PRELIMINARYPRINCIPLES

CHAP. I.—PRELIMINARY PRINCIPLES.

SECT. I.—Of the Latitude and Longitude of a Place.

The situation of a place on the surface of the earth is estimated by its distance from two imaginary lines intersecting each other at right angles. The one of these is called the equator, and the other the first meridian. The situation of the equator is fixed; but that of the first meridian is arbitrary, and therefore different nations assume different first meridians. In Great Britain, we esteem that to be the first meridian which passes through the Royal Observatory at Greenwich.

The equator divides the earth into two equal parts, called the northern and southern hemispheres; and the latitude of a place is its distance from the equator, reckoned on a meridian in degrees and parts of a degree, being either north and south, according as it is in the northern or southern hemisphere.

The first meridian being continued round the globe, divides it into two equal parts, called the eastern and western hemispheres; and the longitude of a place is that portion of the equator contained between the first meridian and the meridian of the given place, and is either east or west, according as it is in the eastern or western hemisphere respectively to the first meridian.

PROB. I. The latitudes of two places being given, to find the difference of latitude.

RULE. Subtract the less latitude from the greater, if the latitudes be of the same name, but add them if of contrary; and the remainder or sum will be the difference of latitude.

Example 1. Required the difference of latitude between the Lizard, in latitude $49^\{\backslash circ\}\; 58\text{'}\; N.$ and Cape St Vincent, in latitude $37^\{\backslash circ\}\; 3\text{'}\; N.$

Ex. 2. What is the difference of latitude between Funchal, in latitude $32^\{\backslash circ\}\; 38\text{'}\; N.$, and the Cape of Good Hope, in latitude $34^\{\backslash circ\}\; 29\text{'}\; S.$?

PROB. II. Given the latitude of one place, and the difference of latitude between it and another place; to find the latitude of that place.

RULE. If the given latitude and the difference of latitude be of the same name, add them; but if of different names, subtract them, and the sum or remainder will be the latitude required of the same name with the greater.

Ex. 1. A ship from latitude $39^\{\backslash circ\}\; 22\text{'}\; N.$ sailed due north 560 miles. Required the latitude come to.

Latitude come to,

Ex. 2. A ship from latitude $7^\{\backslash circ\}\; 19\text{'}\; N.$ sailed 854 miles south. Required the latitude come to.

Latitude come to,

PROB. III. The longitudes of two places being given, to find their difference of longitude.

RULE. If the longitudes of the given places are of the same name, subtract the less from the greater, and the remainder is the difference of longitude; but if the longitudes are of contrary names, their sum is the difference of longitude. If this exceeds $180^\{\backslash circ\}$, subtract it from $360^\{\backslash circ\}$, and the remainder is the difference of longitude.

Preliminary Principles. Ex. 1. Required the difference of longitude between Edinburgh and New York, their longitudes being 3° 12' W. and 74° 2' W. respectively.

Longitude of New York, 74° 2' W.
Longitude of Edinburgh, 3 12 W.

Difference of longitude, 70 50

Ex. 2. What is the difference of longitude between Maskelyne's Isles, in longitude 167° 59' E. and Olinda, in longitude 34° 54' W.

Longitude of Maskelyne's Isles, 167° 59' E.
Longitude of Olinda, 35 54 W.

Sum, 203 53

Subtract from, 360 0

Difference of longitude, 156 2

PROB. IV. Given the longitude of a place, and the difference of longitude between it and another place, to find the longitude of the latter place.

RULE. If the given longitude and the difference of longitude be of a contrary name, subtract the less from the greater, and the remainder is the longitude required, of the same name with the greater quantity; but if they are of the same name, add them, and the sum is the longitude sought, of the same name with that given. If this sum exceeds 180°, subtract it from 360°; the remainder is the required longitude, of a contrary name to that given.

Ex. 1. A ship from longitude 9° 54' E. sailed westerly till the difference of longitude was 23° 18'. Required the longitude come to.

Longitude sailed from,
Difference of longitude,

9° 54' E. Preliminary Principles.
23 18 W.

Longitude come to, 13 24 W.
Ex. 2. The longitude sailed from is 25° 9' W. and difference of longitude 18° 46' W. Required the longitude come to.

Longitude left, 25° 9' W.
Difference of longitude, 18 46 W.

Longitude in, 43 55 W.

SECT. II.—Of the Tides.

The theory of the tides has already been explained under the article ASTRONOMY, and will again be further illustrated under that of TIDES. In this place, therefore, it remains only to explain the method of calculating the time of high water at a given place.

As the tides depend upon the joint actions of the sun and moon, and therefore upon the distance of these objects from the earth and from each other; and as, in the method generally employed to find the time of high water, whether by the mean time of new moon, or by the epacts, or tables deduced therefrom, the moon is supposed to be the sole agent, and to have an uniform motion in the periphery of a circle, whose centre is that of the earth; it is hence obvious that this method cannot be accurate, and by observation the error is sometimes found to exceed two hours. This method is therefore rejected, and another given, in which the error will seldom exceed a few minutes, unless the tides are greatly influenced by the winds.

TABLE I.—For determining the Time of High Water.

RULE. Let the approximate time of high water be found, by taking the corrections for the moon's horizontal parallax for the nearest noon or midnight from Table I. Again, to this time and the given longitude take from the Nautical Almanack the moon's horizontal parallax. Also to the time of the moon's transit over the meridian of Greenwich apply the variation answering to the longitude and daily variation between the given and preceding day if the longitude is east. Subtract it from the transit over the meridian of Greenwich, and the remainder will be the time of transit over the meridian of the given place. But if the longitude be west, the correction answering to the longitude and daily variation of transit between the given and following day must be added to the time of transit over the meridian of Greenwich to obtain the time of transit over the meridian of the given place. To the time of high water, if new and full moon at the given place, add the reduced time of transit over the meridian of the same place, and to the sum apply the equation from the

table answering to the time of transit and horizontal parallax formerly found; the result will be the true mean time of high water required. The apparent time may be found by applying the equation of time, with its proper sign.

Ex. 1. Required the time of high water at Leith on Wednesday the 10th of May 1837, in longitude 3° 11' W.

By the rule, the time of high water will be about six o'clock in the evening. In this case, the moon's horizontal parallax will be 54' 16", and the time of transit 4^h 50^m mean time, or 4^h 54^m apparent time by applying the equation of time 3^m 50^s by addition.

Preliminary Principles. If the sum exceed 12^h 25^m, subtract this number from it; if it exceed 24^h 50^m, subtract as before, and the remainder will be the time of high water in the afternoon of the given day nearly. The time of high water of the tide preceding may be found nearly by subtracting 25^m from it, and the succeeding tide by adding 25^m to it. In cases of great accuracy, however, a computation should be made for each tide in a manner similar to that above.

Ex. 2. Required the time of high water at Aberdeen on the 21st of June 1837, in longitude 2° 6' W.

As before, the time of high water will readily be found to be about three o'clock.

Here the horizontal parallax of the moon will be 60° 30', and the mean time of transit on the given day 15^h 32^m. But as this transit exceeds 12^h, it will be necessary to take the time of transit over the under meridian, or, what comes to the same thing, half the sum of the transits on the given and preceding days, or $\backslash frac\{1\}\{2\}(14^h\; 32^m\; +\; 15^h\; 32^m)$,
Correction from the table, . . . . . = 15^h 2^m
Remainder, . . . . . = 0 44

High water at new and full moon, . . . . . + 1 10

Sum exceeding 12^h . . . . . = 15 28
By rule, subtract . . . . . = 12 25

Apparent time of high water, . . . . . = 3 3
Equation of time, . . . . . + 0 1

Mean time, . . . . . = 3 4

Ex. 3. Required the depth at Aberdeen at the same time, the rise of spring tides being 19 feet, denoted by a in Table II. part 1, and that of the neap 14 feet, by b.

Now, by Table II. part 1, to transit 15^h 2^m, and horizontal parallax 60° 30', will be obtained $0.917\; \backslash times\; 19\; +\; 0.242\; \backslash times\; 14\; =\; 20.8$ feet.

Ex. 4. Required the height of the tide at 3^h 15^m after high water.
By part 2, $20.8\; \backslash times\; 0.5\; =\; 10.4$ feet.

In this manner, the time and rise of the tide may be readily obtained nearly, unless both are much influenced by the strength and direction of the wind.

SECT. III.—Of measuring a Ship's Run in a given Time.

The method commonly used at sea to find the distance sailed in a given time is by means of a log-line and half-minute glass. A description of these is given under the articles Log and Log-LINE, which see.

It has been already observed, that the interval between each knot on the line ought to be fifty feet, in order to adapt it to a glass that runs thirty seconds. But although the line and glass be at any time perfectly adjusted to each other, yet as the line shrinks after being wet, and as the weather has a considerable effect upon the glass, it will therefore be necessary to examine them from time to time; and the distance given by them must be corrected accordingly. The distance sailed may, therefore, be affected by an error in the glass, or in the line, or in both. The true distance may, however, be found as follows:

PROB. I. The distance sailed by the log, and the seconds run by the glass, being given, to find the true distance, the line being supposed right.

RULE. Multiply the distance given by the log by 30, and divide the product by the seconds run by the glass, the quotient will be the true distance.

Ex. 1. The hourly rate of sailing by the log is nine knots, and the glass is found to run out in 35 seconds. Required the true rate of sailing.

Ex. 2. The distance sailed by the log is 73 miles, and the glass runs out in 26 seconds. Sought the true distance.

PROB. II. Given the distance sailed by the log, and the measured interval between two adjacent knots on the line; to find the true distance, the glass running exactly 30 seconds.

RULE. Multiply twice the distance sailed by the measured length of a knot, point off two figures to the right, and the remainder will be the true distance.

Ex. 1. The hourly rate of sailing by the log is five knots, and the interval between knot and knot measures 53 feet. Required the true rate of sailing.

Ex. 2. The distance sailed is 64 miles, by a log-line which measures 42 feet to a knot. Required the true distance.

PROB. III. Given the length of a knot, the number of seconds run by the glass in half a minute, and the distance sailed by the log; to find the true distance.

RULE. Multiply the distance sailed by the log by six times the measured length of a knot, and divide the product by the seconds run by the glass; the quotient, pointing off one figure to the right, will be the true distance.

Ex. The distance sailed by the log is 159 miles, the measured length of a knot is 42 feet, and the glass runs 33 seconds in half a minute. Required the true distance.

CHAP. II.—OF PLANE SAILING.

Plane sailing is the art of navigating a ship upon principles deduced from the notion of the earth's being an extended plane. On this supposition the meridians are esteemed as parallel right lines. The parallels of latitude are at right angles to the meridians; the lengths of the degrees on the meridians, equator, and parallels of latitude, are everywhere equal; and the degrees of longitude are reckoned on the parallels of latitude as well as on the equator. In this sailing four things are principally concerned, namely, the course, distance, difference of latitude, and departure.

The course is the angle contained between the meridian and the line described by the ship, and is usually expressed in points of the compass.

The distance is the number of miles a ship has sailed on a direct course in a given time.

The difference of latitude is the portion of a meridian contained between the parallels of latitude sailed from and come to; and is reckoned either north or south, according as the course is in the northern or southern hemisphere.

The departure is the distance of the ship from the meridian of the place she left, reckoned on a parallel of latitude. In this sailing, the departure and difference of longitude are esteemed equal.

In order to illustrate the above, let A represent the position of any given place, and AB the meridian passing through that place; also let AC represent the line described by a ship, and C the point arrived at. From C draw CB perpendicular to AB. Now in the triangle ABC, the angle BAC represents the course, the side AC the distance, AB the difference of latitude, and BC the departure.

In constructing a figure relating to a ship's course, let the upper part of what the figure is to be drawn on represent the north, then the lower part will be south, the right-hand side east, and the left-hand side west.

A north and south line is to be drawn to represent the meridian of the place from which the ship sailed; and the

upper or lower part of this line, according as the course is southerly or northerly, is to be marked as the position of that place. From this point as a centre, with the chord of 60°, an arch is to be described from the meridian towards the right or left, according as the course is easterly or westerly; and the course, taken from the line of chords if given in degrees, but from the line of rhumbs if expressed in points of the compass, is to be laid upon this arch, beginning at the meridian. A line drawn through this point and that sailed from will represent the distance, which, if given, must be laid thereon, beginning at the point sailed from. A line is to be drawn from the extremity of the distance perpendicular to the meridian; and hence the difference of latitude and departure will be obtained.

If the difference of latitude is given, it is to be laid upon the meridian, beginning at the point representing the place the ship left; and a line drawn from the extremity of the difference of latitude perpendicular to the meridian, till it meets the distance produced, will limit the figure.

If the departure is given, it is to be laid off on a parallel, and a line drawn through its extremity will limit the distance. When either the distance and difference of latitude, distance and departure, or difference of latitude and departure, are given, the measure of each is to be taken from a scale of equal parts, and laid off on its respective line, and the extremities connected. Hence the figure will be formed.

Plane Sailing. PROB. I. Given the course and distance, to find the difference of latitude and departure.

Example. A ship from St Helena, in latitude $15^\{\backslash circ\}\; 55\text{'}\; S.$ sailed S. W. by S. 158 miles. Required the latitude come to, and departure.

By Construction.

Draw the meridian AB, and with the chord of $60^\{\backslash circ\}$ describe the arch mn, and make it equal to the rhumb of three points, and through n draw AC equal to 158 miles; from C, draw CB perpendicular to AB; then AB applied to the scale from which AC was taken will be found to measure 131.4, and BC 87.8.

Fig. 2.

By Calculation.

To find the difference of latitude.

To find the departure.

By Inspection.

In the traverse table, the difference of latitude answering to the course 3 points, and distance 158 miles, in a distance column, is 131.4, and departure 87.8.

By Gunter's Scale.

The extent from 8 points to 5 points, the complement of the course on the line of sine rhumbs (marked S. R.) will reach from the distance 158 to 131.4, the difference of latitude on the line of numbers; and the extent from 8 points to 3 points on sine rhumbs will reach from 158 to 87.8, the departure on numbers.

PROB. II. Given the course and difference of latitude, to find the distance and departure.

Example. A ship from St George's, in latitude $38^\{\backslash circ\}\; 45\text{'}$ north, sailed S. E. $\backslash frac\{1\}\{2\}\; S.$; and the latitude by observation was $35^\{\backslash circ\}\; 7\text{'}\; N.$ Required the distance run, and departure.

By Construction.

Draw the portion of the meridian AB equal to 218 m.: from the centre A with the chord of $60^\{\backslash circ\}$ describe the arch mn, which make equal to the rhumb of $3\backslash frac\{1\}\{2\}$ points: through n draw the line AC, and from B draw BC perpendicular to AB, and let it be produced till it meets AC in C. Then the distance AC, being applied to the scale, will measure 282 m. B to the distance 282 m. and the departure BC 179 miles.

Fig. 3.

By Calculation.

To find the distance.

To find the departure.

By Inspection.

Find the given difference of latitude 218 m. in a latitude column, under the course of $3\backslash frac\{1\}\{2\}$ points; opposite to which, in a distance column, is 282 miles; a departure column is 178.9 m. the distance and departure required.

By Gunter's Scale.

Extend the compass from $4\backslash frac\{1\}\{2\}$ points, the complement of the course, to 8 points on sine rhumbs; that extent will reach from the difference of latitude 218 miles to the distance 282 miles on numbers; and the extent from 4 points to the course $3\backslash frac\{1\}\{2\}$ points on the line of tangent rhumbs (marked T. R.) will reach from 218 miles to 178.9, the departure on numbers.

PROB. III. Given course and departure, to find the distance and difference of latitude.

Example. A ship from Palma, in latitude $28^\{\backslash circ\}\; 37\text{'}\; N.$ sailed N. W. by W. and made 192 miles of departure. Required the distance run, and latitude come to.

By Construction.

Make the departure BC equal to 192 miles, draw BA perpendicular to BC, and from the centre C, with the chord of $60^\{\backslash circ\}$, describe the arch mn, which make equal to the rhumb of 3 points, the complement of the course; draw a line through C n, which produce till it meet BA in A: then the distance AC being measured, will be equal to 231 m. and the difference of latitude AB will be 128.3 miles.

Fig. 4.

By Calculation.

To find the distance.

To find the difference of latitude.

By Inspection.

Find the departure 192 m. in its proper column above the given course 5 points; and opposite thereto is the distance 231 miles, and difference of latitude 128.3, in their respective columns.

¹ For the method of resolving the various problems in navigation by the sliding gunter, the reader is referred to Dr Mackay's Treatise on the Description and Use of that Instrument.

The extent from 5 points to 8 points on the line of sine rhumbs, being laid from the departure 192 on numbers, will reach to the distance 231 on the same line; and the extent from 5 points to 4 points on the line of tangent rhumbs will reach from the departure 192, to the difference of latitude 128.3 on numbers.

PROB. IV. Given the distance and difference of latitude, to find the course and departure.

Example. A ship from a place in latitude 43° 13' N., sails between the north and east 285 miles; and is then by observation found to be in latitude 46° 31' N. Required the course and departure.

Draw the portion of the meridian AB equal to 198 miles; from B draw BC perpendicular to AB; then take the distance 285 miles from the scale, and with one foot of the compass in A describe an arch intersecting BC in C, and join AC. With the chord of 60° describe the arch mn, the portion of which, contained between the distance and difference of latitude, applied to the line of chords, will measure 46°, the course; and the departure BC being measured on the line of equal parts, will be found equal to 205 miles.

Find the given distance in the table in its proper column; and if the difference of latitude answering thereto is the same as that given, namely, 198, then the departure will be found in its proper column, and the course at the top or bottom of the page, according as the difference of latitude, is found in a column marked lat. at top or bottom. If the difference of latitude thus found does not agree with that given, turn over till the nearest thereto is found to answer to the given distance. This is in the page marked 46 degrees at the bottom, which is the course, and the corresponding departure is 205 miles.

The extent from the distance 285 to the difference of latitude 198 on numbers, will reach from 90° to 44°, the

complement of the course on sines; and the extent from 90° to the course 46° on the line of sines being laid from the distance 285, will reach to the departure 205 on the line of numbers.

PROB. V. Given the distance and departure, to find the course and difference of latitude.

Example. A ship from Fort-Royal in the island of Grenada, in latitude 12° 9' N., sailed 260 miles between the south and west, and made 190 miles of departure. Required the course and latitude come to.

Draw BC perpendicular to AB, and equal to the given departure 190 miles; then from the centre C, with the distance 260 miles, sweep an arch intersecting AB in A, and join AC. Now describe an arch from the centre A with the chord of 60°, and the portion mn of this arch, contained between the distance and difference of latitude, measured on the line of chords, will be 47°, the course; and the difference of latitude AB, applied to the scale of equal parts, measures 177½ miles.

Seek in the traverse table until the nearest to the given departure is found in the same line with the given distance 260. This is found to be in the page marked 47° at the bottom, which is the course; and the corresponding difference of latitude is 177.5.

The extent of the compass, from the distance 260 to the departure 190 on the line of numbers, will reach from 90° to 47°, the course on the line of sines; and the extent from 90° to 43°, the complement of the course on sines, will reach from the distance 260 to the difference of latitude 177½ on the line of numbers.

PROB. VI. Given difference of latitude and departure, sought course and distance.

Example. A ship from a port in latitude 7° 56' S. sailed between the south and east till her departure was 132 miles, and was then by observation found to be in latitude 12° 3' S. Required the course and distance.

By Construction.

Draw the portion of the meridian AB equal to the difference of latitude 247 miles; from B draw BC perpendicular to AB, and equal to the given departure 132 miles, and join AC: then with the chord of 60° describe an arch from the centre A; and the portion mn of this arch, being applied to the line of chords, will measure about 28°; and the distance AC, measured on the line of equal parts, will be 280 miles.

Fig. 7.

By Calculation.

To find the course.

To find the distance.

By Inspection.

Seek in the table till the given difference of latitude and departure, or the nearest thereto, are found together in their respective columns, which will be under 28°, the required course; and the distance answering thereto is 280 miles.

By Gunter's Scale.

The extent from the given difference of latitude 247 to the departure 132 on the line of numbers, will reach from 45° to 28°, the course on the line of tangents; and the extent from 62°, the complement of the course, to 90° on sines, will reach from the difference of latitude 247 to the distance 280 on numbers.

CHAP. III.—OF TRAVERSE SAILING.

If a ship sail upon two or more courses in a given time, the irregular tract she describes is called a traverse; and to resolve a traverse, is the method of reducing these several courses, and the distances run, into a single course and distance. The method chiefly used for this purpose at sea is by inspection, which shall therefore be principally adhered to, and is as follows.

Make a table of a breadth and depth sufficient to contain the several courses, &c. This table is to be divided into six columns; the courses are to be put in the first, and the corresponding distances in the second column; the third and fourth columns are to contain the differences of latitude, and the two last the departures.

Now, the several courses and their corresponding distances being properly arranged in the table, find the difference of latitude and departure answering to each in the traverse table; remembering that the difference of latitude is to be put in a north or south column, according as the course is in the northern or southern hemisphere; and that the departure is to be put in an east column if the course is easterly, but in a west column if the course is westerly; observing also, that the departure is less than the difference of latitude when the course is less than four points, or 45°; otherwise greater.

Add up the columns of northing, southing, easting, and

westing, and set down the sum of each at its bottom; then the difference between the sums of the north and south columns will be the difference of latitude made good, of the same name with the greater; and the difference between the sums of the east and west columns is the departure made good, of the same name with the greater sum.

Now, seek in the traverse table till a difference of latitude and departure are found to agree as nearly as possible with those above; then the distance will be found on the same line, and the course at the top or bottom of the page, according as the difference of latitude is greater or less than the departure.

In order to resolve a traverse by construction, describe a circle with the chord of 60°, in which draw two diameters at right angles to each other, at whose extremities are to be marked the initials of the cardinal points, north being uppermost.

Lay off each course on the circumference, reckoned from its proper meridian; and from the centre to each point draw lines, which are to be marked with the proper number of the course.

On the first radius lay off the first distance from the centre, and through its extremity, and parallel to the second radius, draw the second distance, of its proper length; through the extremity of the second distance, and parallel to the third radius, draw the third distance, of its proper length; and thus proceed until all the distances are drawn.

A line drawn from the extremity of the last distance to the centre of the circle will represent the distance made good; and a line drawn from the same point perpendicular to the meridian, produced if necessary, will represent the departure; and the portion of the meridian intercepted between the centre and departure will be the difference of latitude made good.

Ex. 1. A ship from Fyal, in lat. 38° 32' N., sailed as follows: E. S. E. 163 miles, S. W. $\backslash frac\{1\}\{2\}$ W. 110 miles, S. E. $\backslash frac\{3\}\{4\}$ S. 180 miles, and N. by E. 68 miles. Required the latitude come to, the course, and distance made good.

By Inspection.

By Construction.

With the chord of 60° describe the circle NE, SW (fig. 8), the centre of which represents the place the ship sailed from; draw two diameters NS, EW at right angles to each other, the one representing the meridian, and the other the parallel of latitude of the place sailed from. Take each course from the line of rhumbs, lay it off on the circumference from its proper meridian, and number it in order 1, 2, 3, 4. Upon the first rhumb C1, lay off the first distance 163 miles from C to A; through it draw the second distance AB parallel to C2, and equal to 110 miles;

Traverse through B draw BD equal to 180 miles, and parallel to C3; and draw DE parallel to C4, and equal to 68 miles. Now CE being joined, will represent the distance made good; which, applied to the scale, will measure 281 miles. The arch Sn, which represents the course, being measured on the line of chords, will be found equal to $41\backslash frac\{1\}\{2\}^\backslash circ$. From E draw EF perpendicular to CS produced; then CF will be the difference of latitude, and FE the departure made good; which, applied to the scale, will be found to measure 210 and 186 miles respectively.

As the method by construction is scarcely ever practised at sea, it seems therefore unnecessary to apply it to the solution of the following examples.

Ex. 2. A ship from latitude $1^\backslash circ\; 58\text{'}\; S.$ sailed as under. Required her present latitude, course, and distance made good.

Ex. 3. Yesterday at noon we were in latitude $13^\backslash circ\; 12\text{'}\; N.$, and since then have run as follows: S. S. E. 36 miles, S. 12 miles, N. W. $\backslash frac\{1\}\{2\}$ W. 28 miles, W. 30 miles, S. W. 42 miles, W. by N. 39 miles, and N. 20 miles. Required our present latitude, departure, and direct course and distance.

Ex. 4. The course per compass from Greigness¹ to the May is S. W. $\backslash frac\{1\}\{2\}$ S., distance 58 miles; from the May to the Staples, S. by E. $\backslash frac\{3\}\{4\}$ E., 44 miles; and from the Staples to Flamborough Head, S. by E., 110 miles. Required the course per compass, and distance from Greigness to Flamborough Head.

Hence the course per compass is nearly S. $1^\backslash circ$ W., and distance $192\backslash frac\{1\}\{2\}$ miles.

CHAP. IV.—OF PARALLEL SAILING.

The figure of the earth is spherical, and the meridians gradually approach each other, and meet at the poles. The difference of longitude between any two places is the angle at the pole contained between the meridians of those places, or it is the arc of the equator intercepted between the meridians of the given places; and the meridian distance between two places in the same parallel is the arc thereof contained between their meridians. It hence follows, that the meridian distance, answering to the same difference of longitude, will vary with the latitude of the parallel upon which it is reckoned; and the same difference of longitude will not answer to a given meridian distance when reckoned upon different parallels.

Parallel sailing is, therefore, the method of finding the distance between two places lying in the same parallel whose longitudes are known; or, to find the difference of longitude answering to a given distance run in an east or west direction. This sailing is particularly useful in making low or small islands.

In order to illustrate the principles of parallel sailing, let CABP (fig. 9) represent a section of one fourth part of the earth, the arch ABP being part of a meridian; CA the equatorial, and CP the polar semi-axis. Also let B be the situation of any given place on the earth; and join BC, which will be equal to CA or CP.² The arch AB, or angle ACB, is the measure of the latitude of the place B; and the arch BP, or angle BCP, is that of its complement. If BD be drawn from B perpendicular to CP, it will represent the cosine of latitude to the radius BC or CA.

Now since circles and similar portions of circles are in the direct ratio of their radii; therefore,

As radius

Is to the cosine of latitude,

So is any given portion of the equator

To a similar portion of the given parallel.

But the difference of longitude is an arc of the equa-

¹ Greigness is about $2\backslash frac\{1\}\{2\}$ miles distant from Aberdeen, in nearly a S. E. by E. $\backslash frac\{1\}\{2\}$ E. direction.

² This is not strictly true, as the figure of the earth is that of an oblate spheroid; and therefore the radius of curvature varies with the latitude. The difference between CA and CP, according to Sir Isaac Newton's hypothesis, is about 17 miles.

Parallel Sailing. tor, and the distance between any two places under the same parallel is a similar portion of that parallel.

Hence $R\; :\; \backslash cos\; \backslash text\{\; latitude\}\; ::\; \backslash text\{diff.\; longitude\}\; :\; \backslash text\{distance\}$.
And, by inversion,
$\backslash cos\; \backslash text\{\; latitude\}\; :\; R\; ::\; \backslash text\{distance\}\; :\; \backslash text\{diff.\; of\; longitude\}$.
Also,
$\backslash text\{diff.\; of\; longitude\}\; :\; \backslash text\{distance\}\; ::\; R\; :\; \backslash cos\; \backslash text\{\; latitude\}$.

PROB. I. Given the latitude of a parallel, and the number of miles contained in a portion of the equator, to find the miles contained in a similar portion of that parallel.

Ex. 1. Required the number of miles contained in a degree of longitude in latitude $55^\{\backslash circ\}\; 58\text{'}$.

By Construction.

Draw the indefinite right line AB (fig. 10); make the angle BAC equal to the given latitude $55^\{\backslash circ\}\; 58\text{'}$, and AC equal to the number of miles contained in a degree of longitude at the equator, namely, 60; from C draw CB perpendicular to AB; and AB, being measured on the line of equal parts, will be found equal to 33.5, the miles required.

Fig. 10.

By Calculation.

By Inspection.

To $56^\{\backslash circ\}$, the nearest degree to the given latitude, and distance 60 miles, the corresponding difference of latitude is 33.6, which is the miles required.

By Gunter's Scale.

The extent from $90^\{\backslash circ\}$ to $34^\{\backslash circ\}$, the complement of the given latitude on the line of sines, will reach from 60 to 33.6 on the line of numbers.

There are two lines on the other side of the scale, with respect to Gunter's line, adapted to this particular purpose; one of which is entitled chords, and contains the several degrees of latitude. The other, marked M. L. signifying miles of longitude, is the line of longitudes, and shows the number of miles in a degree of longitude in each parallel. The use of these lines is therefore obvious.

Ex. 2. Required the distance between Treguier in France, in longitude $3^\{\backslash circ\}\; 14\text{'}\; W.$, and Gaspey Bay, in longitude $64^\{\backslash circ\}\; 27\text{'}\; W.$, the common latitude being $48^\{\backslash circ\}\; 47\text{'}\; N.$

PROB. II. Given the number of miles contained in a portion of a known parallel, to find the number of miles in a similar portion of the equator.

Example. A ship from Cape Finisterre, in latitude $42^\{\backslash circ\}\; 52\text{'}\; N.$, and longitude $9^\{\backslash circ\}\; 17\text{'}\; W.$, sailed due west 342 miles. Required the longitude come to.

By Construction.

Draw the straight line AB (fig. 11) equal to the given distance 342 miles, and make the angle BAC equal to $42^\{\backslash circ\}\; 52\text{'}$, the given latitude; from B draw BC perpendicular to AB, meeting AC in C; then AC applied to the scale will measure 466, the difference of longitude required.

Fig. 11.

By Calculation.

PROB. III. Given the number of miles contained in any portion of the equator, and the miles in a similar portion of a parallel; to find the latitude of that parallel.

Example. A ship sailed due east 358 miles, and was found by observation to have differed her longitude $8^\{\backslash circ\}\; 42\text{'}$. Required the latitude of the parallel.

By Construction.

Make the line AB (fig. 12) equal to the given distance; to which let BC be drawn perpendicular, with an extent equal to $522\text{'}$, the difference of longitude; describe an arch from the centre A, cutting BC in C; then the angle BAC, being measured by means of the line of chords, will be found equal to $46\backslash frac\{1\}\{2\}^\{\backslash circ\}$, the required latitude.

Fig. 12.

By Calculation.

PROB. IV. Given the number of miles contained in the portion of a known parallel, to find the length of a similar portion of another known parallel.

Example. From two ports in latitude $33^\{\backslash circ\}\; 58\text{'}\; N.$, distance 348 miles, two ships sail directly north till they are in latitude $48^\{\backslash circ\}\; 23\text{'}\; N.$. Required their distance.

By Construction.

Draw the line CB, CE (fig. 13), making angles with CP equal to the complements of the given latitudes, namely, $56^\{\backslash circ\}\; 2\text{'}$ and $41^\{\backslash circ\}\; 37\text{'}$ 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 278 miles.

Fig. 13.

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° 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.

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°; 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° 53'.

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 sailed upon 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 rectilinear. 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 correspond-

ing 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 rectilinear triangle will, therefore, represent the course, distance, difference of latitude, and departure.

Although the parts HG, GI, and angle G of the rectilinear 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 sailed from or upon 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 sine PT to the sine 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.

By Construction.

Draw the right line AD (fig. 18) to represent the meridian of the May; with the chord of $60^\backslash circ$ describe the arch mn, upon which lay off the chord of $32^\backslash circ\; 59\text{'}$, the complement of the middle latitude from m to n: from D through n draw the line DC equal to $60^\backslash circ$ the difference of longitude, and from C draw CB perpendicular to AD: make BA equal to $98^\backslash circ$ 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 $73^\backslash circ\; 24\text{'}$, the required course.

Fig. 18.

By Calculation.

To find the course.¹

To find the distance.

The true course, therefore, from the Island of May to the Naze of Norway is $N.\; 73^\backslash circ\; 24\text{'}\; E.$, $E.\; N.\; E.\; \backslash frac\{3\}\{4\}\; E.$ nearly; but as the variation at the May is $2\backslash frac\{1\}\{2\}$ points west, therefore the course per compass from the May is $E.\; by\; S.$

PROB. II. Given one latitude, course, and distance sailed, to find the other latitude and difference of longitude.

Example. A ship from Brest, in latitude $48^\backslash circ\; 23\text{'}\; N.$ and longitude $4^\backslash circ\; 30\text{'}\; W.$, sailed $S.\; W.\; \backslash frac\{3\}\{4\}\; W.$ 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 $46^\backslash circ\; 1\text{'}\; N.$, and the middle latitude $47^\backslash circ\; 12\text{'}$. Now make the angle DCB equal to $47^\backslash circ\; 12\text{'}$; and DC, being measured, will be 281, the difference of longitude: hence the longitude come to is $9^\backslash circ\; 11\text{'}\; W.$

By Calculation.

To find the difference of latitude.

To find the difference of longitude.²

PROB. III. Given both latitudes and course, required the distance and difference of longitude.

Example. A ship from St Antonio, in latitude $17^\backslash circ\; 0\text{'}\; N.$ and longitude $24^\backslash circ\; 25\text{'}\; W.$, sailed $N.\; W.\; \backslash frac\{3\}\{4\}\; N.$, till by observation her latitude was found to be $28^\backslash circ\; 34\text{'}\; N.$ Required the distance sailed, and longitude come to.

By Construction.

Fig. 19.

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.

To find the difference of longitude.

PROB. IV. Given one latitude, course, and departure, to find the other latitude, distance, and difference of longitude.

Example. A ship from latitude $26^\backslash circ\; 30\text{'}\; N.$, and longitude $45^\backslash circ\; 30\text{'}\; W.$, sailed $N.\; E.\; \backslash frac\{1\}\{2\}\; N.$ till her departure is 216 miles. Required the distance run, and latitude and longitude come to.

By Construction.

Fig. 20.

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 $30^\backslash circ\; 53\text{'}$, and middle latitude $28^\backslash circ\; 42\text{'}$. Now make the angle BCD equal to the middle latitude, and the difference of longitude DC applied to the scale will measure 246'.

¹ For R: cosine mid. lat. :: diff. of long. : departure,
And diff. of lat. : dep. :: R : tangent course;
Hence diff. of lat. : cosine mid. lat. :: diff. of long. : tang. course,
Or diff. of lat. : diff. of long. :: cosine mid. lat. : tang. course.

² This proportion is obvious, by considering the whole figure as an oblique-angled plane triangle.

PROB. V. Given both latitudes and distance to find the course and difference of longitude.

Example. From Cape Sable, in latitude 43° 24' N. and longitude 65° 39' W., a ship sailed 246 miles on a direct course between the south and east, and was then by observation in latitude 40° 48' N. Required the course and longitude in.

Make AB (fig. 21) equal to 156 miles; draw BC perpendicular to AB, and make AC equal to 246 miles. Draw CD, making with CB an angle of 42° 6' the middle latitude. Now DC will be found to measure 256, and the course or angle A will measure 50° 39'.

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 9° 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.

Make AB (fig. 22) 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 36½°, and the distance and difference of longitude, measured on the line of equal parts, will be found to be 1403 and 946 respectively.

PROB. VII. Given one latitude, distance, and departure; to find the other latitude, course, and difference of longitude.

Example. A ship from Bordeaux, in latitude 44° 50' N., and longitude 0° 35' W., sailed between the north and west 37½ miles, and made 210 miles of westing. Required the course, and the latitude and longitude come to.

With the given distance and departure make the triangle ABC (fig. 23). Now the course being measured on the line of chords is about 34½°, and the difference of latitude on the line of numbers is 309 miles; hence the latitude come to is 49° 59' N., and middle 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.

To find the difference of latitude.

To find the difference of longitude.

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 was found to be in longitude 5° 26' E., and has made 220 miles of easting. Required the latitude come to, course, and distance run.

By Construction.

Make BC (fig. 24) equal to the departure 220, and CD equal to the difference of longitude 396: then the middle latitude BCD being measured, will be found equal to 56° 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 to 54½°.

Fig. 24.

By Calculation.

To find the middle latitude.

To find the course.

To find the distance.

PROB. IX. Given the course and distance sailed, and difference of longitude; to find both latitudes.

Example. A ship from a port in north latitude, sailed S. E. ¼ S. 438 miles, and differed her longitude 7° 28'. Required the latitude sailed 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° 58', and the difference of latitude AB 324 miles; hence the latitude left is 51° 40', and that come to 46° 16'.

Fig. 25.

By Calculation.

To find the difference of latitude.

To find the middle latitude.

PROB. X. To determine the difference of longitude made good upon compound courses, by middle latitude sailing.

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 sailing.

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° 18' N., longitude 9° 46' W. sailed as follows:—S. W. by S. 54 miles, W. by N. 63 miles, N. N. W. 48 miles, and N. E. ½ E. 85 miles. Required the latitude and longitude come to.

The above method is that always practised to find the difference of longitude made good in the course of a day's run, and will, no doubt, give the difference of longitude tolerably exact in any probable run a ship may make in that time, especially near the equator. But in a high la-

titude, when the distances are considerable, this method is not to be depended on. To illustrate this, let a ship be supposed to sail from latitude 57° N., as follows: E. 240 miles, N. 240 miles, W. 240 miles, and S. 240 miles; then, by the above method, the ship will be come to the same place she left. It will, however, appear evident from the following consideration, that this is by no means the case; for let two ships, from latitude 61° N., and distant 240 miles, sail directly south till they are in latitude 57° N.; now, their distance, being computed by Problem IV. of Parallel Sailing, will be 269.6 miles; and therefore, if the ship sailed as above, she will be 29.6 miles west of the place sailed from, and the error in longitude will be equal to $240\; \backslash times\; \backslash secant\; 61^\backslash circ\; -\; \backslash secant\; 57^\backslash circ\; =\; 29.6\; \backslash times\; \backslash secant\; 57^\backslash circ\; =\; 54.4$.

Theorems might be investigated for computing the 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° 30' N., long. 27° 15' W., sailed as follows: S. S. W. 46 miles, S. W. 61 miles, S. by W. 59 miles, S. E. by E. 86 miles, S. by E. $\backslash frac\{1\}\{2\}$ E. 76 miles. Required the lat. and long. come to.

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 are resolved with the utmost accuracy, by the assistance 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 secants of each minute of latitude.

For by parallel sailing,
$R\; :\; \backslash cos.\; \backslash text\{\; of\; lat.\}\; ::\; \backslash text\{\; part\; of\; equat.\}\; :\; \backslash text\{\; similar\; part\; of\; parallel.\}$
And because the equator and meridian on the globe are equal; therefore,
$R\; :\; \backslash cos.\; \backslash text\{\; lat.\}\; ::\; \backslash text\{\; part\; of\; meridian\}\; :\; \backslash text\{\; similar\; part\; of\; parallel.\}$
Or sec. lat. : $R\; ::\; \backslash text\{\; part\; of\; merid.\}\; :\; \backslash text\{\; similar\; part\; of\; parallel.\}$

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 $1\text{'},\; 2\text{'},\; 3\text{'},\; \backslash dots$ &c. and the corresponding parts of the enlarged meridian be represented by $a,\; b,\; c,\; \backslash dots$; then,

Hence secant $1\text{'}\; :\; \backslash text\{\; part\; of\; mer.\; \}\; a\; ::\; \backslash text\{\; secant\; \}\; 2\text{'}\; :\; \backslash text\{\; part\; of\; mer.\; \}\; b\; ::\; \backslash text\{\; secant\; \}\; 3\text{'}\; :\; \backslash text\{\; part\; of\; mer.\; \}\; c,\; \backslash dots$

Therefore, by 12th V. Euclid,
Secant $1\text{'}\; :\; \backslash text\{\; part\; of\; mer.\; \}\; a\; ::\; \backslash text\{\; secant\; \}\; 1\text{'}\; +\; \backslash text\{\; secant\; \}\; 2\text{'}\; +\; \backslash text\{\; secant\; \}\; 3\text{'},\; \backslash dots\; :\; \backslash text\{\; parts\; of\; \}\; a\; +\; b\; +\; \backslash text\{\; mer.\; \}\; c,\; \backslash dots$

That is, the meridional parts of any given latitude are equal to the sum of the secants of the minutes in that latitude.¹

Since $CD\; :\; LK\; ::\; R\; :\; \backslash text\{secant\; \}\; LD$, fig. 15.

And in the triangle CED,
$ED\; :\; CD\; ::\; R\; :\; \backslash text\{tangent\; \}\; CED;$

Therefore $ED\; :\; LK\; ::\; R^2\; :\; \backslash text\{secant\; \}\; LD\; \backslash times\; \backslash text\{tangent\; \}\; CED.$

But $\backslash frac\{ED\; \backslash times\; \backslash text\{sec.\; \}\; LD\}\{R\}$ is the enlarged portion of the meridian answering to ED. Now the sum of all the quantities $\backslash frac\{ED\; \backslash times\; \backslash text\{secant\; \}\; LD\}\{R\}$ 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.

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 rectilinear triangle, having an angle equal to the course; the adjacent

side equal to the meridional difference of latitude, and the Mercator's 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.

The meridional parts on the terrestrial spheroid of $\backslash frac\{1\}\{60\}$th of compression to any latitude $l$, may be found from the following formula. $P\; =\; 7915\; \backslash cdot\; 705\; \backslash log.\; \backslash tan.\; (45^\backslash circ\; +\; \backslash frac\{1\}\{2\}l)\; -\; 22\; \backslash cdot\; 88\; \backslash sin.\; l\; -\; 0\; \backslash cdot\; 0508\; \backslash sin.\; 3l,\; \backslash dots$ &c.

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 $42^\backslash circ\; 52\text{'}\; N$, longitude $9^\backslash circ\; 17\text{'}\; W$, and Port Praya in the island of St Jago, in latitude $14^\backslash circ\; 54\text{'}\; N$, and longitude $23^\backslash circ\; 29\text{'}\; W$.

1678

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 $23^\backslash circ\; 37\text{'}$.

By Calculation.

To find the course.

To find the distance.

¹ 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.

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 $38^\{\backslash circ\}\; 47\text{'}\; N.$, longitude $75^\{\backslash circ\}\; 4\text{'}\; W.$, sailed 267 miles N. E. by 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 $42^\{\backslash circ\}\; 29\text{'}\; N.$, 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 $71^\{\backslash circ\}\; 48\text{'}\; W.$

By Calculation.

Lat. Cape Hinlopen, = $38^\{\backslash circ\}\; 47\text{'}\; N.$ Mer. parts, 2528
Difference of lat. . . . 3 42 N.

Latitude come to, . . . 42 29 N. Mer. parts, 2821

Meridional difference of latitude, 293

To find the difference of longitude.

to the difference of longitude 195-8 . . . 2-29176

Longitude Cape Hinlopen, . . . $75^\{\backslash circ\}\; 4\text{'}\; W.$
Difference of longitude, . . . 3 16 E.

Longitude come to, . . . 71 48 W.

PROB. III. Given the latitudes and bearing of two places, to find their distance and difference of longitude.

Example. A ship from Port Canso in Nova Scotia, in latitude $45^\{\backslash circ\}\; 20\text{'}\; N.$, longitude $60^\{\backslash circ\}\; 55\text{'}\; W.$, sailed S. E. $\backslash frac\{1\}\{4\}\; S.$, and, by observation, was found to be in latitude $41^\{\backslash circ\}\; 14\text{'}\; N.$ Required the distance sailed and longitude come to.

Lat. Port Canso, . . . $45^\{\backslash circ\}\; 20\text{'}\; N.$ . Mer. parts, 3058
Lat. in, by observation, $41^\{\backslash circ\}\; 14\text{'}\; N.$ . 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 to 338; draw AE, making an angle with AD equal to $3\backslash frac\{1\}\{4\}$ 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.

To find the difference of longitude.

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 $33^\{\backslash circ\}\; 58\text{'}\; N.$, longitude $6^\{\backslash circ\}\; 20\text{'}\; W.$, the corrected course was N.W. by W. $\backslash frac\{1\}\{2\}\; W.$, 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, will be found to be equal to 476 and 224 respectively; hence the latitude come to is $37^\{\backslash circ\}\; 42\text{'}\; N.$, and meridional difference of latitude 276. Make AD equal to 276; and draw DE perpendicular thereto, meeting the distance produced in E; then DE applied to the scale will be found to measure 516. The longitude in is, therefore, $14^\{\backslash circ\}\; 56\text{'}\; W.$

By Calculation.

To find the distance.

To find the difference of latitude.

Latitude in, 37 42 N. Mer. parts, 2445

Mer. difference of latitude, 276

To find the difference of longitude.