NAVIGATION

IS the art of conducting or carrying a ship from one port to another.

HISTORY.

THE poets refer the invention of the art of navigation to Neptune, some to Bacchus, others to Hercules, others to Jason, and others to Janus, who is said to have made the first ship. Historians ascribe it to the Æginetes, the Phœnicians, Tyrians, and the ancient inhabitants of Britain. Some suppose, that the first hint was taken from the flight of the kite; others, as Oppian (De Piscibus, lib. i.), from the fish called nautilus: others ascribe it to accident.—Scripture refers the origin of so useful an invention to God himself, who gave the first specimen thereof in the ark built by Noah under his direction. For the raillery which the good man underwent on account of his enterprise shows evidently enough that the world was then ignorant of any thing like navigation, and that they even thought it impossible.

However, profane history represents the Phœnicians, especially those of their capital Tyre, as the first navigators; being urged to seek a foreign commerce by the narrowness and poverty of a slip of ground they possessed along the coasts; as well as by the conveniency of two or three good ports, and by their natural genius for traffic. Accordingly, Lebanon, and the other neighbouring mountains, furnishing them with excellent wood for ship-building, in a short time they were masters of a numerous fleet; and constantly hazarding new navigations, and settling new trades, they soon arrived at an incredible pitch of opulence and populousness: insomuch as to be in a condition to send out colonies, the principal of which was that of Carthage; which, keeping up their Phœnician spirit of commerce, in time not only equalled Tyre itself, but vastly surpassed it; sending its merchant fleets through Hercules's Pillars, now the straits of Gibraltar, along the western coasts of Africa and Europe; and even, if we believe some authors, to America itself.

Tyre, whose immense riches and power are represented in such lofty terms both by sacred and profane authors, being destroyed by Alexander the Great, its navigation and commerce were transferred by the conqueror to Alexandria, a new city admirably situated for those purposes; proposed for the capital of the empire of Asia, which Alexander then meditated. And thus arose the navigation of the Egyptians; which was afterwards so much cultivated by the Ptolemies, that Tyre and Carthage were quite forgotten.

Egypt being reduced into a Roman province after the battle of Actium, its trade and navigation fell into the hands of Augustus; in whose time Alexandria was only inferior to Rome; and the magazines of the ca-

pital of the world were wholly supplied with merchandise from the capital of Egypt.

At length, Alexandria itself underwent the fate of Tyre and Carthage; being surprised by the Saracens, who in spite of the emperor Heraclius, overspread the northern coasts of Africa, &c. whence the merchants being driven, Alexandria has ever since been in a languishing state, though it still has a considerable part of the commerce of the Christian merchants trading to the Levant.

The fall of Rome and its empire drew along with it not only that of learning and the polite arts, but that of navigation; the barbarians into whose hands it fell, contenting themselves with the spoils of the industry of their predecessors.

But no sooner were the more brave among those nations well settled in their new provinces, some in Gaul, as the Franks; others in Spain, as the Goths; and others in Italy, as the Lombards; but they began to learn the advantages of navigation and commerce, and the methods of managing them, from the people they subdued; and this with so much success, that in a little time some of them became able to give new lessons, and set on foot new institutions for its advantage. Thus it is to the Lombards we usually ascribe the invention and use of banks, book-keeping, exchanges, recharges, &c.

It does not appear which of the European people, after the settlement of their new masters, first betook themselves to navigation and commerce. Some think it began with the French; though the Italians seem to have the justest title to it; and are accordingly regarded as the restorers thereof, as well as of the polite arts, which had been banished together from the time the empire was torn asunder. It is the people of Italy then, and particularly those of Venice and Genoa, who have the glory of this restoration; and it is to their advantageous situation for navigation they in great measure owe their glory. In the bottom of the Adriatic were a great number of marshy islands, only separated by narrow channels, but those well screened, and almost inaccessible, the residence of some fishermen, who here supported themselves by a little trade in fish and salt, which they found in some of these islands. Thither the Veneti, a people inhabiting that part of Italy along the coasts of the gulf, retired, when Alaric king of the Goths, and afterwards Attila king of the Huns, ravaged Italy.

These new islanders, little imagining that this was to be their fixed residence, did not think of composing any body politic; but each of the 72 islands of this little archipelago continued a long time under its several masters, and each made a distinct commonwealth. When their commerce was become considerable enough to give jealousy to their neighbours, they began to think

think of uniting into a body. And it was this union, first begun in the sixth century, but not completed till the eighth, that laid the sure foundation of the future grandeur of the state of Venice. From the time of this union, their fleets of merchantmen were sent to all the ports of the Mediterranean; and at last to those of Egypt, particularly Cairo, a new city built by the Saracen princes on the eastern banks of the Nile, where they traded for the spices and other products of the Indies. Thus they flourished, increased their commerce, their navigation, and their conquests on the terra firma, till the league of Cambrai in 1508, when a number of jealous princes conspired to their ruin; which was the more easily effected by the diminution of their East India commerce, of which the Portuguese had got one part and the French another. Genoa, which had applied itself to navigation at the same time with Venice, and that with equal success, was a long time its dangerous rival, disputed with it the empire of the sea, and shared with it the trade of Egypt and other parts both of the east and west.

Jealousy soon began to break out; and the two republics coming to blows, there was almost continual war for three centuries before the superiority was ascertained; when, towards the end of the 14th century, the battle of Chioza ended the strife; the Genoese, who till then had usually the advantage, having now lost all; and the Venetians, almost become desperate, at one happy blow, beyond all expectation, secured to themselves the empire of the sea, and superiority in commerce.

About the same time that navigation was retrieved in the southern parts of Europe, a new society of merchants was formed in the north, which not only carried commerce to the greatest perfection it was capable of till the discovery of the East and West Indies, but also formed a new scheme of laws for the regulation therefore, which still obtain under the names of Uses and Customs of the Sea. This society is that famous league of the Hanse towns, commonly supposed to have begun about the year 1164. See HANSE TOWNS.

For the modern state of navigation in England, Holland, France, Spain, Portugal, &c. see COMMERCE, COMPANY, &c.

We shall only add, that in examining the causes of commerce passing successively from the Venetians, Genose, and Hanse towns, to the Portuguese and Spaniards, and from these again to the English and Dutch, it may be established as a maxim, that the relation between commerce and navigation, or if we may be allowed to say it, their union, is so intimate, that the fall of the one inevitably draws after it that of the other; and that they will always either flourish or dwindle together. Hence so many laws, ordinances, statutes, &c. for its regulation; and hence particularly that celebrated act of navigation, which an eminent foreign author calls the palladium or tutelary deity of the commerce of England; which is the standing rule, not only of the British among themselves, but also of other nations with whom they traffic.

The art of navigation has been exceedingly improved in modern times, both with regard to the form of the vessels themselves, and with regard to the methods of working them. The use of rowers is now entirely superseded by the improvements made in the

formation of the sails, rigging, &c. by which means ships can not only sail much faster than formerly, but can tack in any direction with the greatest facility. It is also very probable that the ancients were neither so well skilled in finding the latitudes, nor in steering their vessels in places of difficult navigation, as the moderns. But the greatest advantage which the moderns have over the ancients is from the mariner's compass, by which they are enabled to find their way with as great facility in the midst of an immeasurable ocean, as the ancients could have done by creeping along the coast, and never going out of sight of land. Some people indeed contend, that this is no new invention, but that the ancients were acquainted with it. They say, that it was impossible for Solomon to have sent ships to Ophir, Tarshish, and Parvaim, which last they will have to be Peru, without this useful instrument. They insist, that it was impossible for the ancients to be acquainted with the attractive virtue of the magnet, and to be ignorant of its polarity. Nay, they affirm, that this property of the magnet is plainly mentioned in the book of Job, where the loadstone is mentioned by the name of topaz, or the stone that turns itself. But it is certain, that the Romans, who conquered Judea, were ignorant of this instrument; and it is very improbable, that such an useful invention, if once it had been commonly known to any nation, would have been forgot, or perfectly concealed from such a prudent people as the Romans, who were so much interested in the discovery of it.

Among those who admit that the mariner's compass is a modern invention, it has been much disputed who was the inventor. Some give the honour of it to Flavio Gioia of Amalfi in Campania*, who lived about the beginning of the 14th century; while others say that it came from the east, and was earlier known in Europe. But, at whatever time it was invented, it is certain, that the mariner's compass was not commonly used in navigation before the year 1420. In that year the science was considerably improved under the auspices of Henry duke of Visco, brother to the king of Portugal. In the year 1485, Roderick and Joseph, physicians to John II. king of Portugal, together with one Martin de Bohemia, a Portuguese, native of the island of Fayal, and scholar to Regiomontanus, calculated tables of the sun's declination for the use of sailors, and recommended the astrolabe for taking observations at sea. Of the instructions of Martin, the celebrated Christopher Columbus is said to have availed himself, and to have improved the Spaniards in the knowledge of the art; for the farther progress of which a lecture was afterwards founded at Seville by the emperor Charles V.

The discovery of the variation is claimed by Columbus, and by Sebastian Cabot. The former certainly did observe this variation without having heard of it from any other person, on the 14th of September 1492, and it is very probable that Cabot might do the same. At that time it was found that there was no variation at the Azores, where some geographers have thought proper to place the first meridian; though it hath since been observed that the variation alters in time.—The use of the cross staff now began to be introduced among sailors. This ancient instrument is described by John Werner of Nuremberg, in his

his annotations on the first book of Ptolemy's Geography, printed in 1514. He recommends it for observing the distance between the moon and some star, in order thence to determine the longitude.

At this time the art of navigation was very imperfect, on account of the inaccuracies of the plane chart, which was the only one then known, and which, by its gross errors, must have greatly misled the mariner, especially in voyages far distant from the equator. Its precepts were probably at first only set down on the sea charts, as is the custom at this day: but at length there were two Spanish treatises published in 1545; one by Pedro de Medina; the other by Martin Cortes, which contained a complete system of the art, as far as it was then known. These seem to have been the oldest writers who fully handled the art; for Medina, in his dedication to Philip prince of Spain, laments that multitudes of ships daily perished at sea, because there were neither teachers of the art, nor books by which it might be learned; and Cortes, in his dedication, boasts to the emperor that he was the first who had reduced navigation into a compendium, valuing himself much on what he had performed. Medina defended the plane chart; but he was opposed by Cortes, who showed its errors, and endeavoured to account for the variation of the compass, by supposing the needle to be influenced by a magnetic pole (which he called the point attractive), different from that of the world; which notion hath been farther prosecuted by others. Medina's book was soon translated into Italian, French, and Flemish, and served for a long time as a guide to foreign navigators. However, Cortes was the favourite author of the English nation, and was translated in 1561; while Medina's work was entirely neglected, though translated also within a short time of the other. At that time the system of navigation consisted of the following particulars, and others similar: An account of the Ptolemaic hypothesis, and the circles of the sphere; of the roundness of the earth, the longitudes, latitudes, climates, &c. and eclipses of the luminaries; a calendar; the method of finding the prime, epoch, moon's age, and tides; a description of the compass, an account of its variation, for the discovering of which Cortes said an instrument might easily be contrived; tables of the sun's declination for four years, in order to find the latitude from his meridian altitude; directions to find the same by certain stars; of the course of the sun and moon; the length of the days; of time and its divisions; the method of finding the hour of the day and night; and lastly, a description of the sea chart, on which to discover where the ship is, they made use of a small table, that showed, upon an alteration of one degree of the latitude, how many leagues were run in each rhumb, together with the departure from the meridian. Besides, some instruments were described, especially by Cortes; such as one to find the place and declination of the sun, with the days, and place of the moon; certain dials, the astrolabe, and cross staff; with a complex machine to discover the hour and latitude at once.

About the same time were made proposals for finding the longitude by observations of the moon.— In 1530, Gemma Frisius advised the keeping of the time by means of small clocks or watches, then, as he

says, newly invented. He also contrived a new sort of cross staff, and an instrument called the nautical quadrant; which last was much praised by William Cunningham, in his Astronomical Glass, printed in the year 1559.

In 1537 Pedro Nunez, or Nonius, published a book in the Portuguese language, to explain a difficulty in navigation proposed to him by the commander Don Martin Alphonso de Susa. In this he exposes the errors of the plane chart, and likewise gives the solution of several curious astronomical problems; amongst which is that of determining the latitude from two observations of the sun's altitude and intermediate azimuth being given. He observed, that though the rhumbs are spiral lines, yet the direct course of a ship will always be in the arch of a great circle, whereby the angle with the meridians will continually change: all that the steersman can here do for the preserving of the original rhumb, is to correct these deviations as soon as they appear sensible. But thus the ship will in reality describe a course without the rhumb line intended; and therefore his calculations for assigning the latitude, where any rhumb line crosses the several meridians, will be in some measure erroneous. He invented a method of dividing a quadrant by means of concentric circles, which, after being much improved by Dr Halley, is used at present, and is called a nonius.

In 1577, Mr William Bourne published a treatise, in which, by considering the irregularities in the moon's motion, he shows the errors of the sailors in finding her age by the epoch, and also in determining the hour from observing on what point of the compass the sun and moon appeared. He advises, in sailing towards the high latitudes, to keep the reckoning by the globe, as there the plane chart is most erroneous. He despairs of our ever being able to find the longitude, unless the variation of the compass should be occasioned by some such attractive point as Cortes had imagined; of which, however, he doubts: but as he had shown how to find the variation at all times, he advises to keep an account of the observations, as useful for finding the place of the ship; which advice was prosecuted at large by Simon Stevin, in a treatise published at Leyden in 1599; the substance of which was the same year printed at London in English by Mr Edward Wright, entitled the Haven-finding Art. In this ancient tract also is described the way by which our sailors estimate the rate of a ship in her course, by an instrument called the log. This was so named from the piece of wood or log that floats in the water while the time is reckoned during which the line that is fastened to it is veering out. The author of this contrivance is not known; neither was it taken notice of till 1607, in an East India voyage published by Purchas: but from this time it became famous, and was much taken notice of by almost all writers on navigation in every country; and it still continues to be used as at first, though many attempts have been made to improve it, and contrivances proposed to supply its place; many of which have succeeded in quiet water, but proved useless in a stormy sea.

In 1581 Michael Coignet, a native of Antwerp, published a treatise, in which he animadverted on Medina. In this he showed, that as the rhumbs are spirals,

rals, making endless revolutions about the poles, numerous errors must arise from their being represented by straight lines on the sea charts; but though he hoped to find a remedy for these errors, he was of opinion that the proposals of Nonius were scarcely practicable, and therefore in a great measure useless. In treating of the sun's declination, he took notice of the gradual decrease in the obliquity of the ecliptic; he also described the cross staff with three transverse pieces, as it is at present made, and which he owned to have been then in common use among the sailors. He likewise gave some instruments of his own invention; but all of them are now laid aside, excepting perhaps his nocturnal. He constructed a sea table to be used by such as sailed beyond the 60th degree of latitude; and at the end of the book is delivered a method of sailing on a parallel of latitude by means of a ring dial and a 24 hour glass. The same year the discovery of the dipping needle was made by Mr Robert Norman*. In his publication on that art he maintains, in opposition to Cortes, that the variation of the compass was caused by some point on the surface of the earth, and not in the heavens: he also made considerable improvements in the construction of compasses themselves; showing especially the danger of not fixing, on account of the variation, the wire directly under the fleur de luce; as compasses made in different countries have it placed differently. To this performance of Norman's is always prefixed a discourse on the variation of the magnetic needle, by Mr William Burrough, in which he shows how to determine the variation in many different ways. He also points out many errors in the practice of navigation at that time, and speaks in very severe terms concerning those who had published upon it.

All this time the Spaniards continued to publish treatises on the art. In 1585 an excellent compendium was published by Roderico Zamorano; which contributed greatly towards the improvement of the art, particularly in the sea charts. Globes of an improved kind, and of a much larger size than those formerly used, were now constructed, and many improvements were made in other instruments; however, the plane chart continued still to be followed, though its errors were frequently complained of. Methods of removing these errors had indeed been sought after; and Gerard Mercator seems to have been the first who found the true method of doing this, so as to answer the purposes of seamen. His method was to represent the parallels both of latitude and longitude by parallel straight lines, but gradually to augment the former as they approached the pole. Thus the rhumbs, which otherwise ought to have been curves, were now also extended into straight lines; and thus a straight line drawn between any two places marked upon the chart would make an angle with the meridians, expressing the rhumb leading from the one to the other. But though, in 1569, Mercator published an universal map constructed in this manner, it doth not appear that he was acquainted with the principles on which this proceeded; and it is now generally believed, that the true principles on which the construction of what is called Mercator's chart depends, were first discovered by an Englishman, Mr Edward Wright.

Mr Wright supposes, but, according to the general opinion, without sufficient grounds, that this enlarge-

ment of the degrees of latitude was known and mentioned by Ptolemy, and that the same thing had also been spoken of by Cortes. The expressions of Ptolemy alluded to, relate indeed to the proportion between the distances of the parallels and meridians; but instead of proposing any gradual enlargement of the parallels of latitude in a general chart, he speaks only of particular maps; and advises not to confine a system of such maps to one and the same scale, but to plan them out by a different measure, as occasion might require: only with this precaution, that the degrees of longitude in each should bear some proportion to those of latitude; and this proportion is to be deduced from that which the magnitude of the respective parallels bears to a great circle of the sphere. He adds, that in particular maps, if this proportion be observed with regard to the middle parallel, the inconvenience will not be great though the meridians should be straight lines parallel to each other. Here he is said only to mean, that the maps should in some measure represent the figures of the countries for which they are drawn. In this sense Mercator, who drew maps for Ptolemy's tables, understood him; thinking it, however, an improvement not to regulate the meridians by one parallel, but by two; one distant from the northern, the other from the southern extremity of the map by a fourth part of the whole depth; by which means, in his maps, though the meridians are straight lines, yet they are generally drawn inclining to each other towards the poles. With regard to Cortes, he speaks only of the number of degrees of latitude, and not of the extent of them; nay, he gives express directions that they should all be laid down by equal measurement on a scale of leagues adapted to the map.

For some time after the appearance of Mercator's map, it was not rightly understood, and it was even thought to be entirely useless, if not detrimental.— However, about the year 1592, its utility began to be perceived; and seven years after, Mr Wright printed his famous treatise entitled, The Correction of certain Errors in Navigation, where he fully explained the reason of extending the length of the parallels of latitude, and the uses of it to navigators. In 1610, a second edition of Mr Wright's book was published with improvements. An excellent method was proposed of determining the magnitude of the earth; at the same time it was judiciously proposed to make our common measures in some proportion to a degree on its surface, that they might not depend on the uncertain length of a barley corn. Some of his other improvements were, "The table of latitudes for dividing the meridian computed to minutes;" whereas it had been only divided to every tenth minute. He also published a description of an instrument which he calls the sea rings; and by which the variation of the compass, altitude of the sun, and time of the day, may be determined readily at once in any place, provided the latitude is known. He showed also how to correct the errors arising from the eccentricity of the eye in observing by the cross-staff. He made a total amendment in the tables of the declinations and places of the sun and stars from his own observations made with a six foot quadrant in the years 1594, 95, 96, and 97. A sea quadrant to take altitudes by a forward or backward observation; and likewise with a contrivance for the ready finding the latitude

latitude by the height of the pole star, when not upon the meridian. To this edition was subjoined a translation of Zamorano's Compendium above mentioned, in which he corrected some mistakes in the original; adding a large table of the variation of the compass observed in very different parts of the world, to show that it was not occasioned by any magnetic pole.

These improvements soon became known abroad.— In 1658, a treatise entitled, Hypomnemata Mathematica, was published by Simon Stevin, for the use of Prince Maurice. In that part relating to navigation, the author having treated of sailing on a great circle, and shown how to draw the rhumbs on a globe mechanically, sets down Wright's two tables of latitudes and of rhumbs, in order to describe these lines more accurately, pretending even to have discovered an error in Wright's table. But all Stevin's objections were fully answered by the author himself, who showed that they arose from the gross way of calculating made use of by the former.

In 1624, the learned Wellebrordus Snellius, professor of mathematics at Leyden, published a treatise of navigation on Wright's plan, but somewhat obscurely: and as he did not particularly mention all the discoveries of Wright, the latter was thought by some to have taken the hint of all his discoveries from Snellius. But this supposition is long ago refuted: and Wright enjoys the honour of those discoveries which is justly his due.

Mr Wright having shown how to find the place of the ship on his chart, observed that the same might be performed more accurately by calculation: but considering, as he says, that the latitudes, and especially the courses at sea, could not be determined so precisely, he forbore setting down particular examples; as the mariner may be allowed to save himself this trouble, and only mark out upon his chart the ship's way, after the manner then usually practised. However, in 1614, Mr Raphe Handson, among his nautical questions subjoined to a translation of Pitiscus's trigonometry, solved very distinctly every case of navigation, by applying arithmetical calculations to Wright's table of latitudes, or of meridional parts, as it hath since been called. Though the method discovered by Wright for finding the change of longitude by a ship sailing on a rhumb is the proper way of performing it, Handson also proposes two ways of approximation to it without the assistance of Wright's division of the meridian line. The first was computed by the arithmetical mean between the cosines of both latitudes; the other by the same mean between the secants as an alternative, when Wright's book was not at hand; though this latter is wider from the truth than the first. By the same calculations also he showed how much each of these compendiums deviates from the truth, and also how widely the computations on the erroneous principles of the plane chart differ from them all. The method, however, commonly used by our sailors is commonly called the middle latitude; which, though it errs more than that by the arithmetical mean between the two cosines, is preferred on account of its being less opulent: yet in high latitudes it is more eligible to use that of the arithmetical mean between the logarithmic cosines, equivalent to the geometrical mean between the cosines themselves; a method since pro-

posed by Mr John Bassat. The computation by the middle latitude will always fall short of the true change of longitude; that by the geometrical mean will always exceed; but that by the arithmetical mean falls short in latitudes above 45 degrees, and exceeds in lesser latitudes. However, none of these methods will differ much from the truth when the change of latitude is sufficiently small.

About this time logarithms were invented by John Napier, baron of Merchiston in Scotland, and proved of the utmost service to the art of navigation. From which Mr Edmund Gunter constructed a table of logarithmic sines and tangents to every minute of the quadrant, which he published in 1620. In this work he applied to navigation, and other branches of mathematics, his admirable ruler known by the name of Gunter's scale*; on which are described lines of logarithms, of logarithmic sines and tangents, of meridional parts, &c. He greatly improved the sector for the same purposes. He showed also how to take a back observation by the cross staff, whereby the error arising from the eccentricity of the eye is avoided. He described likewise another instrument, of his own invention called the cross bow, for taking altitudes of the sun or stars, with some contrivances for the more ready collecting the latitude from the observation. The discoveries concerning logarithms were carried to France in 1624 by Mr Edmund Wingate, who published two small tracts in that year at Paris. In one of these he taught the use of Gunter's scale; and in the other, of the tables of artificial sines and tangents, as modelled according to Napier's last form, erroneously attributed by Wingate to Briggs.

Gunter's rule was projected into a circular arch by the Reverend Mr William Oughtred in 1633, and its uses fully shown in a pamphlet entitled, The Circles of Proportion, where, in an appendix, are well treated several important points in navigation. It has also been made in the form of a sliding ruler.

The logarithmic tables were first applied to the different cases of sailing by Mr Thomas Addison, in his treatise entitled, Arithmetical Navigation, printed in 1625. He also gives two traverse tables, with their uses; the one to quarter points of the compass, the other to degrees. Mr Henry Gellibrand published his discovery of the changes of the variation of the compass, in a small quarto pamphlet, entitled, A discourse mathematical on the variation of the magnetic needle, printed in 1635. This extraordinary phenomenon he found out by comparing the observations made at different times near the same place by Mr Burrough, Mr Gunter, and himself, all persons of great skill and experience in these matters. This discovery was likewise soon known abroad; for Father Athanasius Kircher, in his treatise entitled, Magnes, first printed at Rome in 1641, informs us, that he had been told it by Mr John Greaves; and then gives a letter of the famous Marinus Mersennus, containing a very distinct account of the same.

As altitudes of the sun are taken on shipboard by observing his elevation above the visible horizon; to obtain from thence the sun's true altitude with correctness, Wright observes it to be necessary that the dip of the visible horizon below the horizontal plane passing through the observer's eye should be brought into the account,

account, which cannot be calculated without knowing the magnitude of the earth. Hence he was induced to propose different methods for finding this; but complains that the most effectual was out of his power to execute; and therefore contented himself with a rude attempt, in some measure sufficient for his purpose: and the dimensions of the earth deduced by him corresponded very well with the usual divisions of the log line; however, as he wrote not an express treatise on navigation, but only for the correcting such errors as prevailed in general practice, the log line did not fall under his notice. Mr Richard Norwood, however, put in execution the method recommended by Mr Wright as the most perfect for measuring the dimensions of the earth, with the true length of the degrees of a great circle upon it; and, in 1635, he actually measured the distance between London and York; from whence, and the summer solstitial altitudes of the sun observed on the meridian at both places, he found a degree on a great circle of the earth to contain 367,196 English feet, equal to 57,300 French fathoms or toises: which is very exact, as appears from many measures that have been made since that time. Of all this Mr Norwood gave a full account in his treatise called The Seaman's Practice, published in 1637. He there shows the reason why Snellius had failed in his attempt: he points out also various uses of his discovery, particularly for correcting the gross errors hitherto committed in the divisions of the log line. But necessary amendments have been little attended to by sailors, whose obstinacy in adhering to established errors has been complained of by the best writers on navigation. This improvement has at length, however, made its way into practice, and few navigators of reputation now make use of the old measure of 42 feet to a knot. In that treatise also Mr Norwood describes his own excellent method of setting down and perfecting a sea reckoning, by using a traverse table; which method he had followed and taught for many years. He shows also how to rectify the course by the variation of the compass being considered; as also how to discover currents, and to make proper allowance on their account. This treatise, and another on trigonometry, were continually reprinted, as the principal books for learning scientifically the art of navigation. What he had delivered, especially in the latter of them, concerning this subject, was contracted as a manual for sailors, in a very small piece called his Epitome; which useful performance has gone through a great number of editions. No alterations were ever made in the Seaman's Practice till the 12th edition in 1676, when the following paragraph was inserted in a smaller character: "About the year 1672, Monsieur Picart has published an account in French, concerning the measure of the earth, a breviate whereof may be seen in the Philosophical Transactions, N^o 112, wherein he concludes one degree to contain 365,184 English feet, nearly agreeing with Mr Norwood's experiment;" and this advertisement is continued through the subsequent editions as late as the year 1732.

About the year 1645, Mr Bond published in Norwood's Epitome a very great improvement in Wright's method, by a property in his meridian line, whereby its divisions are more scientifically assigned than the author himself was able to effect; which was from this theorem, that these divisions are analogous to the excesses of the lo-

garithmic tangents of half the respective latitudes augmented by 45 degrees above the logarithm of the radius. This he afterwards explained more fully in the third edition of Gunter's works, printed in 1653, where, after observing that the logarithmic tangents from 45° upwards increase in the same manner that the secants added together do, if every half degree be accounted as a whole degree of Mercator's meridional line. His rule for computing the meridional parts belonging to any two latitudes, supposed on the same side of the equator, is to the following effect: "Take the logarithmic tangent, rejecting the radius, of half each latitude, augmented by 45 degrees; divide the difference of those numbers by the logarithmic tangent of 45° 30', the radius being likewise rejected; and the quotient will be the meridional parts required, expressed in degrees." This rule is the immediate consequence from the general theorem, That the degrees of latitude bear to one degree (or 60 minutes, which in Wright's table stands for the meridional parts of one degree), the same proportion as the logarithmic tangent of half any latitude augmented by 45 degrees, and the radius neglected, to the like tangent of half a degree augmented by 45 degrees, with the radius likewise rejected. But here was farther wanting the demonstration of this general theorem, which was at length supplied by Mr James Gregory of Aberdeen in his Exercitationes Geometricæ, printed at London in 1668; and afterwards more concisely demonstrated, together with a scientific determination of the divisor, by Dr Halley in the Philosophical Transactions for 1695, N^o 219, from the consideration of the spirals into which the rhumbs are transformed in the stereographic projection of the sphere upon the plane of the equinoctial; and which is rendered still more simple by Mr Roger Cotes, in his Logometria, first published in the Philosophical Transactions for 1714, N^o 388. It is moreover added in Gunter's book, that if $\backslash frac\{1\}\{2\}$th of this division, which does not sensibly differ from the logarithmic tangent of 45° 1' 30" (with the radius subtracted from it), be used, the quotient will exhibit the meridional parts expressed in leagues, and this is the divisor set down in Norwood's Epitome. After the same manner the meridional parts will be found in minutes, if the like logarithmic tangent of 45° 1' 30", diminished by the radius, be taken; that is, the number used by others being 12633, when the logarithmic tables consist of eight places of figures besides the index.

In an edition of the Seaman's Kalender, Mr Bond declared, that he had discovered the longitude by having found out the true theory of the magnetic variation; and to gain credit to his assertion, he foretold, that at London in 1657 there would be no variation of the compass, and from that time it would gradually increase the other way; which happened accordingly. Again, In the Philosophical Transactions for 1668, N^o 40, he published a table of the variation for 49 years to come. Thus he acquired such reputation, that his treatise, entitled, The Longitude Found, was in 1676 published by the special command of Charles II. and approved by many celebrated mathematicians. It was not long, however, before it met with opposition; and in 1678 another treatise, entitled, The Longitude not Found, made its appearance; and as Mr Bond's hypothesis

hypothesis did not in any manner answer its author's sanguine expectations, the affair was undertaken by Dr Halley. The result of his speculation was, that the magnetic needle is influenced by four poles; but this wonderful phenomenon seems hitherto to have eluded all our researches. In 1700, however, Dr Halley published a general map, with curve lines expressing the paths where the magnetic needle had the same variation; which was received with universal applause. But as the positions of these curves vary from time to time, they should frequently be corrected by skilful persons; as was done in 1644 and 1756, by Mr William Mountaine, and Mr James Dodson, F. R. S. In the Philosophical Transactions for 1690, Dr Halley also gave a dissertation on the monsoons; containing many very useful observations for such as sail to places subject to these winds.

After the true principles of the art were settled by Wright, Bond, and Norwood, the authors on navigation became so numerous, that it would be impossible to enumerate them. New improvements were daily made, and every thing relative to it was settled with an accuracy not only unknown to former ages, but which would have been reckoned utterly impossible. The earth being found to be a spheroid, and not a perfect sphere, with the shortest diameter passing through the poles, a tract was published in 1741, by the Rev. Doctor Patrick Murdoch, wherein he accommodated

Wright's sailing to such a figure; and Mr Colin MacLaurin, the same year, in the Philosophical Transactions, N^o 461, gave a rule for determining the meridional parts of a spheroid; which speculation is farther treated of in his book of Fluxions, printed at Edinburgh in 1742.

Among the later discoveries in navigation, that of finding the longitude both by lunar observations and by time-keepers is the principal. It is owing chiefly to the rewards offered by the British parliament that this has attained the present degree of perfection. We are indebted to Dr Maskelyne for putting the first of these methods in practice, and for other important improvements in navigation. The time-keepers, constructed by Harrison for this express purpose, were found to answer so well, that he obtained the parliamentary reward.

The only works that have appeared of late in navigation are those on the longitude and navigation by Dr Mackay, of which the following account is transcribed from the Anti-Jacobin Review for September 1804.

"This publication, (Dr Mackay's Treatise on Navigation) and that on the longitude by the same author, form the most correct and practical system of navigation and nautical science hitherto published in this country; they may be considered not only of individual utility, but of national importance."

THEORY OF NAVIGATION.

THE motion of a ship in the water is well known to depend on the action of the wind upon its sails, regulated by the direction of the helm. As the water is a resisting medium and the bulk of the ship very considerable, it thence follows that there is always a great resistance on her fore-part; and when this resistance becomes sufficient to balance the moving force of the wind upon the sails, the ship attains her utmost degree of velocity, and her motion is no longer accelerated. This velocity is different according to the different strength of the wind; but the stronger the wind, the greater resistance is made to the ship's passage through the water: and hence, though the wind should blow ever so strong, there is also a limit to the velocity of the ship: for the sails and ropes can bear but a certain force of air; and when the resistance on the fore-part becomes more than equivalent to their strength, the velocity can be no longer increased, and the rigging gives way.

The direction of a ship's motion depends on the position of her sails with regard to the wind, combined with the action of the rudder. The most natural direction of the ship is, when she runs directly before the wind, the sails are then disposed, so as to be at right angles thereto. But this is not always the case, both on account of the variable nature of the winds, and the situation of the intended port, or of intermediate headlands or islands. When the wind therefore happens not to be favourable, the sails are placed so as to make an oblique angle both with the direction of the ship and with the wind; and the sails, together with the rudder, must be managed in such a manner, that the direction of the ship may make an acute angle with that of the

wind; and the ship, making boards on different tacks, will by this means arrive at the intended port.

The reason of the ship's motion in this case is, that the water resists the side more than the fore-part, and that in the same proportion as her length exceeds her breadth. This proportion is so considerable, that the ship continually flies off where the resistance is least, and that sometimes with great swiftness. In this way of sailing, however, there is a great limitation: for if the angle made by the keel with the direction of the wind be too acute, the ship cannot be kept in that position; neither is it possible for a large ship to make a more acute angle with the wind than about 6 points; though small sloops, it is said, may make an angle of about 5 points with it. In all these cases, however, the velocity of the ship is greatly retarded; and that not only on account of the obliquity of her motion, but by reason of what is called her lee-way. This is occasioned by the yielding of the water on the lee-side of the ship, by which means the vessel acquires a compound motion, partly in the direction of the wind, and partly in that which is necessary for attaining the desired port.

It is perhaps impossible to lay down any mathematical principles on which the lee-way of a ship could be properly calculated; only we may see in general that it depends on the strength of the wind, the roughness of the sea, and the velocity of the ship. When the wind is not very strong, the resistance of the water on the lee-side bears a very great proportion to that of the current of air; and therefore it will yield but very little: however, supposing the ship to remain

remain in the same place, it is evident, that the water having once begun to yield, will continue to do so for some time, even though no additional force was applied to it; but as the wind continually applies the same force as at first, the lee-way of the ship must go on constantly increasing till the resistance of the water on the lee-side balances the force applied on the other, when it will become uniform, as doth the motion of a ship sailing before the wind. If the ship changes her place with any degree of velocity, then every time she moves her own length, a new quantity of water is to be put in motion, which hath not yet received any momentum, and which of consequence will make a greater resistance than it can do when the ship remains in the same place. In proportion to the swiftness of the ship, then, the lee-way will be the less: but if the wind is very strong, the velocity of the ship bears but a small proportion to that of the current of air; and the same effects must follow as though the ship moved slowly, and the wind was gentle; that is, the ship must make a great deal of lee-way.—The same thing happens when the sea rises high, whether the wind is strong or not; for then the whole water of the ocean, as far as the swell reaches, has acquired a motion in a certain direction, and that to a very considerable depth. The mountainous waves will not fail to carry the ship very much out of her course; and this deviation will certainly be according to their velocity and magnitude. In all cases of a rough sea, therefore, a great deal of lee-way is made.—Another circumstance also makes a variation in the quantity of the lee-way; namely, the lightness or heaviness of the ship; it being evident, that when the ship sinks deep in the water, a much greater quantity of that element is to be put in motion before she can make any lee-way, than when she swims on the surface. As therefore it is impossible to calculate all these things with mathematical exactness, it is plain that the real course of a ship is exceedingly difficult to be found, and frequent errors must be made, which can only be corrected by celestial observations.

In many places of the ocean there are currents, or places where the water, instead of remaining at rest, runs with a very considerable velocity for a great way in some particular direction, and which will certainly carry the ship greatly out of her course. This occasions an error of the same nature with the lee-way: and therefore, whenever a current is perceived, its direction and velocity ought to be determined, and the proper allowances made.

Another source of error in reckoning the course of a ship proceeds from the variation of the compass.

There are few parts of the world where the needle points exactly north; and in those where the variation is known, it is subject to very considerable alterations. By these means the course of the ship is mistaken; for as the sailors have no other standard to direct them than the compass, if the needle, instead of pointing due north, should point north-east, a prodigious error would be occasioned during the course of the voyage, and the ship would not come near the port to which she was bound. To avoid errors of this kind, the only method is, to observe the sun's amplitude and azimuth as frequently as possible, by which the variation of the compass will be perceived, and the proper allowances

can then be made for errors in the course which this may have occasioned.

Errors will arise in the reckoning of a ship, especially when she sails in high latitudes, from the spherical figure of the earth; for as the polar diameter of our globe is found to be considerably shorter than the equatorial one, it thence follows, that the farther we remove from the equator, the longer are the degrees of latitude. Of consequence, if a navigator assigns any certain number of miles for the length of a degree of latitude near the equator, he must vary that measure as he approaches towards the poles, otherwise he will imagine that he hath not sailed so far as he actually hath done. It would therefore be necessary to have a table containing the length of a degree of latitude in every different parallel from the equator to either pole; as without this a troublesome calculation must be made at every time the navigator makes a reckoning of his course. Such a table, however, hath not yet appeared; neither indeed does it seem to be an easy matter to make it, on account of the difficulty of measuring the length even of one or two degrees of latitude in different parts of the world. Sir Isaac Newton first discovered this spheroidal figure of the earth; and showed, from experiments on pendulums, that the polar diameter was to the equatorial one as 229 to 230. This proportion, however, hath not been admitted by succeeding calculators. The French mathematicians, who measured a degree on the meridian in Lapland, made the proportion between the equatorial and polar diameters to be as 1 to 0.9891. Those who measured a degree at Quito in Peru, made the proportion 1 to 0.9924, or 266 to 265. M. Bouguer makes the proportion to be as 179 to 178; and M. Buffon, in one part of his theory of the earth, makes the equatorial diameter exceed the polar one by $\backslash frac\{1\}\{227\}$ of the whole. According to M. du Séjour, this proportion is as 321 to 320; and M. de la Place, in his Memoir upon the Figure of Spheroids, has deduced the same proportion. From these variations it appears that the point is not exactly determined, and consequently that any corrections which can be made with regard to the spheroidal figure of the earth must be very uncertain.

It is of consequence to navigators in a long voyage to take the nearest way to their port; but this is scarcely possible to be done. The shortest distance between any two points on the surface of a sphere is measured by an arch of a great circle intercepted between them; and therefore it is advisable to direct the ship along a great circle of the earth's surface. But this is a matter of considerable difficulty, because there are no fixed marks by which it can be readily known whether the ship sails in the direction of a great circle or not. For this reason the sailors commonly choose to direct their course by the rhumbs, or the bearing of the place by the compass. These bearings do not point out the shortest distance between places; because, on a globe, the rhumbs are spirals, and not arches of great circles. However, when the places lie directly under the equator, or exactly under the same meridian, the rhumb then coincides with the arch of a great circle, and of consequence shows the nearest way. The sailing on the arch of a great circle is called great circle sailing; and the cases of it depend all on the solution of problems in spherical trigonometry.

Containing the Various Methods of Sailing.

THE art of navigation depends upon astronomical and mathematical principles. The places of the sun and fixed stars are deduced from observation and calculation, and arranged in tables, the use of which is absolutely necessary in reducing observations taken at sea, for the purpose of ascertaining the latitude and longitude of the ship, and the variation of the compass. The problems in the various sailings are resolved either by trigonometrical calculation, or by tables or rules formed by the assistance of trigonometry. By mathematics, the necessary tables are constructed, and rules investigated for performing the more difficult parts of navigation. For these several branches of science, and for logarithmic tables, the reader is referred to the respective articles in this work. A few tables are given at the end of this article; but as the other tables necessary for the practice of navigation are to be found in almost every treatise on that subject, it therefore seems unnecessary to insert them in this place.

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

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

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

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

RULE. Subtract the less latitude from the greater, if the latitudes be of the same name, but add them if

of contrary; and the remainder or sum will be the difference of latitude.

Example 1. Required the difference of latitude between the Lizard, in latitude $49^\{\backslash circ\}\; 57\text{'}\; N.$ and Cape St Vincent, in latitude $37^\{\backslash circ\}\; 2\text{'}\; N.$

Difference of latitude $12^\{\backslash circ\}\; 55\text{'}\; =\; 775$ miles.

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

Difference of latitude $67^\{\backslash circ\}\; 7\text{'}\; =\; 4027$ miles.

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

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

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

Latitude come to $48^\{\backslash circ\}\; 42\text{'}\; N.$

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

Latitude come to $6^\{\backslash circ\}\; 55\text{'}\; S.$

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

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

Ex. 1. Required the difference of longitude between Edinburgh and New York, their longitudes being $3^\{\backslash circ\}\; 14\text{'}\; W.$ and $74^\{\backslash circ\}\; 10\text{'}\; W.$ respectively?

Difference of longitude $70^\{\backslash circ\}\; 56\text{'}\; W.$

Ex. 2. What is the difference of longitude between Maskelyne's Isles in longitude $167^\{\backslash circ\}\; 59\text{'}\; E.$ and Olinde, in longitude $35^\{\backslash circ\}\; 5\text{'}\; W.$?

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

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

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

Ex. 2. The longitude sailed from is 25° 9' W. and difference of longitude 18° 46' W.—Required the longitude come to?

SECT. II. Of the Tides.

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

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

PROB. I. To reduce the times of the moon's phases as given in the Nautical Almanac to the meridian of a known place.

RULE. To the time of the proposed phase, as given in the Nautical Almanac, apply the longitude of the place in time, by addition or subtraction, according as it is east or west, and it will give the time of the phase at the given place.

Ex. 1. Required the time of new moon at Salonique in May 1793?

Time of new moon required, in May 9 17 4
Ex. 2. What is the time of the last quarter of the moon at Resolution bay in October 1793?

PROB. II. To find the time of high water at a known place.

RULE. In the Nautical Almanac seek in the given month, or in that immediately preceding or following it, for the time of that phase which happens nearest to the given day: reduce the time of this phase to the meridian of the given place by Prob. I. and take the difference between the reduced time and the noon of the given day.

Find the equation answering to this difference in Table VII. which applied to the time of high water on the day of new or full moon at the given place, according as the table directs, will give the approximate time of high water in the afternoon.

Now, take the interval between the reduced time of the phase and the approximate time of high water; find the corresponding equation, which applied as before to the syzygy time of high water, will give the time of the afternoon high water.

If the time of the morning high water is required, increase the last interval by 12 hours, if the given day falls before the phase, or diminish it by 12 hours when after that phase; and the equation to this time, applied to the syzygy time, gives the morning time of high water.

Ex. 1. Required the morning and afternoon times of high water at Leith, 11th December 1793?

The time of high water found by the common method is about an hour and a half sooner.

Ex. 2. Required the time of high water at Funchal, 15th November 1793?

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

It has been already observed, that the interval between each knot on the line ought to be 50 feet, in order to adapt it to a glass that runs 30 seconds. But although the line and glass be at any time perfectly adjusted to each other, yet as the line shrinks after being wet, and as the weather has a considerable effect upon the glass, it will therefore be necessary to examine them from time to time; and the distance given by them must be corrected accordingly. The distance sailed may, therefore, be affected by an error in the glass, or in the line, or in both. The true distance may, however, be found as follows.

PROB. I. The distance sailed by the log, and the se-

conds run by the glass, being given, to find the true distance, the line being supposed right.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

In order to illustrate the above, let A (fig. 1.) represent the position of any given place, and AB the meridian passing through that place; also let AC represent the line described by a ship, and C the point arrived at. From C draw CB perpendicular to AB. Now in the triangle ABC, the angle BAC represents the course, the side AC the distance, AB the difference of latitude, and BC the departure.

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

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

If the difference of latitude is given, it is to be laid upon the meridian, beginning at the point representing

the place the ship left; and a line drawn from the extremity of the difference of latitude perpendicular to the meridian, till it meets the distance produced, will limit the figure.

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

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

Example. A ship from St Helena, in latitude 15° 55' S. sailed S. W. by S. 158 miles. Required the latitude come to, and departure.

Draw the meridian AB (fig. 2.), and with the chord of 60° describe the arch mn, and make it equal to the rhumb of 3 points, and through n draw AC equal to 158 miles; from C, draw CB perpendicular to AB; then AB applied to the scale from which AC was taken, will be found to measure 131.4 and BC 87.8.

To find the difference of latitude.

to the difference of latitude 131.4 - 2.11851

To find the departure.

to the departure 87.8 - 1.94340

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

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

Latitude come to - 18 6 S.

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

Example.

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

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

By Construction.

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

By Calculation.

To find the distance.

By Inspection.

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

By Gunter's Scale.

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

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

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

By Construction.

Make the departure BC (fig. 4.) equal to 192 miles, draw BA perpendicular to BC, and from the centre C, with the chord of $60^\{\backslash circ\}$, describe the arch $mn$, which make equal to the rhumb of 3 points, the complement of the course; draw a line through C n, which produce till it meets BA in A: then the distance AC being measured, will be equal to 231 m. and the difference of latitude AB will be 128.3 miles.

By Calculation.

To find the distance.

To find the difference of latitude.

By Inspection.

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

By Gunter's Scale.

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

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

Example. A ship from a place in latitude $43^\{\backslash circ\}\; 13\text{'}\; N$, sails between the north and east 285 miles; and is then by observation found to be in latitude $46^\{\backslash circ\}\; 31\text{'}\; N$: Required the course and departure?

By Construction.

Draw the portion of the meridian AB (fig. 5.) equal to 198 miles; from B draw BC perpendicular to AB: then take the distance 285 miles from the scale, and with one foot of the compass in A describe an arch intersecting BC in C, and join AC. With the chord of $60^\{\backslash circ\}$ describe the arch $mn$, the portion of which, contained between the distance and difference of latitude applied to the line of chords, will measure $46^\{\backslash circ\}$, the course; and the departure BC being measured on the line of equal parts, will be found equal to 205 miles.

By Calculation.

To find the course.

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

The extent from the distance 285 to the difference of latitude 198 on numbers, will reach from 90° to 44°, the complement of the course on sines; and the extent from 90° to the course 46° on the line of sines being laid from the distance 285, will reach to the departure 205 on the line of numbers.

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

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

Draw BC (fig. 6.) perpendicular to AB, and equal to the given departure 190 miles; then from the centre C, with the distance 260 miles, sweep an arch intersecting AB in A, and join AC. Now describe an arch from the centre A with the chord of 60°, and the portion m n of this arch, contained between the distance and difference of latitude, measured on the line of chords, will be 47° the course; and the difference of latitude AB applied to the scale of equal parts, measures 177½ miles.

Seek in the traverse table until the nearest to the given departure is found in the same line with the given distance 260. This is found to be in the page marked 47° at the bottom, which is the course; and the corresponding difference of latitude is 177.3.

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

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

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

Draw the portion of the meridian AB (fig. 7.) equal to the difference of latitude 247 miles; from B draw BC perpendicular to AB, and equal to the given departure 132 miles, and join AC: then with the chord of 60° describe an arch from the centre A; and the portion m n of this arch being applied to the line of chords, will measure about 28°; and the distance AC, measured on the line of equal parts, will be 280 miles.

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

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

If a ship sail upon two or more courses in a given time, the irregular track she describes is called a traverse; and to resolve a traverse, is the method of reducing these several courses, and the distances run, in-

to a single course and distance. The method chiefly used for this purpose at sea is by inspection, which shall therefore be principally adhered to; and is as follows.

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

Now, the several courses and their corresponding distances being properly arranged in the table, find the difference of latitude and departure answering to each in the traverse table; remembering that the difference of latitude is to be put in a north or south column, according as the course is in the northern or southern hemisphere; and that the departure is to be put in an east column if the course is easterly, but in a west column if the course is westerly: Observing also, that the departure is less than the difference of latitude when the course is less than 4 points or $45^\backslash circ$; otherwise greater.

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

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

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

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

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

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

EXAMPLES.

I. A ship from Fyal, in lat. $38^\backslash circ\; 32\text{'}\; N$, sailed as follows: ESE 163 miles, SW $\backslash frac\{1\}\{2\}\; W$ 110 miles, SE $\backslash frac\{1\}\{2\}\; S$ 180 miles, and N by E 68 miles. Required the latitude come to, the course, and distance made good.

By Inspection.

By Construction.

With the chord of $60^\backslash circ$ describe the circle NE, SW (fig. 8.), the centre of which represents the place the ship sailed from: draw two diameters NS, EW at right angles to each other; the one representing the meridian, and the other the parallel of latitude of the place sailed from. Take each course from the line of rhumbs, lay it off on the circumference from its proper meridian, and number it in order 1, 2, 3, 4. Upon the first rhumb $C\_1$, lay off the first distance 163 miles from C to A; through it draw the second distance AB parallel to $C\_2$, and equal to 110 miles; through B draw BD equal to 180 miles, and parallel to $C\_3$; and draw DE parallel to $C\_4$, and equal to 68 miles. Now CE being joined, will represent the distance made good; which applied to the scale will measure 281 miles. The arch S n, which represents the course, being measured on the line of chords, will be found equal to $41\backslash frac\{1\}\{2\}^\backslash circ$. From E draw EF perpendicular to CS produced; then CF will be the difference of latitude, and FE the departure made good; which applied to the scale will be found to measure 210 and 186 miles respectively.

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

II. A ship from latitude $1^\backslash circ\; 38\text{'}\; S$. sailed as under. Required her present latitude, course, and distance made good?

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

IV. The course per compass from Greigness (B) to the May is SW $\backslash frac\{1\}\{2\}$ S, distance 58 miles; from the May to the Staples S $\backslash frac\{1\}\{2\}$ E, 44 miles; and from the Staples to Flamborough Head S $\backslash frac\{1\}\{2\}$ E, 110 miles. Required the course per compass, and distance from Greigness to Flamborough Head?

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

CHAP. IV. Of Parallel Sailing.

THE figure of the earth is spherical, and the meridians gradually approach each other, and meet at the poles. The difference of longitude between any two places is the angle at the pole contained between the meridians of those places; or it is the arch of the

equator intercepted between the meridians of the given places; and the meridian distance between two places in the same parallel, is the arch thereof contained between their meridians. It hence follows, that the meridian distance, answering to the same difference of longitude, will be variable with the latitude of the parallel upon which it is reckoned; and the same difference of longitude will not answer to a given meridian distance when reckoned upon different parallels.

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

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

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

As radius

Is to the cosine of latitude;

So is any given portion of the equator

To a similar portion of the given parallel.

But the difference of longitude is an arch of the equator; and the distance between any two places under the same parallel, is a similar portion of that parallel.

Hence R : cosine latitude :: Diff. longitude : Distance.

And by inversion,

Cosine latitude : R :: Distance : Diff. of longitude.

Also,

Diff. of longitude : Distance :: R : cos. latitude.

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

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

By Construction.

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

(B) Greigness is about 2 $\backslash frac\{1\}\{2\}$ miles distant from Aberdeen, in nearly a SE $\backslash frac\{1\}\{2\}$ E direction.

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

To 56°, the nearest degree to the given latitude, and distance 60 miles, the corresponding difference of latitude is 33.6, which is the miles required.

The extent from 90° to 34°, the complement of the given latitude on the line of sines, will reach from 60 to 33.6 on the line of numbers.

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

Ex. 2. Required the distance between Treguier in France, in longitude 3° 14' W, and Gaspey Bay, in longitude 64° 27' W, the common latitude being 48° 47' N?

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

Example. A ship from Cape Finisterre, in latitude 42° 52' N, and longitude 9° 17' W, sailed due west 342 miles. Required the longitude come to?

Draw the straight line AB (fig. 11.) equal to the given distance 342 miles, and make the angle BAC equal to 42° 52', the given latitude: from B draw BC perpendicular to AB, meeting AC in C; then AC applied to the scale will measure 466$\backslash frac\{1\}\{2\}$, the difference of longitude required.

The nearest degree to the given latitude is 43°; under which, and opposite to 171, half the given distance in a latitude column, is 234, in a distance column, which doubled gives 468, the difference of longitude.

If the proportional part answering to the difference

between the given latitude and that used, be applied to the above, the same result with that found by calculation will be obtained.

The extent from 47° 8', the complement of latitude to 60° on the line of sines, being laid the same way from the distance 342, will reach to the difference of longitude 466$\backslash frac\{1\}\{2\}$ on the line of numbers.

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

Example. A ship sailed due east 358 miles, and was found by observation to have differed her longitude 8° 42'. Required the latitude of the parallel?

Make the line AB (fig. 12.) equal to the given distance; to which let BC be drawn perpendicular, with an extent equal to 522', the difference of longitude; describe an arch from the centre A, cutting BC in C; then the angle BAC being measured by means of the line of chords, will be found equal to 46°$\backslash frac\{2\}\{3\}$, the required latitude.

As the difference of longitude and distance exceed the limits of the table, let therefore the half of each be taken; these are 261 and 179 respectively. Now, by entering the table with these quantities, the latitude will be found to be between 46 and 47 degrees. Therefore, to latitude 46°, and distance 261 miles, the corresponding difference of latitude is 181$\backslash frac\{1\}\{3\}$, which exceeds the half of the given distance by 2$\backslash frac\{1\}\{3\}$. Again, to latitude 47°, and distance 261, the difference of latitude is 170.0, being 1$\backslash frac\{1\}\{3\}$ less than the half of that given: therefore the change of distance answering to a change of 1° of latitude is 3$\backslash frac\{1\}\{3\}$.

Now, as 3$\backslash frac\{1\}\{3\}$ : 2$\backslash frac\{1\}\{3\}$ :: 1° : 42'.

Hence the latitude required is 46° 42'.

The extent from 522 to 358 on the line of numbers, will reach from 90° to about 43°$\backslash frac\{1\}\{3\}$, the complement of which 46$\backslash frac\{2\}\{3\}$ is the latitude required.

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

Example. From two ports in latitude 33° 58' N, distance 348 miles, two ships sail directly north till they are in latitude 48° 23' N. Required their distance?

Draw the line CB, CE (fig. 13.), making angles with

with CP equal to the complements of the given latitudes, namely, $56^\{\backslash circ\}\; 2\text{'}$ and $41^\{\backslash circ\}\; 37\text{'}$ respectively: make BD equal to the given distance 348 miles, and perpendicular to CP; now from the centre C, with the radius CB, describe an arch intersecting CE in E; then EF drawn from the point E, perpendicular to CP, will represent the distance required; which being applied to the scale, will measure $278\backslash frac\{1\}\{2\}$ miles.

By Calculation.

By Inspection.

Under $34^\{\backslash circ\}$, and opposite to 174, half the given distance in a latitude column is 210 in a distance column; being half the difference of longitude answering thereto. Now, find the difference of latitude to distance 210 miles over $48^\{\backslash circ\}$ of latitude, which is $140\text{'}.5$; from which $1\text{'}.1$ (the proportional part answering to 23 minutes of latitude) being subtracted, gives $139\text{'}.4$ which doubled is $278\text{'}.8$, the distance required.

By Gunter's Scale.

The extent from $56^\{\backslash circ\}\; 2\text{'}$, the complement of the latitude left, to $41^\{\backslash circ\}\; 37\text{'}$, the complement of that come to, on the line of sines, being laid the same way from 348, will reach to $278\backslash frac\{1\}\{2\}$, the distance sought on the line of numbers.

PROB. V. Given a certain portion of a known parallel, together with a similar portion of an unknown parallel; to find the latitude of that parallel.

Example. Two ships, in latitude $56^\{\backslash circ\}\; 0\text{'}\; N$, distant 180 miles, sail due south; and having come to the same parallel, are now 232 miles distant. The latitude of that parallel is required?

By Construction.

Make DB (fig. 14.) equal to the first distance 180 miles, DM equal to the second 232, and the angle DBC equal to the given latitude $56^\{\backslash circ\}$; from the centre C, with the radius CB, describe the arch BE; and through M draw ME parallel to CD, intersecting the arch BE in E; join EC and draw EF perpendicular to CD: then the angle FEC will be the latitude required; which being measured, will be found equal to $43^\{\backslash circ\}\; 53\text{'}$.

By Calculation.

By Inspection.

To latitude $56^\{\backslash circ\}$, and half the first distance 90 in a latitude column, the corresponding distance is 161, which is half the difference of longitude. Now 161, and 116, half the second distance, are found to agree between 43 and 44 degrees; therefore, to latitude $43^\{\backslash circ\}$ and distance 161, the corresponding difference of latitude is $117\text{'}.7$; the excess of which above $116\text{'}$ is $1\text{'}.7$: and to latitude $44^\{\backslash circ\}$, and distance 161, the differ-

ence of latitude is $115\text{'}.8$: hence $117.7\; -\; 115.8\; =\; 1\text{'}.9$, the change answering to a difference of $1^\{\backslash circ\}$ of latitude. Therefore, $1\text{'}.9\; :\; 1\text{'}.7\; ::\; 1^\{\backslash circ\}\; :\; 53\text{'}$. Hence, the latitude is $43^\{\backslash circ\}\; 53\text{'}$.

By Gunter's Scale.

The extent from 180 to 232 on the line of numbers, being laid in the same direction on the line of sines, from $34^\{\backslash circ\}$, the complement of the latitude sailed from, will reach to $46^\{\backslash circ\}\; 7\text{'}$, the complement of the latitude come to.

CHAP. V. Of Middle Latitude Sailing.

THE earth is a sphere, and the meridians meet at the poles; and since a rhumb-line makes equal angles with every meridian, the line a ship describes is, therefore, that kind of a curve called a spiral.

Let AB (fig. 15.) be any given distance sailed upon an oblique rhumb, PBN, PAM the extreme meridians, MN a portion of the equator, and PCK, PEL, two meridians intersecting the distance AB in the points CE infinitely near each other. If the arches BS, CD, and AR, be described parallel to the equator, it is hence evident, that AS is the difference of latitude, and the arch MN of the equator, the difference of longitude, answering to the given distance AB and course PAB.

Now, since CE represents a very small portion of the distance AB, DE will be the correspondent portion of a meridian: hence the triangle EDC may be considered as rectilinear. If the distance be supposed to be divided into an infinite number of parts, each equal to CE, and upon these, triangles be constructed whose sides are portions of a meridian and parallel, it is evident these triangles will be equal and similar; for, besides the right angle, and hypotenuse which is the same in each, the course or angle CED is also the same. Hence, by the 12th of V. Euc. the sum of all the hypotenuses CE, or the distance AB, is to the sum of all the sides DE, or the difference of latitude AS, as one of the hypotenuses CE is to the corresponding side DE. Now, let the triangle GHI (fig. 16.) be constructed similar to the triangle CDE, having the angle G equal to the course: then as GH : GI :: CE : DC :: AB : AS.

Hence, if GH be made equal to the given distance AB, then GI will be the corresponding difference of latitude.

In like manner, the sum of all the hypotenuses CE, or the distance AB, is to the sum of all the sides CD, as CE is to CD, or as GH to HI, because of the similar triangles.

The several parts of the same rectilinear triangle will, therefore, represent the course, distance, difference of latitude, and departure.

Although the parts HG, GI, and angle G of the rectilinear triangle GHI, are equal to the corresponding parts AB, AS, and angle A, of the triangle ASB upon the surface of the sphere; yet HI is not equal to BS, for HI is the sum of all the arcs CD; but CD is greater than OQ, and less than ZX: therefore HI is greater than BS, and less than AR. Hence the difference of longitude MN cannot be inferred from the departure reckoned either upon the parallel sailed from, or on that come to, but on some intermediate parallel

TV, such that the arch TV is exactly equal to the departure: and in this case, the difference of longitude would be easily obtained. For TV is to MN as the sine PT to the sine PM; that is, as the cosine of latitude is to the radius.

The latitude of the parallel TV is not, however, easily determined with accuracy; various methods have, therefore, been taken in order to obtain it nearly, with as little trouble as possible: first, by taking the arithmetical mean of the two latitudes for that of the mean parallel: secondly, by using the arithmetical mean of the cosines of the latitudes: thirdly, by using the geometrical mean of the cosines of the latitudes: and lastly, by employing the parallel deduced from the mean of the meridional parts of the two latitudes. The first of these methods is that which is generally used.

In order to illustrate the computations in middle latitude sailing, let the triangle ABC (fig. 17.) represent a figure in plane sailing, wherein AB is the difference of latitude, AC the distance, BC the departure, and the angle BAC the course. Also, let the triangle DBC be a figure in parallel sailing in which DC is the difference of longitude, BC the meridian distance, and the angle DCB the middle latitude. In these triangles there is, therefore, one side BC common to both; and that triangle is to be first resolved in which two parts are given, and then the unknown parts of the other triangle will be easily obtained.

PROB. I. Given the latitudes and longitudes of two places, to find the course and distance between them.

Example. Required the course and distance from the island of May, in latitude $56^\{\backslash circ\}\; 12\text{'}\; N$, and longitude $2^\{\backslash circ\}\; 37\text{'}\; W$, to the Naze of Norway, in latitude $57^\{\backslash circ\}\; 50\text{'}\; N$, and longitude $7^\{\backslash circ\}\; 27\text{'}\; E$.

By Construction.

Draw the right line AD (fig. 18.) to represent the meridian of the May; with the chord of $60^\{\backslash circ\}$ describe the arch $mn$, upon which lay off the chord of $32^\{\backslash circ\}\; 59\text{'}$, the complement of the middle latitude from $m$ to $n$: from D through $n$ draw the line DC equal to $604\text{'}$, the difference of longitude, and from C draw CB perpendicular to AD: make BA equal to $98\text{'}$, the difference of latitude, and join AC; which applied to the scale will measure 343 miles, the distance sought: and the angle A being measured by means of the line of

chords, will be found equal to $73^\{\backslash circ\}\; 24\text{'}$, the required course.

By Calculation.

To find the course (D).

to the tangent of the cosine $73^\{\backslash circ\}\; 24\text{'}$ $10.52572$

To find the distance.

to the distance $343$ $2.53534$

By Inspection.

To middle latitude $57^\{\backslash circ\}$, and 151 one-fourth of the difference of longitude in a distance column, the corresponding difference of latitude is 82.2.

Now 24.5, one fourth of the difference of latitude, and 82.2, taken in a departure column, are found to agree nearest in table marked $6\backslash frac\{1\}\{2\}$ points at the bottom, which is the course; and the corresponding distance $85\backslash frac\{1\}\{2\}$ multiplied by 4 gives 343 miles, the distance required.

By Gunter's Scale.

The extent from 98 the difference of latitude, to 604 the difference of longitude on numbers, being laid the same way from $33^\{\backslash circ\}$, the complement of the middle latitude on sines, will reach to a certain point beyond the termination of the line on the scale. Now the extent between this point and $90^\{\backslash circ\}$ on sines, will reach from $45^\{\backslash circ\}$ to $73^\{\backslash circ\}\; 24\text{'}$, the course on the line of tangents. And the extent from $73^\{\backslash circ\}\; 24\text{'}$ the course, to $33^\{\backslash circ\}$ the complement of the middle latitude on the line of sines, being laid the same way from 604 the difference of longitude, will reach to 343 the distance on the line of numbers.

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

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

Example. A ship from Brest, in latitude $48^\{\backslash circ\}\; 23\text{'}\; N$, and longitude $4^\{\backslash circ\}\; 30\text{'}\; W$, sailed $SW\backslash frac\{1\}\{4\}W\; 238$ miles. Required the latitude and longitude come to?

By Construction.

With the course and distance construct the triangle ABC (fig. 17.), and the difference of latitude AB being measured, will be found equal to 142 miles: hence the latitude come to is $46^\{\backslash circ\}\; 1\text{'}\; N$, and the middle latitude $47^\{\backslash circ\}\; 12\text{'}$. Now make the angle DCB equal to

(D) For R. : cosine mid. lat. :: Diff. of long. : Departure;

And diff. of lat. : Dep. :: R. : Tangent course.

Hence diff. of lat. : cosine mid. lat. :: diff. of long. : tang. course;

Or diff. of lat. : diff. of long. :: cosine mid. lat. : tang. course.

middle to $47^\{\backslash circ\}\; 12\text{'}$; and DC being measured, will be 281, the middle difference of longitude: hence the longitude come to is $9^\{\backslash circ\}\; 11\text{'}\; W.$

To the course $4\backslash frac\{1\}\{2\}$ points, and distance 238 miles, the difference of latitude is 141.8, and the departure 191.1. Hence the latitude come to is $46^\{\backslash circ\}\; 1\text{'}\; N$, and middle latitude $47^\{\backslash circ\}\; 12\text{'}$. Then to middle latitude $47^\{\backslash circ\}\; 12\text{'}$, and departure 191.1 in a latitude column, the corresponding distance is 281, which is the difference of longitude.

The extent from 8 points to $3\backslash frac\{1\}\{2\}$ points, the complement of the course on sine rhumbs, being laid the same way from the distance 238, will reach to the difference of latitude 142 on the line of numbers; and the extent from $42^\{\backslash circ\}\; 48\text{'}$ the complement of the middle latitude, to $53^\{\backslash circ\}\; 26\text{'}$, the course on the line of sines, will reach from the distance 238 to the difference of longitude 281 on numbers.

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

Example. A ship from St Antonio, in latitude $17^\{\backslash circ\}\; 0\text{'}\; N$, and longitude $24^\{\backslash circ\}\; 25\text{'}\; W$, sailed NW $\backslash frac\{1\}\{2\}\; N$, till by observation her latitude is found to be $28^\{\backslash circ\}\; 34\text{'}\; N$. Required the distance sailed, and longitude come to?

Construct the triangle ABC (fig. 19.), with the given course and difference of latitude, and make the angle BCD equal to the middle latitude. Now the distance AC and difference of longitude DC being

measured, will be found equal to 864 and 558 respectively.

To course $3\backslash frac\{1\}\{2\}$ points, and difference of latitude 231.3 one third of that given, the departure is 171.6 and distance 288, which multiplied by 3 is 864 miles.

Again to the middle latitude $22^\{\backslash circ\}\; 47\text{'}$, or $23^\{\backslash circ\}$, and departure 271.6 in a latitude column, the distance is 186, which multiplied by 3 is 558, the difference of longitude.

The extent from $4\backslash frac\{1\}\{2\}$ points, the complement of the course, to 8 points on the line of sine rhumbs, will reach from the difference of latitude 694 to the distance 864 on numbers; and the extent from the course $36^\{\backslash circ\}\; 34\text{'}$ to $67^\{\backslash circ\}\; 13\text{'}$, the complement of middle latitude on sines, will reach from the distance 864 to the difference of longitude 558 on numbers.

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

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

With the course and departure construct the triangle ABC (fig. 20.), and the distance and difference of latitude being measured, will be found equal to 340 and 263 respectively. Hence the latitude come to is $30^\{\backslash circ\}\; 53\text{'}$, and middle latitude $28^\{\backslash circ\}\; 42\text{'}$. Now make the angle BCD equal to the middle latitude, and the difference of longitude DC applied to the scale will measure $246\text{'}$.

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

Under the course 3$\backslash frac\{1\}\{2\}$ points, and opposite to 108, half the departure, the distance is 170, and difference of latitude 131$\backslash frac\{1\}\{2\}$; which doubled, give 340 and 263 for the distance and difference of latitude respectively.

Again, to middle latitude 28$^\{\backslash circ\}$ 42', and departure 108, the distance is 123; which doubled is 246 the difference of longitude.

The extent from the course 3$\backslash frac\{1\}\{2\}$ points, on sine rhumbs, to the departure 216 on numbers, will reach from 8 points on sine rhumbs to about 340, the distance on numbers; and the same extent will reach from 4$\backslash frac\{1\}\{2\}$ points, the complement of the course, to 263, the difference of latitude on numbers; and the extent from 61$^\{\backslash circ\}$ 18' the complement of the middle latitude, to 90$^\{\backslash circ\}$ on sines, will reach from the departure 216 to the difference of longitude 246 on numbers.

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

Example. From Cape Sable, in latitude 43$^\{\backslash circ\}$ 24' N, and longitude 65$^\{\backslash circ\}$ 39' W, a ship sailed 246 miles on a direct course between the south and east, and is then by observation in latitude 40$^\{\backslash circ\}$ 48' N. Required the course and longitude in?

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

The distance 246, and difference of latitude 156, are found to correspond above 4$\backslash frac\{1\}\{2\}$ points, and the departure is 190.1. Now, to the middle latitude 42$^\{\backslash circ\}$, and departure 190.1 in a latitude column, the corresponding distance is 256, which is the difference of longitude required.

The extent from 246 miles, the distance, to 156, the difference of latitude on numbers, will reach from 90$^\{\backslash circ\}$ to about 39$^\{\backslash circ\}$$\backslash frac\{1\}\{2\}$, the complement of the course on the line of sines; and the extent from 48$^\{\backslash circ\}$, the complement of the middle latitude, to 50$^\{\backslash circ\}$$\backslash frac\{1\}\{2\}$, the course on sines, will reach from the distance 246m. to the difference of longitude 256m. on numbers.

PROB. VI. Given both latitudes and departure; sought the course, distance, and difference of longitude.

Example. A ship from Cape St Vincent, in latitude 37$^\{\backslash circ\}$ 2' N, longitude 9$^\{\backslash circ\}$ 2' W, sails between the south and west; the latitude come to is 18$^\{\backslash circ\}$ 16' N, and departure 838 miles. Required the course and distance run, and longitude come to?

Make AB (fig. 22.) equal to the difference of latitude 1126 miles, and BC equal to the departure 838, and join AC; draw CD so as to make an angle with CB equal to the middle latitude 27$^\{\backslash circ\}$ 39'. Then the course being measured on chords is about 36$^\{\backslash circ\}$$\backslash frac\{1\}\{2\}$, and the distance and difference of longitude, measured on the line of equal parts, will be found to be 1403 and 946 respectively.

To find the difference of longitude.

To find the difference of longitude.

By Inspection.

One tenth of the difference of latitude 112.6 and of the departure 83.8, are found to agree under 3½ points, and the corresponding distance is 140, which multiplied by 10 gives 1400 miles. And to middle latitude 27° ½, and 209.5 one fourth of the departure in a latitude column, the distance is 236.5; which multiplied by 4 is 946, the difference of longitude.

By Gunter's Scale.

The extent from the difference of latitude 112.6 to the departure 83.8 on numbers, will reach from 45° to 36° ½ the course on tangents; and the extent from 53° ½ the complement of the course to 90° on sines, will reach from 112.6 to 140.3 the distance on numbers. Lastly, the extent from 62° ½ the complement of the middle latitude, to 90° on sines, will reach from the departure 83.8 to the difference of longitude 946 on numbers.

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

Example. A ship from Bourdeaux, in latitude 44° 50' N, and longitude 0° 35' W, sailed between the north and west 374 miles, and made 210 miles of westing. Required the course, and the latitude and longitude come to?

By Construction.

With the given distance and departure make the triangle ABC (fig. 23.). Now the course being measured on the line of chords is about 34° ½, and the difference of latitude on the line of numbers is 309 miles: hence the latitude come to, is 49° 59' N, and middle latitude 47° 25'. Then make the angle BCD equal to 47° 25', and DC being measured will be 310 miles, the difference of longitude.

By Calculation.

To find the course.

To find the difference of latitude.

By Inspection.

The half of the distance 187, and of the departure 105, are found to agree nearest under 34°, and the difference of latitude answering thereto is 155; which doubled is 310 miles.

Again, to middle latitude 47° 25', and departure 105 in the latitude column, the corresponding distance is 155 miles, which doubled is 310 miles, the difference of longitude.

By Gunter's Scale.

The extent from the distance 374 miles to the departure 210 miles on the line of numbers, will reach from 90° to 34° 10', the course on the line of sines; and the extent from 90° to 55° 50', the complement of the course on sines, will reach from the distance 374 to the difference of latitude 309 miles on numbers.

Again, the extent from 42° 35', the complement of the middle latitude, to 90° on sines, will reach from the departure 210 to the difference of longitude 310 on numbers.

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

Example. A ship from latitude 54° 56' N, longitude 1° 10' W, sailed between the north and east, till by observation she is found to be in longitude 5° 26' E, and has made 220 miles of easting. Required the latitude come to, course, and distance run?

By Construction.

Make BC (fig. 24.) equal to the departure 220, and Fig. 24. CD equal to the difference of longitude 396:—then the middle latitude BCD being measured, will be found equal to 59° 15': hence the latitude come to is 57° 34', and difference of latitude 158°. Now make AB equal to 158, and join AC, which applied to the scale, will measure 271 miles. Also the course BAC being measured on chords will be found equal to 54° ½.

By Calculation.

To find the middle latitude.

As the differences of longitude and departure exceed the limits of the tables, let, therefore, their halves be taken; these are 198 and 110 respectively. Now these are found to agree exactly in the page marked 5 points at the bottom. Whence the middle latitude is 56° 15', and difference of latitude 158 miles.

Again, the difference of latitude 158 and departure 220 will be found to agree nearly above 54° the course, and the distance on the same line is 271 miles.

The extent from the difference of longitude 396 to the departure 220 on numbers, will reach from 90° to 33° 45', the complement of the middle latitude on sines; and hence the difference of latitude is 158 miles. Now the extent from 158 to 220 on numbers, will reach from 45° to 54° 19' on tangents; and the extent from the complement of the course 35° 41' to 90° on sines, will reach from the difference of latitude 158 to the distance 271 on numbers.

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

Example. A ship from a port in north latitude, sailed S E $\backslash frac\{1\}\{2\}$ S 438 miles, and differed her longitude 7° 28'. Required the latitude sailed from, and that come to?

With the course and distance construct the triangle ABC (fig. 25.), and make DC equal to 448 the given difference of longitude. Now the middle latitude BCD will measure 48° 58', and the difference of latitude AB 324 miles; hence the latitude left is 51° 40', and that come to 46° 16'.

To the course 3$\backslash frac\{1\}\{2\}$ points, and half the distance 219 miles, the departure is 147.0, and difference of latitude 162.2; which doubled is 324.4. Again, to half the difference of longitude 224 in a distance column, the difference of latitude is 149.9 above 48°, and 146.9 over 49°.

Now, as 30 : 29 :: 60' : 58'.

Hence the middle latitude is 48° 58': the latitude sailed from is therefore 51° 40', and latitude come to 46° 16'.

The extent from 8 points to 4$\backslash frac\{1\}\{2\}$ points, the complement of the course on sine rhumbs, will reach from the distance 438 miles to the difference of latitude 324.5 on numbers. And the extent from the difference of longitude 448, to the distance 438 on numbers, will reach from the course 42° 11' to the complement of the middle latitude 41° 2' on sines. Hence the latitude left is 51° 40', and that come to 46° 16'.

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

RULE I. With the several courses and distances find the difference of latitude and departure made good, and the ship's present latitude, as in traverse sailing.

Now enter the traverse table with the given middle latitude, and the departure in a latitude column, the corresponding distance will be the difference of longitude, of the same name with the departure.

Example. A ship from Cape Clear, in latitude 51° 18' N, longitude 9° 46' W, sailed as follows:—SW $\backslash frac\{1\}\{2\}$ S 34 miles, W $\backslash frac\{1\}\{2\}$ N 63 miles, NNW 48 miles, and NE $\backslash frac\{1\}\{2\}$ E 85 miles. Required the latitude and longitude come to?

The above method is not always practised to find the difference of longitude made good in the course of

of a day's run; and will, no doubt, give the difference of longitude tolerably exact in any probable run a ship may make in that time, especially near the equator. But in a high latitude, when the distances are considerable, this method is not to be depended on.—To illustrate this, let a ship be supposed to sail from latitude $57^\{\backslash circ\}N$, as follows: E 240 miles, N 240 miles, W 240 miles, and S 240 miles: then, by the above method, the ship will be come to the same place she left. It will, however, appear evident from the following consideration, that this is by no means the case; for let two ships, from latitude $61^\{\backslash circ\}N$, and distant 240 miles, sail directly south till they are in latitude $57^\{\backslash circ\}N$; now their distance being computed by Problem IV. of Parallel Sailing, will be 269.6 miles; and, therefore, if the ship sailed as above, she will be 29.6 miles west of the place sailed from; and the error in longitude will be equal to $240\; \backslash times\; \backslash secant\; 61^\{\backslash circ\}\; -\; \backslash secant\; 57^\{\backslash circ\}\; =\; 29.6\; \backslash times\; \backslash secant\; 57^\{\backslash circ\}\; =\; 54.4$.

Theorems might be investigated for computing the Mercator's errors to which the above method is liable. These corrections may, however, be avoided, by using the following method.

RULE 11. Complete the traverse table as before, to which annex five columns; the first column is to contain the several latitudes the ship is in at the end of each course and distance; the second, the sums of each following pair of latitude; the third, half the sums, or middle latitudes; and the fourth and fifth columns are to contain the differences of longitude.

Now find the difference of longitude answering to each middle latitude and its corresponding departure, and put them in the east or west difference of longitude columns, according to the name of the departure. Then the difference of the sums of the east and west columns will be the difference of longitude made good, of the same name with the greater.

Example. A ship from Halliford in Iceland, in lat. $64^\{\backslash circ\}\; 30\text{'}\; N$, long. $27^\{\backslash circ\}\; 15\text{'}\; W$, sailed as follows: SSW 46 miles, SW 61 miles, S$\backslash frac\{1\}\{2\}$W 59 miles, SE$\backslash frac\{1\}\{2\}$E 86 miles, S$\backslash frac\{1\}\{2\}$E$\backslash frac\{1\}\{2\}$E 79 miles. Required the lat. and long. come to?

CHAP. VI. Of Mercator's Sailing.

It was observed in Middle Latitude Sailing, that the difference of longitude made upon an oblique rhumb could not be exactly determined by using the middle latitude. In Mercator's sailing, the difference of longitude is very easily found, and the several problems of sailing resolved with the utmost accuracy, by the assistance of Mercator's chart or equivalent tables.

In Mercator's chart, the meridians are straight lines parallel to each other; and the degrees of latitude, which at the equator are equal to those of longitude, increase with the distance of the parallel from the equator. The parts of the meridian thus increased are called meridional parts. A table of these parts was first constructed by Mr Edward Wright, by the continual addition of the secants of each minute of latitude.

For by parallel sailing,
$R\; :\; \backslash cos.\; \backslash text\{\; of\; lat.\}\; ::\; \backslash text\{\; part\; of\; equat.\}\; :\; \backslash text\{\; similar\; part\; of\; parallel.\}$
4 P 2 And

And because the equator and meridian on the globe are equal; therefore,

$R\; :\; \backslash cos.\; \backslash text\{\; lat.\}\; ::\; \backslash text\{\; part\; of\; meridian\}\; :\; \backslash text\{\; similar\; part\; of\; parallel.\}$
Or $\backslash sec.\; \backslash text\{\; lat.\}\; :\; R\; ::\; \backslash text\{\; part\; of\; merid.\}\; :\; \backslash text\{\; similar\; part\; of\; parallel.\}$

Hence, $\backslash frac\{\backslash secant\; \backslash text\{\; latitude\}\}\{\backslash text\{part\; of\; meridian\}\}\; =\; \backslash frac\{R\}\{\backslash text\{part\; of\; parallel\}\}$

But in Mercator's chart the parallels of latitude are equal, and radius is a constant quality. If, therefore, the latitude be assumed successively equal to and the corresponding parts of the enlarged meridian be represented by ; then,

Hence

Therefore by 12th V. Euclid.

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

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

And in the triangle CED,

$ED\; :\; CD\; ::\; R\; :\; \backslash text\{tangent\; CED\};$

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

Hence $LK\; =\; \backslash frac\{ED\; \backslash times\; \backslash sec.\; LD\; \backslash times\; \backslash text\{tang.\; CED\}\}\{R^2\}$

$\backslash frac\{ED\; \backslash times\; \backslash sec.\; LD\}\{R\}\; \backslash times\; \backslash frac\{\backslash text\{tang.\; CED\}\}\{R\}$

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

of all the ED's contained in AS, will be the meridional parts answering to the difference of latitude AS; and MN is the sum of all the corresponding portions of the equator LK.

Whence $MN\; =\; \backslash text\{mer.\; diff.\; of\; lat.\}\; \backslash times\; \backslash text\{tangent\; \}\; \backslash frac\{CED\}\{R\}$.

That is, the difference of longitude is equal to the meridional difference of latitude multiplied by the tangent of the course, and divided by the radius.

This equation answers to a right-angled rectilinear triangle, having an angle equal to the course; the adjacent side equal to the meridional difference of latitude, and the opposite side the difference of longitude. This triangle is, therefore, similar to a triangle constructed, with the course and difference of latitude, according to the principles of plane sailing, and the homologous sides will be proportional. Hence, if, in fig. 26 the angle A represents the course, AB the difference of latitude, and if AD be made equal to the meridional difference of latitude; then DE, drawn perpendicular to AD, meeting the distance produced to E, will be the difference of longitude.

It is scarcely necessary to observe, that the meridional difference of latitude is found by the same rules as the

proper difference of latitude; that is, if the given latitudes be of the same name, the difference of the corresponding meridional parts will be the meridional difference of latitude; but if the latitudes are of a contrary denomination, the sum of these parts will be the meridional difference of latitude.

PROB. I. Given the latitudes and longitudes of two places, to find the course and distance between them.

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

Difference of lat. = $27^\backslash circ\; 58\text{'}$ Mer. diff. lat. 1948

Diff. longitude = $14^\backslash circ\; 12\text{'}\; =\; 852$.

By Construction.

Draw the straight line AD (fig. 26.) to represent the Fig. 26. meridian of Cape Finisterre, upon which lay off AB, AD equal to 1678, and 1948, the proper and meridional differences of latitude; from D draw DE perpendicular to AD, and equal to the difference of longitude 852, join AE, and draw BC parallel to DE; then the difference AC will measure 1831 miles, and the course BAC $23^\backslash circ\; 37\text{'}$.

By Calculation.

To find the course.

By Inspection.

As the meridian difference of latitude and difference of longitude are too large to be found in the tables, let the tenth of each be taken; these are 194.8 and 85.2 respectively. Now these are found to agree nearest under $24^\backslash circ$; and to 167.8, one-tenth of the proper difference of latitude, the distance is about 183 miles, which multiplied by 10 is 1830 miles.

By Gunter's Scale.

The extent 1948, the meridional difference of latitude, to 852, the difference of longitude on the line of numbers, will reach from $45^\backslash circ$ to $23^\backslash circ\; 37\text{'}$, the course

(E) This is not strictly true; for instead of taking the sum of the secants of every minute in the distance of the given parallel from the equator, the sum of the secants of every point of latitude should be taken.

Mer. parts on the line of tangents. And the extent from $66^\{\backslash circ\}\; 23\text{'}$, the complement of the course to $90^\{\backslash circ\}$ on sines, will reach from 1678, the proper difference of latitude, to 1831, the distance on the line of numbers.

PROB. II. Given the course and distance, sailed from a place whose situation is known, to find the latitude and longitude of the place come to.

Example. A ship from Cape Hinlopen in Virginia, in latitude $38^\{\backslash circ\}\; 47\text{'}\; N$, longitude $75^\{\backslash circ\}\; 4\text{'}\; W$, sailed 267 miles NEbN. Required the ship's present place?

By Construction.

With the course and distance sailed, construct the triangle ABC (fig. 27.); and the difference of latitude AB being measured, is 222 miles: hence the latitude come to is $42^\{\backslash circ\}\; 29\text{'}\; N$, and the meridional difference of latitude 293. Make AD equal to 293; and draw DE perpendicular to AD, and meeting AC produced in E: then, the difference of longitude DE being applied to the scale of equal parts will measure 196; the longitude come to is therefore $71^\{\backslash circ\}\; 48\text{'}\; W$.

By Calculation.

To find the difference of latitude.

To find the difference of longitude.

By Inspection.

To the course 3 points, and distance 267 miles, the difference of latitude is 222 miles: hence the latitude in is $42^\{\backslash circ\}\; 29\text{'}$, and the meridional difference of latitude 293. Again, to course three points, and 146.5 half the mer. difference of latitude, the departure is 97.9, which doubled is 195.8, the difference of longitude.

By Gunter's Scale.

The extent from 8 points to the complement of the course 5 points, on sine rhumbs, will reach from the distance 267 to the difference of latitude 222 on numbers; and the extent from 4 points to 3 points on tangent rhumbs, will reach from the meridional difference of latitude 293 to the difference of longitude 196 on numbers.

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

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

By Construction.

Make AB (fig. 28.) equal to 246, and AD equal Fig. 28. to 338; draw AE, making an angle with AD equal to $3\backslash frac\{1\}\{4\}$ points, and draw BC, DE perpendicular to AD. Now AC being applied to the scale, will measure 332, and DE 306.

By Calculation.

To find the distance.

To find the difference of longitude.

By Inspection.

Under the course $3\backslash frac\{1\}\{4\}$ points, and opposite to half the difference of latitude 123 in a latitude column is 166 in a distance column, which doubled is 332 the distance; and opposite to 169, half the meridional difference of latitude in a latitude column, is 153 in a departure column, which doubled is 306, the difference of longitude.

By Gunter's Scale.

The extent from the complement of the course $4\backslash frac\{1\}\{4\}$ points to 8 points on sine rhumbs, will reach from the difference of latitude 246 m. to the distance 332 on numbers; and the extent from 4 points, to the course $3\backslash frac\{1\}\{4\}$ points on tangent rhumbs, will reach from the meridional difference of latitude 338 to the difference of longitude 306 on numbers.

PROB. IV. Given the latitude and longitude of the place sailed from, the course and departure; to find the distance, and the latitude and longitude of the place come to.

Example. A ship sailed from Sallee in latitude $33^\{\backslash circ\}\; 58\text{'}\; N$, longitude $6^\{\backslash circ\}\; 20\text{'}\; W$, the corrected course was NWbW $\backslash frac\{1\}\{4\}\; W$, and departure 420 miles. Required the distance run, and the latitude and longitude come to?

By Construction.

With the course and departure construct the triangle ABC (fig. 29.); now AC and AB being measured, Fig. 29. will be found to be equal to 476 and 224 respectively: hence

Mercurator's Sailing. Hence the latitude come to is $37^\{\backslash circ\}\; 42\text{'}\; N$, and meridional difference of latitude 276. Make AD equal to 276; and draw DE perpendicular thereto, meeting the distance produced in E; then DE applied to the scale will be found to measure 516'. The longitude in is, therefore, $14^\{\backslash circ\}\; 56\text{'}\; W$.

Above $5\backslash frac\{1\}\{2\}$ points the course, and opposite to 210 half the departure, are 238 and 112; which doubled, we have 476 and 224, the distance and difference of latitude respectively. And to the same course, and opposite to 138, half the meridional difference of latitude, in a latitude column, is 258 in a departure column; which being doubled is 516, the difference of longitude.

The extent from $5\backslash frac\{1\}\{2\}$ points, the course on sine rhumbs, to the departure 420 on numbers, will reach from eight points on sine rhumbs to the distance 476 on numbers; and from the complement of the course $2\backslash frac\{1\}\{2\}$ points on sine rhumbs, to the difference of latitude 224 on numbers.

Again, the extent from difference of latitude 224 to the meridional difference of latitude 276 on numbers, will reach from the departure 420 to the difference of longitude 516 on the same line.

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

Example. A ship from St Mary's, in latitude $36^\{\backslash circ\}\; 57\text{'}\; N$, longitude $25^\{\backslash circ\}\; 9\text{'}\; W$, sailed on a direct course between the north and east 1162 miles, and is then by observation in latitude $49^\{\backslash circ\}\; 57\text{'}\; N$. Required the course steered, and longitude come to?

Make AB (fig. 30.) equal to 780, and AD equal to 1081; draw BC, DE perpendicular to AD; make AC equal to 1162', and through AC draw ACE. Then the course or angle A being measured, will be found equal to $47^\{\backslash circ\}\; 50\text{'}$, and the difference of longitude DE will be 1194.

Because the distance and difference of latitude exceed the limits of the table, take the tenth of each; these are 116.2 and 78.0: Now these are found to agree nearest above $4\backslash frac\{1\}\{2\}$ points, which is therefore the course; and to this course, and opposite to 108.1, one tenth of the meridional difference of latitude, in a latitude column, is 119.3 in a departure column, which multiplied by 10 is 1193, the difference of longitude.

The extent from the distance 1162 m. to the difference of latitude 780 m. on numbers, will reach from $90^\{\backslash circ\}$ to $42^\{\backslash circ\}\; 10\text{'}$ in the line of sines. And the extent $45^\{\backslash circ\}$, to the course $47^\{\backslash circ\}\; 50\text{'}$ on the line of tangents, will reach from the meridional difference of latitude 1081 to the difference of longitude 1194 on numbers.

PROB. VI. Given the latitudes of two places, and the departure, to find the course, distance, and difference of longitude.

Example. From Aberdeen, in latitude $57^\{\backslash circ\}\; 9\text{'}\; N$, longitude $2^\{\backslash circ\}\; 8\text{'}\; W$, a ship sailed between the south and east till her departure is 146 miles, and latitude come to $53^\{\backslash circ\}\; 32\text{'}\; N$. Required the course and distance run, and longitude come to?

With the difference of latitude 217 m. and departure 146 m. construct the triangle ABC (fig. 31.), make AD

Mercator's Sailing. AD equal to 382, draw DE parallel to BC, and produce AC to E: Then the course BAC will measure $33^\{\backslash circ\}\; 56\text{'}$, the distance AC 261, and the difference of longitude DE 257.

to the tangent of the course $33^\{\backslash circ\}\; 56\text{'}$ - 9.82789

to the distance - 261.5 - 2.41755

to the difference of longitude 257 - 2.40995

Longitude of Aberdeen - - - $2^\{\backslash circ\}\; 8\text{'}\; W$

Difference of longitude - - - 4 17 E

Longitude come to - - - 2 9 E

The difference of latitude 217, and departure 146, are found to agree nearest under $34^\{\backslash circ\}$, and the corresponding distance is 262 miles. To the same course, and opposite to 190.7, the nearest to 191, half the meridional difference of latitude, is 128.6 in a departure column, which doubled is 257, the difference of longitude.

The extent from the difference of latitude 217, to the departure 146 on numbers, will reach from $45^\{\backslash circ\}$ to about $34^\{\backslash circ\}$, the course on the line of tangents; and the same extent will reach from the meridional difference of latitude 382 to 257, the difference of longitude on numbers.—Again, the extent from the course $34^\{\backslash circ\}$ to $90^\{\backslash circ\}$ on sines, will reach from the departure 146 to the distance 261 on numbers.

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

Example. A ship from Naples, in latitude $40^\{\backslash circ\}\; 51\text{'}\; N$, longitude $14^\{\backslash circ\}\; 14\text{'}\; E$, sailed 252 miles on a direct course between the south and west, and made 173 miles of westing. Required the course made good, and the latitude and longitude come to?

With the distance and departure make the triangle ABC (fig. 32.) as formerly.—Now the course BAC being measured by means of a line of chords will be found equal to $43^\{\backslash circ\}\; 21\text{'}$, and the difference of latitude applied to the scale of equal parts will measure 183: hence the latitude come to is $37^\{\backslash circ\}\; 48\text{'}\; N$, and meridional difference of latitude 237.—Make AD equal to 237, and complete the figure, and the difference of

longitude DE will measure 224: hence the longitude in is $10^\{\backslash circ\}\; 30\text{'}\; E$.

to the sine of the course $43^\{\backslash circ\}\; 21\text{'}$ - 9.83665

to the difference of latitude 183.2 - 2.26304

Latitude of Naples $40^\{\backslash circ\}\; 51\text{'}\; N$. Mer. parts 2690

Difference of latitude 3 3 S.

Latitude come to - 37 48 N. Mer. parts 2453

to the difference of longitude 223.7 - 2.34972

Longitude of Naples - - - $14^\{\backslash circ\}\; 14\text{'}\; E$

Difference of longitude - - - 3 44 W

Longitude in - - - $10^\{\backslash circ\}\; 30\text{'}\; E$

Under $43^\{\backslash circ\}$ and opposite to the distance 252 m. the departure is 171.8, and under $44^\{\backslash circ\}$, and opposite to the same distance, the departure is 175.0.

Then as $3.2\; :\; 1.2\; ::\; 60\text{'}\; :\; 22\text{'}$.

Hence the course is $43^\{\backslash circ\}\; 21\text{'}$.

Again, under $43^\{\backslash circ\}$ and opposite to 118.5, half the meridional difference of latitude in a latitude column, is 110.5 in a departure column; also under $44^\{\backslash circ\}$ and opposite to 118.5 is 114.4.

Then as $3.2\; :\; 1.2\; ::\; 3.9\; :\; 1.5$.

And $110.5\; +\; 1.5\; =\; 112$, which doubled is 224, the difference of longitude.

The extent from the distance 252 on numbers, to $90^\{\backslash circ\}$ on sines, will reach from the departure 173 on numbers, to the course $43^\{\backslash circ\}\; 21\text{'}$ on sines; and the same extent that will reach from the complement of the course $46^\{\backslash circ\}\; 39\text{'}$ on sines will reach to the difference of latitude 183 on numbers.—Again, the extent from $45^\{\backslash circ\}$ to $43^\{\backslash circ\}\; 21\text{'}$ on tangents will reach from the meridional difference of latitude 237, to the difference of longitude 224, on numbers.

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

Example. A ship from Tercera, in latitude $38^\{\backslash circ\}\; 45\text{'}\; N$, longitude $27^\{\backslash circ\}\; 6\text{'}\; W$, sailed on a direct course, which, when corrected, was N $32^\{\backslash circ\}\; E$, and is found by observation to be in longitude $18^\{\backslash circ\}\; 24\text{'}\; W$. Required the latitude come to, and distance sailed?

Longitude

Fig. 33. Make the right-angled triangle ADE (fig. 33.) having the angle A equal to the course 32°, and the side DE equal to the difference of longitude 522: then AD will measure 835, which added to the meridional parts of the latitude left, will give those of the latitude come to 48° 46'; hence, the difference of latitude is 601: make AB equal thereto, to which let BC be drawn perpendicular; then AC applied to the scale will measure 708 miles.

To find the meridional difference of latitude.

Difference of latitude 10 = 601 miles.

To find the distance.

To course 32°, and opposite to 130.5, one fourth of the given difference of longitude in a departure column, the difference of latitude is 208.8, which multiplied by 4 is 835, the meridional difference of latitude; hence the latitude in is 48° 46' N, and difference of latitude 601.

Again, to the same course, and opposite to 200, one third of the difference of latitude, the distance is 236, which multiplied by 3 gives 708 miles.

The extent from the course 32°, to 45° on tangents, will reach from the difference of longitude 522 to the meridional difference of latitude 835 on numbers.— And the extent from the complement of the course 58° to 90° on sines, will reach from the difference of latitude 601, to the distance 708 miles on numbers.

PROB. IX. To find the difference of longitude made good upon compound courses.

RULE. With the several courses and distances, complete the Traverse Table, and find the difference of latitude, departure, and course made good, and the latitude come to as in Traverse Sailing. Find also the meridional difference of latitude.

Now to the course and meridional difference of latitude, in a latitude column, the corresponding departure will be the difference of longitude, which applied to the longitude left will give the ship's present longitude.

Example. A ship from port St Julian, in latitude 49° 10' S, longitude 68° 44' W, sailed as follows; Mercator's Sailing. ESE 53 miles, SE by S 74 miles, E by N 68 m. SE by E 47 miles, and E 84 miles. Required the ship's present place?

Although the above method is that usually employed at sea to find the difference of longitude, yet as it has been already observed, it is not to be depended on, especially in high latitudes, long distances, and a considerable variation in the courses, in which case the following method becomes necessary.

RULE II. Complete the Traverse Table as before, to which annex five columns. Now with the latitude left, and the several differences of latitude, find the successive latitudes, which are to be placed in the first of the annexed columns; in the second, the meridional parts corresponding to each latitude is to be put; and in the third, the meridional differences of latitude.

Then to each course, and corresponding meridional difference of latitude, find the difference of longitude, by PROB. VI. which place in the fourth or fifth columns, according as the coast is easterly or westerly; and the difference between the sums of these columns will be the difference of longitude made good upon the whole, of the same name with the greater.

1. When the course is north or south, there is no difference of longitude.

2. When the course is east or west, the difference of longitude cannot be found by Mercator's Sailing; in this case the following rule is to be used.

To the nearest degree to the given latitude taken as a course, find the distance answering to the departure in a latitude column; this distance will be the difference of longitude.

Ex. 1. Four days ago we took our departure from Faro-head, in latitude $58^\{\backslash circ\}\; 40\text{'}\; N$, and longitude $4^\{\backslash circ\}\; 50\text{'}\; W$, Mercator's Sailing. and since have sailed as follows: NW 32 miles, W 69 miles, WNW 93 miles, W$\backslash frac\{1\}\{2\}$S 77 miles, SW 58 miles, and W$\backslash frac\{1\}\{2\}$S 49 miles.—Required our present latitude and longitude?

Ex. 2. A ship from latitude $78^\{\backslash circ\}\; 15\text{'}\; N$, longitude $28^\{\backslash circ\}\; 14\text{'}\; E$, sailed the following courses and distances. The latitude come to is required, and the longitude, by both methods: the bearing and distance of Hacluit's head-land, in latitude $79^\{\backslash circ\}\; 55\text{'}\; N$, longitude $11^\{\backslash circ\}\; 55\text{'}\; E$, is also required?

Now to 78.5 half the meridional difference of latitude, and 185.0 half the difference of longitude, the course $67^\{\backslash circ\}$, and opposite to the difference of latitude 27, the distance is 69 miles.—Hence Hacluit's head-land bears S $67^\{\backslash circ\}\; E$, distant 69 miles.

CHAP. VII. Containing the Method of resolving the several Problems of Mercator's Sailing, by the Assistance of a Table of Logarithmic Tangents.

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

RULE. To the arithmetical complement of the logarithm of the distance, add the logarithm of the difference of longitude in minutes, and the log. cosine of the given latitude; the sum rejecting radius will be the log. sine of the approximate course.

To the given latitude taken as a course in the traverse table, add half the difference of longitude in a distance column; the corresponding departure will be

the first correction of the course, which is subtractive Method if the given latitude is the least of the two; otherwise, resolve the Problems of Mercator's Sailing.

In Table A, under the complement of the course, and opposite to the first correction in the side column, is the second correction. In the same table find the number answering to the course at the top, and difference of longitude in the side column; and such part of this number being taken as is found in Table B opposite to the given latitude, will be the third correction. Now these two corrections, subtracted from the course corrected by the first correction, will give the true course.

Now the course and distance being known, the difference of latitude is found as formerly.

Example. From latitude 50° N, a ship sailed 290 miles between the south and west, and differed her longitude 5°. Required the course, and latitude come to?

To find the difference of latitude.

This problem was proposed, and resolved, by Mr Robert Hues in his Treatise on the Globes, printed at London in the year 1639, p. 181.

It was afterwards proposed by Dr Halley, in the second volume of the Miscellanea Curiosa, p. 34. in the following words:

A ship sails from a given latitude, and, having run a certain number of leagues, has altered her longitude by a given angle; it is required to find the course steered. And he then adds—The solution hereof would be very acceptable, if not to the public, at least to the author of this tract, being likely to open some further light into the mysteries of geometry.

Since that time, this problem has been solved in an indirect manner, by several writers on navigation, and others:—As Monsieur Bouguer, in his Nouveau Traité de Navigation; Mr Robertson, in the second volume of his Elements of Navigation; Mr Emerson, in his Theory of Navigation, which accompanies his Mathematical Principles of Geography; Mr Israel Lyons, in the Nautical Almanack for 1772; and Monsieur Bezout, in his Traité de Navigation; and lately, Baron Maseres, with the assistance of Mr Attwood, has given the first direct solution of this problem. For a comparison of the various solutions which have hitherto been made of this problem, the reader is referred to that by Dr Mackay, in the fourth and sixth volumes of Baron Maseres's Scriptores Logarithmici.

It was intended in this place to have given rules, to make allowance for the spheroidal figure of the earth: but as the ratio of the polar to the equatorial semi-axis is not as yet determined with sufficient accuracy, neither is it known if both hemispheres be similar figures; therefore these rules would be grounded on assumption only, and might probably err more from the truth

Oblique truth than those adapted to the spherical hypothesis. This therefore is supposed to be a sufficient apology for not inserting them.

CHAP. VIII. Of Oblique Sailing.

OBLIQUE sailing is the application of oblique-angled plane triangles to the solution of problems at sea. This sailing will be found particularly useful in going along shore, and in surveying coasts and harbours, &c.

Ex. 1. At 11h A. M. the Girdle Ness bore WNW, and at 2h P. M. it bore NWbN: the course during the interval SSW, five knots an hour. Required the distance of the ship from the Ness at each station?

By Construction.

Describe the circle NE, SW (fig. 34.), and draw the diameters NS, EW, at right angles to each other: from the centre C, which represents the first station, draw the WNW line CF; and from the same point draw CH, SSW, and equal to 15 miles the distance sailed.—From H draw HF in a NWbN direction, and the point F will represent the Girdle Ness. Now the distances CF, HF will measure 19.1 and 26.5 miles respectively.

By Calculation.

In the triangle FCH are given the distance CH 15 miles, the angle FCH equal to 9 points, the interval between the SSW and WNW points, and the angle CHF equal to 4 points, being the supplement of the angle contained between the SSW and NWbN points; hence CFH is 3 points: to find the distances CF, HF.

To find the distance CF.

To find the distance FH.

Ex. 2. The distance between the SE point of the island of Jersey and the island of Brehaut is 13 leagues: and the correct bearing and distance of Cape Frehel from the island of Brehaut is SEbE 26 miles. It is also known that the SE point of Jersey bears NNE from Cape Frehel: from whence the distance of these two is required, together with the bearing of the same point from the island of Brehaut?

By Construction.

Describe a circle, (fig. 35.) and draw two diameters at right angles, the extremities of which will represent the cardinal points, north being uppermost.—Let the centre B represent Brehaut, from which draw the SEbE line BF equal to 26 miles, and the point F will represent Cape Frehel, from which draw the NNE line FI; make BI equal to 39 miles: Then FI applied to the scale will measure 34½ miles, and the inclination of BI to the meridian will be found equal to 63°½.

By Calculation.

In the triangle BIF are given BI and BF equal to 39 miles, and 26 miles respectively; and the angle BFI equal to 7 points: To find the side FI, and angle FBI.

To find the angle BIF.

Bearing of Jersey from Brehaut N 63° 20' E.

To find the distance FI.

Ex. 3. At noon Dungeness bore per compass NW, distance 5 leagues; and having run NWbW 7 knots an hour, at 5 P. M. we were up with Beachyhead. Required the bearing and distance of Beachyhead from Dungeness?

By Construction.

Describe a circle (fig. 36.) to represent the horizon; from the centre C draw the NW line CD equal to 15 miles; and the NWbW line CB equal to 35 miles; join DB, which applied to the scale will measure about 26½ miles; and the inclination of DB to the meridian will be found equal to N 79°½ W.

By Calculation.

In the triangle DCB are given the distances CD, CB equal to 15 and 35 miles respectively; and the angle BCD equal to 4 points; to find the angles B and D, and the distance BD.

To find the angles.

Distance CB=35, sum of the ang. 16 points.

CD=15, angle C - 4

Magnetic bearing - N 79° 45' W. Or by allowing 2½ points of westerly variation, the true bearing of Beachyhead from Dungeness will be W½S nearly.

Ex. 4. Running up Channel E&S per compass at the rate of 5 knots an hour. At 11^h A. M. the Eddystone lighthouse bore N6E$\backslash frac\{1\}\{2\}$E, and the Start point NEbE$\backslash frac\{1\}\{2\}$E; and at 4 P. M. the Eddystone bore NWbN, and the Start N$\backslash frac\{1\}\{2\}$E. Required the distance and bearing of the Start from the Eddystone, the variation being 2$\backslash frac\{1\}\{2\}$ points W?

By Construction.

Fig. 37. Let the point C (fig. 37.) represent the first station, from which draw the N6E$\backslash frac\{1\}\{2\}$E line CA, the NEbE$\backslash frac\{1\}\{2\}$E line CB, and the E&S line CD, which make equal to 25 miles the distance run in the elapsed time; then from D draw the NEbN line DA intersecting CA in A, which represents the Eddystone; and from the same point draw the N$\backslash frac\{1\}\{2\}$E line DB cutting CB in B, which therefore represents the Start. Now the distance AB applied to the scale will measure 22.9, and the bearing per compass BAF will measure 73° 7'.

By Calculation.

In the triangle CAD are given CD equal to 25 miles, the angle CAD equal to 4$\backslash frac\{1\}\{2\}$ points, the distance between N6E$\backslash frac\{1\}\{2\}$E and NWbN; and the angle ADC equal to 4 points, the distance between the NWbN and WbN points; to find the distance CA.

In the triangle BCD, are given the distance CD 25 miles, the angle CBD 4$\backslash frac\{1\}\{2\}$ points the interval between NEbE$\backslash frac\{1\}\{2\}$E and N$\backslash frac\{1\}\{2\}$E, and CDB 7$\backslash frac\{1\}\{2\}$ points, the distance between WbN and N$\backslash frac\{1\}\{2\}$E; to find the distance CD.

In the triangle CAB, the distances CA, CB, are given, together with the included angle ACB, equal to 4 points, the distance between N6E$\backslash frac\{1\}\{2\}$E and NEbE$\backslash frac\{1\}\{2\}$E; to find the angle CAB and distance AB.

Bearing per compass S 73 22 E or ESE$\backslash frac\{1\}\{2\}$E; and

the variation 2$\backslash frac\{1\}\{2\}$ points being allowed to the left of ESE$\backslash frac\{1\}\{2\}$E, gives E$\backslash frac\{1\}\{2\}$N, the true bearing of the Start from the Eddystone.

To find the distance.

Ex. 5. A ship from a port in latitude 57° 9' N, longitude 2° 9' W, sailed 82 miles on a direct course, and spoke a ship that had run 100 miles from a port in latitude 56° 21' N, longitude 2° 50' W. Required the course of each ship, and the latitude and longitude come to?

By Construction.

With the meridional difference of latitude, the difference of longitude, and difference of latitude, construct the triangles ADE, ABC (fig. 38.) as in Mer. Fig. 31. cator's Sailing; then A will represent the northernmost, and C the southernmost port. The distance AC applied to the scale will measure 53 miles, and the bearing BCA will be 25° 14'. From the points A and C, with distances equal to 82 and 100 miles respectively, describe arches intersecting each other in M, which will therefore be the place of meeting. Now the angle ABM, the ship's course from the southernmost port, will measure N 80° 1$\backslash frac\{1\}\{2\}$E; and the other ship's course, or angle BAM, will be 67° 4$\backslash frac\{1\}\{2\}$, or ESE. From M draw the parallel MNP, and AN will be the difference of latitude made by the one ship, and CP that by the other ship: hence either of these being measured and applied to its correspondent latitude, will give 56° 38', the latitude in. Make AF equal to 57, the meridional difference of latitude between the northernmost port and latitude in: from F draw FG perpendicular to AF, and produce AM to G, then FG will be the difference of longitude, which applied to the scale will measure 139: hence the longitude in, is 0° 10' E.

By Calculation.

In the triangle ADE, ABC, are given AD equal to 87, DE equal to 41, and AB equal to 48; to find the angle BAC and distance AC.

In the triangle AMC, the three sides are given to find the angles.

In the right-angled triangle AMN, given AM, and the angle MAN, to find the differences of latitude AN.

WINDWARD sailing is, when a ship by reason of a contrary wind is obliged to sail on different tacks in order to gain her intended port; and the object of this sailing is to find the proper course and distance to be run on each tack.

Ex. 1. A ship is bound to a port 48 miles directly to the windward, the wind being SSW, which it is intended to reach on two boards; and the ship can lie

within six points of the wind. Required the course and distance on each tack?

Draw the SSW line CB (fig. 39.) equal to 48 miles. Fig. 39. Make the angles ACB, ABC, each equal to 6 points. Hence the first course will be W, and the second SE: also the distance CA, or AB, applied to the scale, will measure 62½ miles, the distance to be sailed on each board.

From A draw AD perpendicular to BC; then in the triangle ADC are given CD equal to 24 miles; and the angle ACD, equal to 6 points, to find the distance AC.

Ex. 2. The wind at NW, a ship bound to a port 64 miles to the windward, proposes to reach it on three boards; two on the starboard, and one on the larboard tack, and each within 5 points of the wind. Required the course and distance on each tack?

Draw the NW line CA (fig. 40.) equal to 64 miles; Fig. 40. from C draw CB WbS, and from A draw AD parallel thereto, and in an opposite direction; bisect AC in E, and draw BED parallel to the N6E rhumb, meeting CB, AD in the points B and D: then CB=AD applied to the scale will measure 36½ miles, and BD=2CB=72½ miles.

From B draw BF perpendicular to AC; then, in the triangle BFC are given the angle BCF equal to 5 points, and CF equal to one-fourth of CA=16 m. to find CB.

Ex. 3. A ship which can lie within 5½ points of the wind, is bound to a port 36 miles to the windward, the wind being NEbN, which it is intended to reach on four boards, the first being on the larboard tack. Required the course and distance on each?

Draw the NEbN line CA (fig. 41.) equal to 36 miles, and bisect it in B; from C and B draw lines parallel to the E½S rhumb; and from A and B draw lines parallel to the SSE½E point, meeting the former in the points D and E. Now the distances AD, BD, BE, and CE, are equal; and any one of them applied to the scale will measure 19.1 miles.

From E draw EF perpendicular to AC; and in the triangle CFE are given CF=9 m. and the angle FCE=5½ points, to find CE.

Ex. 4. A ship bound to a port bearing N$\backslash frac\{1\}\{2\}$W distant 40 miles, with the wind at N$\backslash frac\{1\}\{2\}$E$\backslash frac\{1\}\{2\}$E, intends to reach it on two boards. Required the course and distance on each tack, the ship lying within 5$\backslash frac\{1\}\{2\}$ points of the wind?

By Construction.

Fig. 42. Draw the N$\backslash frac\{1\}\{2\}$W line CA (fig. 42.) equal to 40 miles; and because the wind is N$\backslash frac\{1\}\{2\}$E$\backslash frac\{1\}\{2\}$E, and the ship can lie within 5$\backslash frac\{1\}\{2\}$ points of the wind, the course on the larboard tack will be E$\backslash frac\{1\}\{2\}$N, and on the starboard NW. Therefore, from the centre C draw the E$\backslash frac\{1\}\{2\}$N line CB, and from it draw the NW line AB, meeting CB in B; then CB and AB applied to the scale will measure 26.7 and 48.1 m. respectively.

By Calculation.

In the triangle ACB, given AC=40 miles, and the angles A, B, and C, equal to 3, 5, and 8 points respectively, to find AB and BC.

To find the distance CB.

Ex. 5. A ship close hauled within 5 points of the wind, and making one point of leeway, is bound to a port bearing SSW distant 54 miles, the wind being S$\backslash frac\{1\}\{2\}$E: It is intended to make the port at three boards, the first of which must be on the larboard tack in order to avoid a reef of rocks. Required the course and distance on each tack?

By Construction.

Fig. 43. Draw the SSW line CA (fig. 43.) equal to 54 m. and as the wind is S$\backslash frac\{1\}\{2\}$E, and the ship makes her course good within 6 points of the wind, therefore the course on the larboard tack will be SW$\backslash frac\{1\}\{2\}$W, and on the starboard E$\backslash frac\{1\}\{2\}$S: hence from C draw the SW$\backslash frac\{1\}\{2\}$W line CB, and from A draw AD parallel thereto; bisect CA in E, and draw BED parallel to the E$\backslash frac\{1\}\{2\}$S line; then will CB and AD be the distances on the larboard tack, which applied to the scale, each will be found to measure 37.4; and the distance on the starboard tack BD will measure 42.4 miles.

By Calculation.

The triangles CBE, EAD are equal and similar: hence in the first of these are given CE, equal to 27 miles, half the distance between the ship and port; the angles C, B, and E, equal to 3, 4, and 9 points respectively, to find CB and BE.

Ex. 6. A ship plying to the windward, with the wind at NNE, after sailing 51 miles on each of two tacks, is found by observation to have made 36 miles of difference of latitude. How near the wind did she make her way good?

By Construction.

Make CA (fig. 44.) equal to 36 miles; draw AB perpendicular to CA, and draw the NNE line CB, meeting AB in B; make CD, BD each equal to 51 miles, and these being measured, will be found equal to 6 points.

By Calculation.

In the triangles CAB, BCD, are given AB equal to 36 m. CD=BD=51, and the angle ACB equal to 2 points; to find the angle BCD.

CHAP. X. Of Current Sailing.

THE computations in the preceding chapters have been performed upon the assumption that the water has no motion. This may no doubt answer tolerably well in those places where the ebbings and flowings are regular, as then the effect of the tide will be nearly counterbalanced. But in places where there is a constant current or setting of the sea towards the same point, an allowance for the change of the ship's place arising therefrom must be made: And the method of resolving these problems, in which the effect of a current, or heave of the sea, is taken into consideration, is called current sailing.

In a calm, it is evident a ship will be carried in the direction and with the velocity of the current. Hence, if a ship sails in the direction of the current, her rate will be augmented by the rate of the current; but if sailing directly against it, the distance made good will be equal to the difference between the ship's rate as given by the log and that of the current. And the absolute motion of the ship will be a-head, if her rate exceeds that of the current; but if less, the ship will make sternway. If the ship's course be oblique to the current, the distance made good in a given time will be represented by the third side of a triangle, whereof the distance given by the log, and the drift of the current in the same time, are the other sides; and the true course will be the angle contained between the meridian and the line actually described by the ship.

Current Sailing. Ex. 1. A ship sailed NNE at the rate of 8 knots an hour, during 18 hours, in a current setting NW$\backslash frac\{1\}\{2\}$W$\backslash frac\{1\}\{2\}$ miles an hour. Required the course and distance made good?

Plate CCLXXV. Fig. 45. By Construction. Draw the NNE line CA (fig. 45.) equal to $18\; \backslash times\; 8\; =\; 144$ miles; and from A draw AB parallel to the NW$\backslash frac\{1\}\{2\}$W rhumb, and equal to $18\; \backslash times\; \backslash frac\{1\}\{2\}\; =\; 45$ miles: now BC being joined will be the distance, and NCB the course. The first of these will measure 159 miles, and the second $6^\{\backslash circ\}\; 23\text{'}$.

By Calculation. In the triangle ACB, are given AC=144 miles, AB=45 miles, and the angle CAB=9 points, to find BAC and BC.

Ex. 2. A ship from a port in latitude $42^\{\backslash circ\}\; 52\text{'}\; N$, sailed S$\backslash frac\{1\}\{2\}$W$\backslash frac\{1\}\{2\}$W 17 miles in 7 hours, in a current setting between the north and west; and then the same port bore ENE, and the ship's latitude by observation was $42^\{\backslash circ\}\; 42\text{'}\; N$. Required the setting and drift of the current?

By Construction. Draw the S$\backslash frac\{1\}\{2\}$W$\backslash frac\{1\}\{2\}$W line CA (fig. 46.) equal to 17 miles, and make CB equal to 10 miles, the difference of latitude: through B draw the parallel of latitude BD, and draw the WSW line CD, intersecting BD in D: AD being joined, will represent the drift of the current, which applied to the scale will measure 20.2, and the angle DAE will be its setting, and will be found equal to $72^\{\backslash circ\}$.

By Calculation. In the triangle CBD, given CB=10 miles, and the angle BCD=6 points; to find the distance CD.

Again, In the triangle ACD are given the distance AC=17 miles, CD=26.13, and the angle ACD $4\backslash frac\{1\}\{2\}$ points; to find the remaining parts.

To find the setting of the current.
Distance DC=26.13 Angle ACD=$4\backslash frac\{1\}\{2\}$ points.
Distance AC=17.0 CAD+CDA $11\backslash frac\{1\}\{2\}$

Setting of the current EAD = $71^\{\backslash circ\}\; 55\text{'}$.
To find the drift of the current.
As the sine of CAD - $88^\{\backslash circ\}\; 47\text{'}$ - 9.99990
is to the sine of ACD - $4\backslash frac\{1\}\{2\}$ points - 9.88819
so is the distance CD - 26.13 - 1.41710

Ex. 3. A ship, from latitude $38^\{\backslash circ\}\; 20\text{'}\; N$, sailed 24 hours in a current setting NW$\backslash frac\{1\}\{2\}$N, and by account is in latitude $38^\{\backslash circ\}\; 42\text{'}\; N$, having made 44 miles of easting; but the latitude by observation is $38^\{\backslash circ\}\; 58\text{'}\; N$. Required the course and distance made good, and the drift of the current.

By Construction. Make CE (fig. 47.) equal to 22 miles, the difference of latitude by D, R, and EA=54 miles, the departure, and join CA; make CD=38 miles, the difference of latitude by observation; draw the parallel of latitude DB, and from A draw the NW$\backslash frac\{1\}\{2\}$N line AB, intersecting DB in B, and AB will be the drift of the current in 24 hours: CB being joined, will be the distance made good, and the angle DCB the true course. Now, AB and CB applied to the scale, will measure 19.2 and 50.5 respectively: and the angle DCB will be $41^\{\backslash circ\}\; \backslash frac\{1\}\{2\}$.

By Calculation. From B draw BF perpendicular to AE, then in the triangle AFB are given BF=16 miles, and the angle ABF=3 points; to find AB and AF.

True departure EF=DB=33.31
Now,

Now, in the triangle CDB are given the difference of latitude and departure; to find the course and distance.

To find the course.

to the tangent of the course - $41^\{\backslash circ\}\; 14\text{'}$ - 9.94279

To find the distance.

to the distance - - - 50.53 - 1.70354

Ex. 4. In the Straits of Sunda, at 2 P. M. steering SE $\backslash frac\{1\}\{2\}$ S at the rate of 5 knots an hour, I passed close by the small islands off Hog point. At 6, not having changed our course, came to anchor on the Java shore. Upon setting the said island from this anchoring place, I find it bears due north, its distance by the chart being 22 miles. It follows from hence, that our course has been affected by a current. Required its velocity and direction?

By Construction.

Fig. 48. From A (fig. 48.) draw the SE $\backslash frac\{1\}\{2\}$ S line AB=20, which will represent the ship's apparent track through the water; draw AC equal to 22 miles south, and C will be the ship's real place; and BC being joined will be the current's drift in four hours; which applied to the scale will measure 12.3; from A draw AD parallel to BC, and the angle CAD will be the direction of the current, and will be found to measure $64^\{\backslash circ\}\; \backslash frac\{1\}\{2\}$.

By Calculation.

In the triangle ABC, given AB=20 m. AC=22 m. and the included angle A=3 points; to find the remaining parts.

To find the setting of the current.

Distance AC=22 m. Included angle =3 points.

Setting of the current S $64^\{\backslash circ\}\; 12\text{'}\; W$, or SW $W\backslash frac\{1\}\{2\}\; W$.

To find the drift of the current.

to the velocity of cur. BC 12 34 - 1.09137

and $\backslash frac\{12.34\}\{4\}=3.1$, its hourly rate.

Example. 5. A ship bound from Dover to Calais, lying 21 miles to the SE $\backslash frac\{1\}\{2\}$ E, and the flood tide setting NE $\backslash frac\{1\}\{2\}$ E $2\backslash frac\{1\}\{2\}$ miles an hour. Required the course

she must steer, and the distance run by the log at 6 knots an hour to reach her port?

By Construction.

In the position of the SE $\backslash frac\{1\}\{2\}$ E rhumb, draw DC = 21 miles (fig. 49.); draw DE NE $\backslash frac\{1\}\{2\}$ E = $2\backslash frac\{1\}\{2\}$ miles; from E with 6 miles cut DC in F; draw DB parallel to EF, meeting CB drawn parallel to DE: then the distance DB applied to the scale will measure 19.4, and Fig. 49. the course SDB will be SE $\backslash frac\{1\}\{2\}$ S.

By Calculation.

In the triangle DBF, given DE=2$\backslash frac\{1\}\{2\}$ miles, EF=6 miles and the angle EDF=6 points; to find the angle DFE=CBD.

to the sine of DFE - - - 22 38 - 9.58541

Angle - SDC = $5\backslash frac\{1\}\{2\}$ points = 61 52

Course SDB - - - 39 14 = SE $\backslash frac\{1\}\{2\}$ S.

In the triangle DBC, given DC=21 miles, the angle BDC=DFE=22$^\{\backslash circ\}$ 38', and the angle DCB=DEF=6 points; to find the distance DB.

to the distance by the log DB. 21 m. 1.28785

CHAP. XI. Instruments proposed to solve the various Problems in Sailing, independent of Calculation.

VARIOUS methods, besides those already given, have been proposed to save the trouble of calculation.— One of these methods is by means of an instrument composed of rulers, so disposed as to form a right-angled triangle, having numbers in a regular progression marked on their sides. These instruments are made of different materials, such as paper, wood, brass, &c. and are differently constructed, according to the fancy of the inventor. Among instruments of this kind, that by John Cooke, Esq. seems to be the best. A number of other instruments, very differently constructed, have been proposed for the same purpose; of these, however, we shall only take notice of the rectangular instrument, by A. Mackay, LL.D. F.R.S.E. &c.

I. Of COOKE'S Triangular Instrument.

Description. The stock a b c d (fig. 50.) is a parallelopiped: The length from a to b is two feet, the breadth from a to d two inches, and the depth is one inch and a half. The stock is perforated longitudinally, so as to be capable of containing within it e f, a cylindrical piece of wood one inch diameter; g h is an aperture on the surface of the stock about a quarter of an inch wide, which discloses one-twelfth part of the surface of the cylinder contained; the edge d c is divided into twelve parts, each of these is subdivided into six parts, and each of these again into ten parts. The surface of the cylinder is divided longitudinally into twelve parts, and on each of them is engraved a portion

portion of a line of meridional parts 22 feet long, which contains the meridional parts for every minute from the equator as far towards the pole as navigation is practicable; and the smallest division on it is not less than $\backslash frac\{1\}\{12\}$th of an inch. By rolling and sliding this cylinder, any part of any line on it may be brought into any position which may be required; the box $i$ is engraved into the edge of the stock $a\; b$, so that it may move freely from $a$ to $b$; a limb from this box extends to $k$, which serves to mark that degree of the perpendicular $i\; l$ which is parallel to the centre of the semicircle $m$; $i\; l$ is two feet long, and graduated on both edges as the stock; it is perpendicular to the stock, and is fixed in the box $i$, by which it may be moved from $a$ to $b$; $o\; p\; n$ is a semicircle of six inches radius, engraved, as appears in the plate, which slides freely from $c$ to $d$ in a groove in the edge of the stock $c\; d$; $m\; q$ is the index moving on the centre $m$, the edge of which marks the course on the semicircle; it is two feet long, and divided into 72 parts; and these are subdivided in the same manner as those on the stock and perpendicular, to which they are equal; $r$ is a vernier attached to the index to show minutes; $S$ is a vernier composed of concentric semicircles, which slides along the edge $q\; m$, to the intersection of the perpendicular and index, where it serves as a vernier to both; below $x$ is a small piece of ivory, with a mark on it to point out the degree of the line $d\; c$, which is perpendicularly under the centre of the semicircle. Fig. 51. is a view of the back part of the instrument.

Use. The method of working every case which occurs in navigation, is to make the instrument similar to that ideal triangle which is composed of the difference of latitude, departure, and distance; or, to that composed of the meridional difference of latitude, difference of longitude, and enlarged distance; or, to that composed of the difference of longitude, departure, and sine of the middle latitude; which is done by means of the data procured from the compass, log-line, and quadrant: whence it follows, from the nature of similar triangles, or from the relation which exists between the sides of triangles and the sines of their opposite angles, that the parts of the instrument become proportional to those which they represent; and will ascertain the length of the lines, or the extent of the angles sought, by its gradations.

In the practice of this instrument, a small square is necessary in order to bring the centre of the semicircle perpendicularly over the meridional degree corresponding to the latitude.

Plane Sailing.

PROB. I. The course and distance sailed being given, to find the difference of latitude and departure.

Example. A ship from latitude $24^\{\backslash circ\}\; 18\text{'}\; N$, sailed $NW\backslash frac\{1\}\{2\}N$ 168 miles. Required the latitude come to, and departure?

Set the centre of the semicircle perpendicularly over the given latitude $24^\{\backslash circ\}\; 18\text{'}$, and the index to the course 3 points; move the perpendicular until it cut the index at the given distance 168; then at the point of intersection on the perpendicular is 93.3 miles, the departure, and on the base, by the edge of the box, is $26^\{\backslash circ\}\; 38\text{'}$, the latitude come to.

PROB. II. Both latitudes and course given, to find the distance and departure.

Example. Let the latitude sailed from be $43^\{\backslash circ\}\; 50\text{'}\; N$, that come to $47^\{\backslash circ\}\; 8\text{'}\; N$, and the course $NNE$. Required the distance and departure?

Move the centre of the semicircle to the latitude left $43^\{\backslash circ\}\; 50\text{'}$, and the edge of the box to the latitude come to $47^\{\backslash circ\}\; 8\text{'}$; fix the index at the given course 2 points; then at the point of intersection of the index and perpendicular is the distance 214 miles on the index, and the departure 82 miles on the perpendicular.

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

Example. Let the latitude sailed from be $32^\{\backslash circ\}\; 38\text{'}\; N$, the course $SW\backslash frac\{1\}\{2\}S$, and the departure 200 miles. Required the distance and latitude come to?

Move the centre of the semicircle to the latitude left $32^\{\backslash circ\}\; 38\text{'}$, set the index to the given course 3 points, and move the perpendicular till the given departure 200 cuts the index; at this point on the index is 360 miles, and the edge of the box will cut the latitude come to $27^\{\backslash circ\}\; 39\text{'}\; N$.

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

Example. Let the latitude left be $17^\{\backslash circ\}\; 10\text{'}\; N$, the latitude come to $21^\{\backslash circ\}\; 40\text{'}\; N$, and the distance sailed on a direct course between the north and west 300 miles. Required the course and departure?

Move the semicircle and box to the given latitudes, and the index until the distance found thereon meets the perpendicular; then at the point of contact on the perpendicular is 130.8, the departure, and on the semicircle by the index is $25^\{\backslash circ\}\; 50\text{'}$, the course.

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

Example. The distance sailed is 246 miles between the south and east, the departure is 138 miles, and the latitude left $51^\{\backslash circ\}\; 10\text{'}\; N$. Required the course and latitude come to?

Set the centre of the semicircle to $51^\{\backslash circ\}\; 10\text{'}$, the latitude sailed from; find the distance 246 on the index, and the departure 138 on the perpendicular; then move both till these points meet, and the course $34^\{\backslash circ\}\; 10\text{'}$ will be found on the semicircle by the index, and the latitude in $47^\{\backslash circ\}\; 47\text{'}\; N$, by the edge of the box.

PROB. VI. Both latitudes and departure given, to find the course and distance.

Example. A ship from latitude $43^\{\backslash circ\}\; 10\text{'}\; N$, sailed between the north and west till she was in latitude $47^\{\backslash circ\}\; 14\text{'}\; N$, and has made 170 miles of departure. Required the course and distance?

Move the centre of the semicircle over $43^\{\backslash circ\}\; 10\text{'}$, and the edge of the box to $47^\{\backslash circ\}\; 14\text{'}$; find the departure on the perpendicular, and bring the edge of the index thereto; now at the point of intersection is the distance 297.4 miles on the index, and the course $34^\{\backslash circ\}\; 52\text{'}$ on the semicircle.

Traverse Sailing.

Example. A ship from latitude $46^\{\backslash circ\}\; 48\text{'}\; N$, sailed $SSW\backslash frac\{1\}\{2\}W$ 24 miles, $S\backslash frac\{1\}\{2\}W$ 36 miles, and $S\backslash frac\{1\}\{2\}E$ 40 miles.

miles. Required the latitude in, together with the direct course and distance?

Set the semicircle to the latitude sailed from $46^\{\backslash circ\}\; 48\text{'}$, and the index to the course $SSW\backslash frac\{1\}\{2\}W$; mark the distance 24 on the index, and bring the perpendicular to meet it; then the index will cut the departure 11.3 on the perpendicular, and the perpendicular will cut the latitude $46^\{\backslash circ\}\; 27\text{'}\; N$ on the base. For the next course and distance, bring the semicircle to the latitude marked by the perpendicular, and lay down the course $SW$: if it be towards the first meridian, move the last marked departure until it meets the index, and the limb of the box will mark the present departure; but if the course be from the first meridian, bring the last departure 11.3 to the limb of the box, the index will mark the departure made good 18.3 on the perpendicular, and the latitude arrived at $45^\{\backslash circ\}\; 52\text{'}$ will be marked on the base by the perpendicular: proceed in the same manner with all the courses of which the traverse consists, then the difference of latitude $1^\{\backslash circ\}\; 36\text{'}$ will be intercepted between the latitude sailed from $46^\{\backslash circ\}\; 48\text{'}$, and the latitude come to $45^\{\backslash circ\}\; 12\text{'}$ last marked by the perpendicular; and also the departure made good will be intercepted between that point on the perpendicular where the first departure commenced, and that where the last terminated. Now, with the difference of latitude $1^\{\backslash circ\}\; 36\text{'}$ and the departure, the course will be $S\; 8^\{\backslash circ\}\; 30\text{'}\; W$, and distance 97 miles, by last problem in Plane Sailing.

Parallel Sailing.

PROB. I. The difference of longitude between two places in one parallel of latitude given, to find the distance between them.

Example. Let the common latitude be $49^\{\backslash circ\}\; 30\text{'}\; N$, and the difference of longitude $3^\{\backslash circ\}\; 30\text{'}$. Required the distance?

Set the index to $49^\{\backslash circ\}\; 30\text{'}$, the complement of the latitude on the semicircle; mark the difference of longitude in miles on the index; then move the perpendicular until it meets the termination of the difference of longitude on the index, and the part of the perpendicular intercepted between the limb of the box and the point of intersection will be the distance 136.4 miles.

PROB. II. The distance between two places in one parallel of latitude given, to find the difference of longitude between them.

Example. Let the latitude of the given parallel be $49^\{\backslash circ\}\; 30\text{'}\; N$, the distance sailed 136.4 E. Required the difference of longitude?

Set the index to the complement of the latitude $49^\{\backslash circ\}\; 30\text{'}$, and mark the distance sailed on the perpendicular; then move it until it meets the index, and the point of intersection will show the difference of longitude $210^\{\backslash circ\}$ or $3^\{\backslash circ\}\; 30\text{'}$ on the index.

PROB. III. Given the distance sailed on a parallel, and the difference of longitude, to find the latitude of that parallel.

Example. The distance sailed due east is 136.4, and the difference of longitude $3^\{\backslash circ\}\; 30\text{'}$. Required the latitude of the parallel?

Find the difference of longitude $210^\{\backslash circ\}$ on the index, and the distance 136.4 on the perpendicular, and move

both until these numbers meet, and the complement of the latitude $4^\{\backslash circ\}\; 30\text{'}$ will be shown by the index on the semicircle.

Mercator's and Middle Latitude Sailing.

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

Example. Required the course and distance between two places whose latitudes and longitudes are $50^\{\backslash circ\}\; 30\text{'}\; N$, $19^\{\backslash circ\}\; 0\text{'}\; W$, and $54^\{\backslash circ\}\; 30\text{'}\; N$, $15^\{\backslash circ\}\; 30\text{'}\; W$, respectively?

By Mercator's Sailing.

To find the course.

Move the centre of the semicircle perpendicularly over the meridional degree answering to latitude $50^\{\backslash circ\}\; 30\text{'}\; N$, then move the box until the edge of the perpendicular cuts the meridional parts of the other latitude $54^\{\backslash circ\}\; 30\text{'}\; N$, and move the index until it cuts the difference of longitude $3^\{\backslash circ\}\; 30\text{'}$ on the perpendicular, and the index will mark the course $30^\{\backslash circ\}\; 10\text{'}$, or $NNE\backslash frac\{1\}\{2\}E$ nearly on the semicircle.

To find the distance.

Screw the index to this course, and move the centre of the semicircle to the latitude $50^\{\backslash circ\}\; 30\text{'}\; N$, and the edge of the perpendicular to the latitude $54^\{\backslash circ\}\; 30\text{'}\; N$, then the perpendicular will cut the distance 254.7 on the index.

By Middle Latitude Sailing.

To find the departure.

Move the centre of the semicircle to the latitude $50^\{\backslash circ\}\; 30\text{'}$, and the edge of the index to the complement of the middle latitude $37^\{\backslash circ\}\; 20\text{'}$ on the semicircle; then move the box until the edge of the perpendicular intersects the termination of the difference of longitude $210$ miles on the index, which point of intersection will mark the departure 128 on the perpendicular.

To find the course and distance.

Move the edge of the perpendicular to the other latitude $54^\{\backslash circ\}\; 30\text{'}$, and the index until it cuts the departure 128 on the perpendicular; then will the perpendicular mark the distance on the index 254.7 miles, and the index will mark the course on the semicircle $30^\{\backslash circ\}\; 10\text{'}$, or $NNE\backslash frac\{1\}\{2\}E$ nearly.

PROB. II. Both latitudes and course given, to find the distance and difference of longitude.

Example. A ship from latitude $50^\{\backslash circ\}\; 30\text{'}\; N$, longitude $19^\{\backslash circ\}\; 0\text{'}\; W$, sailed $N\; 30^\{\backslash circ\}\; 10\text{'}\; E$, until she is in latitude $54^\{\backslash circ\}\; 30\text{'}\; N$. Required the distance and difference of longitude?

By Mercator's Sailing.

To find the difference of longitude.

Move the box and semicircle as in the former problem to the meridional parts of the given latitudes, then set the index to the course, and it will mark the difference of longitude $3^\{\backslash circ\}\; 30\text{'}$ on the perpendicular: Hence the longitude in is $15^\{\backslash circ\}\; 30\text{'}\; W$.

To find the distance.

Move the perpendicular and semicircle to the given latitudes, and put the index to the given course; then the perpendicular will cut the distance 254.7 miles on the index.

Move the semicircle and perpendicular to the given latitudes, and the index to the course; then the perpendicular will show the departure 128 miles, and the index the distance 254.7 miles at the point of intersection.

Set the index to the complement of the middle latitude on the semicircle, and move the box until the termination of the departure on the perpendicular meets the index, which will mark the difference of longitude thereon 210 m. or $3^\{\backslash circ\}\; 30\text{'}$.

PROB. III. Both latitudes and distance given, to find the course and difference of longitude.

Example. From latitude $50^\{\backslash circ\}\; 50\text{'}\; N$, longitude $19^\{\backslash circ\}\; 0\text{'}\; W$, a ship sailed 254.7 miles between the north and east, and by observation is in latitude $54^\{\backslash circ\}\; 30\text{'}\; N$. Required the course and difference of longitude?

Move the perpendicular and semicircle to the given latitudes, and the index until the distance sailed marked on it meets the perpendicular; then the index will mark the course $N\; 30^\{\backslash circ\}\; 10\text{'}\; E$ on the semicircle.

Screw the index to the course, move the perpendicular and semicircle to the meridional parts of the given latitudes, and the space intercepted between the limb of the box and the index will be the difference of longitude $3^\{\backslash circ\}\; 30\text{'}$.

Move the semicircle and perpendicular to the given latitudes, and the index until the distance sailed on it cuts the perpendicular; then the perpendicular will show the departure 128 miles, and the semicircle the course $N\; 30^\{\backslash circ\}\; 10\text{'}\; E$.

Set the index to $37^\{\backslash circ\}\; 20\text{'}$, the complement of the middle latitude on the semicircle, and move the perpendicular until the termination of the departure on it cuts the index: then the point of intersection will mark the difference of longitude 210 miles on the index.

PROB. IV. Both latitudes and departure given, to find the course, distance, and difference of longitude.

Example. Let the latitude and longitude sailed from be $56^\{\backslash circ\}\; 40\text{'}\; S$ and $28^\{\backslash circ\}\; 55\text{'}\; E$ respectively, the latitude come to $61^\{\backslash circ\}\; 20\text{'}\; S$, and departure 172 miles. Required the course, distance, and difference of longitude?

Move the perpendicular and semicircle to the given latitude (H); then move the index till it meets the extremity of the departure on the perpendicular; the

distance will be marked on the index 329, and the course $S\; 31^\{\backslash circ\}\; 35\text{'}\; E$, or $SSE\backslash frac\{1\}\{4\}E$ nearly, on the semicircle.

Move the perpendicular and semicircle to the meridional parts of the given latitudes, and the index will cut the difference of longitude on the perpendicular $5^\{\backslash circ\}\; 35\text{'}$.

Set the index to $31^\{\backslash circ\}$, the complement of the middle latitude on the semicircle, and move the perpendicular until the departure marked on it cuts the index, and this point of intersection will mark the difference of longitude on the index 335 m. or $5^\{\backslash circ\}\; 35\text{'}$.

PROB. V. One latitude, course, and distance given, to find the difference of latitude and difference of longitude.

Example. Let the latitude left be $56^\{\backslash circ\}\; 40\text{'}\; S$, longitude $28^\{\backslash circ\}\; 55\text{'}\; E$, the course $S\; 31^\{\backslash circ\}\; 35\text{'}\; E$, and distance 329 m. Required the latitude and longitude come to?

Set the semicircle to the latitude sailed from, and the index to the course, and bring the perpendicular to the distance, which at the same time will mark the latitude come to $61^\{\backslash circ\}\; 20\text{'}\; S$.

Screw the index to the course, and move the semicircle and perpendicular to the meridional parts of both latitudes; then the index will cut the difference of longitude on the perpendicular $5^\{\backslash circ\}\; 35\text{'}$.

The semicircle and perpendicular being set to both latitudes, and the index to the course, it will show the departure 172.7 on the perpendicular.

Set the index to $31^\{\backslash circ\}$ the complement of the middle latitude on the semicircle, and move the perpendicular until the departure marked on it cuts the index, and the division on the index at the point of intersection will be the difference of longitude 335.

PROB. VI. One latitude, course, and departure given, to find the distance, difference of latitude, and difference of longitude.

Example. Let the latitude sailed from be $56^\{\backslash circ\}\; 40\text{'}\; N$, longitude $28^\{\backslash circ\}\; 35\text{'}\; W$, the course $N\; 31^\{\backslash circ\}\; 35\text{'}\; W$, and departure 172.7. Required the distance, and the latitude and longitude come to?

Move the semicircle to the latitude left, and the index to the course; mark the departure on the perpendicular,

(H) In southern latitudes, the end of the cylinder where the numbers begin must be turned towards the north pointed out by the semicircle; and in northern latitudes, it must be reversed.

Instruments to solve Problems in Sailing, without Calculation.

pendicular, and move it until the termination thereof meets the index, then the point of intersection will show the distance 329 miles on the index, and the perpendicular will show the latitude arrived at $61^\{\backslash circ\}\; 20\text{'}\; N$ on the base.

To find the difference of longitude.

Screw the index, and move the perpendicular and semicircle to the meridional parts of both latitudes, then the index will cut the difference of longitude $5^\{\backslash circ\}\; 35\text{'}$ on the perpendicular.

By Middle Latitude Sailing.

Find the distance sailed and latitude in as above, and the difference of longitude as in Problem IV. by middle latitude sailing.

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

Example. The latitude sailed from is $48^\{\backslash circ\}\; 30\text{'}\; N$, and longitude $14^\{\backslash circ\}\; 40\text{'}\; W$; the distance run is 345 miles between the south and east, and the departure 200 miles. Required the course, and the latitude and longitude come to?

By Mercator's Sailing.

To find the course and latitude come to.

Move the semicircle to the latitude left, mark the distance on the index, and the departure on the perpendicular, move both until these points meet; then will the index show the course $S\; 35^\{\backslash circ\}\; 26\text{'}\; E$ on the semicircle, and the latitude come to $43^\{\backslash circ\}\; 49\text{'}$ on the base.

The difference of longitude is found as in the preceding problem.

By Middle Latitude Sailing.

The course and latitude come to are found as above, and the difference of longitude as in Problem IV. by middle latitude sailing.

II. OF DR MACKAY'S Rectangular Instrument.

Plate Description. Fig. 52. is a representation of this instrument, of about one-third of the original size.—
Fig. 52. The length CA is divided into 100 equal parts, and the breadth CB into 70; but in this plate every second division only is marked, in order to avoid confusion; through these divisions parallels are drawn, terminating at the opposite sides of the instrument. Upon the upper and right hand sides are two scales; the first contains the degrees of the quadrant, and the other the points and quarters of the compass. M is an index moveable about the centre C, and divided in the same manner as the sides (1). Fig. 53. is a portion of the enlarged meridian, so constructed that the first degree is equal to three divisions on the instrument; and therefore, in the use of this line, each division on the instrument is to be accounted 20 minutes. The size of the plate would not admit of the continuation of the line.

Use. From a bare inspection of this instrument, it

is evident that any triangle whatever may be formed on it. In applying it to nautical problems, the course is to be found at top, or right-hand side, in the column of degrees or points, according as it is expressed; the distance is to be found on the index, the difference of latitude at either side column, and the departure at the head or foot of the instrument. The numbers in these columns may represent miles, leagues, &c.; but when used in conjunction with the enlarged meridional line, then 10 is to be accounted 100 miles, 20 is to be esteemed 200 miles, and so on, each number being increased in a tenfold ratio; and the intermediate numbers are to be reckoned accordingly.

Plane Sailing.

PROB. I. The course and distance sailed given, to find the difference of latitude and departure.

Example. Let the course be $NE\backslash frac\{1\}\{2\}N$, distance 44 miles. Required the difference of latitude and departure?

Move the index until the graduated edge be over $3\backslash frac\{1\}\{2\}$ points, and find the given distance 44 miles on the index: this distance will be found to cut the parallel of 34 miles, the difference of latitude in the side column, and that of 28 miles, the departure at the top.

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

Example. Required the distance and departure answering to the course $28^\{\backslash circ\}$, and difference of latitude 60 miles?

Lay the index over the given course $28^\{\backslash circ\}$: find the difference of latitude 60 miles in the side column; its parallel will cut the index at 68 miles, the distance and the corresponding departure at the top is 32 miles.

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

Example. Let the course be $SSW$ and the departure 36 miles. Required the distance and difference of latitude?

Lay the index over two points; find the departure at the top, and its parallel will cut the index at 94 miles the distance, and the difference of latitude on the side column is 87 miles.

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

Example. The distance is 35 leagues, and the difference of latitude 30 leagues. Required the course and departure?

Bring 35 leagues on the index to the parallel of 30 leagues in the side; then the departure at the top is 18 leagues, and the course by the edge of the index on the line of rhumbs is $2\backslash frac\{1\}\{4\}$ points.

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

Example. Let the distance be 58 miles, and the departure

(1) In the original instrument are two slips, divided like the side and end of the instrument. One of these slips is moveable in a direction parallel to the side of the instrument, and the other parallel to the end.

Instrument to solve Problems in Sailing without Calculation.

Departure 15 miles. Required the course and difference of latitude?

Move the index until 58 found thereon cuts the parallel of 15 from the top: this will be found to intersect the parallel of 56 miles, the difference of latitude; and the course by the edge of the ruler is $15^\{\backslash circ\}$.

PROB. VI. The difference of latitude and departure being given, to find the course and distance.

Example. Let the difference of latitude be 30 miles, the departure 28 miles. Required the course and distance?

Bring the index to the intersection of the parallels of 30 and 28; then the distance on the index is 41 miles, and the course by its edge is $43^\{\backslash circ\}$.

Traverse Sailing.

Find the difference of latitude and departure answering to each course and distance by Problem I. of Plane Sailing, and from thence find the difference of latitude and departure made good; with which find the course and distance by the last problem.

An example is unnecessary.

Parallel Sailing.

PROB. I. Given the difference of longitude between two places on the same parallel, to find the distance between them.

Example. Let the latitude of a parallel be $48^\{\backslash circ\}$, and the difference of longitude between two places on it $3^\{\backslash circ\}\; 40\text{'}$. Required their distance?

Put the index to $48^\{\backslash circ\}$, the given latitude, and find the difference of longitude 220 on the index, and the corresponding parallel from the side will be 147, the distance required.

PROB. II. The latitude of a parallel, and the distance between two places on that parallel, being given, to find the difference of longitude between them.

Example. The latitude of a parallel is $56^\{\backslash circ\}$, and the distance between two places on it 200 miles. Required their difference of longitude?

Put the index to the given latitude, and find the distance in the side column, and the intersection of its parallel with the index will give 358, the difference of longitude sought.

PROB. III. Given the distance and difference of longitude between two places on the same parallel, to find the latitude of that parallel.

Example. The number of miles in a degree of longitude is 46.5. Required the latitude of the parallel?

Bring 60 on the index to cut the parallel of 46.5 from the side, then the edge of the index will give $39^\{\backslash circ\}\; 11\text{'}$, the latitude required.

Middle Latitude and Mercator's Sailing.

PROB. I. The latitudes and longitudes of two places being given, to find the course and distance between them.

Example. Required the course and distance between Genoa, in latitude $44^\{\backslash circ\}\; 25\text{'}\; N$, longitude $8^\{\backslash circ\}\; 36\text{'}\; E$, and Palermo, in latitude $38^\{\backslash circ\}\; 10\text{'}\; N$, longitude $13^\{\backslash circ\}\; 38\text{'}\; E$?

By Mercator's Sailing.

Take the interval between $38^\{\backslash circ\}\; 10\text{'}$ and $44^\{\backslash circ\}\; 25\text{'}$ on

the enlarged meridian, which laid off from C upwards will reach to 500; now find the difference of longitude 302 at the top, and bring the divided edge of the index to the intersection of the corresponding parallels, and the index will show the course $31^\{\backslash circ\}\; 8\text{'}$ on the line of degrees; then find the difference of the latitude 375 on the side column, and its parallel will intersect the index at 438, the distance.

By Middle Latitude Sailing.

Put the index to $41^\{\backslash circ\}\; 18\text{'}$, the complement of the middle latitude on degrees, and the difference of longitude 302 on the index will intersect the parallel of 227, the departure, in the side column. Now move the index to the intersection of the parallels of 375 and 227, the first being found in the side column, and the other at top or bottom; then the distance answering thereto on the index will be 438, and the course on the scale of degrees is $41^\{\backslash circ\}\; 10\text{'}$.

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

Example. Let the latitude and longitude sailed from be $39^\{\backslash circ\}\; 22\text{'}\; N$, and $12^\{\backslash circ\}\; 8\text{'}\; W$ respectively, the course $NNW\backslash frac\{1\}\{2\}W$, and distance 500 miles. Required the latitude and longitude come to?

By Mercator's Sailing.

Put the index to the course $2\backslash frac\{1\}\{2\}$ points, and find the distance 500 miles thereon; then the corresponding difference of latitude will be 441 miles, and the departure $235\backslash frac\{1\}\{2\}$ miles, hence the latitude in is $46^\{\backslash circ\}\; 43\text{'}\; N$. Now take the interval between the latitudes of $39^\{\backslash circ\}\; 22\text{'}$ and $46^\{\backslash circ\}\; 43\text{'}$ on the enlarged meridian, which laid off from C will reach to about 605, the parallel of which will intersect the vertical parallel of the difference of longitude 323 at the edge of the index: hence the longitude in is $17^\{\backslash circ\}\; 31\text{'}\; W$.

By Middle Latitude Sailing.

Find the difference of latitude and departure as before, and hence the latitude in is $46^\{\backslash circ\}\; 43\text{'}\; N$, and the middle latitude $43^\{\backslash circ\}\; 3\text{'}$. Now put the index to $43^\{\backslash circ\}\; 3\text{'}$, and the horizontal parallel of the departure $235\backslash frac\{1\}\{2\}$ will intersect the index at 322, the difference of longitude.

PROB. III. Both latitudes and course given, to find the distance and difference of longitude.

Example. The latitude sailed from is $22^\{\backslash circ\}\; 54\text{'}\; S$, and longitude $42^\{\backslash circ\}\; 40\text{'}\; W$, the course is SE by E, and latitude come to $26^\{\backslash circ\}\; 8\text{'}\; S$. Required the distance sailed, and longitude in?

By Mercator's Sailing.

Bring the index to 5 points, the given course, and the parallel of 194, the difference of latitude found in the side column, will intersect the index at 349, the distance; and it will cut the vertical parallel of 290, the departure.

Take the interval between the given latitudes $22^\{\backslash circ\}\; 54\text{'}$ and $26^\{\backslash circ\}\; 8\text{'}$ on the enlarged meridian; lay off that extent from the centre on the side column, and it will reach to 213: the parallel of this number will intersect the vertical parallel of 319, the difference of longitude. Hence the longitude in is $37^\{\backslash circ\}\; 21\text{'}\; W$.

By

By Middle Latitude Sailing.

With the given course and difference of latitude find the distance and departure as before; then bring the index to the middle latitude $24^\{\backslash circ\}\; 31\text{'}$; find the departure 290 in the side column, and its parallel will intersect the index at 319, the difference of longitude.

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

Example. The latitude and longitude left are $20^\{\backslash circ\}\; 30\text{'}\; N$, and $49^\{\backslash circ\}\; 17\text{'}\; W$, respectively; the course is $NE\backslash frac\{1\}\{4\}N$, and departure 212 miles. Required the latitude and longitude come to, and distance sailed?

By Mercator's Sailing.

Put the index to the given course $3\backslash frac\{1\}\{4\}$ points, and the vertical parallel of 212 will cut the index at 356, the distance, and the horizontal parallel of 286, the difference of latitude; the latitude come to is therefore $25^\{\backslash circ\}\; 16\text{'}\; N$.

Now take the interval between the latitudes $20^\{\backslash circ\}\; 30\text{'}$, and $25^\{\backslash circ\}\; 16\text{'}$ on the enlarged meridian, which laid off from the centre C will reach to 311; and this parallel will intersect the vertical parallel of the difference of longitude 230, at the edge of the index. Hence the longitude in is $45^\{\backslash circ\}\; 27\text{'}\; W$.

By Middle Latitude Sailing.

Find the distance and difference of latitude as directed above; then bring the index to $22^\{\backslash circ\}\; 53\text{'}$, the middle latitude, and the horizontal parallel of 212, the departure, will intersect the index at 230, the difference of longitude.

PROB. V. Both latitudes and distance given, to find the course and difference of longitude.

Example. The distance sailed is 500 miles between the north and west; the latitude and longitude left are $40^\{\backslash circ\}\; 10\text{'}\; N$, and $9^\{\backslash circ\}\; 20\text{'}\; W$ respectively, and the latitude in is $46^\{\backslash circ\}\; 40\text{'}\; N$. Required the course and longitude in?

By Mercator's Sailing.

Bring the distance 500 on the index to intersect the horizontal parallel of the difference of latitude 390; then the course $38^\{\backslash circ\}\; 44\text{'}$ is found on the line of degrees by the edge of the index, and the vertical parallel of the above point of intersection is that answering to 313, the departure.

Take the interval between the latitudes $40^\{\backslash circ\}\; 10\text{'}$, and $46^\{\backslash circ\}\; 40\text{'}$, which lay off from the centre C, and its horizontal parallel will intersect the vertical parallel of 431, the difference of longitude, by the edge of the index, it being in the same position as before. Hence the longitude in is $16^\{\backslash circ\}\; 31\text{'}\; W$.

By Middle Latitude Sailing.

The course and departure are found as formerly, and the middle latitude is $43^\{\backslash circ\}\; 25\text{'}$, to which bring the edge of the index, and the horizontal parallel of 313, the departure, will intersect the index at 431, the difference of longitude.

PROB. VI. Both latitudes and departure given, to find the course, distance, and difference of longitude.

Example. Let the latitude sailed from be $42^\{\backslash circ\}\; 52\text{'}\; N$,

long. $9^\{\backslash circ\}\; 17\text{'}\; W$, the departure 250 miles W, and the Sea Charts latitude come to $36^\{\backslash circ\}\; 18\text{'}\; N$. Required the course and distance sailed, and the longitude come to?

By Mercator's Sailing.

Find the point of intersection of the horizontal parallel of 394, the difference of latitude, and the vertical parallel of 250, the departure; to this point bring the index, and the corresponding division thereon will be 467 miles, and the course on the scale of degrees by the edge of the index will be $32^\{\backslash circ\}\; 24\text{'}$.

Take the interval between the latitudes on the enlarged meridian; which being laid off from the centre will reach to 512: now the horizontal parallel of 512 will cut the vertical parallel of 325, the difference of longitude, at the edge of the index. The longitude come to is therefore $14^\{\backslash circ\}\; 42\text{'}\; W$.

By Middle Latitude Sailing.

The course and distance are to be found in the same manner as above. Then bring the index to $39^\{\backslash circ\}\; 35\text{'}$, the middle latitude, and the horizontal parallel of 250 will intersect the edge of the index at 324, the difference of longitude.

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

Example. A ship from latitude $32^\{\backslash circ\}\; 38\text{'}\; N$, longitude $17^\{\backslash circ\}\; 6\text{'}\; W$, sailed 586 miles between the south and west, and made 336 miles of departure:—Required the course, and the latitude and longitude come to?

By Mercator's Sailing.