Parallel Sailing.
With CP equal to the complements of the given latitudes, namely, $56^\circ 2'$ and $41^\circ 37'$ respectively; make BD equal to the given distance 348 miles, and perpendicular to CP; now from the centre C, with the radius CB, describe an arch intersecting CE in E; then EF drawn from the point E, perpendicular to CP, will represent the distance required; which being applied to the scale, will measure 278½ miles.
By Calculation.
As the cofine of the latitude left $35^\circ 58'$ 9.01874 is to the cofine of the lat. come to $48^\circ 23'$ 9.82226 so is the given distance 348 2.54158
to the distance required - 278.6 2.44510
By Inspection.
Under $34^\circ$, and opposite to $174$, half the given distance in a latitude column is 210 in a distance column; being half the difference of longitude answering thereto. Now, find the difference of latitude to distance 210 miles over $48^\circ$ of latitude, which is $140^\circ$.5; from which $1^\circ.1$ (the proportional part answering to 23 minutes of latitude) being subtracted, gives $139^\circ.4$ which doubled is $278^\circ.8$, the distance required.
By Gunter's Scale.
The extent from $56^\circ 2'$, the complement of the latitude left, to $41^\circ 37'$, the complement of that come to, on the line of fines, being laid the same way from 348, will reach to 278½, the distance sought on the line of numbers.
Prob. V. Given a certain portion of a known parallel, together with a similar portion of an unknown parallel; to find the latitude of that parallel.
Example. Two ships, in latitude $56^\circ 0'$ N., distant 180 miles, sail due south; and having come to the same parallel, are now 232 miles distant. The latitude of that parallel is required?
By Construction.
Make DB (fig. 14.) equal to the first distance 180 miles, DM equal to the second 232, and the angle DBC equal to the given latitude $56^\circ$; from the centre C, with the radius CB, describe the arch BE; and through M draw ME parallel to CD, intersecting the arch BE in E; join EC and draw EF perpendicular to CD: then the angle FEC will be the latitude required; which being measured, will be found equal to $43^\circ 53'$.
By Calculation.
As the distance on the known parallel 180 2.25527 is to the distance on that required 232 2.36549 so is the cofine of the latitude left $56^\circ 0'$ 9.74756
to the cofine of the latitude come to 43 53 9.85778
By Inspection.
To latitude $56^\circ$, and half the first distance 90 in a latitude column, the corresponding distance is 161, which is half the difference of longitude. Now 161, and 116, half the second distance, are found to agree between 43 and 44 degrees; therefore, to latitude $43^\circ$ and distance 161, the corresponding difference of latitude is $17^\circ.7$; the excess of which above $116'$ is $1^\circ.7$; and to latitude $44^\circ$, and distance 161, the difference of latitude is $115'.8$: hence $117.7 - 115.8 = 1.9$, the change answering to a difference of $1^\circ$ of latitude.
Therefore, $1^\circ.9 : 1^\circ.7 :: 1^\circ : 53'$
Hence, the latitude is $43^\circ 53'$.
By Gunter's Scale.
The extent from 180 to 232 on the line of numbers, being laid in the same direction on the line of fines, from $34^\circ$, the complement of the latitude failed from, will reach to $46^\circ 7'$, the complement of the latitude come to.
Chap. V. Of Middle Latitude Sailing.
The earth is a sphere, and the meridians meet at the poles; and since a rhumb-line makes equal angles with every meridian, the line a ship describes is, therefore, that kind of a curve called a spiral.
Let AB (fig. 15.) be any given distance failed upon Fig. 15., an oblique rhumb, PBN, PAM the extreme meridians, MN a portion of the equator, and PCK, PEL two meridians intersecting the distance AB in the points CE infinitely near each other. If the arches BS, CD, and AR, be described parallel to the equator, it is hence evident, that AS is the difference of latitude, and the arch MN of the equator, the difference of longitude, answering to the given distance AB and course PAB.
Now, since CE represents a very small portion of the distance AB, DE will be the correspondent portion of a meridian: hence the triangle EDC may be considered as rectilineal. If the distance be supposed to be divided into an infinite number of parts, each equal to CE, and upon these, triangles be constructed whose sides are portions of a meridian and parallel, it is evident these triangles will be equal and similar; for, besides the right angle, and hypotenuse which is the same in each, the course or angle CED is also the same. Hence, by the 12th of V. Euc. the sum of all the hypotenuses CE, or the distance AB, is to the sum of all the sides DE, or the difference of latitude AS, as one of the hypotenuses CE is to the corresponding side DE. Now, let the triangle GIH (fig. 16.) be constructed similar to the triangle CDE, having the angle G equal to the course: then as GH : GI :: CE : DC :: AB : AS.
Hence, if GH be made equal to the given distance AB, then GI will be the corresponding difference of latitude.
In like manner, the sum of all the hypotenuses CE, or the distance AB, is to the sum of all the sides CD, as CE is to CD, or as GH to HI, because of the similar triangles.
The several parts of the same rectilineal triangle will, therefore, represent the course, distance, difference of latitude, and departure.
Although the parts HG, GI, and angle G of the rectilineal triangle GIH, are equal to the corresponding parts AB, AS, and angle A, of the triangle ASB upon the surface of the sphere; yet HI is not equal to BS, for HI is the sum of all the arcs CD; but CD is greater than OQ, and less than ZX: therefore HI is greater than BS, and less than AR. Hence the difference of longitude MN cannot be inferred from the departure reckoned either upon the parallel failed from, or on that come to, but on some intermediate parallel. TV, such that the arch TV is exactly equal to the departure: and in this case, the difference of longitude would be easily obtained. For TV is to MN as the fine PT to the fine PM; that is, as the cosine of latitude is to the radius.
The latitude of the parallel TV is not, however, easily determined with accuracy; various methods have, therefore, been taken in order to obtain it nearly, with as little trouble as possible: first, by taking the arithmetical mean of the two latitudes for that of the mean parallel; secondly, by using the arithmetical mean of the cosines of the latitudes; thirdly, by using the geometrical mean of the cosines of the latitudes; and lastly, by employing the parallel deduced from the mean of the meridional parts of the two latitudes. The first of these methods is that which is generally used.
In order to illustrate the computations in middle latitude sailing, let the triangle ABC (fig. 17) represent a figure in plane sailing, wherein AB is the difference of latitude, AC the distance, BC the departure, and the angle BAC the course. Also, let the triangle DBC be a figure in parallel sailing, in which DC is the difference of longitude, BC the meridian distance, and the angle DCB the middle latitude. In these triangles there is, therefore, one side BC common to both; and that triangle is to be first resolved in which two parts are given, and then the unknown parts of the other triangle will be easily obtained.
**Prob. I.** Given the latitudes and longitudes of two places, to find the course and distance between them.
**Example.** Required the course and distance from the island of May, in latitude 56° 12' N, and longitude 2° 37' W, to the Naze of Norway, in latitude 57° 50' N, and longitude 7° 27' E.
| Latitude Isle of May | 56° 12' N | |----------------------|-----------| | Latitude Naze of Norway | 57° 50' N |
Difference of latitude = 1° 38' = 98' Middle latitude = 57° 1 Longitude Isle of May = 2° 37' W Longitude Naze of Norway = 7° 27' E
Difference of longitude = 10° 4' = 60'
**By Construction.**
Draw the right line AD (fig. 18) to represent the meridian of the May; with the chord of 60° describe the arc mn, upon which lay off the chord of 32° 59', the complement of the middle latitude from m to n; from D through n draw the line DC equal to 60°; the difference of longitude, and from C draw CB perpendicular to AD; make BA equal to 98', the difference of latitude, and join AC; which applied to the scale will measure 343 miles, the distance sought; and the angle A being measured by means of the line of chords, will be found equal to 73° 24', the required course.
**By Calculation.**
To find the course (d).
As the difference of latitude = 98' = 1.99123 is to the difference of longitude = 60° = 2.78164 so is the cosine of middle latitude = 57° 1' = 9.73591
to the tangent of the cosine = 73° 24' = 10.52572
To find the distance.
As radius = 10.00000 is to the secant of the course = 73° 24' = 10.54411 so is the difference of latitude = 98' = 1.99123
to the distance = 343 = 2.53534
**By Inspection.**
To middle latitude = 57°, and 1° 51' one-fourth of the difference of longitude in a distance column, the corresponding difference of latitude is 8.2.2.
Now 2.4.5, one-fourth of the difference of latitude, and 8.2.2, taken in a departure column, are found to agree nearest in table marked 64 points at the bottom, which is the course; and the corresponding distance 8.5.1 multiplied by 4 gives 343 miles, the distance required.
**By Gunter's Scale.**
The extent from 98 the difference of latitude, to 60° the difference of longitude on numbers, being laid the same way from 33°, the complement of the middle latitude on lines, will reach to a certain point beyond the termination of the line on the scale. Now the extent between this point and 92° on lines, will reach from 45° to 73° 24', the course on the line of tangents. And the extent from 73° 24' the course, to 33° the complement of the middle latitude on the line of lines, being laid the same way from 60° the difference of longitude, will reach to 343 the distance on the line of numbers.
The true course, therefore, from the island of May to the Naze of Norway is N 73° 24' E, ENE 3° E nearly; but as the variation at the May is 25 points west, therefore, the course per compass from the May is E 8 S.
**Prob. II.** Given one latitude, course, and distance failed, to find the other latitude and difference of longitude.
**Example.** A ship from Brest, in latitude 48° 23' N, and longitude 4° 30' W, sailed SW 1/4 W 238 miles. Required the latitude and longitude come to?
**By Construction.**
With the course and distance construct the triangle ABC (fig. 17), and the difference of latitude AB being measured, will be found equal to 142 miles; hence the latitude come to is 46° 1' N, and the middle latitude 47° 12'. Now make the angle DCB equal to... Practice.
To find the difference of latitude.
As radius - 10.00000 is to the cosine of the course - 3\(\frac{1}{4}\) points - 9.77593 so is the distance - 238 - 2.37658
To the difference of latitude - 141.8 - 2.15161 Latitude of Breft, 48° 23' N - 48° 23' N Difference of lat. 2 22 S half 1 11 S
Lat. come to 46° 1' N. Mid.lat. 47 12
To find the difference of longitude (E).
As the cosine of Mid. Lat. 47° 12' - 9.83215 is to the fine of the course - 3\(\frac{1}{4}\) points - 9.9483 so is the distance - 238 - 2.37658
To the difference of longitude - 281.3 - 2.44926 Longitude of Breft - 4° 30' W Difference of longitude - 4° 41 W
Longitude come to - 9 11 W.
By Inspection.
To the course 3\(\frac{1}{4}\) points, and distance 238 miles, the difference of latitude is 141.8, and the departure 191.1. Hence the latitude come to is 46° 1' N, and middle latitude 47° 12'. Then to middle latitude 47° 12', and departure 191.1 in a latitude column, the corresponding distance is 281', which is the difference of longitude.
By Gunter's Scale.
The extent from 8 points to 3\(\frac{1}{4}\) points, the complement of the course on fine rhumbs, being laid the same way from the distance 238, will reach to the difference of latitude 142 on the line of numbers; and the extent from 42° 48' the complement of the middle latitude, to 53° 26', the course on the line of fines, will reach from the distance 238 to the difference of longitude 281 on numbers.
Prob. III. Given both latitudes and course, required the distance and difference of longitude?
Example. A ship from St Antonio, in latitude 17° 0' N, and longitude 24° 25' W, sailed NW, \(\frac{3}{4}\) N, till by observation her latitude is found to be 28° 34' N. Required the distance sailed, and longitude come to?
Latitude St Antonio 17° 0' N - 17° 0' N Latitude by observation 28 34 N - 28 34 N
Difference of lat. 11 34 = 694 m. 45 34 Middle lat. 22 47
By Construction.
Construct the triangle ABC (fig. 19.), with the given course and difference of latitude, and make the angle BCD equal to the middle latitude. Now the distance AC and difference of longitude DC being measured, will be found equal to 864 and 558 respectively.
By Calculation.
To find the distance.
As radius - 10.00000 is to the secant of the course - 3\(\frac{1}{4}\) points - 10.00517 so is the difference of lat. 694 - 2.84136
To the distance - 864 - 2.93653
To find the difference of longitude.
As the cosine of middle latitude 22° 47' - 9.96472 is to the tangent of the course - 3\(\frac{1}{4}\) points - 9.87020 so is the difference of latitude 694 - 2.84136
To the difference of longitude - 558.3 - 2.74684 Longitude of St Antonio - 24° 25' W Difference of longitude - 9 18 W
Longitude come to - 33 43 W
By Inspection.
To course 3\(\frac{1}{4}\) points, and difference of latitude 231.3 one third of that given, the departure is 171.6 and distance 288, which multiplied by 3 is 864 miles.
Again to the middle latitude 22° 47', or 23°, and departure 171.6 in a latitude column, the distance is 186, which multiplied by 3 is 558, the difference of longitude.
By Gunter's Scale.
The extent from 4\(\frac{3}{4}\) points, the complement of the course, to 8 points on the line of fine rhumbs, will reach from the difference of latitude 694 to the distance 864 on numbers; and the extent from the course 36° 34' to 67° 13', the complement of middle latitude on fines, will reach from the distance 864 to the difference of longitude 558 on numbers.
Prob. IV. Given one latitude, course, and departure, to find the other latitude, distance, and difference of longitude.
Example. A ship from latitude 26° 30' N, and longitude 45° 30' W, sailed NE, till her departure is 216 miles. Required the distance run, and latitude and longitude come to?
By Construction.
With the course and departure construct the triangle ABC (fig. 20.), and the distance and difference of latitude, being measured, will be found equal to 340 and 263 respectively. Hence the latitude come to is 30° 53', and middle latitude 28° 42'. Now make the angle BCD equal to the middle latitude, and the difference of longitude DC applied to the scale will measure 246'.
By Calculation.
To find the distance.
As the fine of the course - 3\(\frac{1}{4}\) points - 9.80236 is to radius - 10.00000 so is the departure - 216 - 2.33445
To the distance - 340.5 - 2.53209
(e) This proportion is obvious, by considering the whole figure as an oblique-angled plane triangle. To find the difference of latitude.
As the tangent of the course \(3\frac{1}{2}\) points \(9.91417\) is to radius \(10.00000\) so is the departure \(216\) \(2.33445\)
to the difference of lat. \(263.2\) \(2.42028\) Latitude sailed from \(26°30'N\) \(26°30'N\) Difference of latitude \(4°23'N\) half \(2°12'N\)
Latitude come to \(30°53'N\). Mid. lat. \(28°42'\)
To find the difference of longitude.
As radius \(10.00000\) is to the secant of the mid. lat. \(28°42'\) \(10.05693\) so is the departure \(216\) \(2.33445\)
to the difference of longitude \(246.2\) \(2.39138\) Longitude left \(45°35'W\) Difference of longitude \(4°6'E\)
Longitude come to \(41°24'W\)
By Inspection.
The distance \(246\), and difference of latitude \(156\), are found to correspond above \(4\frac{1}{2}\) points, and the departure is \(190.\). Now, to the middle latitude \(42°\), and departure \(190.\) in a latitude column, the corresponding distance is \(256\), which is the difference of longitude required.
By Gunter's Scale.
The extent from \(246\) miles, the distance, to \(156\), the difference of latitude on numbers, will reach from \(90°\) to about \(39°\), the complement of the course on the line of sines; and the extent from \(48°\), the complement of the middle latitude, to \(50°\), the course on sines, will reach from the distance \(246\)m. to the difference of longitude \(256\)m. on numbers.
Prob. VI. Given both latitudes and departure; sought the course, distance, and difference of longitude.
Example. A ship from Cape St Vincent, in latitude \(37°2'N\), longitude \(6°2'W\), sails between the south and west; the latitude come to is \(18°16'N\), and departure \(838\) miles. Required the course and distance run, and longitude come to?
Latitude Cape St Vincent, \(37°2'N\) Latitude come to \(18°16'N\) Difference of latitude \(18°46'=2126\) sum \(55°18'\) Middle latitude \(17°39'\)
By Construction.
Make \(AB\) (fig. 21.) equal to the difference of latitude \(1126\) miles, and \(BC\) equal to the departure \(838\), and join \(AC\); draw \(CD\) so as to make an angle with \(CB\) equal to the middle latitude \(27°39'\). Then the course being measured on chords is about \(36°39'\), and the distance and difference of longitude, measured on the line of equal parts, will be found to be \(1403\) and \(946\) respectively.
By Calculation.
To find the course.
As the difference of latitude \(1126\) \(3.05154\) is to the departure \(838\) \(2.92324\) so is radius \(10.00000\)
to the tangent of the course \(36°39'\) \(9.87170\)
To find the distance.
As radius \(10.00000\) is to the secant of the course \(36°39'\) \(10.09566\) so is the difference of latitude \(1126\) \(3.05154\)
to the distance \(1403\) \(3.14720\)
To To find the difference of longitude.
As radius - 10,00000 is to the secant of mid. lat. 27° 39' so is the departure - 838
to the difference of longitude - 946 Longitude Cape St Vincent - 9° 2'W Difference of longitude - 15 46W
Longitude come to - 24 48W
By Inspection.
One tenth of the difference of latitude 112.6 and of the departure 83.8, are found to agree under 34 points, and the corresponding distance is 142, which multiplied by 10 gives 1420 miles. And to middle latitude 27° 3', and 209.5 one fourth of the departure in a latitude column, the distance is 236.5; which multiplied by 4 is 946, the difference of longitude.
By Gunter's Scale.
The extent from the difference of latitude 1126 to the departure 838 on numbers, will reach from 45° to 36° 1/2 the course on tangents; and the extent from 53° 1/2 the complement of the course to 90° on lines, will reach from 1126 to 1403 the distance on numbers. Lastly, the extent from 62° 1/2 the complement of the middle latitude, to 90° on lines, will reach from the departure 838 to the difference of longitude 946 on numbers.
PROB. VII. Given one latitude, distance, and departure, to find the other latitude, course, and difference of longitude.
Example. A ship from Bourdeaux, in latitude 44° 50' N, and longitude 5° 35' W, sailed between the north and west 374 miles, and made 210 miles of sailing. Required the course, and the latitude and longitude come to?
By Construction.
With the given distance and departure make the triangle ABC (fig. 23). Now the course being measured on the line of chords is about 34° 10', and the difference of latitude on the line of numbers is 309 miles; hence the latitude come to, is 49° 50' N, and middle latitude 47° 25'. Then make the angle BCD equal to 47° 25', and DC being measured will be 310 miles, the difference of longitude.
By Calculation.
To find the course.
As the distance - 374 2.37287 is to the departure - 210 2.32222 so is radius - 10,00000
to the sine of the course - 34° 10' 9.74935
To find the difference of latitude.
As radius - 10,00000 is to the cosine of the course 34° 10' 9.91772 so is the distance - 374 2.57287
to the difference of latitude - 309.4 2.49059 Latitude of Bourdeaux 44° 50'N Difference of latitude - 5° 9 N half 2 33
Latitude come to - 49° 59 N Mid. lat. 47° 25'
Middle Latitude Sailing.
To find the difference of longitude.
As radius - 10,00000 is to the secant of mid. lat. 47° 25' so is the departure - 210 2.32222
to the difference of longitude - 310.3 2.49185 Longitude of Bourdeaux - 0° 35'W Difference of longitude - 5° 10 W
Longitude in - 5° 45 W
By Inspection.
The half of the distance 187, and of the departure 105, are found to agree nearest under 34°, and the difference of latitude answering thereto is 155; which doubled is 310 miles.
Again, to middle latitude 47° 25', and departure 105 in a latitude column, the corresponding distance is 155 miles, which doubled is 310 miles, the difference of longitude.
By Gunter's Scale.
The extent from the distance 374 miles to the departure 210 miles on the line of numbers, will reach from 90° to 34° 10', the course on the line of lines; and the extent from 90° to 55° 50', the complement of the course on lines, will reach from the distance 374 to the difference of latitude 309 miles on numbers.
Again, the extent from 42° 35', the complement of the middle latitude, to 90° on lines, will reach from the departure 210 to the difference of longitude 310 on numbers.
PROB. VIII. Given one latitude, departure, and difference of longitude, to find the other latitude, course, and distance.
Example. A ship from latitude 54° 56' N, longitude 1° 10' W, sailed between the north and east, till by observation she is found to be in longitude 5° 26' E, and has made 220 miles of sailing. Required the latitude come to, course, and distance run?
Longitude left - 1° 10'W Longitude come to - 5° 26'E Difference of longitude - 6° 36'=396
By Construction.
Make BC (fig. 24.) equal to the departure 220, and Fig. 24. CD equal to the difference of longitude 396;—then the middle latitude BCD being measured, will be found equal to 59° 15': hence the latitude come to is 57° 34', and difference of latitude 158°. Now make AB equal to 158, and join AC, which applied to the scale, will measure 271 miles. Also the course BAC being measured on chords will be found equal 54° 1'.
By Calculation.
To find the middle latitude.
As the departure - 220 2.34242 is to the diff. of longitude - 396 2.59769 so is radius - 10,00000
To the secant of mid. lat. - 56° 15' 10.25527
Double, mid. lat. - 112° 30' Latitude left - 54° 56' Latitude come to - 57° 34' Diff. of latitude - 2° 38'=158 miles To find the course.
As the difference of latitude \(158\) is to the departure \(220\), so is radius \(2.19866\).
To find the distance.
As radius \(10.00000\) is to the secant of the course \(54^\circ 19'\), so is the difference of latitude \(158\) to the distance \(270.9\).
By Inspection.
As the difference of longitude and departure exceed the limits of the tables, let, therefore, their halves be taken; these are \(108\) and \(110\) respectively. Now these are found to agree exactly in the page marked \(5\) points at the bottom. Whence the middle latitude is \(56^\circ 15'\), and difference of latitude \(158\) miles.
Again, the difference of latitude \(158\) and departure \(220\) will be found to agree nearly above \(54^\circ\) the course, and the distance on the same line is \(271\) miles.
By Gunter's Scale.
The extent from the difference of longitude \(396\) to the departure \(220\) on numbers, will reach from \(90^\circ\) to \(33^\circ 45'\), the complement of the middle latitude on fines; and hence the difference of latitude is \(158\) miles. Now the extent from \(158\) to \(220\) on numbers, will reach from \(45^\circ\) to \(54^\circ\) on tangents; and the extent from the complement of the course \(35^\circ 3'\) to \(90^\circ\) on fines, will reach from the difference of latitude \(158\) to the distance \(271\) on numbers.
Prob. IX. Given the course and distance failed, and difference of longitude; to find both latitudes.
Example. A ship from a port in north latitude, sailed \(SSE_4S\) \(438\) miles, and differed her longitude \(7^\circ 28'\). Required the latitude failed from, and that come to?
By Construction.
With the course and distance construct the triangle \(ABC\) (fig. 25.), and make \(DC\) equal to \(448\) the given difference of longitude. Now the middle latitude \(BCD\) will measure \(48^\circ 38'\), and the difference of latitude \(AB\) \(324\) miles; hence the latitude left is \(51^\circ 40'\), and that come to \(46^\circ 16'\).
By Calculation.
To find the difference of latitude.
As radius \(10.00000\) is to the cosine of the course \(3\frac{1}{2}\) pts., so is the distance \(438\) \(2.64147\).
To find the middle latitude.
As the difference of longitude \(448\) is to the distance \(438\), so is the fine of the course \(3\frac{1}{2}\) pts., \(9.82708\).
To the cosine of mid. latitude \(48^\circ 58'\) half difference of latitude \(2^\circ 42'\) \(9.81727\).
Latitude failed from \(51^\circ 40'\)
Latitude come to \(46^\circ 16'\)
By Inspection.
To the course \(3\frac{1}{2}\) points, and half the distance \(219\) miles, the departure is \(147.2\), and difference of latitude \(152.2\); which doubled is \(323.4\). Again, to half the difference of longitude \(224\) in a distance column, the difference of latitude is \(149.9\) above \(48^\circ\), and \(146.9\) over \(49^\circ\).
Now, as \(30 : 29 :: 60' : 58'\).
Hence the middle latitude is \(48^\circ 58'\); the latitude failed from is therefore \(51^\circ 40'\), and latitude come to \(46^\circ 16'\).
By Gunter's Scale.
The extent from \(8\) points to \(4\frac{1}{2}\) points, the complement of the course on fine rhumbs, will reach from the distance \(438\) miles to the difference of latitude \(324\) on numbers. And the extent from the difference of longitude \(448\), to the distance \(438\) on numbers, will reach from the course \(48^\circ 11'\) to the complement of the middle latitude \(41^\circ 2'\) on fines. Hence the latitude left is \(51^\circ 40'\), and that come to \(46^\circ 16'\).
Prob. X. To determine the difference of longitude made good upon compound courses, by middle latitude failing.
Rule I. With the several courses and distances find the difference of latitude and departure made good, and the ship's present latitude, as in traverse failing.
Now enter the traverse table with the given middle latitude, and the departure in a latitude column, the corresponding distance will be the difference of longitude, of the same name with the departure.
Example. A ship from Cape Clear, in latitude \(51^\circ 18'\) N, longitude \(9^\circ 46'\) W, sailed as follows:—SWbS \(34\) miles, WbN \(63\) miles, NNW \(48\) miles, and NE\(E\) \(85\) miles. Required the latitude and longitude come to?
| Courses | Diff. | Latitude of Cape Clear | |---------|------|------------------------| | SWbS | 54 | 44.9 | | WbN | 63 | 12.3 | | NNW | 48 | 44.4 | | NE\(E\) | 85 | 53.9 |
| Latitude come to | Sum | Middle latitude | |------------------|-----|----------------| | \(52^\circ 24'\)N | 103 | \(51^\circ 51'\) |
Now, to middle latitude \(51^\circ 51'\) or \(52^\circ\), and departure \(44.5\) in a latitude column, the difference of longitude is \(72\) in a distance column.
Longitude of Cape Clear \(9^\circ 46'\) W
Difference of longitude \(1^\circ 12'\) W
Longitude come to \(10^\circ 58'\) W
The above method is that always practised to find the difference of longitude made good in the course of Theorems might be investigated for computing the Mercator's errors to which the above method is liable. These corrections may, however, be avoided, by using the following method.
**Rule II.** Complete the traverse table as before, to which annex five columns: the first column is to contain the several latitudes the ship is in at the end of each course and distance; the second, the sums of each following pair of latitude; the third, half the sums, or middle latitudes; and the fourth and fifth columns are to contain the differences of longitude.
Now find the difference of longitude answering to each middle latitude and its corresponding departure, and put them in the east or west difference of longitude columns, according to the name of the departure. Then the difference of the sums of the east and west columns will be the difference of longitude made good, of the same name with the greater.
**Example.** A ship from Halliford in Iceland, in lat. $64^\circ 30' N$, long. $27^\circ 15' W$, sailed as follows: SSW $46$ miles, SW $61$ miles, SbW $59$ miles, SEbE $86$ miles, SbE½E $76$ miles. Required the lat. and long. come to?
| Traverse Table. | Longitude Table. | |-----------------|------------------| | **Courses.** | **Diff. of Lat.** | **Departure.** | **Successive Latitudes.** | **Sums.** | **Middle Latitudes.** | **Diff. of Longitude** | | | **N** | **S** | **E** | **W** | **N** | **S** | **E** | **W** | **E** | **W** | | SSW | 46 | 42.5 | 17.6 | 64° 30' | 63 48 | 128° 18' | 64° 9' | 40.4 | | SW | 61 | 43.1 | 43.1 | 63 5 | 126 53 | 63 27 | 96.4 | | SbW | 59 | 57.9 | 11.5 | 62 7 | 125 12 | 62 36 | 25.0 | | SEbE | 86 | 47.8 | 71.5 | 61 19 | 123 26 | 61 43 | 150.9 | | SbE½E | 76 | 72.7 | 22.0 | 60 6 | 121 25 | 60 43 | 45.0 |
By Rule I.
Latitude Halliford - $64^\circ 30' N$ Difference of latitude - $4^\circ 24' S$
Latitude in - $60^\circ 6' N$ Sum - $124^\circ 36'$ Middle latitude - $62^\circ 18'$
Now, to middle lat. $62^\circ 18'$, and departure $21.3$, the difference of long. is $46^\circ E$.
Long. Halliford - $27^\circ 15' W$
Longitude in - $16^\circ 29'$ The error of comm. method, in this Ex. is $12'$.
**Chap. VI. Of Mercator's Sailing.**
It was observed in Middle Latitude Sailing, that the difference of longitude made upon an oblique rhumb could not be exactly determined by using the middle latitude. In Mercator's sailing, the difference of longitude is very easily found, and the several problems of sailing resolved with the utmost accuracy, by the affluence of Mercator's chart or equivalent tables.
In Mercator's chart, the meridians are straight lines parallel to each other; and the degrees of latitude, which at the equator are equal to those of longitude, increase with the distance of the parallel from the equator. The parts of the meridian thus increased are called *meridional parts*. A table of these parts was first constructed by Mr Edward Wright, by the continual addition of the fecants of each minute of latitude.
For by parallel sailing,
\[ R : \text{cof. of lat.} :: \text{part of equat.} : \text{similar part of parallel.} \]
And And because the equator and meridian on the globe are equal; therefore,
\[ R : \text{cof. lat.} :: \text{part of meridian} : \text{similar part of parallel}. \]
Or sec. lat. \( R :: \text{part of merid.} : \text{similar part of parallel}. \)
Hence, \( \frac{\text{secant latitude}}{\text{part of meridian}} = \frac{R}{\text{part of parallel}} \).
But in Mercator's chart the parallels of latitude are equal, and radius is a constant quantity. If therefore, the latitude be assumed successively equal to \( 1^\circ, 2^\circ, 3^\circ \), &c., and the corresponding parts of the enlarged meridian be represented by \( a, b, c, \) &c.; then,
\[ \frac{\text{secant } 1^\circ}{\text{part of mer. } a} = \frac{\text{secant } 2^\circ}{\text{part of mer. } b} = \frac{\text{secant } 3^\circ}{\text{part of mer. } c}, \] &c.
Hence secant \( 1^\circ : \text{part of mer. } a :: \text{secant } 2^\circ : \text{part of mer. } b :: \text{secant } 3^\circ : \text{part of mer. } c, \) &c.
Therefore by 12th V. Euclid,
Secant \( 1^\circ : \text{part of mer. } a + \text{secant } 2^\circ + \text{secant } 3^\circ, \) &c.: parts of \( a + b + \text{mer. } c, \) &c.
That is, the meridional parts of any given latitude are equal to the sum of the secants of the minutes in that latitude (E).
Since \( CD : LK :: R : \text{secant } LD, \) fig. 15.
And in the triangle CED,
\[ ED : CD :: R : \text{tangent } CED; \]
Therefore, \( ED : LK :: R^2 : \text{secant } LD \times \text{tangent } CED \)
\[ \frac{ED \times \text{sec. } LD}{R} \times \frac{\text{tang. } CED}{R}. \]
But \( \frac{ED \times \text{secant } LD}{R} \) is the enlarged portion of the meridian answering to \( ED. \) Now the sum of all the quantities \( \frac{ED \times \text{secant } LD}{R} \) corresponding to the sum of all the ED's contained in AS, will be the meridional parts answering to the difference of latitude AS; and MN is the sum of all the corresponding portions of the equator LK.
Whence \( MN = \text{mer. diff. of lat.} \times \text{tangent } \frac{CED}{R}. \)
That is, the difference of longitude is equal to the meridional difference of latitude multiplied by the tangent of the course, and divided by the radius.
This equation answers to a right-angled rectilineal triangle, having an angle equal to the course; the adjacent side equal to the meridional difference of latitude, and the opposite side the difference of longitude. This triangle is, therefore, similar to a triangle constructed, with the course and difference of latitude, according to the principles of plane sailing, and the homologous sides will be proportional. Hence, if, in fig. 26, the angle A represents the course, AB the difference of latitude, and if AD be made equal to the meridional difference of latitude; then DE, drawn perpendicular to AD, meeting the distance produced to E, will be the difference of longitude.
It is scarcely necessary to observe, that the meridional difference of latitude is found by the same rules as the proper difference of latitude; that is, if the given latitudes be of the same name, the difference of the corresponding meridional parts will be the meridional difference of latitude; but if the latitudes are of a contrary denomination, the sum of these parts will be the meridional difference of latitude.
**Prob. I.** Given the latitudes and longitudes of two places, to find the course and distance between them.
Ex. Required the course and distance between Cape Finisterre, in latitude \( 42^\circ 52' N, \) longitude \( 9^\circ 17' W, \) and Port Praya in the island of St Jago, in latitude \( 14^\circ 54' N, \) and longitude \( 23^\circ 29' W? \)
| Lat. Cape Finisterre | Mer. parts | |---------------------|-----------| | Latitude Port Praya | Mer. parts |
Difference of lat. \( = 27^\circ 58' \) Mer. diff. lat. \( = 1948 \)
Longitude Cape Finisterre \( = 9^\circ 17' W \) Longitude Port Praya \( = 23^\circ 29' W \)
Diff. longitude \( = 14^\circ 12' = 852. \)
**By Construction.**
Draw the straight line AD (fig. 26.) to represent the meridian of Cape Finisterre, upon which lay off AB, AD equal to 1678, and 1948, the proper and meridional differences of latitude; from D draw DE perpendicular to AD, and equal to the difference of longitude 852, join AE, and draw BC parallel to DE; then the difference AC will measure 1831 miles, and the course BAC \( 23^\circ 37'. \)
**By Calculation.**
To find the course.
As the meridian difference of lat. \( 1948 : 3.28959 \) is to the difference of longitude \( 852 : 2.93044 \) so is radius \( 10.00000 \)
to the tangent of the course \( 23^\circ 37' : 9.64085 \)
To find the distance.
As radius \( 10.00000 \) is to the secant of the course \( 23^\circ 37' : 10.03798 \) so is the difference of latitude \( 1678 : 3.22479 \)
to the distance \( 1831 : 3.26277 \)
**By Inspection.**
As the meridian difference of latitude and difference of longitude are too large to be found in the tables, let the tenth of each be taken; these are 194.8 and 85.2 respectively. Now these are found to agree nearest under \( 24^\circ \); and to 167.8, one-tenth of the proper difference of latitude, the distance is about 183 miles, which multiplied by 10 is 1830 miles.
**By Gunter's Scale.**
The extent 1948, the meridional difference of latitude, to 852, the difference of longitude on the line of numbers, will reach from \( 45^\circ \) to \( 23^\circ 37', \) the course on
---
(e) This is not strictly true; for instead of taking the sum of the secants of every minute in the distance of the given parallel from the equator, the sum of the secants of every point of latitude should be taken. Practice.
Mercator's on the line of tangents. And the extent from $66^\circ 23'$, sailing, the complement of the course to $90^\circ$ on fines, will reach from $1678$, the proper difference of latitude, to $1831$, the distance on the line of numbers.
**Prob. II.** Given the course and distance, sailed from a place whose situation is known, to find the latitude and longitude of the place come to.
*Example.* A ship from Cape Hinlopen in Virginia, in latitude $38^\circ 47'$ N, longitude $75^\circ 4'$ W, sailed 267 miles NE&N. Required the ship's present place?
*By Construction.*
With the course and distance sailed, construct the triangle ABC (fig. 27.) and the difference of latitude AB being measured, is 222 miles; hence the latitude come to is $42^\circ 29'$ N, and the meridional difference of latitude 293. Make AD equal to 293; and draw DE perpendicular to AD, and meeting AC produced in E; then, the difference of longitude DE being applied to the scale of equal parts will measure 196; the longitude come to is therefore $71^\circ 48'$ W.
*By Calculation.*
To find the difference of latitude.
| As radius | 10.00000 | |-----------|----------| | is to the cosine of the course, - 3 points | 9.91985 | | so is the distance | 267 | 2.42651 |
To the difference of latitude - 222 - 2.34636
Lat. Cape Hinlopen = $38^\circ 47'$ N. Mer. parts 2528
Difference of lat. - 3 42' N.
Latitude come to - 42 29' N. Mer. parts 2828
Meridional difference of latitude 293
To find the difference of longitude.
| As radius | 10.00000 | |-----------|----------| | is to tangent of the course, - 3 points | 9.82489 | | so is the mer. diff. of latitude | 293 | 2.46687 |
To the difference of longitude - 195.8 - 2.29176
Longitude Cape Hinlopen - 75° 4' W
Difference of longitude - 3 16' E
Longitude come to - 71 48' W
*By Inspection.*
To the course 3 points, and distance 267 miles, the difference of latitude is 222 miles; hence the latitude in is $42^\circ 29'$, and the meridional difference of latitude 293. Again, to course 3 points, and 146.5 half the mer. difference of latitude, the departure is 97.9, which doubled is 195.8, the difference of longitude.
*By Gunter's Scale.*
The extent from 8 points to the complement of the course 5 points, on fine rhumbs, will reach from the distance 267 to the difference of latitude 222 on numbers; and the extent from 4 points to three points on tangent rhumbs, will reach from the meridional difference of latitude 293 to the difference of longitude 196 on numbers.
**Prob. III.** Given the latitudes and bearing of two places; to find their distance and difference of longitude.
*Example.* A ship from Port Canfo in Nova Scotia, in latitude $45^\circ 20'$ N, longitude $60^\circ 55'$ W, sailed SE $\frac{1}{4}$ S, and by observation is found to be in latitude $41^\circ 14'$ N. Required the distance sailed, and longitude come to?
Lat. Port Canfo - $45^\circ 20'$ N - Mer. parts - 3058 Lat. in by observation $41^\circ 14'$ N - Mer. parts - 2720
Difference of lat. - 4 6 = 246 Mer. diff. lat. 338
*By Construction.*
Make AB (fig. 28.) equal to 246, and AD equal Fig. 28. to 338; draw AE, making an angle with AD equal to 3½ points, and draw BC, DE perpendicular to AD. Now AC being applied to the scale, will measure 332, and DE 306.
*By Calculation.*
To find the distance.
As radius - 10.00000 is to the secant of the course, - 3½ points - 10.13021 so is the difference of latitude - 246 - 2.39093
To the distance - 332 - 2.52114
To find the difference of longitude.
As radius - 10.00000 is to the tangent of the course, - 3½ points - 9.95729 so is the mer. diff. of latitude - 338 - 2.52892
To the difference of longitude - 306.3 - 2.48621
Longitude Port Canfo - $60^\circ 55'$ W
Difference of longitude - 5 6 E
Longitude in - 53 49' W
*By Inspection.*
Under the course 3½ points, and opposite to half the difference of latitude 123 in a latitude column is 166 in a distance column, which doubled is 332 the distance; and opposite to 169, half the meridional difference of latitude in a latitude column, is 153 in a departure column, which doubled is 306, the difference of longitude.
*By Gunter's Scale.*
The extent from the complement of the course 4½ points to 8 points on fine rhumbs, will reach from the difference of latitude 246 m. to the distance 332 on numbers; and the extent from 4 points, to the course 3½ points on tangent rhumbs, will reach from the meridional difference of latitude 338 to the difference of longitude 306 on numbers.
**Prob. IV.** Given the latitude and longitude of the place sailed from, the course and departure; to find the distance, and the latitude and longitude of the place come to.
*Example.* A ship sailed from Sallee in latitude $33^\circ 8'$ N, longitude $6^\circ 20'$ W, the corrected course was NWbW $\frac{1}{2}$ W, and departure 420 miles. Required the distance run, and the latitude and longitude come to?
*By Construction.*
With the course and departure construct the triangle ABC (fig. 29.) now AC and AB being measured, Fig. 29. will be found to be equal to 476 and 224 respectively: By Calculation.
To find the distance.
As radius - 10.00000 is to the cotangent of the course, 5½ pts - 10.05457 so is the departure - 420 - 2.62325
to the distance - 476.2 - 2.67782
To find the difference of latitude.
As radius - 10.00000 is to the cotangent of the course, 5½ pts - 9.72796 so is the departure - 420 - 2.62325
to the difference of latitude - 224.5 - 2.35121
Lat. of Sallee 33° 58' N Mer. parts 2169 Diff. of lat. 3 44 N
Lat. in 37° 42' N Mer. parts 2445
Mer. difference of latitude - 276
To find the difference of longitude.
As radius - 10.00000 is to the tangent of the course, 5½ pts - 10.27204 so is the mer. diff. of latitude - 276 - 2.44091
to the difference of longitude - 516.3 - 2.71295
Longitude of Sallee - 6° 20' W Difference of longitude - 8 36 W
Longitude in - 14° 56' W
By Inspection.
Above 5½ points the course, and opposite to 2½ half the departure, are 238 and 112; which doubled, we have 476 and 224, the distance and difference of latitude respectively. And to the same course, and opposite to 138, half the meridional difference of latitude, in a latitude column, is 258 in a departure column; which being doubled is 516, the difference of longitude.
By Gunter's Scale.
The extent from 5½ points, the course on fine rhumbs, to the departure 420 on numbers, will reach from 8 points on fine rhumbs to the distance 476 on numbers; and from the complement of the course 2½ points on fine rhumbs, to the difference of latitude 224 on numbers.
Again, the extent from difference of latitude 224 to the meridional difference of latitude 276 on numbers, will reach from the departure 420 to the difference of longitude 516 on the same line.
Prob. V. Given the latitudes of two places, and their distance, to find the course and difference of longitude.
Example. A ship from St Mary's, in latitude 36° 57' N, longitude 25° 9' W, sailed on a direct course between the north and east 1162 miles, and is then by observation in latitude 49° 57' N. Required the course steered, and longitude come to?
Latitude of St Mary's - 36° 57' N Mer. parts 3470 Lat. come to - 49° 57' N Mer. parts 2389
Difference of lat. - 13° 0 Mer. diff. lat. 1081
By Construction.
Make AB (fig. 30.) equal to 780, and AD equal Fig. 32. to 1081; draw BC, DE perpendicular to AD; make AC equal to 1162, and through AC draw ACE. Then the course or angle A being measured, will be found equal to 47° 50', and the difference of longitude DE will be 1194.
By Calculation.
To find the course.
As the distance - 1162 - 3.06521 is to the difference of latitude, 780 - 2.89209 so is radius - - 10.00000
to the cosine of the course - 47° 50' - 9.82688
To find the difference of longitude.
As radius - 10.00000 is to the tangent of the course, 47° 50' - 10.04302 so is the mer. diff. of latitude - 1081 - 3.03383
to the difference of longitude - 1194 - 3.07685
Longitude of St Mary's - 25° 9' W Difference of longitude - 19° 54' E
Longitude in - 5° 15' W
By Inspection.
Because the distance and difference of latitude exceed the limits of the table, take the tenth of each; these are 116.2 and 78.0; Now these are found to agree nearest above 4½ points, which is therefore the course; and to this course, and opposite to 108.1, one tenth of the meridional difference of latitude, in a latitude column, is 119.3 in a departure column, which multiplied by 10 is 1193, the difference of longitude.
By Gunter's Scale.
The extent from the distance 1162 m. to the difference of latitude 780 m. on numbers, will reach from 90° to 42° 10' in the line of fines. And the extent 45°, to the course 47° 50' on the line of tangents, will reach from the meridional difference of latitude 1081 to the difference of longitude 1194 on numbers.
Prob. VI. Given the latitudes of two places, and the departure, to find the course, distance, and difference of longitude.
Example. From Aberdeen, in latitude 57° 9' N, longitude 2° 8' W, a ship sailed between the south and east till her departure is 146 miles, and latitude come to 53° 32' N. Required the course and distance run, and longitude come to?
Latitude Aberdeen 57° 9' N Mer. parts 4199 Latitude come to 53° 32' N Mer. parts 3817
Difference of latitude 3° 37' = 217' Mer. diff. lat. 382
By Construction.
With the difference of latitude 217 m. and departure Fig. 31., construct the triangle ABC (fig. 31.), make AD Practice.
Mercator's AD equal to 382, draw DE parallel to BC, and produce AC to E: Then the course BAC will measure 33° 56', the distance AC 261, and the difference of longitude DE 257.
By Calculation.
To find the course.
As the difference of latitude 217 - 2.33464 is to the departure 146 - 2.16435 so is radius 10.00000
to the tangent of the course 33° 56' - 9.82780
To find the distance.
As radius 10.00000 is to the secant of the course 33° 56' - 10.08109 so is the difference of latitude 217 - 2.33464
to the distance 261.5 - 2.41755
To find the difference of longitude.
As the difference of latitude 217 - 2.33464 is to the mer. diff. of latitude 382 - 2.18206 so is the departure 146 - 2.16435
to the difference of longitude 257 - 2.40995
Longitude of Aberdeen 2° 8' W Difference of longitude 4° 17' E
Longitude come to 2° 9' E
By Inspection.
The difference of latitude 217, and departure 146, are found to agree nearest under 34°, and the corresponding distance is 262 miles. To the same course, and opposite to 190.7, the nearest to 191, half the meridional difference of latitude, is 128.6 in a departure column, which doubled is 257, the difference of longitude.
By Gunter's Scale.
The extent from the difference of latitude 217, to the departure 146 on numbers, will reach from 45° to about 34°, the course on the line of tangents; and the same extent will reach from the meridional difference of latitude 382 to 257, the difference of longitude on numbers.—Again, the extent from the course 34° to 90 on fines, will reach from the departure 146 to the distance 261 on numbers.
PROB. VII. Given one latitude, distance, and departure; to find the other latitude, course, and difference of longitude.
Example. A ship from Naples, in latitude 40° 51' N, longitude 14° 14' E, sailed 252 miles on a direct course between the south and west, and made 173 miles of sailing. Required the course made good, and the latitude and longitude come to?
By Construction.
With the distance and departure make the triangle ABC (fig. 32.), as formerly.—Now the course BAC being measured by means of a line of chords will be found equal to 43° 21', and the difference of latitude applied to the scale of equal parts will measure 183': hence the latitude come to is 37° 48' N, and meridional difference of latitude 237.—Make AD equal to 237, and complete the figure, and the difference of longitude DE will measure 224': hence the longitude Mercator's in is 10° 30' E.
By Calculation.
To find the course.
As the distance 252 - 2.40140 is to the departure 173 - 2.23825 so is radius 10.00000
to the fine of the course 43° 21' - 9.83665
To find the difference of latitude.
As radius 10.00000 is to the cosine of the course 43° 21' - 9.86164 so is the distance 252 - 2.40140
to the difference of latitude 183.2 - 2.26304
Latitude of Naples 40° 51' N. Mer. parts 2693 Difference of latitude 3° 3' S.
Latitude come to 37° 48' N. Mer. parts 2453
Meridional difference of latitude 237
To find the difference of longitude.
As radius 10.00000 is to the tangent of the course 43° 21' - 9.97497 so is the mer. diff. of latitude 237 - 2.37475
to the difference of longitude 223.7 - 2.34972
Longitude of Naples 14° 14' E Difference of longitude 3° 44' W
Longitude in 10° 30' E
By Inspection.
Under 43° and opposite to the distance 252 m. the departure is 171.8, and under 44°, and opposite to the same distance, the departure is 175.0.
Then as 3.2 : 1.2 :: 65' : 22'.
Hence the course is 43° 22'.
Again, under 43° and opposite to 118.5, half the meridional difference of latitude in a latitude column, is 110.5 in a departure column; also under 44° and opposite to 118.5 is 114.4.
Then as 3.2 : 1.2 :: 3.9 : 1.3.
And 110.5 + 1.5 = 112, which doubled is 224, the difference of longitude.
By Gunter's Scale.
The extent from the distance 252 on numbers, to 90° on fines, will reach from the departure 173 on numbers, to the course 43° on fines; and the same extent that will reach from the complement of the course 46° on fines will reach to the difference of latitude 183 on numbers.—Again, the extent from 45° to 43° on tangents will reach from the meridional difference of latitude 237, to the difference of longitude 224, on numbers.
PROB. VIII. Given one latitude, course, and difference of longitude: to find the other latitude and distance.
Example. A ship from Tercera, in latitude 38° 45' N, longitude 27° 6' W, failed on a direct course, which, when corrected, was N 32° E, and is found by observation to be in longitude 18° 24' W. Required the latitude come to, and distance failed? By Construction.
Make the right-angled triangle ADE (fig. 33.) having the angle A equal to the course 32°, and the side DE equal to the difference of longitude 522; then AD will measure 835, which added to the meridional parts of the latitude left, will give those of the latitude come to 48° 46′; hence, the difference of latitude is 601: make AB equal thereto, to which let BC be drawn perpendicular; then AC applied to the scale will measure 708 miles.
By Calculation.
To find the meridional difference of latitude.
As radius 10.000000 is to the co-tangent of the course 32° 0′ 10.20421 so is the difference of longitude 5 22 2.71767
to the mer. difference of latitude 8352 2.92188 Latitude of Tercera 35° 45′ N Mer. parts 2526 Mer. diff. of lat. 835
Latitude come to - 48° 46′ N Mer. parts 3361
Difference of latitude 10 = 601 miles.
To find the distance.
As radius 10.000000 is to the secant of the course 32° 0′ 10.07158 so is the difference of latitude 601 2.77887
to the distance 707.7 2.85045
By Inspection.
To course 32°, and opposite to 130.5, one fourth of the given difference of longitude in a departure column, the difference of latitude is 208.8, which multiplied by 4 is 835, the meridional difference of latitude; hence the latitude in is 48° 46′ N, and difference of latitude 601.
Again, to the same course, and opposite to 200, one third of the difference of latitude, the distance is 236, which multiplied by 3 gives 708 miles.
By Gunter's Scale.
The extent from the course 32°, to 45° on tangents, will reach from the difference of longitude 522 to the meridional difference of latitude 835 on numbers.—And the extent from the complement of the course 58° to 90° on sines, will reach from the difference of latitude 601, to the distance 708 miles on numbers.
Prob. IX. To find the difference of longitude made good upon compound courses.
Rule. With the several courses and distances, complete the Traverse Table, and find the difference of latitude, departure, and course made good, and the latitude come to as in Traverse Sailing. Find also the meridional difference of latitude.
Now to the course and meridional difference of latitude, in a latitude column, the corresponding departure will be the difference of longitude, which applied to the longitude left will give the ship's present longitude.
Example. A ship from port St Julian, in latitude Mercator's 49° 10′ S, longitude 68° 44′ W, sailed as follows; Sailing. ESE 53 miles, SEbS 74 miles, E by N 68 m. SEbE½E 47 miles, and E 84 miles. Required the ship's present place?
| Courses | Diff. | Diff. of Lat. | Departure | |---------|------|--------------|-----------| | ESE | 53 | 20.3 | 49.0 | | SEbN | 74 | 61.5 | 41.1 | | EbyN | 68 | 13.3 | 66.7 | | SEbE½E | 47 | 22.1 | 41.5 | | E | 84 | | 84.0 |
| Latitude left | 13.3 | 103.9 | 282.3 | |---------------|------|-------|-------| | Latitude come to | 50° 41′ S m. pt. 3397 |
Mer. difference of latitude - 142
Now to course 72°, and opposite to 71, half the mer. difference of latitude in a latitude column, is 218.7 in a departure column, which doubled is 437, the difference of longitude.
Longitude of Port St Julian - 68° 44′ W Difference of longitude - 7 17 E Longitude come to - 61 27 W
Although the above method is that usually employed at sea to find the difference of longitude, yet as it has been already observed, it is not to be depended on, especially in high latitudes, long distances, and a considerable variation in the courses, in which case the following method becomes necessary.
Rule II. Complete the Traverse Table as before, to which annex five columns. Now with the latitude left, and the several differences of latitude, find the successive latitudes, which are to be placed in the first of the annexed columns; in the second, the meridional parts corresponding to each latitude is to be put; and in the third, the meridional differences of latitude.
Then to each course, and corresponding meridional difference of latitude, find the difference of longitude, by Prob. IV, which place in the fourth or fifth columns, according as the coast is easterly or westerly; and the difference between the sums of these columns will be the difference of longitude made good upon the whole, of the same name with the greater.
Remarks.
1. When the course is north or south, there is no difference of longitude. 2. When the course is east or west, the difference of longitude cannot be found by Mercator's Sailing; in this case the following rule is to be used. 3. To the nearest degree to the given latitude taken as a course, find the distance answering to the departure in a latitude column: this distance will be the difference of longitude.
Ex. Ex. 1. Four days ago we took our departure from Faro-head, in latitude $58^\circ 40' N$, and longitude $4^\circ 50' W$. Mercator's Sailing, and since have failed as follows: NW $32$ miles, W $69$ miles, WNW $93$ miles, WB $77$ miles, SW $58$ miles, and W $49$ miles.—Required our present latitude and longitude?
| Traverse Table | Longitude Table | |---------------|-----------------| | Courses | Diff. of Lat. | Departure | Succesive Latitudes | Merid. Parts | Merid. Diff. Lat. | Diff. of Longitude | | | N | S | E | W | N | S | E | W | N | S | E | W | | NW | 32 | 22.6| 22.6| 58° 40' | 437° | 45 | 45° | | W | 69 | | 69.0| 441.5 | 45 | 134° | | WNW | 93 | 35.6| 85.9| 4484 | 69 | 166.5| | WBs | 77 | 15.0| 75.5| 4454 | 30 | 151.0| | SW | 58 | 41.0| 41.0| 4374 | 80 | 80.0 | | W$^4$S | 49 | 7.2 | 48.5| 4361 | 13 | 88.0 | | W$^1$S | | | | 342.5 | | | | W$^1$S | 343 | 5.0 | | | | |
Longitude of Faro head $4^\circ 50' W.$ Difference of longitude $11^\circ 4 W.$ Longitude in $15^\circ 54 W.$
Ex. 2. A ship from latitude $78^\circ 15' N$, longitude $28^\circ 14' E$, sailed the following courses and distances. The latitude come to is required, and the longitude, by both methods: the bearing and distance of Hacluit's head-land, in latitude $79^\circ 55' N$, longitude $11^\circ 55' E$, is also required?
| Traverse Table | Longitude Table | |---------------|-----------------| | Courses | Diff. of Latitude | Departure | Succesive Latitudes | Merid. Parts | Merid. Diff. Lat. | Diff. of Longitude | | | N | S | E | W | N | S | E | W | N | S | E | W | | WNW | 154 | 58.9| | 142.3| 79 | 14 | 8120 | 303 | | | | 731.7| | SW | 96 | | 67.9| 67.9 | 78 | 6 | 7774 | 346 | | | | 346.0| | NW$^4$W | 89 | 56.4| | 68.8 | 79 | 2 | 8056 | 282 | | | | 343.6| | NbE | 110 | 107.9| | 21.5 | 80 | 50 | 8676 | 620 | | | | 218.0| | NW$^4$N | 56 | 45.0| | 33.4 | 81 | 35 | 8970 | 294 | | | | 218.0| | SbE$^4$E | 78 | | 73.4| 26.3 | 80 | 22 | 8504 | 466 | | | | 166.7|
By Rule I. Latitude left $78^\circ 15' N.$ Mer. pts. $7817$ Diff. of latitude $2^\circ 7 N.$ Lat. come to $80^\circ 22 N.$ Mer. pts. $8504$
Meridional diff. of latitude $687$ As difference of lat. $126.9$ is to mer. diff. of lat. $687$ fo is the departure $264.6$ to diff. of longit. $1432$ Longitude left $28^\circ 14' E.$ Longitude in $4^\circ 22' E.$
The error of this method, in the present example, is therefore $1^\circ 23'$.
To find the bearing and distance of Hacluit's head-land. Lat. H. H. = $79^\circ 55' N$. M. P. $8347$ Lon. $11^\circ 55' E.$ Lat. ship. = $80^\circ 22 N$. M. P. $8504$ Lon. $5^\circ 45' E.$ Diff. lat. $0^\circ 27$ M. D. L. $157$ D. L. $6^\circ 10'$ Now to $78.5$ half the meridional difference of latitude, and $185.0$ half the difference of longitude, the course $67^\circ$, and opposite to the difference of latitude $27$, the distance is $69$ miles.—Hence Hacluit's head-land bears $S 67^\circ E.$ distant $69$ miles. CHAP. VII. Containing the Method of resolving the several Problems of Mercator's Sailing, by the Assistance of a Table of Logarithmic Tangents.
Prob. I. Given one latitude, distance, and difference of longitude; to find the course, and other latitude.
Rule. To the arithmetical complement of the logarithm of the distance, add the logarithm of the difference of longitude in minutes, and the log. cosine of the given latitude; the sum rejecting radius will be the log. sine of the approximate course.
To the given latitude taken as a course in the traverse table, add half the difference of longitude in a distance column; the corresponding departure will be the first correction of the course, which is subtractive if the given latitude is the least of the two; otherwise, additive.
In Table A, under the complement of the course, and opposite to the first correction in the side column, is the second correction. In the same table find the number answering to the course at the top, and difference of longitude in the side column; and such part of this number being taken as is found in Table B opposite to the given latitude, will be the third correction. Now these two corrections, subtracted from the course corrected by the first correction, will give the true course.
Now the course and distance being known, the difference of latitude is found as formerly.
| Arc. | 1° | 2° | 3° | 4° | 5° | 6° | 7° | 8° | |------|----|----|----|----|----|----|----|----| | 1° | 3' | 1' | 1' | 1' | 1' | 1' | 1' | 1' | | 2 | 12 | 6 | 4 | 2 | 2 | 2 | 2 | 2 | | 3 | 27 | 13 | 8 | 6 | 4 | 4 | 4 | 4 | | 4 | 47 | 23 | 14 | 10 | 7 | 7 | 7 | 7 | | 5 | 74 | 36 | 23 | 16 | 11 | 11 | 11 | 11 | | 6 | 107| 52 | 33 | 22 | 16 | 11 | 11 | 11 | | 7 | 145| 70 | 44 | 30 | 21 | 15 | 15 | 15 | | 8 | 190| 92 | 58 | 40 | 28 | 19 | 19 | 19 |
Example. From latitude 50° N, a ship sailed 290 miles between the south and west, and differed her longitude 5°. Required the course, and latitude come to?
Distance - 290. ar. co. log. 7.53760 Diff. of longitude - 300. log. 2.47712 Latitude - 50° 0' co. - 9.80807
Approximate course - 41° 41' fine - 9.82279 To lat. 50°, and half diff. long. 150 in a dist. col. the first corr. in a dep. col. is 115 + 1 55 Approximate course - 41° 41' Cor. - 1 55 In Table A to co. course 48° and 1st corr. 1° 55' the second direction is To course 41° and diff. long. 5°, the number is 15, of which \( \frac{5}{8} \) (Tab. B) being taken, gives True course - S. 43° 31' W
To find the difference of latitude. As radius - 10.00000 is to the cosine of the course 43° 33' - 9.86020 so is the distance - 290 - 2.46240 to the difference of latitude - 210.2 - 2.32260 Latitude left - 50° 0' N Difference of latitude - 3° 30' S
Latitude come to - 46° 30' N
This problem was proposed, and resolved, by Mr Robert Hues in his Treatise on the Globes, printed at London in the year 1639, p. 181.
It was afterwards proposed by Dr Halley, in the second volume of the Miscellanea Curiosa, p. 35, in the following words:
A ship sails from a given latitude, and, having run a certain number of leagues, has altered her longitude by a given angle; it is required to find the course steered.
And he then adds—The solution hereof would be very acceptable, if not to the public, at least to the author of this treatise, being likely to open some further light into the mysteries of geometry.
Since that time, this problem has been solved in an indirect manner, by several writers on navigation, and others—As Monsieur Bouguer, in his Nouveau Traité de Navigation; Mr Robertson, in the second volume of his Elements of Navigation; Mr Emerson, in his Theory of Navigation, which accompanies his Mathematical Principles of Geography; Mr Israel Lyons, in the Nautical Almanack for 1772; and Monsieur Bezout, in his Traité de Navigation; and lately, Baron Maferes, with the assistance of Mr Attwood, has given the first direct solution of this problem. For a comparison of the various solutions which have hitherto been made of this problem, the reader is referred to that by Dr Mackay, in the fourth and fifth volumes of Baron Maferes's Scriptores Logarithmici.
It was intended in this place to have given rules, to make allowance for the spheroidal figure of the earth; but as the ratio of the polar to the equatorial semi-axis is not as yet determined with sufficient accuracy, neither is it known if both hemispheres be similar figures; therefore these rules would be grounded on assumption only, and might probably err more from the truth. **NAVIGATION**
**By Calculation.**
In the triangle BIF are given BI and BF equal to 39 miles, and 26 miles respectively; and the angle BFI equal to 7 points; To find the side FI, and angle FBI.
To find the angle BIF.
As the distance BI - 39 - 1.59106 is to the distance BF - 26 - 1.41497 so is the fine of BFI - 78° 45' - 9.99157
to the fine of BIF - 40 50 - 9.81548 Sum - 119 35
Angle FBI - 60 25 EBF - 33 45
Difference, or EBI - 26 40
Bearing of Jersey from Brehaut N 63 20 E.
To find the distance FI.
As the fine of BFI - 78° 45' - 9.99157 is to the fine of FBI - 60 25 - 9.99394 so is the distance BI - 39 miles - 1.59106
to the distance FI - 34° 58' - 1.53883
**Ex. 3.** At noon Dungeness bore per compass NW, distance 5 leagues; and having run NW by W 7 knots an hour, at 5 P.M. we were up with Beachyhead. Required the bearing and distance of Beachyhead from Dungeness?
**By Construction.**
Describe a circle (fig. 36.) to represent the horizon; Fig. 36., from the centre C draw the NW line CD equal to 15 miles; and the NW by W line CB equal to 35 miles; join DB, which applied to the scale will measure about 26½ miles; and the inclination of DB to the meridian will be found equal to N 79° 24' W.
**By Calculation.**
In the triangle DCB are given the distances CD, CB equal to 15 and 35 miles respectively; and the angle BCD equal to 4 points; to find the angles B and D, and the distance BD.
To find the angles.
Distance CB = 35, sum of the ang. 16 points CD = 15, angle C - 4
Sum - 50, angles B and D 12 Difference 20, half sum - 6 pts. = 67° 30' As the sum of the distances - 50 - 1.09897 is to their difference - 20 - 1.30103 so is the tangent of half sum angles 67° 30' - 10.38378
to the tangent of half their diff. 44° - 9.98484
Angle CDB - 111° 30' Supplement - 68° 30' Angle, NCD - 111° 15'
Magnetic bearing - N 79° 45' W. Or by allowing 2½ points of westerly variation, the true bearing of Beachyhead from Dungeness will be W ¼ S nearly.
---
**Chap. VIII. Of Oblique Sailing.**
Oblique sailing is the application of oblique-angled plane triangles to the solution of problems at sea. This sailing will be found particularly useful in going along shore, and in surveying coasts and harbours, &c.
**Ex. 1.** At 11th A.M. the Girdle Nefs bore WNW, and at 2h P.M. it bore NW by N: the course during the interval SbW five knots an hour. Required the distance of the ship from the Nefs at each station?
**By Construction.**
Describe the circle NE, SW (fig. 34.), and draw the diameters NS, EW, at right angles to each other: from the centre C, which represents the first station, draw the WNW line CF; and from the same point draw CH, SbW, and equal to 15 miles the distance sailed.—From H draw HF in a NW by N direction, and the point F will represent the Girdle Nefs. Now the distances CF, HF will measure 19.1 and 26.5 miles respectively.
**By Calculation.**
In the triangle FCH are given the distance CH 15 miles, the angle FCH equal to 9 points, the interval between the SbW and WNW points, and the angle CHF equal to 4 points, being the supplement of the angle contained between the SbW and NW by N points; hence CFH is 3 points: to find the distances CF, HF.
To find the distance CF.
As the fine of CFH - 3 points - 9.74474 is to the fine of CHF - 4 points - 9.84948 so is the distance CH 15 miles - 1.17609
to the distance CF - 19.07 - 1.28083
To find the distance FH.
As the fine of CFH - 3 points - 9.74474 is to the fine of FCH - 4 points - 9.99157 so is the distance CH - 15 miles - 1.17609
to the distance FH - 26.48 - 1.42292
**Ex. 2.** The distance between the SE point of the island of Jersey and the island of Brehaut is 13 leagues; and the correct bearing and distance of Cape Frehel from the island of Brehaut is SE by E 26 miles. It is also known that the SE point of Jersey bears NNE from Cape Frehel: from whence the distance of these two is required, together with the bearing of the same point from the island of Brehaut?
**By Construction.**
Describe a circle, (fig. 35.) and draw two diameters at right angles, the extremities of which will represent the cardinal points, north being uppermost.—Let the centre B represent Brehaut, from which draw the SE by E line BF equal to 26 miles, and the point F will represent Cape Frehel, from which draw the NNE line FI; make BI equal to 39 miles; Then FI applied to the scale will measure 34½ miles, and the inclination of BI to the meridian will be found equal to 63° 17'. To find the distance.
As the fine of CDB - 111°30' - 9.96868 is to the fine of BCD - 45° - 9.84948 so is the distance BC - 35 - 1.54497
to the distance BD - 26.6 - 1.42487
Ex. 4. Running up Channel EbS per compass at the rate of 5 knots an hour. At 11 A.M., the Eddifton lighthouse bore N4E1/2E, and the Start point NEbE1/2E; and at 4 P.M., the Eddifton bore NWbN, and the Start N1/2E. Required the distance and bearing of the Start from the Eddifton, the variation being 24 points W?
By Construction.
Let the point C (fig. 37.) represent the first station, from which draw the N4E1/2E line CA, the NEbE1/2E line CB, and the EbS line CD, which make equal to 25 miles the distance run in the elapsed time; then from D draw the NEbN line DA intersecting CA in A, which represents the Eddifton; and from the same point draw the N1/2E line DB cutting CB in B, which therefore represents the Start. Now the distance AB applied to the scale will measure 22.9, and the bearing per compass BAF will measure 73°01'.
By Calculation.
In the triangle CAD are given CD equal to 25 miles, the angle CAD equal to 4½ points, the distance between N4E1/2E and NWbN; and the angle ADC equal to 4 points, the distance between the NWbN and WBn points; to find the distance CA.
As the fine of CAD 4½ points - 9.86979 is to the fine of CDA 4 points - 9.84948 so is the distance CD 25 miles - 1.39794
to the distance CA 23.86 - 1.37763
In the triangle BCD, are given the distance CD 25 miles, the angle CBD 4½ points the interval between NEbE1/2E and N1/2E, and CDB 7½ points, the distance between WBn and N1/2E; to find the distance CD.
As the fine of CBD 4½ points - 9.88819 is to the fine CDB 7½ points - 9.99947 so is the distance CD 25 miles - 1.39794
to the distance CB 32.3 - 1.50922
In the triangle CAB, the distances CA, CB, are given, together with the included angle ACB, equal to 4 points, the distance between N4E1/2E and NEbE1/2E; to find the angle CAB and distance AB.
Distance CB 32.3 Angle ACB = 45°0'
Distance CA 23.86 Sum of CAB and ABC 135°
Sum 56.16 Half - - 67 30 Difference 8.44 As the sum of the distances 56.16 - 1.74943 is to their difference 8.44 - 0.92634 so is the tangent of half } 67 30 - 10.38278 sum angles to the tangent of half } 19 56 - 9.55969 diff. angles
Angle CAB - - 87 26 Angle CAF - - 14 4
Bearing per compass S 73°22' E or ESE1/2E; and the variation 2½ points being allowed to the left of ESE1/2E, gives E1/2N, the true bearing of the Start from the Eddifton.
To find the distance.
As the fine of CAB 87°26' - - 9.99956 is to the fine of ACB 45° - - 9.84948 so is the distance CB 32.3 - 1.50922
to the distance AB 22.86 - 1.35914
Ex. 5. A ship from a port in latitude 57°9' N, longitude 2°9' W, sailed 82 miles on a direct course, and spoke a ship that had run 100 miles from a port in latitude 56°21' N, longitude 2°50' W.—Required the course of each ship, and the latitude and longitude come to?
Lat. 57°9' N Mer. parts 4199 Lon. 2°9' W 56°21' N ------------------ 4112 - 2°50' W
Diff. of lat. 48 Mer. diff. lat. 87 Diff. lon. 41
By Construction.
With the meridional difference of latitude, the difference of longitude, and difference of latitude, construct the triangles ADE, ABC (fig. 38.) as in Mer.-Fig. 32, cator's Sailing; then A will represent the northernmost, and C the southernmost port. The distance AC applied to the scale will measure 53 miles, and the bearing BCA will be 25°0'. From the points A and C, with distances equal to 82 and 100 miles respectively, describe arches intersecting each other in M, which will therefore be the place of meeting.—Now the angle ABM, the ship's course from the southernmost port, will measure N80°2'E; and the other ship's course, or angle BAM, will be 67°2', or ESE. From M draw the parallel MNP, and AN will be the difference of latitude made by the one ship, and CP that by the other ship: hence either of these being measured and applied to its correspondent latitude, will give 56°38', the latitude in. Make AF equal to 57, the meridional difference of latitude between the northernmost port and latitude in: from F draw FG perpendicular to AF, and produce AM to G, then FG will be the difference of longitude, which applied to the scale will measure 139: hence the longitude in, is 0°10' E.
By Calculation.
In the triangle ADE, ABC, are given AD equal to 87, DE equal to 41, and AB equal to 48; to find the angle BAC and distance AC.
To find the bearing of the ports.
As the meridional diff. of lat. 87 - - 1.93952 is to the diff. of long. - - 41 - - 1.61278 so is radius - - - - 10.00000
to the tangent of the bearing 25°14' - - 9.67326
To find the distance of the ports.
As radius - - - - 10.00000 is to the secant of the } 25°14' - - 10.04355 bearing fo is the diff. of latitude 48 - - 1.68124
to the distance 53.06 - 1.72479
In the triangle AMC, the three sides are given to find the angles. To find the angle ACM.
| AM | 82 | | MC | 100 ar. co. log. | 8.00000 | | AC | 53.06 ar. co. log. | 8.27523 |
Sum | 235.06 log. | 2.07015 | Half | 117.53 log. | 1.55059 | Difference | 53.53 |
| 27 29 | cosine | 9.94798 |
Angle ACM | 54 58 | Angle BAC | 25 14 |
Southernmost ship's course | N 80 12 E |
To find the angle MAC.
As AM | 82 | 1.91381 | is to MC | 100 | 2.00000 | fo is the sine of ACM | 54 58 | 9.91319 |
to the sine of MAC | 93 3 | 9.99938 | Angle BAC | 25 14 |
Northernmost ship's course | S 67 49 E, or ESE |
In the right-angled triangle AMN, given AM, and the angle MAN, to find the differences of latitude AN.
As radius | 10.00000 | is to the cosine of | 67° 49' | 9.57700 | the course | 82 | 1.91381 |
to the diff. of lat. | 30.96 | 1.49081 | Latitude of northernmost port | 57 9 Mer. parts | 4199 | Latitude in | 56 38 Mer. parts | 4142 |
Meridional difference of latitude | 57 |
To find the difference of longitude FG.
As radius | 10.00000 | is to the tangent of | 67° 49' | 10.38960 | the course | 57 | 1.75587 |
to the diff. of long. | 139.8 | 2.14547 | Longitude left | 2° 9' W | Difference of longitude | 2 20 E |
Longitude in | 0 11 |
**CHAP. IX. Of Windward Sailing.**
Windward sailing is, when a ship by reason of a contrary wind is obliged to sail on different tacks in order to gain her intended port; and the object of this sailing is to find the proper course and distance to be run on each tack.
Ex. 1. A ship is bound to a port 48 miles directly to the windward, the wind being SSW, which it is intended to reach on two boards; and the ship can lie within six points of the wind. Required the course Windward Sailing and distance on each tack?
**By Construction.**
Draw the SSW line CB (fig. 39.) equal to 48 miles, Fig. 39. Make the angles ACB, ABC, each equal to 6 points. Hence the first course will be W, and the second SE; also the distance CA, or AB, applied to the scale, will measure 62½ miles, the distance to be sailed on each board.
**By Calculation.**
From A draw AD perpendicular to BC; then in the triangle ADC are given CD equal to 24 miles; and the angle ACD, equal to 6 points, to find the distance AC.
As radius | 10.00000 | is to the secant of C | 6 points | 10.41716 | fo is CD | 24 miles | 1.38021 | to CA | 62.7 | 1.79737 |
Ex. 2. The wind at NW, a ship bound to a port 64 miles to the windward, proposes to reach it on three boards; two on the starboard, and one on the larboard tack, and each within 5 points of the wind. Required the course and distance on each tack?
**By Construction.**
Draw the NW line CA (fig. 40.) equal to 64 miles; Fig. 40. from C draw CB WB S, and from A draw AD parallel thereto, and in an opposite direction; bisect AC in E, and draw BED parallel to the NBE rhumb, meeting CB, AD in the points B and D: then CB = AD applied to the scale will measure 36½ miles, and BD = 2CB = 72½ miles.
**By Calculation.**
From B draw BF perpendicular to AC; then, in the triangle BFC are given the angle BCF, equal to 5 points, and CF equal to one-fourth of CA = 16 m. to find CB.
As radius | 10.00000 | is to the secant of BCF | 5 points | 10.25526 | fo is CF | 16 m. | 1.20412 | to CB | 36.25 | 1.55938 |
Ex. 3. A ship which can lie within 5½ points of the wind, is bound to a port 36 miles to the windward, the wind being NEbN, which it is intended to reach on four boards, the first being on the larboard tack. Required the course and distance on each?
**By Construction.**
Draw the NEbN line CA (fig. 41.) equal to 36 Fig. 41. miles, and bisect it in B; from C and B draw lines parallel to the E½S rhumb; and from A and B draw lines parallel to the SSE½E point, meeting the former in the points D and E. Now the distances AD, BD, BE, and CE, are equal; and any one of them applied to the scale will measure 19.1 miles.
**By Calculation.**
From E draw EF perpendicular to AC; and in the triangle CFE are given CF = 9 m. and the angle FCE = 5½ points, to find CE. Ex. 4. A ship bound to a port bearing NNE distant 40 miles, with the wind at NNE, intends to reach it on two boards. Required the course and distance on each tack, the ship lying within 5 points of the wind?
By Construction.
Draw the NNE line CA (fig. 42.) equal to 40 miles; and because the wind is NNE, and the ship can lie within 5 points of the wind, the course on the larboard tack will be ENE, and on the starboard NW. Therefore, from the centre C draw the ENE line CB, and from it draw the NW line AB, meeting CB in B; then CB and AB applied to the scale will measure 26.7 and 48.1 m. respectively.
By Calculation.
In the triangle ACB, given AC = 40 miles, and the angles A, B, and C, equal to 3, 5, and 8 points respectively, to find AB and BC.
To find the distance CB.
As the fine of B - 5 points 9.91985 is to the fine of A - 3 points 9.74474 so is the distance CA - 40 miles 1.60206
to the distance CB - 26.73 1.42695
To find the distance AB.
As the fine of B - 5 points 9.91985 is to the fine of C - 8 points 10.00000 so is the distance CA - 40 miles 1.60206
to the distance AB - 48.11 1.68221
Ex. 5. A ship close hauled within 5 points of the wind, and making one point of leeway, is bound to a port bearing SSW, distant 54 miles, the wind being SSE. It is intended to make the port at three boards, the first of which must be on the larboard tack in order to avoid a reef of rocks. Required the course and distance on each tack?
By Construction.
Draw the SSW line CA (fig. 43.) equal to 54 m. and as the wind is SSE, and the ship makes her course good within 6 points of the wind, therefore the course on the larboard tack will be SWB, and on the starboard ESE; hence from C draw the SWB line CB, and from A draw AD parallel thereto; bisection CA in E, and draw BED parallel to the ESE line; then will CB and AD be the distances on the larboard tack, which applied to the scale, each will be found to measure 37.4; and the distance on the starboard tack BD will measure 42.4 miles.
By Calculation.
The triangles CBE, EAD are equal and similar; hence in the first of these are given CE, equal to 27 miles, half the distance between the ship and port; the angles C, B, and E, equal to 3, 4, and 9 points respectively, to find CB and BE.
To find CB, the distance on the larboard tack.
As the fine of B - 4 points 9.84948 is to the fine of E - 9 points 9.99157 so is the distance CE - 27 miles 1.43136
to the distance BC - 37.45 1.57345
To find BE, half the distance on the starboard tack.
As the fine of B - 4 points 9.84948 is to the fine of C - 3 points 9.74474 so is the distance CE - 27 miles 1.43136
to the distance BE - 21.21 1.32662
Whole distance AC - 42.42
Ex. 6. A ship plying to the windward, with the wind at NNE, after sailing 51 miles on each of two tacks, is found by observation to have made 36 miles of difference of latitude. How near the wind did she make her way good?
By Construction.
Make CA (fig. 44.) equal to 36 miles; draw AB perpendicular to CA, and draw the NNE line CB, meeting AB in B; make CD, BD each equal to 51 miles, and these being measured, will be found equal to 6 points.
By Calculation.
In the triangles CAB, BCD, are given AB equal to 36 m., CD = BD = 51, and the angle ACB equal to 2 points; to find the angle BCD.
As the distance CD - 51 1.70757 is to the diff. of latitude CA - 18 1.25527 so is the secant of ACB - 2 points 10.03438
to the cosine of BCD - 67° 32' 9.58208
Chap. X. Of Current Sailing.
The computations in the preceding chapters have been performed upon the assumption that the water has no motion. This may no doubt answer tolerably well in those places where the ebbs and flows are regular, as then the effect of the tide will be nearly counterbalanced. But in places where there is a constant current or setting of the sea towards the same point, an allowance for the change of the ship's place arising therefrom must be made: And the method of resolving these problems, in which the effect of a current, or heave of the sea, is taken into consideration, is called current sailing.
In a calm, it is evident a ship will be carried in the direction and with the velocity of the current. Hence, if a ship sails in the direction of the current, her rate will be augmented by the rate of the current; but if sailing directly against it, the distance made good will be equal to the difference between the ship's rate as given by the log and that of the current. And the absolute motion of the ship will be a-head, if her rate exceeds that of the current; but if less, the ship will make sternway. If the ship's course be oblique to the current, the distance made good in a given time will be represented by the third side of a triangle, whereof the distance given by the log, and the drift of the current in the same time, are the other sides; and the true course will be the angle contained between the meridian and the line actually described by the ship. Ex. 1. A ship sailed NNE at the rate of 8 knots an hour, during 18 hours, in a current setting NWbW2½ miles an hour. Required the course and distance made good?
By Construction.
Draw the NNE line CA (fig. 45.) equal to 18 × 8 = 144 miles; and from A draw AB parallel to the NWbW rhumb, and equal to 18 × 2½ = 45 miles: now BC being joined will be the distance, and NCB the course. The first of these will measure 159 miles, and the second 6° 23'.
By Calculation.
In the triangle ACB, are given AC = 144 miles, AB = 45 miles, and the angle CAB = 9 points, to find BAC and BC.
To find the course made good.
| Dist. AC | 144 | Ang. BAC = 9 pts = 101° 15' | | Dist. AB | 45 |
Sum - 189
As the sum of the sides - 189 - 2.27646 is to the difference of the sides - 99 - 1.99563 so is the tan. of half sum angles - 39 22½ - 9.91417
to the tan. of half diff. angles - 23 15½ - 9.63334
Angle ACB - 16° 7' Angle ACN - 22° 30'
Course made good N 6 23
To find the distance.
As the fine of ACB - 16° 7' - 9.44341 is to the fine of CAB - 101° 15' - 9.99157 so is the distance AB - 45 - 1.65321
to the distance CB - 159 - 2.20137
Ex. 2. A ship from a port in latitude 42° 52' N, sailed SbW½W 17 miles in 7 hours, in a current setting between the north and west; and then the same port bore ENE, and the ship's latitude by observation was 42° 42' N. Required the setting and drift of the current?
By Construction.
Draw the SbW½W line CA (fig. 46.) equal to 17 miles, and make CB equal to 10 miles, the difference of latitude: through B draw the parallel of latitude BD, and draw the WSW line CD, intersecting BD in D: AD being joined, will represent the drift of the current, which applied to the scale will measure 20.2, and the angle DAE will be its setting, and will be found equal to 72°.
By Calculation.
In the triangle CBD, given CB = 10 miles, and the angle BCD = 6 points; to find the distance CD.
As radius - 10.00000 is to the secant of BCD - 6 points - 10.41710 so is the diff. of lat. CB - 10 miles - 1.00000
to the distance CD - 26.13 - 1.41710
Again, in the triangle ACD are given the distance AC = 17 miles, CD = 26.13, and the angle ACD 4½ points; to find the remaining parts.
To find the setting of the current.
Distance DC = 26.13 Angle ACD = 4½ points. Distance AC = 17.0 CAD + CDA 11½
Sum - 43.13 CAD + CDA 2 = 5½ = 64 41'
Difference - 9.13
As the sum of the sides - 43.13 - 1.63478 is to the differ. of the sides - 9.13 - 0.96047 so is tang. half sum angles - 64° 41' - 10.32509
to tang. half diff. angles - 24 6' - 9.65078
Angle CAD - 88° 47' Angle CAE = ACB = r½ pt. = 16° 52'
Setting of the current EAD = 71° 55'
To find the drift of the current.
As the fine of CAD - 88° 47' - 9.99990 is to the fine of ACD - 4½ points - 9.88819 so is the distance CD - 26.13 - 1.41710
to the drift of current AD 20.2 - 1.30539
Hence the hourly rate of the current is 20.2 = 2.9 knots.
Ex. 3. A ship, from latitude 38° 20' N, sailed 24 hours in a current setting NWbN, and by account is in latitude 38° 42' N, having made 44 miles of easting; but the latitude by observation is 38° 58' N. Required the course and distance made good, and the drift of the current.
By Construction.
Make CE (fig. 47.) equal to 22 miles, the difference Fig. 47. of latitude by D, R, and EA = 54 miles, the departure, and join CA; make CD = 38 miles, the difference of latitude by observation; draw the parallel of latitude DB; and from A draw the NWbN line AB, intersecting DB in B, and AB will be the drift of the current in 24 hours: CB being joined, will be the distance made good, and the angle DCB the true course. Now, AB and CB applied to the scale, will measure 19.2 and 50.5 respectively: and the angle DCB will be 41° 2'.
By Calculation.
From B draw BF perpendicular to AE, then in the triangle AFB are given BF = 16 miles, and the angle ABF = 3 points; to find AB and AF.
To find the drift of the current AB:
As radius - 10.00000 is to the secant of ABF - 3 points - 10.08015 so is BF - 16 miles - 1.20412
to the drift of the current AB - 19.24 - 1.28427
Hence the hourly rate = 19.24 = 0.8.
To find AF.
As radius - 10.00000 is to the tangent of ABF - 3 points - 9.82459 so is BF - 16 miles - 1.20412
to AF - 10.69 - 1.02901.
Departure by account EA - 44.
True departure EF = DB = 33.31. Now, in the triangle CDB are given the difference of latitude and departure; to find the course and distance.
To find the course. As the difference of latitude CD - 38° - 1.57978 is to the departure DB - 33°31' - 1.52257 so is radius - - - - 10.00000
to the tangent of the course - 41°14' - 9.94279
To find the distance. As radius - - - - 10.00000 is to the secant of the course - 41°14' - 10.12376 so is the difference of latitude - 38° - 1.57978
to the distance - - - - 50.53 - 1.70354
Ex. 4. In the Straits of Sunda, at 2 P.M. steering SE6S at the rate of 5 knots an hour, I passed close by the small islands off Hog point. At 6, not having changed our course, came to anchor on the Java shore. Upon setting the said island from this anchoring place, I find it bears due north, its distance by the chart being 22 miles. It follows from hence, that our course has been affected by a current. Required its velocity and direction?
By Construction. From A (fig. 48.) draw the SEbS line AB=20, which will represent the ship's apparent track through the water; draw AC equal to 22 miles south, and C will be the ship's real place; and BC being joined will be the current's drift in four hours; which applied to the scale will measure 12.3; from A draw AD parallel to BC, and the angle CAD will be the direction of the current, and will be found to measure 64°15'.
By Calculation. In the triangle ABC, given AB=20m. AC=22m. and the included angle A=3 points; to find the remaining parts.
To find the setting of the current. Distance AC=22m. Included angle =3 points.
| AB | 20 | |----|----| | B+C | 13 |
Sum - 42 Difference - 2 As the sum of the sides - 42 - 1.62325 is to the diff. of the sides - 0.30103 so is the tang. of half sum angles 73°7'11" - 10.51806
to tang. of half diff. angles 8.5554 - 9.19584
Setting of the current S 64°12W, or SWbW34W.
To find the drift of the current. As the fine of ACB 64°12' - 9.95449 is to the fine of BAC 33°45' - 9.74474 so is the distance AB - 20 - 1.30103
to the velocity of cur. BC 12°34' - 1.09137 and 12°34' = 3.1, its hourly rate.
Example 5. A ship bound from Dover to Calais, lying 21 miles to the SEbE½E, and the flood tide setting NE½E 2½ miles an hour. Required the course she must steer, and the distance run by the log at 6 knots an hour to reach her port?
By Construction. In the position of the SEbE½E rhumb, draw DC = 21 miles (fig. 49.) draw DE NE½E = 2½ miles; from E with 6 miles cut DC in F; draw DB parallel to EF, meeting CB drawn parallel to DE: then the distance DB applied to the scale will measure 19.4, and the course SDB will be SE½S.
By Calculation. In the triangle DBF, given DE=2½ miles, EF =6 miles and the angle EDF=6 points; to find the angle DFE=CBD.
As the hourly rate of failing - 6m. 0.77815 is to the hourly rate of current 2½m. 0.39794 so is the fine of EDF=6 points 67°30' 9.96562
to the fine of DFE - 22°38' 9.58541 Angle SDC=3½ points = 61°52'
Course SDB - - - - 39°14'=SE½S.
In the triangle DBC, given DC=21 miles, the angle BDC=DFE=22°38', and the angle DCB=DEF=6 points; to find the distance DB.
As the fine of DBC - 89°52' - 9.99999 is to the fine of DCB - 67°30' - 9.96562 so is the true distance DC 21 m. 1.32222
to the distance by the log DB. 21 m. 1.28785
Chap. XI. Instruments proposed to solve the various Problems in Sailing, independent of Calculation.
Various methods, besides those already given, have been proposed to save the trouble of calculation.—One of these methods is by means of an instrument composed of rulers, so disposed as to form a right-angled triangle, having numbers in a regular progression marked on their sides. These instruments are made of different materials, such as paper, wood, brass, &c. and are differently constructed, according to the fancy of the inventor. Among instruments of this kind, that by John Cooke, Esq. seems to be the best. A number of other instruments, very differently constructed, have been proposed for the same purpose; of these, however, we shall only take notice of the rectangular instrument, by A. Mackay, LL.D. F.R.S.E. &c.
I. Of Cooke's Triangular Instrument.
Description. The stock abcd (fig. 50.) is a parallelopiped: The length from a to b is two feet, the breadth from a to d two inches, and the depth is one inch and a half. The stock is perforated longitudinally, so as to be capable of containing within it, a cylindrical piece of wood one inch diameter; g h is an aperture on the surface of the stock about a quarter of an inch wide, which discloses one-twelfth part of the surface of the cylinder contained; the edge dc is divided into twelve parts, each of these is subdivided into six parts, and each of these again into ten parts. The surface of the cylinder is divided longitudinally into twelve parts, and on each of them is engraved a portion... Practice.
Instruments portion of a line of meridional parts 22 feet long, which contains the meridional parts for every minute from the equator as far towards the pole as navigation is practicable; and the smallest division on it is not less than \( \frac{1}{6} \)th of an inch. By rolling and sliding this cylinder, any part of any line on it may be brought into any position which may be required; the box \( i \) is engraved into the edge of the stock \( a b \), so that it may move freely from \( a \) to \( b \); a limb from this box extends to \( k \), which serves to mark that degree of the perpendicular \( l \) which is parallel to the centre of the semicircle \( m \); \( l \) is two feet long, and graduated on both edges as the stock; it is perpendicular to the stock, and is fixed in the box \( i \), by which it may be moved from \( a \) to \( b \); \( o p n \) is a semicircle of six inches radius, engraved, as appears in the plate, which slides freely from \( c \) to \( d \) in a groove in the edge of the stock \( c d \); \( m q \) is the index moving on the centre \( m \), the edge of which marks the course on the semicircle; it is two feet long, and divided into 72 parts; and these are subdivided in the same manner as those on the stock and perpendicular, to which they are equal; \( r \) is a vernier attached to the index to show minutes; \( s \) is a vernier composed of concentric semicircles, which slides along the edge \( q m \), to the intersection of the perpendicular and index, where it serves as a vernier to both; below \( x \) is a small piece of ivory, with a mark on it to point out the degree of the line \( d c \), which is perpendicularly under the centre of the semicircle. Fig. 51, is a view of the back part of the instrument.
Use. The method of working every case which occurs in navigation, is to make the instrument similar to that ideal triangle which is composed of the difference of latitude, departure, and distance; or, to that composed of the meridian difference of latitude, difference of longitude, and enlarged distance; or, to that composed of the difference of longitude, departure, and sine of the middle latitude; which is done by means of the data procured from the compass, log-line, and quadrant: whence it follows, from the nature of similar triangles, or from the relation which exists between the sides of triangles and the sines of their opposite angles, that the parts of the instrument become proportional to those which they represent; and will ascertain the length of the lines, or the extent of the angles sought, by its graduations.
In the practice of this instrument, a small square is necessary in order to bring the centre of the semicircle perpendicularly over the meridional degree corresponding to the latitude.
Plane Sailing.
Prob. I. The course and distance being given, to find the difference of latitude and departure.
Example. A ship from latitude \( 24^\circ 18' \) N, sailed NW by N 168 miles. Required the latitude come to, and departure?
Set the centre of the semicircle perpendicularly over the given latitude \( 24^\circ 18' \), and the index to the course 3 points; move the perpendicular until it cut the index at the given distance 168; then at the point of intersection on the perpendicular is 93.3 miles, the departure, and on the base, by the edge of the box, is \( 26^\circ 38' \), the latitude come to.
Vol. XIV. Part II.
Prob. II. Both latitudes and course given, to find the distance and departure.
Example. Let the latitude failed from be \( 43^\circ 50' \) N, that come to \( 47^\circ 8' \) N, and the course NNE. Required the distance and departure?
Move the centre of the semicircle to the latitude left \( 43^\circ 50' \), and the edge of the box to the latitude come to \( 47^\circ 8' \); fix the index at the given course 2 points; then at the point of intersection of the index and perpendicular is the distance 214 miles on the index, and the departure 82 miles on the perpendicular.
Prob. III. Given the course and departure, to find the distance and difference of latitude.
Example. Let the latitude failed from be \( 32^\circ 38' \) N, the course SW by S, and the departure 200 miles. Required the distance and latitude come to?
Move the centre of the semicircle to the latitude left \( 32^\circ 38' \), set the index to the given course 3 points, and move the perpendicular till the given departure 200 cuts the index; at this point on the index is 360 miles, and the edge of the box will cut the latitude come to \( 27^\circ 39' \) N.
Prob. IV. Given the difference of latitude and distance, to find the course and departure.
Example. Let the latitude left be \( 17^\circ 10' \) N, the latitude come to \( 21^\circ 40' \) N, and the distance failed on a direct course between the north and west 300 miles. Required the course and departure?
Move the semicircle and box to the given latitudes, and the index until the distance found thereon meets the perpendicular; then at the point of contact on the perpendicular is 130.8, the departure, and on the semicircle by the index is \( 21^\circ 50' \), the course.
Prob. V. The distance and departure given, to find the course and difference of latitude.
Example. The distance failed is 246 miles between the south and east, the departure is 138 miles, and the latitude left \( 51^\circ 10' \) N. Required the course and latitude come to?
Set the centre of the semicircle to \( 51^\circ 10' \), the latitude failed from; find the distance 246 on the index, and the departure 138 on the perpendicular; then move both till these points meet, and the course \( 34^\circ 10' \) will be found on the semicircle by the index, and the latitude in \( 47^\circ 47' \) N, by the edge of the box.
Prob. VI. Both latitudes and departure given, to find the course and distance.
Example. A ship from latitude \( 43^\circ 10' \) N, sailed between the north and west till she was in latitude \( 47^\circ 14' \) N, and has made 170 miles of departure. Required the course and distance?
Move the centre of the semicircle over \( 43^\circ 10' \), and the edge of the box to \( 47^\circ 14' \); find the departure on the perpendicular, and bring the edge of the index thereunto; now at the point of intersection is the distance 297.4 miles on the index, and the course \( 34^\circ 52' \) on the semicircle.
Traverse Sailing.
Example. A ship from latitude \( 46^\circ 48' \) N, sailed SSW \( \frac{1}{2} \) W 24 miles, SW 36 miles, and S \( \frac{1}{2} \) E 40 miles. Instruments miles. Required the latitude in, together with the direct course and distance?
Problems in Sailing. Set the semicircle to the latitude sailed from $46^\circ 48'$, without the index to the course SSW $^{\circ} W$; mark the distance $24$ on the index, and bring the perpendicular to meet it; then the index will cut the departure $11.3$ on the perpendicular, and the perpendicular will cut the latitude $46^\circ 27'$ N on the base. For the next course and distance, bring the semicircle to the latitude marked by the perpendicular, and lay down the course SW $^{\circ}$; if it be towards the first meridian, move the last marked departure until it meets the index, and the limb of the box will mark the present departure; but if the course be from the first meridian, bring the last departure $11.3$ to the limb of the box, the index will mark the departure made good $18.3$ on the perpendicular, and the latitude arrived at $45^\circ 52'$ will be marked on the base by the perpendicular; proceed in the same manner with all the courses which the traverse consists, then the difference of latitude $1^\circ 36'$ will be intercepted between the latitude sailed from $46^\circ 48'$, and the latitude come to $45^\circ 12'$ last marked by the perpendicular; and also the departure made good will be intercepted between that point on the perpendicular where the first departure commenced, and that where the last terminated.
Now, with the difference of latitude $1^\circ 36'$ and the departure, the course will be S $8^\circ 30'$ W, and distance $97$ miles, by last problem in Plane Sailing.
Parallel Sailing.
Prob. I. The difference of longitude between two places in one parallel of latitude given, to find the distance between them.
Example. Let the common latitude be $49^\circ 30'$ N, and the difference of longitude $3^\circ 30'$. Required the distance?
Set the index to $40^\circ 30'$, the complement of the latitude on the semicircle; mark the difference of longitude in miles on the index; then move the perpendicular until it meets the termination of the difference of longitude on the index, and the part of the perpendicular intercepted between the limb of the box and the point of intersection will be the distance $136.4$ miles.
Prob. II. The distance between two places in one parallel of latitude given, to find the difference of longitude between them.
Example. Let the latitude of the given parallel be $49^\circ 30'$ N, the distance sailed $136.4$ E. Required the difference of longitude?
Set the index to the complement of the latitude $49^\circ 30'$, and mark the distance sailed on the perpendicular; then move it until it meets the index, and the point of intersection will show the difference of longitude $210$ or $3^\circ 30'$ on the index.
Prob. III. Given the distance sailed on a parallel, and the difference of longitude, to find the latitude of that parallel.
Example. The distance sailed due east is $136.4$, and the difference of longitude $3^\circ 30'$. Required the latitude of the parallel?
Find the difference of longitude $210$ on the index, and the distance $136.4$ on the perpendicular, and move both until these numbers meet, and the complement of the latitude $4^\circ 30'$ will be shown by the index on the semicircle.
Mercator's and Middle Latitude Sailing.
Prob. I. The latitudes and longitudes of two places given, to find the direct course and distance between them.
Example. Required the course and distance between two places whose latitudes and longitudes are $50^\circ 30'$ N, $19^\circ 0'$ W, and $54^\circ 30'$ N, $15^\circ 30'$ W, respectively?
By Mercator's Sailing.
To find the course.
Move the centre of the semicircle perpendicularly over the meridional degree answering to latitude $50^\circ 30'$ N, then move the box until the edge of the perpendicular cuts the meridional parts of the other latitude $54^\circ 30'$ N, and move the index until it cuts the difference of longitude $3^\circ 30'$ on the perpendicular, and the index will mark the course $50^\circ 15'$, or NNE $^{\circ} E$ nearly on the semicircle.
To find the distance.
Screw the index to this course, and move the centre of the semicircle to the latitude $50^\circ 30'$ N, and the edge of the perpendicular to the latitude $54^\circ 30'$ N, then the perpendicular will cut the distance $254.7$ on the index.
By Middle Latitude Sailing.
To find the departure.
Move the centre of the semicircle to the latitude $50^\circ 30'$ N, and the edge of the index to the complement of the middle latitude $37^\circ 25'$ on the semicircle; then move the box until the edge of the perpendicular intercepts the termination of the difference of longitude $210$ miles on the index, which point of intersection will mark the departure $128$ on the perpendicular.
To find the course and distance.
Move the edge of the perpendicular to the other latitude $45^\circ 30'$, and the index until it cuts the departure $128$ on the perpendicular; then will the perpendicular mark the distance on the index $254.7$ miles, and the index will mark the course on the semicircle $30^\circ 15'$, or NNE $^{\circ} E$ nearly.
Prob. II. Both latitudes and course given, to find the distance and difference of longitude.
Example. A ship from latitude $50^\circ 50'$ N, longitude $10^\circ 0'$ W, sailed N $32^\circ 10'$ E, until she is in latitude $54^\circ 30'$ N. Required the distance and difference of longitude?
By Mercator's Sailing.
To find the difference of longitude.
Move the box and semicircle as in the former problem to the meridional parts of the given latitudes, then fet the index to the course, and it will mark the difference of longitude $3^\circ 30'$ on the perpendicular; Hence the longitude in is $15^\circ 30'$ W.
To find the distance.
Move the perpendicular and semicircle to the given latitudes, and put the index to the given course; then the perpendicular will cut the distance $254.7$ miles on the index. By Middle Latitude Sailing.
To find the distance and departure.
Move the semicircle and perpendicular to the given latitudes, and the index to the course; then the perpendicular will show the departure 128 miles, and the index the distance 254.7 miles at the point of intersection.
To find the difference of longitude.
Set the index to the complement of the middle latitude on the semicircle, and move the box until the termination of the departure on the perpendicular meets the index, which will mark the difference of longitude thereon 210 m. or 3° 30'.
Prob. III. Both latitudes and distance given, to find the course and difference of longitude.
Example. From latitude 50° 50' N, longitude 190° W, a ship sailed 254.7 miles between the north and east, and by observation is in latitude 54° 30' N. Required the course and difference of longitude?
By Mercator's Sailing.
To find the course.
Move the perpendicular and semicircle to the given latitudes, and the index until the distance marked on it meets the perpendicular; then the index will mark the course N 30° 10' E on the semicircle.
To find the difference of longitude.
Screw the index to the course, move the perpendicular and semicircle to the meridional parts of the given latitudes, and the space intercepted between the limb of the box and the index will be the difference of longitude 3° 30'.
By Middle Latitude Sailing.
To find the departure and course.
Move the semicircle and perpendicular to the given latitudes, and the index until the distance marked on it cuts the perpendicular; then the perpendicular will show the departure 128 miles, and the semicircle the course N 30° 10' E.
To find the difference of longitude.
Set the index to 37° 20', the complement of the middle latitude on the semicircle, and move the perpendicular until the termination of the departure on it cuts the index; then the point of intersection will mark the difference of longitude 210 miles on the index.
Prob. IV. Both latitudes and departure given, to find the course, distance, and difference of longitude.
Example. Let the latitude and longitude sailed from be 56° 40' S and 28° 55' E respectively, the latitude come to 61° 20' S, and departure 172.7 miles. Required the course, distance, and difference of longitude?
By Mercator's Sailing.
To find the course and distance.
Move the perpendicular and semicircle to the given latitude (H); then move the index till it meets the extremity of the departure on the perpendicular; the distance will be marked on the index 329, and the course S 31° 35' E, or SSE 1/2 E nearly, on the semicircle.
To find the difference of longitude.
Move the perpendicular and semicircle to the meridional parts of the given latitudes, and the index will cut the difference of longitude on the perpendicular 5° 35'.
By Middle Latitude Sailing.
The course and distance are found as before.
To find the difference of longitude.
Set the index to 31°, the complement of the middle latitude on the semicircle, and move the perpendicular until the departure marked on it cuts the index, and this point of intersection will mark the difference of longitude on the index 335 m. or 5° 35'.
Prob. V. One latitude, course, and distance given, to find the difference of latitude and difference of longitude.
Example. Let the latitude left be 56° 40' S, longitude 28° 55' E, the course S 31° 35' E, and distance 329 m. Required the latitude and longitude come to?
By Mercator's Sailing.
To find the latitude come to.
Set the semicircle to the latitude sailed from, and the index to the course, and bring the perpendicular to the distance, which at the same time will mark the latitude come to 61° 20' S.
To find the difference of longitude.
Screw the index to the course, and move the semicircle and perpendicular to the meridional parts of both latitudes; then the index will cut the difference of longitude on the perpendicular 5° 35'.
By Middle Latitude Sailing.
The latitude arrived at is found as above.
To find the departure.
The semicircle and perpendicular being set to both latitudes, and the index to the course, it will show the departure 172.7 on the perpendicular.
To find the difference of longitude.
Set the index to 31°, the complement of the middle latitude on the semicircle, and move the perpendicular until the departure marked on it cuts the index, and the division on the index at the point of intersection will be the difference of longitude 335.
Prob. VI. One latitude, course, and departure given, to find the distance, difference of latitude, and difference of longitude.
Example. Let the latitude sailed from be 56° 40' N, longitude 28° 35' W, the course N 31° 35' W, and departure 172.7. Required the distance, and the latitude and longitude come to?
By Mercator's Sailing.
To find the distance and latitude come to.
Move the semicircle to the latitude left, and the index to the course; mark the departure on the perpendicular,
(h) In southern latitudes, the end of the cylinder where the numbers begin must be turned towards the north pointed out by the semicircle; and in northern latitudes, it must be reversed. Instrumental, and move it until the termination thereof meets the index, then the point of intersection will show the distance 329 miles on the index, and the perpendicular will show the latitude arrived at $61^\circ 25' N$.
To find the difference of longitude.
Screw the index, and move the perpendicular and semicircle to the meridional parts of both latitudes, then the index will cut the difference of longitude $5^\circ 33'$ on the perpendicular.
By Middle Latitude Sailing.
Find the distance failed and latitude in as above, and the difference of longitude as in Problem IV. by middle latitude sailing.
**Prob. VII.** One latitude, the distance failed, and departure given, to find the course, difference of latitude, and difference of longitude.
*Example.* The latitude failed from is $48^\circ 30' N$, and longitude $14^\circ 40' W$; the distance run is 345 miles between the fourth and east, and the departure 200 miles. Required the course, and the latitude and longitude come to?
By Mercator's Sailing.
To find the course and latitude come to.
Move the semicircle to the latitude left, mark the distance on the index, and the departure on the perpendicular, move both until these points meet; then will the index show the course $S 35^\circ 26' E$ on the semicircle, and the latitude come to $43^\circ 49'$ on the base.
The difference of longitude is found as in the preceding problem.
By Middle Latitude Sailing.
The course and latitude come to are found as above, and the difference of longitude as in Problem IV. by middle latitude sailing.
II. Of Dr Mackay's Rectangular Instrument.
*Description.* Fig. 52. is a representation of this instrument, of about one-third of the original size.—Fig. 52.
The length CA is divided into 100 equal parts, and the breadth CB into 70; but in this plate every second division only is marked, in order to avoid confusion; through these divisions parallels are drawn, terminating at the opposite sides of the instrument. Upon the upper and right hand sides are two scales; the first contains the degrees of the quadrant, and the other the points and quarters of the compass. M is an index moveable about the centre C, and divided in the same manner as the sides (1). Fig. 53. is a portion of the enlarged meridian, so constructed that the first degree is equal to three divisions on the instrument; and therefore, in the use of this line, each division on the instrument is to be accounted 20 minutes. The size of the plate would not admit of the continuation of the line.
*Use.* From a bare inspection of this instrument, it is evident that any triangle whatever may be formed Instruments on it. In applying it to nautical problems, the course is to be found at top, or right-hand side, in the column of degrees or points, according as it is expressed; the distance is to be found on the index, the difference of latitude at either side column, and the departure at the head or foot of the instrument. The numbers in these columns may represent miles, leagues, &c.; but when used in conjunction with the enlarged meridian line, then 10 is to be accounted 100 miles, 20 is to be esteemed 200 miles, and so on, each number being increased in a tenfold ratio; and the intermediate numbers are to be reckoned accordingly.
Plane Sailing.
**Prob. I.** The course and distance failed given, to find the difference of latitude and departure.
*Example.* Let the course be NE $\frac{1}{2} N$, distance 44 miles. Required the difference of latitude and departure?
Move the index until the graduated edge be over 35 points, and find the given distance 44 miles on the index; this distance will be found to cut the parallel of 34 miles, the difference of latitude in the side column, and that of 28 miles, the departure at the top.
**Prob. II.** Given the course and difference of latitude, to find the distance and departure.
*Example.* Required the distance and departure answering to the course $28^\circ$, and difference of latitude 60 miles?
Lay the index over the given course $28^\circ$: find the difference of latitude 60 miles in the side column; its parallel will cut the index at 68 miles, the distance and the corresponding departure at the top is 32 miles.
**Prob. III.** The course and departure given, to find the distance and difference of latitude.
*Example.* Let the course be SSW and the departure 36 miles. Required the distance and difference of latitude?
Lay the index over two points; find the departure at the top, and its parallel will cut the index at 94 miles the distance, and the difference of latitude on the side column is 87 miles.
**Prob. IV.** Given the distance and difference of latitude, to find the course and departure.
*Example.* The distance is 35 leagues, and the difference of latitude 30 leagues. Required the course and departure?
Bring 35 leagues on the index to the parallel of 30 leagues in the side; then the departure at the top is 18 leagues, and the course by the edge of the index on the line of rhumbs is $2\frac{1}{2}$ points.
**Prob. V.** Given the distance and departure, to find the course and difference of latitude.
*Example.* Let the distance be 58 miles, and the departure
---
(1) In the original instrument are two slips, divided like the side and end of the instrument. One of these slips is moveable in a direction parallel to the side of the instrument, and the other parallel to the end. Practice.
Instruments to solve Problems in Sailing, without Calculation.
Required the course and difference of latitude?
Move the index until it is found thereon cuts the parallel of 15 from the top; this will be found to intersect the parallel of 56 miles, the difference of latitude; and the course by the edge of the ruler is 15°.
Prob. VI. The difference of latitude and departure being given, to find the course and distance.
Example. Let the difference of latitude be 30 miles, the departure 28 miles. Required the course and distance?
Bring the index to the intersection of the parallels of 30 and 28; then the distance on the index is 41 miles, and the course by its edge is 43°.
Traverse Sailing.
Find the difference of latitude and departure answering to each course and distance by Problem I. of Plane Sailing, and from thence find the difference of latitude and departure made good; with which find the course and distance by the last problem.
An example is unnecessary.
Parallel Sailing.
Prob. I. Given the difference of longitude between two places on the same parallel, to find the distance between them.
Example. Let the latitude of a parallel be 48°, and the difference of longitude between two places on it 3° 40′. Required their distance?
Put the index to 48°, the given latitude, and find the difference of longitude 220 on the index, and the corresponding parallel from the side will be 147, the distance required.
Prob. II. The latitude of a parallel, and the distance between two places on that parallel, being given, to find the difference of longitude between them.
Example. The latitude of a parallel is 56°, and the distance between two places on it 200 miles. Required their difference of longitude?
Put the index to the given latitude, and find the distance in the side column, and the intersection of its parallel with the index will give 358, the difference of longitude sought.
Prob. III. Given the distance and difference of longitude between two places on the same parallel, to find the latitude of that parallel.
Example. The number of miles in a degree of longitude is 46.5. Required the latitude of the parallel?
Bring 60 on the index to cut the parallel of 46.5 from the side, then the edge of the index will give 39° 11′, the latitude required.
Middle Latitude and Mercator's Sailing.
Prob. I. The latitudes and longitudes of two places being given, to find the course and distance between them.
Example. Required the course and distance between Genoa, in latitude 44° 25′ N, longitude 8° 36′ E, and Palermo, in latitude 38° 10′ N, longitude 13° 38′ E?
By Mercator's Sailing.
Take the interval between 38° 10′ and 44° 25′ on the enlarged meridian, which laid off from C upwards will reach to 500; now find the difference of longitude 322 at the top, and bring the divided edge of the index to the intersection of the corresponding parallels, and the index will show the course 31° 8′ on the line of degrees; then find the difference of the latitude 375 on the side column, and its parallel will intersect the index at 438, the distance.
By Middle Latitude Sailing.
Put the index to 41° 18′, the complement of the middle latitude on degrees, and the difference of longitude 322 on the index will intersect the parallel of 227, the departure, in the side column. Now move the index to the intersection of the parallels of 375 and 227, the first being found in the side column, and the other at top or bottom; then the distance answering thereto on the index will be 438, and the course on the scale of degrees is 41° 10′.
Prob. II. Given one latitude, course, and distance, to find the other latitude and difference of longitude.
Example. Let the latitude and longitude failed from be 39° 22′ N, and 12° 8′ W respectively, the course NNW 4½ W, and distance 500 miles. Required the latitude and longitude come to?
By Mercator's Sailing.
Put the index to the course 2½ points, and find the distance 500 miles thereon; then the corresponding difference of latitude will be 441 miles, and the departure 235½ miles, hence the latitude in is 46° 43′ N. Now take the interval between the latitudes of 39° 22′, and 46° 43′ on the enlarged meridian, which laid off from C will reach to about 605, the parallel of which will intersect the vertical parallel of the difference of longitude 323 at the edge of the index; hence the longitude in is 17° 31′ W.
By Middle Latitude Sailing.
Find the difference of latitude and departure as before, and hence the latitude in is 46° 43′ N, and the middle latitude 43° 3′. Now put the index to 43° 3′, and the horizontal parallel of the departure 235½ will intersect the index at 322, the difference of longitude.
Prob. III. Both latitudes and course given, to find the distance and difference of longitude.
Example. The latitude failed from is 22° 54′ S, and longitude 42° 40′ W, the course is SE by E, and latitude come to 26° 8′ S. Required the distance failed, and longitude in?
By Mercator's Sailing.
Bring the index to 5 points, the given course, and the parallel of 194; the difference of latitude found in the side column will intersect the index at 349, the distance; and it will cut the vertical parallel of 290, the departure.
Take the interval between the given latitudes 22° 54′ and 26° 8′ on the enlarged meridian; lay off that extent from the centre on the side column, and it will reach to 213; the parallel of this number will intersect the vertical parallel of 319, the difference of longitude. Hence the longitude in is 37° 21′ W. By Middle Latitude Sailing.
With the given course and difference of latitude find the distance and departure as before; then bring the index to the middle latitude $24^\circ 31'$; find the departure $292$ in the side column, and its parallel will intersect the index at $319$, the difference of longitude.
**Prob. IV.** One latitude, course, and departure, given, to find the other latitude, distance, and difference of longitude.
*Example.* The latitude and longitude left are $20^\circ 32'$ N, and $46^\circ 17'$ W, respectively; the course is NE $4N$, and departure $212$ miles. Required the latitude and longitude come to, and distance failed?
**By Mercator's Sailing.**
Put the index to the given course $3\frac{3}{4}$ points, and the vertical parallel of $212$ will cut the index at $356$, the distance, and the horizontal parallel of $286$, the difference of latitude; the latitude come to is therefore $25^\circ 16'$ N.
Now take the interval between the latitudes $20^\circ 30'$, and $25^\circ 16'$ on the enlarged meridian, which laid off from the centre C will reach to $311$; and this parallel will intersect the vertical parallel of the difference of longitude $230$, at the edge of the index. Hence the longitude in is $45^\circ 27'$ W.
**By Middle Latitude Sailing.**
Find the distance and difference of latitude as directed above; then bring the index to $22^\circ 53'$, the middle latitude, and the horizontal parallel of $212$, the departure, will intersect the index at $230$, the difference of longitude.
**Prob. V.** Both latitudes and distance given, to find the course and difference of longitude.
*Example.* The distance failed is $500$ miles between the north and west; the latitude and longitude left are $40^\circ 10'$ N, and $9^\circ 20'$ W respectively, and the latitude in is $46^\circ 40'$ N. Required the course and longitude in?
**By Mercator's Sailing.**
Bring the distance $500$ on the index to intersect the horizontal parallel of the difference of latitude $390$; then the course $38^\circ 44'$ is found on the line of degrees by the edge of the index, and the vertical parallel of the above point of intersection is that answering to $313$, the departure.
Take the interval between the latitudes $40^\circ 10'$, and $46^\circ 40'$, which lay off from the centre C, and its horizontal parallel will intersect the vertical parallel of $431$, the difference of longitude, by the edge of the index, it being in the same position as before. Hence the longitude in is $16^\circ 31'$ W.
**By Middle Latitude Sailing.**
The course and departure are found as formerly, and the middle latitude is $43^\circ 25'$, to which bring the edge of the index, and the horizontal parallel of $313$, the departure, will intersect the index at $431$, the difference of longitude.
**Prob. VI.** Both latitudes and departure given, to find the course, distance, and difference of longitude.
*Example.* Let the latitude failed be $42^\circ 52'$ N, long. $9^\circ 17'$ W, the departure $250$ miles W, and the Sea-Charts, latitude come to $36^\circ 18'$ N. Required the course and distance failed, and the longitude come to?
**By Mercator's Sailing.**
Find the point of intersection of the horizontal parallel of $394$, the difference of latitude, and the vertical parallel of $250$, the departure; to this point bring the index, and the corresponding division thereon will be $467$ miles, and the course on the scale of degrees by the edge of the index will be $32^\circ 24'$.
Take the interval between the latitudes on the enlarged meridian; which being laid off from the centre will reach to $512$; now the horizontal parallel of $512$ will cut the vertical parallel of $325$, the difference of longitude, at the edge of the index. The longitude come to is therefore $14^\circ 42'$ W.
**By Middle Latitude Sailing.**
The course and distance are to be found in the same manner as above. Then bring the index to $39^\circ 35'$, the middle latitude, and the horizontal parallel of $250$ will intersect the edge of the index at $324\frac{1}{2}$, the difference of longitude.
**Prob. VII.** Given one latitude, distance, and departure, to find the other latitude, course, and difference of longitude.
*Example.* A ship from latitude $32^\circ 38'$ N, longitude $17^\circ 6'$ W, sailed $586$ miles between the south and west, and made $336$ miles of departure:—Required the course, and the latitude and longitude come to?
**By Mercator's Sailing.**
Move the index till the distance $586$ intersects the vertical parallel of the departure $336$; then the corresponding horizontal parallel will be $482$, the difference of latitude, and the course $35^\circ$. Hence the latitude in is $24^\circ 38'$ N.
Now take the interval between the latitudes on the enlarged meridian, which laid off from the centre will reach to $547$, the horizontal parallel of which will cut the vertical parallel of $383$, the difference of longitude. The longitude in is therefore $23^\circ 29'$ W.
**By Middle Latitude Sailing.**
Find the course and difference of latitude as before, and hence the middle latitude is $28^\circ 38'$, to which bring the index, and the horizontal parallel of $336$, the departure, will intersect the index at $383$, the difference of longitude.
It seems unnecessary to enlarge any further on the use of this instrument, as the above will make it sufficiently understood.
**Chap. XII. Of Sea-Charts.**
The charts usually employed in the practice of navigation, are of two kinds, namely, *Plane* and *Mercator's Charts*. The first of these is adapted to represent a portion of the earth's surface near the equator; and the last for all portions of the earth's surface. For a particular description of these, reference has already been made from the article *Chart*, to those of *Plane* and *Mercator*: and as these charts are particularly described under the above articles, it is therefore sufficient in this place to describe their use. Use of the Plane Chart.
Prob. I. To find the latitude of a place on the chart.
Rule. Take the least distance between the given place and the nearest parallel of latitude; now this distance applied the same way on the graduated meridian, from the extremity of the parallel, will give the latitude of the proposed place.
Thus the distance between Bonavita and the parallel of 15 degrees, being laid from that parallel upon the graduated meridian, will reach to $16^\circ 3'$, the latitude required.
Prob. II. To find the course and distance between two given places on the chart.
Rule. Lay a ruler over the given places, and take the nearest distance between the centre of any of the compasses on the chart and the edge of the ruler; move this extent along, so as one point of the compass may touch the edge of the rule, and the straight line joining their points may be perpendicular thereto; then will the other point show the course. The interval between the places, being applied to the scale, will give the required distance.
Thus the course from Palma to St Vincent will be found to be about SSW $\frac{1}{4}$ W, and the distance $13^\circ \frac{1}{4}$ or 795 m.
Prob. III. The course and distance failed from a known place being given, to find the ship's place on the chart.
Rule. Lay a ruler over the place failed from, parallel to the rhumb, expressing the given course; take the distance from the scale, and lay it off from the given place by the edge of the ruler; and it will give the point representing the ship's present place.
Thus, suppose a ship had failed SWB W 160 miles from Cape Palmas; then by proceeding as above, it will be found that she is in latitude $2^\circ 57'$ N.
The various other problems that may be resolved by means of this chart require no further explanation, being only the construction of the remaining problems in Plane Sailing on the chart.
Use of Mercator's Chart.
The method of finding the latitude and longitude of a place, and the course or bearing between two given places by this chart, is performed exactly in the manner as in the Plane Chart, which see.
Prob. I. To find the distance between two given places on the chart.
Case I. When the given places are under the same meridian.
Rule. The difference or sum of their latitudes, according as they are on the same or on opposite sides of the equator, will be the distance required.
Case II. When the given places are under the same parallel.
Rule. If that parallel be the equator, the difference or sum of their longitudes is the distance; otherwise, take half the interval between the places, lay it off upwards and downwards on the meridian from the given parallel, and the intercepted degrees will be the distance between the places.
Or, take an equal extent of a few degrees from the meridian on each side of the parallel, and the number of extents, and parts of an extent, contained between the places, being multiplied by the length of an extent, will give the required distance.
Case III. When the given places differ both in latitude and longitude.
Rule. Find the difference of latitude between the given places, and take it from the equator or graduated parallel; then lay a ruler over the two places, and move one point of the compass along the edge of the ruler until the other point just touches a parallel; then the distance between the place where the point of the compass rests by the edge of the ruler, and the point of intersection of the ruler and parallel, being applied to the equator, will give the distance between the places in degrees and parts of a degree, which multiplied by 6c will reduce it to miles.
Prob. II. Given the latitude and longitude in, to find the ship's place on the chart.
Rule. Lay a ruler over the given latitude, and lay off the given longitude from the first meridian by the edge of the ruler, and the ship's present place will be obtained.
Prob. III. Given the course failed from a known place, and the latitude in, to find the ship's present place on the chart.
Rule. Lay a ruler over the place failed from, in the direction of the given course, and its intersection with the parallel of latitude arrived at will be the ship's present place.
Prob. IV. Given the latitude of the place left and the course and distance failed, to find the ship's present place on the chart.
Rule. The ruler being laid over the place failed from, and in the direction of the given course, take the distance failed from the equator, put one point of the compass at the intersection of any parallel with the ruler, and the other point of the compass will reach to a certain place by the edge of the ruler. Now this point remaining in the same position, draw in the other point of the compass until it just touches the above parallel when swept round: apply this extent to the equator, and it will give the difference of latitude. Hence the latitude in will be known, and the intersection of the corresponding parallel with the edge of the ruler will be the ship's present place.
The other problems of Mercator's Sailing may be very easily resolved by this chart; but as they are of less use than those given, they are, therefore, omitted, and may serve as an exercise to the student.
BOOK II.
Containing the method of finding the Latitude and Longitude of a Ship at Sea, and the Variation of the Compass.
Chap. I. Of Hadley's Quadrant.
Hadley's quadrant is the chief instrument in use at present for observing altitudes at sea. The form of this instrument, according to the present mode of construction,
Method of finding the Latitude and Longitude at Sea.
Plate CCCLXVII Fig. 54.
Construction, is an octagonal sector of a circle, and therefore contains 45 degrees; but because of the double reflection, the limb is divided into 90 degrees. See ASTRONOMY and QUADRANT. Fig. 54 represents a quadrant of the common construction, of which the following are the principal parts.
1. ABC, the frame of the quadrant. 2. BC, the arch or limb. 3. D, the index; ab, the subdividing scale. 4. E, the index-glass. 5. F, the fore horizon-glass. 6. G, the back horizon-glass. 7. K, the coloured or dark glasses. 8. HI, the vanes or sights.
Of the Frame of the Quadrant.
The frame of the quadrant consists of an arch BC, firmly attached to the two radii AB, AC, which are bound together by the braces LM, in order to strengthen it, and prevent it from warping.
Of the Index D.
The index is a flat bar of brass, and turns on the centre of the octant: at the lower end of the index there is an oblong opening; to one side of this opening the vernier scale is fixed, to subdivide the divisions of the arch; at the end of the index there is a piece of brass, which bends under the arch, carrying a spring to make the subdividing scale lie close to the divisions. It is also furnished with a screw to fix the index in any desired position. The best instruments have an adjusting screw fitted to the index, that it may be moved more slowly, and with greater regularity and accuracy, than by the hand. It is proper, however, to observe, that the index must be previously fixed near its right position by the above-mentioned screw.
Of the Index Glass E.
Upon the index, and near its axis of motion, is fixed a plane speculum, or mirror of glass quicksilvered. It is set in a brass frame, and is placed so that its face is perpendicular to the plane of the instrument. This mirror being fixed to the index moves along with it, and has its direction changed by the motion thereof; and the intention of this glass is to receive the image of the sun, or any other object, and reflect it upon either of the two horizon-glasses, according to the nature of the observation.
The brass frame with the glass is fixed to the index by the screw c; the other screw serves to replace it in a perpendicular position, if by any accident it has been deranged.
Of the Horizon Glasses F, G.
On the radius AB of the octant are two small speculums: the surface of the upper one is parallel to the index glass, and that of the lower one perpendicular thereto, when o on the index coincides with o on the limb. These mirrors receive the reflected rays, and transmit them to the observer.
The horizon-glasses are not entirely quicksilvered; the upper one F is only silvered on its lower half, or that next the plane of the quadrant, the other half being left transparent, and the back part of the frame cut away, that nothing may impede the sight through the unsilvered part of the glass. The edge of the foil of this glass is nearly parallel to the plane of the instrument, and ought to be very sharp, and without a flaw. The other horizon glass is silvered at both ends. In the middle there is a transparent slit through which the horizon may be seen.
Each of these glasses is set in a brass frame, to which there is an axis passing through the wood work, and is fitted to a lever on the under side of the quadrant, by which the glass may be turned a few degrees on its axis, in order to set it parallel to the index-glass. The lever has a contrivance to turn it slowly, and a button to fix it. To set the glasses perpendicular to the plane of the instrument, there are two sunk screws, one before and one behind each glass: these screws pass through the plate on which the frame is fixed into another plate; so that by loosening one and tightening the other of these screws, the direction of the frame with its mirror may be altered, and set perpendicular to the plane of the instrument.
Of the Coloured Glasses K.
There are usually three coloured glasses, two of which are tinged red and the other green. They are used to prevent the solar rays from hurting the eye at the time of observation. These glasses are set in a frame, which turns on a centre, so that they may be used separately or together as the brightness of the sun may require. The green glass is particularly useful in observations of the moon; it may also be used in observations of the sun, if that object be very faint. In the fore-observation, these glasses are fixed as in fig. 54; but when the back observation is used, they are removed to N.
Of the two Sight Vanes, H, I.
Each of these vanes is a perforated piece of brass, designed to direct the sight parallel to the plane of the quadrant. That which is fixed at I is used for the fore, and the other for the back, observation. The vane I has two holes, one exactly at the height of the silvered part of the horizon-glass, the other a little higher, to direct the sight to the middle of the transparent part of the mirror.
Of the divisions on the Limb of the Quadrant.
The limb of the quadrant is divided from right to left into 90 primary divisions, which are to be considered as degrees, and each degree is subdivided into three equal parts, which are therefore of 30 minutes each: the intermediate minutes are obtained by means of the scale of divisions at the end of the index.
Of the Vernier, or Subdividing Scale.
The dividing scale contains a space equal to 21 divisions of the limb, and is divided into 20 equal parts. Hence the difference between a division on the dividing scale and a division on the limb is one-twentieth of a division on the limb, or one minute. The degree and minute pointed out by the dividing scale may be easily found thus:
Observe what minute on the dividing scale coincides with a division on the limb; this division being added to the degree and part of a degree on the limb, immediately dately preceding the first division on the dividing scale, will be the degree and minute required.
Thus suppose the fourteenth minute on the dividing scale coincided with a division on the limb, and that the preceding division on the limb to \( o \) on the vernier was \( 56^\circ 40' \); hence the division shown by the vernier is \( 56^\circ 54' \). A magnifying glass will assist the observer to read off the coinciding divisions with more accuracy.
Adjustments of Hadley's Quadrant.
The adjustments of the quadrant consist in placing the mirrors perpendicular to the plane of the instrument. The fore horizon-glass must be set parallel to the speculum, and the planes of the speculum and back horizon-glass produced must be perpendicular to each other when the index is at \( o \).
**Adjustment I.** To set the index-glass perpendicular to the plane of the quadrant.
**Method 1.** Set the index towards the middle of the limb, and hold the quadrant so that its plane may be nearly parallel to the horizon; then look into the index-glass; and if the portion of the limb seen by reflection appears in the same plane with that seen directly, the speculum is perpendicular to the plane of the instrument. If they do not appear in the same plane, the error is to be rectified by altering the position of the screws behind the frame of the glass.
**Method 2.** This is performed by means of the two adjusting tools, fig. 55, 56, which are two wooden frames, having two lines on each, exactly at the same distance from the bottom.
Place the quadrant in a horizontal position on a table; put the index about the middle of the arch; turn back the dark glasses; place one of the above-mentioned tools near one end of the arch, and the other at the opposite end, the side with the lines being towards the index-glass; then look into the index-glass, directing the sight parallel to the plane of the instrument, and one of the tools will be seen by direct vision, and the other by reflection. By moving the index a little, they may be brought exactly together. If the lines coincide, the position of the mirror is right; if not, they must be made to coincide by altering the screws behind the frame, as before.
**Adjustment II.** To set the fore horizon-glass perpendicular to the plane of the instrument.
Set the index to \( o \); hold the plane of the quadrant parallel to the horizon; direct the sight to the horizon, and if the horizons seen directly and by reflection are apparently in the same straight line, the fore horizon-glass is perpendicular to the plane of the instrument; if not, one of the horizons will appear higher than the other. Now if the horizon seen by reflection is higher than that seen directly, release the nearest screw in the pedeal of the glass, and screw up that on the farther side, till the direct and reflected horizons appear to make one continued straight line. But if the reflected horizon is lower than that seen directly, unscrew the fartherest, and screw up the nearest screw till the coincidence of the horizons is perfect, observing to leave both screws equally tight, and the fore horizon-glass will be perpendicular to the plane of the quadrant.
**Adjustment III.** To set the fore horizon-glass parallel to the index-glass, the index being at \( o \).
**Method of finding the Latitude and Longitude at Sea.**
Set \( o \) on the index exactly to \( o \) on the limb, and fix it in that position by the screw at the under side; hold the plane of the quadrant in a vertical position, and direct the sight to a well-defined part of the horizon; then if the horizon seen in the silvered part coincides with that seen through the transparent part, the horizon-glass is adjusted; but if the horizons do not coincide, unscrew the milled screw in the middle of the lever on the other side of the quadrant, and turn the nut at the end of the lever until both horizons coincide, and fix the lever in this position by tightening the milled screw.
As the position of the glass is liable to be altered by fixing the lever, it will therefore be necessary to re-examine it, and if the horizons do not coincide, it will be necessary either to repeat the adjustment, or rather to find the error of adjustment, or, as it is usually called, the index-error; which may be done thus:
Direct the sight to the horizon, and move the index until the reflected horizon coincides with that seen directly; then the difference between \( o \) on the limb and \( o \) on the vernier is the index error; which is additive when the beginning of the vernier is to the right of \( o \) on the limb, otherwise subtractive.
**Adjustment IV.** To set the back horizon-glass perpendicular to the plane of the instrument.
Put the index to \( o \); hold the plane of the quadrant parallel to the horizon, and direct the sight to the horizon through the back sight vane. Now if the reflected horizon is in the same straight line with that seen through the transparent part, the glass is perpendicular to the plane of the instrument. If the horizons do not unite, turn the sunk screws in the pedeal of the glass until they are apparently in the same straight line.
**Adjustment V.** To set the back horizon-glass perpendicular to the plane of the index-glass produced, the index being at \( o \).
Let the index be put as much to the right of \( o \) as twice the dip of the horizon amounts to; hold the quadrant in a vertical position, and apply the eye to the back vane; then if the reflected horizon coincides with that seen directly, the glass is adjusted; if they do not coincide, the screw in the middle of the lever on the other side of the quadrant must be released, and the nut at its extremity turned till both horizons coincide. It may be observed, that the reflected horizon will be inverted; that is, the sea will be apparently uppermost and the sky lowermost.
As this method of adjustment is esteemed troublesome, and is often found to be very difficult to perform at sea, various contrivances have therefore been proposed to render this adjustment more simple. Some of these are the following.
1. Mr Dollond's method of adjusting the back horizon-glass.
In this method an index is applied to the back horizon-glass, by which it may be moved so as to be parallel to the index-glass, when \( o \) on the vernier coincides with \( o \) on the limb. When this is effected, the index of the back horizon-glass is to be moved exactly \( 90^\circ \) from its former position, which is known by means of a divided arch for that purpose; and then the plane of the back horizon-glass will be perpendicular to the plane of the index-glass produced. 2. Mr Blair's method of adjusting the back horizon-glas.
All that is required in this method is to polish the lower edge of the index-glas, and expose it to view. The back horizon-glas is adjusted by means of a reflection from this polished edge, in the very same method as the fore horizon-glas is adjusted by the common method.
In order to illustrate this, let RIHE (fig. 57) represent a pencil of rays emitted from the object R, incident on the index-glas I, from which it is reflected to the fore horizon-glas H, and thence to the eye at E. By this double reflection, an image of the object is formed at r. RIHE represents another pencil from the same object R, coming directly through the fore horizon-glas to the eye at E; so that the doubly reflected image r appears coincident with the object R itself, seen directly.
When this coincidence is perfect, and the object R so very distant as to make the angle IRH insensible, the position of the speculums I and H will differ infensibly from parallelism; that is, the quadrant will be adjusted for the fore observation. Now it is from the ease and accuracy with which this adjustment can at any time be made, that the fore-observation derives its superiority over the back observation. But by grinding the edge of the index-glas perpendicular to its reflecting surface, and polishing it, the observation is rendered capable of an adjustment equally easy and accurate as the fore horizon-glas: for by a pencil of rays emitted from the object S, incident on the reflecting edge of the index-glas D, thence reflected to the back horizon-glas B, and from that to the eye at e, an image will be formed at s; which image being made to coincide with the object S itself, seen directly, affords the position of the back horizon-glas relative to the index-glas, with the same precision, and in a manner equally direct, as the former operation does that of the fore horizon-glas.
Directions for adjusting the Back Horizon-Glas.
The method of adjusting the quadrant for the back-observation is this. If it is to be done without making use of the telescope, place the index at o, and, applying the eye to the hole in the sight vane (k), or tube for directing the sight, direct it through the back horizon-glas to the horizon, if that is the object to be used for adjusting. The two horizons are then to be made to coincide, holding the quadrant first in a vertical and then in a horizontal position; by which means both adjustments will be effected as in the fore-observation.
There will be no difficulty in finding the reflected horizon, if the observer first directs his eye to that part of the horizon-glas where he observes the image of the polished edge of the index-glas, which will appear double. When the direct horizon is made to appear in this case, the reflected one will be seen close by it, unless the instrument wants a great adjustment. In this case, a little motion of the back horizon-glas backwards and forwards will presently bring it into view.
When the horizon, or any obscure terrestrial object, is to be made use of for adjusting by means of the reflecting edge, there is a precaution to be taken, without which the observer will sometimes meet with what will appear an unaccountable difficulty; for if the sky, or other object behind him, should happen to be pretty bright, he will not be able to discern the horizon at all. This arises from the image of the object behind him, which is reflected from the silvered surface of the index-glas, appearing to coincide with the horizon; in which case, the bright picture of the former, which is formed in the bottom of the eye, prevents the fainter impression of the latter from being perceived. This will be avoided, either by applying a black screen over the silvered surface of the index-glas, or, without being at this trouble, by standing at a door or window, so that only the dark objects within can be reflected from the index-glas; but if the observation is to be made in the open air, a hat, or any such dark obstacle, held before the silvered surface of the index-glas, will very effectually remove this inconvenience.
It may be remarked, that some observers, instead of making the principal adjustment, place the speculums parallel, by moving the index without altering the position of the horizon-glas: and the difference between o on the vernier and o on the limb is the index error, which must be subtracted from all angles measured by the back-observation, when o on the index, is to the right of o on the limb; and added when to the left.
3. Mr Wright's method of adjusting the back horizon-glas of his improved patent quadrant.
Fig. 58. Is a representation of the quadrant complete in all its parts for use. A, is the reflecting surface of the index-glas, which is made of the usual length, and \( \frac{1}{2} \) of an inch broad. The bottom part is covered in front by the brass frame, and the reflecting surface is \( \frac{1}{2} \) on the back. B, the fore horizon-glas, placed as usual: O, the back horizon-glas, now placed under the fore sight-vane on the first radius of the quadrant I: C, the sight-vane of the fore horizon-glas: D, the sight-vane of the back horizon-glas: E, the coloured glasses in a brass frame, in the proper place for the fore observation: F, a hole in the frame to receive the coloured glasses when an observation is to be taken with the back horizon-glas in the common way, by turning the back to the sun: G, a hole in the frame of the farthest radius K, to receive the coloured glasses when an observation is to be taken by the new method; which is by looking through the lower hole in the sight-vane of the back horizon-glas, directly at
(k) Besides the hole in the sight-vane, commonly made, there must be another nearer to the horizon-glas, and so placed that an eye directed through it to the centre of the horizon glas shall there perceive the image of the polished edge of the index-glas. This hole must not be made small like the other, but equal to the ordinary size of the pupil of the eye, there being on some occasions no light to spare. at the sun in the line of sight DN; the horizon from behind will then be reflected from the back of the index-glasfs to the horizon-glasfs, and from thence to the eye. (See fig. 62.) H, a bras clamp on the upper end of the index, having a milled screw underneath, which fastens the round plate to the index when required. (See fig. 59.) IK, the graduated arch of the quadrant divided into 90 degrees: L, the bras index which moves over the graduated arch: M, the vernier to subdivide the divisions on the arch into single minutes of a degree.
Fig. 59 shows the upper part of the index L on a larger scale, with part of the bras frame that fastens the index-glasfs, and the three adjusting screws D to adjust its axis vertical to the plane of the quadrant: B, the centre on which the milled plate O moves over the index: The dotted line BF is the distance it is required to move: K, the adjusting screw to stop it in its proper place for adjusting the back observation-glasfs: G, a piece of bras fastened to the index opposite to the clamp H, to keep the plate O always close to the index L.
Fig. 60 represents the parallel position of the index and horizon glasfs after adjustment by the sun: BC, a ray from the sun incident on the index-glasfs C, and from thence reflected to the fore horizon-glasfs D, and again to the eye at E, in the line DE, where the eye sees the sun at A by direct vision, and the image by reflection, in one; the parallel lines AE and BC being so near to each other, that no apparent angle can be observed in the planes of the index and horizon-glasfs, when adjusted by a distant object.
In fig. 61, the index glasfs is removed 45 degrees from the plane of the fore horizon-glasfs, and fixed in its proper place for adjusting the back horizon-glasfs parallel to its plane, in the same manner as the fore horizon-glasfs is adjusted.
In fig. 62, the index-glasfs (after the adjustment of the fore and back horizon-glasfs) is carried forward by the index on the arch 90 degrees, and makes an angle of 45° with the plane of the fore horizon-glasfs, and is at right angles to the plane of the back horizon-glasfs. The eye at E now sees the sun in the horizon at H, reflected by the index and horizon glasfs from the zenith at Z, the image and object being 90 degrees distant. The back horizon K is now reflected from the back surface of the index-glasfs C to the horizon-glasfs M, and from thence to the eye at D, in a right line with the fore horizon F. In order to make an exact contact of the fore and back horizons at F, the index must be advanced beyond the 90th degree on the arch, by a quantity equal to twice the dip of the horizon.
The quadrant is adjusted for the fore-observation as usual, having previously fixed the index-glasfs in its proper place by the milled screw at H, as represented in fig. 59.
To adjust the Quadrant for the Back-observation.
Fasten the index to 90° on the limb; loosen the screw H (fig. 59), and turn the plate O by the milled edge until the end of the adjusting screw K touch the edge of the clamp M; and by means of a distant object observe if the glasfs are then parallel, as at fig. 60: if they are, fasten the screw H; if not, with a screw-driver turn the screw K gently to the right or left to make them perfect, and then fasten the screw. Now remove the index back to O on the limb, and the index-glasfs will be parallel to the back horizon glasfs E, fig. 61.; If not, make them so by turning the adjusting screw of the glasfs E, the eye being at the upper hole in the fight-vane D, and the fight directed to the horizon, or any distant object in the direction DN (fig. 58.). Now the index remaining in this position, the index-glasfs is to be returned, to stop at the pin E, and it will be parallel to the fore horizon-glasfs as at first: then the quadrant will be adjusted for both methods of observation.
To observe the Sun's Altitude by the Back-observation.
Remove the coloured glasfses to G (fig. 58.), and look through the lower hole in the fight-vane D, in the line of direction DN, directly to the sun, and move the index forward on the arch exactly in the same manner as in the fore-observation: make the contact of the sun's limb and the back horizon exact, and the degrees and minutes shown by the index on the limb is the sun's zenith distance. It may be observed, that the horizon will be inverted. If the sun's lower limb be observed, the semidiameter is to be subtracted from the zenith distance; but if the upper limb is observed, the semidiameter is to be added.
The observation may be made in the usual manner, by turning the back to the sun. In this case the coloured glasfses are to be shifted to F, and proceed according to the directions formerly given.
Use of Hadley's Quadrant.
The altitude of any object is determined by the position of the index on the limb, when by reflection that object appears to be in contact with the horizon.
If the object whose altitude is to be observed be the sun, and if so bright that its image may be seen in the transparent part of the fore horizon-glasfs, the eye is to be applied to the upper hole in the fight vane; otherwise, to the lower hole: and in this case, the quadrant is to be held so that the sun may be bisected by the line of separation of the silvered and transparent parts of the glasfs. The moon is to be kept as nearly as possible in the same position; and the image of the star is to be observed in the silvered part of the glasfs adjacent to the line of separation of the two parts.
There are two different methods of taking observations with the quadrant. In the first of these the face of the observer is directed towards that part of the horizon immediately under the sun, and is therefore called the fore observation. In the other method, the observer's back is to the sun, and it is hence called the back-observation. This last method of observation is to be used only when the horizon under the sun is obscured, or rendered indistinct by fog or any other impediment.
In taking the sun's altitude, whether by the fore or back observation, the observer must turn the quadrant about upon the axis of vision, and at the same time turn himself about upon his heel, so as to keep the sun always in that part of the horizon glasfs which is at the same distance as the eye from the plane of the quadrant. In this way the reflected sun will describe an arch of a parallel circle round the true sun, whose convex side... will be downwards in the fore-observation and upwards in the back; and consequently, when by moving the index, the lowest point of the arch in the fore-observation, or highest in the back, is made to touch the horizon, the quadrant will stand in a vertical plane, and the altitude above the visible horizon will be properly observed. The reason of these operations may be thus explained: The image of the sun being always kept in the axis of vision, the index will always show on the quadrant the distance between the sun and any object seen directly which its image appears to touch; therefore, as long as the index remains unmoved, the image of the sun will describe an arch everywhere equidistant from the sun in the heavens, and consequently a parallel circle about the sun, as a pole. Such a translation of the sun’s image can only be produced by the quadrant’s being turned about upon a line drawn from the eye to the sun, as an axis. A motion of rotation upon this line may be resolved into two, one upon the axis of vision, and the other upon a line on the quadrant perpendicular to the axis of vision; and consequently a proper combination of these two motions will keep the image of the sun constantly in the axis of vision, and cause both jointly to run over a parallel circle about the sun in the heavens; but when the quadrant is vertical, a line thereon perpendicular to the axis of vision becomes a vertical axis; and as a small motion of the quadrant is all that is wanted, it will never differ much in practice from a vertical axis. The observer is directed to perform two motions rather than the single one equivalent to them on a line drawn from the eye to the sun: because we are not capable, while looking towards the horizon, of judging how to turn the quadrant about upon the elevated line going to the sun as an axis, by any other means than by combining the two motions above mentioned, so as to keep the sun’s image always in the proper part of the horizon-glass. When the sun is near the horizon, the line going from the eye to the sun will not be far removed from the axis of vision; and consequently the principal motion of the quadrant will be performed on the axis of vision, and the part of motion made on the vertical axis will be but small. On the contrary, when the sun is near the zenith, the line going to the sun is not far removed from a vertical line, and consequently the principal motion of the quadrant will be performed on a vertical axis, by the observer’s turning himself about, and the part of the motion made on the axis of vision will be but small. In intermediate altitudes of the sun, the motions of the quadrant on the axis of vision, and on the vertical axis, will be more equally divided.
Observations taken with the quadrant are liable to errors, arising from the bending and elasticity of the index, and the resilience it meets with in turning round its centre: whence the extremity of the index, on being pushed along the arch, will sensibly advance before the index-glass begins to move, and may be seen to recoil when the force acting on it is removed. Mr Hadley seems to have been apprehensive that his instrument would be liable to errors from this cause; and in order to avoid them, gives particular directions that the index be made broad at the end next the centre, and that the centre, or axis itself, have as easy a motion as is consistent with steadiness; that is, an entire freedom from looseness, or shake, as the workmen term it. By strictly complying with these directions the error in question may indeed be greatly diminished; so far, perhaps, as to render it nearly infensible, where the index is made strong, and the proper medium between the two extremes of a thake at the centre on one hand, and too much stiffness there on the other, is nicely hit; but it cannot be entirely corrected. For to more or less of bending the index will always be subject; and some degree of resilience will remain at the centre, unless the friction there could be totally removed, which is impossible.
Of the reality of the error to which he is liable from this cause, the observer, if he is provided with a quadrant furnished with a screw for moving the index gradually, may thus satisfy himself. After finishing the observation, lay the quadrant on a table, and note the angle; then cautiously loosen the screw which fastens the index, and it will immediately, if the quadrant is not remarkably well constructed, be seen to start from its former situation, more or less according to the perfection of the joint and the strength of the index. This starting, which is owing to the index recoiling after being released from the confined state it was in during the observation, will sometimes amount to several minutes; and its direction will be opposite to that in which the index was moved by the screw at the time of finishing the observation. But how far it affects the truth of the observation, depends on the manner in which the index was moved in setting it to o, for adjusting the instrument; or in finishing the observations necessary for finding the index error.
The easiest and best rule to avoid these errors seems to be this: In all observations made by Hadley’s quadrant, let the observer take notice constantly to finish his observations, by moving the index in the same direction which was used in setting it to o for adjusting; or in the observations necessary for finding the index error. If this rule is observed, the error arising from the spring of the index will be obviated. For as the index was bent the same way, and in the same degree in adjusting as in observing, the truth of the observations will not be affected by this bending.
To take Altitudes by the Fore-observation.
I. Of the Sun.
Turn down either of the coloured glasses before the horizon-glass, according to the brightness of the sun; direct the sight to that part of the horizon which is under the sun, and move the index until the coloured image of the sun appear in the horizon-glass; then give the quadrant a slow vibratory motion about the axis of vision; move the index until the lower or upper limb of the sun is in contact with the horizon, at the lowest part of the arch described by this motion; and the degrees and minutes shown by the index on the limb will be the altitude of the sun.
II. Of the Moon.
Put the index to o, turn down the green glass, place the eye at the lower hole in the light vane, and observe the moon in the filigreed part of the horizon-glass; move the index gradually, and follow the moon’s reflected image until the enlightened limb is in contact with with the horizon, at the lower part of the arch described by the vibratory motion as before, and the index will show the altitude of the observed limb of the moon. If the observation is made in the day-time, the coloured glass is unnecessary.
III. Of a Star or Planet.
The index being put to 0, direct the sight to the star through the lower hole in the right-vane and transparent part of the horizon-glass; move the plane of the quadrant a very little to the left, and the image of the star will be seen in the silvered part of the glass. Now move the index, and the image of the star will appear to descend: continue moving the index gradually until the star is in contact with the horizon at the lowest part of the arch described; and the degrees and minutes shown by the index on the limb will be the altitude of the star.
To take Altitudes by the Back-observation.
I. Of the Sun.
Put the stem of the coloured glasses into the perforation between the horizon-glasses, turn down either according to the brightness of the sun, and hold the quadrant vertically; then direct the sight through the hole in the back right-vane, and the transparent slit in the horizon-glass to that part of the horizon which is opposite to the sun; now move the index till the sun is in the silvered part of the glass, and by giving the quadrant a vibratory motion, the axis of which is that of vision, the image of the sun will describe an arch whose convex side is upwards; bring the limb of the sun, when in the upper part of this arch, in contact with the horizon; and the index will show the altitude of the other limb of the sun.
II. Of the Moon.
The altitude of the moon is observed in the same manner as that of the sun, with this difference only, that the use of the coloured glass is unnecessary unless the moon is very bright; and that the enlightened limb, whether it be the upper or lower, is to be brought in contact with the horizon.
III. Of a Star or Planet.
Look directly to the star through the vane and transparent slit in the horizon-glass; move the index until the opposite horizon, with respect to the star, is seen in the silvered part of the glass; and make the contact perfect as formerly. If the altitude of the star is known nearly, the index may be set to that altitude, the sight directed to the opposite horizon, and the observation made as before.
Sect. II. Of finding the Latitude of a Place.
The observation necessary for ascertaining the latitude of a place, is that of the meridional altitude of a known celestial object; or two altitudes when the object is out of the meridian. The latitude is deduced with more certainty and with less trouble from the first of these methods, than from the second; and the sun, for various reasons, is the object most proper for this purpose at sea. It, however, frequently happens, that by the interposition of clouds, the sun is obscured at noon; and by this means the meridian altitude is lost. In this case, therefore, the method by double altitudes becomes necessary. The latitude may be deduced from three altitudes of an unknown object, or from double altitudes, the apparent times of observation being given.
The altitude of the limb of an object observed at sea, requires four separate corrections in order to obtain the true altitude of its centre: these are for semidiameter, dip, refraction, and parallax. (See Astronomy, and the respective articles). The first and last of these corrections vanish when the observed object is a fixed star.
When the altitude of the lower limb of any object is observed, its semidiameter is to be added thereto in order to obtain the central altitude; but if the upper limb be observed, the semidiameter is to be subtracted. If the altitude be taken by the back-observation, the contrary rule is to be applied. The dip is to be subtracted from, or added to, the observed altitude, according as the fore or back-observation is used. The refraction is always to be subtracted from, and the parallax added to, the observed altitude.
Prob. I. To reduce the sun’s declination to any given meridian.
Rule. Find the number in Table IX. answering to the longitude in the table nearest to that given, and to the nearest day of the month. Now, if the longitude is west, and the declination increasing, that is, from the 20th of March to the 22d of June, and from the 22d of September to the 22d of December, the above number is to be added to the declination: during the other part of the year, or while the declination is decreasing, this number is to be subtracted. In east longitude, the contrary rule is to be applied.
Ex. 1. Required the sun’s declination at noon 16th April 1810, in longitude 84° W?
Sun’s declination at noon at Greenwich 9° 59′ 2″ N. Number from Table IX. + 5°
Reduced declination 10° 4′ 2″
Ex. 2. Required the sun’s declination at noon 22d March 1793, in longitude 151° E?
Sun’s declination at noon at Greenwich 0° 53′ N. Equation from Table X. - 10
Reduced declination 0° 43′ N
Prob. II. Given the sun’s meridian altitude, to find the latitude of the place of observation.
Rule. The sun’s semidiameter is to be added to, or subtracted from, the observed altitude, according as the lower or upper limb is observed; the dip answering to the height from Table V. is to be subtracted if the fore-observation is used; otherwise, it is to be added; and the refraction answering to the altitude from Table IV. is to be subtracted: hence the true altitude of the sun’s centre will be obtained. Call the altitude south or north, according as the sun is south or north at the time of observation; which subtracted from 90°, will give the zenith distance of a contrary denomination.
Reduce the sun’s declination to the meridian of the place of observation, by Prob. I.; then the sum or difference Ex. 1. October 19, 1810, in longitude 32° E, the meridian altitude of the sun's lower limb was 48° 53' S, height of the eye 18 feet. Required the latitude?
| Obs. alt. sun's lower limb | 48° 53' S | |---------------------------|-----------| | Sun's dec. Oct. noon | 9° 51' S | | Semidiameter | +0 16 | | Equation Table IX | - |
Dip and refraction = 5
Reduced declination = 9° 49' S
True alt. sun's centre = 49° 4' S
Zenith distance = 40° 56' N
Latitude = 31° 7' N
Ex. 2. November 16, 1812, in longitude 158° W, the meridian altitude of the sun's lower limb was 87° 37' N, height of the eye 10 feet. Required the latitude?
| Obs. alt. sun's lower limb | 87° 37' N | |---------------------------|-----------| | Sun's dec. noon | 18° 48' S | | Semidiameter | +0 16 | | Equation tab. | +0 8 |
Dip and refraction = 3
Reduced declination = 19° 5' S
True alt. sun's centre = 87° 50' N
Zenith distance = 21° 10' S
Latitude = 21° 6' S
Ex. 3. December 19, 1811, being nearly under the meridian of Greenwich, the altitude of the sun's upper limb at noon was 40° 30' S, height of the eye 20 feet. Required the latitude?
| Observed altitude of the sun's upper limb | 4° 30' S | |------------------------------------------|----------| | Sun's semidiameter | +0 16 | | Dip and refraction | +0 15 |
True altitude of the sun's centre = 3° 59' S
Zenith distance = 86° 1' N
Declination = 23° 25' S
Latitude = 62° 36' N
Ex. 4. August 23, 1812, in longitude 107° E, the meridian altitude of the sun's lower limb by the back observation was 61° 8' N, and the height of the eye 14 feet. Required the latitude?
| Observed altitude of the sun's upper limb | 61° 8' N | |------------------------------------------|----------| | Sun's semidiameter | +0 16 | | Dip and refraction | +0 3 | | Refraction | +0 3/4 |
True altitude of the sun's centre = 60° 55' N
Zenith distance = 29° 5' S
Reduced declination = 11° 26' N
Latitude = 17° 39' S
The dip in Table V. answers to an entirely open and unobstructed horizon. It, however, frequently happens, that the sun is over the land at the time of observation, and the ship nearer to the land than the visible horizon would be if unconfined. In this case, the dip will be different from what it would otherwise have been, and is to be taken from Table VI., in which the height is expressed at the top, and the distance from the land in the side column in nautical miles.—Seamen, in general, can estimate the distance of any object from the ship with sufficient exactness for this purpose, especially when that distance is not greater than six miles; which is the greatest distance of the visible horizon from an observer on the deck of any ship.
Prob. III. Given the meridian altitude of a fixed star, to find the latitude of the place of observation.
Rule. Correct the altitude of the star by dip and refraction, and find the zenith distance of the star as formerly; take the declination of the star from Table XI. and reduce it to the time of observation. Now, the sum or difference of the zenith distance and declination of the star, according as they are of the same or of a contrary name, will be the latitude of the place of observation.
Ex. 1. December 1, 1810, the meridian altitude of Sirius was 59° 5' S, height of the eye 14 feet. Required the latitude?
| Observed altitude of Sirius | 59° 5' S | |-----------------------------|----------| | Dip and refraction | +0 4 | | True altitude | 59° 46' S| | Zenith distance | 30° 14' N| | Declination | 16° 28' S| | Latitude | 13° 46' N|
Ex. 2. February 17, 1797, the meridian altitude of Procyon was 71° 15' N, the height of the eye 10 feet. Required the latitude?
| Observed altitude of Procyon | 71° 15' N| |------------------------------|----------| | Dip and refraction | +0 3 | | True altitude | 71° 12' N| | Zenith distance | 18° 48' S| | Declination | 5° 43' N | | Latitude | 13° 5' S |
Prob. IV. Given the meridian altitude of a planet, to find the latitude of the place of observation.
Rule. Compute the true altitude of the planet as directed in last problem (which is sufficiently accurate for altitudes taken at sea); take its declination from the Nautical Almanac, p. iv. of the month, and reduce it to the time and meridian of the place of observation; then the sum or difference of the zenith distance and declination of the planet will be the latitude as before.
Ex. 1. August 7, 1812, the meridian altitude of Saturn was 68° 42' N, and height of the eye 15 feet. Required the latitude?
| Observed altitude of Saturn | 68° 42' N| |-----------------------------|----------| | Dip and refraction | +0 4 | | True altitude | 68° 38' N| | Zenith distance | 21° 22' S| | Declination | 22° 42' S| | Latitude | 44° 6' S |
Ex. 2. October 15, 1812, the meridian altitude of Jupiter was 81° 5' S, height of the eye 18 feet. Required the latitude? **Practice.**
**NAVI GATION.**
| Method of finding the Latitude and Longitude at Sea. | |-----------------------------------------------------| | Observed altitude of Jupiter | $81^\circ 5' S$ | | Dip | $-3$ | | True altitude | $81^\circ 2 S$ | | Zenith distance | $8^\circ 58 N$ | | Declination | $19^\circ 4 S$ | | Latitude | $10^\circ 6 S$ |
**PROB. V.** Given the meridian altitude of the moon, to find the latitude of the place of observation.
**Rule.** Take the number \( \dagger \) answering to the ship's longitude, and daily variation of the moon's passing the meridian; which being applied to the time of passage given in the Nautical Almanac, will give the time of the moon's passage over the meridian of the ship.
Reduce this time to the meridian of Greenwich; and by means of the Nautical Almanac find the moon's declination, horizontal parallax, and semidiameter at the reduced time.
Apply the semidiameter and dip to the observed altitude of the limb, and the apparent altitude of the moon's centre will be obtained; to which add the correction answering to the apparent altitude and horizontal parallax \( \dagger \), and the sum will be the true altitude of the moon's centre; which subtracted from \( 90^\circ \), the remainder is the zenith distance, and the sum or difference of the zenith distance and declination, according as they are of the same or of a contrary name, will be the latitude of the place of observation.
**Ex. I.** July 24, 1812, in longitude \( 15^\circ W \), the altitude of the sun's lower limb at midnight was \( 8^\circ 58' \) height of the eye 18 feet. Required the latitude?
Time of passage over the mer. of Greenwich = \( 9h 19' \)
Equation Table XX. \( +0^\circ 4 \)
Time of passage over mer. ship \( -9^\circ 23 \)
Longitude in time \( -2^\circ 0 \)
Reduced time \( -11^\circ 23 \)
Moon's dec. at midnight, Table IX. \( =14^\circ 53' N \)
Eq. to time from midnight \( -0^\circ 4 \)
Reduced declination \( -14^\circ 49' N \)
Moon's hor. par. \( -53' 25'' \)
Moon's semidiameter \( -15^\circ 6 \)
Augmentation \( +0^\circ 14 \)
Aug. semidiameter \( -15^\circ 20 \)
**Observed altitude of the moon's lower limb** \( 81^\circ 15' N \)
Semidiameter \( +0^\circ 15 \)
Dip \( -0^\circ 3 \)
Apparent altitude of the moon's centre \( 81^\circ 27' N \)
Correction \( +0^\circ 8 \)
True altitude of moon's centre \( 81^\circ 35' N \)
Zenith distance \( 8^\circ 23' S \)
Declination \( 14^\circ 49' N \)
Latitude \( 6^\circ 42' N \)
**Remark.** If the object be on the meridian below the pole at the time of observation, then the sum of the true altitude and the complement of the declination is the latitude, of the same name as the declination or altitude.
**Ex. II.** July 24, 1812, in longitude \( 15^\circ W \), the altitude of the sun's lower limb at midnight was \( 8^\circ 58' \) height of the eye 18 feet. Required the latitude?
Observed altitude sun's lower limb \( -8^\circ 58' \)
Semidiameter \( +0^\circ 16 \)
Dip and refraction \( -0^\circ 10 \)
True altitude of sun's centre \( -9^\circ 4 N \)
Compl. declin. reduced to time and place \( 66^\circ 57' N \)
Latitude \( 76^\circ 1 N \)
**PROB. VI.** Given the latitude by account, the declination and two observed altitudes of the sun, and the interval of time between them, to find the true latitude.
**Rule.** To the log. secant of the latitude by account, add the log. secant of the sun's declination; the sum, rejecting 20 from the index, is the logarithm ratio. To this add the log. of difference of the natural sines of the two altitudes, and the log. of the half elapsed time from its proper column.
Find this sum in column of middle time, and take out the time answering thereto; the difference between which and the half elapsed time will be the time from noon when the greater altitude was observed.
Take the log. answering to this time from column of rising, from which subtract the log. ratio, the remainder is the logarithm of a natural number; which being added to the natural sine of the greater altitude, the sum is the natural cosine of the meridian zenith distance; from which and the sun's declination the latitude is obtained as formerly.
If the latitude thus found differs considerably from that by account, the operation is to be repeated, using the computed latitude in place of that by account (I.).
**Example I.**
(1.) This method is only an approximation, and ought to be used under certain restrictions; namely,
The observations must be taken between nine o'clock in the forenoon and three in the afternoon. If both observations be in the forenoon, or both in the afternoon, the interval must not be less than the distance of the time of observation of the greatest altitude from noon. If one observation be in the forenoon and the other in the afternoon, the interval must not exceed four hours and a half; and in all cases, the nearer the greater altitude is to noon the better.
If the sun's meridian zenith distance be less than the latitude, the limitations are still more contracted. If the latitude be double the meridian zenith distance, the observations must be taken between half past nine in the morning and half past two in the afternoon, and the interval must not exceed three hours and a half. The observations must be taken still nearer to noon, if the latitude exceed the zenith distance in a greater proportion. See Mackelyne's British Mariner's Guide, Dr Mackay's Treatises on the Longitude and Navigation, &c. and Requisite Tables, 2d edit. Example 1. July 9, 1811, in latitude by account 27° N., at 10h 29' A.M. per watch, the corrected altitude of the sun was 65° 24', and at 12h 31', the altitude was 74° 8'. Required the true latitude?
Times per wat Alt. N. Sines Lat. by acc. 37° 0' Secant 0.09765 10h 29' 65° 24' 0.9024 Declination 22° 28' Secant 0.3428
| 12° 31' | 74° 8' | Logarithm ratio | 0.13193 | |---------|--------|-----------------|----------| | 2° 2' | Differ.| 5266 | Logarithm | 3.7148 | | 1° | Half elapsed time | - | 0.57999 | | 31° 10'' | Middle time | - | 4.43349 | | 29° 50' | Rising | - | Log. ratio | 0.13193 | | Natural number | - | 624 | 2.79547 | | Greatest altitude | 74° 8' N. fine | 96190 |
Mer. zenith diff. 14° 30' N cosine 96814 Declination 22° 28' Latitude 26° 58' N.
Ex. 2. October 17, 1812, in latitude 43° 24' N. by account, at 6h 38' P.M. the correct altitude of the sun's centre was 36° 5', and at 2h 46' P.M. the altitude was 24° 49'. Required the latitude?
Times per wat Alt. N. Sines Lat. by acc. 43° 3' Secant 0.13872 ch 38' 36° 5' 58960 Declination 9° 18' Secant 0.00675
| 2° 46' | 24° 49' | Logarithm ratio | 0.14447 | |---------|---------|-----------------|----------| | 2° 8' | Differ.| 16924 | Log. | 4.22850 | | 1° 4' | Half elapsed time | - | 0.55066 | | 1° 41' 20'' | Middle time | - | 4.93633 | | 37° 20' | Rising | - | Log. ratio | 0.14447 | | Natural Number | - | 649 | 2.97737 | | Greatest altitude | 36° 5' N. fine | 58960 |
Mer. zen. distance 53° 15' N cosine 59345 Declination 9° 18' Latitude 43° 57' N.
Ex. 3. In latitude 49° 48' N. by account, the sun's declination being 39° 37' S. at oh 3' 2' P.M. per watch, the altitude of the sun's lower limb was 28° 32', and at 2h 41' it was 19° 25', the height of the eye 12 feet. Required the true latitude?
First observed altit. 28° 32' Second altitude 19° 25' Semidiameter +0° 16' Semidiameter +0° 16' Dip and refraction -0° 5' Dip and refr. -0° 6'
True altitude 28° 43' True altitude 19° 35' Time per wat Alt. N. Sines Lat. by acc. 49° 48' Secant 0.19013 ch 32' 28° 43' 48° 48' Declination 9° 37' Secant 0.0615
| 2° 41' | 19° 35' | Logarithm ratio | 0.09628 | |---------|---------|-----------------|----------| | 2° 9' | Difference | 14530 | Log. | 4.16227 | | 1° 4' 30'' | Half elapsed time | - | 0.55637 | | 1° 37' 0'' | Middle time | - | 4.91492 | | 32° 30' | Rising | - | Log. ratio | 0.09628 | | Natural number | - | 639 | 2.80536 | | Mer. zen. diff. | 60° 52' N. cosine | 48687 | | Declination | 9° 37' S. |
Latitude 51° 15' N.
As the latitude by computation differs 1° 27' from that by account, the operation must be repeated.
Computed latitude 51° 15' Secant 0.20348 Declination 9° 37' Secant 0.0615
Logarithm ratio 0.20348 Difference of nat. fines 14530 Log. 4.16227 Half elapsed time 1h 4' 30'' Log. 0.55637 Middle time 1h 40' 20'' Log. 4.92827 Rising 0° 35' 50'' Log. 3.08630 Natural number 753 2.87667 Gr. altitude 28° 43' N. fine 48048 Mer. zen. diff. 60° 47' N. cosine 48801 Declination 9° 37' Latitude 51° 10' N.
As this latitude differs only 5' from that used in the computation, it may therefore be depended on as the true latitude.
Prob. VII. Given the latitude by account, the sun's declination, two observed altitudes, the elapsed time, and the course and distance run between the observations; to find the ship's latitude at the time of observation of the greater altitude.
Rule. Find the angle contained between the ship's course and the sun's bearing at the time of observation of the least altitude, with which enter the Traverse Table as a course, and the difference of latitude answering to the distance made good will be the reduction of altitude.
Now, if the least altitude be observed in the forenoon, the reduction of altitude is to be applied thereunto by addition or subtraction, according as the angle between the ship's course and the sun's bearing is less or more than eight points. If the least altitude be observed in the afternoon, the contrary rule is to be used.
The difference of longitude in time between the observations is to be applied to the elapsed time by addition or subtraction, according as it is east or west. This is, however, in many cases so inconsiderable as to be neglected.
With the corrected altitudes and interval, the latitude by account and sun's declination at the time of observation of the greatest altitude, the computation is to be performed by the last problem.
Ex. 1. July 6, 1793, in latitude 58° 14' N by account, and longitude 16° E., at 10h 54' A.M. per watch, the altitude of the sun's lower limb was 53° 17', and at 1h 17' P.M. the altitude was 52° 51', and bearing per compass SWbW; the ship's course during the elapsed time was SWbW, and the hourly rate of falling 8 knots, the height of the eye 16 feet. Required the true latitude at the time of observation of the greater altitude?
Sun's bear. at 2d. ob. SWbW, Interval bet. observ. 2h 23' Ship's course SWbW Dist.run=2h 23'x8=19m.
Contained angle 34 points. Now to course 3½ points, and distance 19 miles, the difference of latitude is 14.7 or 15 miles.
First observed alt. 53° 17' Second observed alt. 52° 51' Semidiameter +0 16 Semidiameter +0 16 Dip and refract. -0 4 Dip and refraction -0 4
True altitude 53° 29' Reduction -0 15 Reduced altitude 52° 48'
Time of ob. of gr. alt. 10h 54' A.M. Sun's dec. 22° 39' N. Longitude in time 1 4 Eq. to r. t. +1
Reduced time 9 50 A.M. Red. decl. 22° 40' N. Time per wat. Alt. N. Sines. Lat. by acc. 58° 14' Secant 0.27863 10h 54' 53° 29' 80368 Declination 22° 40' Secant 0.3491
| Logarithm ratio | 0.31354 | |-----------------|----------| | Differ. | 715 | | Half elapsed time | 0.51294 |
Middle time 5 30 Rising 3 6 0
Logarithm ratio 0.31354
Natural number 2001 Greatest altitude 53° 29' N. fine 80368
Mer. zenith dist. 34° 33' N. cof. 82369 Declination 22° 40' N.
Latitude 57° 13' N.
Since the computed latitude differs so much from that by account, it will be necessary to repeat the operation.
Computed latitude 57° 13' Secant 0.26643 Declination 22° 40' Secant 0.3491
Logarithm ratio 0.30134 Difference of natural fines 715 Log. 2.85431 Half elapsed time 1h 11' 30" Log. 0.51294
Middle time 5 20 Rising 1 6 10 Log. 3.66859
Logarithm ratio 0.30134
Natural number 2068 Greatest altitude 53° 29' N. fine 80368
Mer. zen. dist. 34° 29' N. cof. 82436 Declination 22° 40' N.
Latitude 57° 9' N.
As this latitude differs only 4 miles from that used in the computation, it may therefore be depended on as the true latitude.
Remark. If the sun come very near the zenith, the fines of the altitude will vary so little as to make it uncertain which ought to be taken as that belonging to the natural fine of the meridian altitude. In this case, the following method will be found preferable.
To the log. rising of the time from noon found as before, add the log. secant of half the sum of the estimate meridian altitude, and greatest observed altitude; from which subtract the log. ratio, its index being increased by 10, and the remainder will be the log. fine of an arch; which added to the greatest altitude will give the sun's meridian altitude.
Ex. 2. December 21st 1793, in latitude 22° 40' S., by account at 11h 57' the correct altitude of the sun's centre was 89° 10', and at 12h 4' 40", the altitude was 88° 50'. Required the true latitude?
Times per wat. Alt. N. Sines. Lat. by acc. 22° 40' Secant 0.3491 11h 57' 89° 10' 99989 Declination 23° 28' Secant 0.3749
| Logarithm ratio | 0.07240 | |-----------------|----------| | Differ. | 10 | | Half elapsed time | 1.77663 |
Middle time 2.84903 Rising 9.93284
Comp. of lat. by acc. 67° 20' Declination 23° 28'
Sum 90° 48' Estimate mer. altitude 89° 12' 89° 11' sec. 11.84609 Greatest altitude 89° 10'
Logarithm ratio +5 12.77893 Arch 0° 17' fine 7.70653 Greatest altitude 89° 10'
Meridian altitude 89° 27' zen. dist. 0° 33' N. declinat. 23° 28' S.
This differing from the assumed latitude, the work must be repeated.
Latitude 22° 55' fecant 0.03571 Declination 23° 28' fecant 0.03749
Logarithm ratio 0.07320 Difference of natural fines 1° log. 1.00000 Half elapsed time 3' 50" 1.77663
Middle time 2.84983 Rising 0.93284
Comp. of lat. 67° 5' Declination 23° 28'
Sum 90° 33' Mer. alt. 89° 27' 89° 18' sec. 11.91827 Greatest alt. 89° 10'
Log. ratio +5 12.85111 Arch 0° 21' 7.77791 Greatest altitude 89° 10'
Merid. altitude 89° 31' zen. dist. 0° 29' Declination 23° 28'
If the work be repeated with this last latitude, the latter part only may be altered. **NAVIGATION**
| Latitude | Declination | Eft. mer. alt. | Greatest altitude | |----------|-------------|---------------|------------------| | 22° 59' | 23° 28' | 89° 31' | 89° 10' ar. com. |
| fecant | fecant | log. ratio | |----------|-------------|---------------| | 0.03592 | 0.03749 | 0.07341 |
| Interval | Capella's declin. | Sirius's declin. | |----------|-------------------|------------------| | 1h 34' 36" | 45° 46' N | 16° 27' S |
| cofine | cofine | |----------|----------------| | 9.84360 | 9.98185 |
| Sum | 62° 13' N cofine | |----------|------------------| | 46613 | 5599 |
| Arch first | Capella's declin. | Interval | |------------|-------------------|----------| | 24° 13' N fine | 45° 46' N | 1h 34' 36" |
| cofine | fecant | |----------|------------------| | 9.96000 | 0.15640 |
| Arch second | Arch first | Sirius's altitude | |-------------|------------|-------------------| | 1° 11' 28" | 24° 13' | 16° 19' |
| fecant | fecant | |----------|------------| | 0.04000 | 0.01785 |
| Difference | Capella's altitude | |------------|--------------------| | 7° 54N, cofine | 69° 23N, fine |
| 99051 | 93596 |
| Arch third | Arch second | |------------|-------------| | 1h 21' 20" | 1h 11' 28" |
| rising | rising | |-----------|------------| | 3.79464 | 3.73679 |
| Arch fourth | Sirius's declin. | |-------------|------------------| | 9° 52' | 16° 27' | | altitude | 16° 19' |
| rising | cofine | |-----------|--------------| | 1.96708 | 9.98185 | | 9.98215 | |
| Sum | 32° 46' N cofine | |----------|------------------| | 84088 | 85° 1.93008 |
| Latitude | 57° 9 N fine | |----------|--------------| | 84003 | |
---
**CHAP. II. Containing the Method of finding the Longitude at Sea by Lunar Observations.**
**SECT. I. Introduction.**
The observations necessary to determine the longitude by this method are, the distance between the sun and moon, or the moon and a fixed star near the ecliptic, together with the altitude of each. The stars used in the Nautical Almanack for this purpose are the following: namely, α Arietis, Aldebaran, Pollux, Regulus, Spica Virginis, Antares, α Aquilae, Fomalhaut, and α Pegasi; and the distances of the moon’s centre from the sun, and from one or more of these stars, are contained in the viii., ix., x., and xi. pages of the month, at the beginning of every third hour apparent time, by the meridian of Greenwich. The distance between the moon and the sun, or one of these stars, is observed with a sextant; and the altitudes of the objects are taken as usual with a Hadley’s quadrant.
In the practice of this method, it will be found convenient to be provided with three assistants; two of these are to take the altitudes of the sun and moon, or moon and star, at the same time the principal observer is taking the distance between the objects; and the third assistant is to observe the time, and write down... Of finding down the observations. In order to obtain accuracy, it will be necessary to observe several distances, and the corresponding altitudes; the intervals of time between them being as short as possible; and the sum of each divided by the number will give the mean distance and mean altitudes; from which the time of observation at Greenwich is to be computed by the rules to be explained.
If the sun or star from which the moon's distance is observed be at a proper distance from the meridian, the time at the ship may be inferred from the altitude observed at the same time with the distance: in this case, the watch is not necessary; but if that object be near the meridian, the watch is absolutely necessary, in order to connect the observations for ascertaining the apparent time at the ship and at Greenwich with each other.
An observer without any assistants may very easily take all the observations, by first taking the altitudes of the objects, then the distance, and again their altitudes, and reduce the altitudes to the time of observation of the distance; or, by a single observation of the distance, the apparent time being known, the longitude may be determined.
A set of observations of the distance between the moon and a star, and their altitudes, may be taken with accuracy during the time of the evening or morning twilight; and the observer, though not much acquainted with the stars, will not find it difficult to distinguish the star from which the moon's distance is to be observed. For the time of observation nearly, and the ship's longitude by account being known, the estimate time at Greenwich may be found; and by entering the Nautical Almanac with the reduced time, the distance between the moon and given star will be found nearly. Now let the index of the sextant to this distance, and hold the plane of the instrument so as to be nearly at right angles to the line joining the moon's culps, direct the sight to the moon, and by giving the sextant a slow vibratory motion, the axis of which being that of vision, the star, which is usually one of the brightest in that part of the heavens, will be seen in the transparent part of the horizon glass.
**Sect. II. Of the Sextant.**
This instrument is constructed for the express purpose of measuring with accuracy the angular distance between the sun and moon, or between the moon and a fixed star, in order to ascertain the longitude of a place by lunar observations. It is, therefore, made with more care than the quadrant, and has some additional appendages that are wanting in that instrument.
Fig. 63. represents the sextant, so framed as not to be liable to bend. The arch AA is divided into 120 degrees; each degree is divided into three parts; each of these parts, therefore, contains 20 minutes, which are again subdivided by the vernier into every half minute or 30 seconds. The vernier is numbered at every fifth of the lunar divisions, from the right towards the left, with 5, 10, 15, and 20; the first division to the right being the beginning of the scale.
In order to observe with accuracy, and make the images come precisely in contact, an adjusting screw B is added to the index, which may thereby be moved with greater accuracy than it can be by hand; but this forew does not act until the index is fixed by the finger screw C. Care should be taken not to force the adjusting forew when it arrives at either extremity of its adjustment. When the index is to be moved any considerable quantity, the screw C at the back of the sextant must be loosened; but when the index is brought nearly to the division required, this back screw should be tightened, and then the index may be moved gradually by the adjusting screw.
There are four tinged glasses D, each of which is set in a separate frame that turns on a centre. They are used to defend the eye from the brightness of the solar image and the glare of the moon, and may be used separately or together as occasion requires.
There are three more such glasses placed behind the horizon glass at E, to weaken the rays of the sun or moon when they are viewed directly through the horizon glass. The paler glass is sometimes used in observing altitudes at sea, to take off the strong glare of the horizon.
The frame of the index glass I is firmly fixed by a strong cock to the centre plate of the index. The horizon glass F is fixed in a frame that turns on the axes or pivots, which move in an exterior frame; the holes in which the pivots move may be tightened by four screws in the exterior frame. G is a screw by which the horizon glass may be set perpendicular to the plane of the instrument: should this screw become loose, or move too easy, it may be easily tightened by turning the capstan headed screw H, which is on one side of the socket through which the stern of the finger screw passes.
The sextant is furnished with a plain tube (fig. 64.) Fig. 64. without any glasses; and to render the objects still more distinct, it has two telescopes, one (fig. 65.) representing the objects erect, or in their natural position: the longer one (fig. 66.) shows them inverted; Fig. 66. it has a large field of view, and other advantages, and a little use will soon accustom the observer to the inverted position, and the instrument will be as readily managed by it as by the plain tube alone. By a telescope the contact of the images is more perfectly distinguished; and by the place of the images in the field of the telescope, it is easy to perceive whether the sextant is held in the proper place for observation. By sliding the tube that contains the eye-glasses in the inside of the other tube, the object is suited to different eyes, and made to appear perfectly distinct and well defined.
The telescopes are to be screwed into a circular ring at K; this ring rests on two points against an exterior ring, and is held thereto by two screws: by turning one or other of these screws, and tightening the other, the axis of the telescope may be set parallel to the plane of the sextant. The exterior ring is fixed on a triangular brass item that slides in a socket, and by means of a screw at the back of the quadrant may be raised or lowered so as to move the centre of the telescope to point to that part of the horizon glass which shall be judged the most fit for observation. Fig. 67. is a circular head, with tinged glasses to screw on the eye end of either of the telescopes or the plain tube. The glasses are contained in a circular plate which Of finding which has four holes; three of these are fitted with tinged glasses, the fourth is open. By pressing the finger against the projecting edge of this plate, and turning it round, the open hole, or any of the tinged glasses, may be brought between the eye-glass of the telescope and the eye.
Fig. 68. is a magnifying glass, to assist the observer to read off the angle with more accuracy: and fig. 69. a screw-driver.
Mr Hoppe of Church-street, Minories, London, has lately contrived a sextant, with two arches, which is, therefore, preferable to the common sextant.
Adjustments of the Sextant.
The adjustments of a sextant are, to set the mirrors perpendicular to its plane and parallel to each other when the index is at zero, and to set the axis of the telescope parallel to the plane of the instrument. The three first of these adjustments are performed nearly in the same manner as directed in the section on the quadrant: as however the sextant is provided with a set of coloured glasses placed behind the horizon glass, the index error may be more accurately determined by measuring the sun's diameter twice, with the index placed alternately before and behind the beginning of the divisions: half the difference of these two measures will be the index error, which must be added to, or subtracted from, all observations, according as the diameter measured with the index to the left of o is less or greater than the diameter measured with the index to the right of the beginning of the divisions.
Adjustment IV. To set the Axis of the Telescope parallel to the Plane of the Instrument.
Turn the eye end of the telescope until the two wires are parallel to the plane of the instrument; and let two distant objects be selected, as two stars of the first magnitude, whose distance is not less than 90° or 100°; make the contact of these objects as perfect as possible at the wire nearest the plane of the instrument; fix the index in this position; move the sextant till the objects are seen at the other wire, and if the same points are in contact, the axis of the telescope is parallel to the plane of the sextant; but if the objects are apparently separated, or do partly cover each other, correct half the error by the screws in the circular part of the supporter, one of which is above and the other between the telescope and sextant: turn the adjusting screw at the end of the index till the limbs are in contact; then bring the objects to the wire next the instrument; and if the limbs are in contact, the axis of the telescope is adjusted; if not, proceed as at the other wire, and continue till no error remains.
It is sometimes necessary to know the angular distance between the wires of the telescope; to find which, place the wires perpendicular to the plane of the sextant, hold the instrument vertical, direct the sight to the horizon, and move the sextant in its own plane till the horizon and upper wire coincide; keep the sextant in this position, and move the index till the reflected horizon is covered by the lower wire: and the division shown by the index of the limb, corrected by the index error, will be the angular distance between the wires. Other and better methods will readily occur to the observer at land.
Use of the Sextant.
When the distance between the moon and the sun or a star is to be observed, the sextant must be held so that its plane may pass through the eye of the observer and both objects; and the reflected image of the most luminous of the two is to be brought in contact with the other seen directly. To effect this, therefore, it is evident, that when the brightest object is to the right of the other, the face of the sextant must be held upwards; but if to the left, downwards. When the face of the sextant is held upwards, the instrument should be supported with the right hand, and the index moved with the left hand. But when the face of the sextant is from the observer, it should be held with the left hand, and the motion of the index regulated by the right hand.
Sometimes a fitting posture will be found very convenient for the observer, particularly when the reflected object is to the right of the direct one; in this case, the instrument is supported by the right hand, the elbow may rest on the right knee, the right leg at the same time resting on the left knee.
If the sextant is provided with a ball and socket, and a staff, one of whose ends is attached thereto, and the other rests in a belt fastened round the body of the observer, the greater part of the weight of the instrument will by this means be supported by his body.
To observe the Distance between the Moon and any Celestial Object.
1. Between the Sun and Moon.
Put the telescope in its place, and the wires parallel to the plane of the instrument; and if the sun is very bright, raise the plate before the silvered part of the speculum; direct the telescope to the transparent part of the horizon glass, or to the line of separation of the silvered and transparent parts according to the brightness of the sun, and turn down one of the coloured glasses; then hold the sextant so that its plane produced may pass through the sun and moon, having its face either upwards or downwards according as the sun is to the right or left of the moon; direct the sight through the telescope to the moon, and move the index till the limb of the sun is nearly in contact with the enlightened limb of the moon; now fasten the index, and by a gentle motion of the instrument make the image of the sun move alternately past the moon; and, when in that position where the limbs are nearest each other, make the coincidence of the limbs perfect by means of the adjusting screw: this being effected, read off the degrees and parts of a degree shown by the index on the limb, using the magnifying glass; and thus the angular distance between the nearest limbs of the sun and moon is obtained.
2. Between the Moon and a Star.
Direct the middle of the field of the telescope to the line of separation of the silvered and transparent parts of the horizon glass; if the moon is very bright, turn down the lightest coloured glass; and hold the sextant so that its plane may be parallel to that passing through the eye of the observer and both objects; its face being upwards if the moon is to the right of the star, but if to the left, the face is to be held from the observer; now direct the sight through the telescope to the star, and... move the index till the moon appears by the reflection to be nearly in contact with the star; fasten the index, and turn the adjusting screw till the coincidence of the star and enlightened limb of the moon is perfect: and the degrees and parts of a degree shown by the index will be the observed distance between the moon's enlightened limb and the star.
The contact of the limbs must always be observed in the middle between the parallel wires.
It is sometimes difficult for those not much accustomed to observations of this kind, to find the reflected image in the horizon glass: it will perhaps in this case be found more convenient to look directly to the object, and, by moving the index, to make its image coincide with that seen directly.
Sect. III. Of the Circular Instrument of Reflection.
This instrument was proposed with a view to correct the errors to which the sextant is liable; particularly the error arising from the inaccuracy of the divisions on the limb. It consists of the following parts; a circular ring or limb, two moveable indices, two mirrors, a telescope, coloured glasses, &c.
The limb of this instrument is a complete circle of metal, and is connected with a perforated central plate by six radii: it is divided into 720 degrees; each degree is divided into three equal parts; and the division is carried to minutes by means of the index scale as usual.
The two indices are moveable about the same axis, which passes exactly through the centre of the instrument:—the first index carries the central mirror, and the other the telescope and horizon glass; each index being provided with an adjusting screw for regulating its motion, and a scale for showing the divisions on the limb.
The central mirror is placed on the first index immediately above the centre of the instrument, and its plane makes an angle of about 30° with the middle line of the index. The four screws in its pedestal for making its plane perpendicular to that of the instrument have square heads, and are therefore easily turned either way by a key for that purpose.
The horizon glass is placed on the second index near the limb, so that as few as possible may be intercepted of the rays proceeding from the reflected object when to the left. The perpendicular position of this glass is rectified in the same manner as that of the horizon glass of a sextant, to which it is similar. It has another motion, whereby its plane may be disposed so as to make a proper angle with the axis of the telescope, and a line joining its centre, and that of the central mirror.
The telescope is attached to the other end of the index. It is an achromatic astronomical one, and therefore inverts objects; it has two parallel wires in the common focus of the glasses, whose angular distance is between two and three degrees; and which, at the time of observation, must be placed parallel to the plane of the instrument. This is easily done, by making the mark on the eye-piece coincide with that on the tube. The telescope is moveable by two screws in a vertical direction with regard to the plane of the instrument, but is not capable of receiving a lateral motion.
There are two sets of coloured glasses, each set containing four, and differing in shade from each other. The glasses of the larger set, which belongs to the central mirror, should have each about half the degree of shade with which the correspondent glasses of the set belonging to the horizon mirror is tinged. These glasses are kept tight in their places by small pressing screws, and make an angle of about 85° with the plane of the instrument; by which means the image from the coloured glasses is not reflected to the telescope. When the angle to be measured is between 5° and 34°, one of the glasses of the largest set is to be placed before the horizon glass.
The handle is of wood, and is screwed to the back of the instrument, immediately under the centre, with which it is to be held at the time of observation.
Fig. 70. is a plan of the instrument, wherein the limb Fig. 70a. is represented by the divided circular plate; A is the central mirror; a a, the places which receive the stems a a of the glasses, fig. 73.; EF, the first or central index with its scale and adjusting screw; MN, the second or horizon index; GH, the telescope; IK, the screws for moving it towards or from the plane of the instrument; C, the plane of the coloured glasses, fig. 72.; and D, its place in certain observations.
Fig. 71. is a section of the instrument, wherein the Fig. 71a. several parts are referred to by the same letters as in fig. 70.; Fig. 72. represents one of the horizon coloured glasses; and fig. 73. one of the central coloured glasses: Fig. 74. is the key for turning the adjusting screws of the mirrors: Fig. 75. is the handle: Fig. 76. a section of one of the radii towards its middle: Fig. 77. is used in some terrestrial observations for diminishing the light of the direct object, whose place at the time of observation is D: Fig. 78. is the tool for adjusting the central mirror; and for rectifying the position of the telescope with regard to the plane of the instrument, there is another tool exactly of the same size. The height of these is nearly equal to that of the middle of the central mirror.
Adjustments of the Circular Instrument.
I. To set the horizon glass so that none of the rays from the central mirror shall be reflected to the telescope from the horizon mirror, without passing through the coloured glasses belonging to this last mirror.—Place the coloured glasses before the horizon mirror; direct the telescope to the silvered part of that mirror, and make it nearly parallel to the plane of the instrument; move the first index; and if the rays from the central mirror to the horizon glass, and from thence to the telescope, have all the same degree of shade with that of the coloured glasses used, the horizon glass is in its proper position; otherwise the pedestal of the glass must be turned until the uncoloured images disappear.
II. Place the two adjusting tools on the limb, about 35° of the instrument distant, one on each side of the division on the left, answering to the plane of the central mirror produced: then the eye being placed at the upper edge of the nearest tool, move the central index till one half only of the reflected image of this tool is seen in the central mirror towards the left, and move the other tool till its half to the right is hid by the same edge of the mirror; then, if the upper edges of both tools are apparently in the same straight line, Of finding the central mirror is perpendicular to the plane of the instrument; if not, bring them into this position by the screws in the pedestal of the mirror.
III. To set the horizon mirror perpendicular to the plane of the instrument.—The central mirror being previously adjusted, direct the sight through the telescope to any well-defined distant object; then if, by moving the central index, the reflected image passes exactly over the direct object, the mirror is perpendicular; if not, its position must be rectified by means of the screws in the pedestal of the glass.
A planet, or star of the first magnitude, will be found a very proper object for this purpose.
IV. To make the line of collimation parallel to the plane of the instrument.—Lay the instrument horizontally on a table; place the two adjusting tools on the limb, towards the extremities of one of the diameters of the instrument; and at about 15 or 20 feet distant let a well-defined mark be placed, so as to be in the same straight line with the tops of the tools; then raise or lower the telescope till the plane, passing through its axis and the tops of the tools, is parallel to the plane of the instrument, and direct it to the fixed object; turn either or both of the screws of the telescope till the mark is apparently in the middle between the wires; then is the telescope adjusted; and the difference, if any, between the divisions pointed out by the indices of the screws will be the error of the indices. Hence this adjustment may be easily made.
In this process the eye tube must be so placed as to obtain distinct vision.
V. To find that division to which the second index being placed the mirrors will be parallel, the central index being at zero.—Having placed the first index exactly to 0, direct the telescope to the horizon mirror, so that its field may be affected by the line of separation of the silvered and transparent parts of that mirror; hold the instrument vertically, and move the second index until the direct and reflected horizons agree; and the division shown by the index will be that required.
This adjustment may be performed by measuring the sun's diameter in contrary directions, or by making the reflected and direct images of a star or planet to coincide.
Use of the Circular Instrument.
To observe the Distance between the Sun and Moon.
I. The sun being to the right of the moon.
Set a proper coloured glass before the central mirror, if the distance between the objects is less than 35°; but if above that quantity, place a coloured glass before the horizon mirror; make the mirrors parallel, the first index being at 0, and hold the instrument so that its plane may be directed to the objects, with its face downwards, or from the observer; direct the sight through the telescope to the moon; move the second index, according to the order of the divisions on the limb, till the nearest limbs of the sun and moon are almost in contact; fasten that index, and make the coincidence of the limbs perfect by the adjusting screw belonging thereto; then invert the instrument, and move the central index towards the second by a quantity equal to twice the arch passed over by that index; direct the plane of the instrument to the objects; look directly to the moon, and the sun will be seen in the field of the telescope; fasten the central index, and the longitude at sea by Lunar Observations.
II. The sun being to the left of the moon.
Hold the instrument with its face upwards, so that its plane may pass through both objects; direct the telescope to the moon, and make its limb coincide with the nearest limb of the sun's reflected image, by moving the second index; now put the instrument in an opposite position; direct its plane to the objects, and the sight to the moon, the central index being previously moved towards the second by a quantity equal to twice the measured distance; and make the same two limbs that were before observed coincide exactly, by turning the adjusting screw of the first index; then half the angle shown by the first index will be the angular distance between the observed limbs of the sun and moon. This instrument has of late been greatly improved by Captain Mendoza.
To observe the Angular Distance between the Moon and a Fixed Star or Planet.
I. The star being to the right of the moon.
In this case the star is to be considered as the direct object; and the enlightened limb of the moon's reflected image is to be brought in contact with the star or planet, both by a direct and inverted position of the instrument, exactly in the same manner as described in the last article. If the moon's image is very bright, the lightest tinged glass is to be used.
II. The star being to the left of the moon.
Proceed in the same manner as directed for observing the distance between the sun and moon, the sun being to the right of the moon, using the lightest tinged glass, if necessary.
Sect. IV. Of the Method of determining the Longitude from Observation.
Prob. I. To convert degrees or parts of the equator into time.
Rule. Multiply the degrees and parts of a degree by 4, beginning at the lowest denomination, and the product will be the corresponding time. Observing that minutes multiplied by 4 produce seconds of time, and degrees multiplied by 4 give minutes.
Ex. 1. Let 26° 45' be reduced to time.
\[ \begin{array}{c} 26° 45' \\ \times \quad 4 \\ \hline 1h 47' 0'' = \text{time required.} \end{array} \]
Ex. 2. Reduce 8° 37' to time.
\[ \begin{array}{c} 8° 37' \\ \times \quad 4 \\ \hline 34° 28' \end{array} \]
Corresponding time = 5° 34' 28"
Prob. II. To convert time into degrees.
Rule. Multiply the given time by 10, to which add the half of the product. The sum will be the corresponding degrees.
Ex. Ex. 1. Let 3h 4' 28" be reduced to degrees.
\[ \begin{align*} 3h & \quad 4' \quad 28'' \\ & \quad 10 \\ \hline 30 & \quad 44 \quad 40 \\ \text{Half} & = \quad 15 \quad 22 \quad 20 \\ \end{align*} \]
Corresponding deg.: 46° 7'
Ex. 2. Reduce 8h 42' 36" to degrees.
\[ \begin{align*} 8h & \quad 42' \quad 36'' \\ & \quad 10 \\ \hline 87 & \quad 6 \quad 0 \\ 43 & \quad 33 \quad 0 \\ \end{align*} \]
Answer. 13° 39'
Prob. III. Given the time under any known meridian, to find the corresponding time at Greenwich.
Rule. Let the given time be reckoned from the preceding noon, to which the longitude of the place in time is to be applied by addition or subtraction, according as it is east or west; and the sum or difference will be the corresponding time at Greenwich.
Ex. 1. What time at Greenwich answers to 6h 15' at a ship in longitude 76° 45' W?
Time at ship - 6h 15' Longitude in time - 5° 7W
Time at Greenwich - 11° 22'
Ex. 2. Required the time at Greenwich answering to 5h 46' 39" of May 1st, at Canton, whose longitude is 113° 2' 15" E?
Time at Canton, May 1st, - 5h 46' 39" Longitude in time - 7° 32' 9E
Time at Greenwich, April 30 - 14° 30'
Prob. IV. To reduce the time at Greenwich to that under any given meridian.
Rule. Reckon the given time from the preceding noon, to which add the longitude in time if east, but subtract it if west; and the sum or remainder will be the corresponding time under the given meridian.
Ex. 1. What is the expected time of the beginning of the lunar eclipse of February 25, 1793, at a ship in longitude 109° 48' E?
Beg. of eclipse at Greenwich per Naut. Alm. 9h 23' 45' Ship's longitude in time - 7° 19' 12"
Time of beginning of eclipse at ship, - 16° 42' 57"
Ex. 2. At what time may the immersion of the first satellite of Jupiter be observed at Port St Julian, in longitude 68° 44' W, which, by the Nautical Almanack, happens at Greenwich 24th March 1792, at 12h 53' 1"?
App. time of immersion at Greenwich - 17h 53' 1" Longitude of Port St Julian in time - 4° 34' 56" W.
App. time of immer. at Port St Julian - 13° 18' 5"
Prob. V. To find the equation of equal altitudes.
Rule. To the cotangent of half the interval of time in degrees add the tangent of the latitude, and to the cotangent of half the interval add the tangent of the declination. Now if the latitude and declination be of a contrary name, add the corresponding natural numbers; but if of the same name, subtract them.—Then to the ar. co.log. of this sum or difference add the proportional logarithm of one-fourth of the interval expressed in time, and the proportional logarithm of the daily variation of declination; the sum will be the proportional logarithm of the equation of equal altitudes in minutes and seconds, which are to be esteemed seconds and thirds.
Example. Let the latitude of the place of observation be 37° 9' N, the interval of time between the observations of the equal altitudes 17° 17', the fun's declination 17° 48' S, and the daily change of declination 16' 19" 1': Required the equation of equal altitudes?
Half the interval = 2h 38' 15" = 39° 37'. \[ \frac{1}{2} \text{ int.} = 39° 37' \quad \text{coft.} \quad 0.19542 \quad \text{cotang.} \] Lat. 57° 9 tan. 0.18997 dec. 17° 48' ta. 9.50659
\[ \begin{align*} 0.38539 & \quad 2.4288 \\ & \quad 3879 \\ \end{align*} \]
Sum - 2.8167 ar.co.lo.9.5503
One fourth interval - 1h 10' 15" P.L.O.3563
Daily variation of declination - 16' 19" 1/2 P.L.O.424
Equation of equal altitudes - 20' 14" P.L.O.9490
Prob. VI. To find the error of a watch by equal altitudes of the sun.
Rule. In the morning, when the sun is more than two hours distant from the meridian, let a set of observations be taken, confining, for the sake of greater accuracy, of at least three altitudes, which, together with the corresponding times per watch, are to be wrote regularly, the time of each observation being increased by 12 hours. In the afternoon, observe the instants when the sun comes to the same altitudes, and write down each opposite to its respective altitude. Now half the sum of any two times answering to the same altitude will be the time of noon per watch uncorrected. Find the mean of all the times of noon thus deduced from each corresponding pair of observations, to which the equation of equal altitudes is to be applied by addition or subtraction according as the sun is receding from or approaching to the elevated pole, and the sum or difference will be the time per watch of apparent noon, the difference between which and noon will be the error of the watch for apparent time; and the watch will be fast or slow according as the time of noon thereby is more or less than 12 hours.
Example. January 29, 1786, in lat. 57° 9' N, the following equal altitudes of the sun were observed: Required the error of the watch?
Alt. = 8° 5' Time 21h 35' 8" A.M. 2h 55' 43" P.M.
| Alt. | Time | |------|------| | 8° 10' | 36° 8' | | 8° 20' | 38° 9' | | 8° 25' | 39° 12.5' |
\[ \begin{align*} 37.5 & \quad 4.2 \\ 21 & \quad 37 \quad 9.37 \\ \end{align*} \]
Sum - 24° 30' 50.42
Time of noon per watch uncorrected - 12° 15' 25.2
Equation of equal altitudes - 0° 0' 20.2
Time per watch of apparent noon - 12° 15' 5.
Watch fast - 15' 5". The mean time of noon per watch is found by applying the equation of time with a contrary fine.
In practice it will be found convenient to put the index of the quadrant to a certain division, and to wait till either limb of the sun attains that altitude.
**Prob. VII.** Given the latitude of a place, the altitude and declination of the sun, to find the apparent time, and the error of the watch.
*Ex. 1.* September 15, 1792, in latitude 33° 56' S, and longitude 18° 22' E, the mean of the times per watch was 8h 12' 10'' A.M., and that of the altitudes of the sun's lower limb 24° 48'; height of the eye 24 feet. Required the error of the watch?
| Obs. alt. Sun's lower limb | 24° 48' | |---------------------------|---------| | Semidiameter | + 16.0 | | Dip | - 4.7 | | Correction | - 1.9 |
True altitude Sun's centre = 24° 57.4
| Latitude | 33° 56' | | Declination | 2° 45.4 | | Sum | 36° 41.4 | | Sun's altitude | 24° 57.4 |
| nat. cosine 80188 | | nat. fine 42193 |
Difference 37995 - log. 4.57973
Sun's meridian distance = 3h 48' 51" rising
Apparent time = 8h 11' 9"
Time per watch = 8h 12' 10"
Watch fast = 1'
*Ex. 2.* May 6, 1793, in latitude 56° 4' N, and longitude 38° 30' W, at 4h 37' 4" P.M. per watch the altitude of the sun's lower limb was 25° 6'.1, and height of the eye 18 feet. Required the error of the watch for apparent time?
| Altitude sun's lower limb | = 25° 6.1 | |---------------------------|-----------| | Semidiameter | + 15.9 | | Dip | - 4.1 | | Correction | - 1.9 |
True alt. sun's centre = 25° 16.0
| Latitude | 56° 4.0 N. | | Declination | 16° 49.5 N. |
| Difference | 39° 14.5 | | nat. cosine 77448 | | nat. fine 42683 |
Difference 34765 - log. 4.54114
Apparent time = 4h 38' 12" rising
Time per watch = 4h 37' 4"
Watch slow = 1'
**Prob. VIII.** Given the latitude of a place, the altitude of a known fixed star, and the sun's right ascension, to find the apparent time of observation and error of the watch.
**Rule.** Correct the observed altitude of the star, and reduce its right ascension and declination to the time of observation.
With the latitude of the place, the true altitude and declination of the star, compute its horary distance from the meridian by last problem; which being added to, or subtracted from, its right ascension, according as it was observed in the western or eastern hemisphere, the sum or remainder will be the right ascension of the meridian.
From the right ascension of the meridian subtract the sun's right ascension, as given in the Nautical Almanack. Practice.
Of finding back for the noon of the given day; and the remainder of the longitude at sea will be the approximate time of observation; from which subtract the proportional part of the daily variation of right ascension answering thereto, and let the proportional part answering to the longitude be added or subtracted, according as the longitude is east or west, and the result will be the apparent time of observation; and hence the error of the watch will be known.
Ex. 1. December 12, 1792, in lat. 37° 46' N., and longitude 21° 15' E., the altitude of Arcturus east of the meridian was 34° 6'.4, the height of the eye 10 feet. Required the apparent time of observation?
Observed alt. of Arcturus 34° 6'.4 Dip and refraction - 4° 4'
True altitude - 34° 2.0 Latitude - 37° 46.0 N. sec. 0.10209 Declination - 20° 14.4 N. sec. 0.02778
Difference 17° 31.6 N. co. 95358 Altitude of Arcturus 34° 2.0 N. fine 55968
Difference 393924.59539
Arcturus's merid. dist. 4h 8' 10" rising 4.72526 right af. - 14° 6' 13"
Right af. of merid. 9° 58' 3" Sun's right af. - 17° 21' 59"
Approximate time 16° 36' 4" Eq. to approx. time - 3° 3' Eq. to longitude + 16
Ap. time of obs. 16° 33' 17"
Ex. 2. January 29, 1792, in latitude 53° 24' N., and longitude 25° 18' W., by account, at 14h 58' 38", the altitude of Procyon west of the meridian was 19° 58'; height of the eye 20 feet. Required the error of the watch?
Example. March 3, 1792, in latitude 51° 38' N., at 11h 29' 7" P.M. per watch, the altitude of the moon's lower limb was 37° 31', the height of the eye being 10 feet, and the time at Greenwich 13h 43'. Required the error of the watch?
Altitude of the moon's lower limb = 37° 31' Semidiometer + 15 Dip - 3 Correction + 42 Corrected alt. of moon's centre 38° 25' Latitude - 51° 38' N. secant 0.20712 Declination - 17° 0' N. secant 0.01949
Difference 34° 38' Moon's altitude 38° 25'
Moon's meridian distance 3° 14' 36" rising right ascension 7° 22' 54" Right ascension of meridian 10° 37' 30" Sun's right ascension 23° 2' 0"
Apparent time at ship 11° 35' 30" Time per watch 11° 29' 7"
Watch slow 23
Prob. IX. Given the altitude of the moon, the latitude of a place, and the apparent time at Greenwich; to find the apparent time at the place of observation.
Rule. Correct the altitude of the moon's limb by Problem V. p. 731, and reduce its right ascension and declination, and the sun's right ascension, to the Greenwich time of observation. Now with the latitude of the place, the declination and altitude of the moon, compute its meridian distance as before: Which being applied to its right ascension by addition or subtraction, according as it is in the western or eastern hemisphere, will give the right ascension of the meridian. Then the sun's right ascension subtracted from the right ascension of the meridian, will give the apparent time of observation. **Prob. X.** Given the apparent distance between the moon and the sun or a fixed star, to find the true distance.
**Rule.** To the logarithmic difference answering to the moon's apparent altitude and horizontal parallax, add the logarithmic fines of half the sum, and half the difference of the apparent distance and difference of the apparent altitudes; half the sum will be the logarithmic cosine of an arch: now add the logarithm fines of the sum and difference of this arch, and half the difference of the true altitudes, and half the sum will be the logarithmic cosine of half the true distance.
**Example.** Let the apparent altitude of the moon's centre be $48^\circ 22'$, that of the sun's $27^\circ 43'$, the apparent central distance $81^\circ 23' 40''$, and the moon's horizontal parallax $58' 45''$. Required the true distance?
| Apparent altitude moon's centre | $48^\circ 22'$ | |-------------------------------|---------------| | Correction | + $38 26$ |
| Sun's true altitude | $27 41 20$ | | Sun's apparent altitude | $27 43$ | | Moon's apparent altitude | $48 22$ | | Difference | $21 19 6$ | | Half Logarithmic difference | $10 39 33$ | | Apparent distance | $81 23 40$ | | Sum | $102 2 40$ | | Difference | $60 44 40$ | | Half difference true altitudes | $10 39 33$ | | Arch | $51 27 29$ | | Sum | $62 7 2$ | | Difference | $49 47 56$ | | True distance | $81 4 32$ |
**Prob. XI.** To find the time at Greenwich answering to a given distance between the moon and the sun, or one of the stars, used in the Nautical Almanack.
**Rule.** If the given distance is found in the Nautical Almanack opposite to the given day of the month, or to that which immediately precedes or follows it, the time is found at the top of the page. But if this distance is not found exactly in the ephemeris, subtract the prop. log. of the difference between the distances which immediately precede and follow the given distance from the prop. log. of the difference between the given and preceding distances; the remainder will be the prop. log. of the excess of the time corresponding to the given distance, above that answering to the preceding distance: And hence the apparent time at Greenwich is known.
**Example.** September 21, 1792, the true distance between the centres of the sun and moon was $68^\circ 13' 8''$. Required the apparent time at Greenwich?
| Given distance | $65^\circ 13' 5''$ | |----------------|------------------| | Dist. at ix. hours | $67 53 27$ | | Diff. = $-0^\circ 19' 41''$ | P. log. $0612$ | | Dist. at xii. hours | $69 30 6$ | | Diff. = $+1^\circ 36' 39''$ | P. log. $2701$ |
| Excess | $0 36 39$ | P. log. $0911$ | | Preceding time | $9 0 0$ |
App. time at Greenwich | $9 36 39$ |
**Ex. I.** March 17, 1792, in latitude $34^\circ 53' N$, and longitude by account $27^\circ W$, about 9h. A. M. the distance between the nearest limbs of the sun and moon was $68^\circ 3' \frac{1}{2}$; the altitude of the sun's lower limb $33^\circ 18'$; that ### NAVIGATION
**Of finding the Longitude at Sea by Lunar Observations.**
| Time at ship | 9h 0' A.M. | |--------------|------------| | Longitude in time | 1° 48' |
| Reduced time | 10° 48' A.M. | |--------------|-------------| | Altitude moon's upper limb | 31° 3' 0" | | Aug. semidiameter | -16° 10' | | Dip | -3° 18' | | Apparent altitude | 30° 43' 23" | | Correction | +49° 26' | | Moon's true altitude | 31° 32' 49" |
| Sun's apparent altitude | 33° 30' 48" | | Moon's apparent altitude | 30° 43' 23" | | Difference | 2° 47' 25" | | Apparent distance | 68° 35' 40" |
| Sum | 71° 23' 5" | | Difference | 65° 48' 15" |
| Half difference true altitudes | 0° 58' 20" | | Arch | 55° 54' 12" | | Sum | 56° 52' 32" | | Difference | 54° 55' 52" |
| Half true distance | 34° 6' 53" |
| True distance | 68° 13' 46" | | Distance at XXI hours | 69° 11' 20" | | Distance at noon | 67° 32' 38" |
| Proportional part | 1° 45' 0" | | Preceding time | 21° 0' 0" |
| Apparent time at Greenwich | 22° 45' 0" | | Latitude | 34° 53' .0 N | | Declination | 0° 57' .9 S | | Secant | 0.08602 | | Secant | 0.00006 |
| Sum | 35° 50' .9 | | Nat. cosine | 81057 | | Sun's altitude | 33° 29' .5 | | Nat. sine | 55181 |
| Difference | 25° 76' | | Time from noon | 3h 7' 13" |
| Apparent time | 20° 52' 47" | | App. time at Green. | 22° 45' 0" |
| Longitude in time | 1° 52' 13" = 28° 3' W. |
| Diff. sun and moon's nearest limbs | 68° 3' 15" | | Sun's semidiameter | +16° 6' | | Moon's semidiameter | +16° 10' | | Augmentation | +0° 9' | | Apparent central distance | 68° 35' 40" | | Altitude sun's lower limb | 33° 18' | | Sun's semidiameter | +16° 6' | | Dip | -3° 18' | | Sun's apparent altitude | 33° 30' 48" | | Correction | -0° 1' 19" | | Sun's true altitude | 33° 29' 29" | | Moon's true altitude | 31° 32' 49" |
| Difference | 1° 56' 40" | | Half | 0° 58' 20" |
| Logarithmic difference | 9.996336 | | Half | 35° 41' 32" | | Half | 32° 54' 7" | | Sine | 9.765991 | | Sine | 9.734964 | | Cofine | 19.497291 | | Cofine | 9.748645 | | Sine | 9.922977 | | Sine | 9.912998 | | Cofine | 19.835975 | | Cofine | 9.917987 |
| Difference | C° 57' 34" | | P. log. | 4951 | | Difference | 1° 38' 42" | | P. log. | 2610 | | Per log. | 2341 |
| Rising | 4.41291 | | Rising | 4.49899 |
Ex. 2. Example 2. September 2, 1792, in latitude $13^\circ 57'$ N, and longitude by account $56^\circ$ E, several observations of the moon and Altair were taken; the mean of the times per watch was $1^h 18' 59''$ A.M. that of the distance between Altair and the moon's nearest limb $58^\circ 45' 26''$; the mean of the altitude of the moon's lower limb $70^\circ 33'$; and that of Altair $25^\circ 27'$. height of the eye 13 feet. Required the true longitude?
| Time per watch | $1^h 18' 59''$ A.M | |----------------|-------------------| | Longitude in time | $3^\circ 44'$ |
| Distance moon and Altair | $58^\circ 45' 26''$ | | Augmented semidiameter | $+0^\circ 16' 28''$ |
| Reduced time | $9^\circ 34' 59''$ | | Apparent central distance | $59^\circ 1' 54''$ | | Altitude moon | $70^\circ 33'$ | | Altitude of Altair | $25^\circ 27' 4''$ | | Semidiameter and dip | $-0^\circ 13'$ | | Dip | $-0^\circ 3' 4''$ |
| Apparent alt. moon | $70^\circ 20'$ | | Apparent altitude Altair | $25^\circ 24'$ | | Correction | $+0^\circ 19' 40''$ | | Refraction | $-0^\circ 2' 0''$ |
| True altitude moon | $70^\circ 39' 40''$ | | True altitude Altair | $25^\circ 22'$ | | Moon's apparent alt. | $70^\circ 20'$ | | Moon's true altitude | $70^\circ 39' 40''$ |
| Altair's apparent alt. | $25^\circ 24'$ | | Difference | $45^\circ 17' 40''$ |
| Difference | $44^\circ 56'$ | | Apparent distance | $59^\circ 1' 54''$ | | Logarithmic difference | $9.993101$ |
| Sum | $103^\circ 57' 54''$ | | Half | $51^\circ 58' 57''$ | | Sine | $9.806428$ | | Difference | $14^\circ 5' 54''$ | | Half | $7^\circ 2' 57''$ | | Sine | $9.889919$ |
| Half diff. true alt. | $22^\circ 38' 50''$ | | Arch | $72^\circ 1' 57''$ | | Cosine | $9.489224$ |
| Sum | $94^\circ 40' 47''$ | | Difference | $49^\circ 23' 7''$ | | Sine | $9.998548$ | | Sine | $9.880301$ |
| Half true distance | $29^\circ 33' 48''$ | | Cofine | $9.939424$ |
| True distance | $59^\circ 7' 37''$ | | Distance at IX hours | $58^\circ 51' 17''$ | | Difference | $0^\circ 16' 20''$ | | P. log. | $1.0422$ | | at XII hours | $60^\circ 24' 34''$ | | Difference | $1^\circ 33' 17''$ | | P. log. | $0.2855$ |
| Proportional part | $0^\circ 31' 31''$ | | Preceding time | $9^\circ 0' 0''$ | | P. log. | $0.7567$ |
| Apparent time at Greenwich | $9^\circ 31' 31''$ | | Latitude | $13^\circ 57' N$ | | Secant | $0.01300$ | | Declination | $8^\circ 19.8' N$ | | Secant | $0.00461$ |
| Difference | $5^\circ 37.2'$ | | Nat. cofine | $99519$ | | Altitude Altair | $25^\circ 22'$ | | Nat. fine | $42841$ |
| Difference | $56678$ | | Altair's meridian distance | $4^\circ 23' 14''$ | | Right ascension | $19^\circ 40' 40''$ | | Rifing | $4.77102$ |
| Right ascension meridian | $0^\circ 3' 54''$ | | Sun's right ascension | $10^\circ 46' 17''$ |
| Apparent time at ship | $13^\circ 17' 37''$ | | Apparent time at Greenwich | $9^\circ 31' 31''$ |
| Longitude in time | $4^\circ 46' 6'' = 56^\circ 31' \frac{1}{2} E$ |
For various other methods of determining the longitude of a place, the reader is referred to Dr Mackay's Treatise on the Theory and Practice of finding the Longitude at Sea or Land.
Chap. **NAVIGATION**
**Practice.**
**Variation of the Compass.**
**CHAP. III. Of the Variation of the Compass.**
The variation of the compass is the deviation of the points of the mariner's compass from the corresponding points of the horizon; and is denominated east or west variation, according as the north point of the compass is to the east or west of the true north point of the horizon.
A particular account of the variation, and of the several instruments used for determining it from observation, may be seen under the articles Azimuth, Compass, and Variation; and for the method of communicating magnetism to compass needles, see Magnetism.
**Prob. I.** Given the latitude of a place, and the sun's magnetic amplitude, to find the variation of the compass.
**Rule.** To the log. secant of the latitude, add the log. sine of the sun's declination, the sum will be the log. cosine of the true amplitude; to be reckoned from the north or south according as the declination is north or south.
The difference between the true and observed amplitudes, reckoned from the same point, and if of the same name, is the variation; but if of a different name, their sum is the variation.
If the observation be made in the eastern hemisphere, the variation will be east or west according as the observed amplitude is nearer to or more remote from the north than the true amplitude. The contrary rule holds good in observations taken in the western hemisphere.
**Ex. 1.** May 15, 1794, in latitude 33° 10' N, longitude 18° W, about 5th A.M., the sun was observed to rise E by N. Required the variation?
Sun's decl. May 15, at noon 18° 58' N.
Equation to 7th from noon - 4
to 18° W + 1
Reduced declination 18° 55' Sine 9.51080
Latitude 33° 10' Secant 0.07723
True amplitude N 67° 13' E Cosine 9.58803
**Ex. 2.** December 20, 1793, in latitude 31° 38' S, longitude 83° W, the sun was observed to set SW. Required the variation?
Latitude 31° 38' Secant 0.06985
Declination 23° 28' Sine 9.60012
True amplitude S 62° 7' W Cosine 9.66997
Observed ampl. S 45° W
Variation 17° 7' which is east, as the observed amplitude is farther from the north than the true amplitude, the observation being made at unfavouring.
It may be remarked, that the sun's amplitude ought to be observed at the instant the altitude of its lower limb is equal to the sum of 15 minutes and the dip of the horizon. Thus, if an observer be elevated 18 feet above the surface of the sea, the amplitude should be taken at the instant the altitude of the sun's lower limb is 19 minutes.
**Prob. II.** Given the magnetic azimuth, the altitude and declination of the sun, together with the latitude of the place of observation; to find the variation of the compass.
**Rule.** Reduce the sun's declination to the time and place of observation, and compute the true altitude of the sun's centre.
Find the sum of the sun's polar distance and altitude and the latitude of the place, take the difference between the half of this sum and the polar distance.
To the log. secant of the altitude add the log. secant of the latitude, the log. cosine of the half sum, and the log. cosine of the difference; half the sum of these will be the log. sine of half the sun's true azimuth, to be reckoned from the south in north latitude, but from the north in south latitude.
The difference between the true and observed azimuths will be the variation as formerly.
**Ex. 1.** November 18, 1793, in latitude 50° 22' N, longitude 24° 30' W, about three quarters past eight A.M., the altitude of the sun's lower limb was 8° 10', and bearing per compass S, 23° 18' E; height of the eye 20 feet.
Required the variation of the compass?
Sun's decl. 18th Nov. at noon 19° 25' S
Equation to 3rd from noon - 2
to 24° 30' W + 1
Reduced declination 19° 24'
Polar distance 109° 24'
Altitude 8° 16'
Latitude 50° 22'
Sum 168° 2
Half 84° 1
Difference 25° 23'
Half true azimuth 22° 43'
Observed alt. sun's lower limb = 8° 10'
Semidiameter + 16
Dip and refraction - 10
True altitude 8° 16'
Secant 0.00454
Secant 0.19527
Cosine 9.01803
Cosine 9.95591
Sine 19.17375
Sine 9.58687 Half true azimuth 22° 43' 2 True azimuth S 45° 26' E. Observed azimuth S 23° 18' E.
Variation 22° 8' W.
Ex. 2. January 3, 1794, in latitude 33° 52' N, 53° 15' E longitude, about half past three the altitude of the sun's lower limb 41° 18', and azimuth S 50° 25' W, the height of the eye being 20 feet. Required the variation?
Sun's declination at noon 21° 24' S. Equation to time from noon 2 to longitude + 2
Reduced declination 21° 24' S.
Polar distance 111° 24' Altitude 41° 28' Latitude 33° 52'
Sum 186° 44' Half 93° 22' Difference 18° 2'
True altitude 41° 28'
Secant 0.12532 Secant 0.08075
Cofine 8.76883 Cofine 9.97558
Sine 18.95048 Sine 2.47524
True azimuth S 34° 46' W. Observed azimuth S 50° 25' W.
Variation 15° 39' W.
Chap. IV. Of a Ship's Journal.
A journal is a regular and exact register of all the various transactions that happen aboard a ship whether at sea or land, and more particularly that which concerns a ship's way, from whence her place at noon or any other time may be justly ascertained.
That part of the account which is kept at sea is called sea work; and the remarks taken down while the ship is in port are called harbour work.
At sea, the day begins at noon, and ends at the noon of the following day: the first 12 hours, or those contained between noon and midnight, are denoted by P.M. signifying after mid day; and the other 12 hours, or those from midnight to noon, are denoted by A.M. signifying before mid day. A day's work marked Wednesday March 6. began on Tuesday at noon, and ended on Wednesday at noon. The days of the week are usually represented by astronomical characters. Thus O represent Sunday; D Monday; T Tuesday; W Wednesday; Th Thursday; F Friday; and S Saturday.
When a ship is bound to a port so situated that she will be out of sight of land, the bearing and distance of the port must be found. This may be done by Mercator's or Middle-latitude Sailing; but the most expeditious method is by a chart. If islands, capes, or headlands intervene, it will be necessary to find the several courses and distances between each successively. The true course between the places must be reduced to the course per compass, by allowing the variation to the right or left of the true course, according as it is west or east.
At the time of leaving the land, the bearing of some known place is to be observed, and its distance is usually found by estimation. As perhaps the distance thus found will be liable to some error, particularly in hazy or foggy weather, or when that distance is considerable, it will therefore be proper to use the following method for this purpose.
Let the bearing be observed of the place from which the departure is to be taken; and the ship having run a certain distance on a direct course, the bearing of the same place is to be again observed. Now having one side of a plain triangle, namely the distance sailed, and all the angles, the other distances may be found by Prob. I. of Oblique Sailing.
The method of finding the course and distance sailed in a given time is by the compass, the log-line, and half-minute-glass. These have been already described. In the royal navy, and in ships in the service of the East India Company, the log is hove once every hour; but in most other trading vessels only every two hours.
The several courses and distances sailed in the course of 24 hours, or between noon and noon, and whatever remarks are thought worthy of notice, are set down with chalk on a board painted black, called the log-board, which is usually divided into five columns: the first column on the left hand contains the hours from noon to noon; the second and third the knots and parts of a knot sailed every hour, or every two hours, according as the log is marked; the fourth column contains the courses steered; the fifth the winds; and in the sixth the various various remarks and phenomena are written. The log-board is transferred every day at noon into the log-book, which is ruled and divided after the same manner.
The courses steered must be corrected by the variation of the compass and leeway. If the variation is west, it must be allowed to the left hand of the course steered; but if east, to the right hand, in order to obtain the true course. The leeway is to be allowed to the right or left of the course steered, according as the ship is on the larboard or starboard tack. The method of finding the variation, which should be determined daily if possible, is given in the preceding chapter; and the leeway may be understood from what follows.
When a ship is close hauled, that part of the wind which acts upon the hull and rigging, together with a considerable part of the force which is exerted on the sails, tends to drive her to the leeward. But since the bow of a ship exposes less surface to the water than her side, the resistance will be less in the first case than in the second; the velocity in the direction of her head will therefore in most cases be greater than the velocity in the direction of her side; and the ship's real course will be between the two directions. The angle formed between the line of her apparent course and the line she really describes through the water is called the angle of leeway, or simply the leeway.
There are many circumstances which prevent the laying down rules for the allowance of leeway. The construction of different vessels, their trim with regard to the nature and quantity of their cargo, the position and magnitude of the sail set, and the velocity of the ship, together with the swell of the sea, are all susceptible of great variation, and very much affect the leeway. The following rules, are, however, usually given for this purpose:
1. When a ship is close hauled, has all her sails set, the water smooth, with a light breeze of wind, she is then supposed to make little or no leeway. 2. Allow one point when the top-gallant sails are handed. 3. Allow two points when under close reefed topsails. 4. Allow two points and a half when one top-sail is handed. 5. Allow three points and a half when both top-sails are handed. 6. Allow four points when the fore course is handed. 7. Allow five points when under the main-sail only. 8. Allow six points when under balanced mizen. 9. Allow seven points when under bare poles.
These allowances may be of some use to work up the day's work of a journal which has been neglected; but a prudent navigator will never be guilty of this neglect.
A very good method of estimating the leeway is to observe the bearing of the ship's wake as frequently as may be judged necessary; which may be conveniently enough done by drawing a small semicircle on the taffrail, with its diameter at right angles to the ship's length, and dividing its circumference into points and quarters. The angle contained between the semidiameter which points right aft, and that which points in the direction of the wake, is the leeway. But the best and most rational way of bringing the leeway into the day's log is to have a compass or semicircle on the taffrail, as before described, with a low crutch or swivel in its centre; after heaving the log, the line may be slipped into the crutch just before it is drawn in, and the angle it makes on the limb with the line drawn right aft will show the leeway very accurately; which as a necessary article, ought to be entered into a separate column against the hourly distance on the log-board.
In hard blowing weather, with a contrary wind and a high sea, it is impossible to gain any advantage by failing. In such cases, therefore, the object is to avoid as much as possible being driven back. With this intention it is usual to lie to under no more sail than is sufficient to prevent the violent rolling which the vessel would otherwise acquire, to the endangering her masts, and straining her timbers, &c. When a ship is brought to, the tiller is put close over to the leeward, which brings her head round to the wind. The wind having then very little power on the sails, the ship loses her way through the water; which ceasing to act on the rudder, her head falls off from the wind, the sail which has set fills, and gives her fresh way through the water; which acting on the rudder brings her head again to the wind. Thus the ship has a kind of vibratory motion, coming up to the wind and falling off from it again alternately. Now the middle point between those upon which the comes up and falls off is taken for her apparent course, and the leeway and variation is to be allowed from thence, to find the true course.
The setting and drift of currents, and the heave of the sea are to be marked down. These are to be corrected by variation only.
The computation made from the several courses corrected as above, and their corresponding distances, is called a day's work; and the ship's place as deduced therefrom, is called her place by account, or dead reckoning.
It is almost constantly found that the latitude by account does not agree with that by observation. From an attentive consideration of the nature and form of the common log, that its place is alterable by the weight of the line, by currents, and other causes, and also the errors to which the course is liable, from the very often wrong position of the compass in the binnacle, the variation not being well ascertained, an exact agreement of the latitudes cannot be expected.
When the difference of longitude is to be found by dead reckoning, if then the latitudes by account and observation disagree, several writers on navigation have proposed to apply a conjectural correction to the departure or difference of longitude. Thus, if the course be near the meridian, the error is wholly attributed to the distance, and the departure is to be increased or diminished accordingly: if near the parallel, the course only is supposed to be erroneous; and if the course is towards the middle of the quadrant, the course and distance are both assumed wrong. This last correction will, according to different authors, place the ship upon opposite sides of her meridian by account. As these corrections are, therefore, no better than guessing, they should be absolutely rejected.
If the latitudes are not found to agree, the navigator ought to examine his log-line and half-minute-glass, and correct the distance accordingly. He is then to consider if the variation and leeway have been properly ascertained; if not, the courses are to be again corrected, and and no other alteration whatever is to be made on them. He is next to observe if the ship's place has been affected by a current or heave of the sea, and to allow for them according to the best of his judgment. By applying these corrections, the latitudes will generally be found to agree tolerably well; and the longitude is not to receive any farther alteration.
It will be proper, however, for the navigator to determine the longitude of the ship from observation as often as possible; and the reckoning is to be carried forward in the usual manner from the last good observation; yet it will perhaps be very satisfactory to keep a separate account of the longitude by dead reckoning.
General Rules for working a Day's Work.
Correct the several courses for variation and leeway; place them, and the corresponding distances, in a table prepared for that purpose. From whence, by Traverse Sailing, find the difference of latitude and departure made good; hence the corresponding course and distance, and the ship's present latitude, will be known.
Find the middle latitude at the top or bottom of the Traverse Table, and the distance, answering to the departure found in a latitude column, will be the difference of longitude: Or, the departure answering to the course made good, and the meridional difference of latitude in a latitude column, is the difference of longitude. The sum, or difference of which, and the longitude left, according as they are of the same or of a contrary name, will be the ship's present longitude of the same name with the greater.
Compute the difference of latitude between the ship and the intended port, or any other place whose bearing and distance may be required: find also the meridional difference of latitude and the difference of longitude. Now the course answering the meridional difference of latitude found in a latitude column, and the difference of longitude in a departure column, will be the bearing of the place, and the distance answering to the difference of latitude will be the distance of the ship from the proposed place. If these numbers exceed the limits of the Table, it will be necessary to take aliquot parts of them; and the distance is to be multiplied by the number by which the difference of latitude is divided.
It will sometimes be necessary to keep an account of the meridian distance, especially in the Baltic or Mediterranean trade, where charts are used in which the longitude is not marked. The meridian distance on the first day is that day's departure; and any other day it is equal to the sum or difference of the preceding day's meridian distance and the day's departure, according as they are of the same or of a contrary denomination. ### NAVIGATION
**A Journal of a Voyage from London to Funchal in Madeira, in his Majesty's Ship the Resolution, A—— M—— Commander, anno 1793.**
| Days of month | Winds | Remarks on board his Majesty's ship Resolution, 1793 | |---------------|-------|-----------------------------------------------------| | 2 Sept. 28 | SW | Strong gales and heavy rain. At 3 P.M. sent down topgallant yards; at 11 A.M. the pilot came on board. | | 3 Sept. 29 | SW | Moderate and cloudy, with rain. At 10 A.M. cast loose from the sheer hulk at Deptford; got up topgallant yards, and made sail down the river. At noon running through Blackwall reach. | | 4 Sept. 30 | SW Variable | The first part moderate, the latter squally, with rain. At half past one anchored at the Galleons, and moored ship with near a whole cable each way in 5 fathoms; a quarter of a mile off shore. At 3 A.M. strong gales: got down topgallant yards. A.M. the people employed working up junk. Bent the sheet cable. | | 5 Octob. 1 | SSW SW | Fresh gales and squally. P.M. received the remainder of the boatswain's and carpenter's stores on board. The clerk of the cheque mustered the ship's company. | | 6 Octob. 2 | Variable | Variable weather with rain. At noon weighed and made sail; at 5 anchored in Long-reach in 8 fathoms. Received the powder on board. At 6 A.M. weighed and got down the river. At 10 A.M. past the Nore; brought too and hoisted in the boats: double reefed the topsails, and made sail for the Downs. At noon running for the flats of Margate. | | 7 Octob. 3 | NEbE N | First part stormy weather; latter moderate and clear. At 4 P.M. got through Margate Roads. At 5 run through the Downs; and at 6 anchored in Dover Road, in 10 fathoms muddy ground. Dover Castle bore north, and the South Foreland NEbE off shore 1¼ miles. Discharged the pilot. Employed making points, &c., for the sails. Scaled the guns. | | 8 Octob. 4 | NNE | Moderate and fair. Employed working up junk. Received from Deal a cutter of 17 feet, with materials. A.M. strong gales and squally, with rain; got down topgallant yards. |
| Hours | Kn. | Fa. | Courses | Winds | Remarks, 2 Octob. 5, 1793 | |-------|-----|-----|---------|-------|--------------------------| | 1 | | | | NNE | Fresh gales with rain. | | 2 | | | | | Hove short. | | 3 | | | | | Weighed and made sail. | | 4 | | | WSW | | Shortened sail.—Dungeness light NEbE. | | 5 | | | WNW | NE | Fresh breezes and cloudy. | | 6 | | | | | Ditto weather. | | 7 | | | | | Got up topgallant yards. | | 8 | | | | | Set studding sails. | | 9 | | | | | Ditto weather. | | 10 | | | | | St Alban's Head NE. |
---
*Vol. XIV. Part II.* ### Navigation
**A Journal from England towards Madeira**
| Hours | Kn. | Fa. | Course | Winds | Remarks | |-------|-----|-----|--------|-------|---------| | 1 | 8 | | WbN | NE | A fresh steady gale. | | 2 | 8 | | | | Do. weather. | | 3 | 8 | | | | Spoke the Ranger of London, from Carolina. Took in studding sails. | | 4 | 8 | | | | Do. weather. | | 5 | 8 | | | | Eddystone light NbW. | | 6 | 8 | | | | Do. weather. | | 7 | 8 | | | | Eddystone light NE. | | 8 | 8 | | | | Do. weather. | | 9 | 8 | | | | Set lower studding sails. | | 10 | 8 | | | | Fresh breeze and clear weather. | | 11 | 8 | | | | Do. weather. |
**N. Latitude by**
| Course | Dist. | D.L. Dep. | Acc. | Obf. | D. Long. | Acc. | Obf. | W. Lon. by | W. Var. | |--------|-------|-----------|------|------|----------|------|------|------------|---------| | S 52° 5' W. | 93 | 57 | 74 | 49° 11' | 49° 9' | 114° W | 6° 18' | 2½ pts. |
As there is no land in sight this day at noon, and from the course and distance run since the last bearing of the Eddystone light was taken, it is not to be supposed that any part of England will be seen, the departure is therefore taken from the Eddystone; and the distance of the ship from that place is found by resolving an oblique-angled plane triangle, in which all the angles are given, and one side, namely, the distance run (16 miles) between the observations. Hence the distance of the Eddystone at the time the last bearing of the light was taken will be found equal to 18 miles; and as the bearing of the Eddystone from the ship at that time was NE, the ship's bearing from the Eddystone was SE. Now the variation 2½ points W, being allowed to the left of SW, gives SSW ½ W, the true course. The other courses are in like manner to be corrected, and inserted in the following table, together with their respective distances, beginning at 10 o'clock A.M., the time when the last bearing of the Eddystone was taken. The difference of latitude, departure, course, and distance made good, are to be found by Traverse Sailing.
| Courses | Dist. | Diff. of Lat. | Departure | |---------|-------|--------------|-----------| | SSW ½ W | 18 | | | | WSS ½ S | 22 | | | | SW ½ W | 58 | | | | S 52° 5' W | 93 | 56.9 = 57m. | 74° 0' |
Latitude by account - 49 11 N. Sum - 99 19 Middle latitude - 49 40
Now to middle latitude as a course, and the departure 74m. in a latitude column, the difference of long. in a distance column is 114 = 1° 54' W. Longitude of Eddystone - 4 24 W. Longitude in by account - 6 18 W. ### NAVIGATION
**A Journal from England towards Madeira**
| Hours | Kn. | Fa. | Courses | Winds | Remarks, D October 7, 1793 | |-------|-----|-----|---------|-------|---------------------------| | 1 | 6 | 5 | WSW | NE | Fresh breezes. | | 2 | 6 | 5 | | | Sounded 62; fine sand. | | 3 | 6 | 3 | | | Moderate and cloudy. | | 4 | 5 | 3 | | | Unbent the cables, and coiled them. | | 5 | 5 | | N | | Took in fludding sails. | | 6 | 5 | | | | | | 7 | 5 | | | | | | 8 | 4 | 7 | | | Do. weather. | | 9 | 4 | 5 | | | | | 10 | 4 | 5 | | | | | 11 | 4 | | | | | | 12 | 4 | | | | | | 1 | 4 | | | | | | 2 | 4 | | | | | | 3 | 4 | | | | | | 4 | 4 | | | | | | 5 | 3 | | SWNW | NW | Light breeze. | | 6 | 3 | | | | A sail SbE. | | 7 | 3 | | | | Hazy weather. | | 8 | 3 | | | | | | 9 | 3 | | SW | Var. | | | 10 | 3 | | | | | | 11 | 3 | | | | | | 12 | 2 | | | | |
The courses being corrected for variation, and the distances summed up, the work will be as under.
| Course | Diff. | D.L. | Dep. | N. Latitude by Acc. | D. Long. by Acc. | W. Long. by Acc. | W. Var. by acc. | Porto Sancto's Bearing | Distance | |--------|-------|------|------|---------------------|-----------------|-----------------|------------------|------------------------|----------| | S. 38°W. | 99 | 78 | 92 | 47° 51' | 93° W. | 7° 51' | 2½ pts. | S 23° ½ W. | 974 m. |
It is now necessary to find the bearing and distance of the intended port, namely, Funchal; but as that place is on the opposite side of the island with respect to the ship, it is therefore more proper to find the bearing of the east or west end of Madeira; the east end is, however, preferable. But as the small island of Porto Sancto lies a little to the NE of the east end of Madeira, it therefore seems more eligible to find the bearing and distance of that island.
To find the bearing and distance of Porto Sancto.
**Latitude of ship** 47° 51'N. **Mer. parts** 3278 **Lat. of Porto Sancto** 32° 58'N. **Mer. parts** 2097 **Longitude of ship** 7° 51'W. **Lon. of Porto Sancto** 16° 25'W. **Difference of latitude** 14° 53' = 893. **M. D. Lat.** 118° **Difference of long.** 8° 34' = 514.
The course answering to the meridional difference of latitude and difference of longitude is about 23° 5', and the distance corresponding to the difference of latitude is 974 miles. Now as Porto Sancto lies to the southward and westward of the ship, the course is therefore S 23° ½ W.; and the variation, because W., being allowed to the right hand, gives SW¼W nearly, the bearing upon compasses; and which is the course that ought to be steered. ### NAVIGATION
**A Journal from England towards Madeira**
| Hours | Kn. | Fa. | Course | Winds | Remarks, 8 October 8, 1793 | |-------|-----|-----|--------|-------|---------------------------| | 1 | 2 | | SW | NW Variable | Little wind and cloudy. | | 2 | 1 | | | | Tried the current, and found none. | | 3 | | | Ship's head to the SW | Calm. | | 4 | | | | | Calm; a long swell from the SW. | | 5 | | | Ship's head from SW to SSE WSW | S | | 6 | | | | | Light airs and hazy. | | 7 | | | | | Moderate wind and cloudy. | | 8 | | | | | Set top-gallant sails. | | 9 | | | | | By double altitudes of the sun, the latitude was found to be 47° 28' N. |
| Course | Diff. D.L. Dep. | N. Latitude by Acc. | Obf. | D. Long. | W. Long. by Acc. | Obf. | W. Var. | Porto Sancto's Bearing | Distance | |--------|-----------------|---------------------|------|----------|-----------------|------|---------|-----------------------|----------| | S 61° W | 51 25 45 | 47° 28' | 47° 28' | 67° W. | 8° 38' | 2 points. | S 21° W | 932 |
The several courses corrected will be as under.
| Course | Diff. | Diff. of Latit. | Departure | |--------|-------|----------------|-----------| | | | N. | S. | E. | W. | | SSW | 3 | 2.8 | 1.1 | | SW | 23 | 9.2 | 9.2 | | WSW | 22 | 8.4 | 20.3 | | WbSWS | 15 | 4.4 | 14.4 | | S 61° W | 51 | 24.8 = 25 | 45.0 |
Latitude by account 47° 26' Sum - - - 77 Middle latitude 47° 39'
To middle latitude 37° 3', and departure 45' in a latitude column, the difference of longitude in a distance column is 67' = 1° 7' W.
Yesterday's longitude 7° 31' W.
Longitude in by account 8° 58' W.
To find the bearing and distance of Porto Sancto.
| Latitude of ship | 47° 28' N. | Mer. parts | 3244 | Longitude | 8° 58' W. | |------------------|------------|------------|------|-----------|------------| | Lat. of Porto Sancto | 32° 58' N. | Mer. parts | 2097 | Longitude | 16° 25' W. |
Difference of latitude 14° 30' = 870 M. D. lat. 1147 D. longitude 7° 27' = 447'.
Hence the bearing of Porto Sancto is S 21° W., and distance 932 miles. The course per compass is therefore SW nearly. ### NAVIGATION
**A Journal from England towards Madeira**
| Hours | Kn. | Fa. | Courses | Winds | Remarks, Oct 9, 1793 | |-------|-----|-----|---------|-------|---------------------| | 1 | 5 | | WbN | SWbS | Squally with rain. | | 2 | 5 | | | | Handed top-gallant fails. | | 3 | 5 | | | | In first reef topfairs. | | 4 | 5 | 4 | SEbS | | Dark gloomy weather. Tacked ship. | | 5 | 5 | 6 | | | In 2d reef topfairs, and down top gallant yards. | | 6 | 5 | | | | Stormy weather; in fore and mizen top-fails and 3d reef main top-fail. Handed the main top-fail, bent the main stay-fail, and brought to with it and the mizen; reefed the mainail; at 10, wore and lay to under the mainail, got down top-gallant masts; at 12 set the foresail, and balanced the mizen. | | 7 | 4 | | | | The sea stove in several half ports. | | 8 | 4 | | | | The swell abates a little. | | 9 | 3 | | | | The swell abates fast. | | 10 | 3 | | | | Up top-gallant masts. | | 11 | 3 | | | | Set the top-fails. | | 12 | 5 | | | | Clear weather; good observation. |
There is no leeway allowed until two o'clock P.M., when the top-gallant sails are taken in; from 2 to 3 one point is allowed; from 3 to 6, one and a half points are allowed; from 6 to 8, one and three-fourth points are allowed; from 8 to 9, three points; from 9 to 10, four and a half points; from 10 to 12, five points; from 12 to 10 A.M., three and a half points; and from thence to noon two points leeway are allowed. Now the several courses being corrected by variation and leeway will be as under; but as the corrected courses from 2 to 3 P.M. and from 10 to 12 A.M. are the same, namely, west; this, therefore, is inserted in the table, together with the sum of the distances, as a single course and distance. In like manner the courses from 12 to 2, and from 5 to 8, being the same, are inserted as a single course and distance.
| Courses | Diff. | N. Latitude by Acc. | D. Long. by Acc. | W. Long by Acc. | W. Var. | Porto Sancto's Bearing. | Distance. | |---------|-------|---------------------|-----------------|-----------------|--------|------------------------|-----------| | WbN½N | 43 | 12 | 41 | 47° 40' | 47° 39' | 61' | 9° 59' | 2 points |
### Table of Courses, Distances, and Departures
| Courses | Diff. | N. | S. | E. | W. | |---------|-------|----|----|----|----| | WbS | 10 | 2.0| | | 9.8| | W | 15.5 | | | | 15.5| | W½N | 5.4 | 0.5| | | 5.4 | | Eb½S | 10.0 | | 3.1| | 10.1| | Eb½S | 8 | | 1.9| | 7.8 | | E | 3 | | | 3.0| | | NEbE | 1 | 0.6| | | 0.8 | | NW½W | 2 | | 1.1| | 1.7 | | NW½W½W | 17.2 | | 8.1| | 15.2| | NW½W | 11 | | 7.0| | 8.5 | | WbN½N | 7.4 | | 2.1| | 7.1 | | WbN½N | 43 | 12.4| | | 41.5|
*Yeast latitude 47° 28' N.*
Lat. by account 47° 40' N.
To middle latitude 37° 34', and departure 41.5 the difference of longitude is 61° = 1° 1' W.
Yesterday's longitude 8° 58' W.
Longitude in by account 9° 59' W. ### Navigation
**A Journal from England towards Madeira**
| Hours | Kn. | Fa. | Courses | Winds | Remarks, 2 October 10, 1793 | |-------|-----|-----|---------|-------|---------------------------| | 1 | 5 | 3 | W | SSW | Fresh gales with rain. | | 2 | 5 | 7 | | | Do. weather. | | 3 | 6 | | | | Out 3d reef topsails. | | 4 | 6 | | | | Loft a log and line. | | 5 | 6 | | | | Do. weather. | | 6 | 6 | | | | | | 7 | 5 | 6 | | | | | 8 | 5 | 4 | | | | | 9 | 5 | 2 | | | | | 10 | 5 | | | | | | 11 | 5 | | | | | | 12 | 5 | | | | | | 1 | 5 | 5 | | | | | 2 | 5 | | | | | | 3 | 5 | | | | | | 4 | 4 | | | | | | 5 | 4 | | | | | | 6 | 4 | | | | | | 7 | 4 | 3 | | | | | 8 | 4 | 4 | SWbW | SSE | Moderate and cloudy, out all reefs. | | 9 | 4 | 6 | | | Sprung fore topgallant yard, got up another. | | 10 | 5 | 3 | | | Do. weather. | | 11 | 5 | 4 | SEbS | | A sail NE. | | 12 | 5 | | | | Employed working up junk. |
Two points leeway are allowed on the first course, one on the second; and as the ship is 7 points from the wind on the third course, there is no leeway allowed on it. The opposite point to NW, that from which the swell set, with the variation allowed upon it, is the last course in the Traverse Table.
| Course | Dist. | D.L. | Dep. | N. Latitude by | W. Long. by | Porto Sancto's | |--------|-------|------|------|----------------|-------------|---------------| | | | | | Acc. | | Bearing, Distance. | | S 74° W. | 108 | 36 | 104 | 47° 9' | 153° W. | S 12° W., 870 m. |
To find the bearing and distance of Porto Sancto.
- Latitude of ship: 47° 9' - Lat. Port Sancto: 32° 58' - Difference of latitude: 14° 11' = 8° 51' M. D. lat. - Hence the bearing of Porto Sancto is S 12° W., and distance 870 miles; the course per compass is therefore about SWbW. | Hours | Kn. | Fa. | Courses | Winds | Remarks, 2 October 11, 1793 | |-------|-----|-----|---------|-------|---------------------------| | 1 | 4 | | SWbS | ESE | Moderate wind and fair weather. | | 2 | 3 | | | | Shortened sail and set up the topmast rigging, | | 3 | 2 | | | | Do. weather. | | 4 | 3 | | | | Variation per amplitude 21° W. | | 5 | 4 | | | | A fine steady breeze. | | 6 | 4 | | | | By an observation of the moon's distance from α Pegasi, the ship's longitude at half past 8 was 12° 28' W. | | 7 | 4 | | | | Clear weather. | | 8 | 4 | | | | Do. weather. | | 9 | 5 | | | | Set studding sails, &c, | | 10 | 5 | | | | One sail in sight. | | 11 | 5 | | | | Do. weather, good observation. |
| Course | Dift. D.L. Dep. | N. Latitude by Acc. | D. Long. | W. Long. by Acc. | W. Var. Observed. | Porto Sancto's Bearing. | Distance. | |--------|-----------------|---------------------|----------|-----------------|------------------|------------------------|----------| | S 12° 45' W. | 128 125 28 | 45° 4' | 44° 59' | 41° W. | 13° 13' | 12° 59' | 21° | S 12° W. | 737 miles |
The observed variation 21° being allowed to the left of SWbS gives S 12° 45' W, the corrected course, and the distance summed up is 127.9, or 128 miles. Hence the difference of latitude is 124.8, and the departure 28.2. The latitude by account is therefore 45° 4' N, and the middle latitude 46° 6', to which, and the departure 28.2 in a latitude column, the difference of longitude in a distance column is 41° W; which being added to 12° 32' W, the yesterday's longitude gives 13° 13' W, the longitude in by account. But the longitude by observation was 12° 28' W at half past 8 P.M.; since that time the ship has run 96 miles; hence the departure in that interval is 21.2 m. Now half the difference of latitude 47 m. added to 44° 59', the latitude by observation at noon, the sum 45° 46' is the middle latitude; with which and the departure 21.2, the difference of longitude is found to be 3° W; which, therefore, added to 12° 28', the longitude observed, the sum is 12° 59' W, the longitude by observation reduced to noon.
To find the bearing and distance of Porto Sancto.
| Latitude ship | 44° 59' N. | Mer. parts | 3028 | Longitude | 12° 59' W | | Lat. Porto Sancto | 32° 58' N. | Mer. parts | 2997 | Longitude | 16° 25' W | | Difference of latitude | 12° 1° = 721 | M. D. lat. | 931 | D. Longitude | 3° 26' = 206' |
Hence the bearing of Porto Sancto is S 12° W, and distance 737 miles. The course to be steered is therefore S 33° W, or SWbS nearly. | Hours | Kn. | Fa. | Course | Winds | Remarks | |-------|-----|-----|--------|-------|---------| | 1 | 8 | | SWbS | EbN | Fresh gales, and cloudy. | | 2 | 7 | 5 | | | Do, weather. | | 3 | 8 | | | | Hauled down studding sails. | | 4 | 8 | 6 | | | Do, weather. | | 5 | 8 | 4 | | | | | 6 | 8 | | | | | | 7 | 7 | 5 | | | | | 8 | 7 | 3 | | | | | 9 | 7 | 4 | | | | | 10 | 7 | 2 | | | | | 11 | 7 | 6 | | | | | 12 | 7 | 5 | ENE | | A steady gale and fine weather. | | 1 | 7 | 5 | | | Do, weather. | | 2 | 7 | | | | | | 3 | 7 | | | | | | 4 | 7 | 3 | | | | | 5 | 7 | 2 | | | | | 6 | 7 | | | | | | 7 | 7 | 4 | | | | | 8 | 8 | | | | | | 9 | 8 | | | | | | 10 | 8 | | | | | | 11 | 7 | 6 | | | | | 12 | 8 | | | | |
The course corrected by variation is S 23° 31' W, and the distance run is 183 miles; hence the difference of latitude is 177.9, and the departure 42.8.
Yesterday's latitude by observation 44° 59' N. Mer. parts 3028
Difference of latitude 2° 58' S.
Latitude in by account 42° 1' N. Mer. parts 2783
Meridional difference of latitude 245
Now to course 13° 1' W, and meridional difference of latitude 245 in a latitude column, the difference of longitude in a departure column is 59° W; hence the longitudes of yesterday by account and observation, reduced to the noon of this day, will be 14° 12' W and 13° 58' respectively.
To find the bearing and distance of Porto Sancto.
| Latitude ship | 42° 1' N. | Mer. parts | 2783 | Longitude | 13° 58' W | |---------------|-----------|------------|------|-----------|-----------| | Lat. Port Sancto | 32° 58' N. | Mer. parts | 2097 | Longitude | 16° 25' W |
Difference of latitude 9° 3' = 543 M.D. latitude 686 D. Longitude 2° 27' = 147.
The meridional difference of latitude and difference of longitude will be found to agree nearest under 12, the correct bearing of Porto Sancto; and the variation being allowed to the right hand of S 12° W, gives S 32° W, the bearing per compass; and the distance answering to the difference of latitude 543, under 12 degrees, is 555 miles. ### NAVIGATION
**A Journal from England towards Madeira**
| Hours | Kn. | Fa. | Course | Winds | Remarks, Oct. 13, 1793 | |-------|-----|-----|--------|-------|----------------------| | 1 | 8 | SWbS | ENE | A steady gale, and fine weather. | | 2 | 8 | 5 | | At 34 minutes past three, the distance between the nearest limbs of the sun and moon, together with the altitude of each, were observed; from whence the ship's longitude at that time is 14° 1' W. | | 3 | 8 | 6 | | Hauled in the lower studding-falls. | | 4 | 8 | | | At 9h 22', by an observation of the moon's distance from α Pegasi, the longitude was 14° 20' W. | | 5 | 8 | | | Fresh gales, and clear. | | 6 | 8 | | | Do. weather. | | 7 | 8 | | | Variation per amplitude 19° 51' W. | | 8 | 8 | | | Do. per azimuth 19° 28' W. Set studding-falls. | | 9 | 8 | | | Carried away a fore-top-mast-studding-fall boom, got up another. | | 10 | 8 | | | Fresh gales. Took in studding-falls. |
| Course | Diff. | D.L. | Dep. | N. Latitude by | W. Long. by | W. Var. by Obl. | Porto Sancto's | |--------|-------|------|------|----------------|------------|----------------|----------------| | | | | | Acc. | Obs. | | | | SbW½W | 184 | 178 | 45 | 39° 3' | 59° W. | 15° 11' 14° 52' | 1¾ pts. |
The mean of the variation is about 1¾ points W; hence the course corrected is SbW½W; with which and the distance run 184 miles, the difference of latitude is 178.5, and the departure 44.7.
Yesterday's latitude - 42° 1' N. Mer. parts 2783
Difference of latitude - 2° 58' S.
Latitude in by account - 39° 3' N. Mer. parts 2549
Meridional difference of latitude - 234
Now, to course 1¾ points, and meridional difference of latitude 234, the difference of longitude is about 59 m.; which, added to yesterday's longitude by account 14° 12' W, the sum 15° 11' W is the longitude in by account at noon. The longitudes by observation are reduced to noon as follow:
The distance run between noon and 3h 34' P.M. is 29 miles; to which, and the course 1¾ points, the difference of latitude is 28'.
Yesterday's latitude at noon - 42° 1' N.
Latitude at time of observation - 41° 33' N. Mer. parts 2746
Latitude at noon - 39° 3' N. Mer. parts 2549
Meridional difference of latitude - 197
Then, to course 1¾ points, and meridional difference of latitude 197 in a latitude column, the difference of longitude in a departure column is 49' W; which added to 14° 1' W, the longitude by observation, the sum 14° 50' W is the longitude reduced to noon.
Again, the distance run between the preceding noon and 9h 22' P.M. is 75 miles; hence the corresponding difference of latitude is 72.8, or 73 miles; the ship's latitude at that time is therefore 40° 48' N.
Latitude at time of observation - 40° 48' N. Mer. parts 2686
Latitude at noon - 39° 3' N. Mer. parts 2549
Meridional difference of latitude - 137
Now, with the corrected course, and meridional difference of latitude, the difference of longitude is 34' W; which added to 40° 20' W, the sum is 14° 54' W, the reduced longitude. The mean of which and the former reduced longitude is 14° 52' W, the correct longitude.
VOL. XIV. Part II. ### NAVIGATION
_A Journal from England towards Madeira_
| Hours | Kn. | Fa. | Courses | Winds | Remarks, 1 October 14, 1793 | |-------|-----|-----|---------|-------|---------------------------| | 1 | 8 | | SWBS | Ebs | Fresh gales and hazy, single reefed topsails. | | 2 | 5 | | | | Got down topgallant yards. | | 3 | 7 | 5 | SSW | | Do. weather, and a confused swell running. | | 4 | 7 | | | | More moderate. | | 5 | 7 | 4 | | | Do. with lightning all round the compasses. | | 6 | 7 | 1 | Variable| | Squally, with rain. | | 7 | 6 | | | | Moderate weather; out reefs, and up topgallant yards. | | 8 | 5 | | SWBS | SEBS | At 11h 10' A.M. the latitude from double altitudes of the sun was 37° 10'. Clear weather. |
As the ship is close hauled from 2 o'clock A.M., 1½ points leeway are allowed upon that course and a point on the two following courses.
| Course | Diff. | D.L. | Dep. | N. Latitude by | D. Long. | W. Long. by | Porto Santo's | |--------|-------|------|------|----------------|----------|-------------|--------------| | | | | | Acc. | Obf. | Acc. | Bearing. | | S 16° W | 116 | 111 | 32 | 37° 12' | 37° 8' | 41° W. | 15° 52' | | | | | | | | | 15° 33' | | | | | | | | | 1¼ pts. | | | | | | | | | S 10° W. | | | | | | | | | 254 m. |
The latitude by observation at 11h 10' A.M. is 37° 10', and from that time till noon the ship has run about 4 miles. Hence the corresponding difference of latitude is 2 miles, which subtracted from the latitude observed gives 37° 8', the latitude reduced to noon.
To find the bearing and distance of Porto Santo,
- Latitude of ship: 37° 8' N. - Latitude Porto Santo: 32° 58' N. - Difference of latitude: 4° 10' = 250 M.D. lat. - Longitude: 15° 33' W. - Longitude Porto Santo: 16° 25' W. - Diff. longitude: 52
Hence the bearing of Porto Santo is S 15° W, or SSW ¼ W nearly, per compass, and the distance is 254 miles. ### NAVIGATION
**A Journal from England towards Madeira**
| Hours | Kn. | Fa. | Courses | Winds | Remarks | |-------|-----|-----|---------|-------|---------| | 1 | 4 | | WbS | SbW | Moderate and clear weather. | | 2 | 4 | | | | Employed working points and rope-bands. Ditto weather. | | 3 | 3 | 6 | | | Fine clear weather. | | 4 | 3 | | | | Ditto weather. | | 5 | 3 | 4 | | | Variable. | | 6 | 3 | | WbN | SWbS | | | 7 | 3 | | | | Variation per mean of several azimuths 18° 0' W. Ditto weather. Tacked ship. Sail-makers making wind-fails. | | 8 | 3 | 2 | | | A fine steady breeze. Cloudy. |
---
Half a point of leeway is allowed on each course; but as the variation is expressed in degrees, it will be more convenient and accurate to reduce the several courses into one, leeway only being allowed upon them. The course thus found is then to be corrected for variation, with which and the distance made good the difference of latitude and departure are to be found.
| Course | Diff. | D.L. | Dep. | Acc. | Obf. | D. Long. | W. Long. by Account | W. Var. by Obf. | Porto Sancto's Bearing | Distance | |--------|-------|------|------|------|------|----------|-------------------|-----------------|------------------------|----------| | S 66°W | 56 | 21 | 52 | 36° 47' | | 65' W | 16° 57' | 16° 38' | S 1/2 E | 229 |
To find the bearing and distance of Porto Sancto,
| Latitude ship | 36° 47' N. | Mer. pts | 2376 | Longitude | 16° 38' W. | | Lat. of Porto Sancto | 32° 58' N. | Mer. pts | 2297 | Longitude | 16° 25' W. | | Diff of latitude | -3° 49' = 229 | M. D. Lat. | 279 | D. Longitude | 0° 13' |
Hence the course is S 1/2 E, distance 229 miles; and the course per compass is SbW 1/2 W nearly.
---
**Practice**
**Ship's Journal**
---
**Ship's Journal**
---
**Ship's Journal** ### NAVIGATION
**A Journal from England towards Madeira**
| Hours | Kn. | Fa. | Courses | Winds | Remarks, 8 October 16, 1793 | |-------|-----|-----|---------|-------|-----------------------------| | 1 | 6 | SbE | SWbW | Fresh gales. | | 2 | 6 | 4 | S | W | Do. and cloudy. | | 3 | 7 | | | | | | 4 | 7 | | | | | | 5 | 7 | | | | | | 6 | 7 | | | | | | 7 | 7 | | | | | | 8 | 7 | | | | | | 9 | 8 | | | | | | 10 | 8 | | | | | | 11 | 8 | | | | | | 12 | 8 | | | | | | 1 | 8 | | | | | | 2 | 8 | | | | | | 3 | 8 | | | | | | 4 | 9 | | | | | | 5 | 9 | | | | | | 6 | 9 | | | | | | 7 | 9 | | | | | | 8 | 9 | | | | | | 9 | 9 | | | | | | 10 | 9 | | | | | | 11 | 7 | | | | | | 12 | 8 | | | | |
Variation per amplitude 1\(\frac{1}{2}\) points W.
People employed occasionally.
Do. weather. Observed sun’s meridian altitude.
| Course. | Diff. | D.L. Dep. | N. Latit. by | W. Long. by | W. Var. | Porto Sancto’s | |---------|-------|-----------|--------------|-------------|---------|---------------| | S 8° E | 186 | 185 | 33° 42' | 33° 46' | 31' E. | 16° 26' | | | | | | | | 1\(\frac{1}{2}\) pts. | | | | | | | | S 17° W. | | | | | | | | 50 miles. |
Half a point of leeway is allowed on the first course; which, and the others, are corrected for variation as usual.
| Courses | Diff. | Diff. of latit. | Departure | |---------|-------|-----------------|-----------| | SEbS | 12.4 | | 10.3 | | SbE\(\frac{1}{2}\)E. | 43. | | 41.2 | | S\(\frac{1}{2}\)E. | 65. | | 64.7 | | S. | 68.5 | | 68.5 | | S8° E. | 18.6 | | 184.7 | | | | | 25.8 |
Yesterday’s latitude - 36° 47' N.
Latitude by account - 33° 42' N.
Sum - 70° 29'
Middle latitude - 35° 15'
To the middle latitude and the departure, the difference of longitude in a distance column is 31' E.
Yesterday’s long. by acc. 16° 57' W. by obl. 16° 38' W.
Difference of long. 0° 31' E.
Longitude in 16° 26' W. 16° 7' W.
To find the bearing and distance of Porto Sancto.
| Latitude ship | 33° 46' N. | Mer. parts | 2155 | Longitude | 16° 7' W. | | Lat. Porto Sancto | 32° 58' N. | Mer. parts | 2097 | Longitude | 16° 25' W. | | Difference of latitude | 48 | Mer. diff. lat. | 58 | Diff. long. | 18 |
Hence the bearing of Porto Sancto is S 17° W. distance 50 miles. This journal is performed by inspection agreeable to the precepts given. Other methods might have been used for the same purpose; for which the two instruments already described and explained seem well adapted. We cannot, however, omit recommending the sliding gunter, which will be found very expeditious, not only in performing a day's work, but also in resolving most other nautical problems. See SLIDING-Gunter.
It will be found very satisfactory to lay down the ship's place on a chart at the noon of each day, and her situation with respect to the place bound to, and the nearest land, will be obvious. The bearing and distance of the intended or any other port, and other requisites, may be easily found by the chart as already explained; and indeed, every day's work may be performed on the chart; and thus the use of tables superfluous.
EXPLANATION OF THE TABLES.
TABLE I. To reduce points of the compass to degrees, and conversely.
The two first and two last columns of this table contain the several points and quarter-points of the compass; the third column contains the corresponding number of points and quarters; and the fourth, the degrees &c. answering thereto. The manner of using this table is obvious.
TABLE II. The miles and parts of a mile in a degree of longitude at every degree of latitude.
The first column contains degrees of latitude, and the second the corresponding miles in a degree of longitude; the other columns are a continuation of the first and second. If the given latitude consists of degrees and minutes, a proportional part of the difference between the miles answering to the given and following degrees of latitude is to be subtracted from the miles answering to the given degree.
Example. Required the number of miles in a degree of longitude, in latitude $57^\circ 9'$?
The difference between the miles answering to the latitudes of $57^\circ$ and $58^\circ$ is $0.89$.
Then as $60' : 9' :: 0.89 : x$
$x = \frac{0.89 \times 60}{9} = 5.93$
Miles answering to $57^\circ = 32.68$
Miles answering to $57^\circ 9' = 32.55$
This table may be used in Parallel and Middle Latitude Sailing.
TABLE III. Of the Sun's Semidiameter.
This table contains the angle subtended by the sun's semidiameter at the earth, for every sixth day of the year. The months and days are contained in the first column, and the semidiameter expressed in minutes and seconds in the second column. It is useful in correcting altitudes of the sun's limb, and distances between the sun's limb and the moon. TABLE IV. Of the Refraction in Altitude.
The refraction is necessary for correcting altitudes and distances observed at sea; it is always to be subtracted from the observed altitude, or added to the zenith distance. This table is adapted to a mean state of the atmosphere in Britain, namely, to 29.6 inches of the barometer, and 50° of the thermometer. If the height of the mercury in these instruments be different from the mean, a correction is necessary to reduce the tabular to the true refraction. See REFRACTION.
TABLES V. VI. Of the Dip of the Horizon.
The first of these tables contains the dip answering to a free or unobstructed horizon; and the numbers therein, as well as in the other table, are to be subtracted from the observed altitude when the fore-observation is used; but added, in the back-observation.
When the sun is over the land, and the ship nearer it than the visible horizon when unconfined: in this case, the sun's limb is to be brought in contact with the line of separation of the sea and land; the distance of that place from the ship is to be found by estimation or otherwise; and the dip answering thereto, and the height of eye, is to be taken from Table VI.
TABLE VII. Of the Correction to be applied to the time of high water at full and change of the moon, to find the time of high water on any other day of the moon.
The use of this table is fully explained at Section II. Chap. I. Book I. of this article.
TABLES VIII. IX. X. Of the Sun's Declination, &c.
The first of these tables contains the sun's declination, expressed in degrees, minutes, and tenths of a minute, for four successive years, namely, 1793, 1794, 1795, and 1796; and by means of Table X, may easily be reduced to a future period; observing that, after the 28th of February 1800, the declination answering to the day preceding that given is to be taken.
Ex. I. Required the sun's declination May 1. 1816?
May 1. 1812 is four years after the same day in 1812. Sun's declination May 1. 1812 = 15° 6'.7 N Equation from Table X. = +0 0.6 Sun's declination May 1. 1799 = 15° 7.3 N
Ex. II. Required the sun's declination August 20. 1805?
The given year is 12 years after 1793, and the time is after the end of February 1800. Now, Sun's dec. August 19. 1793 = 12° 34'.6 Equation from Table X. to 12 years = -0 1.9 Sun's declination August 20. 1805 = 12° 32.7
The declination in Table VIII. is adapted to the meridian of Greenwich, and Table IX. is intended to reduce it to any other meridian, and to any given time of the day under that meridian. The titles at the top and bottom of this table direct when the reduction is to be added or subtracted.
TABLE XI. Of the Right Ascensions and Declinations of Fixed Stars.
This table contains the right ascensions and declinations of 60 principal fixed stars, adapted to the beginning of the year 1793. Columns fourth and fifth contain the annual variation arising from the precession of the equinoxes, and the proper motion of the stars; which serves to reduce the place of a star to a period a few years after the epoch of the table with sufficient accuracy. When the place of a star is wanted, after the beginning of 1793, the variation in right ascension is additive; and that in declination is to be applied according to its sign. The contrary rule is to be used when the given time is before 1793.
Example. Required the right ascension and declination of Bellatrix, May 1. 1798?
Right ascension January 1. 1793 = 5h 14' 3" Variation = 3".21 × 5½ y. = +0 0 17 Right Ascension, May 1. 1798 = 5h 14' 20" Declination = 6° 8' 53" N Variation = 4" × 5½ y. = +0 0 21 Declination May 1. 1798 = 6° 9' 14" N
The various other tables necessary in the practice of navigation are to be found in most treatises on that subject. Those used in this article are in Mackay's Treatises on the Longitude and Navigation. ### Table I. To reduce Points of the Compass to Degrees, and conversely.
| North-east Quadrant | South-east Quadrant | Points | D. M. S. | South-west Quadrant | North-west Quadrant | |---------------------|---------------------|--------|----------|---------------------|---------------------| | North. | South. | 0° | 0° | South. | North. | | N½E | S½E | 0° | 48° | S½W | N½W | | N½E | S½E | 0° | 37° | S½W | N½W | | N½E | S½E | 0° | 26° | S½W | N½W | | NbE | SbE | 1° | 11° | SbW | NbW | | NbE½E | SbE½E | 1° | 4° | SbW½W | NbW½W | | NbE½E | SbE½E | 1° | 16° | SbW½W | NbW½W | | NbE½E | SbE½E | 1° | 19° | SbW½W | NbW½W | | NNE | SSE | 2° | 22° | SSE½W | NNW½W | | NNE½E | SSE½E | 2° | 18° | SSE½W | NNW½W | | NNE½E | SSE½E | 2° | 28° | SSE½W | NNW½W | | NNE½E | SSE½E | 2° | 30° | SSE½W | NNW½W | | NE½N | SE½S | 3° | 33° | SW½S | NW½N | | NE½N | SE½S | 3° | 36° | SW½S | NW½N | | NE½N | SE½S | 3° | 39° | SW½S | NW½N | | NE½N | SE½S | 3° | 42° | SW½S | NW½N | | NE½E | SE½E | 4° | 45° | SW½E | NW½E | | NE½E | SE½E | 4° | 48° | SW½E | NW½E | | NE½E | SE½E | 4° | 50° | SW½E | NW½E | | NE½E | SE½E | 4° | 53° | SW½E | NW½E | | NE½E | SE½E | 5° | 56° | SW½E | NW½E | | NE½E | SE½E | 5° | 59° | SW½E | NW½E | | NE½E | SE½E | 5° | 61° | SW½E | NW½E | | NE½E | SE½E | 5° | 64° | SW½E | NW½E | | ENE | ESE | 6° | 67° | WSW | WNW | | E½N½N | E½S½S | 6° | 70° | W½S½S | W½N½N | | E½N½N | E½S½S | 6° | 73° | W½S½S | W½N½N | | E½N½N | E½S½S | 6° | 75° | W½S½S | W½N½N | | E½N | E½S | 7° | 78° | W½S | W½N | | E½N | E½S | 7° | 81° | W½S | W½N | | E½N | E½S | 7° | 84° | W½S | W½N | | E½N | E½S | 7° | 87° | W½S | W½N | | Eaft. | Eaft. | 8° | 90° | Weft. | Weft. |
### Table II. The Miles and Parts of a Mile in a Degree of Longitude at every Degree of Latitude.
| D.L. Miles. | D.L. Miles. | D.L. Miles. | D.L. Miles. | D.L. Miles. | D.L. Miles. | |-------------|-------------|-------------|-------------|-------------|-------------| | 1 | 59.99 | 16 | 57.67 | 31 | 51.43 | | 2 | 59.97 | 17 | 57.36 | 32 | 50.88 | | 3 | 59.92 | 18 | 57.06 | 33 | 50.32 | | 4 | 59.86 | 19 | 56.73 | 34 | 49.74 | | 5 | 59.77 | 20 | 56.38 | 35 | 49.15 | | 6 | 59.67 | 21 | 56.01 | 36 | 48.54 | | 7 | 59.56 | 22 | 55.63 | 37 | 47.92 | | 8 | 59.44 | 23 | 55.23 | 38 | 47.28 | | 9 | 59.26 | 24 | 54.81 | 39 | 46.62 | | 10 | 59.08 | 25 | 54.38 | 40 | 45.95 | | 11 | 58.80 | 26 | 53.93 | 41 | 45.28 | | 12 | 58.68 | 27 | 53.46 | 42 | 44.95 | | 13 | 58.46 | 28 | 52.97 | 43 | 43.88 | | 14 | 58.22 | 29 | 52.47 | 44 | 43.16 | | 15 | 57.95 | 30 | 51.96 | 45 | 42.43 |
### Table III. Sun's Semidiam.
| Mon. Day | Sun's Semidiam. | |----------|-----------------| | January | 16° 19' | | February | 16° 16' | | March | 16° 10' | | April | 16° 18' | | May | 16° 17' | | June | 16° 12' | | July | 16° 10' | | August | 16° 15' | | September| 16° 16' | | October | 16° 14' | | November | 16° 15' | | December | 16° 19' | ### Table IV. **Refraction in Altitude.**
| App. Alt. | Refra. | App. Alt. | Refra. | |-----------|--------|-----------|--------| | D. M. | M. S. | D. M. | M. S. | | 0 | 33 | 6 | 30 | | 5 | 32 | 10 | 6 | | 10 | 31 | 22 | 7 | | 15 | 30 | 35 | 8 | | 20 | 29 | 49 | 9 | | 25 | 28 | 64 | 10 | | 30 | 27 | 79 | 11 | | 35 | 26 | 94 | 12 | | 40 | 25 | 109 | 13 | | 45 | 24 | 124 | 14 |
### Table V. **Dip of the Horizon.**
| Height of eye | Dip of horizon | Height of eye | Dip of horizon | |---------------|----------------|---------------|----------------| | M. S. Feet | M. S. Feet | M. S. Feet | M. S. Feet | | 1 | 0 | 11 | 3 | | 2 | 1 | 12 | 4 | | 3 | 2 | 13 | 5 | | 4 | 3 | 14 | 6 | | 5 | 4 | 15 | 7 | | 6 | 5 | 16 | 8 | | 7 | 6 | 17 | 9 | | 8 | 7 | 18 | 10 | | 9 | 8 | 19 | 11 | | 10 | 9 | 20 | 12 | | 11 | 10 | 21 | 13 | | 12 | 11 | 22 | 14 | | 13 | 12 | 23 | 15 | | 14 | 13 | 24 | 16 | | 15 | 14 | 25 | 17 | | 16 | 15 | 26 | 18 | | 17 | 16 | 27 | 19 | | 18 | 17 | 28 | 20 | | 19 | 18 | 29 | 21 | | 20 | 19 | 30 | 22 | | 21 | 20 | 31 | 23 | | 22 | 21 | 32 | 24 | | 23 | 22 | 33 | 25 | | 24 | 23 | 34 | 26 | | 25 | 24 | 35 | 27 | | 26 | 25 | 36 | 28 | | 27 | 26 | 37 | 29 | | 28 | 27 | 38 | 30 | | 29 | 28 | 39 | 31 | | 30 | 29 | 40 | 32 |
### Table VI. **Dip of the Sea at different distances from the Observer.**
| Height of the eye above the sea in feet. | |------------------------------------------| | 5 | | 10 | | 15 | | 20 | | 25 | | 30 | | 35 | | 40 |
### Table VII. **The Correction to be applied to the time of High-water at Full and Change of the Moon, to find the time of High-water on any other day.**
| Interval of Time | After New or Full Moon | Before 1st or 3rd Quarter | After 1st or 3rd Quarter | Before New or Full Moon | |------------------|------------------------|---------------------------|--------------------------|-------------------------| | | Additive | Additive | Additive | Subtractive | | D H | H. M | H. M | H. M | H. M | | 0 | 0 | 5 | 5 | 0 | | 1 | 8 | 4 | 5 | 2 | | 2 | 17 | 4 | 5 | 1 | | 3 | 26 | 4 | 5 | 0 | | 4 | 35 | 4 | 5 | 0 | | 5 | 44 | 4 | 5 | 0 | | 6 | 53 | 4 | 5 | 0 | | 7 | 62 | 4 | 5 | 0 | | 8 | 71 | 4 | 5 | 0 | | 9 | 80 | 4 | 5 | 0 | | 10 | 89 | 4 | 5 | 0 | | 11 | 98 | 4 | 5 | 0 | | 12 | 107 | 4 | 5 | 0 | | 13 | 116 | 4 | 5 | 0 | | 14 | 125 | 4 | 5 | 0 | | 15 | 134 | 4 | 5 | 0 | | 16 | 143 | 4 | 5 | 0 | | 17 | 152 | 4 | 5 | 0 | | 18 | 161 | 4 | 5 | 0 | | 19 | 170 | 4 | 5 | 0 | | 20 | 179 | 4 | 5 | 0 | | 21 | 188 | 4 | 5 | 0 | | 22 | 197 | 4 | 5 | 0 | | 23 | 206 | 4 | 5 | 0 | | 24 | 215 | 4 | 5 | 0 | | 25 | 224 | 4 | 5 | 0 | | 26 | 233 | 4 | 5 | 0 | | 27 | 242 | 4 | 5 | 0 | | 28 | 251 | 4 | 5 | 0 | | 29 | 260 | 4 | 5 | 0 | | 30 | 269 | 4 | 5 | 0 | | 31 | 278 | 4 | 5 | 0 | | 32 | 287 | 4 | 5 | 0 | | 33 | 296 | 4 | 5 | 0 | | 34 | 305 | 4 | 5 | 0 | | 35 | 314 | 4 | 5 | 0 | | 36 | 323 | 4 | 5 | 0 | | 37 | 332 | 4 | 5 | 0 | | 38 | 341 | 4 | 5 | 0 | | 39 | 350 | 4 | 5 | 0 | | 40 | 359 | 4 | 5 | 0 | | 41 | 368 | 4 | 5 | 0 | | 42 | 377 | 4 | 5 | 0 | | 43 | 386 | 4 | 5 | 0 | | 44 | 395 | 4 | 5 | 0 | | 45 | 404 | 4 | 5 | 0 | | 46 | 413 | 4 | 5 | 0 | | 47 | 422 | 4 | 5 | 0 | | 48 | 431 | 4 | 5 | 0 | | 49 | 440 | 4 | 5 | 0 | | 50 | 449 | 4 | 5 | 0 | | 51 | 458 | 4 | 5 | 0 | | 52 | 467 | 4 | 5 | 0 | | 53 | 476 | 4 | 5 | 0 | | 54 | 485 | 4 | 5 | 0 | | 55 | 494 | 4 | 5 | 0 | | 56 | 503 | 4 | 5 | 0 | | 57 | 512 | 4 | 5 | 0 | | 58 | 521 | 4 | 5 | 0 | | 59 | 530 | 4 | 5 | 0 | | 60 | 539 | 4 | 5 | 0 | | 61 | 548 | 4 | 5 | 0 | | 62 | 557 | 4 | 5 | 0 | | 63 | 566 | 4 | 5 | 0 | | 64 | 575 | 4 | 5 | 0 | | 65 | 584 | 4 | 5 | 0 | | 66 | 593 | 4 | 5 | 0 | | 67 | 602 | 4 | 5 | 0 | | 68 | 611 | 4 | 5 | 0 | | 69 | 620 | 4 | 5 | 0 | | 70 | 629 | 4 | 5 | 0 | | 71 | 638 | 4 | 5 | 0 | | 72 | 647 | 4 | 5 | 0 | | 73 | 656 | 4 | 5 | 0 | | 74 | 665 | 4 | 5 | 0 | | 75 | 674 | 4 | 5 | 0 | | 76 | 683 | 4 | 5 | 0 | | 77 | 692 | 4 | 5 | 0 | | 78 | 701 | 4 | 5 | 0 | | 79 | 710 | 4 | 5 | 0 | | 80 | 719 | 4 | 5 | 0 | | 81 | 728 | 4 | 5 | 0 | | 82 | 737 | 4 | 5 | 0 | | 83 | 746 | 4 | 5 | 0 | | 84 | 755 | 4 | 5 | 0 | | 85 | 764 | 4 | 5 | 0 | | 86 | 773 | 4 | 5 | 0 | | 87 | 782 | 4 | 5 | 0 | | 88 | 791 | 4 | 5 | 0 | | 89 | 800 | 4 | 5 | 0 | | 90 | 809 | 4 | 5 | 0 | | 91 | 818 | 4 | 5 | 0 | | 92 | 827 | 4 | 5 | 0 | | 93 | 836 | 4 | 5 | 0 | | 94 | 845 | 4 | 5 | 0 | | 95 | 854 | 4 | 5 | 0 | | 96 | 863 | 4 | 5 | 0 | | 97 | 872 | 4 | 5 | 0 | | 98 | 881 | 4 | 5 | 0 | | 99 | 890 | 4 | 5 | 0 | | 100 | 900 | 4 | 5 | 0 | ### Table VIII. Sun's Declination for 1809, being the first after leap year.
| Days | January | February | March | April | May | June | July | August | September | October | November | December | |------|---------|----------|-------|-------|-----|------|------|--------|-----------|---------|----------|----------| | 1 | 23° 1'6" | 17° 7'5" | 7° 37'1" | 4° 29'7" | 1° 5'2" | 22° 2'5" | 25° 8'8" | 19° 5'7" | 8° 2'19" | 3° 7'49" | 1° 24'3" | 1° 48'7" | | 2 | 22° 56'4" | 16° 50'3" | 7° 14'3" | 4° 52'8" | 1° 20'2" | 22° 10'6" | 23° 4'7" | 17° 50'3" | 8° 0'0" | 3° 30'7" | 1° 43'3" | 1° 57'9" | | 3 | 22° 50'8" | 16° 32'8" | 6° 51'4" | 5° 15'8" | 1° 37'9" | 22° 18'3" | 23° 0'1" | 17° 35'0" | 7° 38'1" | 3° 54'0" | 1° 24'2" | 2° 2'6" | | 4 | 22° 44'7" | 16° 15'0" | 6° 28'4" | 5° 38'7" | 1° 55'4" | 22° 25'6" | 22° 55'2" | 17° 19'2" | 7° 16'0" | 4° 17'3" | 1° 21'1" | 2° 14'9" | | 5 | 22° 38'2" | 15° 56'9" | 6° 5'2" | 6° 1'5" | 1° 16'12" | 22° 32'4" | 22° 49'9" | 17° 3'2" | 6° 53'8" | 4° 40'4" | 1° 39'6" | 2° 22'8" | | 6 | 22° 31'2" | 15° 38'6" | 5° 42'0" | 6° 24'2" | 1° 26'20" | 22° 38'9" | 22° 44'1" | 16° 46'8" | 6° 31'4" | 5° 3'6" | 1° 57'7" | 2° 32'3" | | 7 | 22° 23'8" | 15° 19'9" | 5° 18'8" | 6° 40'8" | 1° 46'5" | 22° 45'0" | 22° 38'0" | 16° 30'2" | 6° 9'0" | 5° 26'7" | 1° 15'7" | 2° 37'3" | | 8 | 22° 15'9" | 15° 1'1" | 4° 55'4" | 7° 9'3" | 1° 17'29" | 22° 50'7" | 22° 31'5" | 16° 13'3" | 5° 46'4" | 5° 49'7" | 1° 33'3" | 2° 43'9" | | 9 | 22° 7'6" | 14° 41'9" | 4° 32'0" | 7° 31'7" | 1° 17'19" | 22° 56'0" | 22° 24'6" | 15° 50'2" | 5° 23'8" | 6° 12'6" | 1° 30'7" | 2° 50'0" | | 10 | 21° 53'9" | 14° 22'6" | 4° 8'6" | 7° 53'9" | 1° 17'35" | 22° 53'0" | 22° 17'3" | 15° 38'8" | 5° 1'0" | 6° 35'4" | 1° 7'8" | 2° 55'7" |
### Table VIII. The Sun's Declination for 1810, being the second after leap year.
| Days | January | February | March | April | May | June | July | August | September | October | November | December | |------|---------|----------|-------|-------|-----|------|------|--------|-----------|---------|----------|----------| | 1 | 23° 2'7" | 17° 11'6S | 7° 42'7S | 4° 24'0N | 14° 5'5N | 22° 0'5N | 23° 9'7N | 18° 9'3N | 8° 2'17N | 3° 1'8S | 1° 19'6S | 2° 46'4S | | 2 | 22° 57'7" | 16° 54'5" | 7° 19'9" | 4° 47'1" | 1° 15'6" | 22° 8'6" | 23° 5'7" | 17° 54'2" | 8° 5'3" | 3° 2'1" | 1° 38'8" | 2° 55'7" | | 3 | 22° 52'1" | 16° 37'0" | 6° 57'0" | 5° 10'1" | 1° 33'5" | 22° 16'4" | 23° 1'3" | 17° 38'7" | 7° 43'4" | 3° 4'8" | 1° 37'8" | 2° 4'5" | | 4 | 22° 46'2" | 16° 19'3" | 6° 34'0" | 5° 33'1" | 1° 51'1" | 22° 23'7" | 22° 56'4" | 17° 23'0" | 7° 21'3" | 4° 11'7" | 1° 16'6" | 2° 12'9" | | 5 | 22° 39'7" | 16° 1'3" | 6° 10'9" | 5° 55'9" | 1° 6'8" | 22° 30'7" | 22° 51'2" | 17° 0'7" | 6° 59'1" | 4° 34'9" | 1° 35'1" | 2° 20'9" | | 6 | 22° 32'9" | 15° 43'0" | 5° 47'7" | 6° 18'7" | 1° 25'6" | 22° 37'3" | 22° 45'5" | 16° 50'7" | 6° 36'8" | 4° 58'0" | 1° 53'3" | 2° 28'5" | | 7 | 22° 25'6" | 15° 24'4" | 5° 24'5" | 6° 41'3" | 1° 42'4" | 22° 43'5" | 22° 39'5" | 16° 34'2" | 6° 14'4" | 5° 21'1" | 1° 16'11" | 2° 35'6" | | 8 | 22° 17'8" | 15° 5'6" | 5° 1'1" | 7° 3'9" | 1° 58'9" | 22° 49'3" | 22° 33'0" | 16° 17'4" | 5° 51'9" | 5° 44'1" | 1° 29'0" | 2° 42'3" | | 9 | 22° 9'6" | 14° 46'5" | 5° 37'7" | 7° 26'3" | 1° 15'2" | 22° 54'7" | 22° 26'7" | 16° 0'3" | 5° 29'3" | 6° 7'0" | 1° 46'5" | 2° 48'5" | | 10 | 22° 1'0" | 14° 27'2" | 5° 14'2" | 7° 48'5" | 1° 31'1" | 22° 59'7" | 22° 19'0" | 15° 43'0" | 5° 6'6" | 6° 29'9" | 1° 36'3" | 2° 54'3" |
---
**Note:** The tables provide the Sun's declination for each day in the specified years, with the declination given in degrees, minutes, and seconds. ### Table VIII. Sun's Declination for 1811, being the third after leap year.
| Days | January | February | March | April | May | June | July | August | September | October | November | December | |------|---------|----------|-------|-------|-----|------|------|--------|-----------|---------|----------|----------| | 1 | 23° 3'9"S | 19° 15'7"S | 7° 48'2"S | 4° 18'4"N | 14° 53'1"N | 21° 58'5"N | 23° 10'6"N | 18° 12'8"N | 8° 3'2"3 | 2° 56'1"S | 14° 14'8"S | 21° 44'0"S | | 2 | 22° 58.9 | 16° 58.6 | 7° 25.4 | 4° 41.5 | 15° 11.3 | 22° 6.7 | 23° 6.7 | 17° 57.8 | 8° 10.6 | 3° 19.4 | 14° 34.1 | 21° 53.4 | | 3 | 22° 53.5 | 16° 41.3 | 7° 2.6 | 5° 4.6 | 15° 20.2 | 22° 14.6 | 23° 2.3 | 17° 42.4 | 7° 48.7 | 3° 42.7 | 14° 53.1 | 22° 2.3 | | 4 | 22° 47.7 | 16° 23.6 | 6° 39.6 | 5° 27.5 | 15° 46.9 | 22° 22.0 | 22° 57.6 | 17° 26.8 | 7° 26.6 | 4° 5.9 | 15° 11.9 | 22° 10.9 | | 5 | 22° 41.3 | 15° 5.7 | 6° 16.5 | 5° 50.4 | 16° 4.3 | 22° 29.1 | 22° 52.4 | 17° 10.9 | 7° 4.5 | 4° 29.2 | 15° 30.5 | 22° 18.9 | | 6 | 22° 34.6 | 15° 47.5 | 5° 53.4 | 6° 13.1 | 16° 21.4 | 22° 35.8 | 22° 46.9 | 16° 54.7 | 6° 42.2 | 4° 52.3 | 15° 48.8 | 22° 26.6 | | 7 | 22° 27.4 | 15° 29.0 | 5° 30.1 | 6° 35.8 | 16° 38.3 | 22° 42.0 | 22° 41.0 | 16° 38.2 | 6° 19.9 | 5° 15.4 | 16° 6.9 | 22° 33.9 | | 8 | 22° 19.7 | 15° 10.3 | 5° 6.8 | 6° 58.3 | 16° 54.9 | 22° 47.9 | 22° 34.6 | 16° 21.5 | 5° 57.4 | 5° 38.4 | 16° 24.6 | 22° 40.7 | | 9 | 22° 11.6 | 14° 51.3 | 4° 43.5 | 7° 20.7 | 17° 11.2 | 22° 53.4 | 22° 27.9 | 16° 4.5 | 5° 34.8 | 6° 1.3 | 16° 42.1 | 22° 47.0 | | 10 | 22° 3.1 | 14° 32.0 | 4° 20.4 | 7° 43.0 | 17° 27.3 | 22° 58.5 | 22° 20.8 | 15° 47.2 | 5° 12.1 | 6° 24.2 | 16° 59.3 | 22° 52.0 |
### Table VIII. Sun's Declination for 1812, being leap year.
| Days | January | February | March | April | May | June | July | August | September | October | November | December | |------|---------|----------|-------|-------|-----|------|------|--------|-----------|---------|----------|----------| | 1 | 23° 4'9"S | 17° 10'8"S | 7° 31'0"S | 4° 35'7"N | 15° 6'7"N | 22° 4'6"N | 23° 7'6"N | 18° 1'5"N | 8° 15'9"N | 3° 13'7"S | 14° 25'8"S | 21° 51'0"S | | 2 | 22° 0.0 | 17° 2.8 | 7° 8.2 | 4° 58.8 | 15° 24.7 | 22° 12.5 | 23° 3.3 | 17° 46.2 | 7° 54.0 | 3° 37.0 | 14° 48.5 | 22° 0.1 | | 3 | 22° 54.8 | 16° 45.3 | 6° 45.3 | 5° 21.8 | 15° 42.4 | 22° 20.1 | 22° 8.7 | 17° 30.6 | 7° 32.0 | 4° 0.3 | 15° 7.4 | 22° 8.7 | | 4 | 22° 49.0 | 16° 27.9 | 6° 22.2 | 5° 44.7 | 15° 59.9 | 22° 27.3 | 22° 33.6 | 17° 14.7 | 7° 9.9 | 4° 23.5 | 15° 26.0 | 22° 17.0 | | 5 | 22° 42.9 | 16° 10.1 | 5° 59.1 | 6° 7.5 | 16° 17.1 | 22° 34.0 | 22° 48.2 | 16° 58.6 | 6° 47.7 | 4° 40.7 | 15° 44.4 | 22° 24.7 | | 6 | 22° 36.2 | 15° 51.9 | 5° 35.9 | 6° 30.2 | 16° 34.0 | 22° 40.4 | 22° 42.3 | 16° 42.2 | 6° 25.3 | 5° 9.8 | 16° 2.5 | 22° 32.0 | | 7 | 22° 29.1 | 15° 33.5 | 5° 12.6 | 6° 52.8 | 16° 50.8 | 22° 46.4 | 22° 36.1 | 16° 25.5 | 6° 2.8 | 5° 32.9 | 16° 20.3 | 22° 39.0 | | 8 | 22° 21.6 | 15° 14.8 | 4° 49.2 | 7° 15.2 | 17° 2.7 | 22° 52.0 | 22° 29.5 | 16° 8.6 | 5° 40.3 | 5° 55.8 | 16° 37.9 | 22° 4.5 | | 9 | 22° 13.6 | 14° 55.9 | 4° 25.8 | 7° 37.6 | 17° 23.3 | 22° 57.2 | 22° 22.4 | 15° 51.4 | 5° 17.6 | 6° 18.7 | 16° 55.2 | 22° 51.4 | | 10 | 22° 5.2 | 14° 36.5 | 4° 2.3 | 7° 59.8 | 17° 39.1 | 22° 2.0 | 22° 15.0 | 15° 33.9 | 4° 54.8 | 6° 41.5 | 17° 12.2 | 22° 57.0 |
*Table* | Table IX. To reduce the Sun's Declination to any other Meridian, and to any given Time under that Meridian. | |---|---|---|---|---|---|---|---|---|---|---|---|---|---|---| | **Longitude** | 10° | 20° | 30° | 40° | 50° | 60° | 70° | 80° | 90° | 100° | 110° | 120° | 130° | 140° | 150° | 160° | 170° | 180° | | **Add.in W.** | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | | **Sub.in W.** | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | | **Add.in E.** | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | | **Add.in N.** | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | | **Sub.aft.N.** | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | | **Add.aft.N.** | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° | 0° |
Time from Noon. ### Table X. Change of Sun's Declination
| Month | Days | Complete Years | |-------|------|----------------| | | | 4 8 12 16 | | January | | | | | | 1° -1° -3° -4° | | | | 7° 2° 4° 7° | | | | 13° 3° 6° 9° | | | | 19° 4° 7° 11° | | | | 25° 4° 8° 11° |
### Table XI. The Right Ascensions and Declinations of the Principal fixed Stars, adapted to the beginning of the Year 1810.
| Names of Stars | Mag. | Right Ascension in Time | Ann. Var. | Declination | Ann. Var. | |----------------|------|------------------------|-----------|-------------|-----------| | Pegasi | 2 | 0° 3' 57" | 3° .06 | 14° 7' 35" N | +20° .0 | | β Ceti | 2 | 0° 34' 2 | 3° .01 | 19° 3' 36" S | -19° .8 | | Alrucabah, pole star | 2.3 | 0° 54' 15 | 12° .42 | 88° 17' 41" N | +19° .6 | | Mirach | 2 | 0° 59' 7 | 3° .31 | 34° 30' 45" N | +19° .4 | | Achernar | 1 | 1° 30' 38 | 2° .25 | 58° 11' 19" S | -18° .5 | | Almach | 2 | 1° 52' 16 | 3° .62 | 41° 24' 47" N | +17° .7 | | Menkar | 2 | 1° 52' 20 | 3° .11 | 3° 20' 27" N | +14° .7 | | Algol | Var. | 2° 56' 0 | 3° .85 | 40° 12' 53" N | +14° .5 | | Algenib | 1 | 3° 10' 49 | 4° .21 | 49° 9' 31" N | +13° .6 | | Aldabaran | 2 | 4° 25' 1 | 3° .42 | 16° 7' 6" N | +13° .2 | | Capella | 1 | 5° 2' 40 | 4° .41 | 45° 47' 41" N | +5° .1 | | Rigel | 1 | 5° 5' 19 | 2° .87 | 8° 25' 48" S | -4° .8 | | β Tauri | 2 | 5° 14' 17 | 3° .78 | 28° 26' 10" N | +4° .1 | | Bellatrix | 2 | 5° 14' 57 | 3° .21 | 6° 10' 1" N | +4° .0 | | Orionis | 2 | 5° 22' 20 | 3° .07 | 0° 26' 53" S | -3° .4 | | ζ Orionis | 2 | 5° 26' 35 | 3° .04 | 1° 19' 57" S | -3° .0 | | ξ Orionis | 2 | 5° 31' 11 | 3° .03 | 2° 3' 8" S | -2° .6 | | α Columbae | 2 | 5° 32' 45 | 2° .17 | 34° 10' 54" S | -2° .4 | | Betelgeuse | 1 | 5° 44' 53 | 3° .24 | 7° 21' 40" N | +1° .4 | | β Canis Majoris| 2.3 | 6° 14' 22 | 2° .65 | 17° 52' 16" S | +1° .2 | | Canopus | 1 | 6° 19' 43 | 1° .33 | 51° 35' 44" S | +1° .7 | | Sirius | 1 | 6° 36' 46 | 2° .65 | 15° 27' 48" S | +4° .3 | | β Canis Majoris| 2 | 7° 9' 40 | 2° .44 | 26° 6' 0" S | +5° .2 | | Caftor | 1.2 | 7° 22' 27 | 3° .85 | 32° 17' 33" N | -6° .9 | | Procyon | 1.2 | 7° 29' 20 | 3° .14 | 5° 42' 56" N | -7° .5 | | Pollux | 2.3 | 7° 32' 48 | 3° .09 | 28° 27' 28" S | -7° .9 | | η Navis | 2 | 7° 56' 44 | 2° .11 | 39° 28' 20" S | +9° .7 | | θ Navis | 2 | 8° 3' 41 | 1° .85 | 46° 46' 39" S | +10° .3 | | Acubens | 3 | 8° 48' 4 | 3° .30 | 12° 35' 19" N | -13° .4 | | θ Navis | 1 | 9° 11' 6 | 0° .75 | 68° 56' 13" S | +14° .8 | | Alphard | 2 | 9° 18' 14 | 2° .93 | 7° 50' 25" S | +15° .2 | | Regulus | 1 | 9° 58' 14 | 3° .20 | 12° 53' 29" N | -17° .2 | | η Navis | 2 | 10° 37' 43 | 2° .30 | 58° 41' 0" N | +18° .7 | | β Ursae Majoris| 2 | 10° 50' 17 | 3° .71 | 57° 23' 53" N | -19° .1 | | Dubhe | 2 | 10° 51' 54 | 3° .85 | 62° 46' 43" N | -19° .1 | | β Leonis | 2 | 11° 39' 21 | 3° .06 | 15° 37' 9" N | -19° .9 | | β Ursae Majoris| 2 | 11° 47' 45 | 3° .22 | 54° 44' 7" N | -20° .0 | | α Crucis | 1 | 12° 16' 9 | 3° .24 | 62° 2' 46" S | +20° .0 | | γ Crucis | 2 | 12° 20' 42 | 3° .24 | 52° 2' 42" S | +20° .0 | | β Crucis | 2 | 12° 36' 44 | 3° .41 | 58° 38' 55" S | +19° .8 | | Aliath | 2 | 12° 45' 36 | 2° .67 | 57° 10' 48" N | -19° .7 | | Spica Virginis | 1 | 13° 15' 11 | 3° .13 | 10° 9' 54" S | +19° .0 | | β Ursae Majoris| 2 | 13° 16' 23 | 2° .43 | 55° 54' 8" N | -19° .0 | | Benetnach | 2 | 13° 40' 4 | 2° .40 | 50° 15' 58" N | -18° .2 | | β Centauri | 1.2 | 13° 50' 32 | 4° .11 | 59° 26' 51" N | +17° .8 | | Arcturus | 1 | 14° 6' 59 | 2° .72 | 25° 10' 34" N | -19° .1 | | α Centauri | 1 | 14° 27' 16 | 4° .45 | 60° 3' 17" S | +16° .1 | | Alphacca | 2 | 15° 26' 38 | 2° .33 | 27° 21' 44" N | -12° .5 | | β Scorpii | 2 | 15° 54' 26 | 3° .47 | 10° 16' 29" S | +10° .5 | | Antares | 1 | 16° 17' 45 | 3° .04 | 25° 59' 49" S | +8° .7 | | Ras. Algethi | 2 | 17° 5' 59 | 2° .73 | 14° 36' 57" N | -4° .8 | | Ras. Allague | 2 | 17° 26' 7 | 2° .77 | 12° 42' 37" N | -3° .0 | | Vega | 1 | 18° 30' 30 | 2° .02 | 38° 30' 35" N | +2° .6 | | Altair | 1.2 | 19° 41' 30 | 2° .92 | 8° 22' 13" N | +8° .5 | | Deneb | 2 | 20° 34' 56 | 2° .03 | 44° 36' 13" N | +12° .5 | | Gruis | 2 | 21° 56' 11 | 3° .85 | 48° 1' 58" S | -17° .1 | | Fomalhaut | 1 | 22° 47' 6 | 3° .33 | 30° 48' 14" S | -19° .0 | | Scheat | 2 | 22° 54' 34 | 2° .87 | 27° 3' 7" N | +19° .2 | | Markab | 2 | 22° 55' 17 | 2° .96 | 27° 3' 7" N | +19° .2 | | α Andromedae | 2 | 23° 58' 34 | 3° .07 | 28° 2' 31" N | +20° .0 |
NAVIGATION Navigation, Navigation of the Ancients. See Phoenicia and Inland Navigation.
Inland Navigation, the method of conveying commodities from one part of a country to another by means of rivers, lakes, canals, or arms of the sea. See Canal.
We have already, under Canal, taken notice of a method proposed by Dr. Anderson of raising and lowering vessels by means of mechanical powers, instead of dams and locks. We shall describe another mechanical contrivance proposed by Mr. Leach for the same purpose. This machinery is compounded of an inclined plane and wheel in axis. The inclined plane is a parallelogram whose length reaches from the end of one canal to the beginning of another, or to the sea or navigable river, to which the vessel is next to be conveyed; the breadth ought to be 22 feet. It may be made of good oak or deal plank, and sufficiently strong to bear the weight to be laid upon it; and it must be very strongly supported by beams of oak or other wood. It ought to be divided in the middle by a ledge or rib of 12 inches square, the side ribs being nine by 12 inches. The elevation must depend upon particular circumstances. Fig. 1 shows the inclined part of the machine; AB being the wooden part just described, placed between the side of the hill W and the navigable river F. According to the dimensions already given, the two paths A and B on which the vessels move are exactly ten feet wide. G represents the canal, brought perhaps from the distance of several miles to the top of the precipice WW. At the end of the canal, and quite across from R to R, must be built a very strong wall; in which are two sluices with flood gates at K and L, to let out the water occasionally. Between the head of the plane AB, and the end of the canal G, is a horizontal platform divided into two parts, as is represented in the figure by the letters HI. At the end of the canal are six rollers M and N, of use in carrying the boats and lighters in and out of the canal. Near the end of the canal, at S, and T, are two other sluices, with their flood-gates, for letting out a quantity of fluid to drive the other part of the machine. O and P represent the two ends of the towing paths, one on each side of the canal.
Fig. 2 shows the vehicle by which the lighters are conveyed up and down the inclined plane, by the two paths A and B, fig. 1. AA (fig. 2) represents part of the inclined plane, B the vehicle in the position in which it rolls up and down the two paths. C is the body of the vehicle, which is made hollow, to contain a quantity of water occasionally used as a counterbalance for its corresponding vehicle. DDD are three rollers between the bottom of the vehicle and the plane, for the purpose of rolling the boats up and down. HHH are six rollers; four by the horizontal part of the vehicle on which the boat E is to rest in its passage up and down the plane; the other two rollers are in a moveable part, which is fastened to the body of the vehicle with a pair of very strong hinges; and in the passage of the vehicle up and down the plane, it turns up between the head of the boat and the plane, preventing the former from rubbing against the plane. When the vehicle gets up to the top, this moveable part falls down on the platform marked HI, becoming parallel with the horizontal part of the vehicle; after which it serves for a launch and passage to place the boat upon the rollers MN (fig. 1) at the end of the canal. This passage part of the vehicle, together with the three rollers at the end of the canal, is likewise of great use in towing a boat out of the canal, in order to place it on the horizontal part. At the bottom of the cavity of the vehicle is a large hole F, with a valve opening inwardly. Through this hole the water enters when the vehicle sinks into the navigable river F, for the purpose of receiving a boat on the top or horizontal part of the vehicle till it is quite full and then will sink entirely under water, while the boat is towed in on the horizontal part. A small rope K is fastened to the valve, on purpose to lift it up and to keep it so, while the vehicle and boat are ascending up the plane out of the canal; so that the water may discharge itself till as much as is necessary be got out, or till it becomes an equal balance for the corresponding vehicle and its contents, which are descending by the other path. Hence we see, that every machine must have two of these vehicles furnished with rollers as already described, and so constructed that one may be as nearly as possible a counterbalance to the other. As it is necessary that the vehicles should be watertight, the inside of them must be caulked very tight; and they should be capacious enough to hold as much water as will balance the largest boat with its contents. Here it may be observed, that every vessel will be balanced by as many cubic feet of water as it displaces by being put into the water when loaded. The quantity may easily be known, by observing how far the boat sinks in the water, and calculating the bulk of the part immersed.
The machine which puts the vehicles in motion, may either be constructed with an under-shot or breast-water wheel: or by an over-shot water-wheel: or by two walking-wheels, for men to walk in as in cranes, &c.
Fig. 3 shows a front view of the under-shot water-wheel movement; where A is the end of the axis or cylinder of the cog or spur wheel; the diameter of which axis is four feet, and its length not less than 22 feet, as it must be extended quite across the canal from one side to the other, and placed on the top of very strong supports on each side of the canal, about seven feet above the surface of the water, as the loaded boat is to pass backwards and forwards under the cylinder, and at a convenient distance from the wall RR (fig. 1); and placed between the two sluices S and T; on the end of which cylinder is the cog-wheel B (fig. 3). The wheel B is supposed to be 20 feet of diameter, having on its edge 120 cogs; and underneath the cog-wheel is the breast-water one C, 24 feet in diameter, from the tip of one aller-board to the tip of its opposite. On the end of the axis of the water-wheel D is a trundle two feet and a half in diameter, with 15 rounds and shaves contained therein. This must be placed between the two sluices S and T, to let the water out of the canal; which, falling on the float-boards, will turn the wheel round from the right hand towards the left, when the sluice on the left hand of the wheel is opened; but the contrary way when that of the right is opened.—The water falling Island Na-falling upon the boards passes along with the wheel in vigation.
To the axis or cylinder of this machine, which must always be horizontal, are fixed a pair of strong ropes; the ends of each pair fastened to the upper part of the cylinder; it being necessary that they should act in contrary directions. Each must extend the whole length of the plane, and their strength must be proportioned to the weight necessary to be sustained. The two vehicles already mentioned are fastened to the other ends of the ropes; so that one pair of the ropes are wound up by the cylinder turning one way, and the other by its turning the contrary way. Thus when one of the vehicles is at the upper part of the path A, ready to discharge its boat and cargo into the upper canal, the other boat will be at the foot of the path B, all under water in the lower canal, and ready for the reception of a boat to be towed in on the horizontal part of it; so that as one vehicle rolls up on one side of the plane, the other will roll down on the other side, and vice versa.
Fig. 4 shows the movement by means of an over-shot water-wheel. It consists of a water-wheel C, and two spur or cog-wheels A and B. The water-wheel is 18 feet in diameter, and has two rows of buckets placed contrariwise to one another, that it may turn round in contrary directions, according as the one or the other sluice, S or T, is opened. On its axis F is a trundle of three feet diameter, having 18 rounds or staves which fall into the cogs of the second spur-wheel B, causing it to turn round in a direction contrary to that of the water-wheel. This second wheel is likewise 18 feet in diameter, with a trundle of three feet having 18 rounds or staves.—The diameter of the upper spur-wheel A is also 18 feet, but the diameter of its axis is six feet. On the edge of the wheel are 108 cogs. These fall in between the staves of the axis of the other spur-wheel; and thus the third wheel turns round the same way with the water wheel C. The cylinder of this upper spur-wheel must be placed across the canal betwixt the two sluices, on very strong supports, as explained in the former movement, and the two pair of ropes in the same manner.
The movement of the walking wheel is shown (fig. 5). A1 and A2 are two wheels for men to walk in, each of them 24 feet in diameter. B1 and B2 are the axes or cylinders of the two wheels, of equal lengths; viz. 11 feet each, and four in diameter.—At one end of each of the two cylinders C1 and C2, is a wheel of the same diameter with the cylinder. On the edges of these wheels are teeth of an equal number in each wheel; and as the teeth of the wheels mutually fall into each other, the revolutions of both must be performed in the same time. By this contrivance also the cylinders will turn different ways; and the ropes on the two different cylinders will constantly one pair be wound up, and the other wound down, by the natural moving of the machine. DDD is the frame that supports the whole, which must be made very firm and secure.
Let us now suppose, that there is a boat in the upper canal to be brought down, but none to go up for a balance. In this case, as one of the vehicles must be at the top to receive the boat, the other will be at the bottom to take in water. Let then any of the movements just described be set to work, and it is plain, that as the upper vehicle with its boat descends, the under vehicle will ascend with the water; the valve being in the mean time lifted up till a sufficient quantity of water has flowed out, to make the one nearly a counterbalance to the other; so that the vessel may slide down gently and without any violence.
If it happens that a boat is to go up while none is to come down, one of the vehicles being at the foot of the plane under water, and in readiness to have the boat towed upon its horizontal part, one of the sluices at K or L is to be opened, and a quantity of water let into the cistern of the upper vehicle sufficient to counterbalance the boat with its contents which is to ascend. This being done, the machine is set to work, the valve of the under vehicle kept open till the water is all discharged; and then the boat will roll up to the top of the plane.
From this description of the canal and machinery for raising and lowering the vessels, the reader can be at no loss to understand the principles on which it depends. It would be superfluous to adduce examples, or follow our author through his calculations relative to particular cases. We shall only observe, that the difference of time in which vessels may be raised or lowered by the machinery just described, in comparison with what can be done in the common way by dams and locks, must give a very favourable idea of the new method. According to Mr Leach’s computations, a boat with its cargo weighing 10 tons might be raised by the walking machine in 12 or 14 minutes, by the under-shot wheel in 15 minutes, and by the over-shot wheel in 30 minutes; and that through a space of no less than 30 fathoms measured on the inclined plane, or 114 feet perpendicular.