NAVIGATION

N A V I G A T I O N.

NAVIGATION, is the art of conducting or carrying a ship from one port to another. In order to understand this science, particularly the theoretical parts of it, it is necessary that the student be acquainted with the general principles of GEOMETRY, ASTRONOMY, and TRIGONOMETRY. See these articles.

SECT. I. Of the LOG-LINE and COMPASS.

1. THE method commonly made use of for measuring a ship's way at sea, or how far she runs in a given space of time, is by the log-line, and half-minute glass.

2. The log is a flat piece of wood, in shape like a flounder, having a piece of lead fastened to its bottom, which makes it stand or swim upright in the water: to this log is tied or fastened a long line, which is called the log-line; and this is commonly divided into certain spaces, each of which is, or ought to be, such a proportional part of a nautical mile (60 of which make a degree of a great circle on the earth) as half a minute (the time allowed for the experiment) is of an hour.

3. These spaces are called knots, because at the end of each of them there is a piece of twine with knots in it, interceded between the strands of the line, which shews how many of these spaces or knots are run out during the half minute. They commonly begin to be counted at the distance of about 10 fathom or 60 feet from the log; that so the log, when it is hove over board, may be out of the eddy of the ship's wake before they begin to count; and for the more ready discovery of this point of commencement, there is commonly fastened at it a piece of red rag.

4. The log being thus prepared, and hove over board from the poop, and the line veered out (by the help of a reel, that turns easily, and about which it is wound) as fast as the log will carry it away, or rather as the ship fails from it, will shew, according to the time of veering, how far the ship has run in a given time, and consequently her rate of sailing.

5. A degree of a meridian, according to the exactest measures, contains about 69.545 English miles; and each mile by the statute being 5280 feet, therefore a degree

of a meridian will be about 367200 feet; whence the $\backslash frac\{1\}\{60\}$ of that, viz. 2 minute, or nautical mile, must contain 6120 standard feet; consequently, since $\backslash frac\{1\}\{60\}$ minute is the $\backslash frac\{1\}\{120\}$ part of an hour, and each knot being the same part of a nautical mile, it follows, that each knot will contain the $\backslash frac\{1\}\{120\}$ of 6120 feet, viz. 51 feet.

6. Hence it is evident, that whatever number of knots the ship runs in half a minute, the same number of miles she will run in one hour, supposing her to run with the same degree of velocity during that time; and therefore it is the general way to heave the log every hour to know her rate of sailing: but if the force or direction of the wind vary, and not continue the same during the whole hour; or if there has been more sail set, or any sail handed, that so the ship has run swifter or slower in any part of the hour than she did at the time of heaving the log; then there must be an allowance made accordingly for it, and this must be according to the discretion of the artist.

7. Sometimes when the ship is before the wind, and there is a great sea setting after her, it will bring home the log, and consequently the ship will sail faster than is given by the log. In this case it is usual, if there be a very great sea, to allow one mile in ten, and less in proportion, if the sea be not so great. But for the generality, the ship's way is really greater than that given by the log; and therefore, in order to have the reckoning rather before than behind the ship, (which is the safest way,) it will be proper to make the space on the log-line between knot and knot to consist of 50 feet instead of 51.

8. If the space between knot and knot on the log-line should happen to be too great in proportion to the half-minute glass, viz. greater than 50 feet, then the distance given by the log will be too short; and if that space be too small, then the distance run (given by the log) will be too great; therefore to find the true distance run in either case, having measured the distance between knot and knot, we have the following proportion, viz.

As the true distance, 50 feet, is to the measured distance; so are the miles of distance given by the log, to the true distance in miles that the ship has run.

EXAMPLE

EXAMPLE I. Suppose a ship runs at the rate of 6½ knots in half a minute; but measuring the space between knot and knot, I find it to be 56 feet: Required the true distance in miles.

Making it, As 50 feet is to 56 feet, so is 6.25 knots to 7 knots; I find that the true rate of sailing is 7 miles in the hour.

EXAMPLE II. Suppose a ship runs at the rate of 6½ knots in half a minute; but measuring the space between knot and knot, I find it to be only 44 feet: Required the true rate of sailing.

Making it, As 50 feet is to 44 feet, so is 6.5 knots to 5.72 knots; I find that the true rate of sailing is 5.72 miles in the hour.

9. Again, supposing the distance between knot and knot on the log-line to be exactly 50 feet, but that the glass is not 30 seconds; then, if the glass require longer time to run than 30 seconds, the distance given will be too great, if estimated by allowing 1 mile for every knot run in the time the glass runs; and, on the contrary, if the glass requires less time to run than 30 seconds, it will give the distance sailed too small. Consequently, to find the true distance in either case, we must measure the time the glass requires to run out (by the method in the following article;) then we have the following proportion, viz.

As the number of seconds the glass runs, is to half a minute, or 30 seconds; so is the distance given by the log, to the true distance.

EXAMPLE I. Suppose a ship runs at the rate of 7½ knots in the time the glass runs; but measuring the glass, I find it runs 34 seconds: Required the true distance sailed.

Making it, As 34 seconds is to 30 seconds, so is 7.5 to 6.6; I find that the ship sails at the rate of 6.6 miles an hour.

EXAMPLE II. Suppose a ship runs at the rate of 6½ knots; but measuring the glass, I find it runs only 25 seconds: Required the true rate of sailing.

Making it, As 25 seconds is to 30 seconds, so is 6.5 knots to 7.8 knots; I find that the true rate of sailing is 7.8 miles an hour.

10. In order to know how many seconds the glass runs, you may try it by a watch or clock, that vibrates seconds; but if neither of these be at hand, then take a line, and to the one end fastening a plummet, hang the other upon a nail or peg, so as the distance from the peg to the centre of the plummet be 39½ inches: Then this put into motion will vibrate seconds; i. e. every time it passes the perpendicular, you are to count one second; consequently, by observing the number of vibrations that it makes during the time the glass is running, we know how many seconds the glass runs.

11. If there be an error both in the log line and half-minute glass, viz. if the distance between knot and knot and the log-line be either greater or less than 50 feet, and the glass runs either more or less than 30 seconds; then the finding out the ship's true distance will be somewhat more complicate, and admit of three cases, viz.

CASE I. If the glass runs more than 30 seconds, and the distance between knot and knot be less than 50 feet, then the distance given by the log-line, viz. by allowing

1 mile for each knot the ship sails while the glass is running, will always be greater than the true distance, since either of these errors give the distance too great. Consequently, to find the true rate of sailing in this case, we must first find (by Art. 8.) the distance, on the supposition that the log-line is only wrong, and then with this (by Art. 9.) we shall find the true distance.

EXAMPLE. Suppose a ship is found to run at the rate of 6 knots; but examining the glass, I find it runs 35 seconds; and measuring the log-line, I find the distance between knot and knot to be but 46 feet: Required the true distance run.

First, (by Art. 8.) We have the following proportion, viz. As 50 feet : 46 feet :: 6 knots : 5.52 knots. Then (by Art. 9.) As 35 seconds : 30 seconds :: 5.52 knots : 4.73 knots. Consequently the true rate of sailing is 4.73 miles an hour.

CASE II. If the glass be less than 30 seconds, and the place between knot and knot be more than 50 feet; then the distance given by the log will always be less than the true distance, since either of these errors lessen the true distance.

EXAMPLE. Suppose a ship is found to run at the rate of 7 knots; but examining the glass, I find it runs only 25 seconds; and measuring the space between knot and knot on the log line, I find it is 54 feet: Required the true rate of sailing.

First, (by Art. 9.) As 25 seconds : 30 seconds :: 7 knots : 8.4 knots. Then (by Art. 8.) As 50 feet : 54 feet :: 8.4 knots : 9.072 knots. Consequently the true rate of sailing is 9.072 miles an hour.

CASE III. If the glass runs more than 30 seconds, and the space between knot and knot be greater than 50 feet; or if the glass runs less than 30 seconds, and the space between knot and knot be less than 50 feet; then, since in either of these two cases the effects of the errors are contrary, it is plain the distance will sometimes be too great, and sometimes too little, according as the greater quantity of the error lies; as will be evident from the following examples.

EXAMPLE I. Suppose a ship is found to run at the rate of 9½ knots per glass; but examining the glass, it is found to run 36 seconds; and by measuring the space between knot and knot, it is found to be 58 feet: Required the true rate of sailing.

First, (by Art. 8.) As 50 feet : 58 feet :: 9.5 knots : 11.02 knots. Then (by Art. 9.) As 36 seconds : 30 seconds :: 11.02 knots : 8.7 knots. Consequently the ship's true rate of sailing is 8.7 miles an hour.

EXAMPLE II. Suppose a ship runs at the rate of 6 knots per glass; but examining the glass, it is found to run only 20 seconds; and by measuring the log-line, the distance between knot and knot is found to be but 38 feet: Required the true rate of sailing.

First, (by Art. 8.) As 50 feet : 38 feet :: 6 knots : 4.56 knots. Then (by Art. 9.) As 20 seconds : 30 seconds :: 4.56 knots : 6.84 knots. Consequently the true rate of sailing is 6.84 miles an hour.

But if in this case it happen, that the time the glass takes to run be to the distance between knot and knot, as 30, the seconds in half a minute, is to 50, the true distance

distance between knot and knot; then it is plain, that whatever number of seconds the glass consists of, and whatever number of feet is contained between knot and knot; yet the distance given by the log-line, will be the true distance in miles.

12. Though the method of measuring the ship's way by the log-line, described in the foregoing articles, be that which is now commonly made use of; yet it is subject to several errors, and these very considerable. For first, the half-minute or quarter-minute glasses (by which, and the log, the ship's way is determined) are seldom or never true, because dry and wet weather have a great influence on them; so that at one time they may run more, and at another time fewer than 30 seconds, and it is evident that a small error in the glass will cause a sensible one in the ship's way. Again, the chief property of the log is to have it swim upright, or perpendicular to the horizon: but this is too often wanting in logs, because few seamen examine whether it is so or not, and generally take it upon trust, being satisfied if it weigh a little more at the stern than the head: and from this there flows an error in the reckoning; for if the log does not swim upright, it will not hold water, nor remain steady in the place where it is heaved, since the least check in the hand in veering the line will make it come up several feet: this repeated will make the errors become fathoms, and perhaps knots, which, how insignificant so ever they appear, are miles and parts of miles, and amount to a good deal in a long voyage. Another inconvenience attending the log line is its stretching and shrinking; for when a new line is first used, let it be ever so well stretched upon the deck, and measured as true as possible, yet after wetting it shrinks considerably; and consequently to be the better assured of the ship's way by the log-line, we ought to measure and alter the knots on it every time before we use it; but this is seldom done oftener than once a week, and sometimes not above once or twice in a whole voyage; also when the line is measured to its greatest degree of shrinking, it is generally left there; and when, by much use, it comes to stretch again, it is seldom or never mended, though it will stretch beyond what it first shrunk. These, and many other errors, too well known, attending that method of measuring the ship's way by the log-line, plainly answers for a great many errors committed in reckonings. So it is to be wished, that either this method were improved or amended, or that some other method less subject to error were found out.

13. The meridian and prime vertical of any place cuts the horizon in 4 points, at 90 degrees distance from one another, viz. North, South, East, and West; that part of the meridian which extends itself from the place to the north point of the horizon is called the north line; that which tends to the south point of the horizon, is called the south line; and that part of the prime vertical which extends towards the right hand of the observer, when his face is turned to the north, is called the east line; and lastly, that part of the prime vertical which tends towards the left hand, is called the west line; the four points in which these lines meet the horizon, are called the cardinal points.

VOL. III. N^o. 84.

2

14. In order to determine the course of the winds, and to discover their various alterations or shiftings; each quadrant of the horizon intercepted between the meridian and prime vertical, is usually divided into eight equal parts, and consequently the whole horizon into thirty-two; and the lines drawn from the place on which the observer standeth, to the points of division in his horizon, are called rumb lines, the four principal of which are those described in the preceding article, each of them having its name from the cardinal point in the horizon towards which it tends; the rest of the rumb lines have their names compounded of the principal lines on each side of them, as in the figure; (see Navigation Plates, No. 1.) and over which-soever of these lines the course of the wind is directed, that wind takes its name accordingly.

15. The instrument commonly used at sea for directing the ship's way, is called the mariner's compass; which consists of a card and two boxes. The card is a circle made to represent the horizon, whose circumference is quartered and divided into degrees, and also into thirty-two equal parts, by lines drawn from the centre to the several points of division, called points of the compass. On the back side of the card, and just below the south and north line, is fixed a steel needle with a brass cupola, or hollow center in the middle, which is placed upon the end of a fine pin, upon which the card may easily turn about; the needle is touched with a load stone, by which a certain virtue is infused into it, that makes it (and consequently the south and north line on the card above it) hang nearly in the plane of the meridian, by which means the south and north lines on the card produced would meet the horizon in the south and north points; and consequently all the other lines on the card produced would meet the horizon in the respective points.

16. The card is represented in No. 1. in which you may observe, that the capital letters N, S, E, W, denote the four cardinal points, viz. N the North, S the South, &c. and the small letter b signifies the word by: the Rhomb in the middle between any two of the cardinals are expressed by the letters denoting these cardinals, that which denotes the point lying in the meridian having the precedence; thus the rhomb in the middle between the north and east is expressed N E, which is to be read North-East; also S W denotes the South-West rhomb, &c. the other rhombs are expressed according to their situation with respect to these middle rhombs, and the nearest cardinals, as is plain from the foresaid figure.

17. The card is put into a round box, made for it, having a pin erected in the middle, upon which the hollow centre of the needle is fixed, so as the card may lie horizontal, and easily vibrate according to the motion of the needle: the box is covered over with a smooth glass, and is hung in a brass hoop upon two cylindrical pins, diametrically opposite to one another; and this hoop is hung within another brass circle, upon two pins at right angles with the former. These two circles, and the box, are placed in another square wooden box, so that the innermost box, and consequently the card, may keep horizontal which way soever the ship heels.

18. Since the meridians do all meet at the poles, and there

4 Z

there form certain angles with one another; and since, if we move never so little towards the east or west, from one place to another, we thereby change our meridian, and in every place the east and west line being perpendicular to the meridian; it follows, that the east and west line in the first place will not coincide with the east and west line in the second, but be inclined to it at a certain angle: and consequently all the other rhomb lines at each place will be inclined to each other, they always forming the same angles with the meridian. Hence it follows, that all rhombs, except the four cardinals, must be curves or helispherical lines, always tending towards the pole, and approaching it by infinite gyrations or turnings, but never falling into it. Thus let P (No. 2.) be the pole, EQ an arch of the equator, PE PA, &c. meridians, and EFGHKL any

rhombs: then because the angles PEF, PFG, &c. are by the nature of the rhomb line equal, it is evident that it will form a curve line on the surface of the globe, always approaching the pole P, but never falling into it; for if it were possible for it to fall into the pole, then it would follow, that the same line could cut an infinite number of other lines at equal angles, in the same point; which is absurd.

19. Because there are 32 rhombs (or points in the compass) equally distant from one another, therefore the angle contained between any two of them adjacent will be $11^\{\backslash circ\}\; 15\text{'}$, viz. $\backslash frac\{1\}\{32\}$ part of $360^\{\backslash circ\}$; and so the angle contained between the meridian and the NNE will be $11^\{\backslash circ\}\; 15\text{'}$, and between the meridian and the NNE will be $22^\{\backslash circ\}\; 30\text{'}$; and so of the rest, as in the following table.

A Table of the Angles which every $\backslash frac\{1\}\{32\}$ Point of the Compass makes with the Meridian.

SECT. 2. OF PLAIN SAILING.

1. This method of sailing supposes the earth to be a plane, and the meridians parallel to one another; and likewise the parallels of latitude at equal distance from one another, as they really are upon the globe. Though this method be in itself evidently false; yet in a short run, and especially near the equator, an account of the ship's way may be kept by it tolerably well.

2. The angle formed by the meridian and rhumb that a ship sails upon, is called the ship's course. Thus if a ship sails on the NNE rhumb, then her course will be $22^\{\backslash circ\}\; 30\text{'}$, and so of others.

3. The distance between two places lying on the same parallel counted in miles of the equator, or the distance of one place from the meridian of another counted as above on the parallel passing over that place, is called meridional distance; which, in plain sailing, goes under the name of departure.

4. Let A (No. 3.) denote a certain point on the earth's surface, AC its meridian, and AD the parallel of lati-

tude passing through it; and suppose a ship to sail from A on the NNE rhumb till she arrive at B; and through B draw the meridian BD, (which, according to the principles of plain sailing, must be parallel to CA) and the parallel of latitude BC; then the length of AB, viz. how far the ship has sailed upon the NNE rhumb, is called her distance; AC or BD will be her difference of latitude, or northing; CB will be her departure, or easting; and the angle CAB will be the course. Hence it is plain, that the distance sailed will always be greater than either the difference of latitude or departure; it being the hypothenuse of a right-angled triangle, whereof the other two are the legs; except the ship sails either on a meridian, or a parallel of latitude: for if the ship sails on a meridian, then it is plain, that her distance will be just equal to her difference of latitude, and she will have no departure; but if she sail on a parallel, then her distance will be the same with her departure, and she will have no difference of latitude. It is evident also from the figure, that if the course be less than 4 points, or 45 degrees, its complement, viz. the other oblique angle, will be greater

greater than 45 degrees, and so the difference of latitude will be greater than the departure; but if the course be greater than 4 points, then the difference of latitude will be less than the departure; and lastly, if the course be just 4 points, the difference of latitude will be equal to the departure.

5. Since the distance, difference of latitude, and departure, form a right-angled triangle, in which the oblique angle opposite to the departure is the course, and the other its complement; therefore, having any two of these given, we can (by plain trigonometry) find the rest; and hence arise the cases of plain-failing, which are as follow.

CASE I. Course and distance given, to find difference of latitude and departure.

EXAMPLE. Suppose a ship fails from the latitude of 30° 25' north, NNE, 32 miles. (No. 4.) Required the difference of latitude and departure, and the latitude come to. Then (by right angle trigonometry,) we have the following analogy, for finding the departure, viz.

And since her former latitude was north, and her difference of latitude also north; therefore, To the latitude failed from — — 30° 25' N add the difference of latitude — — 00° 29.57

and the sum is the latitude come to 30° 54.57 N

By this case are calculated the tables of difference of latitude, and departure, to every degree, point, and quarter-point of the compass.

CASE II. Course and difference of latitude given, to find distance and departure

EXAMPLE. Suppose a ship, in the latitude of 45° 25' north, fails NE&N $\backslash frac\{1\}\{2\}$ easterly (No. 5.) till she come to the latitude of 46° 55' north. Required the distance and departure made good upon that course.

Since both latitudes are northerly, and the course also northerly; therefore,

the difference of latitude, equal to 90 miles.

And (by rectangular trigonometry) we have the following analogy, for finding the departure BD, viz.

so is the tangent of course A — 39° 22' 9.91404 to the departure BD — — 73.84 1.86828 so the ship has got 73.84 miles to the eastward of her former meridian.

Again, for the distance AD, we have (by rectangular trigonometry) the following proportion, viz.

CASE III. Difference of latitude and distance given, to find course and departure.

EXAMPLE. Suppose a ship fails from the latitude of 56° 50' north, on a rhomb between south and west, 126 miles, and she is then found by observation to be in the latitude of 55° 40' north. Required the course she failed on, and her departure from the meridian. No. 6.

Since the latitudes are both north, and the ship failing towards the equator; therefore,

and the remainder — — — — — 01° 40' equal to 70 miles, is the difference of latitude.

By rectangular trigonometry we have the following proportion for finding the angle of the course F, viz.

CASE IV. Difference of latitude and departure given, to find course and distance.

EXAMPLE. Suppose a ship fails from the latitude of 44° 50' north, between south and east, till she has made 64 miles of sailing, and is then found by observation to be in the latitude of 42° 56' north. Required the course and distance made good.

Since the latitudes are both north, and the ship failing towards the equator; therefore,

and their remains — — — — — 01° 54' equal to 114 miles, the difference of latitude or southing.

In this case by (rectangular trigonometry) we have the following proportion to find the course KGL (No. 7.) viz.

to the tangent of course G — $29^\{\backslash circ\},\; 19\text{'}$ — 9.74928 which, because the ship is sailing between south and east, will be south $29^\{\backslash circ\},\; 19\text{'}$ east, or SSE $\backslash frac\{1\}\{2\}$ east nearly.

Then for the distance, we shall have (by rectangular trigonometry) the following analogy, viz.

consequently the ship has sailed on a SSE $\backslash frac\{1\}\{2\}$ east course 130.8 miles.

CASE V. Distance and departure given, to find course and difference of latitude.

EXAMPLE. Suppose a ship at sea sails from the latitude of $34^\{\backslash circ\},\; 24\text{'}$ north, between north and west 124 miles, and is found to have made of westing 86 miles. Required the course steered, and the difference of latitude or northing made good.

In this case (by rectangular trigonometry) we have the following proportion for finding the course ADB, (No. 8.) viz.

so the ship's course is north $33^\{\backslash circ\},\; 54\text{'}$ west, or NW & N $\backslash frac\{1\}\{4\}$ west nearly.

Then for the difference of latitude, we have (by rectangular trigonometry) the following analogy, viz.

which is equal to 1 degree and 29 minutes nearly. Hence, to find the latitude the ship is in, since both latitudes are north, and the ship sailing from the equator; therefore,
To the latitude sailed from — — — $34^\{\backslash circ\},\; 24\text{'}$
add the difference of latitude — — — $1^\{\backslash circ\},\; 29\text{'}$
the sum is — — — — — $35^\{\backslash circ\},\; 53\text{'}$
the latitude the ship is in north.

CASE VI. Course and departure given, to find distance and difference of latitude.

EXAMPLE. Suppose a ship at sea, in the latitude of $24^\{\backslash circ\},\; 30\text{'}$ south, sails SE & S, till she has made of easting 96 miles. Required the distance and difference of latitude made good on that course.

In this case, by Rectangular Trigonometry, and by Case 2. we have the following proportion for finding the distance, (No. 9.) viz.

Then, for the difference of latitude, we have (by rectangular trigonometry) the following analogy, viz.

equal to $2^\{\backslash circ\},\; 24\text{'}$ nearly. Consequently, since the latitude the ship sailed from was south, and she sailing still towards the south,

is the latitude she is come to south.

6. When a ship steers several courses in 24 hours, then the reducing all these into one, and thereby finding the course and distance made good upon the whole, is commonly called the resolving of a traverse.

7. At sea they commonly begin each day's reckoning from the noon of that day, and from that time they set down all the different courses and distances steered by the ship till noon next day upon the log-board; then from these several courses and distances had from the compass and log line, they compute the difference of latitude and departure for each course (by Case 1. of Plain Sailing;) and these, together with the courses and distances, are set down in a table called the traverse table; which consists of five columns: in the first of which are placed the courses and distances; in the two next the differences of latitude belonging to these courses, according as they are north or south; and in the two last are placed the departures belonging to these courses, according as they are east or west. Then they sum up all the northings, and all the southings; and taking the difference of these, they know the difference of latitude made good by the ship in the last 24 hours, which will be north or south, according as the sum of the northings or southings is greatest: The same way, by taking the sum of all the eastings, and likewise of all the westings, and subtracting the lesser of these from the greater, the difference will be the departure made good by the ship last 24 hours, which will be east or west according as the sum of the eastings is greater or less than the sum of the westings; then from the difference of latitude and departure made good by the ship last 24 hours, found as above, they find the true course and distance made good upon the whole (by Case 4. of Plain Sailing), as also the course and distance to the intended port.

EXAMPLE. Suppose a ship at sea, in the latitude of $48^\{\backslash circ\},\; 24\text{'}$ north at noon any day, is bound to a port in the latitude of $43^\{\backslash circ\},\; 40\text{'}$ north, whose departure from the ship is 144 miles east; consequently the direct course and distance of the ship is SSE $\backslash frac\{1\}\{2\}$ east 315 miles; but by reason of the shifting of the winds she is obliged to steer the following courses till noon next day, viz. SE & S 56 miles, SSE 64 miles, NW & W 48 miles, SW $\backslash frac\{1\}\{2\}$ west 54 miles, and SE & S $\backslash frac\{1\}\{2\}$ east 74 miles. Required the course and distance made good the last 24 hours, and the bearing and distance of the ship from the intended port.

The solution of this traverse depends entirely on the 1st and 4th cases of Plain Sailing; and first we must (by Case 1.) find the difference of latitude and departure for each course. Thus,

1 Course SE & S distance 56 miles.

For departure.

fo is the fine of the course 33°, 45' — 9.74474
to the departure — 31.11 — 1.49293

For difference of latitude.

As radius — — — — — 10.00000
is to the distance — — — — — 1.74819

fo is the co-line of the course 33°, 45' — 9.91985
to the diff. of latitude — 46.57 — 1.66804

2. Course SSE and distance 64 miles.

For departure.

As radius — — — — — 10.00000
is to the distance — — — — — 1.80618

fo is the fine of the course 22°, 30' — 9.58284
to the departure — — — — — 1.38902

For difference of latitude.

As radius — — — — — 10.00000
is to the distance — — — — — 1.80618

fo is the co-line of the course 22°, 30' — 9.96562
to the difference of latitude 59.13 — 1.77180

3. Course NW&W and distance 48 miles.

For departure.

As radius — — — — — 10.00000
is to the distance — — — — — 1.68124

fo is the fine of the course 56°, 15' — 9.91985
to the departure — — — — — 1.60109

For difference of latitude.

As radius — — — — — 10.00000
is to the distance — — — — — 1.68124

fo is the co-line of the course 56°, 15' — 9.74474
to the difference of latitude 26.67 — 1.42598

4. Course S&W $\backslash frac\{1\}\{2\}$ west and distance 54 miles.

For departure.

As radius — — — — — 10.00000
is to the distance — — — — — 1.73239

fo is the fine of the course 16°, 52' — 9.46262
to the departure — — — — — 1.19501

For difference of latitude.

As radius — — — — — 10.00000
is to the distance — — — — — 1.73239

fo is the co-line of the course 16°, 52' — 9.98090
to the difference of latitude 51.67 — 1.71329

5. Course SE&S $\backslash frac\{1\}\{2\}$ east and distance 74 miles.

For departure.

As radius — — — — — 10.00000
is to the distance — — — — — 1.86923

fo is the fine of the course 39°, 22' — 9.80228
to the departure — — — — — 1.67151

For difference of latitude.

As radius — — — — — 10.00000
is to the distance — — — — — 1.86923

fo is the co-line of the course 39°, 22' — 9.88824
to the difference of latitude 57.21 — 1.75747

Now these several courses and distances, together with the differences of latitude and departures deduced from them, being set down in their proper columns in the traverse table, will stand as on next column.

From that table it is plain, since the sum of the northings is 26.67, and of the southings 214.58, the difference between these, viz. 187.91, will be the southing made good by the ship the last 24 hours; also the sum of the eastings being 102.55, and of the westings 55.58, the difference 46.97 will be the easting or departure made good by the ship's

last 24 hours; consequently, to find the true course and distance made good by the ship in that time, it will be, (by Case 4. of Plain Sailing.)

As the difference of latitude — 187.91 2.27393
is to radius — — — — — 10.00000

fo is the departure — — — — — 46.97 1.67182
to the tangent of the course 14°, 03' — 9.39789

which is SE $\backslash frac\{1\}\{2\}$ east nearly. Then for the distance, it will be,
As radius — — — — — 10.00000

is to the difference of latitude 187.91 2.27393
fo is the secant of the course 14°, 03' 10.01319

to the distance — — — — — 193.7 2.28712
consequently the ship has made good the last 24 hours,
on a SE $\backslash frac\{1\}\{2\}$ east course, 193.7 miles: And since the ship
is sailing towards the equator; therefore,

From the latitude sailed from — — — — — 48°, 24' N
take the diff. of latitude made good — — — — — 3, 08 S

there remains — — — — — 45, 16 N
the latitude the ship is in north. And because the port
the ship is bound for lies in the latitude of 43° 40' north,
and consequently south of the ship; therefore,

From the latitude the ship is in — — — — — 45°, 16' N
take the latitude she is bound for — — — — — 43, 40 N

and there remains — — — — — 1, 36
or 96 miles, the difference of latitude or southing the ship
has to make. Again, the whole easting the ship had to

make being 144 miles, and she having already made
46.97 or 47 miles of easting; therefore the departure or
easting she still has to make will be 97 miles: consequently,
to find the direct course and distance between the ship
and the intended port, it will be (by Case 4. of Plain
Sailing)

As the difference of latitude — 96 — 1.98227
is to radius — — — — — 10.00000

fo is the departure — — — — — 97 — 1.98677
to the tangent of the course — 45°, 19' 10.00450

And

As radius — — — — — 10.00000
is to the difference of latitude 96 — 1.98227

fo is the secant of the course 45°, 19' — 10.15293
to the distance — — — — — 136.5 — 2.13620

whence the true bearing and distance of the intended
port is SE, 136.5 miles.

1. SINCE the parallels of latitude do always decrease the nearer they approach the pole, it is plain a degree on any of them must be less than a degree upon the equator. Now in order to know the length of a degree on any of them, let PB (No. 10.) represent half the earth's axis, PA a quadrant of a meridian, and consequently A a point on the equator, C a point on the meridian, and CD a perpendicular from that point upon the axis, which plainly will be the sine of CP the distance of that point from the pole, or the co-sine of CA its distance from the equator; and CD will be to AB, as the sine of CP, or co-sine of CA, is to the radius. Again, if the quadrant PAB is turned round upon the axis PB, it is plain the point A will describe the circumference of the equator whose radius is AB, and any other point C upon the meridian will describe the circumference of a parallel whose radius is CD.

COR. I. Hence (because the circumference of circles are as their radii) it follows, that the circumference of any parallel is to the circumference of the equator, as the co-sine of its latitude is to radius.

COR. II. And since the wholes are as their similar parts, it will be, As the length of a degree on any parallel is to the length of a degree upon the equator, so is the co-sine of the latitude of that parallel to radius.

COR. III. Hence, as radius is to the co-sine of any latitude, so are the minutes of difference of longitude between two meridians, or their distance in miles upon the equator, to the distance of these two meridians on the parallel in miles.

COR. IV. And as the co-sine of any parallel is to radius, so is the length of any arch on that parallel (intercepted between two meridians) in miles, to the length of a similar arch on the equator, or minutes of difference of longitude.

COR. V. Also, as the co-sine of any one parallel is to the co-sine of any other parallel, so is the length of any arch on

the first, in miles, to the length of the same arch on the other in miles.

2. From what has been said, arises the solution of the several cases of parallel sailing, which are as follow.

CASE I. Given the difference of longitude between two places, both lying on the same parallel; to find the distance between those places.

EXAMPLE I. Suppose a ship in the latitude of $54^\{\backslash circ\}\; 20\text{'}$ north, sails directly west on that parallel till she has differed her longitude $12^\{\backslash circ\}\; 45\text{'}$; required the distance sailed on that parallel.

First, The difference of longitude reduced into minutes, or nautical miles, is 765, which is the distance between the meridian sailed from, and the meridian come to, upon the equator; then to find the distance between these meridians on the parallel of $54^\{\backslash circ\}\; 20\text{'}$, or the distance sailed, it will be, by Cor. 3. of the last article,

As radius — — — — — 10.00000
is to the co-sine of the lat. $54^\{\backslash circ\}\; 20\text{'}$ — — — — — 9.76572
so are the minutes of diff. long. — — — — — 765 — — — — — 2.88366
to the distance on the parallel — — — — — 446.1 — — — — — 2.64938

EXAMPLE II. A degree on the equator being 60 minutes or nautical miles; required the length of a degree on the parallel of $51^\{\backslash circ\}\; 32\text{'}$.

By Cor. 3. of the last article, it will be
As radius — — — — — 10.00000
is to the co-sine of the latitude $51^\{\backslash circ\}\; 32\text{'}$ — — — — — 9.79383
so are the minutes in 1 degree on the equator. 60 — — — — — 1.77815
to — — — — — 37.32 — — — — — 1.57198
the miles answering to a degree on the parallel of $51^\{\backslash circ\}\; 32\text{'}$.

By this problem the following table is constructed, shewing the geographic miles answering to a degree on any parallel of latitude; in which you may observe, that the columns marked at the top with D.L. contain the degrees of latitude belonging to each parallel; and the adjacent columns marked at the top, Miles, contain the geographic miles answering to a degree upon these parallels.

A Table shewing how many Miles answer to a Degree of Longitude, at every Degree of Latitude.

Though

Though this table does only shew the miles answering to a degree of any parallel, whose latitude consists of a whole number of degrees; yet it may be made to serve for any parallel whose latitude is some number of degrees and minutes, by making the following proportion, viz.

As 1 degree, or 60 minutes, is to the difference between the miles answering to a degree in the next greater and next less tabular latitude than that proposed; so is the excess of the proposed latitude above the next tabular latitude, to a proportional part; which, subtracted from the miles answering to a degree of longitude in the next less tabular latitude, will give the miles answering to a degree in the proposed latitude.

EXAMPLE. Required to find the miles answering to a degree on the parallel of $56^\{\backslash circ\}\; 44\text{'}$.

First, The next less parallel of latitude in the table than that proposed, is that of $56^\{\backslash circ\}$, a degree of which (by the table) is equal to 33.55 miles; and the next greater parallel of latitude in the table, than that proposed, is that of $57^\{\backslash circ\}$, a degree of which is (by the table) equal to 32.68 miles; the difference of these is 87, and the distance between these parallels is 1 degree or 60 minutes; also the distance between the parallel of $56^\{\backslash circ\}$, and the proposed parallel of $56^\{\backslash circ\}\; 44\text{'}$, is 44 minutes: then by the preceding proportion it will be, As 60 is to 87, so is 44 to 638, the difference between a degree on the parallel of $56^\{\backslash circ\}$ and a degree on the parallel of $56^\{\backslash circ\}\; 44\text{'}$; which therefore taken from 33.55, the miles answering to a degree on the parallel of $56^\{\backslash circ\}$, leaves 32.912, the miles answering to a degree on the parallel of $56^\{\backslash circ\}\; 44\text{'}$, as was required.

CASE II. The distance failed in any parallel of latitude, or the distance between any two places on that parallel, being given; to find the difference of longitude.

EXAMPLE. Suppose a ship in the latitude of $55^\{\backslash circ\}\; 36\text{'}$ north fails directly east 685.6 miles; required how much she has differed her longitude.

By Cor. 4. Art. 1. of this section, it will be
As the co. sine of the lat. — $55^\{\backslash circ\}\; 36\text{'}$ — 9.75202
is to radius — — — — — 10.00000
so is the distance failed — 685.6 — 2.83607
to min. of diff. of long. — 1213 — 3.08405
which reduced into degrees, by dividing by 60, makes $20^\{\backslash circ\}\; 13\text{'}$, the difference of longitude the ship has made.

This may also be solved by help of the foregoing table, viz. by finding from it the miles answering to a degree on the proposed parallel, and dividing with this the given number of miles, the quotient will be the degrees and minutes of difference of longitude required.

Thus in the last example; I find, from the foregoing table, that a degree on the parallel of $55^\{\backslash circ\}\; 36\text{'}$ is equal to 33.89 miles; by this I divide the proposed number of miles 685.6 and the quotient is 20.13 degrees, i. e. $20^\{\backslash circ\}\; 13\text{'}$, the difference of longitude required.

CASE III. The difference of longitude between two places on the same parallel, and the distance between them, being given; to find the latitude of that parallel.

EXAMPLE. Suppose a ship fails on a certain parallel directly west 624 miles, and then has differed her longi-

tude $18^\{\backslash circ\}\; 46\text{'}$ or 1126 miles: Required the latitude of the parallel she failed upon.

By Cor. 3. Art. 1. of this section, it will be,
As the min. of diff. long. — 1126 — 3.05154
is to the distance failed — 624 — 2.79518
so is radius — — — — — 10.00000
to the co. sine of the lat. $56^\{\backslash circ\}\; 21\text{'}$ — 9.74364
consequently the latitude of the ship or parallel she failed upon was $56^\{\backslash circ\}\; 21\text{'}$.

From what has been said, may be solved the following problems.

PROB. I. Suppose two ships in the latitude of $46^\{\backslash circ\}\; 30\text{'}$ north, distant atunder 654 miles, sail both directly north 256 miles, and consequently are come to the latitude of $50^\{\backslash circ\}\; 46\text{'}$ north: Required their distance on that parallel.

By Cor. 6. Art. 1. of this Section, it will be,
As the co. sine of — $46^\{\backslash circ\}\; 30\text{'}$ — 9.83781
is to the co. sine of — $50^\{\backslash circ\}\; 46\text{'}$ — 9.80105
so is — 654 — 2.81558
to — 601 — 2.77882
the distance between the ships when on the parallel of $50^\{\backslash circ\}\; 46\text{'}$.

PROB. II. Suppose two ships in the latitude of $45^\{\backslash circ\}\; 48\text{'}$ north, distant 846 miles, sail directly north till the distance between them is 624 miles: Required the latitude come to, and the distance failed.

By Cor. 5. Art. 1. of this Section, it will be,
As their first distance — 846 — 2.92737
is to their second distance — 624 — 2.79518
so is the co. sine of — $45^\{\backslash circ\}\; 48\text{'}$ — 9.84334
to the co. sine — $59^\{\backslash circ\}\; 04\text{'}$ — 9.71115
the latitude of the parallel the ships are come to.

Consequently to find their distance failed,
From the latitude come to — — — $59^\{\backslash circ\}\; 04\text{'}$
subtract the latitude failed from, — — — $45^\{\backslash circ\}\; 48\text{'}$
and there remains — — — — — $13^\{\backslash circ\}\; 16\text{'}$
equal to 796 miles, the difference of latitude or distance failed.

SEC. 4. OF MIDDLE-LATITUDE SAILING.

WHEN two places lie both on the same parallel, we shewed in the last section, how, from the difference of longitude given, to find the miles of easting or westing between them, & contra. But when two places lie not on the same parallel, then their difference of longitude cannot be reduced to miles of easting or westing on the parallel of either place: for if counted on the parallel of that place that has the greatest latitude, it would be too small; and if on the parallel of that place having the least latitude, it would be too great. Hence the common way of reducing the difference of longitude between two places, lying on different parallels, to miles of easting or westing, & contra, is by counting it on the middle parallel between the places, which is found by adding the latitudes of the two places together, and taking half the sum, which will be the latitude of the middle parallel required. And hence arises the solution of the following cases.

CASE I. The latitudes of two places, and their difference of longitude, given; to find the direct course and distance.

EXAMPLE,

EXAMPLE. Required the direct course and distance between the Lizard in the latitude of $50^\{\backslash circ\}\; 00\text{'}$ north, and longitude of $5^\{\backslash circ\}\; 14\text{'}$ west, and St Vincent in the latitude of $17^\{\backslash circ\}\; 10\text{'}\; N.$ and longitude of $24^\{\backslash circ\}\; 20\text{'}\; W.$

First, To the latitude of the Lizard — $50^\{\backslash circ\},\; 00\text{'}\; N.$
add the latitude of St Vincent — $17^\{\backslash circ\},\; 10\text{'}$

The sum is — — — $67^\{\backslash circ\},\; 10\text{'}$

Half the sum or latitude of } — $33^\{\backslash circ\},\; 35\text{'}\; N.$
the middle parallel is }

Also the difference of latitude is — $33^\{\backslash circ\},\; 50\text{'}$

equal to 1970 miles of southing. Again,

From the longitude of St Vincent — $24^\{\backslash circ\},\; 20\text{'}\; W.$
take the longitude of the Lizard — $05^\{\backslash circ\},\; 14\text{'}$

there remains — — — $16^\{\backslash circ\},\; 06\text{'}$

equal to 1146 min. of diff. of long. west.

Then for the miles of weeping, or departure, it will be, by Case 1. of Parallel Sailing,

As radius — — — 10.00000

is to the co-sine of the } — $33^\{\backslash circ\}\; 35\text{'}$ — 9.92069
middle parallel }

so is min. diff. of long. — 1146 — 3.05918

to the miles of weeping — 954.7 — 2.97987

And for the course it will be, by Case 4. of Plain Sailing,

As the diff. of lat. — 1970 — 3.29447

is to radius — — — 10.00000

so is the departure — 954.7 — 2.97987

to the tang. of the course $25^\{\backslash circ\},\; 51\text{'}$ — 9.68540

which, because it is between south and west, it will be SSW $\backslash frac\{1\}\{4\}$ west nearly.

For the distance, it will be, by the same case,

As radius — — — 10.00000

is to the diff. of lat. — 1970 — 3.29447

so is the secant of the course $25^\{\backslash circ\},\; 51\text{'}$ — 10.04579

to the distance — 2189 — 3.34026

whence the direct course and distance from the Lizard to St Vincent is SSW $\backslash frac\{1\}\{4\}$ W, 2189 miles.

CASE II. One latitude, course, and distance failed being given, to find the other latitude and difference of longitude.

EXAMPLE. Suppose a ship in the latitude of $50^\{\backslash circ\}\; 00\text{'}$ north, sails south $50^\{\backslash circ\}\; 06\text{'}$ west, 150 miles: Required the latitude the ship has come to, and how much she has differed her longitude.

First, For the difference of latitude, it will be, by Case 1. of Plain Sailing,

As radius — — — 10.00000

is to the distance — 150 — 2.17609

so is the co-sine of the course $50^\{\backslash circ\},\; 06\text{'}$ — 9.80716

to the diff. of latitude — 96.22 — 1.98325

equal to $1^\{\backslash circ\},\; 36\text{'}$. And since the ship is sailing towards the equator; therefore,

From the latitude she was in — $50^\{\backslash circ\},\; 00\text{'}$
take the diff. of latitude — — — $1^\{\backslash circ\},\; 36\text{'}$

and there remains — — — $48^\{\backslash circ\},\; 24\text{'}$

the latitude she has come to north. Consequently the

latitude of the middle parallel will be $49^\{\backslash circ\}\; 12\text{'}$.

Then for departure or weeping it will be, by the same case,

As radius — — — 10.00000

is to the distance — 150 — 2.17609

so is the sine of the course $50^\{\backslash circ\}\; 06\text{'}$ — 9.88489

to the departure — 115.1 — 2.06098

As for the difference of longitude, it will be, by Case 2. of Plain Sailing,

As the co-sine of the middle parallel $49^\{\backslash circ\}\; 12\text{'}$ — 9.81519

is to radius — — — 10.00000

so is the departure — 115.1 — 2.06098

to the min. diff. of longitude 176.1 — 2.24579

equal to $2^\{\backslash circ\}\; 56\text{'}$, which is the difference of longitude the ship has made westerly.

CASE III. Course and difference of latitude given; to find the distance sailed, and difference of longitude.

EXAMPLE. Suppose a ship in the latitude of $53^\{\backslash circ\}\; 34\text{'}$ north, sails SE&S, till by observation she is found to be in the latitude of $51^\{\backslash circ\}\; 12\text{'}$, and consequently has differed her latitude $2^\{\backslash circ\}\; 22\text{'}$ or 142 miles: Required the distance sailed, and the difference of longitude.

First, for the departure, it will be (by Case 2. of Plain Sailing),

As radius — — — 10.00000

is to the diff. of latitude — 142 — 2.15229

so is the tang. of course — $33^\{\backslash circ\}\; 45\text{'}$ — 9.82489

to the departure — 94.88 — 1.97718

And for the distance it will be, by the same Case,

As radius — — — 10.00000

is to the diff. of latitude — 142 — 2.15229

so is the secant of the course $33^\{\backslash circ\}\; 45\text{'}$ — 10.08015

to the distance — 170.8 — 2.23244

Then, since the latitude sailed from was $53^\{\backslash circ\}\; 34\text{'}$ north, and the latitude come to $51^\{\backslash circ\}\; 12\text{'}$ north; therefore the middle parallel will be $52^\{\backslash circ\}\; 23\text{'}$; and consequently, for the difference of longitude, it will be (by Case 2. of Parallel Sailing)

As the co-sine of the mid. parallel $52^\{\backslash circ\}\; 23\text{'}$ — 9.78560

is to the departure — 94.88 — 1.97718

so is radius — — — 10.00000

to min. of diff. of longitude — 155.5 — 2.19158

equal to $2^\{\backslash circ\}\; 35\text{'}$ the difference of longitude easterly.

CASE IV. Difference of latitude and distance sailed, given; to find the course and difference of longitude.

EXAMPLE. Suppose a ship in the latitude of $43^\{\backslash circ\}\; 26\text{'}$ north, sails between south and east, 246 miles, and then is found by observation to be in the latitude of $41^\{\backslash circ\}\; 06\text{'}$ north: Required the direct course and difference of longitude.

First, For the course, it will be, by Case 3. of Plain Sailing,

As the distance — 246 — 2.39094

is to radius — — — 10.00000

so is the diff. of latitude — 140 — 2.14613

to the co-sine of the course $55^\{\backslash circ\}\; 19\text{'}$ — 9.75519

which, because the ship sails between south and east, will be south $55^\{\backslash circ\}\; 19\text{'}$ east, or SE&E nearly.

Then for departure, it will be, by the same Case,

As radius — — — 10.00000

is to the distance — 246 — 2.39094

To the fine of the course $55^\{\backslash circ\}\; 19\text{'}$ — 9.91504
to the departure — 202.3 — 2.30598

Lastly, For the difference of longitude, it will be, by
Case 2. of Parallel Sailing,

As the co-line of the mid. par. $42^\{\backslash circ\}\; 16\text{'}$ — 9.86924
is to the departure — 302.3 — 2.30598
so is radius — — — — 10.00000
to min. of diff. of longitude 273.3 — 2.43674
equal to $4^\{\backslash circ\}\; 33\text{'}$, the difference of longitude easterly.

CASE V. Course and departure given, to find difference of latitude, difference of longitude, and distance sailed.

EXAMPLE. Suppose a ship in the latitude of $48^\{\backslash circ\}\; 23\text{'}$ north, sails SWS, till she has made of westing 123 miles; Required the latitude come to, the difference of longitude, and the distance sailed.

First, For the distance, it will be, by Case 6. of Plain Sailing,

As the fine of the course — $33^\{\backslash circ\}\; 45\text{'}$ — 9.74474
is to the departure — 123 — 2.08991
so is radius — — — — 10.00000
to the distance — 221.4 — 2.34517

And for the difference of latitude it will be, by the same
Case,

As the tang. of course — $33^\{\backslash circ\}\; 45\text{'}$ — 9.82489
is to the departure — 123 — 2.08991
so is radius — — — — 10.00000
to the diff. of latitude — 184 — 2.26502
equal to $3^\{\backslash circ\}\; 04\text{'}$; And since the ship is sailing towards the equator, the latitude come to will be $45^\{\backslash circ\}\; 19\text{'}$ north; and consequently the middle parallel will be $46^\{\backslash circ\}\; 51\text{'}$.

Then to find the difference of longitude, it will be, by
Case 2. of Parallel Sailing,

As the co-line of mid. par. $46^\{\backslash circ\}\; 51\text{'}$ — 9.83500
is to departure — 123 — 2.08991
so is radius — — — — 10.00000
to min. of diff. of longit. — 180 — 2.25491
which is equal to $3^\{\backslash circ\}\; 00\text{'}$, the difference of longitude westerly.

CASE VI. Difference of latitude and departure given, to find course, distance, and difference of longitude.

EXAMPLE. Suppose a ship in the latitude of $46^\{\backslash circ\}\; 37\text{'}$ north, sails between south and east, till she has made of easting 146 miles, and is then found by observation to be in the latitude of $43^\{\backslash circ\}\; 24\text{'}$ north; required the course, distance, and difference of longitude.

First, By Case 4. of Plain Sailing, it will be for the course,

As the diff. of latitude — 193 — 2.28556
is to departure — 146 — 2.16137
so is radius — — — — 10.00000
to the tang. of the course — $36^\{\backslash circ\}\; 55\text{'}$ — 9.87581
which, because the ship is sailing between south and east, will be south $36^\{\backslash circ\}\; 55\text{'}$ east, or SEBS $\backslash frac\{1\}\{2\}$ east nearly.

For the distance, it will be, by the same Case,

As radius — — — — 10.00000
is to the diff. of latitude — 193 — 2.28556
so is the secant of the course $36^\{\backslash circ\}\; 55\text{'}$ — 10.09718
to the distance — 241.4 — 2.38274

Then for the difference of longitude, it will be, by Case 2. of Parallel Sailing,

As the co-line of the mid. par. $45^\{\backslash circ\}\; 00\text{'}$ 9.84949
is to the departure — — — — 146 — 2.16137
so is radius — — — — 10.00000
to min. of diff. of longitude — 205 — 2.31188
equal to $3^\{\backslash circ\}\; 25\text{'}$, the difference of longitude easterly.

CASE VII. Distance and departure given, to find difference of latitude, course, and difference of longitude.

EXAMPLE. Suppose a ship in the latitude of $33^\{\backslash circ\}\; 40\text{'}$ north, sails between south and east 165 miles, and has then made of easting 112.5 miles; required the difference of latitude, course, and difference of longitude.

First, For the course, it will be, by Case 5. of Plain Sailing,

As the distance — 165 — 2.21748
is to radius — — — — 10.00000
so is the departure — 102.5 — 2.05115
to the fine of the course $42^\{\backslash circ\}\; 59\text{'}$ — 9.83367
which, because the ship sails between south and east, will be south $42^\{\backslash circ\}\; 59\text{'}$ east or SESE $\backslash frac\{1\}\{2\}$ east nearly.

And for the difference of latitude, it will be, by the same
Case,

As radius — — — — 10.00000
is to the distance — 165 — 2.21748
so is the co-line of the course $42^\{\backslash circ\}\; 59\text{'}$ — 9.86436
to the difference of lat. — 120.7 — 2.08184
equal to $2^\{\backslash circ\}\; 00\text{'}$; consequently the latitude come to will be $31^\{\backslash circ\}\; 40\text{'}$ north, and the latitude of the middle parallel will be $32^\{\backslash circ\}\; 40\text{'}$. Hence, to find the difference of longitude, it will be, by Case 2. of Parallel Sailing,

As the co-line of the mid. par. $32^\{\backslash circ\}\; 40\text{'}$ — 9.92522
is to the departure — 112.5 — 2.05115
so is radius — — — — 10.00000
to min. of diff. of long. — 133.6 — 2.12593
equal to $2^\{\backslash circ\}\; 13\text{'}$ nearly, the difference of longitude easterly.

CASE VIII. Difference of longitude and departure given, to find difference of latitude, course, and distance sailed.

EXAMPLE. Suppose a ship in the latitude of $50^\{\backslash circ\}\; 46\text{'}$ north, sails between south and west, till her difference of longitude is $3^\{\backslash circ\}\; 12\text{'}$, and is then found to have departed from her former meridian 126 miles; required the difference of latitude, course, and distance sailed.

First, For the latitude she has come to, it will be, by
Case 3. of Parallel Sailing,

As min. of diff. of long. — 192 — 2.28330
is to departure — 126 — 2.10037
so is radius — — — — 10.00000
to the co-line of the mid. par. — $48^\{\backslash circ\}\; 59\text{'}$ — 9.81707

Now since the middle latitude is equal to half the sum of the two latitudes (by Art. 1. of this Sect.) and so the sum of the two latitudes equal to double the middle latitude; it follows, that if from double the middle latitude we subtract any one of the latitudes, the remainder will be the other. Hence from twice $48^\{\backslash circ\}\; 59\text{'}$, viz. $97^\{\backslash circ\}\; 58\text{'}$ taking $50^\{\backslash circ\}\; 46\text{'}$ the latitude sailed from, there remains $47^\{\backslash circ\}\; 12\text{'}$ the latitude come to; consequently the difference of latitude is $3^\{\backslash circ\}\; 34\text{'}$, or 214 minutes.

Then for the course, it will be, by Case 4. of Plain Sailing,

As diff. of lat. — 214 — 2.33045
5 B

is to radius — — — — — 10.00000
so is the departure — — — — — 126 — — — — — 2.10037
to the tang. of the course $30^\{\backslash circ\}\; 29\text{'}$ — — — — — 9.76996
which, because it is between south and west, will be south
$30^\{\backslash circ\}\; 29\text{'}$ west, or SSW $\backslash frac\{1\}\{4\}$ west nearly.

And for the distance, it will be, by the same Case,
As radius — — — — — 10.00000
is to the diff. of lat. — — — — — 214 — — — — — 2.33041
so is the secant of the course $30^\{\backslash circ\}\; 29\text{'}$ — — — — — 10.06461
to the distance — — — — — 248.4 — — — — — 2.39502

2. From what has been said, it will be easy to solve a
traverse, by the rules of Middle Latitude Sailing.

EXAMPLE. Suppose a ship in the latitude of $43^\{\backslash circ\}\; 25\text{'}$
north, sails upon the following courses, viz. SWS 63
miles, SSW $\backslash frac\{1\}\{4\}$ west 45 miles, SSE 54 miles, and SWW
74 miles: Required the latitude the ship has come to,
and how far she has differed her longitude.

First, By Case 2. of this Sect. find the difference of
latitude and difference of longitude belonging to each
course and distance, and they will stand as in the follow-
ing table.

Hence it is plain the ship has differed her latitude 186.2
minutes, or $3^\{\backslash circ\}\; 6\text{'}$ and so has come to the latitude of
$40^\{\backslash circ\}\; 19\text{'}$ north, and has made of difference of longitude
143.8 minutes, or $2^\{\backslash circ\}\; 23\text{'}\; 48\text{'}\text{'}$ westerly.

3. This method of sailing, though it be not strictly
true, yet it comes very near the truth, as will be evident,
by comparing an example wrought by this method with
the same wrought by the method delivered in the next
Section, which is strictly true; and it serves, without any
considerable error, in runnings of 450 miles between the
equator and parallel of 30 degrees, of 300 miles be-
tween that and the parallel of 60 degrees, and of 150
miles as far as there is any occasion, and consequently
must be sufficiently exact for 24 hours run.

SECT. 5. OF MERCATOR'S SAILING.

1. THOUGH the meridians do all meet at the pole,
and the parallels to the equator do continually decrease,
and that in proportion to the co-sines of their latitudes;
yet in old sea charts the meridians were drawn parallel
to one another, and consequently the parallels of latitude
made equal to the equator, and so a degree of longitude
on any parallel as large as a degree on the equator: also
in these charts the degrees of latitude were still represented
(as they are in themselves) equal to each other, and to

those of the equator. By these means the degrees of lon-
gitude being increased beyond their just proportion, and
the more so the nearer they approach the pole, the de-
grees of latitude at the same time remaining the same,
it is evident places must be very erroneously marked down
upon these charts with respect to their latitude and lon-
gitude, and consequently their bearing from one another
very false.

2. To remedy this inconvenience, so as still to keep
the meridians parallel, it is plain we must protract, or
lengthen, the degrees of latitude in the same proportion
as those of longitude are, that so the proportion in casting
and westing may be the same with that of southing and
northing, and consequently the bearings of places from
one another be the same upon the chart as upon the
globe itself.

Let ABD (No. 11.) be a quadrant of a meridian, A
the pole, D a point on the equator, AC half the axis, B
any point upon the meridian, from which draw BF per-
pendicular to AC, and BG perpendicular to CD; then
BG will be the sine, and BF or CG the co-sine of BD
the latitude of the point B; draw D the tangent and CE
the secant of the arch CD. It has been demonstrated in
Sect. 3. that any arch of a parallel is to the like arch of the
equator as the co-sine of the latitude of that parallel is
to radius. Thus any arch as a minute on the parallel de-
scribed by the point B, will be to a minute on the equator
as BF or CG is to CD; but since the triangles CGB
CDE are similar, therefore CG will be to CD as CB is
to CE, i. e. the co-sine of any parallel is to radius as
radius is to the secant of the latitude of that parallel.
But it has been just now shown, that the co-sine of any par-
allel is to radius, as the length of any arch as a minute
on that parallel is to the length of the like arch on the
equator: Therefore the length of any arch as a minute
on any parallel, is to the length of the like arch on the
equator, as radius is to the secant of the latitude of that
parallel; and so the length of any arch, as a minute on
the equator, is longer than the like arch of any parallel
in the same proportion, as the secant of the latitude of
that parallel is to radius. But since in this projection the
meridians are parallel, and consequently each parallel of
latitude equal to the equator, it is plain the length of any
arch as a minute on any parallel, is increased beyond its
just proportion, at such rate as the secant of the latitude
of that parallel is greater than radius; and therefore to
keep up the proportion of northing and southing to that
of easting and westing, upon this chart, as it is upon the
globe itself, the length of a minute upon the meridian at
any parallel must also be increased beyond its just propor-
tion at the same rate, i. e. as the secant of the latitude
of that parallel is greater than radius. Thus to find the
length of a minute upon the meridian at the latitude of
$75^\{\backslash circ\}$ degrees, since a minute of a meridian is every where
equal on the globe, and also equal to a minute upon the
equator, let it be represented by unity: then making it
as radius is to the secant of $75^\{\backslash circ\}$ degrees, so is unity to a
fourth number, which is 3.864 nearly; and consequently,
by whatever line you represent one minute on the equator
of this chart, the length of one minute on the enlarged
meridian

meridian at the latitude of 75 degrees, or the distance between the parallel of 75° 00' and the parallel of 75° 01', will be equal to 3 of these lines, and $\backslash frac\{1\}\{3\}$ of one of them. By making the same proportion, it will be found, that the length of a minute on the meridian of this chart at the parallel of 60°, or the distance between the parallel of 60° 00' and that of 60° 01', is equal to two of these lines. After the same manner, the length of a minute on the enlarged meridian may be found at any latitude; and consequently beginning at the equator, and computing the length of every intermediate minute between that and any parallel, the sum of all these shall be the length of a meridian intercepted between the equator and that parallel; and the distance of each degree and minute of latitude from the equator upon the meridian of this chart, computed in minutes of the equator, forms what is commonly called a table of meridional parts.

If the arch BD (No. 11) represent the latitude of any point B, then (CD being radius) CE will be the secant of that latitude: but it has been shown above, that radius is to the secant of any latitude, as the length of a minute upon the equator is to the length of a minute on the meridian of this chart at that latitude; therefore CD is to CE, as the length of a minute on the equator is to the length of a minute upon the meridian, at the latitude of the point B. Consequently, if the radius CD be taken equal to the length of a minute upon the equator, CE, or the secant of the latitude, will be equal to the length of a minute upon the meridian at that latitude. Therefore, in general, if the length of a minute upon the equator be made radius, the length of a minute upon the enlarged meridian will be every where equal to the secant of the arc contained between it and the equator.

Cor. 1. Hence it follows, since the length of every intermediate minute between the equator and any parallel, is equal to the secant of the latitude (the radius being equal to a minute upon the equator) the sum of all these lengths, or the distance of that parallel on the enlarged meridian from the equator, will be equal to the sum of all the secants, to every minute contained between it and the equator.

Cor. 2. Consequently the distance between any two parallels on the same side of the equator is equal to the difference of the sums of all the secants contained between the equator and each parallel, and the distance between any two parallels on contrary sides of the equator is equal to the sum of the sums of all the secants contained between the equator and each parallel.

5. By the tables of meridional parts, which may be seen in Paton, and other writers on this subject, may be constructed the nautical chart, commonly called Mercator's chart. Thus, for example, let it be required to make a chart that shall commence at the equator, and reach to the parallel of 60 degrees, and shall contain 80 degrees of longitude.

Draw the line EQ representing the equator, (see No. 12.) then take, from any convenient line of equal parts, 4800 (the number of minutes contained in 80 degrees,) which set off from E to Q, and this will determine the breadth of the chart.

Divide the line EQ into eight equal parts, in the points

10, 20, 30, &c. each containing 10 degrees, and each of these divided into 10 equal parts will give the single degrees upon the equator; then through the points E, 10, 20, &c. drawing lines perpendicular to EQ, these shall be meridians.

From the scale of equal parts take 4527.4 (the meridional parts answering to 60 degrees,) and set that off from E to A and from Q to B, and join AB; then this line will represent the parallel of 60, and will determine the length of the chart.

Again, from the scale of equal parts take 603.1 (the meridional parts answering to 10 degrees,) and set that off from E to 10 on the line EA; and through the point 10 draw 10, 10, parallel to EQ; and this will be the parallel of 10 degrees. The same way, setting off from E on the line EA, the meridional parts answering to each degree, &c. of latitude, and through the several points drawing lines parallel to EQ, we shall have the several parallels of latitude.

If the chart does not commence from the equator, but is only to serve for a certain distance on the meridian between two given parallels on the same side of the equator; then the meridians are to be drawn as in the last example; and for the parallels of latitude you are to proceed thus, viz. from the meridional parts answering to each point of latitude in your chart, subtract the meridional parts answering to the least latitude, and set off the differences severally, from the parallel of least latitude, upon the two extreme meridians; and the lines joining these points of the meridians shall represent the several parallels upon your chart.

Thus let it be required to draw a chart that shall serve from the latitude of 20 degrees north to 60 degrees north, and that shall contain 80 degrees of longitude.

Having drawn the line DC to represent the parallel of 20 degrees (see No. 12.) and the meridians to it, as in the foregoing example; set off 663.3 (the difference between the meridional parts answering to 30 degrees, and those of 20 degrees) from D to 30, and from C to 30; then join the points 30 and 30 with a right line, and that shall be the parallel of 30. Also set off 1397.6 (the difference between the meridional parts answering to 40 degrees, and those of 20 degrees) from D to 40, and from C to 40, and joining the points 40 and 40 with a right line, that shall be the parallel of 40. And proceeding after the same way, we may draw as many of the intermediate parallels as we have occasion for.

But if the two parallels of latitude that bound the chart, are on the contrary sides of the equator; then draw a line representing the equator and meridians to it, as in the first example; and from the equator set off on each side of it the several parallels contained between it and the given parallels as above, and your chart is finished.

If Mercator's chart, constructed as above, hath its equator extended on each side of the point E 180 degrees, and if the several places on the surface of the earth be there laid down according to their latitudes and longitudes, we shall have what is commonly called Mercator's map of the earth. This map is not to be considered as a similar and just representation of the earth's surface; for in it the figures

figures of countries are distorted, especially near the poles: but since the degrees of latitude are every where increased in the same proportion as those of longitude are, the bearings between the places will be the same in this chart as on the globe; and the proportions between the latitudes, longitudes, and nautical distances, will also be the same on this chart, as on the globe itself; by which means the several cases of navigation are solved after a most easy manner, and adapted to the meanest capacities.

N. B. Here you must take notice, that in all charts, the upper part is the north side, and the lower part or bottom is the south side; also that part of it towards the right hand is the east, and that towards the left hand the west side of the chart.

6. Since, according to this projection, the meridians are parallel right lines; it is plain, that the rhombs which form always equal angles with the meridians, will be straight lines; which property renders this projection of the earth's surface much more easy and proper for use than any other.

7. This method of projecting the earth's surface upon a plane, was first invented by Mr Edward Wright, but first published by Mercator; and hence the sailing by the chart, was called Mercator's sailing.

8. In No. 13. let A and E represent two places upon Mercator's chart, AC the meridian of A, and CE the parallel of latitude passing through E; draw AE, and set off upon AC the length AB equal to the number of minutes contained in the difference of latitude between the two places, and taken from the same scale of equal parts the chart was made by, or from the equator, or any graduated parallel of the chart, and through B draw BD parallel to CE meeting AE in D. Then AC will be the enlarged difference of latitude, AB the proper difference of latitude, CE the difference of longitude, BD the departure, AE the enlarged distance, and AD the proper distance, between the two places A and E; also the angle BAD will be the course, and AE the rhomb line between them.

9. Now since in the triangle ACE, BD is parallel to one of its sides CE; it is plain the triangles ACE, ABD, will be similar, and consequently the sides proportional. Hence arise the solutions of the several cases in this sailing, which are as follow.

CASE I. The latitudes of two places given, to find the meridional or enlarged difference of latitude between them.

Of this case there are three varieties, viz. either one of the places lies on the equator, or both on the same side of it; or lastly, on different sides.

1. If one of the proposed places lies on the equator, then the meridional difference of latitude is the same with the latitude of the other place, taken from the table of meridional parts.

EXAMPLE. Required the meridional difference of latitude between St Thomas, lying on the equator, and St Antonio in the latitude of $17^\{\backslash circ\}\; 20\text{'}$ north. I look in the tables for the meridional parts answering to $17^\{\backslash circ\}\; 20\text{'}$, and find it to be 1056.2, the enlarged difference of latitude required.

2. If the two proposed places be on the same side of

the equator, then the meridional difference of latitude is found by subtracting the meridional parts answering to the least latitude from those answering to the greatest, and the difference is that required.

EXAMPLE. Required the meridional difference of latitude between the Lizard in the latitude of $50^\{\backslash circ\}\; 00\text{'}$ north, and Antigua in the latitude of $17^\{\backslash circ\}\; 30\text{'}$ north. From the meridional parts of — $50^\{\backslash circ\}\; 00\text{'}$ — 3474.5 subtract the meridional parts of $17^\{\backslash circ\}\; 30\text{'}$ — 1066.7

there remains — — — 2407.8
the meridional difference of latitude required.

3. If the places lie on different sides of the equator, then the meridional difference of latitude is found by adding together the meridional parts answering to each latitude, and the sum is that required.

EXAMPLE. Required the meridional difference of latitude between Antigua in the latitude of $17^\{\backslash circ\}\; 30\text{'}$ north, and Lima in Peru in the latitude of $12^\{\backslash circ\}\; 30\text{'}$ South.

To the merid. parts answering to $17^\{\backslash circ\}\; 30\text{'}$ — 1066.7 add these answering to — $12^\{\backslash circ\}\; 30\text{'}$ — 756.1

the sum is — — — 1822.8
the meridional difference of latitude required.

CASE II. The latitudes and longitudes of two places given, to find the direct course and distance between them.

EXAMPLE. Required to find the direct course and distance between the Lizard in the latitude of $50^\{\backslash circ\}\; 00\text{'}$ north, and Port-Royal in Jamaica in the latitude of $17^\{\backslash circ\}\; 40\text{'}$; differing in longitude $70^\{\backslash circ\}\; 46\text{'}$, Port-Royal lying so far to the westward of the Lizard.

PREPARATION.

From the latitude of the Lizard — $50^\{\backslash circ\}\; 00\text{'}$ subtract the latitude of Port-Royal — $17^\{\backslash circ\}\; 40\text{'}$

and there remains — — — 32, 20
equal to 1940 minutes, the proper difference of latitude, Then from the meridional parts of $50^\{\backslash circ\}\; 00\text{'}$ 3474.5 subtract those of — — — $17^\{\backslash circ\}\; 40\text{'}$ 1077.2

and there remains — — — 2397.3
the meridional or enlarged difference of longitude.

GEOMETRICALLY. Draw the line AC (No 14.) representing the meridian of the Lizard at A, and set off from A, upon that line, AE equal to 1940 (from any scale of equal parts) the proper difference of latitude, also AC equal to 2397.3 (from the same scale) the meridional or enlarged difference of latitude. Upon the point C raise CB perpendicular to AC, and make CB equal to 4246 the minutes of difference of longitude.

Join AB, and through E draw ED parallel to BC: so the case is constructed; and AD applied to the same scale of equal parts the other legs were taken from will give the direct distance, and the angle DAE measured by the line of chords will give the course.

By CALCULATION.

For the angle of the course EAD, it will be, (by Rectangular Trigonometry.)

$AC\; :\; CB\; ::\; R\; :\; T$, BAC, i. e.

As the meridional diff. of lat. 2397.3 — 3 37970

is to the difference of long. — 4246.0 — 3.62798
so is radius — — — — — 10.00000
to the tang. of the direct course $60^\{\backslash circ\}\; 33\text{'}$ — 10.34828
which, because Port Royal is southward of the Lizard, and the difference of longitude westerly, will be south $60^\{\backslash circ\}\; 33\text{'}$ west, or SW $\backslash frac\{1\}\{4\}$ west nearly.

Then for the distance AD, it will be, (by rectangular trigonometry)

As the radius — — — — — 10.00000
is to the proper diff. of lat. — 1940 — 3.28780
so is the secant of the course $60^\{\backslash circ\}\; 33\text{'}$ — 10.30833
to the distance — — — — — 3945.6 — 3.59613
consequently the direct course and distance between the Lizard and Port-Royal in Jamaica, is south $60^\{\backslash circ\}\; 33\text{'}$, 3945.6 miles.

CASE III. Course and distance failed given, to find difference of latitude and difference of longitude.

EXAMPLE. Suppose a ship from the Lizard in the latitude of $50^\{\backslash circ\}\; 00\text{'}$ north, sails south $35^\{\backslash circ\}\; 40\text{'}$ west 156 miles. Required the latitude come to, and how much she has altered her longitude.

GEOMETRICALLY. 1. Draw the line BK (No. 15.) representing the meridian of the Lizard at B; from B draw the line BM, making with BK an angle equal to $35^\{\backslash circ\}\; 40\text{'}$, and upon this line set off BM equal to 156 the given distance, and from M let fall the perpendicular MK upon BK.

Then for BK the proper difference of latitude, it will be, (by rectangular trigonometry)

i. e. As radius — — — — — 10.00000
is to the distance — 156 — 2.19312
so is the co-sine of the course $35^\{\backslash circ\}\; 40\text{'}$ — 9.90978
to the proper difference of lat. 127 — 2.10290
equal to $2^\{\backslash circ\}\; 07\text{'}$; and since the ship is sailing from a north latitude towards the south, therefore the latitude come to will be $47^\{\backslash circ\}\; 53\text{'}$ north. Hence the meridional difference of latitude will be 193.4.

2. Produce BK to D, till BD be equal to 193.4; through D draw DL parallel to MK, meeting DM produced in L; then DL will be the difference of longitude: to find which by calculation, it will be, (by rectangular trigonometry)

i. e. As radius — — — — — 10.00000
is to the meridional diff. of lat. 193.4 — 2.28646
so is the tangent of the course $35^\{\backslash circ\}\; 40\text{'}$ — 9.85594
to minutes of diff. of long. — 138.8 — 2.14240
equal to $2^\{\backslash circ\}\; 18\text{'}\; 48\text{'}\text{'}$ the difference of longitude the ship has made westerly.

CASE IV. Given course and both latitudes, viz. the latitude failed from, and the latitude come to; to find the distance failed, and the difference of longitude.

EXAMPLE. Suppose a ship in the latitude of $54^\{\backslash circ\}\; 20\text{'}$ north, sails south $33^\{\backslash circ\}\; 45\text{'}$ east, until by observation she is found to be in the latitude of $51^\{\backslash circ\}\; 45\text{'}$ north; required the distance failed, and the difference of longitude.

GEOMETRICALLY. Draw AB (No. 16.) to represent the meridian of the ship in the first latitude, and set off from A to B 155 the minutes of the proper difference of

latitude, also AG equal to 257.9 the minutes of the enlarged difference of latitude. Through B and G, draw the lines BC and GK perpendicular to AG; also draw AK making with AG an angle of $33^\{\backslash circ\}\; 45\text{'}$, which will meet the two former lines in the points C and K; so the case is constructed, and AC and GK may be found from the line of equal parts: To find which,

By CALCULATION:

First, For the difference of longitude, it will be, (by rectangular trigonometry)

i. e. As radius — — — — — 10.00000
is to the enlarged diff. of lat. — 257.9 — 2.41145
so is the tang. of the course — $33^\{\backslash circ\}\; 45\text{'}$ — 9.82489
to min. of diff. of longitude — 172.3 — 2.23634
equal to $2^\{\backslash circ\}\; 52\text{'}\; 18\text{'}\text{'}$, the difference of longitude the ship has made easterly.

This might also have been found, by first finding the departure BC (by Case 2. of Plain Sailing,) and then it would be

AB : BC :: AG : GK, the difference of longitude required.

Then for the direct distance AC, it will be, (by rectangular trigonometry)

i. e. As radius — — — — — 10.00000
is to the proper diff. of lat. — 155 — 2.19033
so is the secant of the course — $33^\{\backslash circ\}\; 45\text{'}$ — 10.08015
to the direct distance — 186.4 — 2.27048
consequently the ship has sailed south $33^\{\backslash circ\}\; 45\text{'}$ east 186.4 miles, and has differed her longitude $2^\{\backslash circ\}\; 52\text{'}\; 18\text{'}\text{'}$ easterly.

CASE V. Both latitudes, and distance failed, given; to find the direct course, and difference of longitude.

EXAMPLE. Suppose a ship from the latitude of $45^\{\backslash circ\}\; 26\text{'}$ north, sails between north and east 195 miles, and then by observation she is found to be in the latitude of $48^\{\backslash circ\}\; 6\text{'}$ north; required the direct course and difference of longitude.

GEOMETRICALLY. Draw AB (No. 17.) equal to 160 the proper difference of latitude, and from the point B raise the perpendicular BD; then take 195 in your compasses, and setting one foot of them in A, with the other cross the line BD in D. Produce AB, till AC be equal to 233.6 the enlarged difference of latitude. Thro' C draw CK parallel to BD, meeting AD produced in K; so the case is constructed; and the angle A may be measured by the line of chords, and CK by the line of equal parts: To find which,

By CALCULATION:

First, For the angle of the course BAD it will be, (by rectangular trigonometry)

As the proper diff. of lat. — 160 — 2.20412
is to radius — — — — — 10.00000
so is the distance — 195 — 2.29003
to the secant of the course — $34^\{\backslash circ\}\; 52\text{'}$ — 10.08591
which, because the ship is sailing between north and east, will be north $34^\{\backslash circ\}\; 52\text{'}$ east, or NE $\backslash frac\{1\}\{4\}$ N $1^\{\backslash circ\}\; 7\text{'}$ easterly.

Then for the difference of longitude, it will be, (by rectangular trigonometry)

i. e. As radius — — — — — 10.00000
is to the merid. diff. of lat. — 233.6 — 2.36847
so is the tang. of the course — 34°, 52' — 9.84307
to the min. of diff. of longitude — 162.8 — 2.21154
equal to 2° 42' 48", the difference of longitude easterly.

CASE VI. One latitude, course, and difference of longitude, given; to find the other latitude, and distance sailed.

EXAMPLE. Suppose a ship from the latitude of 48° 50' north, sails south 34° 40' west, till her difference of longitude is 2° 44'; required the latitude come to, and the distance sailed.

GEOMETRICALLY. 1. Draw AE (No. 18) to represent the meridian of the ship in the first latitude, and make the angle EAC equal to 34° 40', the angle of the course; then draw FC parallel to AE, at the distance of 164 the minutes of difference of longitude, which will meet AC in the point C. From C let fall upon AE the perpendicular CE; then AE will be the enlarged difference of latitude. To find which by Calculation, it will be, (by rectangular trigonometry.)

i. e. As the tang. of the course 34°, 40' — 9.83984
is to the radius — — — — — 10.00000
so is min. of diff. longitude — 164 — 2.21484
to the enlarged diff. of latitude — 237.2 — 2.37500
and because the ship is sailing from a north latitude southerly, therefore.

From the merid. parts of } — 48°, 50' — 3366.9
the latitude sailed from }
take the merid. difference of latitude — — — — — 237.2

and there remains — — — — — 3129.7
the meridional parts of the latitude come to, viz. 46° 09'.

Hence for the proper difference of latitude,
From the latitude sailed from — — — — — 48°, 50' N
take the latitude come to — — — — — 46°, 09 N

and there remains — — — — — 2°, 41
equal to 161, the minutes of difference of latitude.

2. Set off upon AE the length AD equal to 161 the proper difference of latitude, and through D draw DB parallel to CE; then AB will be the direct distance. To find which by Calculation, it will be, by rectangular trigonometry,

i. e. As radius — — — — — 10.00000
is to the proper diff. of latitude — 161 — 2.20683
so is the secant of the course 34°, 40' — 10.08488
to the direct distance — — — — — 195.8 — 2.29171

CASE VII. One latitude, course, and departure, given; to find the other latitude, distance sailed, and difference of longitude.

EXAMPLE. Suppose a ship sails from the latitude of 54° 36' north, south 42° 33' east, until she has made of departure 116 miles. Required the latitude she is in, her direct distance sailed, and how much she has altered her longitude.

GEOMETRICALLY. 1. Having drawn the meridian AB, (No. 19.) make the angle BAD equal to 42° 33'. Draw FD parallel to AB at the distance of 116, which will meet AD in D. Let fall upon AB the perpendicular DB. Then AB will be the proper difference of latitude, and AD the direct distance: To find which by calculation, first, for the distance AD it will be (by rectangular trigonometry)

i. e. As the sine of the course 42°, 33' — 9.82010
is to the departure — — — — — 116 — 2.06446
so is radius — — — — — 10.00000
to the direct distance — — — — — 171.5 — 2.23436

Then for the proper difference of latitude, it will be, by rectangular trigonometry.

i. e. As the tang. of the course 42°, 33' — 9.96281
is to the departure — — — — — 116 — 2.06446
so is radius — — — — — 10.00000
to the proper difference of latitude — 126.4 — 2.10165
equal to 2° 6', consequently the ship has come to the latitude of 52° 30' north, and so the meridional difference of latitude will be 212.2.

2. Produce AB to E, till AE be equal to 212.2; and through E draw EC parallel to BD, meeting AD produced in C; then EC will be the difference of longitude; to find which by calculation, it will be, (by rectangular trigonometry)

i. e. As radius — — — — — 10.00000
is to the merid. diff. of latitude — 212.2 — 2.32675
so is the tang. of the course 42°, 33' — 9.96281
to the min. of diff. of longitude — 194.8 — 2.28956
equal to 3° 14' 48", the difference of longitude easterly.

This might have been found otherwise, thus: because the triangles ACE, ADB, are similar; therefore it will be, AB : BD :: AE : EC.

i. e. As the proper diff. of latitude 126.4 — 2.10165
is to the departure — — — — — 116 — 2.06446
so is the enlarged diff. of latitude — 212.2 — 2.32675
to min. diff. of longitude — — — — — 194.8 — 2.28956

CASE VIII. Both latitudes and departure given, to find course, distance, and difference of longitude.

EXAMPLE. Suppose a ship from the latitude of 46° 20' N. sails between south and west, till she has made of departure 126.4 miles; and is then found by observation to be in the latitude of 43° 25' north. Required the course and distance sailed, and difference of longitude.

GEOMETRICALLY. Draw AK (No. 20.) to represent the meridian of the ship in her first latitude; set off upon it AC, equal to 165, the proper difference of latitude. Draw BC perpendicular to AC, equal to 126.4 the departure, and join AB. Set off from A, AK equal to 233.3, the enlarged difference of latitude; and through K draw KD parallel to BC, meeting AB produced in D; so the case is constructed, and DK will be the difference of longitude, AB the distance, and the angle A the course; to find which

First, For DC the difference of longitude, it will be, AC:

i. e. As the proper diff. of latitude 165 2.21748
is to the departure — — 126.4 2.10175
so is the enlarged diff. of latitude — 233.3 2.36791
to min. of diff. longitude — 178.7 2.25218
equal to 2° 58' 42", the difference of longitude westerly.

Then for the course it will be, (by rectangular trigonometry.)

i. e. As the proper diff. of latitude 165 2.21748
is to departure — — 126.4 2.10175
so is radius — — — 10.00000
to the tangent of the course 37°, 27' 9.88427
which, because the ship sails between south and west,
will be south 37° 27' west, or SW&S 6° 30' westerly.

Lastly, For the distance AB; it will be, (by rectangular trigonometry.)

i. e. As the sine of the course 37°, 27' — 9.78395
is to the departure — — 126.4 — 2.10175
so is radius — — — 10.00000
to the direct distance — 207.9 — 2.31780

CASE IX One latitude, distance sailed, and departure given; to find the other latitude, difference of longitude, and course.

EXAMPLE. Suppose a ship in the latitude of 48° 33' north, sails between south and east 138 miles, and has then made of departure 112.6. Required the latitude come to, the direct course, and difference of longitude.

GEOMETRICALLY. 1st, Draw BD (No. 21.) for the meridian of the ship at B; and parallel to it draw FE, at the distance of 112.6, the departure. Take 138, the distance, in your compasses, and fixing one point of them in B, with the other cross the line FE in the point E; then join B and E, and from E let fall upon BD the perpendicular ED; so BD will be the proper difference of latitude, and the angle B will be the course; to find which, by calculation,

First, for the course it will be, (by rectangular trigonometry.)

i. e. As the distance — — 138 — 2.13988
is to radius — — — 10.00000
so is the departure — — 112.6 — 2.05154
to the sine of the course — 54° 41' 9.91166
which, because the ship sails between south and east, will be south 54° 41' east, or SE 9° 41' easterly.

Then for the difference of latitude, it will be, (by rectangular trigonometry.)

i. e. As radius — — — 10.00000
is to the distance — — 138 — 2.13988
so is the co-sine of the course 54° 41' 9.76200
to the difference of latitude 79.8 — 1.90188
equal to 1° 19'. Consequently the ship has come to the latitude of 47° 13'. Hence the meridional difference of latitude will be 117.7.

2^ndly, Produce B to A, till BA be equal to 117.7; and through A draw AC parallel to DE, meeting BE produced in C; then AC will be the difference of longitude; to find which by calculation, it will be,

i. e. As the proper diff. of latitude 79.8 1.90188
is to the departure — — 112.6 2.05154
so is the enlarged diff. of latitude 117.7 2.07078
to the diff. of longitude — 166.1 — 2.22044
equal to 2° 46' 06", the difference of longitude easterly.

9. From what has been said, it will be easy to solve a traverse according to the rules of Mercator's sailing.

EXAMPLE. Suppose a ship at the Lizard in the latitude 30° 00' north, is bound to the Madeira in the latitude of 32°, 20' north, the difference of longitude between them being 11° 40', the west end of the Madeira lying so much to the westward of the Lizard, and consequently the direct course and distance (by Case 2. of this Section) is south 26° 15' west 1181.9 miles; but by reason of the winds she is forced to sail on the following courses (allowance being made for lee-way and variation, &c.) viz. SSW 44 miles, SSW $\backslash frac\{1\}\{2\}$ west 36 miles, SW&S 56 miles, and S&E 28 miles. Required the latitude the ship is in, her bearing and distance from the Lizard, and her direct course and distance from the Madeira, at the end of these courses.

The geometrical construction of this traverse is performed by laying down the two ports according to construction of Case 2. of this Section, and the several courses and distances according to Case 3. by which we have the following solution by calculation.

As radius — — — 10.00000
is to the distance — — 44 — 1.64345
so is the co-sine of the course 22°, 30' 9.96562
to the difference of latitude — 40.65 — 1.60907
and since the course is southerly, therefore the latitude come to will be 49° 20' north, and consequently the meridional difference of latitude will be 61.8. Then

As radius — — — 10.00000
is to the enlarged diff. of lat. 61.8 — 1.79099
so is the tan. of the course 22°, 30' — 9.61722
to min. of diff. of longitude 25.6 — 1.40821

As radius — — — 10.00000
is to the distance — — 36 — 1.55630
so is the co-sine of the course 16°, 52' 9.98090
to the difference of latitude 34.46 — 1.53720
and since the course is southerly, therefore the latitude come to will be 48° 45'. Hence the meridional difference of latitude will be 53.4. Then,

As radius — — — 10.00000
is to the enlarged diff. of lat. 53.4 — 1.72754
so is the tan. of the course 16°, 52' — 9.48171
to the difference of longitude 16.19 — 1.20925

As radius — — — 10.00000
is to the distance — — 56 — 1.74819
so is the co-sine of the course 33°, 45' 9.91985

to the difference of latitude — 46.56 1.66804
consequently the latitude come to is 47° 59', and therefore the enlarged difference of latitude will be 69.2.

Then,

For difference of longitude:

4. Course S&E, distance 28 miles.

For difference of latitude:

Then,

For difference of longitude:

Now these several courses and distances, together with the difference of latitude and longitude belonging to each of them, being set down in their proper columns in the Traverse Table, will stand as follow.

Hence it is plain that the ship has made of southing 149.13 minutes, and consequently has come to the latitude of 47° 31' north, and so the meridional difference of latitude between that and her first latitude will be 226.1; and since she has made of difference of longitude 79.44 minutes westerly; therefore for the direct course and distance between the lizard and the ship it will be, (by Case 2. of this Section)

For the direct course:

which because the difference of latitude is southerly, and the difference of longitude westerly, will be south 19° 22' west, or S&W 8° 7' westerly. Then,

For the direct distance:

and there remains — — — 15, 11
equal to 911 minutes, the proper difference of latitude between the ship and the Madera.

and there remains — — — 1196.4
the enlarged difference of latitude between the ship and the Madera.

and there remains — — — 10, 20.56 W
equal to 620.56 min. of difference of longitude between the ship and the Madera westerly.

Then for the direct course and distance between the ship and the Madera, it will be,

For the direct course:

For the direct distance:

10. It is very common, in working a day's reckoning at sea, to find the difference of latitude and departure to each course and distance; and adding all the departures together, and all the differences of latitudes for the whole departure, and difference of latitude made good that day, from thence (by Case 8. of this Section) to find the difference of longitude, &c. made good that day. Now that this method is false, will evidently appear, if we consider that the same departure reckoned on two different parallels will give unequal differences of longitude; and consequently, when several departures are compounded together and reckoned on the same parallel, the difference of longitude resulting from that cannot be the same with the sum of the differences of longitude resulting from the several departures on different parallels; and therefore we have chosen, in the last example of a traverse, to find the difference of longitude answering to each particular course and distance, the sum of which must be the true difference of longitude made good by the ship on these several courses and distances.

11. We shewed, at Art. 5. of this Section, how to construct a Mercedor's chart; and now we shall proceed to its several uses, contained in the following problems.

PROB. 1. Let it be required to lay down a place upon

the chart, its latitude, and the difference of longitude between it and some known place upon the chart being given.

EXAMPLE. Let the known place be the Lizard lying on the parallel of $50^\{\backslash circ\}\; 00\text{'}$ north, and the place to be laid down St Katharines on the east coast of America, differing in longitude from the Lizard $42^\{\backslash circ\}\; 36\text{'}$, lying so much to the westward of it.

Let L represent the Lizard on the chart, (see No. 12.) lying on the parallel of $50^\{\backslash circ\}\; 00\text{'}$ north, its meridian. Set off AE from E upon the equator $EQ$, $42^\{\backslash circ\}\; 36\text{'}$, towards Q, which will reach from E to F. Through F draw the meridian FG, and this will be the meridian of St Katharines; then set off from Q to H upon the graduated meridian QB, $28$ degrees; and through H draw the parallel of latitude HM, which will meet the former meridian in K, the place upon the chart required.

PROB. II. Given two places upon the chart, to find their difference of latitude and difference of longitude.

Through the two places draw parallels of latitude; then the distance between these parallels numbered in degrees and minutes upon the graduated meridian will be the difference of latitude required; and through the two places drawing meridians, the distance between these, counted in degrees and minutes on the equator or any graduated parallel, will be the difference of longitude required.

PROB. III. To find the bearing of one place from another upon the chart.

EXAMPLE. Required the bearing of St Katharines at K (see No. 12.) from the Lizard at L.

Draw the meridian of the Lizard AE, and join K and L with the right line KL; then by the line of chords measuring the angle KLE, and with that entering the tables, we shall have the thing required.

This may also be done, by having compasses drawn on the chart (suppose at two of its corners;) then lay the edge of a ruler over the two places, and let fall a perpendicular, or take the nearest distance from the centre of the compass next the first place, to the ruler's edge; then with this distance in your compasses, slide them along by the ruler's edge, keeping one foot of them close to the ruler, and the other as near as you can judge perpendicular to it, which will describe the rhomb required.

PROB. IV. To find the distance between two given places upon the chart.

This problem admits of four cases, according to the situation of the two places with respect to one another.

CASE I. When the given places lie both upon the equator.

In this case their distance is found by converting the degrees of difference of longitude intercepted between them into minutes.

CASE II. When the two places lie both on the same meridian.

Draw the parallels of those places; and the degrees upon the graduated meridian, intercepted between those parallels, reduced to minutes, give the distance required.

CASE III. When the two places lie on the same parallel.

EXAMPLE. Required to find the distance between the points K and N, (see No. 12.) both lying on the parallel

of $28^\{\backslash circ\}\; 00\text{'}$ north. Take from your scale the chord of $60^\{\backslash circ\}$ or radius in your compasses, and with that extent on KN as a base make the isosceles triangle KPN; then take from the line of sines the co-sine of the latitude, or sine of $72^\{\backslash circ\}$ and set that off from P to S and T. Join S and T with the right line ST, and that applied to the graduated equator will give the degrees and minutes upon it equal to the distance; which, converted into minutes, will be the distance required.

The reason of this is evident from the section of Parallels Sailing; for it has been there demonstrated, that radius is to the co-sine of any parallel, as the length of any arch on the equator, to the length of the same arch on that parallel. Now in this chart KN is the distance of the meridians of the two places K and N upon the equator; and since, in the triangle PNK, ST is the parallel to KN, therefore $PN:PT::NK:TS$. Consequently TS will be the distance of the two places K and N upon the parallel of $28^\{\backslash circ\}$.

If the parallel the two places lie on be not far from the equator, and they not far asunder; then their distance may be found thus. Take the distance between them in your compasses, and apply that to the graduated meridian, so as the one foot may be as many minutes above, as the other is below the given parallel; and the degrees and minutes intercepted, reduced to minutes, will give the distance.

Or it may also be found thus. Take the length of a degree on the meridian at the given parallel, and turn that over on the parallel from the one place to the other, as oft as you can; then as oft as that extent is contained between the places, so many times 60 miles will be contained in the distance between them.

CASE IV. When the places differ both in longitude and latitude.

EXAMPLE. Suppose it were required to find the distance between the two places a and e upon the chart. By

PROB. II. Find the difference of latitude between them; and take that in your compasses from the graduated equator, which set off on the meridian of a, from a to b; then through b draw be parallel to de; and taking ac in your compasses, apply it to the graduated equator, and it will show the degrees and minutes contained in the distance required, which multiplied by 60 will give the miles of distance.

The reason of this is evident from Art. 8. of this S.A. for it is plain ad is the enlarged difference of latitude, and ab the proper; consequently ae the enlarged distance, and ac the proper.

PROB. V. To lay down a place upon the chart, its latitude and bearing from some known place upon the chart being known. or (which is the same) having the course and difference of latitude that a ship has made, to lay down the running of the ship, and find her place upon the chart.

EXAMPLE. A ship from the Lizard in the latitude of $50^\{\backslash circ\}\; 00\text{'}$ north, sails SSW till she has differed her latitude $36^\{\backslash circ\}\; 40\text{'}$. Required her place upon the chart.

Count from the Lizard at L on the graduated meridian downwards (because the course is southerly) $36^\{\backslash circ\}\; 40\text{'}$ to g through

through which draw a parallel of latitude, which will be the parallel the ship is in; then from L draw a SSW line $Lf$, cutting the former parallel in $f$, and this will be the ship's place upon the chart.

PROB. VI. One latitude, course, and distance, failed, given; to lay down the running of the ship, and find her place upon the chart.

EXAMPLE. Suppose a ship at $a$ in the latitude of $20^\{\backslash circ\}\; 00\text{'}$ north, sails north $37^\{\backslash circ\}\; 20\text{'}$, east 191 miles: Required the ship's place upon the chart.

Having drawn the meridian and parallel of the place $a$, set off the rhomb line $ae$, making with $ab$ an angle of $37^\{\backslash circ\}\; 20\text{'}$; and upon it set off 191 from $a$ to $e$; through $e$ draw the parallel $eb$; and taking $ab$ in your compasses, apply it to the graduated equator, and observe the number of degrees it contains; then count the same number of degrees on the graduated meridian from $C$ to $b$, and through $b$ draw the parallel $be$, which will cut $ae$ produced in the point $c$, the ship's place required.

PROB. VII. Both latitudes and distance failed, given; to find the ship's place upon the chart.

EXAMPLE. Suppose a ship sails from $a$, in the latitude of $20^\{\backslash circ\}\; 00\text{'}$ north, between north and east 191 miles, and is then in the latitude of $45^\{\backslash circ\}\; 00\text{'}$ north: Required the ship's place upon the chart.

Draw $de$ the parallel of $45^\{\backslash circ\}$, and set off upon the meridian of $a$ upwards, $ab$ equal to the proper difference of latitude taken from the equator or graduated parallel. Through $b$ draw $be$ parallel to $de$; then with 191 in your compasses, fixing one foot of them in $a$, with the other cross $be$ in $c$. Join $a$ and $c$ with the right line $ac$; which produced will meet $de$ in $e$, the ship's place required.

PROB. VIII. One latitude, course, and difference of longitude, given; to find the ship's place upon the chart.

EXAMPLE. Suppose a ship from the Lizard in the latitude of $50^\{\backslash circ\}\; 00\text{'}$ north, sails SW&W, till her difference of longitude is $42^\{\backslash circ\}\; 36\text{'}$: Required the ship's place upon the chart.

Having drawn $AE$ the meridian of the Lizard at $L$, count from $E$ to $F$ upon the equator $42^\{\backslash circ\}\; 36\text{'}$; and through $F$ draw the meridian $EG$; then from $L$ draw the SW&W line $LK$, and where this meets $FG$, as at $K$, will be the ship's place required.

PROB. IX. One latitude, course, and departure, given; to find the ship's place upon the chart.

EXAMPLE. Suppose a ship at $a$ in the latitude of $20^\{\backslash circ\}\; 00\text{'}$ north, sails north $37^\{\backslash circ\}\; 20\text{'}$ east, till she has made of departure 116 miles: Required the ship's place upon the chart.

Having drawn the meridian of $a$, at the distance of 116, draw parallel to it the meridian $k\; l$. Draw the rhomb line $ac$, which will meet $k\; l$ in some point $c$; then through $c$ draw the parallel $c\; b$, and $a\; b$ will be the proper difference of latitude, and $b\; c$ the departure. Take $a\; b$ in your compasses, and apply it to the equator or graduated parallel; then observe the number of degrees it contains, and count so many on the graduated meridian from $C$ upwards to $b$. Through $b$ draw the parallel $b\; e$, which will meet $a\; c$ produced in some point as $e$, which is the ship's place upon the chart.

PROB. X. One latitude, distance, and departure, given; to find the ship's place upon the chart.

EXAMPLE. Suppose a ship at $a$ in the latitude of $20^\{\backslash circ\}\; 00\text{'}$ north, sails 191 miles between north and east, and then is found to have made of departure 116 miles: Required the ship's place upon the chart.

Having drawn the meridian and parallel of the place $a$, set off upon the parallel $a\; m$ equal to 116, and through $m$ draw the meridian $k\; l$. Take the given distance 191 in your compasses; setting one foot of them in $a$, with the other cross $k\; l$ in $c$. Join $a\; c$, and through $c$ draw the parallel $c\; b$; so $c\; b$ will be the departure, and $a\; b$ the proper difference of latitude; then proceeding with this, as in the foregoing problem, you will find the ship's place to be $e$.

PROB. XI. The latitude failed from, difference of latitude, and departure, given; to find the ship's place upon the chart.

EXAMPLE. Suppose a ship from $a$ in the latitude of $20^\{\backslash circ\}\; 00\text{'}$ north, sails between north and east, till she be in the latitude of $45^\{\backslash circ\}\; 00\text{'}$ north, and is then found to have made of departure 116 miles: Required the ship's place upon the chart.

Having drawn the meridian of $a$, set off upon it, from $a$ to $b$, 25 degrees, (taken from the equator or graduated parallel,) the proper difference of latitude; then thro' $b$ draw the parallel $b\; c$, and make $b\; c$ equal to 116 the departure, and join $a\; c$. Count from the parallel of $a$ on the graduated meridian upwards to $b$ 25 degrees, and through $b$ draw the parallel $b\; e$, which will meet $a\; c$ produced in some point $e$, and this will be the place of the ship required.

12. In the section of Plain Sailing it is plain that the terms meridional distance, departure, and difference of longitude, were synonymous, constantly signifying the same thing: which evidently followed from the supposition of the earth's surface being projected on a plane, in which the meridians were made parallel, and the degrees of latitude equal to one another and to those of the equator. But since it has been demonstrated (in this section) that if, in the projection of the earth's surface upon a plane, the meridians be made parallel, the degrees of latitude must be unequal, still increasing the nearer they come to the pole. It follows that these terms must denote lines really different from one another.

SECT. 6. OF OBLIQUE SAILING.

THE questions that may be proposed on this head being innumerable, we shall only give a few of the most useful.

PROB. I. Coasting along the shore, I saw a cape bear from me NNE; then I stood away NW&W 20 miles, and I observed the same cape to bear from me NE&E. Required the distance of the ship from the cape at each station.

GEOMETRICALLY. Draw the circle NWSE (No. 22) to represent the compass, NS the meridian, and WE the east and west line, and let $C$ be the place of the ship in her first station; then from $C$ set off upon the NW&W line, $CA$ 20 miles, and $A$ will be the place of the ship in her second station.

From

From C draw the NNE line CB, and from A draw AB parallel to the NESE line CD, which will meet CB in B the place of the cape, and CB will be the distance of it from the ship in its first station, and AB the distance in the second: to find which,

In the triangle ABC are given AC, equal to 20 miles; the angle ACB, equal to $78^\{\backslash circ\}\; 45\text{'}$, the distance between the NNE and NWSE lines; also the angle ABC, equal to BCD, equal to $33^\{\backslash circ\}\; 45\text{'}$, the distance between the NNE and NBSE lines; and consequently the angle A, equal to $67^\{\backslash circ\}\; 30\text{'}$.

Hence for CB, the distance of the cape from the ship in her first station, it will be (by oblique trigonometry)

S. ABC : AC :: S. BAC : CB,
i. e. As the sine of the angle B $33^\{\backslash circ\}\; 45\text{'}$ 9.74473
is to the distance run AC 20 — 1.30103
so is the sine of BAC — 67, 30 9.96562
to CB — 33.26 1.52191
the distance of the cape from the ship at the first station.

Then for AB, it will be, by oblique trigonometry,

S. ABC : AC :: S. ACB : AB.
i. e. As the sine of B — $33^\{\backslash circ\}\; 45\text{'}$ 9.74474
is to AC — 20 — 1.30103
so is the sine of C — $78\; -\; 45$ 9.99157
to AB — 35.31 1.54786
the distance of the ship from the cape at her second station.

PROB. II. Coasting along the shore, I saw two headlands; the first bore from me NESE 17 miles, the other SSW miles. Required the bearing and distance of these headlands from one another.

GEOMETRICALLY. Having drawn the compass NWSE (No. 23.) let C represent the place of the ship; set off upon the NESE line CA 17 miles from C to A, and upon the SSW line CB 20 miles from C to B, and join AB: then A will be the first headland, and B the second; also AB will be their distance, and the angle A will be the bearing from the NESE line: to find which

In the triangle ACB are given, AC 17, CB 20, and the angle ACB equal to $101^\{\backslash circ\}\; 15\text{'}$, the distance between the NESE and SSW lines. Hence (by oblique angular trigonometry) it will be

As the sum of the sides AC and CB 37 1.56820
is to their difference — 3 0.47712
so is the tang. of $\backslash frac\{1\}\{2\}$ the sum } $39^\{\backslash circ\}\; 22\backslash frac\{1\}\{2\}\text{'}$ 9.91417
of the angles A and B }
to the tang. of half their diff. 3, 49 8.82309
consequently the angle A will be $43^\{\backslash circ\}\; 11\text{'}$, and the angle B $35^\{\backslash circ\}\; 34\text{'}$; also the bearing of B from A will be SW $1^\{\backslash circ\}\; 49\text{'}$ westerly, and the bearing of A from B will be NE $1^\{\backslash circ\}\; 49\text{'}$ easterly.

Then for the distance AB, it will be, by oblique angular trigonometry,

S. A : CB :: S. C : AB.
i. e. As the sine of A $43^\{\backslash circ\}\; 11\text{'}$ 9.83527
is to CB — 20 — 1.30103
so is the sine of C — $101,\; 15$ — 9.99157
to AB — 28.67 — 1.45733
the distance between the two headlands.

PROB. III. Coasting along the shore, I saw two headlands; the first bore from me NWSE, and the second NNE; then standing away ENE $\backslash frac\{1\}\{2\}$ northerly 20 miles, I found the first bore from me WNW $\backslash frac\{1\}\{2\}$ westerly, and the second NW $\backslash frac\{1\}\{2\}$ westerly. Required the bearing and distance of these two headlands.

GEOMETRICALLY. Having drawn the compass NWSE (No. 24.) let C represent the first place of the ship; from which draw the NWSE line CB, and the NNE line CD, also the ENE $\backslash frac\{1\}\{2\}$ N line CA, which make equal to 20. From A draw AB parallel to the WNW $\backslash frac\{1\}\{2\}$ W line, and AD parallel to the NW $\backslash frac\{1\}\{2\}$ W meeting the two first lines in the points B and D; then B will be the first and D the second headlands. Join the points B and D, and BD will be the distance between them, and the angle CDB the bearing from the NNE line: to find which

1. In the triangle ABC are given the angle BCA, equal to $104^\{\backslash circ\}\; 04\text{'}$, the distance between the NWSE line, and the ENE $\backslash frac\{1\}\{2\}$ E line; the angle BAC, equal to $36^\{\backslash circ\}\; 34\text{'}$, the distance between the WSW $\backslash frac\{1\}\{2\}$ W line and the WNW $\backslash frac\{1\}\{2\}$ W line; the angle ABC equal to $39^\{\backslash circ\}\; 22\text{'}$, the distance between the ESE $\backslash frac\{1\}\{2\}$ E line; and the SWSE line, also the side CA equal to 20 miles; whence for CB, it will be (by oblique trigonometry)

As the sine of CBA — $39^\{\backslash circ\}\; 22\text{'}$ — 9.80228
is to AC — 20 — 1.30103
so is the sine of CAB — $36^\{\backslash circ\}\; 34\text{'}$ — 9.77507
to CB — 18.79 — 1.27382
the distance between the first headland and the ship in her first station.

2. In the triangle ACD, are given the angle ACD, equal to $47^\{\backslash circ\}\; 49\text{'}$, the distance between the ENE $\backslash frac\{1\}\{2\}$ E line, and the NNE line; the angle CAD, equal to $92^\{\backslash circ\}\; 49\text{'}$, the distance between the WSW $\backslash frac\{1\}\{2\}$ W line; and the NW $\backslash frac\{1\}\{2\}$ W line, the angle CDA equal to $39^\{\backslash circ\}\; 22\text{'}$, the distance between the SSW line and the SBE $\backslash frac\{1\}\{2\}$ E line; also the leg CA equal to 20.

Hence for CD, it will be (by oblique trigonometry)
As the sine CAD — $39^\{\backslash circ\}\; 22\text{'}$ — 9.80228
is to AC — 20 — 1.30103
so is the sine of CAD — $92^\{\backslash circ\}\; 49\text{'}$ — 9.99660
to CD — 31.5 — 1.49835
the distance between the second headland and the ship in her first station.

3. In the triangle BCD are given BC 18.79, CD 31.5, and the angle BCD equal to $56^\{\backslash circ\}\; 15\text{'}$, the distance between the NWSE line and the NNE line.

Hence for the angle CDB, it will be (by oblique trigonometry)

As the sum of the sides — 50.29 — 1.70148
is to the difference of sides 12.71 — 1.10415
so is tangent of $\backslash frac\{1\}\{2\}$ sum } — $61^\{\backslash circ\}\; 51\text{'}$ — 10.27189
of the unknown angles }
to tang. of half their diff. 25, 18 — 9.67459
consequently the angle CBD is $87^\{\backslash circ\}\; 10\text{'}$, and the angle CDB $36^\{\backslash circ\}\; 35\text{'}$. Hence the bearing of the first headland from the second will be S $59^\{\backslash circ\}\; 8\text{'}$, W or SW $59\backslash frac\{1\}\{2\}\; W$ nearly; and for the distance between them, it will be,
As the sine of BDC — $36^\{\backslash circ\}\; 35\text{'}$ — 9.77524
is to BC — 18.79 — 1.27382

so is the fine of BCD — $56^\{\backslash circ\},\; 15\text{'}$ — 9.91985
to BD — — — — — 26.21 — — — 1.41843
the distance between the two headlands.

This, and the first problem, are of great use in drawing the plot of any harbour, or laying down any sea coast.

Suppose a ship that makes her way good within $6\backslash frac\{1\}\{2\}$ points of the wind, at north, is bound to a port bearing east 86 miles distance from her: Required the course and distance upon each tack, to gain the intended port.

GEOMETRICALLY. Having drawn the compass NE SW, (No. 25.) let C represent the ship's place, and set off upon the east line CA 86 miles, so A will be the intended port. Draw CD and CB on each side of the north line at $6\backslash frac\{1\}\{2\}$ points distance from it, and through A draw AB parallel to CD meeting CB in B; then the ENE $\backslash frac\{1\}\{2\}$ E line CB, will be the course of the ship upon the starboard tack, and CB its distance on that tack; also the ESE $\backslash frac\{1\}\{2\}$ E line AB, will be the course on the larboard tack, and BA the distance on that tack: to find which

By CALCULATION;

In the triangle ABC are given, the angle ACB, equal to $16^\{\backslash circ\},\; 53\text{'}$, the distance between the east and ENE $\backslash frac\{1\}\{2\}$ E line; the angle CBA, equal to $146^\{\backslash circ\},\; 14\text{'}$, the distance between the ENE $\backslash frac\{1\}\{2\}$ E and the WNW $\backslash frac\{1\}\{2\}$ W lines; the angle BAC equal to $16^\{\backslash circ\},\; 53\text{'}$, the distance between the east and ESE $\backslash frac\{1\}\{2\}$ E lines; also AC 86 miles.

Hence since the angle at A and C are equal, the legs CB and BA will likewise be equal; to find either of which (suppose CB) it will be (by oblique angled trigonometry.)

As the fine of B — $146^\{\backslash circ\},\; 14\text{'}$ — 9.74493
is to AC — — — — — 86 — — — — — 1.93450
so is the fine of A — $16^\{\backslash circ\},\; 53\text{'}$ — 9.46303
to CB — — — — — 44.94 — — — — — 1.65260
the distance the ship must sail on each tack.

There is a great variety of useful questions of this nature that may be proposed; but the nature of them being better understood by practice at sea, we shall leave them, and go on to Current Sailing.

SECT. 7. Concerning CURRENTS, and how to make proper allowances.

1. CURRENTS are certain settings of the stream, by which all bodies (as ships, &c.) moving therein, are compelled to alter their course or velocity, or both; and submit to the motion impressed upon them by the current.

CASE I. If the current sets just with the course of the ship, (i. e.) moves on the same rhomb with it; then the motion of the ship is increased, by as much as is the drift or velocity of the current.

EXAMPLE. Suppose a ship sails SESS at the rate of 6 miles an hour, in a current that sets SESS 2 miles an hour: Required her true rate of sailing.

Here it is evident that the ship's true rate of sailing will be 8 miles an hour.

CASE II. If the current sets directly against the ship's course, then the motion of the ship is lessened by as much as is the velocity of the current.

EXAMPLE. Suppose a ship sails SSW at the rate of 10 miles an hour, in a current that sets NNE 6 miles an hour. Required the ship's true rate of sailing.

Here it is evident that the ship's true rate of sailing will be 4 miles an hour. Hence it is plain,

COR. I. If the velocity of the current be less than the velocity of the ship, then the ship will get so much ahead as is the difference of these velocities.

COR. II. If the velocity of the current be greater than that of the ship, then the ship will fall so much a stern as is the difference of these velocities.

COR. III. Lastly, If the velocity of the current be equal to that of the ship, then the ship will stand still; the one velocity destroying the other.

CASE III. If the current thwarts the course of the ship, then it not only lessens or augments her velocity, but gives her a new direction compounded of the course she steers, and the setting of the current, as is manifest from the following

LEMMA. If a body at A (No. 26.) be impelled by two forces at the same time, the one in the direction AB capable to carry that body from A to B in a certain space of time, and the other in the direction AD capable to carry it from A to D in the same time; complete the parallelogram ABCD, and draw the diagonal AC; then the body at A agitated by these two forces together, will move along the line BC, and will be in the point C at the end of the time in which it would have moved along AD or AB with the forces separately applied.

Hence the solution of the following examples will be evident.

EXAMPLE I. Suppose a ship sails (by the compass) directly south 96 miles in 24 hours, in a current that sets east 45 miles in the same time. Required the ship's true course and distance.

GEOMETRICALLY. Draw AD (see No. 26.) to represent the south and north line of the ship at A, which make equal to 96; from D draw DC perpendicular to AD, equal to 45; and join AC. Then C will be the ship's true place, AC her true distance, and the angle CAD the true course. To find which

By CALCULATION:

First, For the true course DAC, it will be, (by rectangular trigonometry.)

As the apparent distance AD — 96 — 1.98227
is to the current's motion DC — 45 — 1.65321
so is radius — — — — — 10.00000
to the tangent of the true }
course DAC — — — — — 25°, 07' 9.67094
consequently the ship's true course is S $25^\{\backslash circ\},\; 07\text{'}$ E, or
SSE $2^\{\backslash circ\},\; 37\text{'}$, easterly.

Then for the true distance AC, it will be, (by rectangular trigonometry.)

As the fine of the course A — $25^\{\backslash circ\},\; 07\text{'}$ 9.67094
is to the departure DC — — — — — 45 — 1.65321
so is radius — — — — — 10.00000
to the true distance AC — — — — — 106 — 2.02537

EXAMPLE. Suppose a ship sails SE 120 miles in 20 hours, in a current that sets W&N at the rate of 2 miles an hour: Required the ship's true course and distance sailed in that time.

GEOMETRICALLY. Having drawn the compass NESW (No 27.) let C represent the place the ship failed from; draw the SE line CA, which make equal to 120; then will A be the place the ship caped at.

From A draw AB parallel to the W&N line CD, equal to 40, the motion of the current in 20 hours, and join CB; then B will be the ship's true place at the end of 20 hours, CB her true distance, and the angle SCB her true course. To find which

By CALCULATION;

In the triangle ABC, are given CA 120, AB 40, and the angle CAB equal to $34^\{\backslash circ\}\; 45\text{'}$, the distance between the E&S and SE lines, to find the angles B and C, and the side CB.

First, For the angles C and B, it will be, (by oblique trigonometry)

As the sum of the sides CA and AB 160 — 2.20412
is to their difference — 80 — 1.90309

so is the tang. of half the sum }
of the angles B and C — } $73^\{\backslash circ\},\; 07\text{'}$ 10.51783
to the tang. of half their diff. — } $59,\; 45$ 10.21680
consequently the angle B will be $131,\; 52$, and the angle ACB $14^\{\backslash circ\}\; 23\text{'}$. Hence the true course is S $30^\{\backslash circ\},\; 27\text{'}\; E$, or SSE $2^\{\backslash circ\}\; 07\text{'}$ easterly.

Then for the true distance CB, it will be, (by oblique trigonometry)

As the sine of B — $131^\{\backslash circ\},\; 52\text{'}$ — 9.87198
is to AC — 120 — 2.07918
so is the sine of A — $33^\{\backslash circ\},\; 45\text{'}$ — 9.74474
to the true distance CB — 89.53 — 1.95194

EXAMPLE III. Suppose a ship coming out from sea in the night, has sight of Scilly light, bearing NE&N distance 4 leagues, it being then flood tide setting ENE 2 miles an hour, and the ship running after the rate of 5 miles an hour. Required upon what course and how far she must sail to hit the Lizard, which bears from Scilly E&S distance 17 leagues.

GEOMETRICALLY. Having drawn the compass NESW (No. 28.) let A represent the ship's place at sea, and draw the NE&N line AS, which make equal to 12 miles, so S will represent Scilly.

From S draw SL equal to 51 miles, and parallel to the E&S line; then L will represent the Lizard.

From L draw LC parallel to the ENE line, equal to 2 miles, and from C draw CD equal to 5 miles meeting AL in D; then from A draw AB parallel to CD meeting LC produced in B; and AB will be the required distance, and SAB the true course. To find which

By CALCULATION;

In the triangle ASL are given the side AS equal to 12 miles, the side SL equal to 51, and the angle ASL equal to $118^\{\backslash circ\}\; 07\text{'}$, the distance between the NE&N and W&N lines; to find the angles SAL and SLA. Consequently, (by oblique trigonometry,) it will be,
As the sum of the sides AS and SL — 63 1.79934
is to their difference — 39 1.59106
so is the tang. of half the sum }
of the angles SAL and SLA } $30^\{\backslash circ\},\; 56\text{'}$ 9.77763
to the tang. of half their diff. — $20^\{\backslash circ\},\; 21\text{'}$ 9.56935

consequently the angle SAL, will be $51^\{\backslash circ\}\; 17\text{'}$, and so the direct bearing of the Lizard from the ship will be N $85^\{\backslash circ\}\; 02\text{'}\; E$, or E & N $6^\{\backslash circ\}\; 17\text{'}\; E$; and for the distance AL, it will be (by oblique trigonometry.)

the distance between the ship and the Lizard.

Again, in the triangle DLC, are given the angle L equal to $17^\{\backslash circ\}\; 32\text{'}$, the distance between the ENE and N $85^\{\backslash circ\}\; 02\text{'}\; E$ lines; the side LC, equal to 2 miles, the current's drift in an hour; and the side CD, equal to 5 miles, the ship's run in the same time. Hence for the angle D, it will be (by oblique trigonometry.)

consequently since by construction the angle LAB is equal to the angle LDC, the course the ship must steer is S $88^\{\backslash circ\}\; 03\text{'}\; E$.

Then for the distance AB, it will be (by oblique trigonometry.)

consequently, since the ship is sailing at the rate of 5 miles an hour, it follows, that in sailing 8^h 24^m S $88^\{\backslash circ\}\; 03\text{'}\; E$, she will arrive at the Lizard.

EXAMPLE IV. A ship from a certain headland in the latitude of $34^\{\backslash circ\}\; 00\text{'}$ north, sails SE&S 12 miles in three hours, in a current that sets between north and east; and then the same headland is found to bear WNW, and the ship to be in the latitude of $33^\{\backslash circ\}\; 52\text{'}$ north. Required the setting and drift of the current.

GEOMETRICALLY. Having drawn the compass NESW (No. 29.) let A represent the place of the ship, and draw the SE&S line AB equal to 12 miles, also the ESE line AC.

Set off from A upon the meridian AD, equal to 8 miles, the difference of latitude, and through D draw DC parallel to the east and west line WE, meeting AC in C. Join C and B with the right line BC; then C will be the ship's place, the angle ABC the setting of the current from the SE&S line, and the line BC will be the drift of the current in 3 hours. To find which

By CALCULATION;

In the triangle ADC right angled at D, are given the difference of latitude AD equal to 8 miles, the angle DAC equal to $67^\{\backslash circ\}\; 30\text{'}$. Whence for AC, the distance the ship has sailed, it will be
As radius — — — — — 10.00000
is to the diff. of latitude AD — 8 — 0.90309
so is the secant of the course }
DAC — — — — — } $67^\{\backslash circ\},\; 30\text{'}$ 10.41716
to the distance run AC — — — — — 20.9 — 1.32025

Again, in the triangle ABC, are given AB equal to 5 E 12

12 miles, AC equal to 20.9, and the angle BAC equal to $33^\{\backslash circ\}\; 45\text{'}$, the distance between the SEBS and ESE lines. Whence for the angle at B, it will be,

As the sum of the sides AC and AB $32.9$ $1.51720$
is to their difference $—$ $8.9$ $—$ $0.94930$

so is the tang. of half the sum of the angles B and C $\backslash left\backslash \{\; 73^\{\backslash circ\},\; 07\text{'}\; \backslash right\backslash \}$ $—$ $10.51806$

to tang. of $\backslash frac\{1\}\{2\}$ their diff. $—$ $41^\{\backslash circ\},\; 43\text{'}\; \backslash frac\{1\}\{2\}$ $—$ $9.95025$
consequently the angle B is $114^\{\backslash circ\}\; 51\text{'}$, and so the setting of the current will be N $81^\{\backslash circ\}\; 06\text{'}\; E$ or ESE $2^\{\backslash circ\}\; 21\text{'}\; E$. Then for BC the current's drift in 3 hours, it will be,

As the sine of B $—$ $114^\{\backslash circ\},\; 51\text{'}$ $—$ $9.92700$
is to the distance run AC 20.9 $—$ $1.32025$

so is the sine of A $—$ $33^\{\backslash circ\},\; 45\text{'}$ $—$ $9.74474$
to BC $—$ $12.8$ $—$ $1.10719$

the current's drift in 3 hours, and consequently the current sets ESE $2^\{\backslash circ\}\; 21\text{'}\; E$ 4.266 miles an hour.

SECT. 8. Concerning the VARIATION of the COMPASS, and how to find it from the true and observed AMPLITUDES or AZIMUTHS of the sun.

1. The variation of the compass is how far the north or south point of the needle stands from the true south or north point of the horizon towards the east or west; or it is an arch of the horizon intercepted between the meridian of the place of observation and the magnetic meridian.

2. It is absolutely necessary to know the variation of the compass at sea, in order to correct the ship's course; for since the ship's course is directed by the compass, it is evident that if the compass be wrong the true course will differ from the observed, and consequently the whole reckoning differ from the truth.

3. The sun's true amplitude is an arch of the horizon comprehended between the true east or west point thereof, and the centre of the sun at rising or setting; or it is the number of degrees, &c. that the centre of the sun is distant from the true east or west point of the horizon, towards the south or north.

4. The sun's magnetic amplitude is the number of degrees that the centre of the sun is from the east or west point of the compass, towards the south or north point of the same at rising or setting.

5. Having the declination of the sun, together with the latitude of the place of observation, we may from thence find the sun's true amplitude, by the following astronomical proposition, viz.

As the co-sine of the latitude

is to the radius

So is the sine of the sun's declination

to the sine of the sun's true amplitude

which will be north or south according as the sun's declination is north or south.

EXAMPLE. Required the sun's true amplitude in the latitude of $41^\{\backslash circ\}\; 50\text{'}$ north, on the 23d day of April 1731.

First, I find (from the tables of the sun's declination) that the sun's declination the 23d of April is $15^\{\backslash circ\}\; 54\text{'}$ north; then for the true amplitude, it will be, by the former analogy.

As the co-sine of the lat. $41^\{\backslash circ\}\; 50\text{'}$ $—$ $9.87221$
is to radius $—$ $—$ $—$ $10.00000$

so is the sine of the decl. $15^\{\backslash circ\},\; 54\text{'}$ $—$ $9.43760$
to the sine of the amplit. $21^\{\backslash circ\},\; 35\text{'}$ $—$ $9.56548$
which is north, because the declination is north at that time; and consequently, in the latitude of $41^\{\backslash circ\}\; 50\text{'}$ north, the sun rises on the 23d of April $21^\{\backslash circ\}\; 35\text{'}$ from the east part of the horizon towards the north, and sets so much from the west the same way.

6. The sun's true azimuth is the arch of the horizon intercepted between the meridian and the vertical circle passing through the centre of the sun at the time of observation.

7. The sun's magnetic azimuth is the arch of the horizon intercepted between the magnetic meridian and the vertical, passing through the sun.

8. Having the latitude of the place of observation, together with the sun's declination and altitude at the time of observation, we may find his true azimuth after the following method, viz.

Make it,

As the tangent of half the complement of the latitude is to the tangent of half the sum of the distance of the sun from the pole and complement of the altitude

So is the tangent of half the difference between the distance of the sun from the pole and complement of the altitude

To the tangent of a fourth arch

which fourth arch added to half the complement of the latitude will give a fifth arch, and this fifth arch lessened by the complement of the latitude will give a sixth arch.

Then make it

As the radius

is to the tangent of the altitude

so is the tangent of the sixth arch

to the co-sine of the sun's azimuth

which is to be counted from the south or north, to the east or west, according as the sun is situated with respect to the place of observation.

If the latitude of the place and declination of the sun be both north or both south, then the declination taken from $90^\{\backslash circ\}$ will give the sun's distance from the pole; but if the latitude and declination be on contrary sides of the equator, then the declination added to $90^\{\backslash circ\}$ will give the sun's distance from the nearest pole to the place of observation.

EXAMPLE. In the latitude of $51^\{\backslash circ\}\; 32\text{'}$ north, the sun having $19^\{\backslash circ\}\; 39\text{'}$ north declination, his altitude was found by observation to be $38^\{\backslash circ\}\; 18\text{'}$: Required the azimuth.

By the first of the foregoing analogies, it will be

As the tangent of $\backslash frac\{1\}\{2\}$ the complement of the latitude $\backslash left\backslash \{\; 19^\{\backslash circ\},\; 14\text{'}\; 9.54269\; \backslash right\backslash \}$

is to the tangent of $\backslash frac\{1\}\{2\}$ the sum of the distance of the sun from the pole and complement of the altitude $\backslash left\backslash \{\; 61^\{\backslash circ\},\; 01\text{'}\; 10.25655\; \backslash right\backslash \}$

so is the tangent of half their difference $\backslash left\backslash \{\; 9^\{\backslash circ\},\; 19\text{'}\; 7.21499\; \backslash right\backslash \}$

to the tang. of a 4th arch $40^\{\backslash circ\},\; 20\text{'}\; 9.92885$

which fourth arch $40^\{\backslash circ\}\; 20\text{'}$, added to $19^\{\backslash circ\}\; 14\text{'}$ half the complement of the latitude, give a fifth arch $59^\{\backslash circ\}\; 34\text{'}$; and this fifth arch lessened by $38^\{\backslash circ\}\; 28\text{'}$, the complement of the latitude, gives the sixth arch $21^\{\backslash circ\}\; 06\text{'}$; then for the

the azimuth, it will be, by the second of the preceding analogies,

As radius — — — — — 10.00000
is to the tang. of the altitude $38^\{\backslash circ\},\; 18\text{'}$ 9.89749
so is the tang. of the sixth arch $21^\{\backslash circ\},\; 06\text{'}$ — 9.58644
to the co-sine of the azimuth $72^\{\backslash circ\},\; 15\text{'}$ — 9.48393
which, because the latitude is north and the sun south of the place of observation, must be counted from the south towards the east or west; and consequently, if the altitude of the sun was taken in the morning, the azimuth will be $S\; 72^\{\backslash circ\}\; 15\text{'}\; E$, or $ESE\; 4^\{\backslash circ\}\; 45\text{'}\; E$; but if the altitude was taken in the afternoon, the azimuth will be $S\; 72^\{\backslash circ\}\; 15\text{'}\; W$, or $WSW\; 4^\{\backslash circ\}\; 45\text{'}$ westerly.

9. Having found the sun's true amplitude or azimuth by the preceding analogies, and his magnetic amplitude or azimuth by observation, it is evident, if they agree, there is no variation; but if they disagree, then if the true and observed amplitudes at the rising or setting of the sun be both of the same name, i. e. either both north, or both south, their difference is the variation; but if they be of different names, i. e. one north and the other south, their sum is the variation. Again, if the true and observed azimuth be both of the same name, i. e. either both east or both west, their difference is the variation; but if they be of different names, their sum is the variation: And to know whether the variation is easterly, observe this general rule, viz.

Let the observer's face be turned to the sun: then if the true amplitude or azimuth be to the right hand of the observed, the variation is easterly; but if it be to the left, westerly.

To explain which, let NESW (No. 30.) represent a compass, and suppose the sun is really EBS at the time of observation, but the observer sees him off the east point of the compass, and so the true amplitude or azimuth of the sun is to the right of the magnetic or observed; here it is evident that the EBS point of the compass ought to lie where the east point is, and so the north where the N&W is; consequently the north point of the compass is a point too far east, i. e. the variation in this case is easterly. The same will hold when the amplitude or azimuth is taken on the west side of the meridian.

Again, let the true amplitude or azimuth be to the left hand of the observed. Thus, suppose the sun is really EBN at the time of observation, but the observer sees him off the east point of the compass, and so the true amplitude or azimuth to the left of the observed: Here it is evident that the EBN point of the compass ought to stand where the east point is, and so the north where the N&E point is; consequently the north point of the compass lies a point too far westerly, so in this case the variation is west. The same will hold when the sun is observed on the west side of the meridian.

EXAMPLE I. Suppose the sun's true amplitude at rising is found to be $E\; 14^\{\backslash circ\}\; 20\text{'}\; N$, but by the compass it is found to be $E\; 26^\{\backslash circ\}\; 12\text{'}$: Required the variation, and which way it is.

Since they are both the same way, therefore
From the magnetic amplitude — — — — — $E\; 26^\{\backslash circ\},\; 12\text{'}\; N$,

take the true amplitude — — — — — $E\; 14^\{\backslash circ\},\; 20\text{'}\; N$.

and there remains the variation — — — — — $11^\{\backslash circ\},\; 52\text{'}\; E$, which is easterly, because in this case the true amplitude is the right of the observed.

EXAMPLE II. Suppose the sun's true amplitude at setting is $W\; 34^\{\backslash circ\}\; 26\text{'}\; S$, and his magnetic amplitude $W\; 23^\{\backslash circ\}\; 13\text{'}\; S$: Required the variation, and which way it is.

Since they lie both the same way, therefore
From the sun's true amplitude — — — — — $W\; 34^\{\backslash circ\},\; 26\text{'}\; S$,
take his magnetic amplitude — — — — — $W\; 23^\{\backslash circ\},\; 13\text{'}\; S$.

there remains the variation — — — — — $11^\{\backslash circ\},\; 13\text{'}\; W$, which is westerly, because the true amplitude, in this case, is to the left hand of the observed.

EXAMPLE III. Suppose the sun's true altitude at rising is found to be $13^\{\backslash circ\}\; 24\text{'}\; N$, and his magnetic $E\; 12^\{\backslash circ\}\; 32\text{'}\; S$: Required the variation, and which way it lies.

Since the true and observed amplitudes lie different ways, therefore

To the true amplitude — — — — — $E\; 13^\{\backslash circ\},\; 24\text{'}\; N$,
add the magnetic amplitude — — — — — $E\; 12^\{\backslash circ\},\; 32\text{'}\; S$.

the sum is the variation — — — — — $25^\{\backslash circ\},\; 56\text{'}\; W$, which is westerly, because the true amplitude is, in this case, to the left of the observed.

EXAMPLE IV. Suppose the sun's true altitude at setting is found to be $W\; 8^\{\backslash circ\}\; 24\text{'}\; N$, but his magnetic amplitude is $W\; 10^\{\backslash circ\}\; 13\text{'}\; S$: Required the variation.

To the true amplitude — — — — — $W\; 8^\{\backslash circ\},\; 24\text{'}\; N$,
add the magnetic — — — — — $W\; 10^\{\backslash circ\},\; 13\text{'}\; S$.

the sum is the variation — — — — — $18^\{\backslash circ\},\; 37\text{'}\; E$, which is easterly, because the true amplitude is to the right of the observed.

EXAMPLE V. Suppose the sun's true azimuth at the time of observation, is found to be $N\; 86^\{\backslash circ\}\; 40\text{'}\; E$, but by the compass it is $N\; 73^\{\backslash circ\}\; 24\text{'}\; E$: Required the variation, and which way it lies.

From the sun's true azimuth, — — — — — $N\; 86^\{\backslash circ\},\; 40\text{'}\; E$,
take the magnetical, — — — — — $N\; 73^\{\backslash circ\},\; 24\text{'}\; E$.

there remains the variation, — — — — — $13^\{\backslash circ\},\; 16\text{'}\; E$, which is easterly, because the true azimuth is to the right of the observed.

EXAMPLE VI. Suppose the sun's true azimuth is $S\; 3^\{\backslash circ\}\; 24\text{'}\; E$, and the magnetical $S\; 4^\{\backslash circ\}\; 36\text{'}\; W$: Required the variation, and which way it lies.

To the true azimuth — — — — — $S\; 3^\{\backslash circ\},\; 24\text{'}\; E$,
add the magnetical azimuth — — — — — $S\; 4^\{\backslash circ\},\; 36\text{'}\; W$.

the sum is the variation — — — — — $8^\{\backslash circ\},\; 60\text{'}\; W$, which is westerly, because the true azimuth is, in this case, to the left of the observed.

10. The variation of the compass was first observed at London, in the year 1580, to be $11^\{\backslash circ\}\; 15\text{'}$ easterly, and in the year 1622 it was $6^\{\backslash circ\}\; 0\text{'}\; E$; also in the year 1634, it was $4^\{\backslash circ\}\; 05\text{'}\; E$, still decreasing, and the needle approaching the true meridian, till it coincided with it, and then there was no variation; after which, the variation

tion began to be westerly; and in the year 1672, it was observed to be $2^\{\backslash circ\}\; 30\text{'}\; W$; also in the year 1683, it was $4^\{\backslash circ\}\; 30\text{'}\; W$; and since that time the variation still continues at London to increase westerly; but how far it will go that way, time and observations will probably be the only means to discover.

Again, at Paris, in the year 1640, the variation was $3^\{\backslash circ\}\; 00\text{'}\; E$; and in the year 1666, there was no variation; but in the year 1681, it was $2^\{\backslash circ\}\; 30\text{'}\; W$, and still continues to go westerly.

In short, from observations made in different parts of the world, it appears, that in different places the variation differs both as to its quantity and denomination, it being east in one place, and west in another; the true cause and theory of which, for want of a sufficient number of observations, has not as yet been fully explained.

SECT. 9. The METHOD of keeping a JOURNAL at sea; and how to correct it, by making proper allowances for the leeway, variation, &c.

1. LEE-WAY is the angle that the rhomb line, upon which the ship endeavours to sail, makes with the rhomb she really sails upon. This is occasioned by the force of the wind or surge of the sea, when she lies to the windward, or is close hauled, which causes her to fall off and glide side-ways from the point of the compass she capes at. Thus let NESW (No. 31.) represent the compass; and suppose a ship at C capes at, or endeavours to sail upon, the rhomb Ca; but by the force of the wind, and surge of the sea, she is obliged to fall off, and make her way good upon the rhomb Cb; then the angle $aCb$ is the leeway; and if that angle be equal to one point, the ship is said to make one point lee way; and if equal to two points, the ship is said to make two points lee way, &c.

2. The quantity of this angle is very uncertain, because some ships, with the same quantity of sail, and with the same gale, will make more lee-way than others; it depending much upon the mould and trim of the ship, and the quantity of water that she draws. The common allowances that are generally made for the lee-way, are as follow.

1. If a ship be close hauled, has all her sails set, the water smooth, and a moderate gale of wind, she is then supposed to make little or no lee-way.

2. If it blow so fresh as to cause the small sails be handed, it is usual to allow one point.

3. If it blow so hard that the top sails must be close reefed, then the common allowance is two points for lee-way.

4. If one top sail must be handed, then the ship is supposed to make between two and three points lee way.

5. When both top-sails must be handed, then the allowance is about four points for lee-way.

6. If blows so hard, as to occasion the fore-course to be handed, the allowance is between $5\backslash frac\{1\}\{2\}$ and 6 points.

7. When both main and fore-courses must be handed, then 6 or $6\backslash frac\{1\}\{2\}$ points are commonly allowed for lee-way.

8. When the mizen is handed, and the ship is trying a hull, she is then commonly allowed about 7 points for lee-way.

3. Though these rules are such as are generally made

use of, yet since the lee-way depends much upon the mould and trim of the ship, it is evident that they cannot exactly serve to every ship; and therefore the best way is to find it by observation: Thus, let the ship's wake be set by a compass in the poop, and the opposite rumb is the true course made good by the ship; then the difference between this and the course given by the compass in the battle, is the lee-way required. If the ship be within sight of land; then the lee-way may be exactly found by observing a point on the land which continues to bear the same way, and the distance between the point of the compass it lies upon and the point the ship capes at will be the lee-way. Thus suppose a ship at C, is lying up NW, towards A; but instead of keeping that course, she is carried on the NNE line CB, and consequently the point B continues to bear the same way from the ship: Here it is evident that the angle ACB, or the distance between the NW line that the ship capes at, and the NNE line that the ship really sails upon, will be the lee-way.

4. Having the course steered, and the lee-way, given; we may from thence find the true course by the following method, viz. Let your face be turned directly to the windward; and if the ship have her larboard tacks on board, count the lee way from the course steered towards the right hand; but if the starboard tacks be on board, then count it from the course steered towards the left hand. Thus, suppose the wind at north, and the ship lies up within 6 points of the wind, with her larboard tacks on board, making one point lee-way; here it is plain, that the course steered is ENE, and the true course EBN; also suppose the wind is at NNW, and the ship lies up within $6\backslash frac\{1\}\{2\}$ points of the wind with her starboard tack on board, making $1\backslash frac\{1\}\{2\}$ point lee way; it is evident that the true course, in this case, is WSW.

5. We have shewed, in the last section, how to find the variation of the compass; and from what has been said there, we have this general rule for finding the ship's true course, having the course steered and the variation given, viz. Let your face be turned towards the point of the compass upon which the ship is steered; and if the variation be easterly, count the quantity of it from the course steered towards the right hand; but if westerly, towards the left hand; and the course thus found is the true course steered. Thus, suppose the course steered is NNE, and the variation one point easterly; then the true course steered will be NNE: Also suppose the course steered is NESE, and the variation one point westerly; then in this case, the true course will be NE; and so of others.

Hence, by knowing the lee-way variation, and course steered, we may from thence find the ship's true course; but if there be a current under foot, then that must be tried, and proper allowances made for it, as has been shown in the section concerning Currents, from thence to find the true course.

6. After making all the proper allowances for finding the ship's true course, and making as just an estimate of the distance as we can; yet by reason of the many accidents that attend a ship in a day's running, such as different rates of sailing between the times of heaving the log, the

N.1.

N.2.

N.3.

N.4.

N.5.

N.6.

N.7.

N.8.

N.9.

N.12.

N.10.

N.13.

N.11.

N.14.

N. 15.

N. 16.

N. 17.

N. 18.

N. 19.

N. 20.

N. 21.

N. 22.

N. 23.

N. 24.

N. 25.

N. 26.

N. 27.

N. 28.

N. 29.

N. 30.

N. 31.

N. 33.

the want of due care at the helm by not keeping her steady, but suffering her to yaw and fall off; sudden storms, when no account can be kept, &c.: the latitude by account frequently differs from the latitude by observation; and when that happens, it is evident there must be some error in the reckoning; to discover which, and where it lies, and also how to correct the reckoning, you may observe the following rules.

1^st, If the ship sail near the meridian, or within 2 or 2½ points thereof; then if the latitude by account disagrees with the latitude by observation, it is most likely that the error lies in the distance run; for it is plain that in this case it will require a very sensible error in the course to make any considerable error in the difference of latitude, which cannot well happen if due care be taken at the helm, and proper allowances be made for the lee-way, variation, and currents. Consequently if the course be pretty near the truth, and the error in the distance run regularly through the whole, we may, from the latitude obtained by observation, correct the distance and departure by account, by the following analogies, viz. As the difference of latitude by account is to the true difference of latitude, so is the departure by account to the true departure, and so is the direct distance by account to the true direct distance.

The reason of this is plain: for let AB (No. 33.) denote the meridian of the ship at A, and suppose the ship sails upon the rhomb AE near the meridian, till by account she is found in C, and consequently her difference of latitude by account is AB; but by observation she is found in the parallel ED, and so her true difference of latitude is AD, her true distance AE, and her true departure DE; then since the triangles ABC ADE are similar, it will be AB : AD :: BC : DE, and AB : AD :: AC : AE.

EXAMPLE. Suppose a ship from the latitude of 45° 20' north, after having sailed upon several courses near the meridian for 24 hours, her difference of latitude is computed to be upon the whole 95 miles southerly, and her departure 34 miles easterly; but by observation she is found to be in the latitude of 43° 10' north, and consequently her true difference of latitude is 130 miles southerly; then for the true departure, it will be, As the difference of latitude by account 95 is to the true difference of latitude 130, so is the departure by account 34 to the true departure 46.52, and so is the distance by account 100.9 to the true distance 138.

2^dy, If the courses are for the most part near the parallel of east and west, and the direct course be within 5½ or 6 points of the meridian; then if the latitude by account differs from the observed latitude, it is most probable that the error lies in the course, or distance, or perhaps both; for in this case it is evident, the departure by account will be very nearly true; and thence by the help of this, and the true difference of latitude, may the true course and direct distance be readily found by Case 4. of Plain Sailing.

EXAMPLE. Suppose a ship from the latitude of 43° 50' north, after having sailed upon several courses near the parallel of east and west, for the space of 24 hours, is found by dead reckoning to be in the latitude of 42° 45' north, and to have made 160 miles of westing; but by a good observation the ship is found to be in the latitude of 42° 35' north: Required the true course, and direct distance sailed.

With the true difference of latitude 75 miles, and departure 160 miles, we shall find (by Case 4. of Plain Sailing) the true course to be S 64° 53' W, and the direct distance 176.7 miles.

3^dy, If the courses are for the most part near the middle of the quadrant, and the direct course within 2 and 6 points of the meridian; then the error may be either in the course, or in the distance, or in both, which will cause an error both in the difference of latitude and departure; to correct which, having found the true difference of latitude by observation, with this, and the direct distance by dead reckoning, find a new departure (by Case 3. of Plain Sailing;) then half the sum of this departure, and that by dead reckoning, will be nearly equal to the true departure; and consequently with this, and the true difference of latitude, we may (by Case 4. of Plain Sailing) find the true course and distance.

EXAMPLE. Suppose a ship from the latitude of 44° 38' north, sails between south and east upon several courses, near the middle of the quadrant, for the space of 24 hours, and is then found by dead reckoning to be in the latitude of 42° 15' north, and to have made of easting 136 miles; but by observation she is found to be in the latitude of 42° 04' north: Required her true course and distance.

With the true distance of latitude 154 miles, and the direct distance by dead reckoning 197.4, you will find (by Case 3. of Plain Sailing) the new departure to be 123.4, and half the sum of this and the departure by dead reckoning will be 123.7 the true departure; then with this, and the true difference of latitude, you will find (by Case 4. of Plain Sailing) the true course to be S 39° 00' E, and the direct distance 198.2 miles.

7. In keeping a ship's reckoning at sea, the common method is to take from the log-board the several courses and distances stemmed by the ship last 24 hours, and to transfer these together with the most remarkable occurrences into the log-book, into which also are inserted the courses corrected, and the difference of latitude and difference of longitude made good upon each; then the whole day's work being finished in the log-book, if the latitude by account agree with the latitude by observation, the ship's place will be truly determined; if not, then the reckoning must be corrected according to the preceding rules, and placed in the journal.

The form of the Log-book and Journal, together with an example of 2 days work, you have here subjoined.

Note, To express the days of the week, they commonly use the characters by which the sun and planets are expressed, viz. ☉ denotes Sunday, ☾ Monday, ♀ Tuesday, ♃ Wednesday, ♄ Thursday, ♀ Friday, and ♄ denotes Saturday.

The FORM of the
L O G - B O O K,

With the Manner of working Days Works at SEA.

Hence the ship, by account, has come to the latitude of 47° 46' north, and has differed her longitude 2° 5' westerly; so this day I have made my way good S 31° 31' W, distance 157.4 miles.

At noon the Lizard bore from me N 31° 31' E, distance 157.4 miles; and having observed the latitude, I found it agreed with the latitude by account.

Hence the ship, by account, has come to the latitude of 47° 17' north, and has differed her longitude 49' easterly; consequently she has got 1° 16' to the westward of the Lizard, and has made her way good the last 24 hours 849° 08' E, distance 44.3 miles.

At noon the Lizard bore from me north 17° 7' east, distance 170.6 miles.

This day I had an observation, and found the latitude by account to disagree with the latitude by observation by 11 minutes, I being so much further to the southward than by dead reckoning, which by the third of the preceding rules I correct as in the Journal.

A JOURNAL from the Lizard towards Jamaica in the ship Neptune, J. M. commander.