The first measurement was kindly provided by Angeline Pendergrass during a research voyage on the R/V Thomas G Thompson off the coast of WA state. She took the data from the aft deck at a height of eye estimated to be10 ft.
Short of having an accurate watch at hand, the procedure was to take a cell phone picture of the sunset, just as the upper limb dropped below the visible horizon, which marked that time in her phone. Then she proceeded to the wheelhouse and took a picture of the GPS screen, which showed the UTC and the location of the vessel which was drifting at the time. The cell phone time showed the delay was 1m 7s, so she could then figure an accurate time of sunset with the associated position.
The results were:
Sunset (upper limb crossing the visible horizon)
03:17:49 UTC (4/23)
Lat 48º 48º 16.33’ N
Lon 123º 59.00’ W
Height of eye = 10 ft
(Though it is far beyond specs related to this discussion, this vessel has special props and instrumentation to hold a steady position in rough conditions. The actual position was known in principle to sub-meter accuracy during this period. The above numbers are rounded down.)
At this stage there are several ways to analyze this information.
PART 1. COMPUTER SOLUTION
The easiest and likely most accurate is to not go for an actual Lon, but rather just compute a regular cel nav LOP. To do this, we just interpret this as a “sextant sight” with Upper Limb Hs = 0º 0’ from an HE = 10 ft at the given time.
A key point to keep in mind for emergency determination of Lon is we are free to assume we know our latitude precisely. The reason is we have many ways to find accurate Lat without time, and with time we can get it very easily from a noon sight. Not to mention that we also get accurate Lat from any set of star sights, even if the watch used is wrong. The Lat will be right, but the Lon will be wrong by an amount directly related to the watch error at the rate of 15’ Lon error for each 1m of time error.
In this example we will assume we know our Lat and are just going for our Lon. So for now we just choose some random value of what we might have thought our Lon was before we did the sight, Let us say our DR Lon before the sight was 124º 05’ W.
So we do a normal sight reduction with this data:
HE=10 ft. IC = 0 ZD = 0, WE = 0
Hs UL sun = 0º 0’, UTC = 03h 17m 49s on April 23, 2013
DR position = 48º 16.33’ N, 124º 05.00’ W
You can solve this with a Nautical Almanac and Sight Reduction Tables, or with a computer or calculator program. We used the StarPilot program developed at Starpath and from this you will find an altitude intercept of a = 4.3’ A 290.1 T.
This result is plotted in the picture below.
With the plot expanded to 30’ parallels we find a Lon of 124º 01’ W, which is off by only 2’, well within the uncertainties of this process, due primarily to refraction uncertainties at the surface.
You can get better results by expanding the plot to 6’ parallels, and you can check the sight reduction with this data from USNO (see www.starpath.com/usno).
Celestial Navigation Data for 2013 Apr 23 at 3:17:49 UT
SUN GHA 229 52.2 Dec N12 33.5
So the summary so far is a timed sun set can get you a Lon from a known Lat using a simple sight reduction. The easiest way is to use a calculator for the process, but it is all doable by hand. Must be careful in figuring the a-value as you will be subtracting two negative numbers. (We will fill in that detail later. Students in our Advanced Ocean Nav Course get to slug that out in practice exercises.)
We should note that this level of accuracy was fortuitous. We should expect an uncertainty in this type of measurement of about ± 5’ ...even with everything well known. In this case, the HE could have been off some. If it was really 15 ft and not 10 ft, then this would change the a-value to 5.2’ A 290.1, which makes the final Lon about spot on correct. (In fact, if the deck height were 10 ft, then HE was closer to 15 ft, but again, the uncertainty to expect is as noted. We need to check that with the ship specs.)
PART 2. TABLES SOLUTION
Now you get a neat interpolation problem: find xx:xx
50º 00’ N 19:06
48º 16.3 N xx:xx
45º 00’ N 18:56
and the answer is 19:02:33 for UTC sunset (observed at Greenwich)
The other option you might have are tables from the back of the Tide Tables, as shown below.
And now you have a really neat double interpolation problem. Note we were lucky with the Almanac since the date happened to be in the middle of the 3 days covered on each page. If that were not the case, then even the Almanac calls for a double interpolation. You can do this by hand, or use tools we present at www.starpath.com/calc :
Apr 21 19:03
Apr 23 xx:xx
Apr 26 19:11
and we find xx:xx = 19:06:12 for 50 N on apr 23
Apr 21 18:59
Apr 23 xx:xx
Apr 26 19:06
and we find xx:xx = 19:01:48 for 48 N on apr 23
and now interpolate for Lat
50º 00’ N 19:06:12
48º 16.3 N xx:xx:xx
48º 00’ N 19:01:48
and we find xx:xx: = 19:02:24 for 48º 16.3’N on Apr 23.
The first thing we see is the two tables do not yield the same results. the almanac gave us 19:02:33, and reason is simple, all times are rounded to nearest whole minute, so we have 29s floating around on each end of each number. There is no way around that. We must just accept right now that if you are using tables, we have an extra uncertainty which we might estimate at ± 30s, but that would be too harsh. That is, we are getting our result from at least 2 numbers and more likely 4 numbers, so the uncertainty in each of those could sometimes cancel each other out. Thus are extra uncertainty is more like one fourth of square root of the sum of the squares, which would be sqrt(4x900)/4 = ± 15s.
In this example we might then say the best value we have is the average of the two ± 15s, which is 19:02:29 ± 15s. The 15s automatically adds a Lon uncertainty of about ± 4’ of Lon.
Note too that if we had nautical almanac, we could compute the actual time of sunrise very accurately, but with those tools, i.e. almanac and calculator, we are better off just doing a sight reduction, which has all those steps buried in it.
So now we have this information: Observed from Greenwich (Lon) = 0º the sunset would occur at 19:02:29 UTC but we observed it at 03:17:49 UTC. Therefore what must our longitude be?
Observed time of sunset (UTC) = Greenwich sunset time (UTC) + Lon (West)
so Lon West = Observed time - Greenwich sunset time = 03 17 49 − 19 02 24, which is negative, so we must add 24h:
03h 17m 49s
−19h −02m −24s
= 8h 15m 25s
then convert angles to time at the rate of 1h= 15º, 1m = 15’, and 4s = 1’
8h = 120º, 15m = 225’ = 3º 45’ and 25s = 6’, so Lon = 123º 51’ W, compared to known Lon of 123º 59’, or off by 8’, well within our known uncertainties.
So it all works. We have about 8 sights taken from the JRH rowboat which we are still evaluating. We have the correct times, but have to go back to the logged positions to make the comparisons. We will add that data shortly.