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A Lunar Example: Sun and Moon vertically aligned
From: Frank Reed
Date: 2017 Dec 18, 15:53 -0800

A few days ago, I suggested in a post that you can learn a lot about lunars by considering a case where the Sun and Moon are vertically aligned. I'm going to modify the details just slightly, as marked in square brackets [ ] in the quoted section here. I wrote:

To puzzle this out for yourself, you'll learn an enormous amount by considering a case where the Moon and Sun are aligned on the same azimuth --one right above the other. Suppose, for example, I am in the tropics and I look in the [western] sky in the [afternoon] and see the moon low above the horizon and the Sun high up, right on the same azimuth. I have an assistant with a cheap plastic sextant who will shoot the altitudes of the two bodies, and I myself will measure the angle between the Sun and Moon with a fine, properly-adjusted Plath sextant. We approximately measure the altitude of the Moon (Upper Limb) as 20° 10' and the approximate altitude of the Sun (LL) as 75° 30'. In addition, I have with exceeding care and with the best accuracy I can manage measured the limb-to-limb distance between the Sun and Moon as 55° 17.3'. Also, assume 4' dip [17 feet height of eye] and zero index correction on both instruments. What, given these observations, would be the corrected lunar distance angle? You'll need one bit of almanac data to do the work, namely the Moon's HP. Let's suppose that's [54.5']. [Also suppose it's early January so you can look up the Sun's SD]. See if you can work it out.

Did any of you try the original version? With some slight modifications (primarily the given HP value), you should be able to work out this case. And remember, you don't need any complicated tables or tools to do this. I'll lay out the steps in outline:

• Pre-clear: for this you need the SD of the Sun and Moon. For the Moon, the SD is always 0.2724·HP so in this case that's 14.8'. For the altitudes you need to subtract 4' of dip from both and subtract the Moon's SD from its altitude and add the Sun's SD to its altitude. Next we add both SD's to the lunar distance, and in this case working carefully to get every tenth of a minute.
• Now look up the altitude correction for the Moon's center (if you're using the N.A. tables, you follow the rules for a bubble sextant altitude correction) and the altitude correction for the Sun's center (you can just use the standard refraction for stars here).
• The Moon is lifted by the clearing proces, while the Sun is lowered. And they're aligned vertically so the corrections just add/subtract directly from the pre-cleared distance to give you the cleared lunar distance.
• When you're done, your cleared distance should be nearly 54°59.7', plus or minus a tenth or two.

That's the setup to get started understanding this concept of "clearing lunars". If the math of lunars is new to you, you may want to stop here.

For folks with a little more experience messing around with lunars, this case presents an interesting mathematical issue. While it's a perfectly valid example which would produce an excellent value for GMT --dead-on really, it may break certain methodologies that head down the wrong road. In particular, the standard "direct triangle" solutions (however implemented) depend on the constancy of the difference in azimuth between the two bodies, called ΔZ or dZ (sometimes called, idiosyncratically, RBA). But mathematically we only need cos(dZ). If your method goes beyond this necessity and computes dZ itself, you'll be in trouble sometimes. The math is robust and can handle some degree of approximation in the altitudes so long as you calculate cos(dZ) and stop, but it become less robust, and in fact may give a hard, fatal error, if you calculate dZ as an angle. And if you get a fatal error, you'll be thrown back on your dead reckoning... which would be a shame because it's not a real problem with the math of lunars --it's a mistake that results from being too literal-minded about the math.

Finally, if you want to try this out from a "real world" context, visit the middle of the Pacific at 20°00' S, 176°30' W (sailing from Tonga to Fiji) on January 12, 2018 at 00:55:00 UT. There you will find the Sun and Moon aligned vertically with details just as given above. If you don't want to roll your own calculation here, you can just use my online web app to Clear a Lunar at http://reednavigation.com/lunars/. You can enter the inputs in less than 30 seconds, and you'll have the estimated error in your lunar in no time. While you're there, notice the value for "Cos Diff Azm". That's evidence of the mathematical issue I mentioned above.

Frank Reed
Clockwork Mapping / ReedNavigation.com
Conanicut Island USA

PS: I'll insert a lunars-relevant advert here. Read no further if that makes your eyes burn!!
I currently have for sale on ebay an unusual 7x35 scope that's really ideal for lunars and fits most Tamaya sextants and all of the Tamaya-likes, including e.g. "International Nautical" and "M.A.C." sextants. Naturally it's also a nice star scope with good "eye relief", much more comfortable to use than the standard scope on those sextants. This is a relatively rare item. Link to that auction: https://www.ebay.com/itm/142627596058. I also currently have up for sale the standard Celestaire 7x35 sccope for an Astra IIIB sextant and separately an Astra IIIB sextant itself in excellent condition. The full list: https://www.ebay.com/sch/clockwork-mapping/m.html.

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