NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Attempting to decipher a "time sight"
From: Stan K
Date: 2016 May 30, 16:23 -0400
From: Stan K
Date: 2016 May 30, 16:23 -0400
Frank,
You wrote, transcribed from Hart,
The following work shows the polar distance:—
20°56'00" Dec. Naut. Aim.
....1 44 Correction, Table 5. Bowditch.
20 57 44
....1 53 Correc. 2nd, for 4h. from noon, Tab. 5.
--------
20°59'37" True Dec. at 4 P. M.
90
--------
69°00'23" Polar Distance.
....1 44 Correction, Table 5. Bowditch.
20 57 44
....1 53 Correc. 2nd, for 4h. from noon, Tab. 5.
--------
20°59'37" True Dec. at 4 P. M.
90
--------
69°00'23" Polar Distance.
If your latitude and declination are both North or both South subtract the declination from 90° ; if one is North and the other South, add, for the polar distance.
These matters being ascertained, the rule for the body of the work is as follows :
Add together the sun's altitude, latitude and polar distance ; take half the sum of all; from that half sum subtract the sun's altitude. Then by Table 27 of Bowditch, or by some book of Logarithms, ascertain the secant of the latitude, the co-secant of the polar distance, (rejecting 10 in each index,) the co-sine of the above-mentioned half sum, and the sine of the above remainder. Add them together and take half the sum of that addition; which half sum, found in the line of sines, will answer to the time at the ship, also found in the same table. Subtract the time at the ship, thus found, from the time given by the chronometer (which you have already noted) and the difference, turned into longitude by Table 21, will be the longitude your ship is in. Or, to obtain the last result, multiply the difference of time by 15, (or 3 x 5=15.) The figures for this day will then stand thus :—(See following page.)
Add together the sun's altitude, latitude and polar distance ; take half the sum of all; from that half sum subtract the sun's altitude. Then by Table 27 of Bowditch, or by some book of Logarithms, ascertain the secant of the latitude, the co-secant of the polar distance, (rejecting 10 in each index,) the co-sine of the above-mentioned half sum, and the sine of the above remainder. Add them together and take half the sum of that addition; which half sum, found in the line of sines, will answer to the time at the ship, also found in the same table. Subtract the time at the ship, thus found, from the time given by the chronometer (which you have already noted) and the difference, turned into longitude by Table 21, will be the longitude your ship is in. Or, to obtain the last result, multiply the difference of time by 15, (or 3 x 5=15.) The figures for this day will then stand thus :—(See following page.)
Sun's Alt...37°06'...Log.
Lat.........40 .......11575 Sec. Tab. 27.
Polar Dist..69 .......02985 Co-Sec. Tab. 27.
Sum........146 06
Half-sum....73°03'...9.46469 Co-Sine. Tab. 27.
Sun's Alt...37 06
Remainder...35°57'...9.76870 Sine. Tab. 27.
....................19.37899 Sum
H.M.S.
3 54 16 Eq.to Sine...9.68849 Half-Sum
...3 28 Equation of Time, (Tab. 4, A.)
-------
3 50 48 Time at Ship.
7 35 58 Time by Chronometer.
-------
3 45 10 Equal to 56°17'30" Long, of Ship. Tab. 21.
Lat.........40 .......11575 Sec. Tab. 27.
Polar Dist..69 .......02985 Co-Sec. Tab. 27.
Sum........146 06
Half-sum....73°03'...9.46469 Co-Sine. Tab. 27.
Sun's Alt...37 06
Remainder...35°57'...9.76870 Sine. Tab. 27.
....................19.37899 Sum
H.M.S.
3 54 16 Eq.to Sine...9.68849 Half-Sum
...3 28 Equation of Time, (Tab. 4, A.)
-------
3 50 48 Time at Ship.
7 35 58 Time by Chronometer.
-------
3 45 10 Equal to 56°17'30" Long, of Ship. Tab. 21.
I note that there is a typo in either Hart's original or your transcription. It shows the sum correctly as 19.37899. Half of this is 9.689495, so, to five decimal places, the half-sum should be either 9.68949 or 9.68950, but not 9.68849. The time shown of 3-54-16 is correct for 9.68949 (or so).
Stan
