A meridian is a longitude line. Your meridian is the longitude you
are on. Meridian transit—also called meridian passage (mer pass)—is the time a
celestial body crosses your meridian. The sextant height of the body above the
horizon at that moment provides your latitude with a just a few steps of
paperwork, and thus this has been a popular exercise in cel nav since earliest
days.
When it is the sun crossing your meridian, it is called Local
Apparent Noon (LAN), with the sun at its peak height in the sky, bearing either
due south or due north. Latitude from LAN is part of every text on cel nav
since the mid 1700s, but it has clung to prominence in modern times far beyond
its practical value.
We tend to think most often of stars and other bodies moving east to
west across the sky. The motion is left to right looking south, or right to
left looking north. These paths are called the upper transit of a body across
the horizon, because in both cases the bodies are moving over the top of their
nearest poles. Upper transits reach their highest angle above the horizon
as they cross the meridian bearing due north or south.
But if we look toward the elevated pole in either hemisphere, we
also see circumpolar bodies moving the other direction, west to east, as they
pass under their nearest pole. These bodies, in contrast, reach their lowest
elevation as they cross the meridian. These are called lower transits. We can
find our latitude from either transit.
To get an accurate latitude this way we need to measure the peak
sextant height (upper transit) or minimum sextant height (lower transit), which
defines the transit. This is usually accomplished by estimating the time of the
transit from our DR position and the Nautical Almanac—a process that takes just
minutes—and then we start the sights somewhat before that time, and take a
series of sights until we see that the maximum or minimum has indeed been
captured.
It can happen, however, that our best plans don’t work out. Just
as we approach the peak height in a noon sight, the sun might get covered by
clouds. We end up with a sight near the transit, but not exactly at the transit.
Likewise we might notice at twilight that we have time to take a lower transit
star or planet near the meridian, but discover that it is already going back
up. We missed the lowest point at the true transit. Or we might catch one going
down, but it gets too light out to see it start back up. Again, we have a sight
near the meridian transit, but not right at the transit.
A near miss on the transit, however, does not have to be a miss on a latitude determination. The paths of the bodies are well predicted, so if
we have accurate time and a reasonable DR accuracy we can figure out
what the height of the body would have been had we actually seen it cross the
meridian, and from this we can figure a latitude in the normal meridian transit
method. This type of near miss sight solved for latitude is called an
exmeridian sight.
And that is the process we will discuss here, but I must say up
front that the only reason we do this is because the USCG asks exmeridian
questions on their cel nav exams. We do not cover these in our regular cel nav
course. In fact, it seems that the latest publication of the USCG database of questions
(August, 2015) has more exmeridian questions than it used to, and they are
asking lower levels of licenses to solve them. [Need note here to prove this.]
If it were not for the USCG tests, and likeminded thinking
elsewhere, these methods would have disappeared 50 years ago. I like to think
of this as the way the USCG supports navigation schools, so we remain grateful
to them.
The reason these particular cel nav questions should have
disappeared is they are not needed—meaning they do not add to the
navigation—and they risk presenting a false sense of accuracy in latitude. The
sextant sight alone (regardless of what time it was taken) provides a line of position
(LOP) on the chart that we know we are on. If the time happened to be at mer
pass, the LOP is parallel to the bottom of the chart (i.e. it is a latitude
line), but if the sight is a bit earlier or later, the line is tilted some
degrees, depending on how far the time was from mer pass, as shown in Figure 1.
The navigation information is no different. We are somewhere on the line.
The exmeridian sights originated at a time when it was not easy to do what we now called a sight reduction to find an LOP—in fact, the methodology had not yet been discovered at the time, so this was a way to get something from a sight that would not otherwise be useful. This has long not been the case, so these should be thrown out.

Figure 1. LOPs near meridian passage. Each LOP is perpendicular to the azimuth line pointed to the GP of the sun. 
Indeed, the prominence of even the simple meridian transit
sights—forgetting about exmeridians—is likewise overrated, for exactly the
same reason. Meridian sights take much longer than regular sights, because we
need a longer sequence of sights to identify the peak or minimum value, and
these sights must be done at a specific time of day. In contrast, we do just as
good navigation taking the sights whenever we need them in the most efficient
manner.
The popularity of taking the noon sight is largely tradition; we could just
as well find our position at a comfortable time near midday, then DR to the
time of LAN we did (or will) experience, and record that LAN position in the
logbook for the sake of tradition.
At this point we show a few graphics to
illustrate meridian and exmeridian sights, then jump straight into shop talk
(ie we are going to assume the reader is familiar with standard sight
reduction) to show a trick way to solve exmeridians that is faster and less
prone to error than the conventional solution using Bowditch Tables 24 and 25.
The key concept is the zenith distance, z = 90º  Ho, is equal to
the distance on the globe between the observer’s position (Lat, Lon) and the
geographical position (GP) of the body. As a body circles the earth, z decreases
as it gets closer, while its height in the sky (Ho) increases, as shown in
Figure 2. At mer pass, z is minimum and Ho is maximum for an upper transit. If
the GP is close enough to us (meaning for practical purposes, Lat and Dec are
not more than about 80º apart) then we see the body in the sky throughout its
path around the earth. Viewed in the sky, it is called circumpolar. For these
bodies seen crossing our meridian on the other side of the earth, the situation
is reversed—at lower transit, z is maximum, and Ho is minimum.
A lower transit sight must be of a circumpolar body, which means Lat and Dec are Same Name and the Dec > (90Lat). Figure 3 below shows the simple way we can determine Lat from a
true meridian transit.

Figure 3. Computing Lat from meridian sights of circumpolar stars. Only addition and subtraction of Ho and the Dec is needed for upper or lower transits. The height of the elevated pole is always equal to our latitude. To be circumpolar, a star must be Same Name with Dec > (90  Lat). 
When we miss the true transit, we have the situation shown in
Figure 4.

Figure 4. Body heights at upper and lower transits and the height differences between them. The correction is not precisely the same for upper and lower transits even if they miss the meridian by the same time difference, because the peak height of the path affects its shape, but for either one, the correction is the same on each side of the transit at the same time offset. Since the problems can involve any latitude looking either direction, we recommend plotting a sketch of the LOP before choosing an answer. 
Sample USCG UpperTransit Exmeridian Question
253. (5.1.2.1D11) On 15 August [1981] an exmeridian altitude of the
Sun’s lower limb at upper transit was observed at 1130 ZT. Your DR position is
LAT 26°24.0’S, LONG 155°02.0’E, and your sextant altitude (Hs) is 48° 45.9’.
The index error is 2.6’ on the arc, and your height of eye is 51.5 feet. The
chronometer time of the observation is 01h 27m 38s, and the chronometer error
is 02m 14s slow. Find the latitude at meridian transit from the exmeridian
observation.
o (A) LAT 26°32.6’S
• (B) LAT 26°51.6’S
o (C) LAT 26°57.0’S
o (D) LAT 27°09.9’S
Solution
Step
1. Figure UTC from CT and look up Dec and GHA at UTC of sight time.
DRLon is 155º 02’ E; so zone description (ZD) = 155/15 = 10.33, or ZD = 10h.
Sight at ZT 1130, so UTC is about 0130, so UTC = 01h 27m 38s + 02m 14s = 01h
29m 52s. [Note on USCG time keeping]
From Nautical Almanac,
converted to decimals for later computations.
Dec = N 14º 07.9’ = N 14.132º
GHA = 201º 20.4’ = 201.340º
DR Lat = 26º 24.0’ S = 26.400º S
DR Lon = 155º 2.0’ E = 155.03º E
LHA = GHA + LonE = 356.370º
Step
2. Do a sight reduction by computation from the DR position, just
as if we were going to do normal navigation and plot an LOP… which we are in
fact going to do. We can use computation here (as opposed to Pub 229) because
to be prepared for the full range of USCG great circle sailings questions we
must know the same formula.
Hc = arcsin [ sin Lat * sin Dec + cos Lat * cos Dec * cos LHA ] Z
= arccos [ ( sin Dec  sin Lat * sin Hc ) / cos Lat * cos Hc ]
Sign Rules: enter all angles as positive, but if contrary name,
enter Dec as a negative number… which is the case at hand.
Hc = arcsin [sin (26.400) sin (14.132) + cos (26.400) cos (14.132)
cos (356.370) ]
Now find the azimuth angle Z.
Z = arccos { [ sin (14.132)  sin (26.400) x sin(49.316)] / [cos(26.400)
x cos (49.316)] }
Z = arccos (0.995596) = 174.6
(Had Z ended up negative, we would change it to Z+180, but this
example does not call for that.)
Step
3. Convert Z to Zn and Hs to Ho and figure avalue to plot the LOP.
(This is all standard procedure if we were just plotting an LOP from this sun
sight.) S Lat with LHA > 180, so
The standard rules for converting Z to Zn:
N Lat, LHA > 180, Zn = Z, else Zn = 360  Z
S Lat, LHA > 180, Zn = 180  Z, else Zn = 180 + Z
We have S Lat, with LHA > 180, so
Zn = 180  z = 180  174.6 = 005.4
(The sun was indeed just to the right of our meridian, looking
north.)
Ha = Hs ± IC  dip = 48° 45.9’ 2.6’  7.0’ = 48º 36.3’
Ho = Ha ± alt corr = 48º 36.3’ + 15.1 = 48º 51.4’
a = Hc  Ho = 49º 18.9’  48º 51.4’ = 27.5’ A 005.4
Step
4. Make a sketch of the LOP plot to figure the exmeridian Lat,
i.e. to find what our latitude would be if we assume the DRLon is correct,
which is part of the rash assumptions made in all exmeridian sights. We could
plot this and read it from the scales, but for exmeridian only we can compute
the offset; the sketch just keeps the situation in perspective, ie does the
correction make our exmeridian Lat larger or smaller than the DRLat, as shown
in Figure 5.

Figure 5. A
plot sketch to check the computation of dLat. The scale does not
matter, we just want to show that for S Lat with Zn to the N, an avalue
Away makes the Lat a larger number to the south. Note that in this
fictitious problem made up by the USCG for the exam, we find a latitude
that is near 30' off the DRLat with a method that must assume the
corresponding DRLon is correct. We know the LOP is correct, and we
should in practice just leave it at that.

The offset can be computed from dLat = a / cos Zn, which means the
exmeridian Lat = DRLat + dLat . In this example, dLat = 27.5 / cos 5.4 =
27.6’ and the final answer is 26º 24.0’ + 27.6’ = 26º 51.6’ S, which is answer
B.
Sample USCG Lower Transit Exmeridian Question
164. (5.1.2.1C1) On 16 June [1981] in DR position LAT 50°57.0’S,
LONG 53°03.9’W (ZD+4), you take an exmeridian observation of Acrux at lower
transit. The chronometer time of the sight is 10h 08m 18s, and the chronometer
error is 02m 12s fast. The sextant altitude (hs) is 23°49.0’. The index error
is 1.1’ off the arc, and your height of eye is 26 feet. What is the latitude at
meridian transit?
• (A) 50°41.2’S
o (B) 51°02.2’S
o (C) 51°33.0’S
o (D) 51°41.2’S
Solution
Step 1. Figure
UTC. (It is a pity we have to do this, but we do.)
In contrast to the first example, the ZT of the sight was not given, but with
ZD = +4h we can check CT. That is, CT = 1008, means either UTC = 1008 or 2208.
The first gives sight time = 0608 ZT; the latter gives 1808ZT. Under some circumstances we might be
able to judge from this alone, but not in this case, ie either one could be
twilight without further knowledge. So we need to check which one of these
happens to be in twilight when stars and horizon can both be seen.
Checking the 1981 Nautical
Almanac, and noting that DRLon 53º 3.9’ = 3h 32m (Arc to Time table), we
find that sight time in the morning (ie nautical to civil twilight) = 0646 to
0812 LMT = 1018 to 1144 UTC, and in the evening, 1630 to 1715 LMT = 2002 to
2047 UTC. So CT 1008 is a
morning sight corresponding to 1008 UTC.
UTC = 10h 08m 18s 02m 12s = 10h 06m 06s.
[Note this was about 10m before nautical twilight, but still
within reason as they could have had (in the virtual circumstance of this
fictitious problem) very clear skies that let them see the horizon early.]
Look up Nautical
Almanac data and convert to
decimals.
Dec = S63º 00.0’ = S63.000º
GHA = 229° 43.4’ = 229.723º
DR Lat = 50° 57.0’S = 50.950º S
DR Lon = 53°03.9’ W= 53.075º W
LHA = GHA  LonW = 176.658º
[Acrux is a bright star at the base of the Southern Cross, and now
we know that at sight time it was 176.6º west of us. This is confusing
information as we know we were looking south and would expect—from upper
transit experience—a Zn within a few degrees of 180, which for upper transits
would in turn call for LHA of a few degrees or (360 minus a few degrees).
But this is a lower transit sight, which frankly confuses the
reasoning. But it does not matter. We are just going to do a sight reduction
and plot the LOP. We do not care if it is upper or lower, or even if it is not
anywhere near transit. This note is just to say, do not be confused by the LHA
value.]
Step 2. Do
a sight reduction from the DR position.
Hc = arcsin [sin (50.950) sin (63.000) + cos (50.950) cos (63.000)
cos (176.658) ]
Hc = arcsin (0.406426) = 23.980º
= 23º 58.8’
and find Z from:
Z = arccos { [ sin (63.000)  sin (50.959) x sin(23.980)] / [cos(50.950)
x cos (23.980)] }
Z = arccos (0.99958) = 1.6 and Zn = 181.6
Step
3. Convert Hs to Ho and figure avalue to plot the LOP. (This is
all standard procedure if we were just plotting an LOP from this sun sight.)
Ha = Hs ± IC 
dip = 23° 49.0’ +1.1’  4.9’ = 23º 45.2’
Ho = Ha ± alt
corr = 23º 45.2’ 2.2 = 23º 43.0’
a = Hc  Ho = 23º 58.8’  23º 43.0’ = 15.8’ A 181.6
Step 4. We
could make the sketch, but at Zn = 181.6 the angle off the meridian is just
1.6º, which will not change the avalue, so dLat = avalue. We are S Lat
looking S, so an Away avalue makes the Lat a smaller number.
The offset can be computed from dLat = a / cos Zn, which means the
exmeridian Lat = DRLat + dLat . In this example, dLat = 15.8 / cos 181.6 =
15.8’ and the final answer is 50° 57.0’  15.8’ = 50º 41.2’ S, which is answer
A.
__________________
Final note:
We have used a shortcut method to find a logical value of the Lat,
namely if your DR Lon is right, then this is the right Lat. But this is
not the Lat at meridian transit. It is the Lat at the time of the
sight. These USCG exam questions inevitably ask for the "Latitude at
meridian transit," and then proceed to find the result using two tables
in Bowditch, which are specifically annotated as "...remember that the
value obtained [using Tables 24 and 25] is the latitude at the time of
observation, not at the time of meridian transit." And since course and speed are never given with these problems, this is all we can do.
So I maintain that these problems, along with the ones they have for finding compass error by amplitude, should be removed from the tests. The latter should be removed for the same generic reason, namely there is an easier and more accurate way to get the result... maybe not 80 years ago (before sight reduction tables) but certainly today.