NavList:

Francis you wrote:
" I've successfully.used the Fuller slide rule to do lunars using Letcher's formulae,and the John Karl method.I'll try your "easy" series method soon."

Letcher's method for clearing lunars is just another variant of the series method. It's his own slightly quirky approach, but it's fundamentally the same thing. The idea that there are dozens and dozens of methods of working lunars is a classic "lumpers versus splitters" taxonomy problem. If we highlight every tiny variant, then we can fill our "stamp collection" with endless methods and make a mountain out of a molehill. If we recognize the basic mathematics, then there are only a handful of methods, and the whole subject is much less difficult.

You wrote:
"I get good results when starting from RBA (azimuths difference?) but cannot work out how to get the unknown RBA"

Ditch the expression "RBA". It wasn't used historically, and it's not used anywhere except in John's book. You don't need an acronym to describe something as trivial as azimuth difference. It adds no value. You DO see that this is just the difference in azimuth between the two bodies, right??

As for "good results", I commented on that in the last message. Any results which yield an error of 1' at each triangle solution would be considered mediocre to poor results by the standards of lunars. But as I said, you could use rough results as a sanity check. Of course, it doesn't really matter, since there's no real reason to work out this "Lunars by Bygrave" problem except hobbyist amusement, right?

You also wrote:
"A few weeks ago you mentioned another short clearance method. Did you by chance get time to write that up?"

I've written it up in NavList messages before, and I described the basics of it not long ago. You do realize that it involves no trig, right? So it won't require that Bygrave! :)

Here's another brief version of it:
In a surprisingly large number of cases, primarily in the tropics, you can clear lunars by mapping onto the equivalent "vertical circle" case. The three inputs to the problem, as always, are the observed lunar arc, and the observed altitudes of the Moon and the other body. The trick is to throw out the altitude of the Moon and replace it with the altitude that would force the other two quantities onto a vertical circle which effectively "collapses" the spherical triangle problem.

For a specific example, suppose you're at 5°00'N and 45°00'W on December 9, 2013 at 19:00:00 GMT. The Sun is in the WSW 23°33.0' high (altitude of center, obs'd alt. corrected for SD and dip), the Moon is in the ESE 59°31.2' high (center again), and you measure the lunar distance as 92°22.2'(that's center-to-center after subtracting both SDs). Big complicated spherical trigonometry problem to clear that, right? Or maybe not. Note that the three angles add up to 177°50.7'. If the two objects were on a vertical circle, then the sum would be 180° exactly. So throw out the Moon's altitude and replace it by 64°04.8'. The justification for this step is a story for another day. That fictional altitude is the altitude the Moon would have if it were in the ENE, exactly opposite the Sun in azimuth, on the same vertical circle with the Sun, keeping the other two quantities fixed. Now the sum is 180°00.0' exactly. Ah, but in the case of a vertical circle, the clearing process is trivial. You just add the altitude corrections directly onto the center-to-center distance, and you're done! No spherical trig at all. Note that the Moon's altitude correction is determined from this modified "fictional" altitude for the Moon, but there's no other complication. The clearing equation is nothing more than LDclear=LDobs+corr1+corr2 (with appropriate signs naturally). The only catch is that you need a test to decide whether this technique is applicable; when can we skip the spherical trigonometry? Generally, when the Moon is high in the sky and the distance is close to 90°, it works great. And of course if the two bodies are near a vertical circle in the first place, it also works great. Either a table or a (very brief) calculation can test this out. Since historically navigators frequently shot lunars under just those circumstances where this trick would work, it could have saved a lot of time. And if I get into my Tardis and visit some 18th century lunarians, I can teach them this and they'll all be very grateful, and then it will be part of history and nothing new under the sun.

You concluded:
"I have found a paper by wales 1788 clearly describing lunars taken on the
dolphin, prior to Cook and almanacs and cleared by using 1767 Mariner's Guide. Will send if interested."

That's not necessary unless you have something specific that you want to discuss. I'm familiar with the paper by Wales and I'm familiar with the Mariner's Guide. You had previously asked if Cook and company on their first voyage could have used the Mariner's Guide to generate their own almanac data after the short run of Nautical Almanacs then available had expired. I agreed that they could have done just that. This was fundamentally equivalent to calculating almanac pages on the go. Cook was accompanied by experts in lunar ephemeris calculations on that first voyage, and they had direct access to the complete calculations of the Royal Observatory (as of the date of departure). So they may also have had early versions of the data for the years after the previously published Nautical Almanacs which could have considerably reduced the work. There are a number of possibilities. But in any case, there was no particular difficulty here. I don't see any puzzle to solve.

-FER

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