# NavList:

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

**Re: Clearing lunars**

**From:**Frank Reed

**Date:**2010 Aug 26, 22:48 -0700

George H, you wrote:

"In investigating Janet Taylor's early works, "Luni-solar Tables", and "Navigation Simplified", I have become increasingly disappointed by the many flaws to be found therein. I was, at least, allowing the lady a bit of credit for her perception in discovering that impossibility in a set-problem."

George, how generous. You gonna pat the "lady" on the head, too?? :>

You added:

"But it seems that even that should be denied her, as the error had been publicly aired, half a century earlier."

It's an error to call it an error, without qualification. But definitely it's an example that could confuse some students. This specific problem makes a non-triangle by some 13 degrees, so there's no way that this particular case could represent some modest observational error in the observed altitudes. The general issue stands nonetheless (and there were SEVERAL such examples in the Tables Requisite): you can have cases where the triangle cannot exist --with one side longer than the sum of the other two-- and yet the process of clearing the lunar is still valid. As I noted previously, and indeed as I have described many times in NavList messages, the lunar clearing process, in many cases, is not affected by large errors in the Moon's altitude. If you don't believe me, then try a case: the apparent altitudes of the centers of the Sun and Moon, cleared of dip and corrected for SDs are both recorded as 45 degrees. The measured lunar distance corrected for SDs is 89d 00' exactly (the exact lunar distance matters). Suppose that the Moon's HP is 57' (near the mean value, just for the sake of this example) and the refraction is standard. What is the cleared lunar distance? In this case, the triangle is valid since the sum of the zenith distances of the centers is greater than the observed center-to-center distance. Now work the problem again but under the assumption that the altitude of the Moon was observed rather casually and was recorded as 47 degrees instead of 45. If you want an observational explanation, perhaps the altitude of the Moon was shot as a preliminary, then some lunar distances were measured and averaged, then the altitude of the Sun was observed, but the Moon was found to be hidden by clouds and its altitude couldn't be taken again. What would the navigator do? One option would be to work the problem through by the usual methods (perhaps Lyons or Witchell, as in the examples) and you would find no problem, despite the fact that the numbers are not consistent with any possible spherical triangle! Another option would be to drop the Moon's altitude and calculate it (a long, laborious process if it's calculated from the DR --and quite un-necessary since the math doesn't care, or a short and easy process if a vertical circle lunar is assumed --for which, see below).

Regarding the Maskelyne-Nauticus exchange, you wrote:

"I hope Frank's digging is successful, or he can provide a reference, as that sounds rather interesting."

I'm still looking for my copies... It's off my computer on a CD somewhere. I'm sure I'll find it. It's not on Google Books, but a small book published by "Nauticus" covering the letters is indexed there (no text, just the title). It's something like "Mystical Mathematics applied to Moon-hauling". I think he was trying to be funny...

And you wrote:

"I wonder what was the actual date of that exchange, "back in the 1780's", as Moore seems somewhat lax in reprinting the same problem as late as 1795."

I am fairly sure that the letters were published in 1785. Moore continued to use the example either because Maskelyne answered the complaints to the satisfaction of his audience (assuming Moore read the letters) or because there was no mention of them in the errata implying that Maskelyne considered the issue unworthy of concern (assuming Moore didn't read the letters). It's quite possible that Maskelyne did eventually count that specific example as an error in pedagogy but he did not consider it an error in mathematics or navigation, at least in 1785 --only a transcription error with no consequence. The example still serves to illustrate the methods even if the data are abstract. A student learning the methods from that example would not be led astray. But clearly students who dug a little deeper might become confused by it, and that alone was good enough cause to drop it from later textbooks.

You wondered:

"But is it the RIGHT answer? That question becomes meaningless, if it can't be related to a real-life situation, or set up as a physical or mathematical model."

Please see the above example with the measured distance at 89d 00' and the altitude of the Moon either 45 degrees or 47 degrees, one of which yields a seemingly impossible triangle. THESE OBSERVATIONS CAN HAPPEN IN PRACTICE. As I noted the process of clearing a lunar is robust --unless it is set up wrong-- and error in the altitudes that results in a non-triangle (on paper!) can still yield nearly exact navigational information. By the way, for modern lunarians, you can easily set this up wrong so I should mention what might happen. Suppose you create a spreadsheet to clear lunars and you use the standard direct triangle solution. That is, you take the observed zenith distances of the objects' centers and the observed center-to-center measured lunar distance, and you solve the spherical triangle ZSM (zenith, Sun/star, Moon) for the angle at Z, the difference in azimuth. Then you take that difference in azimuth, replace the measured zenith distances by the z.d.'s corrected for refraction and parallax and reverse the operation to get the corrected lunar distance. Easy as pi. BUT there's a catch. You shouldn't solve for the actual angular difference in azimuth at Z but rather for the cosine of that angle. It's possible in practice for that number, cos(Z), to end up with a value slightly greater than 1.0 in absolute value. If you then ask a calculator or spreadsheet to give you the corresponding angle, you will get an error. That angle does not exist in the real numbers. But you don't need that angle anyway. Just calculate cos(Z) and continue. The math works out fine, and the results are accurate.

You wrote:

"However, I don't see how those words of Moore, that Frank quotes, would provide any answer to her criticism of his impossible example."

Well, he COULDN'T answer her specific complaint since he had was DEAD. Moore died relatively early, in 1807. Among the famous authors of navigation manuals in this period, he was the first to have real commercial success and among the first to die. That seems to have made it quite a bit easier for critics to assault his works. But I found it rather charming that he had already answered her "Euclidean" comment on that lunar problem with his broad "Euclidean" answer to critics, whoever they may be, way back in 1794. Of course it doesn't address her specific question. He was dust by then.

And George, you wrote:

"Now we come to quite different ground, a matter which Frank has raised before. It is quite true that special geometries arise in which longitude can be derived from a lunar distance observation, without involving trig. And it's also true that the processing of lunars in that way may be tolerant of quite wide deviations from those exact geometries. But how tolerant? How much deviation from that geometry can be accepted, and still keep within an error-band of an arc-minute or so, outside which a lunar loses much of its value?"

A whole arc-minute?? Well, quite a large number of cases! I previously focused on cases where the resulting error was only a TENTH of a minute, and even then there are far more cases than you would expect. It's not just small, first-order deviations from the exact vertical circle case that work, but much larger deviations.

You asked:

"So, what criteria does our navigator adopt, in order to decide whether such a short-cut is acceptable, or whether a full-blown trig solution is needed?"

There is a very simple test. You compare the difference between the observed altitude of the Moon and the adjusted altitude (adjusted to force vertical circle geometry). If that difference is smaller than 6*tan(LD)/cos(h_moon) then any error resulting from treating it as a vertical circle case is a tenth of a minute of arc or less. If such a method had been made available historically, this test condition probably would have been made available in a short table of one or two pages. If I had a time machine, that's what I would recommend upon my arrival in 1780... Barring advances in time travel, the fact that such simple methods weren't advocated historically by people like Maskelyne tells us something about their focus and perhaps their lack of vision. Or maybe it's just a historical accident that such a quick method for clearing lunars wasn't more widely recognized.

You asked:

"Or would he be better off, applying the traditional trig technique in ever case, whether or not a short-cut might have sufficed? After all, in the days of lunar distance, trig, logs, and tables were the regular diet of a navigator, and they had no fear of such matters"

Clearing lunars WAS NOT difficult mathematically. The exact same mathematical skills and indeed nearly the same methods were required to clear lunars as were required to work common time sights. But clearing lunars WAS time-consuming and tedious. And if you can do something in less time with the same accuracy, you surely would. Right?

George, you added:

"unlike Frank's students of today."

I think that deserves a reply. Students today --at least those who take an interest in celestial navigation-- aren't dunces when it comes to trig and logs, which, after all, are no more than secondary school mathematics. Students who were in secondary school before about 1980-90 got much more practical experience with paper methods, but the ideas are not alien to those who followed them, and they do understand what's going on. It's the little things that confound students working traditional navigation problems today. They're very happy working on calculators and computers and smartphones, partly because it means that don't have to worry about taking logs from tables and adding them up by hand, which strike the students I've spoken with, not as difficult work, but as "busy work" (interesting historically but otherwise pointless, once you've done it once). So what gets to them? Subtracting angles actually seems to cause more frustration than things like trig and logs. I suppose it's because they know it should be easy and discovering that they get lost with something seemingly basic, right at the beginning of the work, leads to a lingering feeling of frustration.

Generally, there are as many approaches to the study of celestial navigation today as there are students. On one axis, some are intensely interesting in the mathematical underpinnings of the subject and they derive much of their pleasure from the topic in "tricky trigonometry". Others want as little to do with the mathematics as possible. They focus on the physical tasks of observation and sextant use and plotting on charts, and they tend to be quite happy learning relatively rote methods mathematically, which, after all, were the norm historically among navigators. I try to find solutions and tools for the broad range of students across that entire spectrum. There is no "ONE WAY" to do it. Now, you and I and many others on NavList definitely like the interesting, occasionally difficult math. That's our taste. But I also like finding easy, short solutions to some of these very long problems, and perhaps that makes me different from you. It seems like "easy" is almost a four-letter-word for you, but even experts should know easy methods, shouldn't they? And students often get a real kick out of them when they're just getting started.

You wrote:

"What, I wonder, rendered lunars less useful in higher latitudes? That reasoning escapes me."

It's not a big issue. Presumably just the obvious factors: greater probability of continuous overcast, less favorable geometries, etc. in higher latitudes. Whatever the reasons, you'll find more lunars in old logbooks taken within 30 degrees of the equator (yes, a bit broader than the tropics) than outside those limits. They would be equal in number by random chance alone since half of the Earth's surface is within 30 degrees of the equator.

-FER

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