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Wednesday, October 19, 2011

Ex-Meridians for USCG exams

This post is about celestial navigation, however it is most emphatically NOT part of the "Celestial Navigation 101" series. This topic is beyond the scope of even what we normally teach as "advanced celestial navigation". However, esoteric though it is, it shows up on USCG licensing exams for 1600 ton Mate Oceans and above. So this post is mostly intended for Oceans license candidates in the US Merchant Marine; however, it may have some interest for those working with celestial navigation generally. It happens that at work today I stumbled across a very simple way to solve ex-meridian problems, and so I thought I'd pass that along here.

We have not yet discussed Local Apparent Noon sights in the CelNav 101 series, so let me start with a very brief description of this. Local Apparent Noon, or LAN, is the moment when the sun crosses your longitude line. At that moment it is the highest it is going to be all day. Measure the altitude of the sun with a sextant at that moment, and then, comparing that altitude to the known Declination (latitude) of the sun, you can determine your own latitude. I'm going to explain this in more detail in the CelNav 101 series, but that's the gist of it.

The advantage of LAN is that it is a very easy sight to take with a sextant, and it is very easy to compute. That, however, is its only benefit. The fact that the Line of Position derived from it happens to be a latitude line is irrelevant, because it is as inherently flawed as any other LoP derived from a sextant sight. The real reason we teach LAN at all is that it is a good confidence-builder for fledgling navigators. Once the student masters basic sight-reduction, they never again have a need for LAN, because any celestial object at any time can give them just as much information as the sun can only one time each day.

The technique used to derive latitude from LAN can also, theoretically, be applied to any other celestial body when it crosses your meridian. There's no reason to do so, but it's possible.

These days we confidently rely on twilight sights of stars and planets to confirm our GPS only once each day. However, once upon a time, before the invention of GPS, the "noon sight" was a routine part of the navigator's daily work. If the navigator failed to obtain their noon sight, either due to cloud cover or simply not getting up on deck in time, this was a significant loss of navigational data. So methods were developed to allow the navigator to take a sight a few minutes after the sun had crossed the meridian, and still be able to derive a reasonable latitude from it. This is what is meant by an ex-meridian sight. Centuries ago, when computing a sight reduction required that spherical trigonometric calculations be solved long-hand, ANY celestial line of position which did not require that was an obvious benefit. Since the invention of logarithmic tables (and later calculators and computers) to perform the trig for us, this has not been the case. If a navigator happens to miss their noon sight, they simply take a sight of the sun at whatever time is convenient and reduce that into a line of position which is every bit as useful as an LAN derived latitude. Put another way, no modern navigator worthy of the name would ever use an ex-meridian sight, because a simple sunline is easier and far more accurate.

Except, of course, for USCG license candidates, who are expected to.

There is another CelNav technique on USCG exams which is basically useless and can be done more easily and accurately with another technique which is also required on the same exam. It's called an Amplitude, and is basically a means of computing the azimuth to an object without getting into the spherical trig necessary to compute an azimuth. But with sight-reduction tables or calculators, azimuths are simple, and we have to be able to use them anyway. So the easy way to avoid dealing with amplitudes on a USCG exam is to simply work amplitude questions as azimuth problems.

It turns out that you can do exactly the same thing with ex-meridian sights. Simply compute it as a normal sight-reduction. Because the sights are necessarily either nearly north or nearly south, the intercept can be applied directly to the Assumed Latitude or DR Latitude to derive the latitude of the ex-meridian. That's all.

Here's an actual example from the USCG database. In it they're asking to compute your latitude from the meridian transit of the star Dubhe, in the Big Dipper.

-- On 8 May 1981, in DR position LAT 30°26.0'N, LONG 46°55.1'W, you take an ex-meridian observation of Dubhe. The chronometer time of the sight is 11h 10m 54s, and the chronometer error is 01m 18s slow. The sextant altitude (hs) is 58°35.0'. The index error is 1.5' on the arc, and your height of eye is 44 feet. What is the latitude at meridian transit?

a) 30°12.5'N
b) 30°19.8'N
c) 30°27.6'N
d) 30°35.8'N

30°19.8'N (answer b) is correct. However, if I simply do this as a standard sight reduction (cheating and using a StarPilot calculator, but the result is the same regardless), I get 6.2' Away from 358° T, from my DR latitude of 30° 26.0'N. Since my ex-meridian sight is always going to be essentially north or south of me, I simply subtracted my intercept from my AP (actually DR in this case) and get 30° 19.8'.

In this case it works out to be exactly right, but even worst case it can't possibly be enough off to lead me to pick any of the other answers.

So if you are taking a USCG Oceans Master or Mate exam, you can effectively NOT study the Amplitude or Ex-Meridian methods and still do fine on those questions, so long as you know how to compute a simple sight-reduction. Incidentally, you can also use your sight-reduction method to determine great-circle courses and distances.

At some point soon we'll do a comparison of the two or three leading methods of computing a standard sight-reduction.


  1. I am appalled! Refer to NAUTICAL LOG Post "WORTHY OF THE NAME".

    Good Watch

  2. Hi Peter,

    Was unable to find the post, could it be under a different name?

    I'm very curious to hear your rationale for continuing to teach formal Ex-Meridians and Amplitudes for modern routine celestial navigation. Perhaps it is a failing of my own under-active imagination, but I cannot envision any scenario in which I would have the ability to formally compute an amplitude of a celestial body in which I would also not have a means of sight-reducing them to derive an azimuth. I can envision a very limited application for emergency techniques for determining amplitude in the absence of a nautical almanac or the equivalent (we teach these methods at Starpath as part of our Emergency Navigation curriculum), but those techniques would not yield an amplitude of sufficient accuracy for a USCG exam.

    Ex-Meridians, at least as the USCG examines them, are of even less utility. At least with an emergency amplitude I have a method of deriving a rough azimuth without any tables or calculations, but for an ex-meridian to have any utility you have to presume that you managed to jump into a lifeboat without an almanac but managed to grab your 1981 edition of Bowditch volume 2. Which is maybe not an impossible scenario, but not one I'd plan for in an 80-hour celestial navigation course.

  3. Thank Poseidon for the internet! I am the guy you wrote this for!
    I am prepping for my USCG 2/M Oceans, and ran into 3 'Ex Mer' problems today. I suspected I could just solve them the long way, but figured I would research first.
    If I had my V2, maybe I would have figured it out, but as you pointed out, that is a fatal flaw.
    The 2002 ed. on my ereader does not even mention them.
    Thanks again!