Latitude by Meridian Transit, Ex-meridians, and the USCG Cel Nav Exam

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 ex-meridian 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 ex-meridian 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 ex-meridian 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 like-minded 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 ex-meridian 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 ex-meridians—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 ex-meridian 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 ex-meridians 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.

Figure 2. Relationship between observed height, Ho, and zenith distance, z. Adapted from
Celestial Navigation: A Complete Home Study Course.

A lower transit sight must be of a circumpolar body, which means Lat and Dec are Same Name and the Dec > (90-Lat). 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 Upper-Transit Ex-meridian Question

253. (5.1.2.1D-11) On 15 August [1981] an ex-meridian 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 ex-meridian 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. DR-Lon 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) ]

Hc = arcsin (0.75830) = 49.315º = 49º 18.9’ [ See key strokes ]

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 a-value 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 ex-meridian Lat, i.e. to find what our latitude would be if we assume the DR-Lon is correct, which is part of the rash assumptions made in all ex-meridian sights. We could plot this and read it from the scales, but for ex-meridian only we can compute the offset; the sketch just keeps the situation in perspective, ie does the correction make our ex-meridian Lat larger or smaller than the DR-Lat, 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 a-value 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 DR-Lat with a method that must assume the corresponding DR-Lon 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 ex-meridian Lat = DR-Lat + 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 Ex-meridian Question

164. (5.1.2.1C-1) On 16 June [1981] in DR position LAT 50°57.0’S, LONG 53°03.9’W (ZD+4), you take an ex-meridian 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 DR-Lon 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 a-value 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 a-value, so dLat = a-value. We are S Lat looking S, so an Away a-value makes the Lat a smaller number. 

The offset can be computed from dLat = a / cos Zn, which means the ex-meridian Lat = DR-Lat + 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.

The Navigation Triangle

The sight reduction method we use in celestial navigation is based on this triangle drawn on the earth's surface. You do not, however, need to know anything at all about this triangle to do celestial navigation and do it well.  It comes into play when we wish to understand more about the fundamentals. It is also valuable to know for USCG cel nav exams as it can save much time on some problems, and it is essentially required to solve selected great circle sailings problems—ie, in principle you could use Pub 229 for these, but the test room only has vol 2, whereas some of the sailings questions start off in vol 1.

The navigation triangle. Image from Celestial Navigation: A Complete Home Study Course, Second Edition.

The corners of the triangle are the north or south pole (P), the assumed position (AP), and the geographical position (GP).

The sides of the triangle are:

P-AP = (90° - Lat);
P-GP = (90° - dec); and
AP-GP = (90° - Hc), which is the zenith distance.

The angles between the sides that we care about are:

Local hour angle (LHA) = the angle between P-AP and P-GP

The azimuth angle (Z) = the angle between AP-P and AP-GP

The side we care about is the zenith distance (z) which is the length of the line from AP to GP.

For any celestial sight, we know two sides and the included angle: (side, angle, side) figured from (dec, LHA, Lat), and from this we can mathematically figure the rest of the triangle. The parts that are solved for in a sight reduction by tables or calculator are Hc and Z. The formulas are:

Sin(Hc) = Sin(dec) x Sin(Lat) + Cos(dec) x Cos(Lat) x Cos(LHA)

SIGN RULES when solving by direct computation:

(1) All angles are treated as positive, regardless of hemisphere, except for Contrary Name, make Declination a negative angle, regardless of hemisphere.

and

Cos(Z) = [Sin(dec) - Sin(Lat) x Sin(Hc)] / [Cos(Lat) x Cos(Hc)].

SIGN RULES when solving by direct computation:

(1) All angles are treated as positive, regardless of hemisphere, except for Contrary Name, make Declination a negative angle, regardless of hemisphere.

(2) If the final Z is negative, then  Z = Z + 180, regardless of hemisphere.

-----------------

The formula is for Z (azimuth angle) and not Zn (azimuth). The normal rules for converting Z to Zn based on LHA and Lat apply.

Note also, we have two Z's. Upper case Z = azimuth angle; lower case z = zenith distance. This is traditional, so we do not change it.


Above notes and graphics adapted from Celestial Navigation: A Complete Home Study Course, Second Edition.

First pass at cell phone barometer calibration.

Most modern cell phones have a barometer in them that could be a valuable aid to your navigation.  They tend to draw extra power if we try to record pressure history with them, but we can always turn them on, read the pressure and write it in the log book, and turn it off—ie, the way mariners have recorded pressure history for several hundred years.  The only difference now is these phone barometers are generally pretty accurate, whereas many barometers on boats are not very accurate. As noted below, we have easy and free ways to check their calibration online and we should do so. But even without that, chances are, right out of the box they could still be the most accurate barometer on the boat. More than adequate apps are free; in fact the more complex expensive ones are not as functional in many cases. If you have not done so, it will pay to download a free one and play with it.

So it was inevitable that we start a study of their accuracy and indeed we report here a first  step in that direction... thanks in large part to Erik Kristen who brought a few over to the office this afternoon to test.

Below are the results for a Galaxy S5, iPhone 6+, and a Samsung Note 2.






The actual pressures displayed were at best accurate to within in 0.8 mb and the worse to within 2.2 mb. But the bigger question is are these off by the same amount at all pressures we might run into, and the answer is yes, that seems to be the case in the ones we tested, which is not really a surprise.

In short, they are pretty good. They were each linear over the range covered: best at ± 0.1 and the worst ± 0.2, though we should not conclude anything from this first study on relative models. The S5 would not work below 950 mb, but that was likely an issue with the particular app we were using and not the phone barometer.  A different app would most likely go to lower pressures.

The best app is the simplest one. We just need the sensor pressure and a way to offset it. Many apps include confusing ways to set the pressure to local airport values, or they use GPS elevation, or smarter ones could use the National Elevation data set based on Lat-Lon, but in any case these schemes do not work. The errors in these corrections are larger than the inherent sensor error. We want just plain station pressure read from the sensor, and we have to know it is not sea level pressure till we are on the boat.

The best iPhone app I have found is called "hPa," which is the abbreviation for hecto Pascal, a pressure unit the same as a millibar. Maybe the real name is "barometer." It is in the iPhone App Store.

More news on this as we learn more, but the first pass at testing them looks very promising. You can always use the free resource at www.starpath.com/barometers to set your phone to the right pressure.

We have a link to phones with barometers in the list of resources at www.starpath.com/cyc, but you have to do a bit of research on older phones. For example, the baro phone link lists Moto X (Gen 2, 2014)  as having a barometer, and indeed many of the original spec sheets listed one, but it seems this phone does not have a barometer in it... at least one that can be accessed by any 3rd party apps, which are required to read the pressure.