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From: George Huxtable (no email)
Date: Tue Mar 07 2006 - 12:51:17 EST
Two days ago, I accidentally posted an earlier version of this message before it was ready, and it
contained some errors. These, I hope, have since been corrected in the amended text below, but I
don't know how many other errors may remain.
I have tried to include below the two tables copied from Walden's original messages. These were
perfectly
clear when I received them, but have had all their spacings garbled in the course of passage through
my computer, which I can't see how to fix. Sorry about that. If you want to refer to the details of
those tables, it would be best to go back to his earlier messages, preferably that of 3 March.
======================================
First, I would like to add my appreciation (to Frank's) for the fine detective work that D.Walden
has done on 2 March in transcribing and understanding those somewhat cryptic numbers in the Charles
W Morgan workbook of 1897.
He wrote "Corrections, additions, comments are invited.", and I have a few comments.
He states- "First, there seems to be, in fact three separate sight reductions on the page. (The
lines and boxes don't seem to accurately separate the three calculations, so they may have been
added later.)"
I agree completely. The next, similar, sight reduction on that page seems to have been made a day
later, on the afternoon of the 20th. Then there appears to be a gap of 3 days, in which no such
calculations were made (cloudyweather, perhaps?) followed by the final sight-reduction on that
page, on the 24th. That is judging
by the change in declination from day to day.
Now for the upper calculation, the one that D.Walden has analysed. It's been labelled "Feb 19th
1897". I take it to be an afternoon observation, somewhere South of Japan.
Walden writes-
"After some guessing and iterating, using the fact that the Morgan is likely in the North Pacific
at this time, I take 6"34"31 to be GMT (chronometer + correction), 25"18 to be Latitude North,
129"33 to be the final result for Longitude East. Using this date (obvious), time, Latitude, and
Longitude, I go the USNO online navigation site, and get the following:"
___________________________________________________________________________
U.S. Naval Observatory
Astronomical Applications Department
Celestial Navigation Data
Celestial Navigation Data for 1897 Feb 19 at 6:34:31 UT
For Assumed Position: Latitude N 25 18.0
Longitude E 129 33.0
............ Almanac Data............. | Altitude Corrections
Object .GHA... Dec .....Hc ...Zn... |.. Refr ..SD ..PA ..Sum
..... o. '.... o.. '... o .. ' . ....o... |.. '... ' ...' ..'
SUN..275 07.7.. S11 09.2 ..+33 14.0 . ..235.6 |.. -1.5 .16.2.. 0.1..14.8
__________________________________________________________________________
Those numbers do indeed correspond to the Sun coordinates at a Greenwich date and time, by modern
reckoning, of 6 34 31 UT on 19th Feb 1897. By the almanac of Layton's's time, however, that would
have been regarded as a Greenwich time of 6 34 31 am on the morning of 18th Feb, and at Greenwich
the astronomical (almanac) date wouldn't switch to 19th Feb until Greenwich noon, 5 hours 29 minutes
later. Layton was working from Sun coordinates in the almanac for that Greenwich noon, the start of
the 19th. His local time and date, in the Pacific, was mid-afternoon of 19 Feb, so the date quoted
on that page is correct. But it's as well to keep those complications of astronomical time in mind.
It's a matter that usually makes my head hurt when I try to think it out. I often get it wrong, and
hope I have it right now in this case.
Nautical almanac dates didn't change to the present civil-day pattern, with date changing at
midnight rather than noon, until 1925.
==========================
Walden continues-
Taking only the first and uppermost calculation, I first transcribe the handwritten numbers below
for easier reference:
6 32"50..... 33"03 ...............11"04"09
. ...1"41....... 12 .....04379.......... 7". 8
_______............... _____........................................................ ________
6 34"31 ....33"15..... 00825........ 11"11"17
...........25"18 .....24677 .........2"34
.........101"08..... 86116....... ________
... ______ ....._____ ........11"08"43
129"33 ....159"41 ....15997........ 90
....... ______.................... _____.............................________
Feb 19th 1897. 79"50 ............. ..101"08"43
........33"15
....... ______
.......46"35 .....2"58"46
......... .........13"58
................ _________
.................15"12"44
................ 6"34"31
............... _________
............... 8"38"13
Comment from George. I agree with all those transcribed numbers, which seem perfectly consistent.
It's Layton (faithfully copied by Walden) who confusingly uses those " symbols to separate the parts
of an angle, rather than the conventional symbols for degrees, minutes, and seconds.
___________________________________________________________________________
Walden adds- "The USNO's Sun's declination South, of 11 09.2 is close to the 1897 value of 11"08"43
(almanac 11"04"09 + 7"8, correction for 8 hours after tabulated time (=the daily difference /3) -
2"34 correction for 1hr 26min less than 8 hours after tabulated time)."
========================
Comment from George-
Although this process gets to the right answer, I don't think the explanation is quite right. 11deg
04' 09" must correspond to the Sun's tabulated declination at the following Greenwich noon, 5h 25m
29s after the observation. The daily change in declination is a reduction of about 21' 12". Layton
seems to go about the process of interpolating for the change in latitude in a curious way. First he
adds 7' 08", which Walden takes to be a correction for 8 hours, a third of a day, which would take
him back to 8 hours before noon, or 4 am. Then he has to adjust further for the interval from 4 am
to the time of the sight, at 6h 34m 31s, so he has next to allow for the change of dec in 2h 34m
31s, and subtract it, because dec is decreasing. Layton subtracts 2' 34", which I take to be 3
hours' worth (not the 1 hour 26m that Walden refers to, which appears to be an error, and not 2h
34m, as it ought to be). My guess is that Layton is correcting for changing declination only to the
nearest hour, which is (arguably) accurate enough, to the nearest half-minute of arc. But his
procedure for
interpolating seems remarkably ham-handed.
==============================
This next bit of text is for anyone who wants to check Layton's results, but without access to a set
of log trig tables. Layton had to use logs to work out how far from noon his time-sight was.
However, for those with a calculator with trig functions, there's no longer any need to indulge in
logs at all. Without the complications that the use of logs imposes, exactly the same calculation
can be written much more directly as
t (in degrees) = arc cos ( (sin alt - sin lat sin dec) / (cos lat cos dec)) in which lat and dec are
positive if North, negative if South. To get t in hours, divide by 15.
In the formula above,
t is the angle- or time-difference from apparent noon at Greenwich, so corresponds to the longitude.
alt comes from the Sun's measured sextant altitude, after the necessary corrections have been made,
in this case amounting to 33deg 15', or 33.25deg.
lat is given as 25deg 18' North, or +25.3deg. Presumably this was deduced from a noon altitude of
the Sun. taken a few hours before, and subsequently adjusted to allow for ship's run in the
interval. That working isn't shown on this page, but may perhaps have beeen written down elsewhere,
perhaps in a different logbook.
dec is the Sun's declination, taken from the tables for Greenwich noon, and corrected for the time
interval from noon, arriving at 11deg 08' 43" South, or -11.146deg.
That formula gives a result of 44.663 deg, or in hours 2.9775, or 2h 58m 39s. That can be compared
with Layton's result from his log calculation of 2h 58m 46s. The 7-second difference relates to some
roundings-off that he has seen fit to made.
=====================
There are some unusual features about the way Layton shows his log calculation, copied again here
as-
04379 log sec lat, where lat = 25 deg 18'
00825 log cosec polar distance of 101deg 08' 43"
24677 log cos s/2, 79deg 50'
86116 log sin (s - h), 46deg 35'
------ sum the numbers above to give
15997 but how do you search for such a number in the log hav t table, in order to find t as 2h
58m 46s?
That calculation shows a bit of short-cutting. If we write down the numbers in full, in formal
textbook manner, exactly as the log trig table provides them, we would get instead-
0.04379 (some tables would give this as 10.04379, effectively the same thing)
0.00825 (or perhaps 10.00825)
9.24677
9.24677
-------- sum them
19.15997 and now we can adjust that number in front of the decimal point by discarding tens from
it to end up with 9.15997
And now if you look for 9.15997 in the log hav table, there you will find 2h 58m 46s, which was
Layton's result..
What Layton has done is this- Being completely familiar with that calculation, which he has gone
through day after day, he has a good idea of the sort of time interval that will result. He knows,
without doing any calculation, that the log hav of the result will always be somewhere in the range
between about 8.7 and 9.7. So he simply ignores those numbers preceding the decimal point (and the
decimal point itself) in his calculation. He concentrates entirely on the numbers that follow the
decimal point, knowing that he can guess the other part from his experience. It saves a few seconds
in the calculation.
====================
Finally, for convenience, I copy below those parts of Walden's original message which weren't
touched on in my comments above.
"The USNO's calculated altitude, +33 14.0, is close to the 1897 observation of 33"15 (33"03 from
the sextant + "12, the 'universal' refraction+dip+semi-diameter correction described by Frank Reed)
((Close to the USNO value of 14.8' without dip correction))
Now, for the actual meridian angle calculation using the 'time sight' method. The equation used
here is:
hav t = sec L csc p cos s sin(s-h)
where:
hav=haversine (also useful is hav=(1-cos)/2
sec=secant (sec=1/cos)
csc=cosecant (csc=1/sin)
cos=cosine
sin=sine
h=altitude
t=meridian angle
L=Latitude
d=declination
p=90-d if L and d same name
p=90+d if L and d contrary name
s=1/2 (h+L+p)
We recognize 101"08"43 on the far right bottom as p=90+d.
159"41 is 2 times s or (H+L+p) listed just above as: 33"15, 25:18, and 101"08 transferred from the
far right bottom.
79"50 is 159"41 divided by 2 or s.
Below 79"50, 33"15 is repeated from above to facilitate calculation of the needed (s-h).
46"35 is s-h.
Now for some table (Bowditch Table 44 and 45 seem possible) look-ups (but only 4 entries for
natural to log, and 1 for log to natural!) The use of the formula above has the significant
advantage of including NO addition of subtraction, so one only switches to logs once to multiply,
then back for the answer and you're done.
So, starting at the top of the third column, 04379 is 100,000 times the log base 10 of sec L. In
detail by calculator for the first one:
L=25"18=25+18/60=25.30 deg or 25.3*pi/180=.4415683 radians
cos L= 0.90408255
sec L=1/cos L=1.1060937
log10 1.1060934=.0437919
100,000*.0437919=4379 to 4 digits
00825 is 100,000 times log10 csc p.
24677 is 100,000 times [(log10 cos s)+1], the +1, the standard method to avoid negative logs.
86116 is 100,000 times [(log10 sin(s-h)+1].
15997 is the sum of the logs, which gives the log of the product.
Entering the log to natural haversine table with 15997 yields 2"58"46 in hour, minute, second
notation. Useful equation for calculators; if x=hav t, t=acos(1-2x).
13"58 must be the almanac value for the equation of time, needed for mean to apparent sun. USNO
above, gives GHA of Sun as 275-7.7 at 6:34:31 UT 2/19/1897. Convert 275-7.7 degrees to time,
subtract 12+GMT gives equation of time=14-0.
12hr=180deg is added as required by "the rule" (it's not written down; neither the rule nor the
12).
The GMT 6"34"31 transferred from above is subtracted, giving the final answer for longitude,
8"38"13 in hours, minutes, seconds. Times 15 for degrees gives:
8+38/60+13/3600=8.636944
8.636944*15=129.5541deg=129deg-33min QED.
Ref: Bowditch 1962, Cugle 1936 (a great book, 'underappreciated')."
I have a last request for D. Walden. Please tell us more about Cugle's "great book" of 1936. I have
never even heard of it.
George.
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