although lunar distance tables could be calculated from other ephemeris, they are not published any-more;
(ii) they require one to measure the angle between heavenly bodies (rather than between them and the horizon), which is difficult enough in calm conditions and a cloudless sky let alone on a small boat in any kind of sea;
(iii) they therefore require both bodies to be visible and 'in distance' simultaneously. This last is a really significant constraint.

I propose and describe here an alternative method which does not have these disadvantages. It does require a clock which can measure time intervals to a few tens of seconds over a some hours (12 say). But this is easily possible with the cheapest and most trivial clockwork, and would have been within the capability of Slocum's famous tin alarm clock. The clock does not need setting and can just be started from an arbitrary time. A further advantage is that the calculations are just the normal ones for the Marc St. Hilaire method and so should be familiar.

The method:

1.Start the clock. It will have an as yet undetermined error, call it To. You probably have some idea of the time, to within a couple of hours anyway (you maybe know which ocean you are on;-)

2. Determine your Latitude by measuring the altitude of the sun at local mid day (or from Polaris). Move this, if required, by DR to keep a reasonable estimate of Latitude at all times.

3. Sometime, could be a few hours earlier or later, find a star or two, and/or the sun, such that the azimuth is nearly at 90 or 270 degrees, ie more or less due E or W. Take sights of these stars and/or sun, recording time as displayed on the clock. These sights need not be simultaneous and could be spread over some hours. Record the times as T1, T2 etc. They will all be in error by the same amount, To.

4. Take a sight of the moon when it is near E or W, recording the time, Tm, which is still offset by To. This need not be done at the same time as the sun or star sights are taken, which is a key advantage.

5. In the normal way work out the longitude (long_star) which the (collection of) star and sun sights gives you. As normal one has to move PLs by allowing for DR if required. You have an assumed time and you know latitude, so longitude can be derived. This is the familiar 'longitude by chronometer' method.

6. Do the same calculation to derive the longitude but this time using the moon sight (long_moon). In general this will be different from that given by the star sights.

7. Difference these two longitudes to give delta_long (delta_long = long_star - long_moon). Write down the values of To and delta_long.

8. Guess a new value of To, and go back to step 5. Repeat until one has enough points to draw a graph of delta_long against To (or do it algebraically). The value of To when delta_long is zero is the true value of To and thus clock error, and the corresponding value of longitude is the true longitude. Hence a position fix is obtained without chronometer. Fast convergence can be achieved by knowing that the longitude of the moon advances against that of the sun at a rate of 360 degrees in a synodic month, ie ~29.5 days.

I see no reason that this method shouldn't be about the same accuracy as the classic method of lunar distances (which isn't all that good, maybe 15 miles). It does require more, but simpler, calculations than the classic method, but we have calculators nowadays...


Can anyone see a fatal flaw, and if not, has anybody used or seen this method described before?
 

mjcoon

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(ii) they require one to measure the angle between heavenly bodies (rather than between them and the horizon), which is difficult enough in calm conditions and a cloudless sky let alone on a small boat in any kind of sea;...

But such measurements might be possible with a smartphone that has a good camera. Getting absolute angles would require calibration, but that should be possible if other stars were visible in the photo. I believe that amateur astronomical telescopes must use something similar to automate star-finding.

Has anyone heard of such a smartphone app?

Mike.
 

AndrewB

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I've not seen anything like this suggested before. It sounds in principal as if it would work, but to be quite sure we understand it would be helpful to see a worked example.

But, if I have grasped it, it doesn't sound like it would be anything like as accurate as the conventional method, and even with that you'd be lucky to get 100 nm accuracy, let alone 15 nm. Try working through the consequences of being 2' out in your sextant readings, or a second out in your difference timing, given you don't have a chronometer.
 

John Barry

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I see no reason that this method shouldn't be about the same accuracy as the classic method of lunar distances (which isn't all that good, maybe 15 miles). It does require more, but simpler, calculations than the classic method, but we have calculators nowadays...


Can anyone see a fatal flaw, and if not, has anybody used or seen this method described before?

Lunar distances are not a (direct) method of calculating longitude, they're a method of calculating UT.

Accuracy is about 12 seconds per 0'.1 of measurement error (assuming no calculation error, and the calculations are complex). So a measurement error of about 0'.5 will give UT to within about 1 minute. You won't get much better than this anyway, because the Moon's published GHA (in the Nautical Almanac) is only accurate to about 0'.25. You can get the Moon's position better than that, but it's tricky...

UT accurate to 1 minute will give you (depending on the geometry of the sight) a longitude accurate to about 10 miles (for example, from 51°N 0°E today at 1700UT Ha is 15°24'.9 on 262.6°; at 1701UT it's 15°15'.5 on 262.8; at 1702 it's 15°06'.2 on 263.0°. These would give latitude differences of about 9.4 and 9.3 miles.

Which if you don't have a clock (or if you only have one with only an hour hand) is not a bad start...


John
 

jdc

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Lunar distances are not a (direct) method of calculating longitude, they're a method of calculating UT

Thanks, as you rightly noticed I'd not explained adequately how I'd got to my ~15 miles accuracy estimate.

It was from my error in position fixing being 1.5 miles = 1 sigma, a number I have found to be my average after many sights in real conditions. Doubtless a more expert person on a more stable platform would do better, but it does include all errors, including allowing for distance run when transferring PL in addition of the variances from all sources, not just altitude.

I converted this 1.5 miles to time error equivalent at 15 degrees = 1 hour, and then applied this to the moon's relative movement of 360 degrees in 29.5 days, which gives an effective multiplicative factor of 29.5: 29.5 x 1.5 miles = ~45 miles, but then one can average a bit, reducing maybe 3 fold (take 10 measurements perhaps, so a sqrt(10) improvement) thus arriving at 15 miles.

As you say, pretty good if you've no chronometer at all.
 

AndrewB

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Thanks, as you rightly noticed I'd not explained adequately how I'd got to my ~15 miles accuracy estimate.

It was from my error in position fixing being 1.5 miles = 1 sigma, a number I have found to be my average after many sights in real conditions. Doubtless a more expert person on a more stable platform would do better, but it does include all errors, including allowing for distance run when transferring PL in addition of the variances from all sources, not just altitude.

I converted this 1.5 miles to time error equivalent at 15 degrees = 1 hour, and then applied this to the moon's relative movement of 360 degrees in 29.5 days, which gives an effective multiplicative factor of 29.5: 29.5 x 1.5 miles = ~45 miles, but then one can average a bit, reducing maybe 3 fold (take 10 measurements perhaps, so a sqrt(10) improvement) thus arriving at 15 miles.

As you say, pretty good if you've no chronometer at all.
It would be! Sorry, but I just can't accept your premises - you are drawing a false analogy.

Some years ago we discussed here the accuracy of the conventional lunar distance method, and for that an example proved instructive.

The angular distance between the sun and moon is constantly changing: on average by 30’ per hour, though varying through the month. So the distance between them can be related to GMT. The distance can be worked out from the Almanac and a scientific calculator using the equation:

Cos(distance) = sin (Sun DEC) x sin (Moon DEC) + cos (Sun DEC) x cos (Moon DEC) x cos (Sun GHA-Moon GHA)

(Note this equation is very similar but not identical to that tabulated in the sight reduction tables).

Here is an example based on the extract from the Almanac used for the Ocean Yachtmaster theory course:

On June 20th 1980, EP 40N 30W (somewhere near Azores), assumed GMT somewhere around 1500, from sextant observation the angular distance between the centre of the sun and moon (after applying all measurement corrections) is read as 91º41’. What is the actual GMT?

From the Almanac for Jun 20:

GMT .... Sun GHA .... Sun DEC ... Moon GHA ... Moon DEC ... Calc. Distance
1200 ... 359º37.1’ .. N23º26.3’.. 268º28.4’ .. N3º14.7’ ...... 89º45.5’
1300 .... 14º37.0’ .. N23º26.3’.. 283º02.9’ .. N3º04.9’ ...... 90º12.7’
1400 .... 29º36.9’ .. N23º26.3’.. 297º37.3’ .. N2º55.0’ ...... 90º40.0’
1500 .... 44º36.7’ .. N23º26.3’.. 312º11.8’ .. N2º45.1’ ...... 91º07.1’
1600 .... 59º36.6’ .. N23º26.3’.. 326º46.3’ .. N2º35.3’ ...... 91º34.3’
1700 .... 74º36.5’ .. N23º26.3’.. 341º20.8’ .. N2º25.4’ ...... 92º01.5’
1800 .... 89º36.3’ .. N23º26.3’.. 355º55.3’ .. N2º15.5’ ...... 92º28.6’

(Where the final column is calculated using the above formula). Interpolation of the observed figure of 91º41’ in this table gives a GMT of 1615.

If the observed angular distance had been 2’ less, then the estimated GMT would have been 1610: similarly 2’ more gives 1619. Uncertainty over 9 minutes implies 2¼ degrees of longitude or approximately 100nm at this latitude. My experience is that even this level of accuracy is hard to achieve for sights involving the moon using a conventional sextant aboard a yacht at sea.

I cannot see how your method can better this as fundamentally what it is reliant on is the vertical component only of angular distance (assuming I have grasped it correctly) and so inherently is less precise. Your method does however make the application of correction factors for refraction etc simpler, as well as having the advantage it can be used even when the sun and moon are not visible simultaneously or are at an inconvenient angle to one another.
 
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jdc

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It would be! Sorry, but I just can't accept your premises - you are drawing a false analogy...

Umm, I don't quite follow you. I didn't think I was making any analogy, merely trying to estimate, albeit in a slightly hand-wavey manner, the error one might expect.

My estimate of 1 sigma = 45 miles is not so far from your estimate of +/- 50 miles it seems to me!

I then allowed for improvement due to multiple measurements - assuming the errors are random and zero-mean - which I guessed at best to be a factor of 3, giving 1 sigma = 15 miles. I emphasise the 1 sigma because I assume a bell shaped distribution (2nd order chi-squaredI imagine rather than a 1-D Gaussian), whereas I imagine that your estimate was for a truncated distribution of some sort.
 
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