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From: George Huxtable (no email)
Date: Sun Mar 05 2006 - 19:15:58 EST
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 the odd quibble and comment,
if he doesn't mind.
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, similat, sight reduction seems to have been made a day
later. Then there appears to be a gap of 3 days, in which no such calculations were made (cloudy
weather, perhaps?) followed by the final sight-reduction on that page. 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.
The Greenwich time/date of 6h 34m 31s am / Feb 19th, is perfectly plausible, by the reckoning of
those days, in which the almanac used astronomical time, in which a day extended from one noon to
the next. What we would now reckon, in civil time, to be the morning of Feb 20, the almanac would
then have reckoned as the morning of Feb 19th, which had followed on, after the previous midnight,
from the evening of Feb 19th. Only at noon on the 20th, by modern reckoning, would the almanac-day
of 20th Feb commence. When I try combine that with the date line, and the shift in time due to
longitude differences, my head usually starts to hurt somewhat.
I think that the workbook has these timings correct. What jars, though, is Walden's quotation from
the "USNO on-line navigation site, as follows-
___________________________________________________________________________
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
__________________________________________________________________________
By comparing those numbers with my pocket-calculator almanac, it seems to me that they apply to a
time/date, stated in modern civil terms, of 6:34:31 UT on Feb 20 1897, not Feb 19 as stated. Has D.
Walden made the apropriate adjustment for us, without saying so? If so, then the use of a 24-hour
clock giving UT time would be inappropriate. It's a bit confusing.
He 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
==================================
He 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)."
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. But his procedure for
interpolating seems remarkably ham-handed.
==============================
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. The formula
quoted by Walden can be conveniently rearranged for calculator into-
t(in degrees) = arc cos ( 1 - 2 cos s sin (s-h) / (col L sin p)) (though I haven't checked that
out)
To get the time in terms of hours and fractions of an hour, divide t by 15.
=====================
There are some unusual features about the way Layton uses logs for his calculation.
___________________________________________________________________________
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
__________________________________________________________________________
Things are now looking very good. 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).
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').
---------------------------------
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