USCG Chronometer Time — Bless Their Hearts

We have often had occasion to point out how the USCG supports navigation schools, but one of the contenders for the top price is their use of chronometer time (CT). Someone making up the tests must have read the fine print of the Bowditch definition and had an aha moment on how they could support navigation schools.

“chronometer time. The hour of the day as indicated by a chronometer. Shipboard chronometers are generally set to Greenwich mean time. Unless the chronometer has a 24-hour dial, chronometer time is usually expressed on a 12-hour cycle and labeled AM or PM.”

In other words, CT is the same as UTC, except for a possible chronometer error (CE) correction, but it can be kept on a 12-hr watch dial.  The insidious step in the USCG exam preparation was to use this concept of CT on a 12-hr dial, and then not tell the test taker if the chronometer time is AM or PM! That way, every single USCG cel nav exam question must start out with the candidate having to figure out what time they meant using other information in the question—it brings to mind the old computer game called Myst. 

Granted, candidates do need broader cel nav knowledge to figure out the time, but it runs the risk of implying this is, in some universe, a viable way to keep time, which of course it is not. No navigator in the world would use such a system.

So the first step of each USCG cel nav question is to, for example, read the given time of the event as CT = 10h 13m 20s and then use other information in the question to determine if this is 10h 13m 20s UTC (CT was AM) or is it 22h 13m 20s UTC (CT was PM).

Sometimes the process is straightforward, in that we get at least one zone time (ZT) and a DR-lon.  We can figure the zone description (ZD) and from this convert the ZT given to UTC.

The rule for finding ZD is round the Lon to nearest degree, divide by 15, and round the result to nearest whole number. West is + and East is -.  ZD is defined by this equation:

UTC = ZT + ZD.

Below are a few examples of how candidates must figure out the time.

Here is an easy one:
On 15 August your 0512 zone time position was LAT 29°18.0’N, LON57/G 57°24.0’W. Your vessel was steaming on course 262°T at a speed of 20.0 knots. An observation of the Sun’s lower limb was made at 0824 ZT. The chronometer read 00h 22m 24s and was slow 01m 34s.

Find ZD: 57/15 = 3.8, so ZD = +4.  Then UTC = 0824 + 04 = 1224, so CT must be PM, and we get the right UTC = 12h 22m 24s + 01m 34s = 12h 23m 57s.  The chronometer error correction (CE) is normal. If slow you add it; if fast you subtract it.

Another easy one:
On 10 March in DR position LAT 21°42.0’S, LONG 57°28.0’E, you take an ex-meridian observation of the Sun’s lower limb. The chronometer time of the sight is 08h 28m 17s, and the chronometer error is 00m 00s.

No zone time given, and it does not say upper or lower transit, but it has to be upper for the sun viewed from 21S. We also know it must be near midday on ZT.   Find ZD = 57/15 = 3.8, so ZD = -4 and use UTC = ZT + ZD to find ZT = 0828 - (-4) = 1228.  Or it could be ZT = 2028 -(-4) = 2408,  which is near local midnight, so this can’t be right. The first is correct and UTC = 08h 28m 17s   (CE = 0).

Another (sort of) easy one:
On 24 August in DR position LAT 26°49.4’N, LONG 146°19.4’E, you observe an amplitude of the Sun. The Sun’s center is on the celestial horizon and bears 084°psc. The chronometer reads 07h 55m 06s and is 01m 11s fast. Variation in the area is 15°W. What is the deviation of the magnetic compass?
o (A) 8.0°E
• (B) 8.3°E
o (C) 8.5°E
o (D) 8.7°E°.

This is an amplitude problem, so we know the sun is rising or setting. With bearing given as about 069T (084-15), it must be rising, so we know local time (ZT) is early morning.  The ZD = 146/15 = 9.7 or ZD = -10.  So we have two choices for the local ZT, which in turn tells us if the CT is AM or PM. If CT is AM, then ZT = 0755 + 10 = 1755 ZT, or if PM we have ZT = 1955 +10 = 0555 ZT the previous day. That is, going back to the definition of ZD, we have UTC = ZT + ZD = 0555 ZT (Aug 24) + (-10h) = 1955 Aug 23.

Without looking up the sunrise times, we know this must be 0555 ZT (ie CT was PM) and therefore the right UTC of the sight is 19h 55m 06s -(01m 11s) = 19h 53m 55s on Aug 23.

This beauty brings out a couple nuances for test takers to note. For one, the phrase “On 24 August in DR position...” implies that the date is the zone time date.  DR positions are always given in ZT—at least on all the test questions where the date is not ambiguous, and fortunately that is most of them. So we then learn that the CT given could indeed be on a different date. One way to check this is to compute the actual Zn of the sun at that time on each of the two days by standard sight reduction. The result would be:

19 53 55 Aug 23   GHA 117 51.5   dec N11 16.2   Hc = - 0 01.8    Zn = 077.3 (= 092.3 M) -> dev 8.3E
19 53 55 Aug 24,  GHA 117 55.5   dec N10 55.7   Hc = - 0 07.8    Zn = 077.7 (= 092.7 M) -> dev 8.7E

The other point we are reminded of is the wrong answers in USCG exams are not random. It is probably not fair to call the problems “trick questions,” but there are certainly trick answers. Namely, the wrong answers for most USCG exam questions are the result you would get from making a typical error. We see this here by noting the wrong date leads to one of the wrong answers, but the distracting answers in most amplitude questions are more subtle.  The wrong answers are mostly tied to making an error in the conventional amplitude solution (using Bowditch tables 22 and 23), which is the intended way to solve the problem. I have more to say about these amplitude questions elsewhere, but it is beyond the subject of figuring out what CT means.

Here is one a little more involved:
On 16 June 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.

Again, 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 much information alone, but not in this case— either one could be within 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, which corresponds to 1018 to 1144 UTC.  We find the LMTs in the Almanac, and then correct for the longitude to get the UTC, ie 0646 LMT + 0332 (Lon correction) = 0978 = 1018 UTC.

Evening sights are taken between civil and nautical twilight, which in this case would be 1630 to 1715 LMT, which corresponds to 2002 to 2047 UTC. 

We see that an AM CT is very close to morning twilight time, but the PM CT is not close at all. Thus we know then that the CT given must have been AM, and that the UTC we need for the problem is 10h 08m 18s - (2m 12s) = 10h 06m 06s.

And one final example:
On 30 March in DR position LAT 20°26.2’N, LONG 131°17.9’E, you take an ex-meridian observation of the Moon’s lower limb at upper transit. The chronometer time of the sight is 10h 36m 02s, and the chronometer error is 02m 06s slow.

This one is a more challenging CT puzzle, meaning we simply have to know they ask such questions and be prepared with how to solve them.  There are no zone times given for anything, which always makes the solution more interesting.  We can assume ZD is -9 (from 131º/15 = 8.7, which rounds to -9.

So we ask (after applying CE) does UTC = 10 38 08 or UTC = 22 38 08?

Checking to see what the ZT of the sight  would have been for each of these interpretations, we have to choose between: 1938 ZT Mar 30 or 0737 ZT on Mar 31?  But since this is a moon sight, we cannot get any hints from the twilight times—moon sights could occur in either twilight or throughout the day.  So they are handing us a bit more than the average amount of sleuth work. Unlike the sun, when we know mer pass is near midday on ZT,  we have no such hint about when the moon might cross the meridian.

At this point, we might resort to a nuance mentioned in an earlier example, namely the statement  that your sight is from a DR position on a given date essentially implies the ZT date of the sight time is the one given.   There is no zone time given, just the CT, and the CT is UTC (either AM or PM), but without other information, we have to assume the UTC date is the one given. In other words, this is either 1038 UTC on Mar 30 or it is 2238 on Mar 30.  These two different interpretations of the CT would in fact lead to different days on the ZT clock, which could be all we need to know to choose the right UTC. In other words, from the DR date argument alone we would have to choose the PM interpretation of CT leading to UTC = 22 38 08 Mar 30.

But these arguments about the DR and CT dates are just what we have discerned from looking at a lot of USCG problems. That interpretation of the dates is not spelled out in any official definition.  It would be reassuring to have some physical evidence to back this up—navigators do not like to  rely on just one source of information for crucial decisions.  So we should look into this a bit further.

It is an upper transit, so that means the GHA of the moon must be near the DR-Lon at the sight time. The Lon of 131º E corresponds to a GHA of (360 -131) = 229º.  So we can look up the GHA at 1038 UTC on Mar 30 and at 2238  UTC on Mar 30 to see which one puts the moon on or near the meridian, ie has a GHA of about this value.

Using the Nautical Almanac, we find that on Mar 30, 1981:

1038 UTC,  GHA moon =   46º 39.5’ 
2238 UTC,  GHA moon = 220º 21.5’

At 1038z the moon is no where near the meridian (actually not even above the horizon), so the CT was clearly set to PM, and we have the correct UTC = 10h 38m 08s. 

To save time, you could learn this faster by just checking the GHA at the whole hours in the Almanac (1000 and 2200) since one is not even close. 

Looking ahead, with a GHA of 220W and a Lon of “229W” (ie 131 E), when we finish this problem, we should expect the moon to be to the left of the meridian. Its GP (moving west around the globe) has another 9º to go before it gets to the DR meridian.