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 the 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, interwoven 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 sails 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 \( \frac{1}{60} \) of that, viz. a minute, or nautical mile, must contain 6120 standard feet; consequently, since \( \frac{1}{60} \) minute is the \( \frac{1}{24} \) part of an hour, and each knot being the same part of a nautical mile, it follows, that each knot will contain the \( \frac{1}{24} \) 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 $6\frac{1}{2}$ 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 $6\frac{1}{2}$ 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 or 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 them 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 $7\frac{1}{2}$ 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 $6\frac{1}{2}$ 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 $3\frac{1}{2}$ 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 give 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 an 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 $9\frac{1}{2}$ 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 38 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 6.83 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 forever 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 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.
14. In order to determine the course of the winds, and to discover their various alterations or shifting; 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 rhumb 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 rhumb 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 whichsoever 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 Rhombi 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 foregoing 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 forever the ship heels.
18. Since the meridians do all meet at the poles, and 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 heliapherical 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 rhomb: 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 rhumbs (or points in the compass) equally distant from one another, therefore the angle contained between any two of them adjacent will be $11^\circ 15'$, viz. $\frac{1}{3}$ part of $360^\circ$; and so the angle contained between the meridian and the NNE, will be $11^\circ 15'$, and between the meridian and the NNE will be $22^\circ 30'$; and so of the rest, as in the following table.
A Table of the Angles which every $\frac{1}{4}$ Point of the Compass makes with the Meridian.
| North | North | North | North | |-------|-------|-------|-------| | East | East | East | East | | South | South | South | South |
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 $22^\circ 30'$, 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 latitude 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 hypotenuse 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 third; 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 sails 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.
As radius — — — 10.00000 to the distance AC — — — 32 — — 1.50515 so is the sine of the course A 22° 30' — — 9.58284 to the departure BC — — — 12.25 — — 1.08799 so the ship has made 12.25 miles of departure easterly, or has got so far to the easterly of her meridian. Then for the difference of latitude or northing the ship has made, we have (by rectangular trigonometry) the following analogy, viz.
As radius — — — 10.00000 to the distance AC — — — 32 — — 1.50515 so is the cosine of course A 22° 30' — — 9.96562 to the difference of lat. AB — — — 29.57 — — 1.47077 so the ship has differed her latitude, or made of northing, 29.57 minutes.
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 — — — 0° 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, sails NE by N ½ 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,
From the latitude come to — — — 46° 55' subtract the latitude failed from — — — 45° 25' and their remains — — — 01° 30'
the difference of latitude, equal to — — — 90 miles.
And (by rectangular trigonometry) we have the following analogy, for finding the departure BD, viz.
As radius — — — 10.00000 is to the diff. of latitude AB — — — 90 — — 1.95424 so is the tangent of course A — — — 39° 22' 9.91404 to the departure BD — — — 78° 34' 1.8618 so the ship has got 73 84 miles to the easterly of her former meridian.
Again, for the distance AD, we have (by rectangular trigonometry) the following proportion, viz.
As radius — — — 10.00000 is to the secant of the course 39° 22' — — — 10.11176 so is the difference of latitude AB 90 — — — 1.95424 to the distance AD — — — 116.4 — — 2.06600
Case III. Difference of latitude and distance given, to find course and departure.
Example. Suppose a ship sails 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 sailed on, and her departure from the meridian. No 6.
Since the latitudes are both north, and the ship sailing towards the equator; therefore,
From the latitude failed from — — — 56° 50' subtract the observed latitude — — — 55° 10' 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.
As the distance sailed DF — — — 126 — — 2.10037 is to radius — — — 10.00000 so is the diff. of latitude FD — — — 70 — — 1.84510 to the co-sine of the course F 56° 15' — — — 9.74473 which, because she sails between south and west, will be south 56° 15' west, or SW by W. Then, for the departure, we have (by rectangular trigonometry) the following proportion, viz.
As radius — — — 10.00000 is to the distance sailed DF — — — 126 — — 2.10037 so is the sine of the course F — — — 56° 15' 9.91985 to the departure DE — — — 104.8 — — 2.02022 consequently she has made 104.8 miles of departure westerly.
Case IV. Difference of latitude and departure given, to find course and distance.
Example. Suppose a ship sails from the latitude of 44° 50' north, between south and east, till she has made 64 miles of easterly, 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 sailing towards the equator; therefore,
From the latitude failed from — — — 44° 50' N take the latitude come to — — — 42° 56' and their remains — — — 01° 34' 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.
As the diff. of latitude GK 114 — — — 2.05690 is to radius — — — 10.00000 so is the departure KL — — — 64 — — 1.80618 to the tangent of course $G = 29^\circ 19' - 9.74928$ which, because the ship is sailing between south and east, will be south $29^\circ 19'$ east, or SSE $\frac{1}{2}$ east nearly.
Then for the distance, we shall have (by rectangular trigonometry) the following analogy, viz.
As radius $= 10.00000$
is to the diff. of latitude $GK = 114 = 2.05690$
so is the secant of the course $= 29^\circ 19' = 10.05952$
to the distance $GL = 130.8 = 2.11642$
consequently the ship has sailed on a SSE $\frac{1}{2}$ 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 $34^\circ 24'$ 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.
As the distance $AD = 124 = 2.09342$
is to radius $= 10.00000$
so is the departure $AB = 86 = 1.93450$
to the fine of the course $D = 43^\circ 54' = 9.84108$
so the ship's course is north $33^\circ 54'$ west, or NWbN $\frac{1}{4}$ west nearly.
Then for the difference of latitude, we have (by rectangular trigonometry) the following analogy, viz.
As radius $= 10.00000$
is to the distance $AD = 124 = 2.09342$
so is the co-fine of the course $43^\circ 54' = 9.85766$
to the diff. of latitude $BD = 89.35 = 1.95108$
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 $= 34^\circ 24'$
add the difference of latitude $= 1^\circ 29'$
the sum is $= 35^\circ 53'$
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 $24^\circ 30'$ south, sails SEbS, 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 Cafe 2. we have the following proportion for finding the distance, (No. 9.) viz.
As the fine of the course $G = 33^\circ 45' = 9.74474$
is to the departure $HM = 96 = 1.98227$
so is radius $= 10.00000$
to the distance $GM = 172.8 = 2.23753$
Then, for the difference of latitude, we have (by rectangular trigonometry) the following analogy, viz.
As the tangent of course $= 33^\circ 45' = 9.82489$
is to the departure $HM = 96 = 1.98227$
so is radius $= 10.00000$
to the difference of latitude $GH = 143.7 = 2.15738$
equal to $2^\circ 24'$ nearly. Consequently, since the latitude the ship sailed from was south, and she sailing still towards the south,
To the latitude sailed from $= 24^\circ 30'$
add the difference of latitude $= 2^\circ 24'$
and the sum $= 26^\circ 54'$
is the latitude she is come to south.
6. When a ship stems 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 let down all the different courses and distances stemmed 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 Cafe 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 Cafe 4. of Plain Sailing), as also the course and distance to the intended port.
**Example.** Suppose a ship at sea, in the latitude of $48^\circ 24'$ north at noon any day, is bound to a port in the latitude of $42^\circ 40'$ north, whose departure from the ship is $144$ miles east; consequently the direct course and distance of the ship is SSE $\frac{1}{2}$ 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. SEbS $56$ miles, SSE $\cdot 64$ miles, NWbW $48$ miles, SbW $\frac{1}{2}$ west $54$ miles, and SEbS $\frac{1}{2}$ 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 Cafe 1.) find the difference of latitude and departure for each course. Thus,
1 Course SEbS distance $56$ miles.
For departure.
As radius $= 10.00000$
is to the distance $= 56 = 1.74819$ ### NAVIGATION
**The Traverse Table**
| Course | Distances | Diff. of Lat. | Departure | |--------|-----------|---------------|-----------| | | | N | S | E | W | | SEbS | 56 | 46.57 | 31.11 | | | | SSE | 64 | 59.13 | 24.5 | | | | NWbW | 48 | 26.67 | | | | | SbW½ West | 54 | 51.67 | | | | | SEbS½ East | 54 | 57.21 | 46.94 | | |
**Diff. of Lat.** 187° 91' 46.97' Dep.
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
so is the departure = 46.97' 1.67182
to the tangent of the course = 14° 03' 9.39789
which is SbE½ east nearly. Then for the distance, it will be,
As radius = 10.00000
is to the difference of latitude = 187° 91' 2.27393
so is the secant of the course = 14° 03' 10.01319
to the distance = 193° 2.28712
consequently the ship has made good the last 24 hours, on a SAE ¼ 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 the is bound for = 43° 40' N
and there remains = 1° 36'
or 96 miles, the difference of latitude or fouthing 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 the ship 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° 1.68227
is to radius = 10.00000
so 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° 1.68227
so is the secant of the course = 45° 19' 10.15293
to the distance = 136° 2.13620
whence the true bearing and distance of the intended port is SE, 136° 5 miles.
---
**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 214.48, the difference between these, viz. 187° 91', will be the fouthing 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
Vol. III. No 84. Sect. 3. Of Parallel Sailing.
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 cosine of CA its distance from the equator; and CD will be to AB, as the sine of CP, or cosine 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 cosine 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 cosine of the latitude of that parallel to radius.
Cor. III. Hence, as radius is to the cosine 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 cosine 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 cosine of any one parallel is to the cosine 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 $54^\circ 20'$ north, sails directly west on that parallel till she has differed her longitude $12^\circ 45'$; 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 $54^\circ 20'$, or the distance sailed, it will be, by Cor. 3. of the last article,
As radius $100000$ is to the cosine of the lat. $54^\circ 20'$ $976572$ so are the minutes of diff. long. $765$ $288306$ to the distance on the parallel $446.1$ $264938$
Example II. A degree on the equator being 60 minutes or nautical miles; required the length of a degree on the parallel of $51^\circ 32'$.
By Cor. 3. of the last article, it will be
As radius $100000$ is to the cosine of the latitude $51^\circ 32'$ $979383$ so are the minutes in 1 degree on the equa. $60$ $177815$ to $37.32$ $157198$ the miles answering to a degree on the parallel of $51^\circ 32'$
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.
| D.L. | Miles | D.L. | Miles | D.L. | Miles | D.L. | Miles | |------|-------|------|-------|------|-------|------|-------| | 1 | 59.99 | 19 | 56.73 | 37 | 47.92 | 55 | 34.41 | | 2 | 59.97 | 20 | 56.38 | 38 | 47.28 | 56 | 33.53 | | 3 | 59.92 | 21 | 56.01 | 39 | 46.62 | 57 | 32.68 | | 4 | 59.86 | 22 | 55.63 | 40 | 45.95 | 58 | 31.79 | | 5 | 59.77 | 23 | 55.23 | 41 | 45.28 | 59 | 30.90 | | 6 | 59.67 | 24 | 54.81 | 42 | 44.95 | 60 | 30.00 | | 7 | 59.56 | 25 | 54.38 | 43 | 43.88 | 61 | 29.09 | | 8 | 59.42 | 26 | 53.93 | 44 | 43.16 | 62 | 28.17 | | 9 | 59.26 | 27 | 53.46 | 45 | 42.43 | 63 | 27.24 | | 10 | 59.08 | 28 | 52.97 | 46 | 51.68 | 64 | 26.30 | | 11 | 58.89 | 29 | 52.47 | 47 | 40.92 | 65 | 25.36 | | 12 | 58.68 | 30 | 51.96 | 48 | 40.15 | 66 | 24.41 | | 13 | 58.46 | 31 | 51.43 | 49 | 39.36 | 67 | 23.45 | | 14 | 58.22 | 32 | 50.88 | 50 | 38.57 | 68 | 22.48 | | 15 | 57.95 | 33 | 50.32 | 51 | 37.76 | 69 | 21.50 | | 16 | 57.67 | 34 | 49.74 | 52 | 36.94 | 70 | 20.52 | | 17 | 57.37 | 35 | 49.15 | 53 | 36.11 | 71 | 19.54 | | 18 | 57.06 | 36 | 48.54 | 54 | 35.26 | 72 | 18.55 |
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 $56^\circ 44'$.
First, The next less parallel of latitude in the table than that proposed, is that of $56^\circ$, 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 $57^\circ$, 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 $56^\circ$, and the proposed parallel of $56^\circ 44'$, 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 $56^\circ$ and a degree on the parallel of $56^\circ 44'$; which therefore taken from 33.55, the miles answering to a degree on the parallel of $56^\circ$, leaves 32.912, the miles answering to a degree on the parallel of $56^\circ 44'$, as was required.
Case II. The distance sailed 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 $55^\circ 36'$ north sails 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. $55^\circ 36'$ = 9.75202 is to radius = 10.00000 so is the distance sailed = 685.6 = 2.83607 to min. of diff. of long. = 1213 = 3.08405 which reduced into degrees, by dividing by 60, makes $20^\circ 13'$, 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 $55^\circ 36'$ is equal to 33.80 miles; by this I divide the proposed number of miles 685.6 and the quotient is 20.13 degrees, i.e. $20^\circ 13'$, 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 sails on a certain parallel directly west 624 miles, and then has differed her longitude $18^\circ 46'$ or 1126 miles: Required the latitude of the parallel she sailed 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 sailed = 624 = 2.79518 so is radius = 10.00000 to the co-sine of the lat. $56^\circ 21'$ = 9.74364 consequently the latitude of the ship or parallel she sailed upon was $56^\circ 21'$.
From what has been said, may be solved the following problems.
Prob. I. Suppose two ships in the latitude of $46^\circ 30'$ north, distant at first 654 miles, sail both directly north 256 miles, and consequently are come to the latitude of $50^\circ 46'$ north: Required their distance on that parallel.
By Cor. 6. Art. 1. of this Section, it will be,
As the co-sine of $46^\circ 30'$ = 9.83781 is to the co-sine of $50^\circ 46'$ = 9.80105 so is 654 = 2.81558 to 601 = 2.77882 the distance between the ships when on the parallel of $50^\circ 46'$.
Prob. II. Suppose two ships in the latitude of $45^\circ 48'$ north, distant 846 miles, sail directly north till the distance between them is 624 miles: Required the latitude come to, and the distance sailed.
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 $45^\circ 48'$ = 9.84334 to the co-sine $59^\circ 04'$ = 9.71115 the latitude of the parallel the ships are come to.
Consequently to find their distance sailed,
From the latitude come to $59^\circ 04'$ subtract the latitude sailed from, $45^\circ 48'$ and there remains $13^\circ 16'$ equal to 796 miles, the difference of latitude or distance sailed.
Sect. 4. Of Middle-Latitude Sailing.
When two places lie both on the same parallel, we showed in the last section, how, from the difference of longitude given, to find the miles of easting or westing between them, &c. 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, &c. 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° 00' north, and longitude of 5° 14' west, and St Vincent in the latitude of 17° 10' N. and longitude of 24° 20' W.
First, To the latitude of the Lizard — 50°, 00' N. add the latitude of St Vincent — 17°, 10'
The sum is — 67°, 10' Half the sum or latitude of the middle parallel is — 33°, 35' N. Also the difference of latitude is — 33°, 50' equal to 1970 miles of southing. Again, From the longitude of St Vincent — 24°, 20' W. take the longitude of the Lizard — 05°, 14'
there remains — 16°, 06' equal to 1146 min. of diff. of long. west.
Then for the miles of westing, or departure, it will be, by Case 1. of Parallel Sailing,
As radius — 10.00000 is to the co-sine of the middle parallel — 33° 35' — 9.92069 so is min. diff. of long. — 1146 — 3.05918 to the miles of westing — 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°, 51' — 9.68549 which, because it is between south and west, it will be SSW ½ 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°, 51' — 10.04579 to the distance — 2189 — 3.34026
whence the direct course and distance from the Lizard to St Vincent is SSW ¼ 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° 00' north, sails south 50° 06' west, 150 miles: Required the latitude she 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°, 06' — 9.80716 to the diff. of latitude — 96.22 — 1.08325 equal to 1°, 36'. And since the ship is sailing towards the equator; therefore,
From the latitude she was in — 50°, 00' take the diff. of latitude — 1°, 36' and there remains — 48°, 24' the latitude she has come to north. Consequently the latitude of the middle parallel will be 49° 12'.
Then for departure or westing it will be, by the same case,
As radius — 10.00000 is to the distance — 150 — 2.17609 so is the fine of the course — 50°, 06' — 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° 12' — 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° 56', which is the difference of longitude the ship has made westerly.
Case III. Course and difference of latitude given; to find the distance failed, and difference of longitude.
Example. Suppose a ship in the latitude of 53° 34' north, sails SE 85° till by observation she is found to be in the latitude of 51° 12', and consequently has differed her latitude 2° 22' or 142 miles: Required the distance failed, 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° 45' — 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° 45' — 10.08015 to the distance — 170.8 — 2.22244
Then, since the latitude sailed from was 53° 34' north, and the latitude came to 51° 12' north; therefore the middle parallel will be 52° 23'; 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° 23' — 9.78560 is to the departure — 94.88 — 1.97718 so is radius — 10.00000 to min. diff. of longitude — 155.5 — 2.19158 equal to 2° 35' the difference of longitude easterly.
Case IV. Difference of latitude and distance failed, given; to find the course and difference of longitude.
Example. Suppose a ship in the latitude of 43° 26' north, sails between south and east, 246 miles, and then is found by observation to be in the latitude of 41° 06' 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° 19' — 9.75519 which, because the ship sails between south and east, will be south 55° 19' east, or SSE nearly.
Then for departure, it will be, by the same case,
As radius — 10.00000 is to the distance — 246 — 2.39094 As the co-fine of the mid. par. \(45^\circ\), \(90^\circ\) \(9.84949\) is to the departure \(146\) \(2.16137\) fo is radius \(10.00000\) to min. of diff. of longitude \(205\) \(2.31188\) equal to \(3^\circ 25'\), 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 \(33^\circ 40'\) north, fails between south and east \(165\) miles, and has then made of easterly \(112.5\) 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 \(102.5\) \(2.05115\) fo is the departure \(9.83367\) to the fine of the course \(42^\circ, 59'\) which, because the ship fails between south and east, will be south \(42^\circ 59'\) east or SE \(E \frac{1}{2}\) east nearly.
And for the difference of latitude, it will be, by the same **Case**, As radius \(10.00000\) is to the distance \(165\) \(2.21748\) fo is the co-fine of the course \(42^\circ 59'\) \(9.86436\) to the difference of lat. \(120.7\) \(2.08184\) equal to \(2^\circ 00'\); consequently the latitude come to will be \(31^\circ 40'\) north, and the latitude of the middle parallel will be \(32^\circ 40'\). Hence, to find the difference of longitude, it will be, by **Case 2.** of Parallel Sailing,
As the co-fine of the mid. par. \(32^\circ, 40'\) \(9.92522\) is to the departure \(112.5\) \(2.05115\) fo is radius \(10.00000\) to min. of diff. of long. \(133.6\) \(2.12593\) equal to \(2^\circ 13'\) nearly, the difference of longitude easterly.
**Case VIII.** Difference of longitude and departure given, to find difference of latitude, course, and distance failed.
**Example.** Suppose a ship in the latitude of \(50^\circ 46'\) north, fails between south and west, till her difference of longitude is \(3^\circ 12'\), and is then found to have departed from her former meridian \(126\) miles; required the difference of latitude, course, and distance failed.
First, For the latitude she has come to, it will be, by **Case 2.** of Parallel Sailing,
As min. of diff. of long. \(192\) \(2.28330\) is to departure \(126\) \(2.10027\) fo is radius \(10.00000\) to the co-fine of the mid. par. \(48^\circ, 50'\) \(0.81767\)
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 \(48^\circ 59'\), viz. \(97^\circ 58'\) taking \(50^\circ 46'\) the latitude sailed from, there remains \(47^\circ 12'\) the latitude come to; consequently the difference of latitude is \(3^\circ 34'\), or \(214\) minutes.
Then for the course, it will be, by **Case 4.** of Plain Sailing,
As diff. of lat. \(214\) \(2.33041\)
is to radius — — — 10.00000 so is the departure — — — 2.10037 to the tang. of the course 30° 29' — — — 9.76996 which, because it is between south and west, will be south 30° 29' west, or SSW ¼ west nearly.
And for the distance, it will be, by the same Case, As radius — — — 10.00000 is to the diff. of lat. — — — 2.33041 so is the secant of the course 30° 29' — — — 10.06461 to the distance — — — 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° 25' north, sails upon the following courses, viz. SWbS 63 miles, SSW ¼ west 45 miles, SbE 54 miles, and SWbW 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.
| Courses | Distances | Diff. of Lat. | Diff. of Longit. | |---------|-----------|--------------|----------------| | SWbS | 63 | 52.4 | 47.85 | | SSW ¼ W | 45 | 39.7 | 28.62 | | SbE | 54 | 53.0 | 13.75 | | SWbW | 74 | 41.1 | 81.68 |
Diff. of Lat. 186.2 Diff. of Long. 143.80
Hence it is plain the ship has differed her latitude 186.2 minutes, or 3° 6', and so has come to the latitude of 40° 10' north, and has made a difference of longitude 143.8 minutes, or 2° 23' 48" 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-fines 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 easting 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 fine, and BF or CG the cosine 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 cosine 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 equa- tor 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 cosine 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 cosine of any pa- rallel 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 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 degrees, since a minute of a meridian is everywhere 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 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 \(\frac{3}{8}\) 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 everywhere equal to the secant of the arch 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, 20, 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 662.3 (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 figured... figures of countries are distorted, especially near the poles: but since the degrees of latitude are everywhere 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 failing by the chart, was called Mercator's failing.
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 failing, 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° 20' north. I look in the tables for the meridional parts answering to 17° 20', 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° 00' north, and Antigua in the latitude of 17° 30' north.
From the meridional parts of — 50°, 00' — 3474.5 subtract the meridional parts of 17°, 30' — 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° 30' north, and Lima in Peru in the latitude of 12° 30' South.
To the merid. parts answering to 17°, 30' — 1066.7 add these answering to 12°, 30' — 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° 00' north, and Port-Royal in Jamaica in the latitude of 17° 40'; differing in longitude 70° 46', Port-Royal lying so far to the westward of the Lizard.
Preparation.
From the latitude of the Lizard — 50°, 00' subtract the latitude of Port-Royal — 17°, 40' and there remains — — — 32°, 20' equal to 1940 minutes, the proper difference of latitude.
Then from the meridional parts of 50°, 00' — 3474.5 subtract those of 17°, 40' — 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,) \[ AC : CB :: R : T, \text{BAC}, i.e. \] As the meridional diff. of lat. 2397.3 — 337970 is to the difference of long. — 4246.0 — 3.62798 so is radius — 10.00000 to the tang. of the direct course 60° 33' — 10.34828 which, because Port Royal is southward of the Lizard, and the difference of longitude westerly, will be south 60° 33' west, or SWbW ½ west nearly.
Then for the distance AD, it will be, (by rectangular trigonometry)
\[ R : AE :: \text{Sec. A} : AD, i.e. \]
As the radius — 10.00000 is to the proper diff. of lat. — 1940 — 3.28780 so is the secant of the course — 60° 33' — 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° 33', 3945.6 miles.
Case III. Course and distance failed given, to find dif- ference of latitude and difference of longitude.
Example. Suppose a ship from the Lizard in the lati- tude of 50° 00' north, sails south 35° 40' west 156 miles. Required the latitude come to, and how much she has al- tered her longitude.
Geometrically. 1. Draw the line BK (No. 15.) repre- senting the meridian of the Lizard at B; from B draw the line BM, making with BK an angle equal to 35° 40', 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)
\[ R : MB :: S, BMK : BK, \]
i.e. As radius — 10.00000 is to the distance — 156 — 2.19312 so is the cosine of the course — 35° 40' — 9.90978 to the proper difference of lat. — 127 — 2.10290 equal to 2° 07'; and since the ship is sailing from a north latitude towards the south, therefore the latitude come to will be 47° 53' north. Hence the meridional difference of latitude will be 19° 34'.
2. Produce BK to D, till BD be equal to 193 4; through D draw DL parallel to MK, meeting DM pro- duced in L; then DL will be the difference of longitude: to find which by calculation, it will be, (by rectangular trigonometry)
\[ R : BD :: T, LBD : DL, \]
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° 40' — 9.85594 to minutes of diff. of long. — 138.8 — 2.14240 equal to 2° 18' 48" the difference of longitude the ship has made westerly.
Case IV. Given course and both latitudes, viz. the latitude sailed from, and the latitude come to; to find the distance sailed, and the difference of longitude.
Example. Suppose a ship in the latitude of 54° 20' north, sails south 33° 45' east, until by observation she is found to be in the latitude of 51° 45' north; required the distance sailed, and the difference of longitude.
Geometrically. Draw AB (No. 16.) to repre- sent 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° 45', which will meet the two former lines in the points C and K; so the case is con- structed, and AC and GK may be found from the line of e- qual parts: To find which,
By Calculation:
First, For the difference of longitude, it will be, (by rectangular trigonometry.)
\[ R : AG :: T, GAK : GK, \]
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° 45' — 9.82489 to min. of diff. of longitude — 172.3 — 2.22634 equal to 2° 52' 18", 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 rect- angular trigonometry)
\[ R : AB :: \text{Sec. A} : AC, \]
i.e. As radius — 10.00000 is to the proper diff. of lat. — 155 — 2.19033 so is the secant of the course — 33° 45' — 10.08015 to the direct distance — 186.4 — 2.27048 consequently the ship has sailed south 33° 45' east 186.4 miles, and has differed her longitude 2° 52' 18" 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° 26' north, sails between north and east 105 miles, and then by observation she is found to be in the latitude of 48° 6' 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 o- ther cross the line BD in D. Produce AB, till AC be e- qual 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 mea- sured 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)
\[ AB : R :: AD : \text{Sec. A}, \]
i.e. 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° 52' — 10.08591 which, because the ship is sailing between north and east, will be north 34° 52' east, or NEbN 1° 9' easterly.
Then for the difference of longitude, it will be, (by rectangular trigonometry) **NAVIGATION**
\[ R : AC :: T, A : CK. \]
i.e. As radius \( \frac{1}{10} \) is to the merid. diff. of lat. \( = 233.6 \), so is the tang. of the course \( = 34^\circ 52' \), to min. of diff. of longitude \( = 162.8 \), equal to \( 2^\circ 42' 48'' \), 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 \( 48^\circ 50' \) north, sails south \( 34^\circ 40' \) west, till her difference of longitude is \( 2^\circ 44' \); 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 \( 34^\circ 40' \), the angle of the course; then draw FC parallel to AE, at the distance of 164 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)
\[ T, A : R :: CE : AE, \]
i.e. As the tang. of the course \( = 34^\circ 40' \), is to the radius \( = 9.83984 \), so is min. of diff. longitude \( = 164 \), to the enlarged diff. of latitude \( = 237.2 \), and because the ship is sailing from a north latitude southerly, therefore,
From the merid. parts of \( = 48^\circ 50' \), take the merid. difference of latitude \( = 237.2 \), and there remains \( = 3129.7 \), the meridional parts of the latitude come to, viz. \( 46^\circ 09' \).
Hence for the proper difference of latitude,
From the latitude sailed from \( = 48^\circ 50' \) N take the latitude come to \( = 46^\circ 09' \) N and there remains \( = 2^\circ 41' \), 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,
\[ R : AD :: Sec. A : AB. \]
i.e. As radius \( = 10 \), is to the proper diff. of latitude \( = 161 \), so is the secant of the course \( = 34^\circ 40' \), to the direct distance \( = 195.8 \).
**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 \( 54^\circ 36' \) north, south \( 42^\circ 33' \) 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 \( 42^\circ 33' \). 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)
\[ S, A : BD :: R : AD. \]
i.e. As the sine of the course \( = 42^\circ 33' \), is to the departure \( = 116 \), so is radius \( = 10 \), to the direct distance \( = 171.5 \).
Then for the proper difference of latitude, it will be, by rectangular trigonometry,
\[ T, A : BD :: R : AB. \]
i.e. As the tang. of the course \( = 42^\circ 33' \), is to the departure \( = 116 \), so is radius \( = 10 \), to the proper difference of latitude \( = 126.4 \), equal to \( 2^\circ 6' \); consequently the ship has come to the latitude of \( 52^\circ 30' \) 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)
\[ R : AE :: T, A : EC. \]
i.e. As radius \( = 10 \), is to the merid. diff. of latitude \( = 212.2 \), so is the tang. of the course \( = 42^\circ 33' \), to the min. of diff. of longitude \( = 194.8 \), equal to \( 3^\circ 14' 48'' \), 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 \( = 126.4 \), is to the departure \( = 116 \), so is the enlarged diff. of latitude \( = 212.2 \), to min. diff. of longitude \( = 194.8 \).
**CASE VIII.** Both latitudes and departure given, to find course, distance, and difference of longitude.
**EXAMPLE.** Suppose a ship from the latitude of \( 46^\circ 20' \) 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 \( 43^\circ 25' \) 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
**BY CALCULATION:**
First, For DC the difference of longitude, it will be, AC : CB :: AK : KD.
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.)
AC : BC :: R : T, A.
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 SWB 6° 30' westerly.
Lastly, For the distance AB, it will be, (by rectangular trigonometry.)
S, A : BC :: R : AB.
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 failed, 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. (it, 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.)
BE : R :: DE : S, B.
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.)
R : BE :: Co-S, B : BD.
i.e. As radius — 10.00000 is to the distance — 133 — 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' 19". Hence the meridional difference of latitude will be 117.7.
2dly, 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,
BD : DE :: BA : AC.
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 50° 00' north, is bound to the Madera in the latitude of 32°, 20' north, the difference of longitude between them being 110° 40', the west end of the Madera lying so much to the westward of the Lizard, and consequently the direct course and distance (by Cafe, 2. of this Sect.) 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 ½ west 36 miles, SWB 56 miles, and SSE 28 miles. Required the latitude the ship is in, her bearing and distance from the Lizard, and her direct course and distance from the Madera, at the end of these courses.
The geometrical construction of this traverse is performed by laying down the two ports according to construction of Cafe 2. of this Sect., and the several courses and distances according to Cafe 3., by which we have the following solution by calculation.
1. Course SSW, distance 44 miles.
For difference of latitude:
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,
For difference of longitude,
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
2. Course SSW ½ west, distance 36 miles.
For difference of latitude:
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,
For difference of longitude,
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 — 16.79 — 1.20925
3. Course SWB, distance 56 miles.
For difference of latitude:
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°39', and therefore the enlarged difference of latitude will be 69.2.
Then,
For difference of longitude:
As radius —— 10.00000 is to the enlarged diff. of lat. 69.2 1.84011 so is the tang. of the course 33°45' 9.82489 to the difference of longitude 46.24 1.66500
4. Course SbE, distance 28 miles.
For difference of latitude:
As radius —— 10.00000 is to the distance 28 1.44716 so is the co-line of the course 11°15' 9.99157 to the different of latitude 27.46 1.43873 consequently the latitude come to will be 47°31'; and hence the meridional difference of latitude will be 43.2.
Then,
For difference of longitude:
As radius —— 10.00000 is to the enlarged diff. of lat. 43.2 1.63548 so is the tang. of the course 11°15' 9.29866 to the diff. of longitude 8.59 0.93414
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.
| Courses | Distances | Diff. of Lat. | Diff. of Longit. | |---------|-----------|--------------|----------------| | | | N | S | E | W | | SSW | 44 | 40.65 | 25.6 | | SbWbW | 36 | 34.46 | 16.19 | | SWbS | 56 | 46.56 | 49.24 | | SbE | 28 | 27.46 | 8.59 |
Diff. of Lat. 149.13 8.59 88.03 8.59
Diff. of long. 79.44
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:
As the merid. diff. of latitude 1196.4 3.07788 is to radius 10.00000 so is the difference of longitude 620.56 2.79278 to the tang. of the course 27°25' 9.71493
For the direct distance:
As radius —— 10.00000 is to the proper diff. of latitude 911 2.95952 so is the secant of the course 27°25' 10.05174 to the direct distance 1027 3.01126
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.
We shewed, at Art. 5. of this Section, how to construct a Mercator'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° 00' north, and the place to be laid down St Katharines on the east coast of America, differing in longitude from the Lizard 42° 36', 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° 00' north, its meridian. Set off AE from E upon the equator EQ, 42° 36', 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° 00' north. Take from your scale the chord of 60° 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 fines the co-sine of the latitude, or sine of 72° 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 Parallel Sailing: for it has been there demonstrated, that radius is to the cosine 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°.
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.
**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 bc parallel to de; and taking ac in your compasses, apply it to the graduated equator, and it will shew 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 Sect., 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° 00' north, sails SSW till she has differed her latitude 36° 40'. Required her place upon the chart.
Count from the Lizard at L. on the graduated meridian downwards (because the course is southerly) 36° 40' to g; 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, sailed, 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° 00' north, sails north 37° 20', 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 AC, making with AB an angle of 37° 20'; and upon it set off 191 from A to C; through C draw the parallel CB; 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 BC, which will cut AC produced in the point E, the ship's place required.
**Prob. VII.** Both latitudes and distance sailed, given; to find the ship's place upon the chart.
**Example.** Suppose a ship sails from A in the latitude of 20° 00' north, between north and east 191 miles, and is then in the latitude of 45° 00' north: Required the ship's place upon the chart.
Draw DE the parallel of 45°, 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 BC 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° 00' north, sails SWbW; till her difference of longitude is 42° 36': 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° 36'; and through F draw the meridian EG; then from L draw the SWbW 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° 00' north, sails north 37° 20' 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 KL. Draw the rhomb line AC, which will meet KL in some point C; then through C draw the parallel CB, and AB will be the proper difference of latitude, and BC the departure. Take AB 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 BE, which will meet AC 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° 00' 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 AM equal to 116, and through M draw the meridian KL. Take the given distance, 191 in your compasses; setting one foot of them in A, with the other cross KL in C. Join AC, and through C draw the parallel CB; so CB will be the departure, and AB 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 sailed 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° 00' north, sails between north and east, till she be in the latitude of 45° 00' 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 BC, and make BC equal to 116 the departure, and join AC. Count from the parallel of A on the graduated meridian upwards to B 25 degrees, and through B draw the parallel BE, which will meet AC 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 NWbW 20 miles, and I observed the same cape to bear from me NEbE. 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 NWbW line, CA 20 miles, and A will be the place of the ship in her second station. 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,
By Calculation;
In the triangle ABC are given AC, equal to 20 miles; the angle ACB, equal to 78° 45', the distance between the NNE and NWNW lines; also the angle ABC, equal to BCD, equal to 33° 45', the distance between the NNE and NBEB lines; and consequently the angle A, equal to 67° 30'.
Hence for CB, the distance of the cape from the ship in her first station, it will be (by oblique trigonometry)
\[ S. \text{ABC} : AC :: S. \text{BAC} : CB, \]
i.e. As the sine of the angle B \( 33^\circ 45' \) \( 9.74473 \) is to the distance run AC \( 20 \) \( 1.30103 \) so is the sine of BAC \( 67^\circ 30' \) \( 9.96562 \) 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,
\[ S. \text{ABC} : AC :: S. \text{ACB} : AB. \]
i.e. As the sine of B \( 33^\circ 45' \) \( 9.74474 \) is to AC \( 20 \) \( 1.30103 \) so is the sine of C \( 78^\circ 45' \) \( 9.99157 \) 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 compasses 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
By Calculation;
In the triangle ACB are given, AC 17, CB 20, and the angle ACB equal to 101° 15', 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 \( \frac{1}{2} \) the sum \( 39^\circ 22' \) \( 9.91417 \) of the angles A and B \( 3^\circ 49' \) \( 8.82309 \)
consequently the angle A will be 43° 11', and the angle B 35° 34'; also the bearing of B from A will be SSW 1° 49' westerly, and the bearing of A from B will be NESE 1° 49' easterly.
Then for the distance AB, it will be, by oblique-angular trigonometry,
\[ S. \text{A} : CB :: S. \text{C} : AB. \]
i.e. As the sine of A \( 43^\circ 11' \) \( 9.83527 \) is to CB \( 20 \) \( 1.30103 \) so is the sine of C \( 101^\circ 15' \) \( 9.99157 \) 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 NWbN, and the second NNE; then standing away ESN \( \frac{1}{2} \) northerly 20 miles; I found the first bore from me WNW \( \frac{1}{2} \) westerly, and the second NBbW \( \frac{1}{2} \) westerly. Required the bearing and distance of these two headlands.
Geometrically. Having drawn the compasses NWSE (No. 24.) let C represent the first place of the ship; from which draw the NWbN line CB, and the NNE line CD, also the ESN \( \frac{1}{2} \) N line CA, which make equal to 20. From A draw AB parallel to the WNW \( \frac{1}{2} \) W line, and AD parallel to the NNE \( \frac{1}{2} \) 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
By Calculation;
1. In the triangle ABC are given the angle BCA, equal to 104° 04', the distance between the NWbN line, and the ENE \( \frac{1}{2} \) E line; the angle BAC, equal to 36° 34', the distance between the WSW \( \frac{1}{2} \) W line and the WNW \( \frac{1}{2} \) W line; the angle ABC equal to 39° 22', the distance between the ESE \( \frac{1}{2} \) E line; and the SWbS line, also the side CA equal to 20 miles; whence for CB, it will be (by oblique trigonometry)
As the sine of CBA \( 39^\circ 22' \) \( 9.80228 \) is to AC \( 20 \) \( 1.30103 \) so is the sine of CAB \( 36^\circ 34' \) \( 9.77507 \) 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° 49', the distance between the ENE \( \frac{1}{2} \) E line, and the NNE line; the angle CAD, equal to 92° 49', the distance between the WSW \( \frac{1}{2} \) W line; and the NWbW \( \frac{1}{2} \) W line, the angle CDA equal to 39° 22', the distance between the SSW line and the SbE \( \frac{1}{2} \) E line; also the leg CA equal to 20.
Hence for CD, it will be (by oblique trigonometry)
As the sine CAD \( 39^\circ 22' \) \( 9.80228 \) is to AC \( 20 \) \( 1.30103 \) so is the sine of CAD \( 92^\circ 34' \) \( 9.99960 \) 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° 15', the distance between the NWbN 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 the tangent of \( \frac{1}{2} \) sum \( 61^\circ 51' \) \( 10.27189 \)
of the unknown angles \( 25^\circ 18' \) \( 9.67459 \)
consequently the angle CBD is 87° 10', and the angle CDB 36° 35'. Hence the bearing of the first headland from the second will be S 59° 8', W or SWbW \( \frac{1}{2} \) W nearly; and for the distance between them, it will be,
As the sine of BDC \( 36^\circ 35' \) \( 9.77524 \) is to BC \( 18.79 \) \( 1.27382 \) So is the fine of BCD — 56°, 15' — 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 ½ 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 ½ points distance from it, and through A draw AB parallel to CD meeting CB in B; then the ENE ¼ E line CB, will be the course of the ship upon the starboard tack, and CB its distance on that tack; also the ESE ¼ 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°, 53', the distance between the east and ENE ¼ E line; the angle CBA, equal to 146° 14', the distance between the ENE ¼ E and the WNW ¼ W lines; the angle BAC equal to 16° 53', the distance between the east and ESE ¼ 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°, 14' — 9.74493 is to AC — 86 — 1.93450 so is the fine of A — 16, 53 — 9.46303 to CB — 44.94 — 1.05260 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.
**Sec. 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 thumb 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 SEbS at the rate of 6 miles an hour, in a current that sets SEbS 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° 07' E, or SSE 2° 37', easterly.
Then for the true distance AC, it will be, (by rectangular trigonometry,) As the fine of the course A — 25°, 07' 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 WbN at the rate of 2 miles an hour: Required the ship's true course and distance failed in that time. Geometrically. Having drawn the compass NESW (No. 27.) let C represent the place the ship sailed from; draw the SE line CA, which make equal to 120°; then will A be the place the ship sailed at.
From A draw AB parallel to the WNW 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° 45', 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 73° 07' 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° 23'. Hence the true course is S 30° 27' E, or SSE 2° 07' easterly.
Then for the true distance CB, it will be, (by oblique trigonometry)
As the fine of B 131° 52' = 9.87198 is to AC 120° = 2.07918 so is the fine of A 33° 45' = 9.74474 to the true distance CB 89° 53' = 9.51914
Example III. Suppose a ship coming out from sea in the night, has sight of Scilly light, bearing NEN 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 4S distance 17 leagues.
Geometrically. Having drawn the compass NESW (No. 28.) let A represent the ship's place at sea, and draw the NESE 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° 07', the distance between the NEAN and WNW 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 30° 56' 9.77763 to the tang. of half their diff. — 20° 21' 9.56935
Consequently the angle SAL, will be 51° 17', and so the direct bearing of the Lizard from the ship will be N 85° 02' E, or ENE 6° 17' E; and for the distance AL, it will be (by oblique trigonometry.)
As the fine of SAL 51° 17' = 9.89223 is to SL 51° = 1.70757 so is the fine of ASL 118° 07' = 9.94546 to AL 57° 65' = 1.76080 the distance between the ship and the Lizard.
Again, in the triangle DLC, are given the angle L equal to 17° 32', the distance between the ENE and N 85° 02' 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.)
As the ship's run in 1 hour DC 5° = 0.69897 is to the fine of L 17° 32' = 9.47894 so is the current's drift LC 2° = 0.30103 to the fine of D 60° 55' = 9.08100 consequently since by construction the angle LAB is equal to the angle LDC, the course the ship must steer is S 88° 03' E.
Then for the distance AB, it will be (by oblique trigonometry.)
As the fine of B 135° 33' = 9.61689 is to AL 57° 65' = 1.76080 so is the fine of L 17° 32' = 9.47894 to AB 41° 96' = 1.62285 consequently, since the ship is sailing at the rate of 5 miles an hour, it follows, that in failing 8h 24m S 88° 03' E, she will arrive at the Lizard.
Example IV. A ship from a certain headland in the latitude of 34° 00' north, sails SESE 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° 52' 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 SESE 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 SESE 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° 30'. 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 67° 30' = 10.4176 to the distance run AC 20° 9' = 1.32025
Again, in the triangle ABC, are given AB equal to 12 miles, AC equal to 20.9; and the angle BAC equal to 33° 45', the distance between the SE8S 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 = 73° 07' — 10.51806 to tang. of ¼ their diff. — 41° 45' ½ — 9.95025
Consequently the angle B is 114° 51', and so the setting of the current will be N 81° 06' E or EB 2° 21' E.
Then for BC the current's drift in 3 hours, it will be,
As the sine of B = 114° 51' — 9.92700 is to the distance run AC = 20.9 — 1.32025
So is the sine of A = 33° 45' — 9.74474 to BC — 12.8 — 1.10719
The current's drift in 3 hours, and consequently the current sets EB N 2° 21' 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 astronomic 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° 50' north, on the 23rd day of April 1731.
First, I find (from the tables of the sun's declination) that the sun's declination the 23rd of April is 15° 54' north; then for the true amplitude, it will be, by the former analogy.
As the co-sine of the lat. 41° 50' — 9.87221 is to radius — 10.00000
So is the sine of the decl. 15° 54' — 9.43760 to the sine of the ampl. 21° 35' — 9.56548
Which is north, because the declination is north at that time; and consequently, in the latitude of 41° 50' north, the sun rises on the 23rd of April 21° 35' 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 cosine 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° 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° will give the sun's distance from the nearest pole to the place of observation.
Example. In the latitude of 51° 32' north, the sun having 19° 39' north declination, his altitude was found by observation to be 38° 18': Required the azimuth.
By the first of the foregoing analogies, it will be
As the tangent of ¼ the complement of the latitude 19° 14' 9.54269 is to the tangent of ¼ the sum of the distance of the sun from the pole and complement of the altitude 61° 01' 10.25655
So is the tangent of half their difference 9° 19' 7.21499 to the tang. of a 4th arch 40° 20' 9.92885
Which fourth arch 40° 20', added to 19° 14', half the complement of the latitude, gives a fifth arch 59° 34'; and this fifth arch lessened by 38° 28', the complement of the latitude, gives the sixth arch 21° 06'; then for the the azimuth, it will be, by the second of the preceding analogies,
As radius \( \frac{1}{10} \) is to the tang. of the altitude \( 38^\circ 18' 9.89749 \) so is the tang. of the sixth arch \( 21^\circ 06' \) \( = 9.58644 \) to the co-tang. of the azimuth \( 72^\circ 15' \) \( = 9.48393 \)
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 \( S 72^\circ 15' E \), or ESE \( 4^\circ 45' E \); but if the altitude was taken in the afternoon, the azimuth will be \( S 72^\circ 15' W \), or WSW \( 4^\circ 45' \) 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 NWS 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 NBE 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 \( E 14^\circ 20' N \), but by the compass it is found to be \( E 26^\circ 12' \): Required the variation, and which way it is.
Since they are both the same way, therefore
From the magnetic amplitude \( E 26^\circ 12' N \),
take the true amplitude \( E 14^\circ 20' N \).
and there remains the variation \( 11^\circ 52' E \).
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 \( W 34^\circ 26' S \), and his magnetic amplitude \( W 23^\circ 13' S \): Required the variation, and which way it is.
Since they lie both the same way, therefore
From the sun's true amplitude \( W 34^\circ 26' S \),
take his magnetic amplitude \( W 23^\circ 13' S \),
there remains the variation \( 11^\circ 13' W \).
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 \( 13^\circ 24' N \), and his magnetic \( E 12^\circ 32' S \): Required the variation, and which way it lies.
Since the true and observed amplitudes lie different ways, therefore
To the true amplitude \( E 13^\circ 24' N \),
add the magnetic \( E 12^\circ 32' S \),
the sum is the variation \( 25^\circ 56' W \).
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 \( W 8^\circ 24' N \), but his magnetic amplitude is \( W 10^\circ 13' S \): Required the variation.
To the true amplitude \( W 8^\circ 24' N \),
add the magnetic \( W 10^\circ 13' S \),
the sum is the variation \( 18^\circ 37' E \).
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 \( N 86^\circ 40' E \), but by the compass it is \( N 73^\circ 24' E \): Required the variation, and which way it lies.
From the sun's true azimuth \( N 86^\circ 40' E \),
take the magnetic \( N 73^\circ 24' E \),
there remains the variation \( 13^\circ 16' E \).
which is easterly, because the true azimuth is to the right of the observed.
Example VI. Suppose the sun's true azimuth is \( S 3^\circ 24' E \), and the magnetic \( S 4^\circ 36' W \): Required the variation, and which way it lies.
To the true azimuth \( S 3^\circ 24' E \),
add the magnetic \( S 4^\circ 36' W \),
the sum is the variation \( 8^\circ 00' W \).
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 \( 11^\circ 15' \) easterly, and in the year 1622 it was \( 6^\circ 0' E \); also in the year 1634, it was \( 4^\circ 05' E \), still decreasing, and the needle approaching the true meridian, till it coincided with it, and then there was no variation; after which, the varia- tion began to be westerly; and in the year 1672, it was observed to be $2^\circ 30' W$; also in the year 1683, it was $4^\circ 30' W$; 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 $3^\circ 00' E$; and in the year 1666, there was no variation; but in the year 1681, it was $2^\circ 30' W$, 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.
**Section 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 fall, makes with the rhomb she really falls 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 fall 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 $5\frac{1}{2}$ and 6 points.
7. When both main and fore-courses must be handed, then 6 or 6½ 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.
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 let 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 bitts, 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 NWb, 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 falls 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 E&N; also suppose the wind is at NNW, and the ship lies up within 6½ points of the wind with her starboard tack on board, making 1½ 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 NWbE, and the variation one point easterly; then the true course steered will be NNE: Also suppose the course steered is NEbE, 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 failing between the times of heaving the log, the 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.
1st. If the ship fail 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 \( \frac{AB}{AD} = \frac{BC}{DE} \), and \( \frac{AB}{AD} = \frac{AC}{AE} \).
**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 45° 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.
2dly. 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 Cafe 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 wetting; 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 Cafe 4. of Plain Sailing) the true course to be S 64° 53' W, and the direct distance 176.7 miles.
3dly. 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 Cafe 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 Cafe 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 sailing 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 Cafe 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 Cafe 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-book the several courses and distances steamed 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. S denotes Sunday, M Monday, T Tuesday, W Wednesday, Th Thursday, F Friday, and S denotes Saturday. The FORM of the
LOG-BOOK,
With the Manner of working Days Works at Sea.
| The Log-Book. | The Log-Book. | |---------------|---------------| | H.K. | K. Courses. Winds. Observations and Accidents. Day of | Courses Correct | Diff. | Diff. Lat. | Diff. long. | | | | | N | S | E | W | | 1 | | | SSW | 50 | 46.2| 29.4| | 2 | North | | SWbW| 19 | 18.6| 5.5 | | 3 | | | SW | 49 | 29.7| 45.5| | 4 | | | SWbS| 24.5| 20.2| 20.0| | 5 | | | SW½S| 25.5| 19.5| 24.6| | 6 | | | | | | | | 7 | | | | | | | | 8 | | | | | | | | 9 | | | | | | | | 10 | | | | | | | | 11 | | | | | | | | 12 | | | | | | |
Hence the ship, by account, has come to the latitude of 47° 46' north, and has differed her longitude 20° 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.
The variation I reckon to be one point westerly. ### The Log Book
| Day | Course Corrected | Diff. | Dist. | Lat. | D.R. Long. | |-----|-----------------|------|-------|-----|-----------| | | S E b E | | | | | | | E S E | | | | | | | S E b E | | | | |
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.
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**A Journal from the Lizard towards Jamaica in the ship Neptune, J. M. commander.**
| Week Days | Months Years | Month Days | Winds | Direct Course | Diff. Miles | Latitude Correct | Whole Diff. Long. made | Bearing and Diff. from the Lizard | Remarkable Observations and Accidents | |-----------|--------------|------------|-------|---------------|-------------|------------------|------------------------|------------------------------------|-------------------------------------| | D | | | N b E | S 31, 31 W | 157.4 | 47°, 46' | 2°, 5' W | At noon the Lizard bore N. 31° 31' E. Diff. 157.4 miles. | Fair weather at four P.M. I took my departure from the Lizard, it bearing NNE distance 5 leagues. | | | | | E b S | | | | | | | | | | | N N E | | | | | | | | | | | E N E | | | | | | | | | | | N E b E | | | | | | | | D | | | West | S 34, 01 E | 48.2 | 47°, 06' | 1°, 35' W | At noon the Lizard bore S. 17° 55' W. Diff. 183 miles. | Strong gales of wind and variable. | | | | | N W b W | | | | | | | | | | | S W b W | | | | | | |