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 Jafon, and others to Janus, who is said to have made the first ship. Historians ascribe it to the Ægiates, the Phoenicians, Tyrians, and the ancient inhabitants of Britain. Some will have it, the first hint was taken from the flight of the kite; others, as Oppian, De pescibus, 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 the good man underwent on account of his enterprise shows evidently enough the world was then ignorant of anything like navigation, and that they even thought it impossible.

However, profane history represents the Phoenicians, 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 to 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 opulency and populousness; inasmuch 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 in 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 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 capital of the world were wholly supplied with merchandizes from the capital of Egypt.

At length Alexandria itself underwent the fate of Tyre and Carthage; being surprized 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, rechanges, &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 ordinarily looked on as the reformers 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 of 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 gulph, 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 of uniting into a body. And it was this union, first begun in the fifth 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 parts 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 their 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 thereof, which still obtain under the names of Uts 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 Hansetowns.

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 reasons of commerce's passing successively from the Venetians, Genoese, and Hanse towns, to the Portuguese and Spaniards, and from these again to the English and Dutch, is 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 tutelar 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 hath 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 superceded by the improvements made in the formation of the sails, rigging, &c. by which means the 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 a 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 do agree that the mariner's compass is a modern invention, it hath been much disputed who was the inventor. Some give the honour of it to Flavio Gioia of Amalii in Canaania*, 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 Vifco, brother to the king of Portugal. In the year 1487, Roderic 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 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. There 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 perish 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, exact, 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 fort of cross-staff and an instrument called the nautical quadrant; which last was much praised by William Cuningham, 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 Sufa. In this he exposes the errors of the plane chart, and likewise gives the solution of several curious astronomical problems; among 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 affixing 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 epact, 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, intitled 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 inadverted on Medina. In this he showed, that as the rhumbs are spirals, 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 flower-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 till it 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 enlargement 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. lemly 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 bear 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 only been 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 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 1608, a treatise intitled, 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 to 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 confines 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 confines, is preferred on account of its being less operose: yet in high latitudes it is more eligible to use that of the arithmetical mean between the logarithmic co-fines, equivalent to the geometrical mean between the co-fines themselves; a method since proposed by Mr John Baffat. The computation by the middle latitude will always fall short of the true change of longitude; that by the geometrical mean will always ways 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 intitled, 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 intitled, 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, intitled, A discourse mathematical on the variation of the magnetical 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 intitled, 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 Mercennius, 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, 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 tortoises: 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 seafreckoning, 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° 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 logarithmic 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 Geometricae, 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, No 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, No 338. It is moreover added in Gunter's book, that if \( \frac{1}{2} \)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 Kalendar, 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, No 40, he published a table of the variation for 49 years to come. Thus he acquired such reputation, that his treatise, intitled, 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, intitled, The Longitude not Found, made its appearance; and as Mr Bond's hypothesis did not in any manner answer its author's genuine 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 1744 and 1766, by Mr William Mountraine, 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 everything 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 thro' the poles, a tract was published in 1741 by the Rev. Doctor Patrick Murdoch, wherein he accommodated Wright's failing to such a figure; and Mr Colin MacLaurin, the same year, in the Philosophical Transactions, No 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.

**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 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 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, hath 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 only can 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 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 spheroidal 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 0.99624, 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 \( \frac{1}{8} \) of the whole. According to M. du Sejour, 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 can seldom be done without considerable difficulty. 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.

**PRACTICE OF NAVIGATION.**

**BOOK I.**

Containing the various Methods of Sailing.

**INTRODUCTION.**

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.

**CHAP. I. Preliminary Principles.**

**Sect. I. Of the Latitude and Longitude of a Place.**

The situation of a place on the surface of the earth is estimated by its distance from two imaginary lines intersecting each other at right angles: The one of these is called the equator, and the other the first meridian. The situation of the equator is fixed, but that of the first meridian is arbitrary, and therefore different nations assume different first meridians. In 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 I.** Required the difference of latitude between the Lizard, in latitude 49° 57' N., and Cape St Vincent, in latitude 37° 2' N?

| Latitude of the Lizard | 49° 57' N. | |------------------------|------------| | Latitude of Cape St Vincent | 37° 2' N. |

Difference of latitude = 12° 55' = 775 miles.

**Example II.** What is the difference of latitude between Funchal, in latitude 32° 38' N., and the Cape of Good Hope, in latitude 34° 29' S?

| Latitude of Funchal | 32° 38' N. | |---------------------|------------| | Latitude of Cape of Good Hope | 34° 29' S. |

Difference of latitude = 2° 51' = 175 miles. Practice.

Latitude Latitude of Funchal 32° 38' N. and Longi. Lat. of Cape of Good Hope 34° 29' S.

Difference of latitude 67° 7' = 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 dif- ferent names, subtract them, and the sum or remain- der will be the latitude required of the same name with the greater.

Example I. A ship from latitude 39° 22' N. failed due north 560 miles—Required the latitude come to?

Latitude failed from 39° 22' N. Difference of latitude 560' = 9° 20' N.

Latitude come to 48° 42' N.

Example II. A ship from latitude 7° 19' N. failed 854 miles south—Required the latitude come to?

Latitude failed from 7° 19' N. Difference of latitude 854' = 14° 14' S.

Latitude come to 6° 55' S.

Prob. III. The longitudes of two places being gi- ven, 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°, sub- tract it from 360°, and the remainder is the difference of longitude.

Example I. Required the difference of longitude between Edinburgh and New York, their longitude being 3° 14' W and 74° 10' W. respectively?

Longitude of New York 74° 10' W. Longitude of Edinburgh 3° 14' W.

Difference of longitude 70° 56'

Example II. What is the difference of longitude between Markelyne's Isles, in longitude 167° 59' E. and Olinde, in longitude 35° 5' W?

Longitude of Markelyne's Isles 167° 59' E. Longitude of Olinde 35° 5' W.

Sum 203° 4' Subtract from 360°

Difference of longitude 156° 56'

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 requi- red of the same name with the greater quantity; but if they are of different names, 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 con- trary name to that given.

Example I. A ship from longitude 9° 54' E. failed westerly till the difference of longitude was 23° 18'—

Required the longitude come to?

Longitude failed from 9° 54' E. Difference of longitude 23° 18' W.

Longitude come to 13° 24' W.

Example II. The longitude failed from is 25° 9' W. and difference of longitude 18° 46' W.—Required the longitude come to?

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

Longitude in 43° 55' W.

Sect. II. Of the Tides.

The theory of the tides has been explained under the article Astronomy, and will again be farther il- lustrated under that of Times. In this place, there- fore, it remains only to explain the method of calcu- lating 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 me- thod cannot be accurate, and by observation the error is sometimes found to exceed two hours. That me- thod 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.

Example I. Required the time of new moon at Salonique in May 1793?

Time of new moon per Naut. Alm. 9d 15h 31' Longitude of Salonique in time 0° 1° 33' E.

Time of new moon required in May 9° 17' 4"

Example II. What is the time of the last quarter of the moon at Revolution Bay in October 1793?

Time of last quarter per Naut. Alm. 26d 5h 47' Longitude in time 0° 9° 17' W.

Time at Revolution bay of last quar- ter, October 25° 20' 30"

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 wa- ter 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.

**Example I.** Required the morning and afternoon times of high water at Leith, 11th December 1793?

Nearest phase to 11th Dec. is 1st quart. $9^h\ 20^m\ 29^s$

Longitude of Leith in time $-0\ 0\ 13$

Time at Leith of 1st quarter $9\ 20\ 16$

Given day $11\ 0\ 0$

Difference $1\ 3\ 44$

Time of H.W. at Leith-pier on syz. $0\ 2\ 20$

Equat. from Tab. to $1^d\ 3^h\ 44'$ $+0\ 6\ 32$

Approximate time of high water $11\ 8\ 52$

Reduced time of 1st quarter $9\ 20\ 16$

Interval $1\ 12\ 36$

Time of high water at Leith on syz. $0\ 2\ 20$

Equat. from the Tab. to $1^d\ 12^h\ 36'$ $0\ 7\ 0$

Time of high water at Leith $9\ 20$ P.M.

Time of H.W. at Leith at full & change $2\ 20$

Equat. to $1^d\ 12^h\ 36'-12^h=1^d\ 0^h\ 36'$ $6\ 22$

High water at Leith, Dec. 11th, at $8\ 42$ A.M.

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

**Example II.** Required the time of high water at Funchal, 15th November 1793?

The nearest phase to 15th November is that of full moon. $17^d\ 8^h\ 46'$

Longitude of Funchal in time $-0\ 1\ 8$ W.

Time of full moon at Funchal $17^d\ 7^h\ 38'$

Given day, November $15\ 0\ 0$

Difference $2\ 7\ 38$

Time of high water at Funchal at full and change $0\ 12\ 4$

Equation from the Table to $2^d\ 7^h\ 38'$ before full moon $-0\ 1\ 35$

Approx. time of high water, Nov. 15. $0\ 10\ 29$

Reduced time of full moon $17\ 7\ 38$

Interval $1\ 11\ 9$

Time of high water at full and change $0\ 12\ 4$

Equation to $1^d\ 11^h$ before full moon $0\ 0\ 56$

Time of high water $0\ 11\ 8$ PM.

Equation to $1^d\ 11^h+12^h=1^d\ 23^h$ is $1^h\ 15'$

and $12^h\ 4'=1^h\ 15'=10^h\ 49'$ time of high Ship's Run-water in the forenoon.

**Example III.** Required the time of high water at Dulcey Bay, 24th October 1793?

The nearest phase to the 24th October is the last quarter. $26^d\ 5^h\ 47'$

Longitude of Dulcey Bay in time $+0\ 11\ 5$ E.

Reduced time of first quarter of moon $26\ 16\ 52$

Given day $24\ 0\ 0$

Difference $2\ 16\ 52$

Time of high water at full and change $10^h\ 57'$

Equation to $2^d\ 10^h\ 52'$ before last quarter $+2\ 52$

Approximate time of high water $1\ 49$

Change of equation to app. time $1^h\ 49'$ $3$

Time of high water in the afternoon $1\ 52$

Change of equation to 12 hours $20$

Time of high water in the morning $1\ 32$

**Sect. III. Of measuring a Ship's Run in a given Time.**

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

It has been already observed, that the interval between each knot on the line ought to be 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 seconds run by the glass, being given, to find the true distance, the line being supposed right.

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

**Example I.** 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?

$$\frac{9}{35} = \frac{27}{30} = \text{true rate of sailing}.$$

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

$$\frac{73}{30} = \frac{26(219)}{84} = \text{the true distance}.$$ 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.

**Example I.** 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?

\[ \begin{align*} \text{Measured interval} &= 53 \\ \text{Twice hourly rate} &= 10 \\ \text{True rate of sailing,} &= 5.30 \end{align*} \]

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

\[ \begin{align*} \text{Twice given distance,} &= 128 \\ \text{Measured interval,} &= 42 \\ \text{True distance,} &= 53.76 \end{align*} \]

**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.

**Example.** 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?

\[ \begin{align*} \text{Distance by the log,} &= 159 \\ \text{Six times length of a knot} &= 42 \times 6 = 252 \\ \text{Second run by the glass} &= 33 \times 0.068(121.4) = \text{true distance.} \end{align*} \]

**Chap. II. Of Plane Sailing.**

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

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

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

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

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

In order to illustrate the above, let A (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, to be 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' N. sailed S. W. by S. 158 miles. Required the latitude come to, and departure?

By construction.

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 e 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. By Calculation.

To find the difference of latitude.

As radius 10.00000 is to the cosine of the course 3 points 9.91985 so is the distance 158 2.19866

to the difference of latitude 131.4 2.41851

To find the departure.

As radius 10.00000 is to the sine of the course 3 points 9.74474 so is the distance 158 2.19866

to the departure 87.8 1.94340

By Inspection.

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

By Gunter's Scale.

The extent from 8 points to 5 points, the complement of the course on the line of fine 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 fine rhumbs, will reach from 158 to 87.8, the departure on numbers.

Latitude St Helena = 15° 55' N. Difference of latitude = 2 11 S.

Latitude come to = 13 44 N.

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

Example. A ship from St George's, in latitude 38° 45' north, sailed SE 1/2 S.; and the latitude by observation was 35° 7' N. Required the distance, run, and departure?

Latitude St George's = 38° 45' N. Latitude come to = 35° 7' N.

Difference of latitude = 3° 38' = 218 miles.

By Construction.

Draw the portion of the meridian AB (fig. 3.) equal to 218 m. from the centre A with the chord of 60° describe the arch mn, which make equal to the rhumb of 3½ points; through Ac 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.

As radius 10.00000 is to the secant of the course 3½ points 10.11181 so is the difference of latitude 218 m. 2.33846

to the distance 282 2.45027

To find the departure.

As radius 10.00000 is to the tangent of the course 3½ pts. 9.91417 so is the difference of latitude 218 2.33846

to the departure 178.9 2.25253

By Inspection.

Find the given difference of latitude 218 m. in a latitude column, under the course 3½ points; opposite to which, in a distance column, is 282 miles; and in a departure column is 178.9 m.

Extend the compass from 4½ points, the complement of the course to 8 points on fine rhumbs, that extent will reach from the difference of latitude 2.8 miles to the distance 282 miles in numbers; and the extent from 4 points to the course 3½ 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° 37' 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°, describe the arch mn, which make equal to the rhumb of 3 points, the complement of the course; draw a line through C, which produce till it meets BA in A: then the distance AC being measured, will be equal to 2.1 m., and the difference of latitude AB will be 128.3 miles.

By Calculation.

To find the distance.

As the sine of the course 5 points 9.91985 is to radius 10.00000 so is the departure 192 2.28330

to the distance 230.9 2.36345

To find the difference of latitude.

As the tangent of the course 5 points 10.17511 is to radius 10.00000 so is the departure 192 2.28330

to the difference of latitude 128.3 2.10819

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 138.3, in their respective columns.

By Gunter's Scale.

The extent from 5 points to 8 points on the line of fine 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.

Latitude of Palma = 28° 37' N. Difference of latitude = 2 8' N.

Latitude come to = 30° 45' N.

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

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

Latitude sailed from = 43° 13' N. Latitude by observation = 46° 31' N.

Difference of latitude = 3° 18' = 198 miles.

By Construction.

Draw the portion of the meridian AB (fig. 5.) equal to 198 miles; from B draw BC perpendicular to Practice.

Plane Sailing.

To find the difference of latitude.

As radius is to the cosine of the course \(46^\circ 57'\) so is the distance \(260\) to the difference of latitude \(177.5\).

By Inspection.

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 \(47^\circ\) at the bottom, which is the course; and the corresponding difference of latitude is \(177.3\).

By Gunter's Scale.

The extent of the compass, from the distance \(260\) to the departure \(190\) on the line of numbers, will reach from \(90^\circ\) to \(47^\circ\), the course on the line of fines; and the extent from \(90^\circ\) to \(43^\circ\), the complement of the course on fines, will reach from the distance \(260\) to the difference of latitude \(177.5\) on the line of numbers.

Latitude Fort-Royal \(12^\circ 9' N\) Difference of latitude \(177 = 2^\circ 57' S\)

Latitude in \(9^\circ 12' N\)

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

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

Latitude failed from \(7^\circ 56' S\). Latitude in by observation \(12^\circ 3' S\).

Difference of latitude \(4^\circ 7' = 247\).

By Construction.

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 \(60^\circ\) describe an arch from the centre \(A\); and the portion \(mn\) of this arch being applied to the line of chords, will measure about \(280\) miles; and the distance \(AC\), measured on the line of equal parts, will be \(280\) miles.

By Calculation.

To find the course.

As the difference of latitude \(247\) is to the departure \(132\) so is radius \(10,00000\)

to the tangent of the course \(280^\circ 7'\) \(9.72787\)

To find the distance.

As radius \(10,00000\) is to the secant of the course \(280^\circ 7'\) \(10.05454\) so is the difference of latitude \(247\) \(2.39272\)

to the distance \(280\) \(2.44724\)

By Inspection.

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

By Gunter's Scale.

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

**CHAP. III. Of Traverse Sailing.**

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

Make a table of a breadth and depth sufficient to contain the several courses, &c. This table is to be divided into six columns: the several 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°; otherwise greater.

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

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

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

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

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

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

**Examples.**

I. A ship from Fyial, in latitude 38° 32' N., sailed as follows: ESE 163 miles, SW 110 miles, SE 180 miles, and N by E 68 miles. Required the latitude come to, the course, and distance made good?

**By Inspection.**

| Course | Dist. | Diff. of Latitude | Departure | |--------|-------|------------------|-----------| | ESE | 163 | | | | SW 1/2 W | 110 | | | | SE 1/2 S | 180 | | | | NLE | 68 | | |

| Latitude left, | 38° 32' N. | | Difference of latitude, | 3 21 S. | | Latitude come to | 35 11 N. |

**By Construction.**

With the chord of 60° describe the circle NE, SW (fig. 8), the centre of which represents the place the ship sailed from; draw two diameters NS, EW at right angles to each other; the one representing the meridian, and the other the parallel of latitude of the place sailed from. Take each course from the line of rhumbs, lay it off on the circumference from its proper meridian, and number it in order 1, 2, 3, 4. Upon the first rhumb C1, lay off the first distance from C to A; through it draw the second distance AB parallel to C2, and equal to 110 miles; through B draw BD equal to 180 miles, and parallel to C3; and draw DE parallel to C4, and equal to 68 miles. Now CE being joined, will represent the distance made good; which applied to the scale will measure 281 miles. The arch Sn, which represents the course, being measured on the line of chords, will be found equal to 41°. 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 110 and 186 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 $1^\circ 38' S$ sailed as under. Required her present latitude, course, and distance made good?

| Course | Diff. | Diff. of Latitude | Departure | |--------|-------|------------------|-----------| | NW $N$ | 43 | 35.8 | 23.9 | | WNW | 78 | 29.9 | 72.1 | | SE $E$ | 56 | 31.1 | 46.6 | | WSW $W$ | 62 | 18.0 | 59.3 | | N $E$ | 85 | 84.1 | 12.5 | | | | 149.8 | 59.1 | | | | 49.1 | 59.1 | | N $44^\circ W$ | 139 | 100.7 $= 1^\circ 41'$ | 96.2 |

Latitude come to - 0 3 N.

III. Yesterday at noon we were in latitude $13^\circ 12' N$, and since then have run as follows: SSE 36 miles, S 12 miles, NW $W$ 23 miles, W 30 miles, SW 42 miles, W $N$ 39 miles, and N 20 miles. Required our present latitude, departure, and direct course and distance?

| Course | Diff. | Diff. of Latitude | Departure | |--------|-------|------------------|-----------| | SSE | 36 | 33.3 | 13.8 | | S | 12 | 12.0 | | | NW $W$ | 23 | 17.8 | 21.6 | | W | 30 | | 30.0 | | SW | 42 | 29.7 | 29.7 | | W $N$ | 39 | 7.6 | 38.2 | | N | 20 | 20.0 | | | | | 45.4 | 13.8 | | | | 45.4 | 13.8 | | S $74^\circ W$ | 110 | 29.6 $= 0^\circ 30'$ | 105.7 |

Yesterday's latitude - 13 12 N

Present latitude - 12 42 N

IV. The course per compass from Greigfnefs to the May is SW $S$, distance 58 miles; from the May to the Staples S$E$ $E$, 44 miles; and from the Staples to Flamborough Head S$E$, 110 miles. Required the course per compass, and distance from Greigfnefs to Flamborough Head?

Vol. XII. Part II.

(a) 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.

Hence the course per compass is S $E$, and distance 110 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 where 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 part of the earth, the arch ABP being part of a meridian; CA the equatorial; and CP the polar semiaxis. Also let B be the situation of any given place on the earth; and join BC, which will be equal to CA or CP(A). 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 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 \( R : \cosine \text{ latitude} :: \text{Diff. longitude} : \text{Distance} \).

And by inversion,

\[ \cosine \text{ latitude} : R :: \text{Distance} : \text{Diff. of longitude}. \]

Also,

\[ \text{Diff. of longitude} : \text{Distance} :: R : \cosine \text{ latitude}. \]

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

**Example I.** Required the number of miles contained in a degree of longitude in latitude \(55^\circ 58'N\).

*By Construction.*

Draw the indefinite right line \(AB\) (fig. 10); make the angle \(BAC\) equal to the given latitude \(55^\circ 58'\), 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.

*By Calculation.*

As radius \(10.00000\) is to the cosine of latitude \(55^\circ 58'\) \(9.74794\) so is miles in a deg. of long. at eq. \(60\) \(1.77815\)

to the miles in deg. in a given par. \(33.58\) \(1.52609\)

*By Inspection.*

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

*By Gunter's Scale*

The extent from \(90^\circ\) to \(34^\circ\), the complement of the given latitude on the line of lines, 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 intitled chords, and contains the several degrees of latitude; the other, marked M. L. signifying miles of longitude, is the line of longitude, and shows the number of miles in a degree of longitude in each parallel. The use of these lines is therefore obvious.

**Example II.** Required the distance between Treguier in France, in longitude \(3^\circ 14'W\), and Gaspey Bay, in longitude \(64^\circ 27W\), the common latitude being \(48^\circ 47'N\).

Longitude Treguier \(3^\circ 14'W\) Longitude Gaspey Bay \(64^\circ 27W\)

Difference of longitude \(61^\circ 13' = 3673'\)

As radius \(10.00000\) is to the cosine of latitude \(48^\circ 47'\) \(9.81882\) so is the difference of longitude \(3673'\) \(3.56502\)

to the distance \(2420\) \(3.38384\)

**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 \(42^\circ 52'N\), and longitude \(9^\circ 17'W\), sailed due west 342 miles. Required the longitude come to?

*By Construction.*

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

*By Calculation.*

As radius \(10.00000\) is to the secant of latitude \(42^\circ 52'\) \(10.13493\) so is the distance \(342\) \(2.53403\)

to the difference of longitude \(466.6\) \(2.66896\)

*By Inspection.*

The nearest degree to the given latitude is \(43^\circ\), under which, and opposite to 171, half the given distance in a latitude column, is 23.4 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.

*By Gunter's Scale.*

The extent from \(47^\circ 8'\), the complement of latitude to \(90^\circ\) on the line of lines, being laid the same way from the distance 342, will reach to the difference of longitude \(466\frac{1}{2}\) on the line of numbers.

Longitude Cape Finisterre \(9^\circ 17'W\) Difference of longitude \(7^\circ 47'W\)

Longitude come to \(17^\circ 4W\)

**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 \(8^\circ 42'\). Required the latitude of the parallel?

*By Construction.*

Make the line \(AB\) (fig. 12.) equal to the given distance; to which let \(BC\) be drawn perpendicular, with an extent equal to \(522\), 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 \(46^\circ\), the required latitude.

*By Calculation.*

As the distance \(358\) \(2.55388\) is to the difference of longitude \(522\) \(2.71767\) so is radius \(10.00000\)

to the secant of the latitude \(46^\circ 42'\) \(10.16379\)

*By Inspection.*

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 \(46^\circ\), and distance 261 miles, the corresponding difference of latitude is \(181^\circ 3'\), which exceeds the half of the given distance by \(2^\circ 3'\). Again, to latitude \(47^\circ\), and distance 261, the difference of latitude is \(178^\circ\), being \(1^\circ\) less than the half of that given; therefore the change of distance answering to a change of \(1^\circ\) of latitude is \(3^\circ\).

Now, as \(3^\circ : 2^\circ 3' :: 1^\circ : 42'\)

Hence the latitude required is \(46^\circ 42'\). The extent from 522 to 358 on the line of numbers, will reach from 90° to about 43° 1', the complement of which 46° 1' 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 parts in latitude 33° 58' N, distance 348 miles, two ships sail directly north till they are in latitude 48° 23' N. Required their distance?

**By Construction.**

Draw the lines CB, CE (fig. 13.), making angles with CP equal to the complements of the given latitudes, namely, 56° 2' and 41° 37' respectively; make BD equal to the given distance 348 miles, and perpendicular to CP; now from the centre C, with the radius CB, describe an arch intersecting CE in E; then EF drawn from the point E, perpendicular to CP, will represent the distance required; and which being applied to the scale, will measure 278½ miles.

**By Calculation.**

As the cosine of the latitude left 33° 58' 9.91874 is to the cosine of the latitude come to 48° 23' 9.82226 so is the given distance 348 2.54158 to the distance required 278.6 2.44510

**By Inspection.**

Under 34°, 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° of latitude, which is 140° 51' from which 1° 1' (the proportional part answering to 23 minutes of latitude) being abstracted, gives 139° 8', which doubled is 278° 8', the distance required.

**By Gunter's Scale.**

The extent from 56° 2', the complement of the latitude left, to 41° 37', the complement of that come to on the line of fines, being laid the same way from 348, will reach to 278½, the distance sought on the line of numbers.

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

**Example.** Two ships, in latitude 56° 0' N, distant 180 miles, fall 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°; 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

**By Calculation**

As the distance on the known parallel 180 2.25527 is to the distance on that required 232 2.36549

so is the cosine of the latitude left 56° 0' 9.74756 to the cosine of the latitude come to 43° 53' 9.85778

**By Inspection.**

To latitude 56°, 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° and distance 161, the corresponding difference of latitude is 117° 7'; the excess of which above 116° is 1° 7'; and to latitude 44°, and distance 161, the difference of latitude is 115° 8'; hence 117° 7' - 115° 8' = 1° 9', the change answering to a change of 1° of latitude.

Therefore 1° 9' : 1° 7' :: 1° : 53'

Hence the latitude is 43° 53'.

**By Gunter's Scale.**

The extent from 180 to 232 on the line of numbers, being laid in the same direction on the line of fines, from 34°, the complement of the latitude sailed from, will reach to 46° 7', the complement of the latitude come to.

**Chap. V. Of Middle Latitude Sailing.**

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

Let AB (fig. 15.) be any given distance 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 rectilineal. If the distance be supposed to be divided into an infinite number of parts, each equal to CE, and upon these, triangles be constructed whose sides are portions of a meridian and parallel, it is evident these triangles will be equal and similar; for, besides the right angle, and hypotenuse which is the same in each, the course or angle CED is also the same. Hence, by the 12th of V. Euc. the sum of all the hypotenuses CE, or the distance AB, is to the sum of all the sides DE, or the difference of latitude AS, as one of the hypotenuses CE is to the corresponding side DE. Now, let the triangle GH (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 Middle Latitude Sailing.

CD, as CE is to CD, or as GH to HI, because of the similar triangles.

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

Although the parts HG, GI, and angle G of the rectilineal triangle GIH, are equal to the corresponding parts AB, AS, and angle A, of the triangle ASB upon the surface of the sphere; yet HI is not equal to BS, for HI is the sum of all the arcs CD; but CD is greater than OQ, and less than ZX; therefore HI is greater than BS, and less than AR. Hence the difference of longitude MN cannot be inferred from the departure reckoned either upon the parallel failed from or on that come to, but on some intermediate parallel TV, such that the arch TV is exactly equal to the departure; and in this case the difference of longitude would be easily obtained. For TV is to MN as the fine PT to the fine PM; that is, as the cosine of latitude is to the radius.

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

In order to illustrate the computations in middle latitude sailing, let the triangle ABC (fig. 17.) represent a figure in plain 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 there the unknown parts of the other triangle will be easily obtained.

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

Example. Required the course and distance from the island of May, in latitude 56° 12' N, and longitude 2° 37' W, to the Naze of Norway, in latitude 57° 50' N, and longitude 7° 27' E?

| Latitude Isle of May | 56° 12' N | |----------------------|-----------| | Latitude Naze of Norway | 57° 50' N |

Difference of latitude = 1° 38 = 98' Middle latitude = 57°

Longitude Isle of May = 2° 37' W Longitude Naze of Norway = 7° 27' E

Difference of longitude = 10° 4 = 60'

By Construction.

Draw the right line AD (fig. 18.) to represent the meridian of the May; with the chord of 60° describe the arch mn, upon which lay off the chord of 32° 59', the complement of the middle latitude from m to n; from D through n draw the line DC equal to 604', the difference of longitude, and from C draw CB perpendicular to AC; make BA equal to 98', the difference of latitude, and join AC; which applied to the scale will measure 343 miles, the distance sought: and the angle A being measured by means of the line of chords, will be found equal to 73° 24', the required course.

By Calculation.

To find the course (b).

As the difference of latitude = 98' = 1.99123 is to the difference of longitude 604' = 2.78104 so is the cosine of middle latitude 57° 1' = 9.73591

To the tangent of the cosine 73° 24' = 10.52572

To find the distance.

As radius = 10.00000 is to the secant of the course 73° 24' = 10.54411 so is the difference of latitude 98' = 1.99123

To the distance = 343 = 2.53534

By Inspection.

To middle latitude 57°, and 15° 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 on table marked 6½ points at the bottom, which is the cosine; and the corresponding distance 85½ multiplied by 4 gives 343 miles, the distance.

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°, the complement of the middle latitude on lines, will reach to a certain point beyond the termination of the lines on the scale. Now the extent between this point and 90° on lines, will reach from 45° to 73° 24', the course on the line of tangents. And the extent from 73° 24' the course, to 33° the complement of the middle latitude on the line of lines, being laid the same way from 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 73° N, 24° E, or ENE ¼ E nearly; but as the variation at the May is 2½ points west,

(b) 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. west, therefore the course per compass from the May is E 4 S.

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

**Example.** A ship from Breit, in latitude 48° 23' N. and longitude 4° 30' W. sailed SW 1/2 W 238 miles. Required the latitude and longitude come to?

**By Construction.**

With the course and distance construct the triangle ABC (fig. 17.), and the difference of latitude AB being measured, will be found equal to 142 miles; hence the latitude come to is 46° 1' N., and the middle latitude 47° 12'. Now make the angle DCB equal to 47° 12'; and DC being measured, will be the difference of longitude; hence the longitude come to is

**By Calculation.**

To find the difference of latitude,

| As radius | 10.00000 | |-----------|----------| | Is to the cosine of the course | 9.77593 | | So is the distance | 238 |

To the difference of latitude 141.8

Latitude of Breit, 48° 23' N.

Difference of Lat. 2 22 S.

Lat. come to 46° 1' N. Mid. Lat. 47° 12'

To find the difference of longitude (d).

As the cosine of Mid. Lat. 47° 12' | 9.83215 | |-----------------------------|----------| | Is to the sine of the course | 9.90483 | | So is the distance | 238 |

To the difference of longitude 281.3

Longitude of Breit 4° 30' W.

Difference of longitude 4° 41' W.

Longitude come to 9 11 W.

**By Inspection.**

To the course 4 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° 1' N., and middle latitude 47° 12'. Then to middle latitude 47° 12', and departure 191.1 in a latitude column, the corresponding distance is 281', which is the difference of longitude.

**By Gunter's Scale.**

The extent from 8 points to 3 1/2 points, the complement of the course on fine rhumbs, being laid the same way from the distance 238, will reach to the difference of latitude 142 on the line of numbers; and the extent from 42° 48' the complement of the middle latitude, to 53° 26', the course on the line of fines will reach from the distance 238 to the difference of longitude 281 on numbers.

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

**Example.** A ship from St Antonio, in latitude 17° 0' N. and longitude 24° 25' W., sailed NW 1/2 N. till by observation her latitude is found to be 28° 34' N.

Required the distance sailed, and longitude come to?

| Latitude St Antonio | 17° 0' N. | |---------------------|-----------| | Latitude by observation | 28° 34' N. |

Difference of lat. 11° 34' = 694 m

Middle lat. 22° 47'

**By Construction.**

Construct the triangle ABC (fig. 19), with the given course and difference of latitude, and make the angle BCD equal to the middle latitude. Now the distance AC and difference of longitude DC being measured, will be found equal to 864 and 558 respectively.

**By Calculation.**

To find the distance.

| As radius | 10.00000 | |-----------|----------| | Is to the secant of the course | 3 1/2 points | | So is the difference of lat. | 694 |

To the distance 864

To find the difference of longitude.

| As the cosine of middle latitude 22° 47' | 9.96472 | |-----------------------------------------|----------| | Is to the tangent of the course 3 1/2 points | 9.87020 | | So is the difference of latitude 694 | 2.84136 |

To the difference of longitude 558.3

Longitude of St Antonio 24° 25' W.

Difference of longitude 9 18 W.

Longitude come to 33° 43' W.

**By Inspection.**

To course 3 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° 47', or 23°, and departure 171.6 in a latitude column, the distance is 186, which multiplied by 3 is 558, the difference of longitude.

**By Gunter's Scale.**

The extent from 4 1/2 points, the complement of the course, to 8 points on the line of fine rhumbs, will reach from the difference of latitude 694 to the distance 864 on numbers; and the extent from the course 36° 34' to 67° 13', the complement of middle latitude on fines, will reach from the distance 864 to the difference of longitude 558 on numbers.

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

**Example.** A ship from latitude 26° 30' N. and longitude 30° 30' W. sailed NE 1/2 N. till her departure is 216 miles. Required the distance run, and latitude and longitude come to?

**By Construction.**

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

---

(d) This proportion is obvious, by considering the whole figure as an oblique angled triangle. To find the distance.

As the line of the course 31 points 9.80236 Is to radius 10.00000 So is the departure 216 2.33445

To the distance 340.5 To find the difference of latitude. As the tangent of the course 31 points 9.91417 is to radius 10.00000 so is the departure 216 2.33445

to the difference of lat. 263.2 Latitude sailed from 26° 30' N. Difference of latitude 4° 23' N. half 2° 12' N.

Latitude come to 30° 53' N. Mid.lat. 28° 42'

To find the difference of longitude. As radius 10.00000 is to the secant of the mid. lat. 28° 42' 10.05693 so is the departure 216 2.33445

to the difference of longitude 246.2 Longitude left, 45° 30' W. Difference of longitude 4° 6' E.

Longitude come to 41° 24' W.

By Inspection.

Under the course 31 points, and opposite to 108 half the departure, the distance is 170, and difference of latitude 131; which doubled, give 340 and 263 for the distance and difference of latitude respectively.

Again, to middle latitude 28° 42', and departure 108, the distance is 123; which doubled is 246 the difference of longitude.

By Gunter's Scale.

The extent from the course 31 points, on fine rhumbs, to the departure 216 on numbers, will reach from 8 points on fine rhumbs to about 340, the distance on numbers; and the same extent will reach from 4½ points, the complement of the course, to 263, the difference of latitude on numbers; and the extent from 61° 18' the complement of the middle latitude, to 90° on fines, 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° 24' N. and longitude 65° 39' W., a ship sailed 246 miles on a direct course between the south and east, and is then by observation in latitude 40° 48' N. Required the course and longitude in?

Latitude Cape Sable, 43° 24' N. Latitude by observation, 40° 48' N.

Difference of latitude, 2° 36' = 156° 24' 12"

Middle latitude 42° 6'

By Construction.

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

By Calculation.

To find the course. As the distance 246 is to the difference of latitude 156 so is radius 10.00000

to the cosine of the course 50° 39' 9.80219

To find the difference of longitude. As the cosine of middle latitude 42° 6' 9.87039 is to the fine of the course 50° 39' 9.88834 so is the distance 246 2.39993

to the difference of longitude 256.4 2.40888 Longitude Cape Sable, 65° 39' W. Difference of longitude 4° 16' E.

Longitude come to 61° 23' W.

By Inspection.

The distance 246, and difference of latitude 156, are found to correspond above 4½ points, and the departure is 190.1. Now, to the middle latitude 42°, and departure 190.1 in a latitude column, the corresponding distance is 256, which is the difference of longitude required.

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

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

Latitude Cape St Vincent, 37° 2' N. Latitude come to 18° 16' N.

Difference of latitude 18° 46' = 1126° 55' 18"

Middle latitude 27° 39'

By Construction.

Make AB (fig. 22.) equal to the difference of latitude 1126 miles, and BC equal to the departure 838, and join AC; draw CD so as to make an angle with CB equal to the middle latitude 27° 39'. Then the course being measured on chords is about 36° 3', and the distance and difference of longitude, measured on the line of equal parts, are found to be 1403 and 946 respectively.

By Calculation.

To find the course. As the difference of latitude 1126 is to the departure 838 so is radius 10.00000

to the tangent of the course 36° 39' 9.87170 Practice.

To find the distance.

As radius is to the secant of the course \(36^\circ 39'\) fo is the difference of latitude \(1126\)

\[ \begin{align*} \text{to the distance} & = 1403 \\ \text{To find the difference of longitude.} \\ \text{As radius} & = 10.00000 \\ \text{is to the secant of the mid. lat.} & = 27^\circ 39' \\ \text{fo is the departure} & = 833 \\ \text{to the difference of longitude} & = 946 \\ \text{Longitude Cape St Vincent} & = 9^\circ 2W. \\ \text{Difference of longitude} & = 15^\circ 46W. \\ \text{Longitude come to} & = 24^\circ 48W. \end{align*} \]

By Inspection.

One-tenth of the difference of latitude \(112.6\), and of the departure \(83.8\), are found to agree under \(3\frac{1}{2}\) points, and the corresponding distance is \(140\) which multiplied by \(10\) gives \(1400\) miles. And to middle latitude \(27^\circ \frac{3}{4}\), and \(209.5\) one fourth of the departure in a latitude column, the distance is \(236.5\); which multiplied by \(4\) is \(946\), the difference of longitude.

By Gunter's Scale.

The extent from the difference of latitude \(1126\) to the departure \(838\) on numbers, will reach from \(45^\circ\) to \(36^\circ \frac{3}{4}\) the course on tangents; and the extent from \(53^\circ \frac{1}{2}\) the complement of the course to \(90^\circ\) on lines, will reach from \(1126\) to \(1403\) the distance on numbers. Lastly, the extent from \(62^\circ \frac{1}{2}\) the complement of the middle latitude, to \(90^\circ\) on fines, will reach from the departure \(838\) to the difference of longitude \(946\) on numbers.

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

Example. A ship from Bordeaux, in latitude \(44^\circ 50'N\), and longitude \(0^\circ 35'W\), sailed between the north and west \(374\) miles, and made \(210\) miles of sailing. Required the course and latitude and longitude, come to?

By Construction.

With the given distance and departure mark the triangle ABC (fig. 23.) Now the course being measured on the line of chords is about \(34^\circ \frac{1}{2}\), and the difference of latitude on the line of numbers is \(309\) miles; hence the latitude come to, is \(49^\circ 59'N\), and middle latitude \(47^\circ 25'\). Then make the angle BCD equal to \(47^\circ 25'\), and DC measured will be \(310\) miles, the difference of longitude.

By Calculation.

To find the course.

As the distance \(= 374\) \(2.57287\) is to the departure \(= 210\) \(2.32222\) fo is radius \(= 10.00000\)

\[ \begin{align*} \text{to the fine of the course} & = 34^\circ 10' \\ \text{To find the difference of latitude.} \\ \text{As radius} & = 10.00000 \\ \text{is to the cosine of the course} & = 34^\circ 10' \\ \text{fo is the distance} & = 374 \\ \text{to the difference of latitude} & = 309.4 \\ \end{align*} \]

Make BC (fig. 24.) equal to the departure \(220\), and CD equal to the difference of longitude \(396\); then the middle latitude BCD being measured, will be found equal to \(56^\circ 15'\); hence the latitude come to is \(57^\circ 34'\), and difference of latitude \(158^\circ\). Now make AB equal to \(158\), and join AC, which applied to the scale, will measure \(271\) miles. Also the course BAC being measured on chords will be found equal \(54^\circ \frac{1}{2}\).

By Calculation.

To find the middle latitude.

As the departure \(= 220\) \(2.34242\) is to the difference of longitude \(= 396\) \(2.59769\) fo is radius \(= 10.00000\)

Middle Latitude Sailing.

to the secant of the middle latitude \(56^\circ 15'\) \(10.25527\)

Double the middle latitude \(112^\circ 30'\) Latitude left \(54^\circ 56'\)

Latitude come to \(57^\circ 34'\)

Difference of latitude \(2^\circ 38' = 158\)

To find the course.

As the difference of latitude \(158\) \(2.19866\) is to the departure \(220\) \(2.34242\) so is radius \(10.00000\)

to the tangent of the course \(54^\circ 19'\) \(10.14376\)

To find the distance.

As radius \(10.00000\) is to the secant of the course \(54^\circ 19'\) \(10.23410\) so is the difference of latitude \(158\) \(2.19866\)

to the distance \(270.9\) \(2.43276\)

By Inspection.

As the difference 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 \(56^\circ 15'\), and difference of latitude \(158\) miles.

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

By Gunter's Scale.

The extent from the difference of longitude \(396\) to the departure \(220\) on numbers, will reach from \(90^\circ\) to \(33^\circ 45'\), the complement of the middle latitude on fines; and hence the difference of latitude is \(158\) miles. Now the extent from \(158\) to \(210\) on numbers, will reach from \(45^\circ\) to \(54^\circ 4'\) on tangents; and the extent from the complement of the course \(35^\circ 4'\) to \(90^\circ\) on fines, will reach from the difference of latitude \(158\) to the distance \(271\) on numbers.

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

Example. From a port in south latitude a ship sailed SW \(45^\circ W\), and has made \(690\) miles of difference of latitude, and \(20^\circ 38'\) of difference of longitude.—Required both latitudes and distance?

By Construction.

Construct the triangle ABC (fig. 26.) with the given course and difference of latitude, and make CD equal to \(1228\) the difference of longitude. Then AC applied to the scale will measure \(1088\) miles; and the middle latitude BCD will measure \(46^\circ 47'\). Hence the latitude left is \(41^\circ 2'\), and the latitude come to \(52^\circ 32'\).

By Calculation.

To find the distance.

As radius \(10.00000\) is to the secant of the course \(4^\circ 4\) pts. \(10.19764\) so is the difference of latitude \(690\) \(2.83885\)

to the distance \(1088\) \(3.03649\)

To find the middle latitude.

As the difference of longitude \(1228\) \(3.08920\) is to the distance \(1088\) \(3.03649\) so is the fine of the course \(4^\circ 4\) pts. \(9.88819\)

to the cosine of the middle latitude \(46^\circ 47'\) \(9.83548\)

Half difference of latitude \(5^\circ 45'\)

Latitude sailed from \(41^\circ 2'\) Latitude come to \(52^\circ 32'\)

By Inspection.

To the course \(4^\circ 4\) points, and one-fourth of the given difference of latitude \(172.5\) the departure \(210.2\) Practice.

Middle Latitude Sailing.

210.2, and distance 272, which multiplied by 4 is 1088.

Now the middle latitude answering to the difference of longitude 122°, and departure 840.8, or their aliquot parts, will be found as in last problem to be 46° 47'. Hence the latitudes are 41° 2' and 52° 32' respectively.

By Gunter's Scale.

The extent from the complement of the course 3½ points to 8 points on fine rhumbs, will reach from the difference of latitude 600° to the distance 1088 miles on numbers; and the extent from the difference of longitude 122° to the distance 1088 on numbers, will reach from the course 50° 38' to the complement of the middle latitude 43° 13' on the line of rhumbs. Hence both latitudes are found as before.

Prob. XI. Given the distance failed, difference of latitude, and difference of longitude, to find both latitudes and course.

Example. In north latitude, a ship sailed 458 miles on a direct course between the north and west; and then was found to have differed her latitude 296 miles, and longitude 7° 17'. Required both latitudes and course?

By Construction.

With the distance and difference of latitude construct the triangle ABC (fig. 27.) and make CD equal to the difference of longitude 437°; then the course BAC will be found to measure 49° 44', and the middle latitude BCD 36° 54': Hence the latitude left is 34° 26', and that come to 39° 22'.

By Calculation.

To find the course.

As the difference of latitude 296 is to the distance 458 so is radius

\[ \text{radius} = \frac{296}{458} = 0.64608 \]

to the secant of the course 49° 44'

\[ \text{secant} = \frac{1}{\cos(49° 44')} = 1.1018957 \]

To find the middle latitude.

As the difference of longitude 437 is to the distance 458 so is the sine of the course 49° 44'

\[ \text{sine} = \frac{437}{458} = 0.95444 \]

to the cosine of the middle latitude 36° 54'

\[ \text{cosine} = \frac{1}{\cos(36° 54')} = 0.990293 \]

Half difference of latitude 2° 28'

Latitude left 34° 26'

Latitude come to 39° 22'

By Inspection.

To half the distance 229 the difference of latitude is 150.2 at 49°, and 147.2 at 50°.

Then, as 3:0:2.2::60':44'.

Therefore the course is 49° 44'.

Also the departure is 172.8 at 49°, and 175.4 at 50°.

Hence, as 3:0:2.2::2.6:1.9.

And 172.8 + 1.9 = 174.7 = half the departure.

Now to half the difference of longitude 218.5 in a distance column, the difference of latitude is 176.8 at 36°, and 174.5 at 37°.

Then, as 2.3:2.1::60':54'.

Hence the middle latitude 36° 54'; and therefore the latitude failed from is 34° 26', and that come to 39° 22'.

Vol. XII. Part II.

Navigation.

By Gunter's Scale.

The extent from the distance 458 to the difference of latitude 296 on numbers, will reach from 90° to 40° 16' the complement of the course on rhumbs; and the extent from the difference of longitude 437 to the distance 458 on numbers, will reach from the course 49° 44', to the complement of the middle latitude 36° 54' on the line of rhumbs: Hence the latitudes are 34° 26' and 39° 22' respectively.

Prob. XII. Given the distance, middle latitude, and difference of longitude, to find both latitudes and course.

Example. The distance is 384 miles between the south and east, the middle latitude 54° 6', and difference of longitude 6° 36'. Required both latitude and course?

By Construction.

With the middle latitude 54° 6', and difference of longitude 396, construct the triangle BCD (fig. 28.) and make AC equal to the given distance 384 miles. Then the course BAC will be found to measure 37° 12', and the difference of latitude AB 306 miles — Hence the latitude failed from is 56° 39', and that come to 51° 33'.

By Calculation.

To find the course.

As the distance 384 is to the difference of longitude 396 so is the cosine of middle latitude 54° 6'

\[ \text{cosine} = \frac{384}{396} = 0.976817 \]

to the sine of the course 37° 12'

\[ \text{sine} = \frac{1}{\sin(37° 12')} = 0.978153 \]

To find the difference of latitude.

As radius 10,000,000 is to the cosine of the course 37° 12' so is the distance 384

\[ \text{distance} = \frac{10,000,000}{\cos(37° 12')} = 2,58433 \]

to the difference of latitude 305.9

\[ \text{latitude} = \frac{305.9}{\sin(37° 12')} = 2,48553 \]

Middle latitude 54° 6'

Half difference of latitude 2° 33'

Latitude failed from 56° 39' N

Latitude come to 51° 33' N

By Inspection.

To the middle latitude 54°, and half the difference of longitude 198 in a distance column, the number in a latitude column is 116.4. Now half the distance 192 and 116.4 in a departure column, are found to agree nearly under the course 37°, and the corresponding difference of latitude 153; which doubled is 306 miles. Hence the latitude left is 56° 39' N, and latitude come to 51° 33' N.

By Gunter's Scale.

The extent from the distance 384 to the difference of longitude 396 on the line of numbers, will reach from 35° 54', the complement of the middle latitude, to 37° 12', the course on the line of rhumbs: And the extent from 90° to 52° 48' the complement of the course on rhumbs, will reach from the distance 384 to the difference of latitude 306 on numbers. Hence the latitudes are known.

Prob. XIII. 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 $51^\circ 18' N$, longitude $9^\circ 46' W$, sailed as follows:

- SW by S 34 miles, - WNW 63 miles, - NNW 48 miles, - NE by E 85 miles.

Required the latitude and longitude come to?

| Courses | Diff. | Diff. of latitude | Departure | |---------|------|------------------|-----------| | SW by S | 54 | 44.9 | 30.0 | | WNW | 63 | 12.3 | 61.8 | | NNW | 48 | 44.4 | 18.4 | | NE by E | 85 | 53.9 | |

| Latitude of Cape Clear | 51° 18' N | |------------------------|-----------| | Latitude come to | 52° 24' N | | Sum | 103° 42' | | Middle latitude | 51° 51' |

Now, to middle latitude $51^\circ 51'$ or $52^\circ$, and departure $44.5$ in a latitude column, the difference of longitude is $72$ in a distance column.

Longitude of Cape Clear - $9^\circ 46' W$ Difference of longitude - $1^\circ 12' W$

Longitude come to - $10^\circ 58' W$

Ex. A ship from Halliford in Iceland, in lat. $64^\circ 30' N$, long. $27^\circ 15' W$, sailed as follows: - SSW 46 miles, - SW 61 miles, - SSW 59 miles, - SE by E 86 miles, - SE by E 76 miles.

Required the lat. and long. come to?

| Courses | Diff. | Diff. of Lat. | Departure | |---------|------|--------------|-----------| | SSW | 46 | 42.5 | 17.6 | | SW | 61 | 43.1 | 43.1 | | SJW | 59 | 57.9 | 11.5 | | SE by E | 86 | 47.8 | 71.5 | | SE by E | 76 | 72.7 | 22.0 |

| Latitude Halliford | 64° 30' N | |--------------------|-----------| | Difference of latitude | 4° 24' S | | Latitude in | 60° 6' N | | Sum | 124° 36' | | Middle latitude | 62° 18' |

Now, to middle lat. $62^\circ 18'$, and departure $21.3$, the difference of long. is $46^\circ E$.

Longitude Halliford - $27^\circ 15' W$ Longitude in - $16^\circ 29' W$

The Error of comm. method, in this Ex. is $12'$. It was observed in Middle Latitude Sailing, that the difference of longitude made upon an oblique thumb 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,

\[ R : \text{cof. of lat.} :: \text{part of equat.} : \text{similar part of parallel.} \]

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

\[ R : \text{cof. lat.} :: \text{part of meridian} : \text{similar part of parallel.} \]

Or sec. lat. : R :: part of merid. : similar part of parallel.

Hence, fecant latitude \( \frac{\text{R}}{\text{part of meridian}} = \text{part of parallel.} \)

But in Mercator's chart the parallels of latitude are equal, and radius is a constant quantity. If therefore the latitude be assumed successively equal to \( r, r', r'' \), &c., and the corresponding parts of the enlarged meridian be represented by \( a, b, c, \) &c.; then,

\[ \frac{\text{fecant } r'}{\text{fecant } 2'} = \frac{\text{fecant } 2'}{\text{fecant } 3'} \]

part of mer. \( a : \text{fecant } r' : \text{fecant } 2' : \text{fecant } 3' : \text{part of mer. } c, \) &c.

Hence fecant \( r' : \text{part of mer. } a : \text{fecant } 2' : \text{part of mer. } b : \text{fecant } 3' : \text{part of mer. } c, \) &c.

Therefore by 12th V. Euclid,

Secant \( r' : \text{part of mer. } a : \text{fecant } r' + \text{fecant } 2' + \text{fecant } 3' : \) &c. : parts of mer. \( a + b + c, \) &c.

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

Since CD : LK :: R : fecant LD, fig. 15.

And in the triangle CED,

\[ ED : CD :: R : \text{tangent CED}; \]

Therefore, ED : LK :: R : fecant LD × tangent CED

Hence \( LK = \frac{ED \times \text{sec. LD} \times \text{tang. CED}}{R}. \)

But \( \frac{ED \times \text{sec. LD}}{R} \) is the enlarged portion of the meridian answering to ED. Now the sum of all the quantities \( \frac{ED \times \text{fecant LD}}{R} \) corresponding to the sum of all the ED's contained in AS will be the meridional parts answering to the difference of latitude AS; and MN is the sum of all the corresponding portions of the equator LK.

Whence \( MN = \text{mer. diff. of lat.} \times \text{tangent} \frac{CED}{R}. \)

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

This equation answers to a right-angled rectilineal triangle, having an angle equal to the course; the adjacent side equal to the meridional difference of latitude, and the opposite side the difference of longitude. This triangle is therefore similar to a triangle constructed, with the course and difference of latitude, according to the principles of plane sailing, and the homologous sides will be proportional. Hence if, in fig. 29, 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.

**Example.** Required the course and distance between Cape Finisterre, in latitude 42° 52' N, longitude 9° 17' W, and Port Praya in the island of St Jago, in latitude 14° 54' N, and longitude 23° 29' W?

| Lat. Cape Finisterre | 42° 52' | Mer. parts | 2852 | |---------------------|---------|------------|------| | Latitude Port Praya | 14° 54' | Mer. parts | 904 |

Difference of lat. = 27° 58'

Mer. diff. lat. 1948

Longitude Cape Finisterre - 9° 17' W Longitude Port Praya - 23° 29' W

Diff. longitude = 14° 12' = 852

**By Construction.**

Draw the straight line AD (fig. 29.) to represent the meridian of Cape Finisterre, upon which lay off AB, AD equal to 1678 and 1948, the proper and meridional differences of latitude; from D draw DE perpendicular to AD, and equal to the difference of longitude 852 join AE, and draw BC parallel to DE; then the distance AC will measure 1831 miles, and the course BAC 23° 37'.

**By Calculation.**

To find the course.

As the meridian difference of lat. 1948 = 3.28959 is to the difference of longitude 852 = 2.93044 so is radius 10.00000

To the tangent of the course 23° 37' = 9.64085

To find the distance.

As radius 10.00000 is to the fecant of the course, 23° 37' = 10.03798 so is the difference of latitude 1678 = 3.22479 to the distance 1831 = 3.26277

By

---

(d) 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. 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 1948 and 852 respectively. Now these are found to agree nearest under 24°; 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 meridian difference of latitude, to 852, the difference of longitude on the line of numbers, will reach from 45° to 23° 57', the course on the line of tangents. And the extent from 66° 23', the complement of the course to 90° on fines, will reach from 1678, the proper difference of latitude, to 1831, the distance on the line of numbers.

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

Example. A ship from Cape Hinlopen in Virginia, in latitude 38° 47' N, longitude 75° 4' W, sailed 267 miles NEWN. Required the ship's present place?

By Construction.

With the course and distance sailed construct the triangle ABC (fig. 30.), and the difference of latitude AB being measured, is 222 miles; hence the latitude come to is 42° 20' N, and the meridian 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° 48' W.

By Calculation.

To find the difference of latitude.

| As radius | 10.00000 | |-----------|----------| | is to the cosine of the course, 3 points | 9.91985 | | is the distance | 267 | | to the difference of latitude | 222 |

Lat. Cape Hinlopen = 38° 47' N. Mer. parts 2528

Difference of lat. 3° 42' N.

Latitude come to 42° 29' N. Mer. parts 2821

Meridional difference of lat. 293

To find the difference of longitude.

| As radius | 10.00000 | |-----------|----------| | is to the tangent of the course, 3 points | 9.82489 | | is the mer. diff. of latitude | 293 |

to the difference of longitude 195.8

Longitude Cape Hinlopen 75° 4' W

Difference of longitude 3° 16' E

Longitude come to 71° 48' W

By Inspection.

To the course 3 points, and distance 267 miles, the difference of latitude is 222 miles; hence the latitude is 42° 29', and the meridian difference of latitude 293. Again, to course 3 points, and 146.5 half the mer. difference of latitude, the departure is 97.9, which doubled is 195.8, the difference of longitude.

By Gunter's Scale.

The extent from 8 points to the complement of the course 5 points on fine rhumbs, will reach from the distance 267 to the difference of latitude 222 on numbers; and the extent from 4 points to 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 Canfo in Nova Scotia, in latitude 45° 20' N, longitude 60° 55' W, sailed SE 4° S, by observation is found to be in latitude 41° 14' N. Required the distance sailed, and longitude come to?

Lat. Port Canfo 45° 20' N Mer. parts 3058 Lat. in by observation 41° 14' N Mer. parts 2720

Difference of lat. 4° 6' Mer. diff. lat. 338

By Construction.

Make AB (fig. 31.) equal to 246, and AD equal to 338; draw AE, making an angle with AD equal to 3° 3 points, and draw BC, DE perpendicular to AD. Now AC being applied to the scale, will measure 338, and DE 306.

By Calculation.

To find the distance.

As radius 10.00000 is to the secant of the course, 3.4 points 10.13021 so is the difference of latitude 246 2.39993

to the distance 332 2.52114

To find the difference of longitude.

As radius 10.00000 is to the tangent of the course, 3.1 points 9.95729 so is the mer. diff. of latitude 338 2.52892

to the difference of longitude 306.3 2.45621

Longitude Port Canfo 60° 55' W Difference of longitude 5° 6' E

Longitude in 55° 49' W

By Inspection.

Under the course 3.1 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 4.3 points to 8 points on fine rhumbs, will reach from the difference of latitude 246 m. to the distance 332 on numbers; and the extent from 4 points, to the course 3.1 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° 58' N, longitude 6° 20' W, the corrected course was NW 6° W, and departure 420 miles. Required With the course and departure construct the triangle ABC (fig. 32); now AC and AB being measured, will be found equal to 476 and 224 respectively; hence the latitude come to is 37° 42' 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 therefore is 14° 56' W.

By Calculation.

To find the distance.

As radius 10.00000 is to the cosecant of the course, 5½ pts 10.05457 so is the departure 420 2.62325

to the distance 476.2 2.67782

To find the difference of latitude.

As radius 10.00000 is to the co-tangent of the course, 5½ pts 9.72796 so is the departure 420 2.62325

to the difference of latitude 224.5 2.35124

Lat. of Salce 33° 58' N Mer. parts 2169 Diff. of lat. 3° 44' N

Latitude in 37° 42' N Mer. parts 2445

Mer. difference of latitude 276

To find the difference of longitude.

As radius 10.00000 is to the tangent of the course 5½ pts 10.27224 so is the mer. diff. of latitude 276 2.44091

to the difference of longitude 516.3 2.71295

Longitude of Salce 60° 20' W Difference of longitude 8° 36' W

Longitude in 14° 56' W

By Inspection.

Above 5½ 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.

By Gunter's Scale.

The extent from 5½ points, the course on fine rhumbs, to the departure 420 on numbers, will reach from 8 points on fine rhumbs to the distance 476 on numbers; and from the complement of the course 2½ points on fine rhumbs to the difference of latitude 224 on numbers.

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

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

Example. A ship from St Mary's, in latitude 36° 57' N, longitude 25° 9' W, sailed on a direct course between the north and east 1162 miles, and is then by observation in latitude 49° 57' N. Required the course and longitude come to?

Lat. of St Mary's 36° 57' N Mer. parts 3470 Lat. come to 49° 57' N Mer. parts 2389

Difference of lat. 13° 0 Mer. diff. lat. 1081

780

By Construction.

Make AB (fig. 33.) equal to 780, and AD equal to 1081; draw BC, DE perpendicular to AD; make AC equal to 1162 m. and through AC draw ACE. Then the course or angle A being measured, will be found equal to 47° 50', and the difference of longitude DE will be 1194.

By Calculation.

To find the course.

As the distance 1162 3.06521 is to the difference of latitude, 780 2.89209 so is radius 10.00000

to the cosine of the course 47° 50' 9.82688

To find the difference of longitude.

As radius 10.00000 is to the tangent of the course, 47° 50' 10.04302 so is the mer. diff. of latitude 1081 3.03383

to the difference of longitude 1194 3.07685

Longitude of St Mary's 25° 9' W Difference of longitude 19° 54' E

Longitude in 5° 15' W

By Inspection.

Because the distance and difference of latitude exceed the limits of the table, take the tenth of each; these are 116.2 and 78.0; Now these are found to agree nearest above 4½ points, which is therefore the course; and to this course, and opposite to 108.1, one tenth of the meridional difference of latitude, in a latitude column, is 119.3 in a departure column, which multiplied by 10 is 1193, the difference of longitude.

By Gunter's Scale.

The extent from the distance 1162 m. to the difference of latitude 780 m. on numbers, will reach from 90° to 42° 10' in the line of fines. And the extent 45° to the course 47° 50' on the line of tangents, will reach from the meridional difference of latitude 1081 to the difference of longitude 1194 on numbers.

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

Example. From Aberdeen, in latitude 57° 9' N, longitude 20° 9' W, a ship sailed between the south and east till her departure is 145 miles, and latitude come to 53° 32' N. Required the course and distance run, and longitude come to?

Latitude Aberdeen 57° 9' N. mer. parts 4199 Latitude come to 53° 32' N. mer. parts 3817

Difference of latitude 3° 37' mer. diff. of lat. 382

217. With the difference of latitude 217m. and departure 146m., construct the triangle ABC (fig. 34.), make AD equal to 382, draw DE parallel to BC, and produce AC to E: Then the course BAC will measure 33° 56', the distance AC 261, and the difference of longitude DE 257.

By Calculation.

To find the course.

As the difference of latitude 217 is to the departure 146 so is radius

to the tangent of the course 33° 56' 9.82789

To find the distance.

As radius is to the secant of the course 33° 56' so is the difference of latitude 217

to the distance 261.5 2.41755

To find the difference of longitude.

As the difference of latitude 217 is to the mer. diff. of latitude 382 so is the departure 146

to the difference of longitude 257 Longitude of Aberdeen 2° 9 W Difference of longitude 4° 17 E

Longitude come to 2° 8 E.

By Inspection.

The difference of latitude 217, and departure 146, are found to agree nearest under 34°, and the corresponding distance is 262 miles. To the same course, and opposite to 19° 7, the nearest to 19° half the meridional difference of latitude, is 128.6 in a departure column, which doubled is 257, the difference of longitude.

By Gunter's Scale.

The extent from the difference of latitude 217, to the departure 146 on numbers, will reach from 45° to about 34°, the course on the line of tangents; and the same extent will reach from the meridional difference of latitude 382 to 257, the difference of longitude on numbers.—Again, the extent from the course 34° to 90 on fines, will reach from the departure 146 to the distance 261 on numbers.

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

Example. A ship from Naples, in latitude 40° 51' N, longitude 14° 14' E, sailed 252 miles on a direct course between the south and west, and made 173 miles of setting. Required the course made good, and the latitude and longitude come to?

By Construction.

With the distance and departure make the triangle ABC (fig. 35.) as formerly.—Now the course BAC being measured by means of a line of chords will be found equal to 43° 21', and the difference of latitude applied to the scale of equal parts will measure 183': hence the latitude come to is 37° 48' N, and meridional difference of latitude 237.—Make AD equal to 237, and complete the figure, and the difference of longitude will measure 224': hence the longitude in is 10° 30' E.

By Calculation.

To find the course.

As the distance 252 is to the departure 173 so is radius

to the fine of the course 43° 21' 9.83665

To find the difference of latitude.

As radius is to the cosine of the course 43° 21' so is the distance 252

to the difference of latitude 183.2 Latitude of Naples 40° 51' N. Mer. parts 2690 Difference of latitude 3° 3 S

Latitude come to 37° 48' N. Mer. parts 2453

Meridional difference of latitude 237

To find the difference of longitude.

As radius is to the tangent of the course 43° 21' so is the mer. diff. of latitude 237

to the difference of longitude 223.7 Longitude of Naples 14° 14' E Difference of longitude 3° 44' W

Longitude in 10° 30' E.

By Inspection.

Under 43° and opposite to the distance 252 m. the departure is 171.8, and under 44°, and opposite to the same distance, the departure is 175.0.

Then as 3.2 : 1.2 :: 60' : 22'

Hence the course is 43° 22'.

Again, under 43° and opposite to 118.5, half the meridional difference of latitude in a latitude column, is 110.5 in a departure column; also under 44° and opposite to 118.5 is 114.4.

Then as 3.2 : 1.2 :: 3.9 : 1.5

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

By Gunter's Scale.

The extent from the distance 252 on numbers to 90° on fines will reach from the departure 173 on numbers to the course 43° 21' on fines; and the same extent that will reach from the complement of the course 46° 21' on fines will reach to the difference of latitude on numbers.—Again, the extent from 45° to 43° 21' on tangents will reach from the meridional difference of latitude 237 to the difference of longitude 224 on numbers.

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

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

Longitude of Tercera 27° 6' W Longitude in 18° 24' W

Difference of longitude 8° 42' = 522. By Construction.

Make the right angled triangle ADE (fig. 36.) having the angle A equal to the course 32°, and the side DE equal to the difference of longitude 522; then AD will measure 8355, which added to the meridional parts of the latitude left, will give these of the latitude come to 48° 46'; hence the difference of latitude is 601; make AB equal thereto, to which let BC be drawn perpendicular; then AC applied to the scale will measure 708 miles.

By Calculation.

To find the meridional difference of latitude.

As radius 10,00000 is to the co-tangent of the course 32° 0' 10,20421 so is the difference of longitude 5 22 2,71767

to the mer. difference of latitude '8352 2,92188 Latitude of Tercera 38° 45' N Mer parts 2546 Mer. diff. of lat. 835

Latitude come to 48° 46' N. Mer. parts 3361

Difference of latitude 10 = 601 miles.

To find the distance.

As radius 10,00000 is to the secant of the course 32° 0' 10,07158 so is the difference of latitude 601 2,77887

to the distance 707.7 2,85045

By Inspection.

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

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

By Gunter's Scale.

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

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

Rule. With the several courses and distances, complete the Traverse Table, and find the difference of latitude, departure, and course made good, and the latitude come 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, 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; 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, which place in the fourth or fifth columns, according as the course is easterly or westerly; and the difference between the sums of these columns will be the difference of longitude made good upon the whole of the same name with the greater.

Remarks:

1. When the course is north or south, there is no difference of longitude. 2. When the course is east or west, the difference of longitude cannot be found by Mercator's Sailing; in this case the following rule is to be used: 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.

Example I. Four days ago we took our departure from Faro-head, in latitude 58° 40' N. and longitude 4° 50' W, and since have sailed as follows: NW 32 miles, W 60 miles, WNW 93 miles, WB 77 miles, SW 58 miles, and W 49 miles—Required our present latitude and longitude? ### Navigation

#### Traverse Table

| Course | Diff. | N | S | E | W | |--------|-------|-------|-------|-------|-------| | NW | 32 | 22.6 | | 22.6 | | | W | 69 | | 69.0 | | | | WNW | 93 | 35.6 | | 85.9 | | | WSW | 77 | 15.0 | | 75.5 | | | SW | 58 | 41.0 | | 41.0 | | | WSW | 49 | 7.2 | | 48.5 | |

#### Longitude Table

| Successive Latitudes | Merid. Parts | Merid. Diff. Lat. | Diff. of Longitude | |----------------------|--------------|-------------------|--------------------| | | | | E | W | | 58°40' | 4370 | | | | | 59 | 4415 | 45 | 45.0 | | | 59 | 4415 | | | 134.0 | | 59 | 4484 | 69 | | 166.5 | | 59 | 4454 | 30 | | 151.0 | | 58 | 4374 | 80 | | 80.0 | | 58 | 4361 | 13 | | 88.0 |

Longitude of Faro-head = 4° 30' W.

Difference of longitude = 11° 4 W.

Longitude in = 15° 54 W.

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**Example II.** A ship from latitude 78° 15' N, longitude 28° 14' E. sailed the following courses and distances. The latitude come to is required, and the longitude, by both methods: the bearing and distance of Hacluit's head-land, in latitude 79° 55' N, longitude 11° 55' E, is also required?

#### Traverse Table

| Course | Diff. | N | S | E | W | |--------|-------|-------|-------|-------|-------| | WNW | 154 | 58.9 | | 142.3 | | | SW | 96 | | 67.9 | | 67.9 | | NW½W | 89 | 56.4 | | 68.8 | | | NE | 110 | 107.9 | | 21.5 | | | NW¼N | 56 | 45.0 | | 33.4 | | | SE½E | 78 | | 73.4 | 26.3 | |

By Rule 1ft.

Latitude left = 78° 15' N.

Mer. pts = 7817

Diff. of latitude = 2° 7 N.

Lat. come to = 80° 22 N.

Mer. pts = 8504

Meridional diff. of latitude = 687

As difference of lat. = 126.9

To mer. diff. of latit. = 687

So is the departure = 264.6

To difference of longit. = 1432

Longitude left = 28° 14' E.

Longitude in = 4° 22' E.

The error of this method, in the present example, is therefore 1° 23'.

Now to 78.5 half the meridional difference of latitude, and 185.0 half the difference of longitude, the course 67°, and opposite to the difference of latitude 27°, the distance is 69 miles—Hence Hacluit's head-land bears S 67° 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. The constant quantity 12.633114 (G) is to the difference or sum of the logarithmic tangents of half the co-latitudes of two places, according as these latitudes are of the same, or a contrary denomination; as the tangent of the course is to the difference of longitude.

Demonst. Let CABP, Plate CCCXXXVIII. fig 37, be a section of one fourth of the earth in the plane of the meridian; and let AC be the radius of the equator, and B any given place whose latitude is therefore AB. Draw BD perpendicular to AC, and BE parallel to it; and let Bb be a very small portion of the meridian, as one minute. — Now put CA = r, DB

\[ z = \frac{2.302585 \times 180^\circ \times 60'}{3.15^\circ} \times \log_r \sqrt{\frac{r+y}{r-y}} = \frac{1}{.00012633114} \times \log_r \sqrt{\frac{r+y}{r-y}} \]

But log. \( \sqrt{\frac{r+y}{r-y}} = \log_r \sqrt{\frac{r+\text{fine AB}}{r-\text{fine AB}}} \)

And the tang. \( (45 + \frac{1}{2} \text{AB}) = \frac{1}{\tan_r (45 + \frac{1}{2} \text{AB})} \)

Therefore \( z = \frac{1}{.00012633114} \times \log_r \sqrt{\tan_r (45 + \frac{1}{2} \text{AB})^2} = \log_r \tan_r \text{half co-latitude} \)

Hence the meridional parts answering to any given latitude, is found by dividing the difference between the log. of the radius and the log. tangent of half the complement of latitude, by the constant quantity .00012633114, &c.; and the meridional difference of latitude is obtained by dividing the difference or sum of the logarithmic tangents of half the co-latitudes, according as they are of the same or a contrary name, by the above quantity.

And the meridional difference of latitude multiplied by the tangent of the course, is equal to the difference of longitude. Hence the proposition is manifest.

This method shall be illustrated with examples performed by calculation: the other methods of solution are purposely omitted.

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

Example. Required the bearing and distance of Ostend, in lat. 51° 14' N.; long. 2° 56' E from Aberdeen, in latitude 57° 9' N. and longitude 2° 9' W.

Long. Aberdeen, 2° 9 W. Lat. 57° 9' comp. 32° 51' half 16° 25½ tangent 9 46951 Long. Ostend, 2° 56 E. Lat. 51° 14' comp. 38° 46 half 19 23 tangent 9 54033

Diff. longitude 5° 5° 305 Diff. Lat. 5° 55' = 355 Difference 7682

To find the course.

As the difference of the log. tang. 7682 = 3.88547 is to the constant logarithm 11.10151 so is the diff. of longitude 305 = 2.48430

To the tangent of the course 26° 38' = 9.70034

To find the distance.

As radius 10.00000 is to the secant of the course 26° 38' = 10.04871 so is the difference of latitude 355 = 2.55023

to the dist n.e. 397.1 = 2.59894

Vol. XII. Part II.

(g) In this case the tangent is to consist of five figures besides the index; but if the table extends to 6 or 7 figures, the above number will be 126.33, &c. or 1263.3, &c. To find the distance.

As radius

is to the secant of the course

so is the difference of latitude

Sailing.

To find the difference of longitude.

As the constant logarithm

is to the tangent of the course

so is the sum of the log. tangents

to the difference of longitude

Longitude left

Difference of longitude

Longitude come to

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

Example. From Scarborough, in latitude 54° 20' N. longitude 6° 10' W., a ship sailed NE 1/4 E 210 miles. Required the latitude and longitude come to?

To find the difference of latitude.

As radius

is to the cosine of the course

so is the distance

to the difference of latitude

Lat. left,

Lat. in,

Diff. of lat.

To find the course.

As the distance

is to the difference of latitude

so is radius

to the cosine of the course

To find the difference of longitude.

As the constant logarithm

is to the tangent of the course

so is the difference of log. tangents

to the difference of longitude

Longitude left

Difference of longitude

Longitude come to

PROB. IV. Given both latitudes and departure, to find the course, distance, and difference of longitude.

Example. A ship from latitude 18° 24' S., longitude 3° 25' E. sailed between the north and west upon a direct course, till by observation she is in latitude 12° 42' N., and has made 970 miles of departure. Required the course, distance, and longitude come to?

To find the distance.

As radius

is to the secant of the course

so is the difference of latitude

to the distance

To find the difference of longitude.

As the constant logarithm

is to the tangent of the course

so is the difference of log. tangents

to the difference of longitude

Longitude

Longitude come to

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

Example. A ship from latitude 32° 58' N. longitude 16° 28' W. sailed SE 5 S., and made 164 miles of departure. Required the distance run, and latitude and longitude come to?

To find the distance.

As the sine of the course

is to radius

so is the departure

to the distance

To find the difference of latitude.

As the tangent of the course

is to radius

so is the departure

to the difference of latitude

Lat. left,

Diff. of lat.

Lat. come to,

Difference

To find the difference of longitude.

As the constant logarithm

is to the tangent of the course

so is the difference of log. tangents

to the difference of longitude

Longitude Practice.

Method of Longitude left, 16° 28 W. refolving Difference of longitude, 3 12 E. the Problems of Mercator's Longitude in, 13 16 W. Sailing.

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

Example. A ship from Cape Volatas, in latitude 28° 55' S. longitude 15° 53' E. sailed 286 miles between the south and west, and made 238 miles of departure. Required the course, the latitude and longitude come to?

To find the course.

As the distance 286 2.45637 is to the departure 238 2.37658 so is radius 10,00000

to the sine of the course 56° 19' 9.92021

To find the difference of latitude.

As radius 10,00000 is to the cosine of the course 56° 19' 9.74398 so is the distance 286 2.45637

to the difference of latitude 158.6 2.20035

Lat. Cape Volatas, 28° 55' S. comp. 61° 30' half 30° 3' 1/2 tang. 9.77087 Diff. of Lat. 2 39 S.

Latitude in, 31° 34' S. comp. 58° 26' half 29° 13' tang. 9.74762

To find the difference of longitude.

As the constant logarithm 11.10151 is to the tangent of the course 56° 19' 10.17620 so is the diff. of log. tangents 2325 3.36042

to the difference of longitude 276.1 2.44111

Longitude Cape Volatas, 15° 53' E. Difference of longitude, 4° 36' W.

Longitude come to, 11° 17' E.

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

Example. A ship from latitude 16° 54' N. longitude 62° 16' W. sailed upon a NW. by N. course, until her longitude by observation is 68° 10' W. Required the distance run, and latitude come to?

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

| Table A. | Arc. | 10° | 20° | 0° | 40° | 50° | 60° | 70° | 80° | 90° | |----------|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----| | 1° | 3' | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | | 2 | 12 | 4 | 2 | 2 | 1 | 1 | 1 | 1 | 1 | 1 | | 3 | 27 | 13 | 8 | 6 | 4 | 3 | 2 | 1 | 1 | 1 | | 4 | 47 | 23 | 14 | 10 | 7 | 5 | 3 | 1 | 1 | 1 | | 5 | 74 | 36 | 23 | 16 | 11 | 8 | 5 | 2 | 1 | 1 | | 6 | 107 | 52 | 33 | 22 | 16 | 11 | 7 | 3 | 1 | 1 | | 7 | 145 | 70 | 44 | 30 | 21 | 15 | 9 | 4 | 1 | 1 | | 8 | 190 | 92 | 58 | 40 | 28 | 19 | 12 | 6 | 1 | 1 |

Table B.

| Lat. | 0° | 1° | 2° | 3° | 4° | 5° | 6° | 7° | 8° | 9° | |------|----|----|----|----|----|----|----|----|----|----| | | | | | | | | | | | |

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

Distance 290 ar. co. log. 7.53760 Dif. of longitude 300 log. 2.47712

Approximate course 41° 41' fine 9.82279 To lat. 50°, and half dif. long. 15°, the 1st Corr. in a dep. column is 115 + 1 55

4 U 2 I Method of Intable A to co. course 48° and 1st corr. 1° 55′; the second correction is To course 41° and diff. long 5°, the number is 6′, of which \( \frac{1}{7} \) (Tab. B) being taken gives

True course S. 43° 33′ W

To find the difference of latitude.

As radius 10,000,000 is to the cosine of the course 43° 33′ 9.86020 so is the distance 290 2.46240

to the difference of latitude 210.2 2.32260

Latitude left 50° N

Difference of latitude 3° 30′ S

Latitude come to 46° 30′ N

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 semiaxis 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 which might probably err more from the 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.

**Example I.** At 11th A.M. the Girdlenefs bore WNW, and at 2nd P.M. it bore NWN; the course during the interval SW 5 knots an hour. Required the distance of the Ship from the Nefs at each station?

By Construction.

Describe a circle NE.SW (fig. 38.) 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, SW, and equal to 15 miles the distance sailed.—From H draw HF in a NWN direction, and the point F will represent the Girdlenefs. 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 SW and WNW points; and the angle CHF equal to 4 points, being the supplement of the angle contained between the SW and NWN points; hence CFH is 3 points; to find the distances CF, FH.

To find the distance CF.

As the sine of CFH 3 points 9.74474 is to the sine of CHF 4 points 9.84948 so is the distance CH 15 miles 1.17609

to the distance CF 19.07 1.28083

To find the distance FH.

As the sine of CFH 3 points 9.74474 is to the sine of FCH 9 points 9.99157

so is the distance CH 15 miles 1.17609

to the distance FH 26.43 1.42292

**Example II.** 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 SESE 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 said point from the island of Brehaut?

By Construction.

Describe a circle, (fig. 39.) 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 SESE line DF 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 34.5 miles, and the inclination of BI to the meridian will be found equal to 63°.

By Calculation.

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

To find the angle BIF.

As the distance BI 39 1.59106 is to the distance BF 26 1.41497 so is the sine of BFI 78° 45′ 9.99157

to the fine of BIF 40 50 9.81548

Sum 119 35

Angle FBI 60° 25′

EBF 33° 45′

Difference, or EBI 26° 40′

Bearing of Jersey from Brehaut N63° 20′ E

To find the distance FI.

As the sine of BFI 78° 45′ 9.99157 is to the sine of FBI 60° 25′ 9.93934 so is the distance BI 39 miles 1.59106

to the distance FI 34.58 1.53883

**Example III.** At noon Dungenefs bore per compass N6W distance 5 leagues; and having run NWB 7 knots an hour, at 5 P.M. we were up with Beachyhead. Required the bearing and distance of Beachyhead from Dungenefs?

By Construction.

Describe a circle (fig. 40) to represent the horizon; from the centre C draw the NW line CD equal to 15 miles; and the NWB line CB equal to 35 miles; join DB, which applied to the scale will measure about 26.5 miles; and the inclination of DB to the meridian will be found equal to N75° 4′ W.

By Calculation.

In the triangle DBC 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

Sum 50 angle B and D 12 Difference 20 half sum - 6 pts. = 67°30' As the sum of the distances 50 1.69897 is to their difference 20 1.30103 so is the tangent of half sum angles 67°30' 10.38378

to the tangent of half their difference 44° 9.98484

Angle CDB 111°30' Supplement 68°30' Angle NCD 11°15'

Magnetic bearing N 79°45' W. Or by allowing 2½ points of westerly variation, the true bearing of Beachy-head from Dungeness will be W 1° S nearly.

To find the distance.

As the fine of CDB 111°30' 9.96868 is to the fine of BCD 45° 9.49483 so is the distance BC 35 1.54407

to the distance BD 26.6 1.42487

Example IV. Running up Channel E.S.W. per compass at the rate of 5 knots an hour. At 11th A.M. the Eddystone lighthouse bore N 6° E. and the Start point N 6° E.; and at 4 P.M. the Eddystone bore N.W. and the Start N 3° E. Required the distance and bearing of the Start from the Eddystone, the variation being 2½ points W?

By Construction.

Let the point C (fig. 41) represent the first station, from which draw the N 6° E. line CA, the N 6° E. line CB, and the E.S.W. line CD, which make equal to 25 miles, the distance run in the elapsed time; then from D draw the N.W. line DA intersecting CA in A, which represents the Eddystone; and from the same point draw the N 3° 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°47'.

By Calculation.

In the triangle CAD are given CD equal to 25 miles, the angle CAD equal to 4½ points, the distance between N 6° E. and N.W. and the angle ADC equal to 4 points, the distance between the N.W. and N.E. points; to find the distance CA.

As the fine of CAD 4½ points 9.86979 is to the fine of CDA 4 points 9.84948 so is the distance CD 25 miles 1.39794

to the distance CA 23.86 1.37763

In the triangle BCD, are given the distance CD 25 miles, the angle CBD 4½ points the interval between N 6° E. and N 3° E.; and CDB 7½ points, the distance between N.W. and N.E.; to find the distance CD.

As the fine of CBD 4½ points 9.88319 is to the fine of CDB 7½ points 9.99947 so is the distance CD 25 miles 1.39794

to the distance CB 32.3 1.50922

In the triangle CAB, the distances CA, CB, are given, together with the included angle ACB, equal to 4 points, the distance between N 6° E. and N 6° E.; to find the angle CAB and distance AB.

Distance CA 23.86 Sum of CAB and ABC 135°

Sum 56.16 Half 67°30' Difference 8.44 As the sum of the distances 56.16 1.74943 is to their difference 8.44 0.92634 so is the tangent of half sum angles 67°30' 10.38278

to the tangent of half difference 19.56 9.55969

Angle CAB 87°26' Angle CAF 14°4

Bearing per compass S 73°22'E or E.S.W.; and the variation 2½ points being allowed to the left of E.S.W., gives E 1° N., the true bearing of the Start from the Eddystone.

To find the distance.

As the fine of CAB 87°26' 9.99956 is to the fine of ACB 45° 9.84948 so is the distance CB 32.3 1.50922

to the distance AB 22.86 1.3594

Example V. A ship from a port in latitude 57°9' N., longitude 20°9' W., sailed 82 miles on a direct course, and spoke a ship that had run 100 miles from a port in latitude 56°21' N., longitude 2°50' W. Required the course of each ship, and the latitude and longitude come to?

Lat. 57°9' N Mer parts 4.199 Lon. 2°9' W. 56°21' N 4.112 2°50' W.

Diff. of lat. 48 Mer. diff. lat. 87 Diff. lon. 41

By Construction.

With the meridional difference of latitude, the difference of longitude, and difference of latitude, construct the triangles ADE, ABC (fig. 42.) as in Mercator'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°47'. 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°47' E.; and the other ship's course, or angle BAM, will be 67°47', or E.S.E. 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 By Calculation.

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

To find the bearing of the ports.

As the meridional diff. of lat. 87 - 193952 is to the diff. of long. 41 - 161278 so is radius - 10.00000

to the tangent of the bearing 25° 14' - 9.67326

To find the distance of the ports.

As radius - 10.00000 is to the secant of the bearing 25° 14' - 10.04355 so is the diff. of latitude 48 - 1.68124

to the distance - 53.06 - 1.72479

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

To find the angle ACM.

AM - 82 MC - 100 ar. co. log. - 8.00000 AC - 53.06 ar. co. log. - 8.27523

Sum - 235.06 Half - 117.53 log. - 2.0715 Difference - 35.53 log. - 1.55059

27 29 cosine - 9.94798

Angle ACM 54° 58' Angle BAC 25° 14'

Southernmost ship's course N80° 12' E

To find the angle MAC.

As AM - 82 - 1.91381 is to MC - 100 - 2.00000 so is the sine of ACM - 54° 58' - 9.91319

to the sine of MAC - 93° 3 - 9.99938 Angle BAC - 25° 14'

Northernmost ship's course S 67° 49' E, or ESE.

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

As radius - 10.00000 is to the cosine of the course - 67° 49' - 9.57700 so is the distance - 82 - 1.91381

to the diff. of lat. 30.96 - 1.49081 Latitude of northernmost port 57° 9 Mer. parts 4199

Latitude in 56° 38' Mer. parts 4142

Meridional difference of latitude - 57

To find the difference of longitude FG.

As radius - 10.00000 is to the tangent of the course - 67° 49' - 10.38960

so is the mer. diff. of lat. 57 - 1.75587

to the diff. of long. 139.8 - 2.14547 Longitude left 2° 9' W Difference of longitude 2° 20' E

Longitude in 0° 11' E

CHAP. IX. Of Windward Sailing.

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

Example I. 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 6 points of the wind. Required the course and distance on each tack?

By Construction.

Draw the SSW line CB (fig. 43.) equal to 48 miles. Make the angles ACB, A'B'C', each equal to 6 points. Hence the first course will be W, and the second SE; also the distance CA, or AB, applied to the scale will measure 62½ miles, the distance to be sailed on each board.

By Calculation.

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

As radius - 10.00000 is to the secant of C - 6 points - 10.47116 so is CD - 24 miles - 1.38021

to CA - 62.7 - 1.79737

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

By Construction.

Draw the NW line CA (fig. 44.) equal to 64 miles; from C draw CB W&5, and from A draw AD parallel thereto, and in an opposite direction; bisect AC in E, and draw BED parallel to the N&E rhumb, meeting CB, AD in the points B and D; then CB = AD applied to the scale will measure 36½ miles, and BD = 2CB = 72½ miles

By Calculation.

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

As radius - 10.00000 is to the secant of BCF, - 5 points - 10.25526 so is CF - 16 m. - 1.20412

to CB - 36.25 - 1.53938

Example III. A ship which can lie within 5½ points of the wind, is bound to a port 76 miles to the windward, the wind being N&N, which it is intended to reach Practice.

Windward reach on four boards, the first being on the larboard tack. Required the course and distance on each?

By Construction.

Plate XXXVIII.

Draw the NE'N line CA (fig. 45.) equal to 36 miles, and bisect it in B; from C and B draw lines parallel to the E'S rhumb; and from A and B draw lines parallel to the SSE'E point, meeting the former in the points D and E. Now the distances AD, BD, BE, and CE, are equal; and any one of them applied to the scale will measure 19.1 miles.

By Calculation.

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

As radius

is to the secant of FCE

so is CF

to the distance CE

Example IV. A ship bound to a port bearing NW distant 40 miles, with the wind at NNE, intends to reach it on two boards. Required the course and distance on each tack, the ship lying within 5½ points of the wind?

By Construction.

Draw the NW line CA (fig. 46.) equal to 40 miles; and because the wind is NNE, and the ship can lie within 5½ points of the wind, the course on the larboard tack will be ENE, and on the starboard NW. Therefore, from the centre C draw the ENE line CB, and from it draw the NW line AB, meeting CB in B; then CB and AB applied to the scale will measure 26.7 and 48.1 m. respectively.

By Calculation.

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

To find the distance CB.

As the fine of B

is to the fine of A

so is the distance CA

to the distance CB

To find the distance AB.

As the fine of B

is to the fine of C

so is the distance CA

to the distance AB

Example V. A ship close hauled within 5 points of the wind, and making one point of lee-way, is bound to a port bearing SSW, distant 54 miles, the wind being SSW. 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 or rocks. Required the course and distance on each tack?

By Construction.

Draw the SSW line CA (fig. 47.) equal to 54 m. and as the wind is SSW, and the ship makes her course good within 6 points of the wind, therefore the course on the larboard tack will be SWNW, and on the starboard ESE; hence from C draw the SWNW line CB, and from A draw AD parallel thereto; bisect CA in E, and draw BED parallel to the ESE line; then will CB and AD be the distances on the larboard tack, Windward which applied to the scale, each will be found to measure 37.4; and the distance on the starboard tack BD will measure 42.4 miles.

By Calculation.

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

To find CB, the distance on the larboard tack.

As the fine of B

is to the fine of E

so is the distance CE

to the distance BC

To find BE half the distance on the starboard tack.

As the fine of B

is to the fine of C

so is the distance CE

to the distance BE

Whole distance AC

Example VI. A ship plying to the windward, with the wind at NNE, after sailing 51 miles on each of two tacks, 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. 48.) equal to 36 miles; draw AB perpendicular to CA, and draw the NNE line CB, meeting AB in B; make CD, BD each equal to 51 miles; and these being measured, will be found equal to 6 points.

By Calculation.

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

As the distance CD

is to the diff. of latitude CA

so is the secant of ACB

to the cosine of BCD

Example VII. A ship that makes her way good within 6½ points of the wind, reaches her port on two boards; the first being on the larboard tack 25 miles, and the other on the starboard tack 38 miles; and the difference of latitude is 21 miles north. Required the bearing of the port, and direction of the wind?

By Construction.

With the given distances 25 and 38 miles, and the included angle equal to 16 - 2 × 6½ = 3 points, construct the triangle BCD (fig. 49.); hence CB will be known. Draw CA equal to 21 miles, the given difference of latitude; from A draw AB perpendicular to CA, and make CB equal to what it was before determined; make DE = DC, and draw the line CE, which will represent the direction of the wind, and the angle ACB is the bearing of the port: now ACE will be found equal to 52 ½, and ACB 18°.

By Calculation.

In the triangle BCD are given BC = 25 m., BD = 38 m., and the angle D = 3 points, to find the angle BCD, and distance CB. To find the angle BCD.

Distance BD = 38 Angle BDC 33°45' BC = 25

Sum 63 BCD + CBD = 146°15'

Difference 13 BCD + CBD = 73°7½

As the sum of the sides 63 1.79934 is to the difference of the sides 13 1.11394 so is the tan. of half sum angles 73°7½ 10.51806

to the tan. of half diff. angles 34°13½ 9.83266

Angle BCD 107°21'

To find the distance BC.

As the sine of BCD 107°21' 9.97978 is to the sine of BDC 33°45' 9.74474 so is the distance BD 38 1.57978

to the distance BC 22.12 1.34474

To find the angle ACB, the bearing of the port.

As the distance BC 22.12 1.34474 is to the distance AC 21.1 1.32222 so is radius 10.00000

to the cosine of ACB 18°17' 9.97748

Angle BCD 107°21'

ACD 125°38' DCE 73°7'

Direction of the wind N 52°31'E

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 ebbsings 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 in 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.

Example I. A ship sailed NNE at the rate of 8 knots an hour during 18 hours, in a current setting NW/W 2½ miles an hour—Required the course and current distance made good?

By Construction.

Draw the NNE-line CA (fig. 50), equal to 18×8 = 144 miles, and from A draw AB parallel to the NW/W rhumb, and equal to 18×2½ = 45 miles: now BC being joined will be the distance, and NCB the course. The first of these will measure 159 miles, and the second 6°23'.

By Calculation.

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

To find the course made good.

Diff. AC 144 Angr. BAC = 9 pts = 101°15' Diff. AB .45

Sum 189 B+C 78°45' Diff. 99 B+C 39°22½

As the sum of the sides 189 2.27646 is to the difference of the sides 99 1.99503 so is the tan. of half sum angles 39°22½ 9.91417

to the tan. of half diff. angles 23°15½ 9.63334

Angle ACB 16°7' Angle ACN 22°30'

Course made good N 6°23'E

To find the distance.

As the sine of ACB 16°7' 9.44341 is to the sine of CAB 101°15' 9.99157 so is the distance AB 45 1.05321

to the distance CB 159 2.20137

Example II. A ship from a port in latitude 42°52'N, sailed SSW ¾W 17 miles in 7 hours, in a current setting between the north and west; and then the same port bore ENE, and the ship's latitude by observation was 42°42'N. Required the setting and drift of the current?

By Construction.

Draw the SSW ¾W line CA (fig. 51.) 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 7°2'.

By Calculation.

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

As radius 10.00000 is to the secant of BCD 6 points 10.41710 so is the diff. of lat. CB 10 miles 1.00000

to the distance CD 26.13 1.47110

Again, in the triangle ACD are given the distance AC = 17 miles, CD = 26.13, and the angle ACD 4½ points; to find the remaining parts. To find the setting of the current.

Distance DC = 26.13 Distance AC = 17.0

Angle ACD = 4½ points. CAD + CDA = 11½

Sum = 43.13

Difference = 9.13

As the sum of the sides is to the difference of the sides so is the tang. of half sum angles

64° 41' = 10.32509

to the tang. of half diff. angles

24° 6' = 9.65078

Angle CAD = 88° 47' Angle CAE = ACB = 1 pt. = 16° 52'

Setting of the current EAD = 71° 55'

To find the drift of the current.

As the fine of CAD = 88° 47' = 9.99990 is to the fine of ACD = 4½ points = 9.88819 so is the distance CD = 26.13 = 1.41710

to the drift of curr. AD = 20.2 = 1.30539

Hence the hourly rate of the current is \( \frac{20.2}{7} = 2.9 \) knots.

Example III. A ship, from latitude 38° 20' N., sailed 24 hours in a current setting NW by N., and by account is in latitude 38° 42' N., having made 44 miles of sailing; but the latitude by observation is 38° 58' N. Required the course and distance made good, and the drift of the current?

By Construction.

Make CE (fig. 52.) equal to 22 miles, the difference of latitude by D, R, and EA = 44 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 by 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° 1'.

By Calculation.

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

To find the drift of the current AB.

As radius = 10.00000 is to the secant of ABF = 3 points = 10.08015 so is BF = 16 miles = 1.20412

to the drift of the current AB = 19.24 = 1.28427

Hence the hourly rate \( \frac{19.24}{24} = 0.8 \).

To find AF.

As radius = 10.00000 is to the tangent of ABF = 3 points = 9.84289 so is BF = 16 = 1.20412

to AF = 10.69 = 1.02901

Departure by account EA = 44.

True departure EF = DB = 33.31

Vol. XII. Part II.

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

To find the course.

As the difference of latitude CD = 38. is to the departure DB = 33.31 so is radius = 10.00000

to the tangent of the course = 41° 14' = 9.94279

To find the distance.

As radius = 10.00000 is to the secant of the course = 41° 14' = 10.12376 so is the difference of latitude = 38 = 1.57978

to the distance = 50° 53' = 1.70354

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

By Construction.

From A (fig. 53.) draw the SE by S. line AB = 20, which will represent the ship's apparent tract 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° 1'.

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.

AB = 20 B + C = 13

Sum = 42 Difference = 2

As the sum of the sides is to the difference of the sides so is the tangent of half sum angles

73° 7' = 10.51806

to the tangent of half diff. angles

8.554 = 9.19584

Setting of the current S 64° 12' W., or SW by W.

To find the drift of the current.

As the fine of ACB = 64° 12' = 9.95446 is to the fine of BAC = 33° 45' = 9.74474 so is the distance AB = 20 = 1.30103

to the velocity of current BC = 12° 34' = 1.09137

and \( \frac{12.34}{4} = 3.1 \), its hourly rate.

Example V. A ship bound from Dover to Calais, lying

Current lying 21 miles to the SEbE½E, and the flood tide setting NE½E 2½ miles an hour. Required the course she must steer, and the distance run by the log at 6 knots an hour to reach her port?

By Construction.

In the position of the SEbE½E rhumb, draw DC = 21 miles (fig. 54); draw DE NE½E = 2½ miles; from E with 6 miles cut DC in F; draw DB parallel to EF, meeting CB drawn parallel to DE; then the distance DB applied to the scale will measure 19.4; and the course SDB will be SE½S.

By Calculation.

In the triangle DBF, given DE = 2½ miles, EF = 6 miles, and the angle EDF = 6 points; to find the angle DFE = CBD.

As the hourly rate of sailing 6 m. 0.77815 is to the hourly rate of the current 2½ m. 0.39794 so is the fine of EDF = 6 points 67°30' 9.96562

to the fine of DFE 22 38 9.58541

Angle SDC = 5½ points = 61°52'

Course SDB 39°14'SE½S.

In the triangle DBC, given DC = 21 miles, the angle BDC = DFE = 22°38', and the angle DCB = DEF = 6 points; to find the distance DB.

As the fine of DBC 89°52' 9.99999 is to the fine of DCB 67°30' 9.96562 so is the true distance DC 21 m. 1.32222

to the distance by the log DB. 19.4 m. 1.28785

Example VI. A ship at sea in the night has sight of Scilly light, bearing NE½N, distant 4 leagues, it being then flood tide, setting ENE 2 miles an hour. What course and distance must the ship sail to make the Lizard, which bears from Scilly E½S, distance 17 leagues?

By Construction.

Draw the NE½N line AS = 12 miles (fig. 55); hence S will represent Scilly; from S draw SL = 51 miles, and parallel to the E½S rhumb, then L will represent the Lizard; draw LC parallel to the ENE rhumb, and equal to 2 miles, and make CD = 5 miles; from A draw AB parallel to CD, meeting LC produced in B; then AB will be the distance, and the angle SAB the course: the first of these applied to the scale will measure 41.9 miles, and the course will be S 88° E.

By Calculation.

In the triangle SAL are given the sides AS, SL = 12 and 51 miles respectively, and the angle ASL = 10½ points; to find the other parts.

To find the angles.

Distance SL = 51 m. Angle ASL = 10½ points.

AS = 12 m. SAL + SLA = 5½

Sum 63 m. SAL + SLA = 2½ = 30°56'

Difference 39 m. 2

As the sum of the sides 63 1.79934 is to the diff. of the sides 39 1.59106 so is the tangent of half their sum angles 30°56' 9.77763

to the tan. of half their difference 20 21 9.56935

Angle SAL = 51°17'

NAS = 3 points 33°45' NAL - 85°2 LAE = FLA 4°58' FLB = 2 points 22°30' ALB = DLC 17°32'

To find the distance AL.

As the fine of SAL 51°17' 9.89223 is to the fine of ASL 10½ points 9.94543 so is the distance SL 51 miles 1.70757

to the distance AL 57.64 1.76077

Again, in the triangle DLC, are given the side DC = 5 miles, the ship's run in an hour; LC = 2 miles, the current's drift in the same time; and the angle DLC = 17°32'; to find the angle LDC = LAB.

As the dist. DC = 5 miles 0.69897 is to the distance LC = 2 miles 0.30103 so is the fine of DLC = 17°32' 9.47894

to the fine of LDC 6°55' 9.08100

Angle NAL = 85°2

NAB = 91°57'

Course S 88°3 E

In the triangle ABL, the side AL, together with the angles, are given, to find the distance AB.

As the fine of ABL 155°33' 9.61689 is to the fine of ALB 17°32' 9.47894 so is the distance AL 57.64 1.76077

to the distance AB 41.96 1.62282

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

Various methods, beside 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 And. Mackay, A. M. F. R. S. E.

I. Of Cooke's Triangular Instrument.

Description. The stock abcd (fig. 56.) 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 ef, a cylindrical piece of wood one inch diameter; gb is an aperture on the surface of the stock about a quarter of an inch wide, which discloses one-twelfth part

The surface of the cylinder contained; the edge \(d\) 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 of a line of meridional parts 22 feet long, which contains the meridional parts for every minute from the equator as far towards the pole as navigation is practicable; and the smallest division on it is not less than \(\frac{1}{75}\)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 engrooved into the edge of the block \(a\), so that it may move freely from \(a\) to \(b\); a limb from this box extends to \(k\), which serves to mark that degree of the perpendicular \(l\) which is parallel to the centre of the semicircle \(m\); \(l\) is two feet long, and graduated on both edges as the block; it is perpendicular to the stock, and is fixed in the box \(i\), by which it may be moved from \(a\) to \(b\); \(o\) 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 block \(e\); \(m\) 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 block 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 \(q\), 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 \(d\), which is perpendicularly under the centre of the semicircle. Fig. 57. 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 fine, of the middle latitude; which is done by means of the data procured from the compass, log-line, and quadrant: whence it follows, from the nature of similar triangles, or from the relation which exists between the sides of triangles and the sines of their opposite angles, that the parts of the instrument become proportional to those which they represent; and will ascertain the length of the lines, or the extent of the angles sought, by its graduations.

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

Plane Sailing.

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

Example. A ship from latitude 24° 18' N failed NW \(b\) N 168 miles. Required the latitude come to, and departure?

Set the centre of the semicircle perpendicularly over the given latitude 24° 18', and the index to the course 3 points; move the perpendicular until it cut the index at the given distance 163; then at the point of intersection on the perpendicular is 93.3 miles, the departure, and on the bale, by the edge of the box, is 26° 8', the latitude come to.

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

Example. Let the latitude failed from be 43° 50' N, that come to 47° 8' N, and the course NNE. Required the distance and departure?

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

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

Example. Let the latitude failed from be 32° 38' N, the course SW \(b\) S, and the departure 200 miles. Required the distance and latitude come to?

Move the centre of the semicircle to the latitude left 32° 38', set the index to the given course 3 points, and move the perpendicular till the given departure 200 cuts the index; at this point on the index is 365 miles, and the edge of the box will cut the latitude come to 27° 39' N.

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

Example. Let the latitude left be 17° 10' N, the latitude come to 21° 40' N, and the distance failed on a direct course between the north and west 300 miles. Required the course and departure?

Move the semicircle and box to the given latitudes, and the index until the distance found thereon meets the perpendicular, then at the point of contact on the perpendicular is 130.8, the departure, and on the semicircle by the index is 25° 50', the course.

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

Example. The distance failed is 246 miles between the south and east, the departure is 138 miles, and the latitude left 51° 10' N. Required the course and latitude come to?

Set the centre of the semicircle to 51° 10', the latitude failed from; find the distance 246 on the index, and the departure 138 on the perpendicular; then move both till the points meet, and the course 34° 10' will be found on the semicircle by the index, and the latitude in, 47° 47' N, by the edge of the box.

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

Example. A ship from latitude 43° 10' N, sailed between the north and west till she is in latitude 47° 14' N, and has made 170 miles of departure. Required the course and distance?

Move the centre of the semicircle over 43° 10', and the edge of the box to 47° 14'; 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° 52' on the semicircle.

Traverse Sailing.

Example. A ship from latitude 46° 48' N fail- Instrument: SSW 24 miles, SW 36 miles, and SE 40 miles. Required the latitude in, together with the direct course and distance?

Set the semicircle to the latitude sailed from 46° 48', and the index to the course SSW, 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° 27' N on the base. For the next course and distance, bring the semicircle to the latitude marked by the perpendicular, and lay down the course SW: 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° 52' 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° 36' will be intercepted between the latitude sailed from 46° 48', and the latitude come to 45° 12', last marked by the perpendicular; and also the departure made good will be intercepted between that point on the perpendicular where the first departure commenced, and that where the last terminated. Now, with the difference of latitude 1° 36' and the departure, the course will be S 8° 30' W, and distance 97 miles, by last problem in Plane Sailing.

Parallel Sailing.

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

Example. Let the common latitude be 49° 30' N, and the difference of longitude 3° 30'. Required the distance?

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

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

Example. Let the latitude of the given parallel be 49° 30' N, the distance sailed 136.4 E. Required the difference of longitude?

Set the index to the complement of the latitude 49° 30', and mark the distance sailed on the perpendicular; then move it until it meets the index, and the point of intersection will show the difference of longitude 210' or 3° 30' on the index.

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

Example. The distance sailed due east is 136.4, and the difference of longitude 3° 30'. Required the latitude of the parallel?

Find the difference of longitude 210 on the index, and the distance 136.4 on the perpendicular, and move both until these numbers meet, and the complement of the latitude 49° 30' will be shown by the index on the semicircle.

Mercator's and Middle Latitude Sailing.

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

Example. Required the course and distance between two places whose latitudes and longitudes are 50° 50' N, 19° 0' W, and 54° 30' N, 15° 30' W, respectively?

By Mercator's Sailing.

To find the course.

Move the centre of the semicircle perpendicularly over the meridional degree answering to latitude 50° 50' N, then move the box until the edge of the perpendicular cuts the meridional parts of the other latitude 54° 30' N, and move the index until it cuts the difference of longitude 3° 30' on the perpendicular, and the index will mark the course 30° 10', or NNE 4° 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° 50' N, and the edge of the perpendicular to the latitude 54° 30' N, then the perpendicular will cut the distance 254.7 on the index.

By Middle Latitude Sailing.

To find the departure.

Move the centre of the semicircle to the latitude 50° 50', and the edge of the index to the complement of the middle latitude 37° 20' 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° 30', and the index until it cuts the departure 128 on the perpendicular; then will the perpendicular mark the distance on the index 254.7 miles, and the index will mark the course on the semicircle 30° 10', or NNE 4° E nearly.

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

Example. A ship from latitude 50° 50' N, longitude 19° 0' W, sailed N 30° 10' E, until she is in latitude 54° 30' N. Required the distance and difference of longitude?

By Mercator's Sailing.

To find the difference of longitude.

Move the box and semicircle as in the former problem to the meridional parts of the given latitudes, then set the index to the course, and it will mark the difference of longitude 3° 30' on the perpendicular: Hence the longitude in is 15° 30' W.

To find the distance.

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

By Middle Latitude Sailing.

To find the distance and departure.

Move the semicircle and perpendicular to the given latitudes, and the index to the course; then the perpendicular will show the departure 128 miles, and the Practice.

Instruments to solve problems in sailing, independent of calculation.

To find the difference of longitude.

Set the index to the complement of the middle latitude on the semicircle, and move the box until the termination of the departure on the perpendicular meets the index, which will mark the difference of longitude thereon 210 m. or 3° 30'.

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

Example. From latitude 50° 50' N, longitude 19° W, a ship sailed 254.7 miles between the north and east, and by observation is in latitude 54° 30' N. Required the course and difference of longitude?

By Mercator's Sailing.

To find the course.

Move the perpendicular and semicircle to the given latitudes, and the index until the distance marked on it meets the perpendicular; then the index will mark the course N 30° 10' E on the semicircle.

To find the difference of longitude.

Screw the index to the course, move the perpendicular and semicircle to the meridional parts of the given latitudes, and the space intercepted between the limb of the box and the index will be the difference of longitude 3° 30'.

By Middle Latitude Sailing.

To find the departure and course.

Move the semicircle and perpendicular to the given latitudes, and the index until the distance marked on it cuts the perpendicular; then the perpendicular will show the departure 128 miles, and the semicircle the course N 30° 10' E.

To find the difference of longitude.

Set the index to 37° 20', the complement of the middle latitude on the semicircle, and move the perpendicular until the termination of the departure on it cuts the index; then the point of intersection will mark the difference of longitude 210 miles on the index.

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

Example. Let the latitude and longitude sailed from be 56° 40' S and 28° 55' E respectively, the latitude come to 61° 20' S, and departure 172.7 miles. Required the course, distance, and difference of longitude?

By Mercator's Sailing.

To find the course and distance.

Move the perpendicular and semicircle to the given latitudes (N); then move the index till it meets the extremity of the departure on the perpendicular; the distance will be marked on the index 329, and the course S 31° 35' E or SSE 4° nearly on the semicircle.

To find the difference of longitude.

Move the perpendicular and semicircle to the meridional parts of the given latitudes, and the index will cut the difference of longitude on the perpendicular 5° 35'.

By Middle Latitude Sailing.

The course and distance are found as before.

To find the difference of longitude.

Set the index to 31°, the complement of the middle latitude on the semicircle, and move the perpendicular until the departure marked on it cuts the index, and this point of intersection will mark the difference of longitude on the index 335 m. or 5° 35'.

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

Example. Let the latitude left be 56° 40' S, longitude 28° 55' E, the course S 31° 35' E, and distance 329 m. Required the latitude and longitude come to?

By Mercator's Sailing.

To find the latitude come to.

Set the semicircle to the latitude sailed from, and the index to the course, and bring the perpendicular to the distance, which at the same time will mark the latitude come to 61° 20' S.

To find the difference of longitude.

Screw the index to the course, and move the semicircle and perpendicular to the meridional parts of both latitudes; then the index will cut the difference of longitude on the perpendicular 5° 35'.

By Middle Latitude Sailing.

The latitude arrived at is found as above.

To find the departure.

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

To find the difference of longitude.

Set the index to 31°, the complement of the middle latitude on the semicircle, and move the perpendicular until the departure marked on it cuts the index, and the division on the index at the point of intersection will be the difference of longitude 335.

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

Example. Let the latitude sailed from be 56° 40' N, longitude 28° 35' W, the course N 31° 35' W, and departure 172.7. Required the distance, and the latitude and longitude come to?

By Mercator's Sailing.

To find the distance and latitude come to.

Move the semicircle to the latitude left, and the index to the course; mark the departure on the perpendicular, 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° 20' 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° 35' on the perpendicular.

By Middle Latitude Sailing.

Find the distance sailed and latitude in as above.

---

(n) 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. and the difference of longitude as in Problem IV. by middle latitude sailing.

**Prob. VII.** One latitude, the distance failed, and departure given, to find the course, difference of latitude, and difference of longitude.

**Example.** The latitude failed from is $48^\circ 30' N$, and longitude $14^\circ 40' W$, the distance run is 345 miles between the 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 $8^\circ 35' E$ on the semicircle, and the latitude come to $43^\circ 49'$ on the base.

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

*By Middle Latitude Sailing.*

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

**II. Of Mackay's Rectangular Instrument.**

**Description.** Fig. 58. is a representation of this instrument, of about one-third of the original size.—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. 59. 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 deemed 200 miles, and so on, each number being increased in a tenfold ratio; and the intermediate numbers are to be reckoned accordingly.

**Plane Sailing.**

**Prob. I.** The course and distance failed given, to find the difference of latitude and departure.

**Practice.**

**Example.** Let the course be NE $\frac{1}{4} N$, distance 44 miles. Required the difference of latitude and departure?

Move the index until the graduated edge be over 34 independent points, and find the given distance 44 miles on the index of calculation; this distance will be found to cut the parallel of calculation, 34 miles, the difference of latitude in the side column, and that of 28 miles, the departure at the top.

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

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

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

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

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

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

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

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

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

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

**Example.** Let the distance be 58 miles, and the departure 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 $15^\circ$.

**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 $43^\circ$.

**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.

---

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

By Mercator's Sailing.

Instruments to solve Problems in Sailing, independent of Calculation.

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

Example. Let the latitude of a parallel be 48°, and the difference of longitude between two places on it 3° 40', required their distance?

Put the index to 48°, the given latitude, and find the difference of longitude 220 on the index, and the corresponding parallel from the side will be 147, the distance required.

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

Example. The latitude of a parallel is 56°, and the distance between two places on it 209 miles. Required their difference of longitude?

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

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

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

Bring 60 on the index to cut the parallel of 46.5 from the side, then the edge of the index will give 39° 11', the latitude required.

Middle Latitude and Mercator's Sailing.

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

Example. Required the course and distance between Genoa, in latitude 44° 25' N, longitude 8° 36 E, and Palermo, in latitude 38° 10' N, longitude 13° 38' E?

By Mercator's Sailing.

Take the interval between 38° 10' and 44° 25' on the enlarged meridian, which laid off from C upwards will reach to 500; now find the difference of longitude 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° 8' on the line of degrees; then find the difference of latitude 375 on the side column, and its parallel will intersect the index at 43°, the distance.

By Middle Latitude Sailing.

Put the index to 41° 18', the complement of the middle latitude on degrees, and the difference of longitude 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° 10'.

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° 22' N. and 12° 8' W., respectively, the course NNW ¼ W. and distance 500 miles. Required the latitude and longitude come to?

Put the index to the course 2½ points, and find the distance 500 miles thereon; then the corresponding difference of latitude will be 4.41 miles, and the departure 235½ miles; hence the latitude is 46° 43' N. Now take the interval between the latitudes of 39° 22' and 46° 43' on the enlarged meridian, which laid off from C will reach to about 605, the parallel of which will intersect the vertical parallel of the difference of longitude 323 at the edge of the index; hence the longitude in is 17° 31' W.

By Middle Latitude Sailing.

Find the difference of latitude and departure as before, and hence the latitude is in 46° 43' N, and the middle latitude 43° 3'. Now put the index to 43° 3', and the horizontal parallel of the departure 235½ will intersect the index at 322, the difference of longitude.

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

Example. The latitude sailed from is 22° 54' S, and longitude 42° 45' W, the course is SE by E, and latitude come to 26° 8' 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° 54' and 26° 8' on the enlarged meridian; lay off that extent from the centre on the side column, and it will reach to 213; the parallel of this number will intersect the vertical parallel of 319, the difference of longitude. Hence the longitude in is 37° 21' W.

By Middle Latitude Sailing.

With the given course and difference of latitude find the distance and departure as before; then bring the index to the middle latitude 24° 31'; 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° 30' N. and 46° 17' W. respectively; the course is NE ¼ 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 3½ 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° 16' N.

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

By Middle Latitude Sailing.

Find the distance and difference of latitude as directed above; then bring the index to 22° 53', the middle latitude, and the horizontal parallel of 212, the departure. Instruments 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° 10' N, and 9° 20' W, respectively, and the latitude in is 46° 40' N. Required the course and longitude in?

*By Mercator's Sailing.*

Bring the distance 500 on the index to intersect the horizontal parallel of the difference of latitude 390; then the course 38° 44' is found on the line of degrees by the edge of the index, and the vertical parallel of the above point of intersection is that answering to 31° 3', the departure.

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

*By Middle Latitude Sailing.*

The course and departure are found as formerly, and the middle latitude is 43° 25', to which bring the edge of the index, and the horizontal parallel of 313, the departure, will intersect the index at 431, the difference of longitude.

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

**Example.** Let the latitude sailed from be 42° 52' N, long. 9° 17' W, the departure 250 miles W, and the latitude come to 36° 18' 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° 24'.

Take the interval between the latitudes on the enlarged meridian; which being laid off from the centre will reach to 312; now the horizontal parallel of 512 will cut the vertical parallel of 32° 3', the difference of longitude, at the edge of the index. The longitude come to is therefore 14° 42' W.

*By Middle Latitude Sailing.*

The course and distance are to be found in the same manner as above. Then bring the index to 39° 35', the middle latitude, and the horizontal parallel of 250 will intersect the edge of the index at 324° 3', 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° 38' N, longitude 17° 6' W, sailed 586 miles between the south and No. 238.

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(k) This problem may be expressed thus:—Two places lying on the same parallel, and of these four, the latitude, difference of longitude, distance, and angle of position, any two being given to find the other two.—Now this problem contains four different cases, the most useful of which is given above. The others serve rather as exercises in spherical trigonometry than of any real utility in navigation, and are therefore omitted. The same is to be understood of the following problems.

---

**Great Circle Sailing.**

Move the index till the distance 586 intersects the vertical parallel of the departure 336; then the corresponding horizontal parallel will be 480, the difference of latitude, and the course 35°. Hence the latitude in is 24° 38' N.

Now take the interval between the latitudes on the enlarged meridian, which laid off from the centre will reach to 547, the horizontal parallel of which will cut the vertical parallel of 383, the difference of longitude. The longitude in is therefore 23° 29' W.

*By Middle Latitude Sailing.*

Find the course and difference of latitude as before, and hence the middle latitude is 28° 38', to which bring the index, and the horizontal parallel of 336, the departure, will intersect the index at 383, the difference of longitude.

It seems unnecessary to enlarge any further on the use of this instrument, as the above will make it sufficiently understood.

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**Chap. XII. Of Great Circle Sailing.**

The application of spherical trigonometry to the solution of triangles formed upon the surface of the earth, is called Great Circle Sailing.

The earth being supposed an exact sphere, the shortest distance between two places is the arch of a great circle intercepted between them; and therefore the distance sailed upon a direct course from one place to another, will always be longer than the arch of a great circle contained between them, except when the rhumb line coincides with a great circle, which can only happen when the ship sails on a meridian or on the equator.

Although it is impossible to make a ship describe an arch of a great circle, yet she may be kept so near it as to make the error almost insensible.

The terms that enter into this sailing are, the latitudes of the places, their difference of longitude and distance, and the angles contained between the distance and the meridians of the places, called the angles of position.

**Prob. I.** Given the common latitude of two places on the same parallel, and their difference of longitude, to find the distance and angle of position (x).

**Example.** Required the distance between St Mary's, in latitude 36° 57' N, longitude 25° 9' W, and Cape Henry, in latitude 36° 57' N, and longitude 76° 27' W.

*By Construction.*

Describe the circle EPQS (fig. 60.) to represent the meridian of one of the places; draw the equator EQ and the earth's axis PS at right angles thereto; make ED, QA, each equal to the chord of 36° 57', the the given latitude, and draw the parallel of latitude ABD, the radius of which is the tangent of $53^\circ 3'$; the co-latitude; describe the meridian PBS with the secant of $51^\circ 18'$; the difference of longitude; then A and B will be the two places. Draw the diameter AF, and through the points ABF describe a great circle; then the arch AB will be the distance, and the angle PAB the angle of position. Now these being measured by the rules given in *spherics*, will be found equal to $45^\circ 28'$ and $73^\circ 54'$ respectively.

**By Calculation.**

From P draw PG perpendicular to AB, by describing the arch with the secant of half the difference of longitude; then, in the right-angled spherical triangle AGP, are given AP = $53^\circ 3'$, the complement of latitude, and the angle APG = $25^\circ 39'$, half the difference of longitude; to find AG half the distance, and PAG the angle of position.

1. To find the distance.

| As radius | 10.00000 | |-----------|----------| | is to the fine of AP | $53^\circ 3'$ | | fo is the fine of APG | $25^\circ 39'$ |

to the fine of AG

Distance AB

2. To find the angle of position.

| As radius | 10.00000 | |-----------|----------| | is to the cotangent of AP | $53^\circ 3'$ | | fo is the cotangent of APG | $25^\circ 39'$ |

to the cotangent of PAG

**Prob. II.** Given the latitude of a place, and the difference of longitude between it and a place on the equator, to find the distance between them, and the angles of position.

**Example.** Required the shortest distance between the island of St Thomas, in latitude $0^\circ 0'$, longitude $10^\circ 0$ E, and Port St Julian, in latitude $48^\circ 51'$ S, and longitude $65^\circ 10'$ W?

**By Construction.**

Describe the circle EPQS (fig. 61.), to represent the meridian of one of the places; draw the equator EQ, and axis PS; make EB equal to the chord of $48^\circ 51'$, and B will represent Port St Julian; make CA equal to the semitangent of the complement of the difference of longitude; draw the diameter BF, and through the points BAF draw the great circle BAF; then AB will be the distance, ABE the angle of position at Port St Julian, and BAE the complement of that at St Thomas. These being measured by the rules given in *spherics*, will be found equal to $74^\circ 35'$, $71^\circ 36'$, and $51^\circ 22'$, respectively.

**By Calculation**

In the right-angled spherical triangle AEB, AE, and EB are given, to find AB and the angles A and B.

1. To find the distance.

| As radius | 10.00000 | |-----------|----------| | is to the cosine of AE | $66^\circ 10'$ | | fo is the cosine of EB | $48^\circ 51'$ |

to the cosine of AB

Now $74^\circ 35' = 4475$ miles, which is 57 miles less than the distance found by Mercator's Sailing.

**Vol. XII. Part. II.**

2. To find the angle of position at St Thomas.

| As radius | 10.00000 | |-----------|----------| | is to the fine of AE | $66^\circ 10'$ | | fo is the cotangent of EB | $48^\circ 51'$ |

to the tangent, ang. of position

3. To find the angle of position at Port St Julian.

| As radius | 10.00000 | |-----------|----------| | is to the fine EB | $48^\circ 51'$ | | fo is the cotangent of AE | $66^\circ 10'$ |

to the cotangent of ABE

Hence a ship from St Thomas to Port St Julian must first steer $S 38^\circ 38' W$, and then by constantly altering her course towards the west, so as to arrive at Port St Julian on a course $S 71^\circ 36' W$, she will have sailed the shortest distance between those places.

**Prob. III.** Given the latitudes and longitudes of two places, to find the distance between them, and the angles of position.

**Example.** What is the shortest distance between the Lizard, in latitude $49^\circ 57' N$, longitude $5^\circ 15' W$, and Bermudas, in latitude $32^\circ 35' N$, and longitude $63^\circ 28' W$?

**By Construction.**

Describe the primitive circle (fig. 62.) to represent the meridian of one of the places; make EA = $32^\circ 35'$, and A will represent Bermudas; make EB, QC each equal to $49^\circ 57'$; then with the tangent of the co-latitude $49^\circ 57'$, draw the parallel of latitude of the Lizard, and with the secant of $58^\circ 13'$, the given difference of longitude, draw the oblique circle PBS, intersecting the parallel of latitude in B; which will be the position of the Lizard. Draw the diameter AF, and through the points A, B, F, describe a circle; and the arch AB will be the distance, and the angles A and B the angles of position, which are measured as before.

**By Calculation.**

In the oblique-angled spherical triangle APB are AP, BP, the co-latitudes, and the angle APB the difference of longitude; to find the distance AB, and the angles of position PAB, PBA.

1. To find the distance.

| Difference of long. | $58^\circ 13'$ | |---------------------|---------------| | verified fine | $9.67515$ | | AP | $57^\circ 25'$ | | fine | $9.92563$ | | BP | $40^\circ 3'$ | | fine | $9.80852$ |

Difference

Distance AB

2. To find the angle of position at the Lizard.

| As the fine of AB | $45^\circ 45'$ | |------------------|---------------| | is to the fine of AP | $57^\circ 25'$ | | fo is the fine of P | $58^\circ 13'$ |

to the fine of B

3. To find the angle of position at Bermudas.

| As the fine of AB | $45^\circ 45'$ | |------------------|---------------| | is to the fine of BP | $40^\circ 3'$ | | fo is the fine of P | $58^\circ 13'$ |

to the fine of A

The shortest distance between the Lizard and Bermudas is $45^\circ 45'$ or $2745$ miles, which is 56 miles less than... than the distance found by Mercator's sailing. And a ship to describe the shortest tract must sail from the Lizard S. 80° 20' W., and gradually lessen the course, so as to arrive at Bermuda on the rhumb bearing S. 49° 47' W. The direct course by Mercator's sailing is S. 68° 10' W.

From the preceding examples, it is evident that in order to sail on the arch of a great circle, the ship must continually alter her course. But as this is a difficulty too great to be admitted into the practice of navigation, it has been thought sufficiently ex- act to effect this by a kind of approximation; the principle of which is, that in small arches the diffe- rence between the arch and its chord or tangent is so small, that the one may be substituted for the other in any nautical operations.

Upon this principle, the great circles on the earth are supposed to be made up of short right lines, each of which is a segment of a rhumb line; and on this supposition the solution to the following problem is founded.

**Prob. IV.** Given the latitudes and longitudes of two places, to find the several points in a great circle passing through them, which alter in longitude from either of the places by a given quantity; together with the courses and distances between those points.

**Rule.** Compute the distance of the places, and their angles of position, by one of the preceding pro- blems; find also the perpendicular from the pole to the great circle, passing through the given places, and the several angles at the pole made by the given al- terations of longitude between the perpendicular and the successive meridians come to.

With this perpendicular, and the polar angles seve- rally, find as many corresponding latitudes by the following analogy:

As rad. : co-tan. perp. :: col. 1st pol. ang. : tang. 1st lat. :: col. 2nd pol. ang. : tang. 2nd lat. &c. &c.

Now having the latitudes of the several points in the great circle, and the difference of longitude be- tween each, find the several courses and distances be- tween them; and these will be the courses and dis- tances the ship must run to keep nearly on the arch of a great circle.

**Example I.** A ship from a place in latitude 37° 0' N., longitude 23° 0' W., bound to a place in the same latitude, and in longitude 76° 27' W., intends to sail as near the arch of a great circle as she can, by altering her course at every five degrees of longitude. Required the latitude of each point where the course is proposed to be altered, and also the courses and di- stances between those points?

The triangle APB (fig. 63.) being described, and CCCXLII. the computation made as in Problem I. the distance will be found equal to 42° 6', and the angle of posi- tion A or B = 73° 0'.—Now the triangle APB being isosceles, the perpendicular PI falls in the middle of AB; and the latitudes, courses, and distances being known in the half BI, those in the half IA will also be known.

Let the points a, b, c, d, &c. be the points arrived at on each alteration of five degrees of longitude; then will the arches Pa, Pb, Pc, Pd, &c. be the respective co-latitudes of those places, and are the hypotenuses of the right-angled spherical triangles Pla, Plb, Plc, Pld, &c.

Now in the triangle PIB, given PB = 53° 0', the angle PBI = 73° 9'; to find PI.

As radius is to the sine of PBI so is the sine of PB

| Polar Angles | Successive Diff. Long. | Successive Diff. Lat. | Diff. Lat. | Meridian Parts. | Merid. diff.lat. | Courses. | Distances | |--------------|------------------------|----------------------|-----------|----------------|----------------|----------|-----------| | IPB = 26° 43' | 23° 0' | 37° 0' | 2392.6 | | | | | | IPa = 21° 43' | 28° 0' | 38° 5' | 2474.6 | | | | | | IPb = 16° 43' | 33° 0' | 38° 5' | 2539.8 | | | | | | IPc = 11° 43' | 38° 0' | 39° 3' | 2587.6 | | | | | | IPd = 6° 43' | 43° 0' | 39° 5' | 2618.8 | | | | |

The courses, and the first distance, are found by Mer- cator's Sailing; but as the other courses are near the parallel, the distances cannot be very exactly found by this method; another method is therefore used. The sum of the distances is 1263.1, which doubled is 2526.2, agreeing with the distance found as before. It may be observed, that the distance found by this method cannot be less than the last distance, or that given by Great Circle Sailing, as some authors have found it.

**Example II.** A ship from the Lizard, in latitude 49° 57' N., longitude 5° 15' W., bound to a place in latitude 32° 25' N., and longitude 66° 39' W., proposes to sail on a great circle, and to alter her course at every Practice.

To find the polar angle API.

As radius is to the cosine of AP so is the tangent of PAB

| Polar Angles | Successive Longs. | Diff. Longs. | Successive Lats. | Diff. Lats. | Meridian Parts. | Merid. diff. lat. | Courses. Distances. | |--------------|------------------|-------------|-----------------|------------|----------------|------------------|-------------------| | IPB = 3° 35' | 5° 14' | 8 49 | 215 | 49° 57' | 3 | 3469.8 | | | IPa = 2 | 49 | 11 33 | 169 | 49 58 | 2 | 3474.5 | 4.7 | | IPb = 7 | 49 | 16 38 | 300 | 49 45 | 13 | 3451.2 | 20.2 | | IPc = 12 | 49 | 21 38 | 300 | 49 18 | 27 | 3409.6 | 41.6 | | IPd = 17 | 49 | 26 38 | 300 | 49 37 | 41 | 3347.2 | 62.4 | | IPe = 22 | 49 | 31 38 | 300 | 49 42 | 55 | 3264.7 | 82.5 | | IPf = 27 | 49 | 36 38 | 300 | 49 31 | 71 | 3160.4 | 104.3 | | IPg = 32 | 49 | 41 38 | 300 | 49 45 | 88 | 3034.2 | 126.2 | | IPh = 37 | 49 | 46 38 | 300 | 49 17 | 106 | 2886.4 | 147.8 | | IPi = 42 | 49 | 51 38 | 300 | 49 10 | 127 | 2714.9 | 171.5 | | IPk = 47 | 49 | 56 38 | 300 | 49 38 | 149 | 2520.5 | 194.4 | | IPL = 52 | 49 | 61 38 | 300 | 49 46 | 175 | 2300.7 | 219.8 |

As the four first courses are near the parallel, the corresponding distances were not found by Mercator's Sailing. The sum of the distances 2874.5 agrees very well with, and is not less than, 47° 54', or 2874 miles, the shortest distance between the places.

CHAP. XIII. Of Sea-Charts.

The charts usually employed in the practice of navigation are of two kinds, namely, Plane and Mercator's Charts. The first of these is adapted to represent a portion of the earth's surface near the equator; and the last for all portions of the earth's surface. For a particular description of these, reference has already been made from the article Chart, to those of Plane and Mercator; and as these charts are particularly described under the above articles, it is therefore sufficient in this place to describe their use.

Use of the Plane Chart.

PROB. I. To find the latitude of a place on the chart.

RULE. Take the least distance between the given place and the nearest parallel of latitude; now this distance applied the same way on the graduated meridian, from the extremity of the parallel, will give the latitude of the proposed place.

Thus the distance between Bonavista and the parallel of 15 degrees, being laid from that parallel upon the graduated meridian, will reach to 16° 5', the latitude required.

PROB. II. To find the course and distance between two given places on the chart.

RULE. Lay a ruler over the given places, and take the nearest distance between the centre of any of the compasses on the chart and the edge of the ruler; move this extent along, so as one point of the compass may touch the edge of the rule, and the straight line joining their points may be perpendicular thereto; then will the other point show the course. The interval between the places, being applied to the scale, will give the required distance.

Thus the course from Palma to St Vincent will be found to be about SSW 4 W. and the distance 13° 4' or 795 m.

PROB. III. The course and distance failed from a known place being given, to find the ship's place on the chart.

RULE. Lay a ruler over the place failed from, pa- NAVI GATION.

Rule. The ruler being laid over the place sailed from and in the direction of the given course, take the distance sailed from the equator, put one point of the compass at the intersection of any parallel with the ruler, and the other point of the compass will reach to a certain place by the edge of the ruler. Now this point remaining in the same position, draw in the other point of the compass until it just touch the above parallel when swept round; apply this extent to the equator, and it will give the difference of latitude. Hence the latitude in will be known, and the intersection of the corresponding parallel with the edge of the ruler will be the ship's present place.

The other problems of Mercator's Sailing may be very easily resolved by this chart; but as they are of less use than those given, they are therefore omitted, and may serve as an exercise to the student.

BOOK II.

Containing the method of finding the Latitude and Longitude of a Ship at Sea, and the Variation of the Compass.

CHAP. I. Of Hadley's Quadrant.

Hadley's quadrant is the chief instrument in use at present for observing altitudes at sea. The form of this instrument, according to the present mode of construction, is an octagonal sector of a circle, and therefore contains 45 degrees; but because of the double reflection, the limb is divided into 90 degrees. See Astronomy and Quadrant. Fig. 65 represents a quadrant of the common construction, of which the following are the principal parts.

1. ABC, the frame of the quadrant. 2. BC, the arch or limb. 3. D, the index; ab, the subdividing scale. 4. E, the index-glass. 5. F, the fore horizon-glass. 6. G, the back horizon-glass. 7. K, the coloured or dark glasses. 8. HI, the vanes or sights.

Of the Frame of the Quadrant.

The frame of the quadrant consists of an arch BC, firmly attached to the two radii AB, AC, which are bound together by the braces L M, in order to strengthen it, and prevent it from warping.

Of the Index D.

The index is a flat bar of brass, and turns on the centre of the octant; at the lower end of the index there is an oblong opening; to one side of this opening the vernier scale is fixed, to subdivide the divisions of the arch; at the end of the index there is a piece of brass, which bends under the arch, carrying a spring to make the subdividing scale lie close to the divisions. It is also furnished with a screw to fix the index in any desired position. The best instruments have an adjusting screw fitted to the index, that it may be moved more slowly, and with greater regularity and accuracy, than by the hand. It is proper, however, to observe, that the index must be previously fixed near its right position by the above-mentioned screw. Of the Index Glass E.

Upon the index, and near its axis of motion, is fixed a plane speculum, or mirror of glass quicksilvered. It is set in a brass frame, and is placed so that its face is perpendicular to the plane of the instrument. This mirror being fixed to the index moves along with it, and has its direction changed by the motion thereof; and the intention of this glass is to receive the image of the sun, or any other object, and reflect it upon either of the two horizon-glasses, according to the nature of the observation.

The brass frame with the glass is fixed to the index by the screw c; the other screws serve to replace it in a perpendicular position, if by any accident it has been deranged.

Of the Horizon Glasses F, G.

On the radius A.B of the octant are two small speculums: the surface of the upper one is parallel to the index glass, and that of the lower one perpendicular thereto, when o on the index coincides with o on the limb. These mirrors receive the reflected rays, and transmit them to the observer.

The horizon-glasses are not entirely quicksilvered; the upper one F is only silvered on its lower half, or that next the plane of the quadrant, the other half being left transparent, and the back part of the frame cut away, that nothing may impede the sight through the unslivered part of the glass. The edge of the foil of this glass is nearly parallel to the plane of the instrument, and ought to be very sharp, and without a flaw. The other horizon-glass is silvered at both ends. In the middle there is a transparent slit, through which the horizon may be seen.

Each of these glasses is set in a brass frame, to which there is an axis passing through the woodwork, and is fitted to a lever on the under side of the quadrant, by which the glass may be turned a few degrees on its axis, in order to set it parallel to the index-glass. The lever has a contrivance to turn it slowly, and a button to fix it. To set the glasses perpendicular to the plane of the instrument, there are two sunk screws, one before and one behind each glass: these screws pass through the plate on which the frame is fixed into another plate; so that by loosening one and tightening the other of these screws, the direction of the frame with its mirror may be altered, and set perpendicular to the plane of the instrument.

Of the Coloured Glasses K.

There are usually three coloured glasses, two of which are tinged red and the other green. They are used to prevent the solar rays from hurting the eye at the time of observation. These glasses are set in a frame, which turns on a centre, so that they may be used separately or together as the brightness of the sun may require. The green glass is particularly useful in observations of the moon; it may also be used in observations of the sun, if that object be very faint. In the fore-observation, these glasses are fixed as in fig. 65, but when the back-observation is used, they are removed to N.

Of the two Sight Vanes H, I.

Each of these vanes is a perforated piece of brass, designed to direct the sight parallel to the plane of the quadrant. That which is fixed at I is used for the fore, and the other for the back, observation. The vane I has two holes, one exactly at the height of the silvered part of the horizon-glass, the other a little higher, to direct the sight to the middle of the transparent part of the mirror.

Of the Divisions on the Limb of the Quadrant.

The limb of the quadrant is divided from right to left into 90 primary divisions, which are to be considered as degrees, and each degree is subdivided into three equal parts, which are therefore of 20 minutes each: the intermediate minutes are obtained by means of the scale of divisions at the end of the index.

Of the Vernier or Subdividing Scale.

The dividing scale contains a space equal to 21 divisions of the limb, and is divided into 20 equal parts. Hence the difference between a division on the dividing scale and a division on the limb is one-twentieth of a division on the limb, or one minute. The degree and minute pointed out by the dividing scale may be easily found thus.

Observe what minute on the dividing scale coincides with a division on the limb; this division being added to the degree and part of a degree on the limb, immediately preceding the first division on the dividing scale, will be the degree and minute required.

Thus suppose the fourteenth minute on the dividing scale coincided with a division on the limb, and that the preceding division on the limb to o on the vernier was 56° 40′; hence the division shown by the vernier is 56° 54′. A magnifying glass will assist the observer to read off the coinciding divisions with more accuracy.

Adjustments of Hadley's Quadrant.

The adjustments of the quadrant consist in placing the mirrors perpendicular to the plane of the instrument. The fore horizon-glass must be set parallel to the speculum, and the planes of the speculum and back horizon-glass produced must be perpendicular to each other when the index is at o.

Adjustment I. To set the index-glass perpendicular to the plane of the quadrant.

Method I. Set the index towards the middle of the limb, and hold the quadrant so that its plane may be nearly parallel to the horizon; then look into the index-glass; and if the portion of the limb seen by reflection appears in the same plane with that seen directly, the speculum is perpendicular to the plane of the instrument. If they do not appear in the same plane, the error is to be rectified by altering the position of the screws behind the frame of the glass.

Method II. This is performed by means of the two adjusting tools fig. 66, 67, which are two wooden frames, having two lines on each, exactly at the same distance from the bottom.

Place the quadrant in an horizontal position on a table; put the index about the middle of the arch; turn back the dark glasses; place one of the above-mentioned tools near one end of the arch, and the other at the opposite end, the side with the lines being towards the index-glass; then look into the index-glass, directing the sight parallel to the plane of the instrument, and one of the tools will be seen by direct vision, and the other by reflection. By moving the index a little, they may be brought exactly together. If the lines coincide, the position of the mirror is right; if not, they must be made to coincide by altering the screws behind the frame, as before.

**Adjustment II.** To set the fore horizon-glass perpendicular to the plane of the instrument.

Set the index to o; hold the plane of the quadrant parallel to the horizon; direct the sight to the horizon, and if the horizons seen directly and by reflection are apparently in the same straight line, the fore horizon-glass is perpendicular to the plane of the instrument; if not, one of the horizons will appear higher than the other. Now if the horizon seen by reflection is higher than that seen directly, release the nearest screw in the pedestal of the glass, and screw up that on the farther side, till the direct and reflected horizons appear to make one continued straight line. But if the reflected horizon is lower than that seen directly, unscrew the farthest, and screw up the nearest screw till the coincidence of the horizons is perfect, observing to leave both screws equally tight, and the fore horizon-glass will be perpendicular to the plane of the quadrant.

**Adjustment III.** To set the fore horizon-glass parallel to the index-glass, the index being at o.

Set o on the index exactly to o on the limb, and fix it in that position by the screw at the under side; hold the plane of the quadrant in a vertical position, and direct the sight to a well-defined part of the horizon; then if the horizon seen in the silvered part coincides with that seen through the transparent part, the horizon-glass is adjusted; but if the horizons do not coincide, unscrew the milled screw in the middle of the lever on the other side of the quadrant, and turn the nut at the end of the lever until both horizons coincide, and fix the lever in this position by tightening the milled screw.

As the position of the glass is liable to be altered by fixing the lever, it will therefore be necessary to re-examine it, and if the horizons do not coincide, it will be necessary either to repeat the adjustment, or rather to find the error of adjustment, or, as it is usually called, the index-error; which may be done thus:

Direct the sight to the horizon, and move the index until the reflected horizon coincides with that seen directly; then the difference between o on the limb and o on the vernier is the index error; which is additive when the beginning of the vernier is to the right of o on the limb, otherwise subtractive.

**Adjustment IV.** To set the back horizon-glass perpendicular to the plane of the instrument.

Put the index to o; hold the plane of the quadrant parallel to the horizon, and direct the sight to the horizon through the back sight vane. Now if the reflected horizon is in the same straight line with that seen through the transparent part, the glass is perpendicular to the plane of the instrument: If the horizons do not unite, turn the sunk screws in the pedestal of the glass until they are apparently in the same straight line.

**Adjustment V.** To set the back horizon-glass perpendicular to the plane of the index-glass produced, the index being at o.

Let the index be put as much to the right of o as twice the dip of the horizon amounts to; hold the quadrant in a vertical position, and apply the eye to the back vane: then if the reflected horizon coincides with that seen directly, the glass is adjusted; if they do not coincide, the screw in the middle of the lever on the other side of the quadrant must be released, and the nut at its extremity turned till both horizons coincide. It may be observed, that the reflected horizon will be inverted; that is, the sea will be apparently uppermost and the sky lowermost.

As this method of adjustment is deemed troublesome, and is often found to be very difficult to perform at sea, various contrivances have therefore been proposed to render this adjustment more simple. Some of these are the following.

1. Mr Dollond's method of adjusting the back horizon-glass.

In this method an index is applied to the back horizon-glass, by which it may be moved so as to be parallel to the index-glass, when o on the vernier coincides with o on the limb. When this is effected, the index of the back horizon-glass is to be moved exactly 90° from its former position, which is known by means of a divided arch for that purpose; and then the plane of the back horizon-glass will be perpendicular to the plane of the index-glass produced.

2. Mr Blair's method of adjusting the back horizon-glass.

All that is required in this method is to polish the lower edge of the index-glass, and expose it to view. The back horizon-glass is adjusted by means of a reflection from this polished edge, in the very same method as the fore horizon-glass is adjusted by the common method.

In order to illustrate this, let R I H E (fig. 68.) represent a pencil of rays emitted from the object R, incident on the index-glass I, from which it is reflected to the fore horizon-glass H, and thence to the eye at E. By this double reflection, an image of the object is formed at r. RHE represents another pencil from the same object R, coming directly through the fore horizon-glass to the eye at E; so that the doubly reflected image r appears coincident with the object R itself, seen directly.

When this coincidence is perfect, and the object R so very distant as to make the angle IRH insensible, the position of the speculums I and H will differ infensibly from parallelism; that is, the quadrant will be adjusted for the fore-observation. Now it is from the ease and accuracy with which this adjustment can at any time be made, that the fore-observation derives its superiority over the back-observation. But by grinding the edge of the index glass perpendicular to its reflecting surface, and polishing it, the back-observation is rendered capable of an adjustment equally easy and accurate as the fore horizon-glass: for by a pencil of rays emitted from the object S, incident on the reflecting edge of the index-glass D, thence reflected to the back horizon-glass B, and from that to the eye at e, an image will be formed at s; which image being made to coincide with the object S itself, seen directly, ascertains the position of the back horizon glass relative to the index-glass with the same precision, and in a manner equally direct, as the former operation does that of the fore horizon-glass.

Directions for adjusting the Back Horizon Glass.

The method of adjusting the quadrant for the back observation Method observation is this. If it is to be done without making use of the telescope, place the index at o, and applying the eye to the hole in the sight vane (x), or tube for directing the sight, direct it through the back horizon-glass to the horizon, if that is the object to be used for adjusting. The two horizons are then to be made to coincide, holding the quadrant first in a vertical and then in an horizontal position; by which means both adjustments will be effected as in the fore-observation.

There will be no difficulty in finding the reflected horizon, if the observer first directs his eye to that part of the horizon-glass where he observes the image of the polished edge of the index-glass, which will appear double. When the direct horizon is made to appear in this space, the reflected one will be seen close by it, unless the instrument wants a great adjustment. In this case, a little motion of the back horizon-glass backwards and forwards will presently bring it in view.

When the horizon, or any obscure terrestrial object, is to be made use of for adjusting by means of the reflecting edge, there is a precaution to be taken, without which the observer will sometimes meet with what will appear an unaccountable difficulty; for if the sky, or other object behind him, should happen to be pretty bright, he will not be able to discern the horizon at all. This arises from the image of the object behind him, which is reflected from the silvered surface of the index-glass, appearing to coincide with the horizon; in which case, the bright picture of the former, which is formed in the bottom of the eye, prevents the fainter impression of the latter from being perceived. This will be avoided, either by applying a black screen over the silvered surface of the index-glass, or, without being at this trouble, by standing at a door or window, so that only the dark objects within can be reflected from the index-glass; but if the observation is to be made in the open air, a hat, or any such dark obstacle, held before the silvered surface of the index-glass, will very effectually remove this inconvenience.

It may be remarked, that some observers, instead of making the principal adjustment, place the speculums parallel, by moving the index without altering the position of the horizon-glass; and the difference between o on the vernier and o on the limb is the index error, which must be subtracted from all angles measured by the back-observation, when o on the index is to the right of o on the limb; and added when to the left.

3. Mr Wright's method of adjusting the back horizon-glass of his improved patent quadrant.

Fig. 69. is a representation of the quadrant complete in all its parts for use. A, is the reflecting surface of the index-glass, which is made of the usual length, and \( \frac{9}{10} \) of an inch broad. The bottom part is covered in front by the brass frame, and the reflecting surface is \( \frac{7}{10} \) on the back. B, the fore horizon-

(k) Besides the hole in the sight vane commonly made, there must be another nearer to the horizon-glass, and so placed that an eye directed through it to the centre of the horizon-glass shall there perceive the image of the polished edge of the index-glass. This hole must not be made small like the other, but equal to the ordinary size of the pupil of the eye, there being on some occasions no light to spare. degrees distant. The back horizon K is now reflected from the back surface of the index-glass C to the horizon-glass M, and from thence to the eye at D, in a right line with the fore horizon F. In order to make an exact contact of the fore and back horizons at F, the index must be advanced beyond the 90th degree on the arch, by a quantity equal to twice the dip of the horizon.

The quadrant is adjusted for the fore-observation as usual, having previously fixed the index-glass in its proper place by the milled screw at H, as represented in fig. 70.

To adjust the Quadrant for the Back-observation.

Fasten the index to 90° on the limb; loosen the screw H (fig. 70.), and turn the plate O by the milled edge until the end of the adjusting screw K touch the edge of the clamp M; and by means of a distant object observe if the glasses are then parallel, as at fig. 71.: if they are, fasten the screw H; if not, with a screw-driver turn the screw K gently to the right or left to make them perfect, and then tighten the screw. Now remove the index back to O on the limb, and the index-glass will be parallel to the back horizon-glass E, fig. 72.: If not, make them so by turning the adjusting screw of the glass E, the eye being at the upper hole in the sight-vane D, and the sight directed to the horizon, or any distant object in the direction DN (fig. 69.) Now the index remaining in this position, the index-glass is to be returned, to flop at the pin E, and it will be parallel to the fore horizon-glass as at first: then the quadrant will be adjusted for both methods of observation.

To observe the Sun's Altitude by the Back-observation.

Remove the coloured glasses to G (fig. 69.), and look through the lower hole in the sight-vane D, in the line of direction DN, directly to the sun, and move the index forward on the arch exactly in the same manner as in the fore-observation: make the contact of the sun's limb and the back horizon exact, and the degrees and minutes shown by the index on the limb is the sun's zenith distance. It may be observed, that the horizon will be inverted. If the sun's lower limb be observed, the semidiameter is to be subtracted from the zenith distance; but if the upper limb is observed, the semidiameter is to be added.

The observation may be made in the usual manner, by turning the back to the sun. In this case the coloured glasses are to be shifted to F, and proceed according to the directions formerly given.

Use of Hadley's Quadrant.

The altitude of any object is determined by the position of the index on the limb, when by reflection that object appears to be in contact with the horizon.

If the object whose altitude is to be observed be the sun, and if so bright that its image may be seen in the transparent part of the fore horizon-glass, the eye is to be applied to the upper hole in the sight-vane; otherwise, to the lower hole: and in this case, the quadrant is to be held so that the sun may be bifected by the line of separation of the silvered and transparent parts of the glass. The moon is to be kept as nearly as possible in the same position; and the image of the star is to be observed in the silvered part of the glass adjacent to the line of separation of the two parts.

There are two different methods of taking observations with the quadrant. In the first of these the face of the observer is directed towards that part of the horizon immediately under the sun, and is therefore called the fore-observation. In the other method, the observer's back is to the sun, and it is hence called the back-observation. This last method of observation is to be used only when the horizon under the sun is obscured, or rendered indistinct by fog or any other impediment.

In taking the sun's altitude, whether by the fore or back observation, the observer must turn the quadrant about upon his heel, so as to keep the sun always in that part of the horizon-glass which is at the same distance as the eye from the plane of the quadrant. In this way the reflected sun will describe an arch of a parallel circle round the true sun, whose convex side will be downwards in the fore-observation and upwards in the back; and consequently, when by moving the index, the lowest point of the arch in the fore-observation, or highest in the back, is made to touch the horizon, the quadrant will stand in a vertical plane, and the altitude above the visible horizon will be properly observed. The reason of these operations may be thus explained: The image of the sun being always kept in the axis of vision, the index will always show on the quadrant the distance between the sun and any object seen directly which its image appears to touch; therefore, as long as the index remains unmoved, the image of the sun will describe an arch everywhere equidistant from the sun in the heavens, and consequently a parallel circle about the sun, as a pole. Such a translation of the sun's image can only be produced by the quadrant's being turned about upon a line drawn from the eye to the sun, as an axis. A motion of rotation upon this line may be resolved into two, one upon the axis of vision, and the other upon a line on the quadrant perpendicular to the axis of vision; and consequently a proper combination of these two motions will keep the image of the sun constantly in the axis of vision, and cause both jointly to run over a parallel circle about the sun in the heavens: but when the quadrant is vertical, a line theron perpendicular to the axis of vision becomes a vertical axis; and as a small motion of the quadrant is all that is wanted, it will never differ much in practice from a vertical axis. The observer is directed to perform two motions rather than the single one equivalent to them on a line drawn from the eye to the sun; because we are not capable, while looking towards the horizon, of judging how to turn the quadrant about upon the elevated line going to the sun as an axis, by any other means than by combining the two motions above mentioned, so as to keep the sun's image always in the proper part of the horizon-glass. When the sun is near the horizon, the line going from the eye to the sun will not be far removed from the axis of vision; and consequently the principal motion of the quadrant will be performed on the axis of vision, and the part of motion made on the vertical axis will be but small. On the contrary, when the sun is near the zenith, the line going to the sun is not far removed from a vertical line, and consequently To take Altitudes by the Fore-observation.

I. Of the Sun.

Turn down either of the coloured glasses before the horizon-glass, according to the brightness of the sun; direct the sight to that part of the horizon which is under the sun, and move the index until the coloured image of the sun appears in the horizon-glass; then give the quadrant a slow vibratory motion about the axis of vision; move the index until the lower or upper limb of the sun is in contact with the horizon, at the lowest part of the arch described by this motion; and the degrees and minutes shown by the index on the limb will be the altitude of the sun.

II. Of the Moon.

Put the index to o, turn down the green glass, place the eye at the lower hole in the sight-vane, and observe the moon in the silvered part of the horizon-glass; move the index gradually, and follow the moon's reflected image until the enlightened limb is in contact with the horizon, at the lower part of the arch described by the vibratory motion as before, and the index will show the altitude of the observed limb of the moon. If the observation is made in the daytime, the coloured glass is unnecessary.

III. Of a Star or Planet.

The index being put to o, direct the sight to the star through the lower hole in the sight-vane and transparent part of the horizon-glass; move the plane of the quadrant a very little to the left, and the image of the star will be seen in the silvered part of the glass. Now move the index, and the image of the star will appear to descend; continue moving the index gradually until the star is in contact with the horizon at the lowest part of the arch described; and the degrees and minutes shown by the index on the limb will be the altitude of the star.

To take Altitudes by the Back-observation.

I. Of the Sun.

Put the stem of the coloured glasses into the perforation between the horizon-glasses, turn down either according to the brightness of the sun, and hold the quadrant vertically; then direct the sight through the hole in the back sight-vane, and the transparent slit in the horizon-glass to that part of the horizon which is opposite to the sun; now move the index till the sun is in the silvered part of the glass, and by giving the quadrant a vibratory motion, the axis of which is that of vision, the image of the sun will describe an arch whose convex side is upwards; bring the limb of the sun, when in the upper part of this arch, in contact with the horizon; and the index will show the altitude of the other limb of the sun.

II. Of the Moon.

The altitude of the moon is observed in the same manner as that of the sun, with this difference only, that the use of the coloured glass is unnecessary unless the moon is very bright; and that the enlightened limb, whether it be the upper or lower, is to be brought in contact with the horizon.

III. Of a Star or Planet.

Look directly to the star through the vane and transparent slit in the horizon-glass, move the index until the opposite horizon, with respect to the star, is seen in the silvered part of the glass; and make the contact perfect as formerly. If the altitude of the star is known nearly, the index may be set to that altitude, the sight directed to the opposite horizon, and the observation made as before.

Sect. II. Of finding the Latitude of a Place.

The observation necessary for ascertaining the latitude of a place, is that of the meridional altitude of a known celestial object; or two altitudes when the object is out of the meridian. The latitude is deduced with more certainty and with less trouble from the first of these methods, than from the second; and the sun, for various reasons, is the object most proper for this purpose at sea. It, however, frequently happens, that by the interposition of clouds, the sun is obscured at noon; and by this means the meridian altitude is lost. In this case, therefore, the method by double altitudes becomes necessary. The latitude may be deduced from three altitudes of an unknown object, or from double altitudes, the apparent times of observation being given.

The altitude of the limb of an object observed at sea, requires four separate corrections in order to obtain the true altitude of its centre; these are for semidiameter, dip, refraction, and parallax. (See Astronomy, and the respective articles.) The first and last of these corrections vanish when the observed object is a fixed star.

When the altitude of the lower limb of any object is observed, its semidiameter is to be added thereto in order to obtain the central altitude; but if the upper limb be observed, the semidiameter is to be subtracted. If the altitude be taken by the back-observation, the contrary rule is to be applied. The dip is to be subtracted from, or added to, the observed altitude, according as the fore or back-observation is used. The refraction is always to be subtracted from, and the parallax added to, the observed altitude.

Prob. I. To reduce the sun's declination to any given meridian.

Rule. Find the number in Table IX, answering to the longitude in the table nearest to that given, and to the nearest day of the month. Now, if the longitude is west, and the declination increasing, that is, from the 20th of March to the 22d of June, and from the 22d of September to the 22d of December, the above number is to be added to the declination; during the other part of the year, or while the declination is decreasing, this number is to be subtracted. In east longitude, the contrary rule is to be applied.

Example I. Required the sun's declination at noon 15th April 1793, in longitude 84° W?

Sun's declination at noon at Greenwich 10° 1' 8 N

Number from Table IX, + 5°

Reduced declination 10° 6.8 N

Example II. Required the sun's declination at noon 22d March 1793, in longitude 151° E?

Sun's declination at noon at Greenwich 0° 56' N

Equation from table - 10

Reduced declination 0° 46' N

Prob. II. Given the sun's meridian altitude, to find the latitude of the place of observation.

Rule. The sun's semidiameter is to be added to, or subtracted from, the observed altitude, according as the lower or upper limb is observed; the dip answering to the height from Table V. is to be subtracted if the fore-observation is used; otherwise, it is to be added; and the refraction answering to the altitude from Table IV. is to be subtracted: hence the true altitude of the sun's centre will be obtained. Call the altitude south or north, according as the sun is south or north at the time of observation; which subtracted from 90°, will give the zenith distance of a contrary denomination.

Reduce the sun's declination to the meridian of the place of observation, by Problem I.; then the sum or difference of the zenith distance and declination, according as they are of the same or of a contrary denomination, will be the latitude of the place of observation, of the same name with the greater quantity.

Example I. October 17th 1792, in longitude 32° E, the meridian altitude of the sun's lower limb was 48° 53' S, height of the eye 18 feet. Required the latitude?

Obs. alt. sun's lower limb 48° 53' 3 Sun's dec. 17.O.A. noon.9° 37' S

Semidiameter + 0° 16 Equation Table IX - 2

Dip and refraction - 0° 5 Reduced declination 9° 35' S

True alt. sun's centre 49° 48' Zenith distance 40° 56' N

Latitude 31° 21' N

Example II. November 16th 1793, in longitude 158° W, the meridian altitude of the sun's lower limb was 87° 37' N, height of the eye 10 feet. Required the latitude?

Obs. alt. sun's lower limb 87° 37' N Sun's dec. noon.18° 57' S

Semidiameter + 0° 16 Equation table + 0° 8

Dip and refraction - 0° 3 Reduced declination 19° 5' S

True alt. sun's centre 87° 50' N Zenith distance 21° 10' S

Latitude 21° 15' S

Example III. December 19th 1793, being nearly under the meridian of Greenwich, the altitude of the sun's upper limb at noon was 4° 30' S, height of the eye 20 feet. Required the latitude?

Observed altitude of the sun's upper limb 4° 30' S

Sun's semidiameter - 0° 16

Dip and refraction - 0° 15

True altitude of the sun's centre 3° 59' S

Zenith distance 86° 1' N

Declination 23° 27' S

Latitude 62° 34' N **NAVIGATION**

**Example IV.** August 23rd 1793, in longitude 107° E, the meridian altitude of the sun's lower limb by the back-observation was 61° 8' N, and the height of the eye 14 feet. Required the latitude?

| Observed altitude sun's upper limb | 61° 8' N | |-----------------------------------|----------| | Sun's semidiameter | -0° 16' | | Dip | +0° 3' | | Refraction | -0° 1' |

True altitude of sun's centre: 60° 55' N Zenith distance: 29° 5' S Reduced declination: 11° 20' N

Latitude: 17° 45' S

The dip in Table V. answers to an entirely open and unobstructed horizon. It, however, frequently happens, that the sun is over the land at the time of observation, and the ship nearer to the land than the visible horizon would be if unconfined. In this case, the dip will be different from what it would otherwise have been, and is to be taken from Table VI., in which the height is expressed at the top, and the distance from the land in the side column in nautical miles.

Seamen, in general, can estimate the distance of any object from the ship with sufficient exactness for this purpose, especially when that distance is not greater than six miles, which is the greatest distance of the visible horizon from an observer on the deck of any ship.

**Prob. III.** Given the meridian altitude of a fixed star, to find the latitude of the place of observation.

**Rule.** Correct the altitude of the star by dip and refraction, and find the zenith distance of the star as formerly; take the declination of the star from Table XI., and reduce it to the time of observation. Now, the sum or difference of the zenith distance and declination of the star, according as they are of the same or of a contrary name, will be the latitude of the place of observation.

**Example I.** December 1st 1793, the meridian altitude of Sirius was 59° 50' S, height of the eye 14 feet. Required the latitude?

| Observed altitude of Sirius | 59° 50' S | |-----------------------------------|-----------| | Dip and refraction | -0° 4' |

True altitude: 59° 46' S Zenith distance: 30° 14' N Declination: 16° 27' S

Latitude: 13° 47' N

**Example II.** February 17th 1797, the meridian altitude of Procyon was 71° 15' N, the height of the eye 10 feet. Required the latitude?

| Observed altitude of Procyon | 71° 15' N | |-----------------------------------|-----------| | Dip and refraction | -0° 3' |

True altitude: 71° 12' N Zenith distance: 18° 48' S Declination: 5° 45' N

Latitude: 13° 3' S

**Prob. IV.** Given the meridian altitude of a planet, to find the latitude of the place of observation.

**Rule.** Compute the true altitude of the planet as directed in last problem (which is sufficiently accurate for altitudes taken at sea); take its declination from the Nautical Almanac, page iv. of the month, and reduce it to the time and meridian of the place of observation; then the sum or difference of the zenith distance and declination of the planet will be the latitude as before.

**Example I.** December 10th 1792, the meridian altitude of Saturn was 68° 42' N, and height of the eye 15 feet. Required the latitude?

| Observed altitude of Saturn | 68° 42' N | |-----------------------------------|-----------| | Dip and refraction | -0° 4' |

True altitude: 68° 38' N Zenith distance: 21° 22' S Declination: 7° 26' N

Latitude: 13° 56' S

**Example II.** April 16th 1793, the meridian altitude of Jupiter was 81° 5' S, height of the eye 18 feet. Required the latitude?

| Observed altitude of Jupiter | 81° 5' S | |-----------------------------------|-----------| | Dip | -0° 3' |

True altitude: 81° 2' S Zenith distance: 8° 58' N Declination: 19° 4' S Latitude: 10° 6' S

**Prob. V.** Given the meridian altitude of the moon, to find the latitude of the place of observation.

**Rule.** Take the number answering to the ship's longitude, and daily variation of the moon's palling over the meridian; which being applied to the time of pallage given in the Nautical Almanac, will give the time of Tab. XX., the moon's palling over the meridian of the ship.

Reduce this time to the meridian of Greenwich; and by means of the Nautical Almanac find the moon's declination, horizontal parallax, and semidiameter, at the reduced time.

Apply the semidiameter and dip to the observed altitude of the limb, and the apparent altitude of the moon's centre will be obtained; to which add the correction answering to the apparent altitude and horizontal parallax, and the sum will be the true altitude of the moon's centre; which subtracted from 90°, the remainder is the zenith distance, and the sum or difference of the zenith distance and declination, according as they are of the same or of a contrary name, will be the latitude of the place of observation.

**Example I.** December 24th 1792, in longitude 30° W, the meridian altitude of the moon's lower limb was 81° 15' N, height of the eye 12 feet. Required the latitude?

| Time of pall. over the mer. of Greenwich = 9h 19m | | Equation Table XX. | +0° 4' | | Time of pall. over mer. ship | 9° 23' | | Longitude in time | 2° 0' | | Reduced time | 11° 23' | | Moon's dec. at midnight, Tab. IX. | 14° 53' N| | Eq. to time from midnight | -0° 4' | | Reduced declination | 14° 49' N| | Moon's hor. par. | 55° 25' | ### Method of finding the Latitude and Longitude at Sea

#### Example I. July 1st 1793, in longitude 15° W.

- **Observed altitude of the sun's lower limb** at midnight was 8° 58', height of the eye 18 feet. Required the latitude? - **Observed altitude sun's lower limb**: 8° 58' - **Semidiameter**: +0 16 - **Dip and refraction**: -0 10

#### True altitude of sun's centre

- **Compl. declin. reduced to time and place**: 66° 57 N

#### Latitude

- **True altitude**: 76° 1 N

#### Example II. December 1st 1798, the altitude of the pole star below the pole was 52° 20' N, height of the eye 12 feet. Required the latitude?

- **Observed altitude, pole star**: 52° 20' N - **Dip and refraction**: -0 4

#### True altitude

- **Complement of declination**: 1° 46

#### Latitude

- **True altitude**: 54° 2 N

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**Rule.** To the log. secant of the latitude by account, add the log. secant of the sun's declination; the sum, rejecting 20 from the index, is the logarithm ratio. To this add the log. of the difference of the natural sines of the two altitudes, and the log. of the half elapsed time from its proper column.

Find this sum in column of middle time, and take out the time answering thereto; the difference between which and the half elapsed time will be the time from noon when the greater altitude was observed.

Take the log. answering to this time from column of rising, from which subtract the log. ratio, the remainder is the logarithm of a natural number; which being added to the natural sine of the greater altitude, the sum is the natural cosine of the meridian zenith distance; from which and the sun's declination the latitude is obtained as formerly.

If the latitude thus found differs considerably from that by account, the operation is to be repeated, using the computed latitude in place of that by account (L).

**Example I.** June 4th 1795, in latitude by account

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(1) This method is only an approximation, and ought to be used under certain restrictions; namely,

The observations must be taken between nine o'clock in the forenoon and three in the afternoon. If both observations be in the forenoon, or both in the afternoon, the interval must not be less than the distance of the time of observation of the greatest altitude from noon. If one observation be in the forenoon and the other in the afternoon, the interval must not exceed four hours and an half; and in all cases, the nearer the greater altitude is to noon the better.

If the sun's meridian zenith distance be less than the latitude, the limitations are still more contracted. If the latitude be double the meridian zenith distance, the observations must be taken between half past nine in the morning and half past two in the afternoon, and the interval must not exceed three hours and an half. The observations must be taken still nearer to noon, if the latitude exceeds the zenith distance in a greater proportion. See Molyneux's British Mariner's Guide, and Requisite Tables, 2d Ed. Practice.

Method of finding the Latitude and Longitude at Sea.

Count 37° N at 10h 29' A.M. per watch, the corrected altitude of the sun was 65° 24', and at 12h 31', the altitude was 74° 8'. Required the true latitude?

Times per wat. Alt. N. Sines. Lat. by acc. 37° 0' Secant 0.09765 10h 29' 65° 24' 90924 Declination 22° 28' Secant 0.03428

| 12 31 | 74 8 96190 Logarithm ratio | 0.13193 | |-------|-----------------------------|----------| | 2 2 | Differ. | | | 1 | Logarithm | 3.73148 | | Half elapsed time | 0.57999 | | 3r 10'' Middle time | 4.43340 | | 29 50 Rising | Log. ratio | 0.13193 |

Natural number 624 Greatest altitude 74° 8' N. fine 96190 Mer. zenith dist. 14° N. cosine 90874 Declination 22° 28'

Latitude 36° 58' N.

Example II. October 17th 1793, in latitude 43° 24' N. by account, at 6h 38' P.M., the correct altitude of the sun's centre was 36° 5', and at 2h 46' P.M. the altitude was 24° 49'. Required the latitude?

Times per wat. Alt. N. Sines. Lat. by acc. 43° 24' Secant 0.33872 ch 38' 36° 5' 58896 Declination 9° 32' Secant 0.0604

| 2 46 | 24 49 41972 Logarithm ratio | 0.14476 | |-------|-----------------------------|----------| | 2 8 | Differ. | | | 1 | Logarithm | 4.22830 | | Half elapsed time | 0.55906 | | 1 41 30'' Middle time | 4.92492 | | 37 30 Rising | Log. ratio | 0.14476 |

Natural number 639 Greatest altitude 36° 5' N. fine 58896 Mer. zen. distance 53° 14' N. cosine 59853 Declination 9° 32'

Latitude 43° 42' N.

Example III. August 25th 1793, in latitude 57° N. by account, in the morning the altitude of the sun's lower limb was 34° 22', and 1h 46' after the altitude of the lower limb was 42° 12', the height of the eye 14 feet. Required the latitude?

First altitude 34° 22' Second altitude 42° 12' Sun's semidiameter +0 16 Semidiameter +0 16 Dip and refraction -0 5 Dip and refract. -0 4

Corrected altitude 34° 33' Corrected altit. 42° 24'

Interval Altit. N. Sines. Lat. by acc. 57° 0' Secant 0.26189 of time. 34° 33' 56713 Declination 10° 33' Secant 0.00741

| 42 24 | 67430 Logarithm ratio | 0.27130 | |-------|------------------------|----------| | 1h 46' Difference 10917 | Log. | 4.03007 | | 53 0'' Half elapsed time | 0.63978 | | 1 43 30 Middle time | 4.94115 | | 50 30 Rising | Log. ratio | 0.27130 |

Natural number 753 Gr. altitude 28° 43' N. fine 48048 Mer. zen. dist. 60° 47' N. cosine 48801 Declination 9° 37'

Latitude 51° 10' N.

As this latitude differs only 5' from that used in the computation, it may therefore be depended on as the true latitude.

Prob. VII. Given the latitude by account, the sun's declination, two observed altitudes, the elapsed time, and the course and distance run between the observations; to find the ship's latitude at the time of observation of the greater altitude.

Rule. Find the angle contained between the ship's course and the sun's bearing at the time of observation of the least altitude, with which enter the Traverse Table as a course, and the difference of latitude answering to the distance made good will be the reduction of altitude.

Now, if the least altitude be observed in the forenoon, the reduction of altitude is to be applied thereto by addition or subtraction, according as the angle between... between the ship's course and the sun's bearing is less or more than eight points. If the least altitude be observed in the afternoon, the contrary rule is to be used.

The difference of longitude in time between the observations is to be applied to the elapsed time by addition or subtraction, according as it is east or west. This is, however, in many cases so inconsiderable as to be neglected.

With the corrected altitudes and interval, the latitude by account and sun's declination at the time of observation of the greatest altitude, the computation is to be performed by the last problem.

**Example I.** July 6th 1793, in latitude 58° 14' N. by account, and longitude 6° E. at 10h 54m A.M. per watch, the altitude of the sun's lower limb was 53° 17', and at 1h 17' P.M. the altitude was 52° 51', and bearing per compass SWNW; the ship's course during the elapsed time was SSWW, and the hourly rate of sailing 8 knots, the height of the eye 16 feet. Required the true latitude at the time of observation of the greater altitude?

Sun's bear. at 2d. ob. SWNW. Interval. observ. 2h 23'. Ship's course SSWW Diff. run = 2h 23 x 8 = 19m.

Contained angle 3½ points.

Now to course 3½ points and distance 19 miles, the difference of latitude is 14.7 or 15 miles.

First observed alt. 53° 17' Second observed alt. 52° 51'

Semidiameter +0 16 Semidiameter +0 16

Dip and refract. -0 4 Dip and refraction -0 4

True altitude 53° 29' Reduction -0 15

Reduced altitude 52° 48'

Time of ob. of gr. alt. 10h 54' A.M. Sun's dec. 22° 39'N.

Longitude in time 1 4 Eq. to r.t. + 1

Reduced time 9 50 A.M. Red. decl. 22° 40'N.

Time per wat. Alt. N. Sines Lat. by acc. 58° 14' Secant 0.27863

10h 54' 53° 29' 80368 Declination 22° 40' Secant 0.03491

Logarithm ratio 0.31354

Difference 715 Log. 2.85431

Half elapsed time 0.51294

Middle time 5 20

Rising 1 6 10 Log. 3.61686

Logarithm ratio 0.30134

Natural number 2068

Greatest altitude 53° 29' N. fine 80368

Mer. zen. dist. 34° 29' N. cosine 82436

Declination 22° 40' N.

Latitude 57° 13' N.

Since the computed latitude differs so much from that by account, it will be necessary to repeat the operation.

Computed latitude 57° 13' Secant 0.26643

Declination 22° 40' Secant 0.03491

Logarithm ratio 0.30134

Method of finding the Latitude and Longitude at Sea.

Difference of natural sines 1h 11' 30" 715 Log. 2.85431

Half elapsed time 0.51294

Middle time 5 20 Log. 3.60859

Rising 1 6 10 Log. 3.61686

Logarithm ratio 0.30134

Natural number 2068

Greatest altitude 53° 29' N. fine 80368

Mer. zen. dist. 34° 29' N. cosine 82436

Declination 22° 40' N.

Latitude 57° 9' N.

As this latitude differs only 4 miles from that used in the computation, it may therefore be depended on as the true latitude.

**Example II.** Sept. 13th 1793, in latitude 38° 12' N. by account, and longitude 14° E. at 9h 28' A.M. per watch, the altitude of the sun's lower limb was 40° 42' and azimuth per compass SESE, at 11h 16' A.M. the altitude was 53° 11'; the ship's course during the elapsed time was WNW at the rate of 9 knots an hour, and height of the eye 12 feet. Required the ship's true latitude at the time of the second observation?

Sun's bear. at first obf. SESE. Elapf time 1h 48'.

Ship's course, WNW Diff. run = 1h 48' x 9 = 16m.

Contained angle 11½ points; supplement 4½ pts.

To course 4½ points, and distance run 16 miles, the difference of latitude is 10.7, or 11 miles.

First observed alt. 40° 42' Second obs. alt. 53° 11'

Sun's semidiameter +0 16 Semidiameter +0 16

Dip and refraction -0 4 Dip and refraction -0 4

True altitude 53° 29' Reduction -0 15

Reduced altitude 52° 48'

Time of ob. of gr. alt. 11h 16' A.M. Sun's dec. at noon 3° 32'N.

Longitude in time 0 56 Eq. to r.t. + 1

Reduced time 10 20 Reduced declin. 3° 34'

Time per Watch. Alt. N. Sine. Lat. by acc. 38° 11' Secant 0.10466

9h 28' 40° 43' 05232 Declination 3° 34' Secant 0.00084

Logarithm ratio 0.10550

Middle time 1 16

Rising 1 43 Log. 4.17702

Half elapsed time 0.63182

Middle time 1 37

Rising 0 43 Log. 3.24437

Natural number 1376

Mer. Zen. dist. 33° 16' N. cosine 81640

Declination 3° 34' Secant 0.00084

Latitude 38° 30' Secant 0.10348

Logarithm ratio 0.10932

Difference of natural sines 15032 Log. 4.17702

Half elapsed time 0h 54' 0' Log. 0.63182

Middle time 1 37 50 Log. 4.91815

Rising 0 43 10 Log. 3.26089

Natural number 1413

Greatest Practice.

Method of finding the Latitude and Longitude at Sea.

Greatest altitude 53° 23' N. fine 80264 Mer. Z'n. dist. 35° 14' N cosine 81682 Declination 3° 34' Latitude 38° 48' N.

This latitude differing only 2 miles from that used in the computation, may therefore be relied on as the true latitude.

Remark. If the sun comes very near the zenith, the fines of the altitude will vary so little as to make it uncertain which ought to be taken as that belonging to the natural fine of the meridian altitude. In this case, the following method will be found preferable.

To the log. rising of the time from noon found as before, add the log. secant of half the sum of the estimated meridian altitude, and greatest observed altitude; from which subtract the log. ratio, its index being increased by 10, and the remainder will be the log. fine of an arch; which added to the greatest altitude will give the sun's meridian altitude.

Example. December 21st 1793, in latitude 22° 40' S., by account, at 11h 57', the correct altitude of the sun's centre was 89° 10', and at 12h 4' 4", the altitude was 88° 50'. Required the true latitude?

Times per W.t. A.t. N. Sines. Lat. by acc. 22° 40' sec. 0.03491 11h 57' c" 89° 10' 99989 Declination 23° 28' sec. 0.03749

Logarithm ratio + 5

Arch 0° 17' fine 7.70653 Greatest altitude 89° 10' Meridian altitude 89° 27' zen. dist. 0° 37' N Declination 23° 28'

Latitude 22° 55' S.

This differing from the assumed latitude, the work must be repeated.

Latitude 22° 55' secant 0.03571 Declination 23° 28' secant 0.03749

Logarithm ratio 0.07320 Difference of natural fines, 1° log. 1.00000 Half elapsed time 3° 50" Middle time 0° 50 Rising 3° 0 Comp. of lat. 67° 5' Declination 23° 28' Sum 90° 33'

Prob. VIII. To find the latitude from double altitudes of the sun and the elapsed time; one of these observations being taken near the east or west points, and the other near the meridian.

Rule. With the latitude by account, the sun's declination and least altitude, compute the apparent time of observation by Problem VII. of next chapter. From whence and the interval of time between the observations, the time from noon when the greatest altitude was observed will be known. To the logarithm rising of which, add the logarithmic cofines of the sun's declination and the latitude of the place by account; the sum will be the logarithm of a natural number, which added to the natural fine of the greater altitude, will give the natural cofine of the meridian zenith distance; and hence the latitude is found as formerly.

Or the time from noon being found, the latitude may be computed by the rule given in the preceding remark.

Example. September 1st 1793, in latitude 40° 0' N. by account, at 6h 5' A.M. per watch, the altitude of the sun's lower limb was 16° 21', and at 11h 41' the altitude was 57° 42'; the height of the eye 18 feet. Required the latitude?

First alt. 16° 21' Second alt. 57° 42' Sun's semidia.+0 16 Sun's semidia.+0 16 Dip and refr.-0 7 Dip and refr.-0 5 True altitude 16° 30' True altitude 57° 53'

Method Lat. by acc. Declination Secant Secant Latitude and Longitude at Sea.

| Difference | Altitude | |------------|----------| | 31° 57 nat. cosine | 84851 | | 16° 30 nat. fine | 28402 |

Difference 56449 475165

Time from noon of first obs. 5h c' 40" rising 487170 Interval of time between obs. 4 45"

Time from noon of 2d obs. 0 15 40 rising 236839 Latitude by acc. 40° 0' cosine 9.88425 Declination 8 3 cosine 9.99670

Natural number 178 Greater altitude 57 53 nat. fine 84697

Merid. zen. dist. 31° 55 nat. cosine 84875 Declination 8 3

Latitude 39° 58 N.

Prob IX. Given the altitudes of two known stars, observed at the same or at different times; and if at different times, the interval between the observations; to find the latitude.

Rule. If both altitudes be observed at the same time, call the difference between their right ascensions the reduced interval.

But if the altitudes be taken at different times, reduce the interval between the observations to sidereal time, by adding thereto the proportional part answering to the interval, and 3° 56", the daily acceleration of the fixed stars. Now to the right ascension of the first observed star, add the interval in sidereal time, and the difference between this sum and the right ascension of the other star will be the reduced interval.

To the logarithm rising of the reduced interval, add the logarithmic cosines of the stars declinations; subtract the natural number answering to the sum of these logarithms from the natural cosine of the difference or sum of the stars declinations, according as they are of the same or of a contrary name, and the remainder will be the natural sine of arch first.

To the logarithmic cosine of arch first add the logarithmic secant of declination of the star having the least polar distance, and the logarithm half elapsed time of the reduced interval, the sum will be the logarithm half elapsed time of arch second.

From the natural cosine of the difference between arch first and the altitude of the star having the greatest polar distance, subtract the natural sine of the altitude of the other star, and find the logarithm of the remainder; to which add the logarithm secant of arch first, and the logarithmic secant of the altitude of the star having the greatest polar distance, the sum will be the logarithm rising of arch third. The difference between arches second and third is arch fourth.

To the logarithm rising of arch fourth add the logarithmic cosines of the declination and altitude of the star having the greatest polar distance; subtract the corresponding natural number from the natural cosine of the difference between the altitude and declination, the polar distance being less than 90°;

No 239 Practice.

Of finding Arch first the Longitude Rigel tude at Sea by Lunar Observations.

Difference Alt. of Menkar

Arch third Arch second Arch fourth Declin. of Rigel Alt. of Rigel

Sum

Latitude

71°59' secant 0.50963 29°51' secant 0.06181 42°8 N cosine 74159 43°38 N sine 69004 Difference 5155 3.71223 2h 24' 27" rising 4.28367 3°19' 36" rising 3.45960 0°55' 9" cosine 9.99526 8°27' S cosine 9.93819 29°51' cosine 9.93819 38°18 N cosine 78478 2472 3.39305 49°28 N fine 76006

In order to connect the observations for ascertaining the apparent time at the ship and the longitude with each other.

An observer without any assistants may very easily take all the observations, by first taking the altitudes of the objects, then the distance, and again their altitudes, and reduce the altitudes to the time of observation of the distance; or, by a single observation of the distance, the apparent time being known, the longitude may be determined.

A set of observations of the distance between the moon and a star, and their altitudes, may be taken with accuracy during the time of the evening or morning twilight; and the observer, though not much acquainted with the stars, will not find it difficult to distinguish the star from which the moon's distance is to be observed. For the time of observation nearly, and the ship's longitude by account being known, the ultimate time at Greenwich may be found; and by entering the Nautical Almanac with the reduced time, the distance between the moon and given star will be found nearly. Now set the index of the sextant to this distance, and hold the plane of the instrument so as to be nearly at right angles to the line joining the moon's culps, direct the sight to the moon, and by giving the sextant a slow vibratory motion, the axis of which being that of vision, the star, which is usually one of the brightest in that part of the heavens, will be seen in the filtered part of the horizon glass.

Sect. II. Of the Sextant.

This instrument is constructed for the express purpose of measuring with accuracy the angular distance between the sun and moon, or between the moon and a fixed star, in order to ascertain the longitude of a place by lunar observations. It is, therefore, made with more care than the quadrant, and has some additional appendages that are wanting in that instrument.

Fig. 74 represents the sextant, so framed as not to be liable to bend (m). The arch AA is divided into 120 degrees, each degree is divided into three parts; each of these parts, therefore, contains 20 minutes, which are again subdivided by the vernier into every half minute or 30 seconds. The vernier is numbered at every fifth of the longer divisions, from the right towards the left, with 5, 10, 15, and 20; the first division to the right being the beginning of the scale.

In order to observe with accuracy, and make the images come precisely in contact, an adjusting screw B is added to the index, which may thereby be moved with greater accuracy than it can be by hand; but this screw does not act until the index is fixed by the finger screw C. Care should be taken not to force the adjusting screw when it arrives at either extremity of its adjustment. When the index is to be moved any considerable quantity, the screw C at the back of the sextant must be loosened; but when the index is brought nearly to the division required, this back screw should be tightened, and then the index may be moved gradually by the adjusting screw.

(m) Troughton's patent double-framed sextants are not liable to bend. There are four tinged glasses D, each of which is set in a separate frame that turns on a centre. They are used to defend the eye from the brightness of the solar image and the glare of the moon, and may be used separately or together as occasion requires.

There are three more such glasses placed behind the horizon-glass at E, to weaken the rays of the sun or moon when they are viewed directly through the horizon-glass. The polar glass is sometimes used in observing altitudes at sea, to take off the strong glare of the horizon.

The frame of the index-glass I is firmly fixed by a strong cock to the centre plate of the index. The horizon-glass F is fixed in a frame that turns on the axes or pivots, which move in an exterior frame; the holes in which the pivots move may be tightened by four screws in the exterior frame. G is a screw by which the horizon glass may be set perpendicular to the plane of the instrument; should this screw become loose or move too easily, it may be easily tightened by turning the capstan headed screw H, which is on one side of the socket through which the stem of the finger screw passes.

The sextant is furnished with a plain tube (fig. 75) without any glasses; and to render the objects still more distinct, it has two telescopes, one (fig. 76) representing the object erect, or in their natural position; the longer one (fig. 77) shows them inverted; it has a large field of view and other advantages, and a little use will soon accustom the observer to the inverted position, and the instrument will be as readily managed by it as by the plain tube alone. By a telescope the contact of the images is more perfectly distinguished; and by the place of the images in the field of the telescope, it is easy to perceive whether the sextant is held in the proper place for observation. By sliding the tube that contains the eye-glasses in the inside of the other tube, the object is suited to different eyes, and made to appear perfectly distinct and well defined.

The telescopes are to be screwed into a circular ring at K; this ring rests on two points against an exterior ring, and is held thereto by two screws; by turning one or other of these screws, and tightening the other, the axis of the telescope may be set parallel to the plane of the sextant. The exterior ring is fixed on a triangular brass stem that slides in a socket, and by means of a screw at the back of the quadrant may be raised or lowered so as to move the centre of the telescope to point to that part of the horizon-glass which shall be judged the most fit for observation. Fig. 78. is a circular head, with tinged glasses to screw on the eye end of either of the telescopes or the plain tube. The glasses are contained in a circular plate which has four holes; three of these are fitted with tinged glasses, the fourth is open. By pressing the finger against the projecting edge of this plate, and turning it round, the open hole, or any of the tinged glasses, may be brought between the eyeglass of the telescope and the eye.

Fig. 79. is a magnifying glass, to assist the observer to read off the angle with more accuracy; and (fig. 80.) a screw-driver.

Adjustments of the Sextant.

The adjustments of a sextant are, to set the mirrors perpendicular to its plane and parallel to each other when the index is at Zenith, and to set the axis of the telescope parallel to the plane of the instrument. The three first of these adjustments are performed nearly in the same manner as directed in the section on the quadrant: as, however, the sextant is provided with a set of coloured glasses placed behind the horizon-glass, the index error may be more accurately determined by measuring the sun's diameter twice, with the index placed alternately before and behind the beginning of the divisions; half the difference of these two measures will be the index error, which must be added to, or subtracted from, all observations, according as the diameter measured with the index to the left of 9 is less or greater than the diameter measured with the index to the right of the beginning of the divisions.

Adjustment IV. To set the Axis of the Telescope parallel to the Plane of the Instrument.

Turn the eye end of the telescope until the two wires are parallel to the plane of the instrument; and let two distant objects be selected, as two stars of the first magnitude, whose distance is not less than 9° or 10°; make the contact of these objects as perfect as possible at the wire nearest the plane of the instrument; fix the index in this position; move the sextant till the objects are seen at the other wire, and if the same points are in contact, the axis of the telescope is parallel to the plane of the sextant; but if the objects are apparently separated, or do partly cover each other, correct half the error by the screws in the circular part of the supporter, one of which is above and the other between the telescope and sextant; turn the adjusting screw at the end of the index till the limbs are in contact; then bring the objects to the wire next the instrument; and if the limbs are in contact, the axis of the telescope is adjusted; if not, proceed as at the other wire, and continue till no error remains.

It is sometimes necessary to know the angular distance between the wires of the telescope; to find which, place the wires perpendicular to the plane of the sextant, hold the instrument vertical, direct the sight to the horizon, and move the sextant in its own plane till the horizon and upper wire coincide; keep the sextant in this position, and move the index till the reflected horizon is covered by the lower wire; and the division shown by the index on the limb, corrected by the index error, will be the angular distance between the wires. Other and better methods will readily occur to the observer at hand.

Use of the Sextant.

When the distance between the moon and the sun or a star is to be observed, the sextant must be held so that its plane may pass through the eye of the observer and both objects; and the reflected image of the most luminous of the two is to be brought in contact with the other seen directly. To effect this, therefore, it is evident, that when the brightest object is to the right of the other, the face of the sextant must be held upwards; but if to the left, downwards. When the face of the sextant is held upwards, the instrument should be supported with the right hand, and the index moved with the left hand. But when the face of the sextant is from the observer, it should be held with Sec. III. Of the Circular Instrument of Reflection.

This instrument was proposed with a view to correct the errors to which the sextant is liable; particularly the error arising from the inaccuracy of the divisions on the limb. It consists of the following parts: a circular ring or limb, two moveable indices, two mirrors, a telescope, coloured glasses, &c.

The limb of this instrument is a complete circle of metal, and is connected with a perforated central plate by six radii: it is divided into 720 degrees; each degree is divided into three equal parts; and the division is carried to minutes by means of the index scale as usual.

The two indices are moveable about the same axis, which passes exactly through the centre of the instrument:—the first index carries the central mirror, and the other the telescope and horizon-glass; each index being provided with an adjusting screw for regulating its motion, and a scale for showing the divisions on the limb.

The central mirror is placed on the first index immediately above the centre of the instrument, and its plane makes an angle of about 30° with the middle line of the index. The four screws in its pedestal for making its plane perpendicular to that of the instrument have square heads, and are therefore easily turned either way by a key for that purpose.

The horizon-glass is placed on the second index near the limb, so that as few as possible may be intercepted of the rays proceeding from the reflected object when to the left. The perpendicular position of this glass is rectified in the same manner as that of the horizon-glass of a sextant, to which it is similar. It has another motion, whereby its plane may be disposed so as to make a proper angle with the axis of the telescope, and a line joining its centre, and that of the central mirror.

The telescope is attached to the other end of the index. It is an achromatic astronomical one, and therefore inverts objects; it has two parallel wires in the common focus of the glasses, whose angular distance is between two and three degrees; and which, at the time of observation, must be placed parallel to the plane of the instrument. This is easily done, by making the mark on the eye-piece coincide with that on the tube. The telescope is moveable by two screws in a vertical direction with regard to the plane of the instrument, but is not capable of receiving a lateral motion.

There are two sets of coloured glasses, each set containing four, and differing in shade from each other. The glasses of the larger set, which belongs to the central mirror, should have each about half the degree of shade with which the correspondent glass of the set belonging to the horizon mirror is tinged. These glasses are kept tight in their places by small pressing screws, and make an angle of about 85° with the plane of the instrument; by which means the image from the coloured glass is not reflected to the telescope. When the angle to be measured is between 5° and 34°, one of the glasses of the largest set is to be placed before the horizon-glass.

The handle is of wood, and is screwed to the back of the instrument, immediately under the centre, with which it is to be held at the time of observation. Fig. 81. is a plan of the instrument, wherein the limb is represented by the divided circular plate; A is the central mirror, a a, the places which receive the stems a a of the glas fig. 84; E F, the first or central index, with its scale and adjusting screw; M N, the second or horizon index; G H, the telescope; I K, the screws for moving it towards or from the plane of the instrument; C, the place of the coloured glass fig. 83; and D, its place in certain observations.

Fig. 82 is a section of the instrument, wherein the several parts are referred to by the same letters as in fig. 81: Fig. 83 represents one of the horizon coloured glasses; and fig. 84, one of the central coloured glasses: Fig. 85, is the key for turning the adjusting screws of the mirrors: Fig. 86, is the handle: Fig. 87, a section of one of the radii towards its middle: Fig. 88, is used in some terrestrial observations for diminishing the light of the direct object, whose place at the time of observation is D: Fig. 89, is the tool for adjusting the central mirror, and for rectifying the position of the telescope with regard to the plane of the instrument; there is another tool exactly of the same size. The height of these is nearly equal to that of the middle of the central mirror.

Adjustments of the Circular Instrument.

I. To set the horizon-glass so that none of the rays from the central mirror shall be reflected to the telescope from the horizon mirror, without passing through the coloured glass belonging to this last mirror.—Place the coloured glass before the horizon mirror; direct the telescope to the silvered part of that mirror, and make it nearly parallel to the plane of the instrument; move the first index; and if the rays from the central mirror to the horizon-glass, and from thence to the telescope, have all the same degree of shade with that of the coloured glass used, the horizon-glass is in its proper position; otherwise the pedestal of the glass must be turned until the uncoloured images disappear.

II. Place the two adjusting tools on the limb, about 35° of the instrument distant, one on each side of the division on the left, answering to the plane of the central mirror produced; then the eye being placed at the upper edge of the nearest tool, move the central index till one half only of the reflected image of this tool is seen in the central mirror towards the left, and move the other tool till its half to the right is hid by the same edge of the mirror; then, if the upper edges of both tools are apparently in the same straight line, the central mirror is perpendicular to the plane of the instrument; if not, bring them into this position by the screws in the pedestal of the mirror.

III. To set the horizon mirror perpendicular to the plane of the instrument.—The central mirror being previously adjusted, direct the sight through the telescope to any well-defined distant object; then if, by moving the central index, the reflected image passes exactly over the direct object, the mirror is perpendicular; if not, its position must be rectified by means of the screws in the pedestal of the glass.

A planet, or star of the first magnitude, will be found a very proper object for this purpose.

IV. To make the line of collimation parallel to the plane of the instrument.—Lay the instrument horizontally on a table; place the two adjusting tools on the limb, towards the extremities of one of the diameters of the instrument; and at about 15 or 20 feet distant let a well-defined mark be placed, so as to be in the same straight line with the tops of the tools; then raise or lower the telescope till the plane passing through its axis and the tops of the tools is parallel to the plane of the instrument, and direct it to the fixed object; turn either or both of the screws of the telescope till the mark is apparently in the middle between the wires; then is the telescope adjusted; and the difference, if any, between the divisions pointed out by the indices of the screws will be the error of the indices. Hence this adjustment may in future be easily made.

In this process the eye tube must be so placed as to obtain distinct vision.

V. To find that division to which the second index being placed the mirrors will be parallel, the central index being at Zero.—Having placed the first index exactly to o, direct the telescope to the horizon mirror, so that its field may be bisected by the line of separation of the silvered and transparent parts of that mirror; hold the instrument vertically, and move the second index until the direct and reflected horizons agree; and the division shown by the index will be that required.

This adjustment may be performed by measuring the sun's diameter in contrary directions, or by making the reflected and direct images of a star or planet to coincide.

Use of the Circular Instrument.

To observe the distance between the sun and moon.

I. The sun being to the right of the moon.

Set a proper coloured glass before the central mirror, if the distance between the objects is less than 35°; but if above that quantity, place a coloured glass before the horizon mirror; make the mirrors parallel, the first index being at o, and hold the instrument so that its plane may be directed to the objects, with its face downwards, or from the observer: direct the sight through the telescope to the moon; move the second index, according to the order of the divisions on the limb, till the nearest limbs of the sun and moon are almost in contact: darken that index, and make the coincidence of the limbs perfect by the adjusting screw belonging thereto: then invert the instrument, and move the central index towards the second by a quantity equal to twice the arch passed over by that index: direct the plane of the instrument to the objects: look directly to the moon, and the sun will be seen in the field of the telescope: fasten the central index, and make the contact of the same two limbs exact by means of the adjusting screw: Then half the angle shown by the central index will be the distance between the nearest limbs of the sun and moon.

II. The sun being to the left of the moon.

Hold the instrument with its face upwards, so that its plane may pass through both objects; direct the telescope to the moon, and make its limb coincide with the nearest limb of the sun's reflected image, by moving the second index: now put the instrument in an opposite position; direct its plane to the objects, and the sight to the moon, the central index being previously moved towards the second by a quantity equal to twice the measured distance; and make the same two limbs... Practice.

Of finding limbs that were before observed coincide exactly, by turning the adjusting screw of the first index; then half the angle shown by the first index will be the angular distance between the observed limbs of the sun and moon.

To observe the Angular Distance between the Moon and a Fixed Star or Planet.

I. The star being to the right of the moon. In this case the star is to be considered as the direct object; and the enlightened limb of the moon's reflected image is to be brought in contact with the star or planet, both by a direct and inverted position of the instrument, exactly in the same manner as described in the last article. If the moon's image is very bright, the lightest tinged glass is to be used.

II. The star being to the left of the moon. Proceed in the same manner as directed for observing the distance between the sun and moon, the sun being to the right of the moon, using the lightest tinged glass, if necessary.

Sect. IV. Of the Method of determining the Longitude from Observation.

Prob. I. To convert degrees or parts of the equator into time.

Rule. Multiply the degrees and parts of a degree by 4, beginning at the lowest denomination, and the product will be the corresponding time. Observing that minutes multiplied by 4 produce seconds of time, and degrees multiplied by 4 give minutes.

Example I. Let $26^\circ 45'$ be reduced to time.

$$\begin{array}{c} 26^\circ 45' \\ \times 4 \\ \hline 104^\circ 60' = \text{time required.} \end{array}$$

Example II. Reduce $83^\circ 37'$ to time.

$$\begin{array}{c} 83^\circ 37' \\ \times 4 \\ \hline 332^\circ 148' = \text{time required.} \end{array}$$

Corresponding time = $5^h 34^m 28^s$

Prob. II. To convert time into degrees.

Rule. Multiply the given time by $\frac{1}{4}$, to which add the half of the product. The sum will be the corresponding degrees.

Example I. Let $3^h 4^m 28^s$ be reduced to degrees.

$$\begin{array}{c} 3^h 4^m 28^s \\ \times \frac{1}{4} \\ \hline 0^h 1^m 22^s \end{array}$$

Half = $15^m 22^s$

Corresponding deg. = $46^\circ 0'$

Example II. Reduce $8^h 42^m 36^s$ to degrees.

$$\begin{array}{c} 8^h 42^m 36^s \\ \times \frac{1}{4} \\ \hline 2^h 10^m 9^s \end{array}$$

Answer $130^\circ 39'$

Prob. III. Given the time under any known meridian, to find the corresponding time at Greenwich.

Rule. Let the given time be reckoned from the preceding noon, to which the longitude of the place in time is to be applied by addition or subtraction, according as it is east or west; and the sum or difference will be the corresponding time at Greenwich.

Example I. What time at Greenwich answers to $6^h 15'$ at a ship in longitude $76^\circ 45'W$?

| Time at ship | $6^h 15'$ | |--------------|-----------| | Longitude in time | $5^\circ 7'W$ |

Time at Greenwich, $11^h 22'$

Example II. Required the time at Greenwich answering to $5^h 46' 39''$ of May 1st, at Canton, whose longitude is $113^\circ 2' 15''E$?

| Time at Canton, May 1st | $5^h 46' 39''$ | |--------------------------|---------------| | Longitude in time | $7^\circ 32' 9''E$ |

Time at Greenwich, April 30th, $14^h 30'$

Prob. IV. To reduce the time at Greenwich to that under any given meridian.

Rule. Reckon the given time from the preceding noon, to which add the longitude in time of east, but subtract it if west; and the sum or remainder will be the corresponding time under the given meridian.

Example I. What is the expected time of the beginning of the lunar eclipse of February 25, 1793, at a ship in longitude $109^\circ 48'E$?

Beg. of eclipse at Greenwich per Naut. Alm. $9^h 23' 45''$ Ship's longitude in time, $7^\circ 19' 12''$

Time of beginning of eclipse at ship, $16^h 42' 57''$

Example II. At what time may the immersion of the first satellite of Jupiter be observed at Port St Julian, in longitude $68^\circ 44'W$, which, by the Nautical Almanac, happens at Greenwich 24th March 1792, at $17^h 53' 1''$?

App time of immersion at Greenwich $17^h 53' 1''$ Longitude of Port St Julian in time, $4^\circ 34' 56''W$

App. time of immer. at Port St Julian, $13^h 18' 5''$

Prob. V. To find the equation of equal altitudes.

Rule. To the cotangent of half the interval of time in degrees add the tangent of the latitude, and to the cotangent of half the interval add the tangent of the declination. Now if the latitude and declination be of a contrary name, add the corresponding natural numbers; but if of the same name, subtract them. Then to the ar. co. log. of this sum or difference add the proportional logarithm of one-fourth of the interval expressed in time, and the proportional logarithm of the daily variation of declination, the sum will be the proportional logarithm of the equation of equal altitudes in minutes and seconds, which are to be esteemed seconds and thirds.

Example. Let the latitude of the place of observation be $57^\circ 9'N$, the interval of time between the observations of the equal altitudes $5^h 17'$, the sun's declination $17^\circ 48'S$, and the daily change of declination $16' 19''$: Required the equation of equal altitudes?

Half the interval = $2^h 38' = 39^\circ 37'$

$$\begin{array}{c} \text{int.} = 39^\circ 37' \cos^i.0.19542 \quad \text{cotang.} \quad 0.08209 \\ \text{Lat.} = 57^\circ 9' \tan.0.18997 \quad \text{dec.} 17^\circ 48' \tan.9.50659 \\ \end{array}$$

$$\begin{array}{c} 0.38539 \quad 2.4288 \\ 3879 \quad 9.58863 \end{array}$$

Sum = $2.8167$ ar.co.lo.$9.5103$ **NAVIGATION**

**Rule VI.** To find the error of a watch by equal altitudes of the sun.

**Rule.** In the morning when the sun is more than two hours distant from the meridian, let a set of observations be taken, confining, for the sake of greater accuracy, at least three altitudes, which, together with the corresponding times per watch, are to be wrote regularly, the time of each observation being increased by 12 hours. In the afternoon, observe the instants when the sun comes to the same altitudes, and write down each opposite to its respective altitude. Now half the sum of any two times answering to the same altitude will be the time of noon per watch uncorrected. Find the mean of all the times of noon thus deduced from each corresponding pair of observations, to which the equation of equal altitudes is to be applied by addition or subtraction according as the sun is receding from or approaching to the elevated pole, and the sum or difference will be the time per watch of apparent noon, the difference between which and noon will be the error of the watch for apparent time; and the watch will be fast or slow according as the time of noon thereby is more or less than twelve hours.

**Example.** January 29th, 1785, in lat. 57° 9' N., the following equal altitudes of the sun were observed: Required the error of the watch?

**Example I.** September 15th 1792, in latitude 33° 56' S., and longitude 18° 22' E., the mean of the times per watch was 8h 12' 10" A.M. and that of the altitudes of the sun's lower limb 24° 48'; height of the eye 24 feet. Required the error of the watch?

| Obs alt. Sun's lower limb | 24° 48' | |---------------------------|---------| | Semidiameter | + 16.0 | | Dip | - 4.7 | | Correction | - 1.9 |

True altitude Sun's centre 24° 57.4 Latitude 33° 56' Declination 2° 45.4

Sum 36° 41.4 Sun's altitude 24° 57.4

Sun's meridian distance 3° 48' 5"

Apparent time 8h 11' 9" Time per watch 8h 12' 10" Watch fast 1'

Reduced declination 2° 45.4 S

**Example II.** May 6th 1793, in latitude 56° 4' N., and longitude 38° 30' W., at 4h 37' 4" P.M. per watch, the altitude of the sun's lower limb was 25° 6.1', and height of the eye 18 feet. Required the error of the watch for apparent time?

| Altitude sun's lower limb | = 25° 6.1' | |---------------------------|------------| | Semidiameter | + 15.9 | | Dip | - 4.1 | | Correction | - 1.9 |

True alt. sun's centre 25° 16.0

Sun's declin. per Nautical Almanac 16° 44' 3" Equation to 4h 37' P.M. + 3.4 to 38° 30' W. + 1.3

Reduced declination 16° 49.5 Practice.

| Latitude | Declination | |----------|-------------| | 56° 40' N | secant | | 16° 49' N | secant |

Difference

Sun's altitude

| nat. cosine | nat. sine | |------------|-----------| | 77448 | 42683 |

Difference

Apparent time

| 4h 38' 12" | rising | |------------|-----------|

Time per watch

| 4h 37' 4" | rising | |------------|-----------|

Watch slow

| 1h 8m | rising | |------------|-----------|

Prob. VIII. Given the latitude of a place, the altitude of a known fixed star, and the sun's right ascension, to find the apparent time of observation and error of the watch.

Rule. Correct the observed altitude of the star, and reduce its right ascension and declination to the time of observation.

With the latitude of the place, the true altitude and declination of the star, compute its horary distance from the meridian by last problem; which being added to or subtracted from its right ascension according as it was observed in the western or eastern hemisphere, the sum or remainder will be the right ascension of the meridian.

From the right ascension of the meridian subtract the sun's right ascension as given in the Nautical Almanac for the noon of the given day, and the remainder will be the approximate time of observation; from which subtract the proportional part of the daily variation of right ascension answering thereto, and let the proportional part answering to the longitude be added or subtracted according as the longitude is east or west, and the result will be the apparent time of observation; and hence the error of the watch will be known.

Example I. December 12th 1792, in latitude 37° 46' N, and longitude 21° 15' E, the altitude of Arcturus east of the meridian was 34° 6'.4, the height of the eye 10 feet. Required the apparent time of observation?

Observed alt. of Arcturus 34° 6'.4 Dip and refraction — 4.4

True altitude 34° 2.0 Latitude 37° 46' N. sec 0.10299 Declination 20° 14.4' N. sec 0.02778

Difference 17° 31.6' N. cos 95358 Altitude of Arcturus 34° 2.0' N. sine 55968

Difference 393904.59539

Arcturus's merid. dist. 4h 8' 10" rising 4.72526 right as. 14° 6' 13"

Right as. of merid. 9° 58' 3" Sun's right as. 17° 21' 59"

Approximate time 16° 36' 4" Eq. to approx. time — 3° 3"

Example. March 3d 1792, in latitude 51° 38' N, at 11h 29' 7" P.M. per watch, the altitude of the moon's lower limb was 37° 31', the height of the eye being 10 feet, and the time at Greenwich 13° 43'.

Required the error of the watch? Of finding Altitude of the moon's lower limb = $37^\circ 31'$

Moon's right ascension at Green. time = $7^\text{h} 22' 54''$

Longitude at Sea = $17^\circ 0' N$

By Lunar Dip = $-3$

Sun's right ascension = $23^\text{h} 2' 0''$

Correction = $+42$

Corrected alt. of moon's centre = $38^\circ 25'$

Latitude = $51^\circ 38' N$

Declination = $17^\circ 0' N$

Secant = $0.20712$

Difference = $34^\circ 38'$

Nat. cosine = $82281$

Moon's altitude = $38^\circ 25'$

Nat. sine = $62138$

Difference = $20143$

Difference = $4.30412$

Moon's meridian distance = $3^\circ 14' 36''$ rising

Right ascension = $7^\text{h} 22' 54''$

Right ascension of meridian = $10^\text{h} 37' 30''$

Sun's right ascension = $23^\text{h} 2' 0''$

Apparent time at ship = $11^\text{h} 35' 30''$

Time per watch = $11^\text{h} 29' 7''$

Watch flow = $6^\text{h} 23'$

**Prob. X.** Given the apparent distance between the moon and the sun or a fixed star, to find the true distance.

**Rule.** To the logarithmic difference answering to the moon's apparent altitude and horizontal parallax, add the logarithmic sines of half the sum, and half the difference of the apparent distance and difference of the apparent altitudes; half the sum will be the logarithmic cosine of an arch: now add the logarithm sines of the sum and difference of this arch, and half the difference of the true altitudes, and half the sum will be the logarithmic cosine of half the true distance.

**Example.** Let the apparent altitude of the moon's centre be $48^\circ 22'$, that of the sun's $27^\circ 43'$, the apparent central distance $81^\circ 23' 40''$, and the moon's horizontal parallax $58' 45''$. Required the true distance?

| Apparent altitude sun's centre | $27^\circ 43' 0''$ | |-------------------------------|------------------| | Apparent altitude moon's centre | $48^\circ 22' 0''$ | | Correction | $+1^\circ 40'$ | | Sun's true altitude | $27^\circ 41' 20''$ | | Sun's apparent altitude | $27^\circ 43'$ | | Moon's apparent altitude | $48^\circ 22'$ | | Difference | $20^\circ 39'$ | | Apparent distance | $81^\circ 23' 40''$ | | Sum | $102^\circ 2' 40''$ | | Difference | $60^\circ 44' 40''$ | | Half difference true altitudes | $10^\circ 39' 33''$ | | Arch | $51^\circ 27' 29''$ | | Sum | $62^\circ 7' 2''$ | | Difference | $40^\circ 47' 56''$ |

True distance = $81^\circ 4' 32''$

**Prob. XI.** To find the time at Greenwich answering to a given distance between the moon and the sun, or one of the stars, used in the Nautical Almanac.

**Rule.** If the given distance is found in the Nautical Almanac opposite to the given day of the month, or to that which immediately precedes or follows it, No. 239.

The time is found at the top of the page. But if this distance is not found exactly in the ephemeris, subtract the prop. log. of the difference between the distances which immediately precede and follow the given distance; from the prop. log. of the difference between the given and preceding distances, the remainder will be the prop. log. of the excess of the time corre...

Of finding corresponding to the given distance, above that answering to the preceding distance; And hence the apparent time at Greenwich is known.

Example. September 21, 1792, the true distance between the centres of the sun and moon was $68^\circ 13' 8''$. Required the apparent time at Greenwich?

Given distance $68^\circ 13' 8''$

Diff. at ix hours $67^\circ 53' 27''$ Diff. $= 0^\circ 19' 41''$ P.log. $9674$ Diff. at xii hours $69^\circ 30' 6''$ Diff. $= 1^\circ 36' 39''$ P.log. $2701$

Excess $0^\circ 36' 39''$ P.log. $6911$

Preceding time $9^\circ 0' 0''$

App. time at Greenwich $9^\circ 36' 39''$

Prob. XII. The latitude of a place and its longitude by account being given, together with the distance between, and the altitude of the moon and the sun, or one of the stars in the Nautical Almanac; to find the true longitude of the place of observation.

Rule. Reduce the estimate time of observation to the meridian of Greenwich by Problem III. and to

Example I. March 17, 1792, in latitude $34^\circ 53' N$, and longitude by account $27^\circ W$, about $9^h A.M.$ the distance between the nearest limbs of the sun and moon was $68^\circ 3' 4''$; the altitude of the sun's lower limb $33^\circ 18'$; that of the moon's upper limb $31^\circ 3' 3''$; and the height of the eye $12$ feet. Required the true longitude of the ship?

Time at ship $9^h 0' A.M.$

Longitude in time $1^\circ 48'$

Reduced time $10^\circ 48' A.M.$

Altitude moon's upper limb $31^\circ 3' 3''$

Aug. semidiameter $+ 16' 19''$

Dip $- 3' 18''$

Apparent altitude $30^\circ 43' 23''$

Correction $+ 49' 26''$

Moon's true altitude $31^\circ 32' 49''$

Sun's apparent altitude $33^\circ 30' 48''$

Moon's apparent altitude $30^\circ 43' 23''$

Difference $2' 47' 25''$

Apparent distance $68^\circ 35' 40''$

Logarithmic difference $9.996336$

Sum $71^\circ 23' 5''$

Difference $65^\circ 48' 15''$

Half difference true altitudes $0^\circ 58' 20''$

Arch $55^\circ 54' 12''$

Sum $56^\circ 52' 32''$

Difference $54^\circ 55' 52''$

Half true distance $34^\circ 6' 53''$

True distance $68^\circ 13' 46''$ | True distance | 68°13'45" | |--------------|-----------| | Distance at XXI hours | 69°11'20" | | Distance at noon | 67°32'38" | | Proportional part | - | | Preceding time | - |

| Apparent time at Greenwich | Latitude 34°53'0 N | |-----------------------------|--------------------| | Declination | 0°57'9 S | | Secant | 0.08602 | | Secant | 0.00006 | | Sum | 35°50'9 | | Nat. cofine | 81°057 | | Sun's altitude | 33°29'5 | | Nat. sine | 55°18'1 | | Difference | 25°876 | | Time from noon | 3h 7'13" | | Apparent time | 20°52'47 | | App. time at Green. | 22°45'0 | | Longitude in time | 1°52'13 = 28°3'W |

**Example II.** September 2, 1792, in latitude 13°57'N, and longitude by account 56°E, several observations of the moon and altair were taken; the mean of the times per watch was 1°18'59" A.M. that of the distance between altair and the moon's nearest limb 58°45'26"; the mean of the altitude of the moon's lower limb 70°33'; and that of altair 25°27'.4; height of the eye 13 feet. Required the true longitude?

| Time per watch | 1°18'59" A.M. | |----------------|---------------| | Longitude in time | 3°44'0 | | Augmented semidiameter | - | | Reduced time | 9°34'59" | | Altitude moon | 70°33' | | Semidiameter and dip | -0°13' | | Apparent alt. moon | 70°20' | | Correction | +0°19'40" | | True altitude moon | 70°39'40" | | Moon's apparent alt. | 70°20' | | Altair's apparent alt. | 25°24' | | Difference | 44°56' | | Apparent distance | 59°1'54" | | Logarithmic difference | 9.993101 | | Sum | 103°57'54" | | Difference | 14°5'54" | | Half | 51°58'57" | | Half | 7°2'57" | | Arch | 72°1'57" | | Sum | 94°40'47" | | Difference | 49°23'7" | | Half true distance | 29°33'48" | | True distance | 59°7'37" | | Distance at IX hours | 58°51'17" | | Distance at XII hours | 60°24'34" | | Proportional part | - | | Preceding time | - | | Apparent time at Greenwich | 9°31'3" | **NAVIGATION**

| Latitude | Declination | |----------|-------------| | 13°57' N | 8°19.8 N |

| Difference | Nat. cosine | Secant | |------------|-------------|--------| | 5°37.2 | 99519 | 0.01300|

| Altitude altair | Nat. sine | |-----------------|-----------| | 25°22' | 4.2841 |

| Difference | Secant | |------------|--------| | | 56678 |

| Altair's meridian distance | Rising | |----------------------------|--------| | 4h 23' 14" | |

| Right ascension meridian | Sun's right ascension | |-----------------------------|-----------------------| | 19°40'40" | 10°46'17" |

| Apparent time at ship | Apparent time at Greenwich | |-----------------------------|---------------------------| | 13°17'37" | 9°31'31" |

| Longitude in time | 3°46'6" = 56°31'1/2 East. |

For various other methods of determining the longitude of a place, the reader is referred to the article **LONGITUDE**.

**CHAP. III. Of the Variation of the Compass.**

The variation of the compass is the deviation of the points of the mariner's compass from the corresponding points of the horizon; and is denominated east or west variation, according as the north point of the compass is to the east or west of the true north point of the horizon.

A particular account of the variation, and of the several instruments used for determining it from observation, may be seen under the articles **AZIMUTH**, **COMPASS**, and **VARIATION**; and for the method of communicating magnetism to compass needles, see **MAGNETISM**.

**PROB. I.** Given the latitude of a place, and the sun's magnetic amplitude, to find the variation of the compass.

**RULE.** To the log. secant of the latitude, add the log. sine of the sun's declination, the sum will be the log. cosine of the true amplitude; to be reckoned from the north or south according as the declination is north or south.

The difference between the true and observed amplitudes, reckoned from the same point, and if of the same name, is the variation; but if of a different name, their sum is the variation.

If the observation be made in the eastern hemisphere, the variation will be east or west according as the observed amplitude is nearer to or more remote from the north than the true amplitude. The contrary rule holds good in observations taken in the western hemisphere.

**EXAMPLE I.** May 15, 1794, in latitude 33°10' N, longitude 18° W, about 5th A.M. the sun was observed to rise E & N. Required the variation?

Sun's dec. May 15, at noon 18°58' N.

Equation to 7th from noon — 0°4.

— to 18° W + 0°1

Reduced declination 18°55' Sine 9.51080

Latitude 33°10' Secant 0.07723

True amplitude N 67°13' E Cosine 9.58803

**Variation** 11°32'; which is east, because the observed amplitude is more distant from the north than the true amplitude; the observation being made in the eastern hemisphere.

**EXAMPLE II.** December 20, 1793, in latitude 31°38' S, longitude 83° W, the sun was observed to set SW. Required the variation?

| Latitude | Declination | |----------|-------------| | 31°38' | 23°28' |

| True amplitude S 62°7 W | Cosine 9.66997 | |-------------------------|---------------|

| Observed ampl. S 45°0 W |

**Variation** 17°7'; which is east, as the observed amplitude is farther from the north than the true amplitude, the observation being made at sun-setting.

It may be remarked, that the sun's amplitude ought to be observed at the instant the altitude of its lower limb is equal to the sum of 15 minutes and the dip of the horizon. Thus, if an observer be elevated 18 feet above the surface of the sea, the amplitude should be taken at the instant the altitude of the sun's lower limb is 19 minutes.

**PROB. II.** Given the magnetic azimuth, the altitude and declination of the sun, together with the latitude of the place of observation; to find the variation of the compass.

**RULE.** Reduce the sun's declination to the time and place of observation, and compute the true altitude of the sun's centre.

Find the sum of the sun's polar distance and altitude and the latitude of the place, take the difference between the half of this sum and the polar distance.

To the log. secant of the altitude add the log. secant of the latitude, the log. cosine of the half sum, and the log. cosine of the difference; half the sum of these will be the log. sine of half the sun's true azimuth, to be reckoned from the south in north latitude, but from the north in south latitude.

The difference between the true and observed azimuths will be the variation as formerly. **NAVIGATION**

**Example I.** November 18, 1793, in latitude 50° 22' N, longitude 24° 30' W, about three quarters past eight A.M., the altitude of the sun's lower limb was 8° 10', and bearing per compass S. 23° 18' E; height of the eye 20 feet. Required the variation of the compass?

Sun's declination at noon 19° 25' S. Equation to 31st from noon + 2 Reduced declination 19° 24'

True altitude 8° 16'

Polar distance 109° 24' Altitude 8° 16' Latitude 50° 22'

Secant 0.00454 Secant 0.19527

Sum 168° 2 Half 84° 1 Difference 25° 23'

Cosine 9.01803 Cosine 9.95591

Half true azimuth 22° 43' True azimuth S. 45° 26' E. Observed azimuth S. 23° 18' E.

Variation 22° 8' W.

**Example II.** January 3, 1794, in latitude 33° 52' N, longitude 53° 15' E, about half past three the altitude of the sun's lower limb 41° 18', and azimuth S. 50° 25' W, the height of the eye being 20 feet. Required the variation?

Sun's declination at noon 21° 24' S. Equation to time from noon + 2 Reduced declination 21° 24' S.

True altitude 41° 28'

Polar distance 111° 24' Altitude 41° 28' Latitude 33° 52'

Secant 0.12532 Secant 0.08075

Sum 186° 44' Half 93° 22' Difference 18° 2'

Cosine 8.76883 Cosine 9.97558

Sine 18.95048 Sine 9.47524

True azimuth S. 34° 46' W. Observed azimuth S. 50° 25' W.

Variation 15° 39' W.

**Chap. IV. Of a Ship's Journal.**

A journal is a regular and exact register of all the various transactions that happen aboard a ship whether at sea or land, and more particularly that which concerns a ship's way, from whence her place at noon or any other time may be justly ascertained.

That part of the account which is kept at sea is called *sea-work*; and the remarks taken down while the ship is in port are called *harbour-work*.

At sea, the day begins at noon, and ends at the noon of the following day; the first 12 hours, or those contained between noon and midnight, are denoted by P.M. signifying *after mid-day*; and the other 12 hours, or those from midnight to noon, are denoted by A.M. signifying *before mid-day*. A day's work marked Wednesday March 6 began on Tuesday at noon, and ended on Wednesday at noon. The days of the week are usually represented by astronomical characters. Thus ☀ represents Sunday; ⚛ Monday; ⚜ Tuesday; ⚝ Wednesday; ⚞ Thursday; ⚟ Friday; and ⚠ Saturday.

When a ship is bound to a port so situated that she will be out of sight of land, the bearing and distance of the port must be found. This may be done by Mercator's or Middle-latitude Sailing; but the most expeditious method is by a chart. If islands, capes, or headlands intervene, it will be necessary to find the several course and distances between each successively. The true course between the places must be reduced to the course per compass, by allowing the variation to the right or left of the true course, according as it is west or east.

At the time of leaving the land, the bearing of some known place is to be observed, and its distance is usually found by estimation. As perhaps the distance thus found will be liable to some error, particularly in hazy or foggy weather, or when that distance is considerable, it will therefore be proper to use the following method for this purpose.

Let the bearing be observed of the place from which the departure is to be taken; and the ship having run a certain distance on a direct course, the bearing of the same place is to be again observed. Now having one side of a plane triangle, namely the distance sailed and all the angles, the other distances may be found by Prob. I. of Oblique Sailing.

The method of finding the course and distance sailed in a given time is by the compass, the log line, and half minute-glass. These have been already described. In the royal navy, and in ships in the service of the East India company, the log is hove once every hour; but in most other trading vessels only every two hours.

The several courses and distances sailed in the course of 24 hours, or between noon and noon, and whatever remarks that are thought worthy of notice, are set down with chalk on a board painted black, called the log-board, which is usually divided into six columns: the first column on the left hand contains the hours from noon to noon; the second and third the knots and parts of a knot sailed every hour, or every two hours, according as the log is marked; the fourth column contains the courses steered; the fifth the winds; and in the sixth the various remarks and phenomena are written. The log-board is transcribed every day at noon into the log-book, which is ruled and divided after the same manner.

The courses steered must be corrected by the variation of the compass and leeway. If the variation is west, it must be allowed to the left hand of the course steered; but if east, to the right hand in order to obtain the true course. The leeway is to be allowed to the right or left of the course steered according as the ship is on the larboard or starboard tack. The method of finding the variation, which should be determined daily if possible, is given in the preceding chapter; and the leeway may be understood from what follows.

When a ship is close hauled, that part of the wind which acts upon the hull and rigging, together with a considerable part of the force which is exerted on the sails, tends to drive her to the leeward. But since the bow of a ship exposes less surface to the water than her side, the resistance will be less in the first case than in the second; the velocity in the direction of her head will therefore in most cases be greater than the velocity in the direction of her side; and the ship's real course will be between the two directions. The angle formed between the line of her apparent course and the line she really describes through the water is called the angle of leeway, or simply the leeway.

There are many circumstances which prevent the laying down rules for the allowance of leeway. The construction of different vessels, their trim with regard to the nature and quantity of their cargo, the position and magnitude of the sail set, and the velocity of the ship, together with the swell of the sea, are all susceptible of great variation, and very much affect the leeway. The following rules are, however, usually given for this purpose.

1. When a ship is close hauled, has all her sails set, the water smooth, with a light breeze of wind, she is then supposed to make little or no leeway. 2. Allow one point when the top-gallant sails are handed. 3. Allow two points when under close reefed topsails. 4. Allow two points and an half when one topsail is handed. 5. Allow three points and an half when both topsails are handed. 6. Allow four points when the fore-course is handed. 7. Allow five points when under the mainail only. 8. Allow six points when under balanced mizen. 9. Allow seven points when under bare poles.

These allowances may be of some use to work up the day's works of a journal which has been neglected; but a prudent navigator will never be guilty of this neglect. A very good method of estimating the leeway is to observe the bearing of the ship's wake as frequently as may be judged necessary; which may be conveniently enough done by drawing a small semicircle on the taffrail, with its diameter at right angles to the ship's length, and dividing its circumference into points and quarters. The angle contained between the semidiameter which points right aft and that which points in the direction of the wake is the leeway. But the best and most rational way of bringing the leeway into the day's log is to have a compass or femicircle on the taffrail, as before described, with a low crutch or swivel in its centre: after heaving the log, the line may be slipped into the crutch just before it is drawn in, and the angle it makes on the limb with the line drawn right aft will show the leeway very accurately; which as a necessary article, ought to be entered into a separate column against the hourly distance on the log-board.

In hard blowing weather, with a contrary wind and a high sea, it is impossible to gain any advantage by sailing. In such cases, therefore, the object is to avoid as much as possible being driven back. With this intention it is usual to lie to under no more sail than is sufficient to prevent the violent rolling which the vessel would otherwise acquire, to the endangering her masts, and straining her timbers, &c. When a ship is brought to, the tiller is put close over to the leeward, which brings her head round to the wind. The wind having then very little power on the sails, the ship loses her way through the water; which ceasing to act on the rudder, her head falls off from the wind, the sail which she has set fills, and gives her fresh way through the water; which acting on the rudder brings her head again to the wind. Thus the ship has a kind of vibratory motion, coming up to the wind and falling off from it again alternately. Now the middle point between those upon which she comes up and falls off is taken... taken for her apparent course; and the leeway and variation is to be allowed from thence, to find the true course.

The setting and drift of currents, and the heave of the sea, are to be marked down. These are to be corrected by variation only.

The computation made from the several courses corrected as above, and their corresponding distances, is called a day's work; and the ship's place, as deduced therefrom, is called her place by account, or dead-reckoning.

It is almost constantly found that the latitude by account does not agree with that by observation. From an attentive consideration of the nature and form of the common log, that its place is alterable by the weight of the line, by currents, and other causes, and also the errors to which the course is liable, from the very often wrong position of the compass in the binnacle, the variation not being well ascertained, an exact agreement of the latitudes cannot be expected.

When the difference of longitude is to be found by dead reckoning, if then the latitudes by account and observation disagree, several writers on navigation have proposed to apply a conjectural correction to the departure or difference of longitude. Thus, if the course be near the meridian, the error is wholly attributed to the distance, and the departure is to be increased or diminished accordingly: if near the parallel, the course only is supposed to be erroneous; and if the course is towards the middle of the quadrant, the course and distance are both assumed wrong. This last correction will, according to different authors, place the ship upon opposite sides of her meridian by account. As these corrections are, therefore, no better than guessing, they should be absolutely rejected.

If the latitudes are not found to agree, the navigator ought to examine his log-line and half-minute glass, and correct the distance accordingly. He is then to consider if the variation and leeway have been properly ascertained; if not, the courses are to be again corrected, and no other alteration whatever is to be made on them. He is next to observe if the ship's place has been affected by a current or heave of the sea, and to allow for them according to the best of his judgment. By applying these corrections, the latitudes will generally be found to agree tolerably well; and the longitude is not to receive any farther alteration.

It will be proper, however, for the navigator to determine the longitude of the ship from observation as often as possible; and the reckoning is to be carried forward in the usual manner from the last good observation: yet it will perhaps be very satisfactory to keep a separate account of the longitude by dead-reckoning.

General Rules for working a Day's Work.

Correct the several courses for variation and leeway; place them, and the corresponding distances, in a table prepared for that purpose. From whence, by Traverse Sailing, find the difference of latitude and departure made good: hence the corresponding course and distance, and the ship's present latitude, will be known.

Find the middle latitude at the top or bottom of the Traverse Table, and the distance answering to the departure found in a latitude column, will be the difference of longitude: Or, the departure answering to the course made good, and the meridional difference of latitude in a latitude column, is the difference of longitude. The sum, or difference of which, and the longitude left, according as they are of the same or of a contrary name, will be the ship's present longitude of the same name with the greater.

Compute the difference of latitude between the ship and the intended port, or any other place whose bearing and distance may be required: find also the meridional difference of latitude and the difference of longitude. Now the course answering the meridional difference of latitude found in a latitude column, and the difference of longitude in a departure column, will be the bearing of the place, and the distance answering to the difference of latitude will be the distance of the ship from the proposed place. If these numbers exceed the limits of the Table, it will be necessary to take aliquot parts of them; and the distance is to be multiplied by the number by which the difference of latitude is divided.

It will sometimes be necessary to keep an account of the meridian distance, especially in the Baltic or Mediterranean trade, where charts are used in which the longitude is not marked. The meridian distance on the first day is that day's departure; and any other day it is equal to the sum or difference of the preceding day's meridian distance and the day's departure, according as they are of the same or of a contrary denomination. ### NAVIGATION

**A Journal of a Voyage from London to Funchal in Madeira, in his Majesty's Ship the Resoluton, A—— M—— Commander, anno 1793.**

| Day of Month | Winds | Remarks on board his Majesty's ship Resoluton, 1793. | |--------------|-------|-----------------------------------------------------| | 5 Sept. 28. | SW | Strong gales and heavy rain. At 3 P.M. sent down top gallant yards; at 11 A.M. the pilot came on board. | | 6 Sept. 29. | SW | Moderate and cloudy, with rain. At 10 A.M. cast loose from the sheer hulk at Deptford; got up top-gallant yards, and made sail down the river. At noon running through Blackwall reach. | | 7 Sept. 30. | SW | The first part moderate, the latter squally with rain. At half past one anchored at the Galleons, and moored ship with near a whole cable each way in 5 fathoms, a quarter of a mile off shore. At 3 A.M. strong gales: got down top-gallant yards. A.M. the people employed working up junk. Bent the sheet cable. | | 8 Octob. 1. | SSW | Fresh gales and squally. P.M. received the remainder of the boatswain's and carpenter's stores on board. The clerk of the cheque muttered the ship's company. | | 9 Octob. 2. | SW | Variable weather, with rain. At noon weighed and made sail; at 5 anchored in Long-reach in 8 fathoms. Received the powder on board. At 6 A.M. weighed and got down the river. At 10 A.M. past the Nore; brought to and hoisted in the boats; double reefed the topsails, and made sail for the Downs. At noon running for the flats of Margate. | | 10 Octob. 3. | NNE | First part stormy weather; latter moderate and clear. At 4 P.M. got through Margate Roads. At 5 run through the Downs; and at 6 anchored in Dover-road, in 10 fathoms muddy ground. Dover Castle bore north, and the south Foreland NESE off shore 1½ miles. Discharged the pilot. Employed making points, &c. for the sails. Scaled the guns. | | 11 Octob. 4. | N | Moderate and fair. Employed working up junk. Received from Deal a cutter of 17 feet, with materials. A.M. strong gales and squally, with rain; got down top-gallant yards. |

| Hours | Kn. | Fa. | Courses | Winds | Remarks, 5 Oct. 5. 1793. | |-------|-----|-----|---------|-------|--------------------------| | 1 | | | | NNE | Fresh gales with rain. | | 2 | | | | | Hove short. | | 3 | | | | | Weighed and made sail. | | 4 | | | WSW | | Shortened sail.—Dungeness light NESE. | | 5 | | | | | Fresh breezes, and cloudy. | | 6 | | | | | Ditto weather. | | 7 | | | | | Got up top-gallant yards. | | 8 | | | | | Set fludding sails. | | 9 | | | | | Ditto weather. | | 10 | | | | | St Alban's Head NSE. | | 11 | | | | | | | 12 | | | | | | ### Navigation

**A Journal from England towards Madeira**

| Hours | Kn. | Fa. | Courses | Winds | Remarks, October 6, 1793 | |-------|-----|-----|---------|-------|--------------------------| | 1 | 8 | | WNW | NE | A fresh steady gale. | | 2 | 8 | | | | Do. weather. | | 3 | 8 | | | | Spoke the Ranger of London, from Carolina. Took in studding sails. | | 4 | 8 | | | | Do. weather. | | 5 | 8 | | | | Eddystone light NW. | | 6 | 8 | | | | Do. weather. | | 7 | 8 | | | | Eddystone light NE. | | 8 | 8 | | | | Do. weather. | | 9 | 8 | | | | Set lower studding sails.| | 10 | 8 | | | | Fresh breeze and clear weather. | | 11 | 8 | | | | Do. weather. | | 12 | 8 | | | | |

| Course | Diff. | D.L. | Dep. | Acc. | Obf. | D. Long. | Acc. | Obf. | W. Lon.by | W.Var. | |--------|-------|------|------|------|------|----------|------|------|-----------|--------| | S 52° ½ W. | 93 | 57 | 74 | 49° 11' | 49° 9' | 114° W. | 6° 18' | 2 ¼ pts. |

As there is no land in sight this day at noon, and from the course and distance run since the last bearing of the Eddystone light was taken, it is not to be supposed that any part of England will be seen; the departure is therefore taken from the Eddystone; and the distance of the ship from that place is found by revolving an oblique angled plane triangle, in which all the angles are given, and one side, namely, the distance run (16 miles) between the observations. Hence the distance from the Eddystone at the time the last bearing of the light was taken will be found equal to 18 miles; and as the bearing of the Eddystone from the ship at that time was NE, the ship's bearing from the Eddystone was SE. Now the variation 2 ¼ points W. being allowed to the left of SW, gives SSW ½ W., the true course. The other courses are in like manner to be corrected, and inferred in the following table, together with their respective distances, beginning at 10 o'clock AM, the time when the last bearing of the Eddystone was taken. The difference of latitude, departure, course, and distance, made good, are to be found by Traverse Sailing.

| Courses | Dist. | Diff. of lat. | Departure | |---------|-------|---------------|-----------| | SSW ½ W. | 18 | N. 17.0 | 6.1 | | WSW ½ S | 22 | S. 5.3 | 21.3 | | SW ½ W. | 58 | E. 34.6 | 46.6 |

S 52° ½ W. Latitude of Eddystone 56.9 = 57m. 74.0

Latitude by account - 49 11N. Sum - 99 19 Middle latitude - 49 40

Now to middle latitude as a course, and the departure 74m. in a latitude column, the difference of long. in a distance column is 114 = 1° 54' W. Longitude of Eddystone - 4 24 W.

Longitude in by account - 6 18 W. ### NAVIGATION

**A Journal from England towards Madeira**

| Hours | Kn. | Fa. | Courses | Winds | Remarks, 2 October 7, 1793 | |-------|-----|-----|---------|-------|---------------------------| | 1 | 6 | 5 | WSW | NE | Fresh breezes. | | 2 | 6 | 5 | | | Sounded 62; fine sand. | | 3 | 6 | | | | Moderate and cloudy. | | 4 | 5 | 3 | | | Unbent the cables, and coiled them. Took in studding sails. | | 5 | 5 | | N | | Do. weather. | | 6 | 5 | | | | Do. weather. | | 7 | 5 | | | | Light breeze. | | 8 | 4 | 7 | SWNW | NW | A sail SE. | | 9 | 4 | 5 | | | Hazy weather. | | 10 | 4 | 5 | | | Do. weather. |

| Course | Dist. | DL. | Dep. | Acc. | Obf. | D. Long. | W. Long. by Acc. | W. Var. by ac. | Porto Sancto's Bearing | Distance | |--------|-------|-----|------|------|------|----------|------------------|-----------------|------------------------|----------| | S 38° W. | 99 | 78 | 62 | 47° 51' | 93° W. | 7° 51' | 2½ pts. | S 23° ½ W. | .974 m. |

The courses being corrected for variation, and the distances summed up, the work will be as under.

| Courses | Diff. | N. Latitude by Obf. | W. Longitude by Acc. | W. Var. by ac. | Porto Sancto's Bearing | Distance | |---------|-------|----------------------|----------------------|----------------|------------------------|----------| | SWzS. | 77 | | | | | | | SSW ½ W.| 12 | | | | | | | SSW ¼ W.| 11 | | | | | | | S 38° W.| 99 | | | | | |

Yesterday's lat. by obser. = 49° 9 N. Latitude by account = 47° 51 N. Sum = 97° Middle latitude = 48° 30'

To middle latitude 48° 30', and departure 61.6 in a latitude column, the corresponding difference of longitude in a distance column is 93°.

Yesterday's longitude = 6° 18 W. Longitude in by account = 7° 51 W.

It is now necessary to find the bearing and distance of the intended port, namely, Funchal; but as that place is on the opposite side of the island with respect to the ship, it is therefore more proper to find the bearing of the east or west end of Madeira; the east end is, however, preferable. But as the small island of Porto Sancto lies a little to the NE of the east end of Madeira, it therefore seems more eligible to find the bearing and distance of that island.

To find the bearing and distance of Porto Sancto.

Latitude of ship = 47° 51 N. Mer. parts = 32° 78 Longitude of ship = 7° 51 W. Lat. of Porto Sancto = 32° 58 N. Mer. parts = 20° 97 Lon. Porto Sancto = 16° 25 W.

Difference of latitude = 14° 53' = 893 M.D. lat. 1181 Difference of long. = 8° 34' = 514

The course answering to the meridional difference of latitude and difference of longitude is about 23° ½, and the distance corresponding to the difference of latitude is 974 miles. Now as Porto Sancto lies to the southward and westward of the ship, the course is therefore S 23° ½ W.; and the variation, because W. being allowed to the right hand, gives SW ¼ W nearly, the bearing per compass; and which is the course that ought to be steered.

Vol. XII. Part II. ### NAVIGATION

**A Journal from England towards Madeira**

| Hours | Kn. | Fa. | Courses | Winds | Remarks, 8 October 8, 1793 | |-------|-----|-----|---------|-------|--------------------------| | 1 | 2 | | SW | NW | Little wind and cloudy. | | 2 | 1 | | | Variable. | Tried the current and found none. | | 3 | | | Ship's head to the SW. | Calm. | | 4 | | | | | Calm; a long swell from the SW. | | 5 | | | | | Light airs and hazy. | | 6 | | | | | Moderate wind and cloudy. | | 7 | | | | | Set top-gallant sails. |

By double altitudes of the sun, the latitude was found to be $47^\circ 28' N$.

| Course | Diff. | D.L. Dep. | N. Latit. by Acc. | Obf. | D. Long. | W. Long. by Acc. | Obf. | W. Var. | Porto Sancto's Bearing. | Distance. | |--------|-------|-----------|-------------------|------|----------|-----------------|------|--------|------------------------|----------| | S 61° W. | 51 | 25 | 45 | $47^\circ 26'$ | $47^\circ 28'$ | 6° W. | $8^\circ 58'$ | 2 points | S 21° W | 932 |

The several courses corrected will be as under.

| Courses | Dist. | Diff. of latitude | Departure | |---------|-------|------------------|-----------| | SSW | 3 | | | | SW | 13 | | | | WSW | 22 | | | | WSSS | 15 | | | | S 61° W.| 51 | 24.8 | = 25 | | Yesterday's latitude | | $47^\circ 51'$ | |

Latitude by account $47^\circ 26'$

Sum $77$

Middle latitude $47^\circ 39'$

To middle latitude $47^\circ 20'$, and departure $45'$ in a latitude column, the difference of longitude in a distance column is $6^\circ 7'$ = $1^\circ 7'$ W

Yesterday's longitude $7^\circ 51'$ W

Longitude in by account $8^\circ 58'$ W

To find the bearing and distance of Porto Sancto.

| Latitude of ship | 47° 28' N | Mer. parts | 3244 | Longitude | 8° 58' W | |------------------|-----------|------------|------|-----------|----------| | Lat. of Porto Sancto | 32° 58' N | Mer. parts | 2097 | Longitude | 16° 25' W |

Difference of latitude $14^\circ 30'$ = 870 M. D. lat. $11^\circ 47'$

D. longitude $7^\circ 27'$ = $44^\circ 7'$

Hence the bearing of Porto Sancto is S 21° W, and distance 932 miles. The course per compass is therefore SW nearly. ### NAVIGATION

**A Journal from England to Madeira**

| Hours | Kn. | Fa. | Courses | Winds | Remarks, 9 October 1793 | |-------|-----|-----|---------|-------|------------------------| | 1 | 5 | | WbN | SWbS | Squally with rain. | | 2 | 5 | | | | Handed top-gallant sails. | | 3 | 5 | | | | In 1st reef topsails. | | 4 | 5 | 4 | | | Dark gloomy weather. Tacked ship. | | 5 | 5 | 6 | SEbS | | In 2nd reef topsails, and down top-gallant yards. | | 6 | 5 | | | | Stormy weather; in fore and mizen topsails and 3rd reef maintop fail. Handed the maintop fail, bent the main-stay sail, and brought to with it and the mizen; reefed the mainsail, at 10, wore and lay to under the mainsail, got down top-gallant masts; at 12 set the foresail, and balanced the mizen. | | 7 | 4 | | | | The sea stove in several half ports. | | 8 | 4 | | | | The swell abates a little. | | 9 | 3 | | | | The fwell abates fast. | | 10 | 3 | | | | Up top-gallant masts. | | 11 | 3 | | | | Set the topsails. | | 12 | 3 | | | | Clear weather; good observation. |

| Course | Diff. | D.L. | Dep. | N. latit. by | W. Long. by | Porto Sancto's | |--------|-------|------|------|--------------|-------------|----------------| | WbN½N | 43 | 12 | 41 | 47° 40' | 9° 59' | 2 points |

There is no leeway allowed until 2 o'clock P.M. when the top-gallant sails are taken in; from 2 to 3 one point is allowed; from 3 to 6, one and a half points are allowed; from 6 to 8, one and three-fourth points are allowed; from 8 to 9, three points; from 9 to 10, four and an half points; from 10 to 12, five points; from 12 to 10 A.M. three and an half points; and from thence to noon two points leeway are allowed. Now the several courses being corrected by variation and leeway will be as under; but as the corrected courses from 2 to 3 P.M. and from 10 to 12 A.M. are the same, namely, west; this, therefore, is inserted in the table, together with the sum of the distances, as a single course and distance. In like manner the courses from 12 to 2, and from 5 to 8 being the same, are inserted as a single course and distance.

| Courses | Diff. of latit. | Departure | |---------|-----------------|-----------| | WbS | | | | W | | | | W½N | | | | E½S½S | | | | E½S½S | | | | NE½E | | | | NWbW | | | | NWbW½W | | | | NW½W | | | | WbN½N | | | | W½N½N | | |

Yeast latitude 47° 28' N

Lat. by account 47° 40' N

To middle latitude 47° 34' and departure 41.5

the difference of longitude is 61' = 1° 1' W

Yesterday's longitude 8° 58' W

Longitude in by account 9° 59' W ### NAVIGATION

**A Journal from England towards Madeira**

| Hours | Kn. | Fa. | Courses | Winds | Remarks, 4 October 10, 1793 | |-------|-----|-----|---------|-------|---------------------------| | 1 | 5 | 3 | W | SSW | Fresh gales with rain. | | 2 | 5 | 7 | | | Do. weather. | | 3 | 6 | | | | Out 3d reef topsails. | | 4 | 6 | | | | Loft a log and line. | | 5 | 5 | | | | Do. weather. | | 6 | 5 | | | | Do. weather. | | 7 | 5 | | | | Moderate and cloudy, out all reefs. | | 8 | 4 | | | | Sprung fore top-gallant yard, got up another. | | 9 | 4 | | | | Do. weather. | | 10 | 5 | | | | A sail NE. | | 11 | 5 | | | | Employed working up junk. | | 12 | 5 | | | | A swell from the NW, which by estimation has set ship 7 miles in the opposite direction. |

| Course | Diff. | D.L. | Dep. | N. Latit. by | D. Long. | W. Long. by | W. Var. | Porto Sancto's | |--------|-------|------|------|--------------|----------|-------------|---------|----------------| | | Acc. | Obf. | | | | | | Bearing. | | | | | | | | | | Distance. | | S 74° W. | 108 | 30 | 104 | 47° 9' | 153° W. | 12° 32' | 2 Points. | S 12° W. | 870 m. |

Two points leeway are allowed on the first course, one on the second; and as the ship is 7 points from the wind on the third course, there is no leeway allowed on it. The opposite point to NW, that from which the swell set, with the variation allowed upon it, is the last course in the Traverse Table.

| Courses | Diff. | Departure | |---------|-------|-----------| | | N. | S. | E. | W. | | W | 86.2 | | | 86.2 | | SWbW | 12.3 | 6.9 | | 10.2 | | SWbS | 24.7 | 20.5 | | 13.7 | | ESE | 7 | 2.7 | 6.5 | | | S 74° W.| 108 | 30.0 | 6.5 | 110.1 |

Yesterday’s latitude 47° 39' Latitude by account 47° 9' Sum 48° Middle latitude 47° 24'

To middle latitude 47° 24', and departure 103.6, the difference of longitude is 153° = 2° 33' W Yesterday’s longitude 9° 59' W

Longitude in 12° 32' W

To find the bearing and distance of Porto Sancto.

Latitude of ship 47° 9' Mer. parts Lat. Porto Sancto 32° 58' Mer. parts Difference of latitude 14° 11' = 85' M. D. lat.

Hence the bearing of Porto Sancto is S 12° W, and distance 870 miles; the course per compass is therefore about SWbW.

A Journal | Hours | Kn. | Fa. | Course | Winds | Remarks, 2 October 11, 1793 | |-------|-----|-----|--------|-------|---------------------------| | 1 | 4 | | SWbS | ESE | Moderate wind and fair weather. | | 2 | 3 | | | | Shortened sail and set up the topmast rigging. | | 3 | 2 | | | | Do. weather. | | 4 | 3 | | | | Variation per amplitude 21° W. | | 5 | 4 | | | | A fine steady breeze. | | 6 | 4 | | | | By an observation of the moon's distance from Pegasi, the ship's longitude at half past 8 was 12° 28' W. | | 7 | 4 | | | | Clear weather. | | 8 | 4 | | | | Do. weather. | | 9 | 5 | | | | Set fludding sails, &c. | | 10 | 5 | | | | One sail in sight. | | 11 | 5 | | | | Do. weather, good observation. | | 12 | 6 | | | | |

| Course | Dist. | D.L | Dep. | N. Latit. by | W. Long. by | Porto Sancto's | |--------|-------|-----|------|--------------|-------------|----------------| | | Acc. | Obs.| Acc. | Obs. | Bearing. | Distance. | | S 12° 45' W. | 128 | 125 | 28 | 45 4 | 44° 59' | 41° W. | 13° 13' | 12° 59' | 21° | S 12° W. | 737 miles. |

The observed variation 21° being allowed to the left of SWbS gives S 12° 45' W, the corrected course, and the distance summed up is 127.9, or 128 miles. Hence the difference of latitude is 124.8, and the departure 28.2. The latitude by account is therefore 45° 4' N, and the middle latitude 46° 6', to which, and the departure 28.2 in a latitude column, the difference of longitude in a distance column is 41° W; which being added to 12° 32' W, the yesterday's longitude, gives 13° 13' W, the longitude in by account. But the longitude by observation was 12° 28' W at half past 8 P.M.; since that time the ship has run 96 miles; hence the departure in that interval is 21.2 m. Now half the difference of latitude 47 m. added to 44° 59', the latitude by observation at noon, the sum 45° 46' is the middle latitude; with which and the departure 21.2, the difference of longitude is found to be 31° W; which therefore added to 12° 28', the longitude observed, the sum is 12° 59' W, the longitude by observation reduced to noon.

To find the bearing and distance of Porto Sancto.

| Latitude ship | 44° 59' N. Mer. parts | 3028 | Longitude | 12° 59' W | | Lat. Porto Sancto | 32° 58' N. Mer. parts | 2097 | Longitude | 16° 25' W | | Difference of latitude | 12° 1° = 721 | M. D. lat. 931 | D. longitude | 3° 26° = 206° |

Hence the bearing of Porto Sancto is S 12° W, and distance 737 miles. The course to be steered is therefore S 33° W, or SWbS nearly. ### Navigation

**A Journal from England towards Madeira**

| Hours | Kn. | Fa. | Course | Winds | Remarks, 2 October 12, 1793 | |-------|-----|-----|--------|-------|-----------------------------| | 1 | 8 | SWsS | ENE | Fresh gales, and cloudy. | | 2 | 5 | | | Do. Weather. | | 3 | 8 | | | Hauled down studding-sails. | | 4 | 6 | | | Do. Weather. | | 5 | 8 | | | A steady gale and fine weather. | | 6 | 8 | | | Do. Weather. | | 7 | 7 | | | Out studding-sails alow and aloft. | | 8 | 7 | | | Variation per azimuth 20° 14' W. | | 9 | 7 | | | A sail in the SW quarter. | | 10 | 7 | | | Sailmaker altering a lower studding-fail. | | 11 | 7 | | | Fine weather, and cloudy. |

**Course. Dist. D.L. Dep. N. Latit. by Acc. Obf. D. Long. W. Long. by Acc. Obf. W. Var. Obf. Porto Sancto's Bearing. Distance.**

| S 13° 31' W. | 183 | 178 | 43 | 42° 1' | 50° W. | 14° 12' | 13° 58' | 20° 14' | S 12° W. | 555 m. |

The course corrected by variation is S 13° 31' W., and the distance run is 183 miles; hence the difference of latitude is 177.9, and the departure 42.8.

Yesterday's latitude by observation - 44° 59' N. Mer. parts - 3028

Difference of latitude - 2° 58' S.

Latitude in by account - 42° 1' N. Mer. parts - 2783

Meridional difference of latitude

Now to course 13° 31', and meridional difference of latitude 245 in a latitude column, the difference of longitude in a departure column is 59' W.; hence the yesterday's longitudes by account and observation, reduced to the noon of this day, will be 14° 12' W and 13° 58' respectively.

To find the bearing and distance of Porto Sancto.

| Latitude ship | 42° 1' N. | Mer. parts | 2783 | Longitude | 13° 58' W. | | Lat. Porto Sancto | 32° 58' N. | Mer. parts | 2097 | Longitude | 16° 25' W. |

Difference of latitude 9° 3' = 543 M.D. latitude 686 D. longitude 2° 27' = 147'.

The meridional difference of latitude and difference of longitude will be found to agree nearest under 12°, the correct bearing of Porto Sancto; and the variation being allowed to the right hand of S 12° W., gives S 32° 1' W., the bearing per compass; and the distance answering to the difference of latitude 543, under 12 degrees, is 555 miles. ### NAVIGATION

**A Journal from England towards Madeira**

| Hours | Kn. | Fa. | Courses | Winds | Remarks, October 13, 1793 | |-------|-----|-----|---------|-------|--------------------------| | 1 | 8 | SWsS | ENE | A steady gale, and fine weather. | | 2 | 8 | 5 | | At 34 minutes past three, the distance between the nearest limbs of the sun and moon, together with the altitude of each, were observed from whence the ship's longitude at that time is 14° 1' W. | | 3 | 8 | 6 | | Hauled in the lower studding-falls. | | 4 | 8 | | | At 9h 22', by an observation of the moon's distance from α Pegasi, the longitude was 14° 20' W. | | 5 | 8 | | | Fresh gales, and clear. | | 6 | 8 | | | Do. weather. | | 7 | 8 | | | Variation per amplitude 19° 51' W. | | 8 | 8 | | | Do. per azimuth 19° 28' W. Set studding-falls. | | 9 | 8 | | | Carried away a fore-top-mast studding-fall boom, got up another. | | 10 | 8 | | | Fresh gales. Took in studding-falls. |

| Course | Diff. D.L. | Dep. | N. latit. by Acc. | Obf. | D. Long. | W. Long. by Acc. | W. Var. by Obf. | Porto Sancto's Bearing | Distance | |--------|------------|------|------------------|------|----------|-----------------|-----------------|------------------------|----------| | SSW 1/4 W. | 184 | 178 | 45 | 39° 3' | 59° W. | 15° 11' | 14° 52 | 1½ pts. |

The mean of the variations is about 1½ points W; hence the course corrected is SSW 1/4 W; with which and the distance run 184 miles, the difference of latitude is 178.5, and the departure 44.7.

Yesterday's latitude 42° 1' N. Mer. parts 2783 Difference of latitude 2° 58' S.

Latitude in by account 39° 3 N. Mer. parts 2549

Meridional difference of latitude

Now, to course 1½ points, and meridional difference of latitude 234, the difference of longitude is about 59 m.; which, added to the yesterday's longitude by account 14° 12' W, the sum 15° 11' W is the longitude in by account at noon. The longitudes by observation are reduced to noon as follow:

The distance run between noon and 3h 34' P. M. is 29 miles; to which, and the course 1½ points, the difference of latitude is 28'.

Yesterday's latitude at noon 42° 1' N.

Latitude at time of observation 41° 33' N. Mer. parts 2746 Latitude at noon 39° 3 N. Mer. parts 2549

Meridional difference of latitude

Then, to course 1½ points, and meridional difference of latitude 197 in a latitude column, the difference of longitude in a departure column is 49' W; which added to 14° 1' W, the longitude by observation, the sum 14° 50' W is the longitude reduced to noon.

Again, the distance run between the preceding noon and 9h 22' P. M. is 75 miles; hence the corresponding difference of latitude is 72.8, or 73 miles; the ship's latitude at that time is therefore 40° 48' N.

Latitude at time of observation 40° 48' N. Mer. parts 2686 Latitude at noon 39° 3 N. Mer. parts 2549

Meridional difference of latitude

Now with the corrected course, and meridional difference of latitude, the difference of longitude is 34' W; which added to 14° 20' W, the sum is 14° 54' W, the reduced longitude. The mean of which and the former reduced longitude is 14° 52' W, the correct longitude.

---

*A Journal* ### Navigation

**A Journal from England towards Madeira**

| Hours | Kn. | Fa. | Courses | Winds | Remarks, 2 October 14, 1793 | |-------|-----|-----|---------|-------|-----------------------------| | 1 | 8 | SWsS | EoS | Fresh gales and hazy, single reefed topsails. | | 2 | 7 | 5 | | Got down top-gallant yards. | | 3 | 7 | 5 | | Do. weather, and a confused swell running; | | 4 | 7 | 4 | SSW | More moderate. | | 5 | 7 | 1 | | Do. with lightning all round the compass. | | 6 | 6 | 5 | Variable. | Squally, with rain. | | 7 | 6 | 5 | | Moderate weather; out reefs, and up top-gallant-yards. | | 8 | 6 | 5 | SWsS | At 11h 10' A.M. the latitude from double altitudes of the sun was 37° 10'. Clear weather. | | 9 | 5 | 5 | SEoS | | | 10 | 5 | 5 | SWsS | | | 11 | 4 | 5 | WSW | | | 12 | 5 | | S | |

As the ship is close hauled from 2 o'clock AM, 1¼ points leeway are allowed upon that course, and 1 point on the two following courses.

| Course | Diff. | DL. | Dep. | Acc. | Obf. | D. Long. | W. Long. by Acc. | Obf. | W. Var. | Bearing | Distance | |--------|-------|-----|------|------|------|----------|------------------|------|--------|---------|----------| | S 16°W.| 116 | 111 | 3.2 | 37° 12' | 37° 8' | 41°W. | 15° 52' | 15° 33' | 1½ pts. | S 10°W. | 254 m. |

The latitude by observation at 11h 10' A.M. is 37° 10', and from that time till noon the ship has run about 4 miles. Hence the corresponding difference of latitude is 2 miles, which subtracted from the latitude observed, gives 37° 8', the latitude reduced to noon.

To find the bearing and distance of Porto Sancto.

- Latitude of ship: 37° 8' N. - Mer. parts: 2403 - Longitude: 15° 33' W. - Latitude Porto Sancto: 32° 58' N. - Mer. parts: 2007 - Longitude: 16° 25' W. - Difference of latitude: 4° 10' = 250 M. D. Lat. 306 - Diff. long.: 52

Hence the bearing of Porto Sancto is S. 10° W., or SSW ¼ W nearly, per compass, and the distance is 254 miles.

No. 239. ### NAVIGATION

**A Journal from England towards Madeira**

| Hours | K.n. | Fa. | Courses | Winds | Remarks, 5 October 15, 1793 | |-------|------|-----|---------|-------|-----------------------------| | 1 | 4 | | WsS | SbW | Moderate and clear weather. | | 2 | 4 | | | | Employed working points and rope-bands. Ditto weather. | | 3 | 6 | | | | Fine clear weather. | | 4 | 3 | | | | Ditto weather. | | 5 | 4 | | | | | | 6 | 3 | | WsN | SWsS | Variation per mean of several azimuths 18° 0' W. Ditto weather. Tacked ship. Sail-makers making wind-falls. A fine steady breeze. Cloudy. | | 7 | 3 | | | | | | 8 | 3 | | | | | | 9 | 4 | | | | | | 10 | 4 | | | | | | 11 | 3 | | Variable.| | | | 12 | 3 | | | | |

| Course | Diff. | D.L. | Dep. | N. Lat by | W. Long. by | W. Var. | Porto Sancto's. | |--------|-------|------|------|-----------|------------|---------|-----------------| | S68° W | 56 | 21 | 52 | 36° 47' | 16° 57' | 16° 38' | S½ E | | | | | | | | | 229 |

Half a point of leeway is allowed on each course; but as the variation is expressed in degrees, it will be more convenient and accurate to reduce the several courses into one, leeway only being allowed upon them. The course thus found is then to be corrected for variation, with which and the distance made good the difference of latitude and departure are to be found.

| Courses | Diff. | Diff. of Latitude | Departure | |---------|-------|-------------------|-----------| | W½ S | 18 | | 1.8 | | W¼ N¼ N| 27 | | 7.8 | | W¼ N | 7 | | 0.7 | | NW¼ W¼ W| 2 | | 0.9 | | NW½ W | 12 | | 7.6 | | S½ E¼ E| 20 | | 19.1 | | | | | 5.8 | | S 86° W| 56 | | 17.0 | | Var. | 18 | | 20.9 | | | | | 17.0 | | | | | 3.9 | | | | | 56.0 |

Tr.cour.S68 W. to which and the distance 56 m. the difference of latitude is 21 m. and the departure 51.9 m. Hence the latitude in at noon is 36° 47' W., and middle latitude 36° 58'; to which and the departure 51.9 in a latitude column, the difference of longitude in a distance column is 65° W.

Yesterday's long. by acc. 15° 52' W. By obs. 15° 33' W. Difference of longitude 1° 5 W. 1° 5 W.

Longitude in 16° 57' 16° 38' W.

To find the bearing and distance of Porto Sancto.

Latitude ship 36° 47' N Mer pts 2376 Longitude 16° 38' W. Lat. of Porto Sancto 32° 58' N Mer. pts. 2097 Longitude 16° 25' W. Diff. of latitude 3° 49' = 229 M.D. Lat. 279 D. Longitude 0° 13'

Hence the course is S½ E, distance 229 miles; and the course per compass is SbW¼ W nearly.

Vol. XII. Part II. ### NAVIGATION

**A Journal from England towards Madeira**

| Hours | Kn. | Fa. | Courses | Winds | Remarks, Oct 16, 1793 | |-------|-----|-----|---------|-------|----------------------| | 1 | 6 | 4 | SSW | SWW | Fresh gales. | | 2 | 6 | 4 | SSW | SWW | Do. and cloudy. | | 3 | 7 | | S | W | A steady fresh gale. | | 4 | 7 | | | | Do. weather. | | 5 | 7 | | | | Do. Weather. | | 6 | 7 | | | | Variation per amplitude 1\(\frac{1}{2}\) points W. | | 7 | 8 | | | | People employed occasionally. | | 8 | 8 | | | | Do. weather. Observed sun's meridian altitude. |

| Course | Diff. D.L. Dep. | N. Latit. by Acc. | Obf. | D.Long. | W. Long. by Acc. | Obf. | W. Var. | Porto Sancto's Bearing | Distance | |--------|-----------------|-------------------|------|---------|------------------|------|---------|------------------------|----------| | S 8° E | 186 | 185 | 26 | 33° 42' | 33° 46' | 31° E | 16° 26' | S 17° W | 50 miles |

Half a point of leeway is allowed on the first course; which, and the others, are corrected for variation as usual.

| Courses | Diff. | Diff. of latit. | Departure | |---------|-------|-----------------|-----------| | SEbS | 12.4 | 10.3 | 6.9 | | SbE+E | 43 | 41.2 | 12.5 | | SbE | 65 | 64.7 | 6.4 | | S | 68.5 | 68.5 | | | S8° E | 186 | 184.7 | 25.8 |

Yesterday's latitude 36° 47' N. Latitude by account 33° 42' N. Sum 70° 29' Middle latitude 35° 15'

To middle latitude and the departure, the difference of longitude in a distance column is 31° E. Yesterday's lon. by acc. 16° 57' W. by obf. 16° 38' W. Difference of long. 0° 31' E.

Longitude in 16° 26' W. 16° 7' W.

To find the bearing and distance of Porto Sancto.

| Latitude ship | 33° 46' N. | Mer. parts | 2155 | Longitude | 16° 7' W. | |---------------|------------|------------|------|-----------|-----------| | Lat. Porto Sancto | 32° 58' N. | Mer. parts | 2097 | Longitude | 16° 25' W. | | Difference of latitude | 48 | Mer. diff. lat. | 58 | Diff. long. | 18 |

Hence the bearing of Porto Sancto is S 17° W, distance 50 miles. ### Table I. To reduce points of the compass to degrees, and conversely.

The two first and two last columns of this table contain the several points and quarter-points of the compass; the third column contains the corresponding number of points and quarters; and the fourth, the degrees &c answering thereto. The manner of using this table is obvious.

| Hours | Kn. | Fa. | Courses | Winds | Remarks | |-------|-----|-----|---------|-------|---------| | 1 | 5 | SSW.| NESE. | Moderate wind and clear. Saw the island of Porto Santo, SWbS. Hauled up to round the east end of Porto Santo. Bent the cables. | | 2 | 5 | S | | Squally weather. Porto Santo WbS. | | 3 | 5 | SWbS.| | Ditto with rain. Porto Santo NE. The Deserters SWbS. | | 4 | 5 | SSW.| | The Deserters WSW. 3 or 4 leagues. | | 5 | 6 | Various.| | Hauled up round the east end of the Deserters. | | 6 | 7 | NNW.| | Violent squalls; clewed up all at times. | | 7 | 8 | NWbN.| | Running into Funchal Roads. | | 8 | 9 | | | Anchored in Funchal Road, with the best bower in 30 fathom black sand and mud. Brazen head EASbS, Loo Rock NW, the Great Church NNE, and the southermost Deferrer SEbS; off shore two-thirds of a mile. Saluted the fort with 13 guns; returned by ditto. Found here his majesty's ship Venus, and 7 English merchant ships. |

### Table II. The miles and parts of a mile in a degree of longitude at every degree of latitude.

The first column contains degrees of latitude, and the second the corresponding miles in a degree of longitude; the other columns are a continuation of the first and second. If the given latitude consists of degrees and minutes, a proportional part of the difference between the miles answering to the given and following degrees of latitude is to be subtracted from the miles answering to the given degree.

**Example.** Required the number of miles in a degree of longitude in latitude $57^\circ 9'$?

The difference between the miles answering to the latitudes of $57^\circ$ and $58^\circ$ is $0.89$.

Then as $60' : 9' :: 0.89 : 0.13$

Miles answering to $57^\circ$ $32.68$

Miles answering to $57^\circ 9'$ $32.55$

This table may be used in Parallel and Middle Latitude Sailing.

### Table III. Of the Sun's Semidiameter.

This table contains the angle subtended by the sun's semidiameter at the earth, for every sixth day of the year. The months and days are contained in the first column, and the semidiameter expressed in minutes and seconds in the second column. It is useful in correcting... ing altitudes of the sun's limb, and distances between the sun's limb and the moon.

**TABLE IV. Of the Refraction in Altitude.**

The refraction is necessary for correcting altitudes and distances observed at sea; it is always to be subtracted from the observed altitude, or added to the zenith distance. This table is adapted to a mean state of the atmosphere in Britain, namely, to 29.6 inches of the barometer, and 50° of the thermometer. If the height of the mercury in these instruments be different from the mean, a correction is necessary to reduce the tabular to the true refraction. See Refraction.

**TABLES V. VI. Of the Dip of the Horizon.**

The first of these tables contains the dip answering to a free or unobstructed horizon; and the numbers therein, as well as in the other table, are to be subtracted from the observed altitude, when the fore-observation is used; but added, in the back-observation.

When the sun is over the land, and the ship nearer it than the visible horizon when unconfined: in this case, the sun's limb is to be brought in contact with the line of separation of the sea and land; the distance of that place from the ship is to be found by estimation or otherwise; and the dip answering thereto, and the height of the eye, is to be taken from Table VI.

**TABLE VII. Of the Correction to be applied to the time of high water at full and change of the moon, to find the time of high water on any other day of the moon.**

The use of this table is fully explained at Section II. Chap I. Book I. of this article.

**TABLES VIII. IX. X. Of the Sun's declination, &c.**

The first of these tables contains the sun's declination, expressed in degrees, minutes, and tenths of a minute, for four successive years, namely, 1793, 1794, 1795, and 1796: and by means of Table X. may easily be reduced to a future period; observing that, after the 28th of February 1800, the declination answering to the day preceding that given is to be taken.

**EXAMPLE I. Required the sun's declination May 1. 1799?**

May 1. 1799 is four years after the same day in 1795.

Sun's declination May 1. 1795 = 15° 9'. 1" N

Equation from Table X. = +0° 0.6"

Sun's declination May 1. 1799 = 15° 9'. 1" N

**EXAMPLE II. Required the sun's declination August 20. 1805?**

The given year is 12 years after 1793, and the time is after the end of February 1800.

Now, Sun's dec. August 19. 1793 = 12° 34'. 6"

Equation from Table X. to 12 years = -0° 1.9"

Sun's declination August 20. 1805 = 12° 32'. 7"

The declination in Table VIII. is adapted to the meridian of Greenwich, and Table IX. is intended to reduce it to any other meridian, and to any given time of the day under that meridian. The titles at the top and bottom of this table direct when the reduction is to be added or subtracted.

**TABLE XI. Of the Right Ascensions and Declinations of Fixed Stars.**

This table contains the right ascensions and declinations of 60 principal fixed stars, adapted to the beginning of the year 1793. Column fourth and fifth contain the annual variation arising from the precession of the equinoxes, and the proper motion of the stars; which serves to reduce the place of a star to a period a few years after the epoch of the table with sufficient accuracy. When the place of a star is wanted, after the beginning of 1793, the variation in right ascension is additive; and that in declination is to be applied according to its sign. The contrary rule is to be used when the given time is before 1793.

**EXAMPLE. Required the right ascension and declination of Bellatrix, May 1. 1798?**

Right ascension January 1. 1793 = 5h 14' 3"

Variation = 3' 21" × 5½ y. = +0° 0.17"

Right ascension, May 1. 1798 = 5h 14' 20"

Declination = 6° 8' 53" N

Variation = 4" × 5½ y. = +0° 0.21"

Declination May 1. 1798 = 6° 9' 14" N

The various other tables necessary in the practice of navigation are to be found in most treatises on that subject. Those used in this article are in Mackay's Treatises on the Longitude and Navigation.

---

As the Author of this Article lives at a distance, several Errata have escaped the Press. They are as follow:

Page 683, col. 1. lines 20. and 22. from bottom. For Markelyne's, read Makelyne's.

685. 2. 10. from bottom. For 55° N, read 55° S.

686. 1. 25. For N. read S.; and in line 28. for 13° 44' N. read 18° 6' S.

687. 1. 7. from bottom. After measures, insert 177.

ib. 2. 23. from bottom. After about, insert 280.

689. 2. 16. from bottom. After one, insert fourth; and in line 21. for where, read whose.

690. 1. 29. For a, read the.

691. 1. 5. from bottom. After to, insert 43° 53'; and in line 29. for abraided, read substracted, and for 8. read 4.

692. 2. 2. from bottom. For 73 N, read N 73°.

693. 1. 15. After be, insert 281; and in line 17. after is, insert 9° 11' W.

695. 1. 21. from bottom. For mark, read make.

698. 2. 19. For on, read in; and in line 20. after 61 dele (°) ### Table I. To reduce Points of the Compass to Degrees, and conversely.

| North-east Quadrant | South-east Quadrant | Points | D. M. S. | South-west Quadrant | North-west Quadrant | |---------------------|---------------------|--------|----------|---------------------|---------------------| | North. | South. | 0° | 0' | South. | North. | | N½E | S½E | 0° | 48' | S½W | N½W | | N¼E | S¼E | 0° | 37' | S¼W | N¼W | | N¾E | S¾E | 0° | 26' | S¾W | N¾W | | N½E | S½E | 1° | 15' | S½W | N½W | | N¼E | S¼E | 1° | 4' | S¼W | N¼W | | N¾E | S¾E | 1° | 14' | S¾W | N¾W | | N½E | S½E | 1° | 16' | S½W | N½W | | N¼E | S¼E | 1° | 52' | S¼W | N¼W | | N¾E | S¾E | 1° | 41' | S¾W | N¾W | | NNE | SSE | 2° | 22' | SSW | NNW | | NNE¼E | SSE¼E | 2° | 18' | SSW¼W | NNW¼W | | NNE¾E | SSE¾E | 2° | 28' | SSW¾W | NNW¾W | | NNE¼E | SSE¼E | 2° | 56' | SSW¼W | NNW¼W | | NE½N | SE½S | 3° | 33' | SW½S | NW½N | | NE¼N | SE¼S | 3° | 36' | SW¼S | NW¼S | | NE¾N | SE¾S | 3° | 39' | SW¾S | NW¾S | | NE¼N | SE¼S | 3° | 42' | SW¼S | NW¼S | | NE½E | SE½E | 4° | 45' | SW½E | NW½E | | NE¼E | SE¼E | 4° | 47' | SW¼E | NW¼E | | NE¾E | SE¾E | 4° | 50' | SW¾E | NW¾E | | NE¼E | SE¼E | 4° | 53' | SW¼E | NW¼E | | NE½E | SE½E | 5° | 56' | SW½E | NW½E | | NE¼E | SE¼E | 5° | 59' | SW¼E | NW¼E | | NE¾E | SE¾E | 5° | 61' | SW¾E | NW¾E | | NE¼E | SE¼E | 5° | 64' | SW¼E | NW¼E | | ENE | ESE | 6° | 67' | WSW | WNW | | ENE¼N | ESE¼S | 6° | 70' | WSW¼S | WNW¼S | | ENE¾N | ESE¾S | 6° | 73' | WSW¾S | WNW¾S | | ENE¼N | ESE¼S | 6° | 75' | WSW¼S | WNW¼S | | E½N | ES½S | 7° | 78' | W½S | W½N | | E¼N | ES¼S | 7° | 81' | W¼S | W¼N | | E¾N | ES¾S | 7° | 84' | W¾S | W¾N | | E¼N | ES¼S | 7° | 87' | W¼S | W¼N | | E½N | ES½S | 8° | 90' | W½S | W½N |

### Table II. The Miles and Parts of a Mile in a Degree of Longitude at every Degree of Latitude.

| D.L. Miles | D.L. Miles | D.L. Miles | D.L. Miles | D.L. Miles | D.L. Miles | D.L. Miles | |------------|------------|------------|------------|------------|------------|------------| | 1 | 59.99 | 16 | 57.67 | 31 | 51.41 | 46 | | 2 | 59.97 | 17 | 57.36 | 32 | 50.88 | 47 | | 3 | 59.92 | 18 | 57.06 | 33 | 50.32 | 48 | | 4 | 59.86 | 19 | 56.73 | 34 | 49.74 | 49 | | 5 | 59.77 | 20 | 56.38 | 35 | 49.15 | 50 | | 6 | 59.67 | 21 | 56.01 | 36 | 48.54 | 51 | | 7 | 59.56 | 22 | 55.63 | 37 | 47.92 | 52 | | 8 | 59.44 | 23 | 55.23 | 38 | 47.28 | 53 | | 9 | 59.26 | 24 | 54.81 | 39 | 46.62 | 54 | | 10 | 59.08 | 25 | 54.38 | 40 | 45.95 | 55 | | 11 | 58.89 | 26 | 53.93 | 41 | 45.28 | 56 | | 12 | 58.68 | 27 | 53.46 | 42 | 44.95 | 57 | | 13 | 58.46 | 28 | 52.97 | 43 | 43.88 | 58 | | 14 | 58.22 | 29 | 52.47 | 44 | 43.16 | 59 | | 15 | 57.95 | 30 | 51.96 | 45 | 42.43 | 60 |

### Table III. The Sun's Semidiam.

| Mon. Day | Sun's Semidiam | |----------|----------------| | January | 1 | 16 19 | | February | 7 | 16 19 | | March | 13 | 16 19 | | April | 19 | 16 18 | | May | 25 | 16 17 | | June | 1 | 16 16 | | July | 7 | 16 15 | | August | 13 | 16 14 | | September| 19 | 16 13 | | October | 25 | 16 12 | | November | 1 | 16 10 | | December | 7 | 16 9 | | | 13 | 16 7 | | | 19 | 16 6 | | | 25 | 16 4 | | | 1 | 16 2 | | | 7 | 16 1 | | | 13 | 15 59 | | | 19 | 15 57 | | | 25 | 15 56 | | | 1 | 15 54 | | | 7 | 15 53 | | | 13 | 15 52 | | | 19 | 15 51 | | | 25 | 15 50 | | | 1 | 15 49 | | | 7 | 15 48 | | | 13 | 15 47 | | | 19 | 15 47 | | | 25 | 15 47 | | | 1 | 15 47 | | | 7 | 15 47 | | | 13 | 15 47 | | | 19 | 15 47 | | | 25 | 15 47 | | | 1 | 15 49 | | | 7 | 15 50 | | | 13 | 15 51 | | | 19 | 15 52 | | | 25 | 15 53 |

**Table.** ### Table IV. **Refraction in Altitude.**

| App. Alt. | Refrac. | App. Alt. | Refrac. | App. Alt. | Refrac. | |-----------|---------|-----------|---------|-----------|---------| | D M | M S | D M | M S | D M | M S | | 0 | 33 | 6 | 30 | 751 | 30 | | 5 | 32 | 10 | 640 | 740 | 31 | | 10 | 31 | 22 | 650 | 730 | 32 | | 15 | 30 | 35 | 7 | 720 | 33 | | 20 | 29 | 50 | 710 | 711 | 34 | | 25 | 28 | 62 | 720 | 735 | 35 | | 30 | 27 | 74 | 740 | 745 | 36 | | 35 | 26 | 86 | 750 | 767 | 37 | | 40 | 25 | 98 | 760 | 789 | 38 | | 45 | 24 | 110 | 770 | 811 | 39 |

### Table V. **Dip of the Horizon.**

| Height of eye above the horizon in feet. | |------------------------------------------| | Dip of horizon in feet. | | M S | | Feet | | 1 | | 2 | | 3 | | 4 | | 5 | | 6 | | 7 | | 8 | | 9 | | 10 | | 11 | | 12 | | 13 | | 14 | | 15 | | 16 | | 17 | | 18 | | 19 | | 20 | | 21 | | 22 | | 23 | | 24 | | 25 | | 26 | | 27 | | 28 | | 29 | | 30 | | 31 | | 32 | | 33 | | 34 | | 35 | | 36 | | 37 | | 38 | | 39 | | 40 |

### Table VI. **Dip of the Sea at different distances from the Observer.**

| Height of the eye above the sea in feet. | |------------------------------------------| | Dip of sea in nautical miles. | | M S | | Feet | | 5 | | 10 | | 15 | | 20 | | 25 | | 30 | | 35 | | 40 |

### Table VII. **The Correction to be applied to the Time of High-water at Full and Change of the Moon, to find the time of High-water in any other day.**

| Interval of Time. | After New or Full Moon | Before 1st or 3rd Quarter | After 1st or 3rd Quarter | Before New or Full Moon | |-------------------|------------------------|---------------------------|--------------------------|-------------------------| | D H. | Additive. | Additive. | Additive. | Subtractive. | | 10 | 36 | 49 | 620 | 37 | | 11 | 45 | 56 | 639 | 47 | | 12 | 54 | 64 | 658 | 57 | | 13 | 63 | 73 | 678 | 67 | | 14 | 72 | 82 | 698 | 77 | | 15 | 81 | 91 | 718 | 87 | | 16 | 90 | 100 | 738 | 97 | | 17 | 99 | 109 | 758 | 107 | | 18 | 108 | 118 | 778 | 117 | | 19 | 117 | 127 | 798 | 127 | | 20 | 126 | 136 | 818 | 137 | | 21 | 135 | 144 | 838 | 147 | | 22 | 144 | 153 | 858 | 157 | | 23 | 153 | 162 | 878 | 167 | | 24 | 162 | 171 | 898 | 177 | | 25 | 171 | 180 | 918 | 187 | | 26 | 180 | 189 | 938 | 197 | | 27 | 189 | 198 | 958 | 207 | | 28 | 198 | 207 | 978 | 217 | | 29 | 207 | 216 | 998 | 227 | | 30 | 216 | 225 | 1018 | 237 | | 31 | 225 | 234 | 1038 | 247 | | 32 | 234 | 243 | 1058 | 257 | | 33 | 243 | 252 | 1078 | 267 | | 34 | 252 | 261 | 1098 | 277 | | 35 | 261 | 270 | 1118 | 287 | | 36 | 270 | 279 | 1138 | 297 | | 37 | 279 | 288 | 1158 | 307 | | 38 | 288 | 297 | 1178 | 317 | | 39 | 297 | 306 | 1198 | 327 | | 40 | 306 | 315 | 1218 | 337 | ### Table VIII. Sun's Declination for 1793, being the first after leap-year.

| Days | January | February | March | April | May | June | July | August | September | October | November | December | |------|---------|----------|-------|-------|-----|------|------|--------|-----------|----------|-----------|----------| | 1 | 22°57'9" | 16°52'8" | 7°17'6" | 4°49'9" | 1°50'11" | 6°27'6" | 11°52'6" | 18°3'N | 3°27'7"S | 14°41'1"S | 21°50'9"S | | 2 | 22°51.5" | 16°34.8" | 6°54.1" | 5°12.9" | 5°35.0" | 12°17.3" | 19°51.9" | 26°42.5" | 3°51.0° | 15.0.1° | 22°5°5.7 | | 3 | 22°45.5" | 16°17.1" | 6°31.1" | 5°35.9" | 5°32.3" | 12°24.7" | 20°15.9" | 26°56.0" | 4°14.3" | 15°18.9" | 22°14.1 | | 4 | 22°39.6" | 15°59.9" | 6°8.0" | 5°58.7" | 16°10.5" | 31.6" | 20°50.7" | 26°45.0" | 4°37.5" | 15°37.4" | 22°22.0 | | 5 | 22°32.7" | 15°40.7" | 5°44.8" | 6°21.4" | 16°27.5" | 31.8" | 22°45.0" | 26°49.1" | 5°0.7 | 15°55.0" | 22°29.5 | | 6 | 22°24.7" | 15°22.1" | 5°21.6" | 6°44.0" | 16°44.3" | 44.3" | 22°43.0" | 26°32.5" | 5°23.8" | 16°13.6" | 22°36.6 | | 7 | 22°16.6" | 14°44.2° | 4°58.2" | 6°6.6" | 17°0.9" | 23°0.0" | 20°32.4" | 26°25.4" | 6°49.4" | 16°43.2" | 22°49.4 | | 8 | 22°8.6" | 14°44.2" | 4°34.8" | 7°28.9" | 17°17.1" | 46.4" | 22°58.5" | 26°25.5" | 5°40.8" | 16°48.7" | 22°49.4 | | 9 | 21°59.7" | 14°24.8° | 4°11.4" | 7°5.1" | 17°33.0° | 3.0" | 22°38.8° | 26°18.3° | 5°0.4" | 16°58.2° | 22°49.4 | | 10 | 21°50.8° | 14°5.2" | 3°47.9" | 8°13.3° | 17°48.7" | 4.9" | 22°10.7° | 26°55.4° | 1°41.2" | 16°32.7° | 22°55.1 | | 11 | 21°41.3" | 13°45.4° | 3°24.3" | 8°35.3° | 18°4.0" | 23°0" | 15°1.7" | 18°3.8° | 1°18.3° | 17°39.1° | 23°5.2 | | 12 | 21°31.9" | 13°25.8° | 3°7.0° | 8°57.2° | 19°31.1° | 15°21.7° | 12°16.4° | 18°47.4° | 3°40.6° | 17°55.3° | 23°9.6 | | 13 | 21°20.0° | 13°1.1° | 2°7.0° | 9°18.9° | 19°33.8° | 16.0" | 20°29.2° | 23°23.3° | 3°32.3° | 18°24.3° | 23°13.5 | | 14 | 20°10.8" | 12°44.6° | 2°13.4° | 9°40.4° | 18°35.5" | 16.0° | 18°33.3° | 23°3.3° | 8°3.1° | 18°11.2° | 23°13.5 | | 15 | 20°56.9" | 12°23.9° | 1°49.7° | 10°18.8° | 19°23.3° | 14.5° | 18°36.5° | 23°15.0° | 8°2.1° | 18°26.3° | 23°10.9 | | 16 | 20°47.3° | 12°3.1° | 1°26.0° | 10°28.1° | 19°36.0° | 13.1° | 18°39.8° | 23°07.9° | 8°1.1° | 18°41.2° | 23°19.9 | | 17 | 20°35.4° | 11°42.0° | 1°28.2° | 10°44.1° | 19°39.7° | 14.1° | 18°43.4° | 23°10.0° | 9°1.0° | 18°57.0° | 23°22.4 | | 18 | 20°23.7° | 11°20.8° | 0°39.9° | 11°5.0° | 19°42.7° | 15.1° | 18°47.0° | 23°13.0° | 9°0.0° | 18°48.4° | 23°24.4 | | 19 | 20°10.2° | 1°39.9° | 0°39.4° | 14°59.8° | 20°42.7° | 15.0° | 18°49.5° | 23°16.2° | 9°56.4° | 18°55.6° | 23°20.0 | | 20 | 19°57.1° | 1°33.8° | 0°37.8° | 15°49.9° | 20°49.2° | 20.0° | 18°52.9° | 23°19.3° | 8°49.7° | 18°53.2° | 23°27.7 | | 21 | 19°43.6° | 0°32.4° | 12°6.5° | 20°20.1° | 23°28.7° | 27.8" | 18°56.1° | 23°22.4° | 8°47.4° | 19°06.8° | 23°27.7 | | 22 | 19°29.7° | 0°36.9° | 12°26.6° | 20°31.9° | 23°27.3° | 27.3" | 18°59.7° | 23°25.1° | 8°44.1° | 19°16.9° | 23°27.5 | | 23 | 19°14.5° | 0°32.9° | 12°14.3° | 20°43.3° | 23°26.1° | 2.3° | 19°03.2° | 23°28.1° | 8°40.8° | 19°24.6° | 23°27.5 | | 24 | 19°09.9° | 0°53.6° | 12°00.3° | 20°44.3° | 23°25.3° | 2.3° | 19°06.4° | 23°31.1° | 8°37.7° | 19°24.6° | 23°27.6 | | 25 | 18°46.0° | 0°57.7° | 11°38.5° | 20°47.0° | 23°13.8° | 2.8° | 19°07.0° | 23°34.6° | 8°34.4° | 19°27.3° | 23°27.6 | | 26 | 18°30.7° | 0°42.5° | 11°22.9° | 20°57.7° | 23°15.0° | 2.7° | 19°08.1° | 23°37.4° | 8°31.3° | 19°27.3° | 23°27.7 | | 27 | 18°15.1° | 0°24.6° | 11°34.0° | 20°59.6° | 23°15.7° | 5.7° | 19°08.7° | 23°40.2° | 8°28.2° | 19°27.3° | 23°27.7 | | 28 | 17°59.2° | 0°8.9° | 11°23.8° | 21°10.3° | 23°13.7° | 5.7° | 19°09.5° | 23°43.1° | 8°25.1° | 19°27.3° | 23°27.7 | | 29 | 17°42.9° | 0°37.4° | 10°58.4° | 21°10.4° | 23°13.1° | 9.1° | 19°09.2° | 23°45.3° | 8°22.0° | 19°27.3° | 23°27.7 | | 30 | 17°23.4° | 0°48.2° | 10°46.6° | 21°10.5° | 23°10.2° | 9.2° | 19°09.2° | 23°47.6° | 8°18.9° | 19°27.3° | 23°27.7 | | 31 | 17°9.5° | 0°6.8° | 10°36.8° | 21°10.6° | 23°10.1° | 9.2° | 19°09.2° | 23°49.7° | 8°15.9° | 19°27.3° | 23°27.7 |

### Table VIII. Sun's Declination for 1794, being the second after leap-year.

| Years | January | February | March | April | May | June | July | August | September | October | November | December | |-------|---------|----------|-------|-------|----|------|------|--------|-----------|----------|----------|----------| | | | | | | | | | | | | | | | | | | | | | | | | | | | | ### Table IX

To reduce the Sun's Declination to any other Meridian, and to any given Time under that Meridian.

| Longitude | 1° | 2° | 3° | 4° | 5° | 6° | 7° | 8° | 9° | 10° | 11° | 12° | 13° | 14° | 15° | 16° | 17° | 18° | |-----------|----|----|----|----|----|----|----|----|----|-----|-----|-----|-----|-----|-----|-----|-----|-----| | 0° | | | | | | | | | | | | | | | | | | | | 1° | | | | | | | | | | | | | | | | | | | | 2° | | | | | | | | | | | | | | | | | | | | 3° | | | | | | | | | | | | | | | | | | | | 4° | | | | | | | | | | | | | | | | | | | | 5° | | | | | | | | | | | | | | | | | | | | 6° | | | | | | | | | | | | | | | | | | | | 7° | | | | | | | | | | | | | | | | | | | | 8° | | | | | | | | | | | | | | | | | | | | 9° | | | | | | | | | | | | | | | | | | | | 10° | | | | | | | | | | | | | | | | | | | | 11° | | | | | | | | | | | | | | | | | | | | 12° | | | | | | | | | | | | | | | | | | | | 13° | | | | | | | | | | | | | | | | | | | | 14° | | | | | | | | | | | | | | | | | | | | 15° | | | | | | | | | | | | | | | | | | | | 16° | | | | | | | | | | | | | | | | | | | | 17° | | | | | | | | | | | | | | | | | | | | 18° | | | | | | | | | | | | | | | | | | |

Time from Noon. ### Table X. Change of Sun's Declination

| Month | Days | Complete Years | |-------|------|---------------| | January | | | | February | | | | March | | | | April | | | | May | | | | June | | | | July | | | | August | | | | September | | | | October | | | | November| | | | December| | |

### Table XI. The Right Ascensions and Declinations of the principal fixed Stars adapted to the beginning of the Year 1793

| Name of Stars | Mag. | Right Ascension in Time | Ann. Var. | Declination | Ann. Var. | |---------------|-----|-------------------------|-----------|-------------|-----------| | Pegasi | 2 | 0h 2' 35" | 3° 06' | 1° 55' N | +20" 0' | | Ceti | 2 | 0 33 11 | 3° 01' | 7 32' S | -19 8' | | Alrucabah, pole star | 2,3 | 0 50 44 | 12° 42' | 88 12' | +19 6' | | Mirach | 2 | 0 58 10 | 3° 31' | 31 13' S | +19 4' | | Achernar | 1 | 1 30 0 | 2° 25' | 58 17' | -18 5' | | Almach | 2 | 1 51 15 | 3° 62' | 41 19' | +17 7' | | Menkar | 2 | 2 51 28 | 3° 11' | 16 17' | +14 7' | | Algol | Var.| 2 54 45 | 3° 85' | 40 8' | +14 5' | | Algenib | 2 | 3 9 38 | 4° 21' | 6 42' | +13 6' | | Aldebaran | 1 | 4 24 3 | 3° 42' | 4 48' N | +8 2' | | Capella | 1 | 5 1 25 | 4° 41' | 45 46' | +5 1' | | Rigel | 1 | 5 4 31 | 2° 37' | 8 27' | +4 8' | | Tauri | 2 | 5 13 13 | 3° 78' | 28 25' | +1 1' | | Bellatrix | 2 | 5 14 3 | 3° 21' | 6 8' | +4 0' | | Orionis | 2 | 5 21 27 | 3° 07' | 0 27' | -3 4' | | Orionis | 2 | 5 25 44 | 3° 04' | 2 20' | -3 0' | | Orionis | 2 | 5 30 20 | 3° 03' | 2 52' | -2 6' | | Columbae | 2 | 5 32 10 | 2° 17' | 34 11' | -2 4' | | Betelgeuse | 1 | 43 58 | 3° 24' | 7 21' | +1 4' | | Canis Majoris | 2,3 | 6 13 37 | 2° 65' | 17 51' | +1 2' | | Canopus | 1 | 6 19 22 | 1° 33' | 52 35' | +1 7' | | Sirius | 1 | 6 36 1 | 2° 05' | 16 26' | +4 3' | | Canis Majoris | 2 | 6 59 59 | 2° 44' | 4 32' | +5 2' | | Caltor | 1,2 | 7 21 22 | 3° 85' | 32 19' | -6 9' | | Procyon | 1,2 | 7 28 27 | 3° 14' | 5 45' | -7 5' | | Pollux | 2,3 | 7 32 37 | 3° 09' | 30 42' | -7 9' | | Navis | 2 | 7 56 19 | 2° 11' | 39 25' | +9 7' | | Navis | 2 | 8 3 10 | 1° 85' | 46 43' | +10 3' | | Acubens | 3 | 8 47 8 | 3° 30' | 12 39' | -13 4' | | Navis | 1 | 9 10 54 | 0° 75' | 68 52' | +14 8' | | Alphard | 2 | 9 17 24 | 2° 93' | 7 46' | +15 2' | | Regulus | 1 | 9 57 20 | 3° 20' | 12 58' | -17 2' | | Navis | 2 | 10 37 4 | 2° 30' | 58 36' | +18 7' | | Ursa Majoris | 2 | 10 49 14 | 3° 71' | 57 29' | -19 1' | | Dubhe | 2 | 10 50 49 | 3° 85' | 62 52' | -19 1' | | Leonis | 2 | 11 38 29 | 3° 06' | 15 43' | -19 9' | | Ursa Majoris | 2 | 11 46 51 | 3° 22' | 54 50' | -20 0' | | Crucis | 1 | 12 15 14 | 3° 24' | 61 57' | -20 0' | | Crucis | 2 | 12 19 47 | 3° 24' | 55 57' | -20 0' | | Crucis | 2 | 12 35 46 | 3° 41' | 58 33' | -19 8' | | Aliath | 2 | 12 44 51 | 2° 67' | 57 14' | -19 7' | | Spica Virginis| 1 | 13 14 18 | 3° 13' | 10 43' | +19 0' | | Ursa Majoris | 2 | 13 15 32 | 2° 43' | 56 01' | -19 0' | | Benetnach | 2 | 13 39 23 | 2° 40' | 50 21' | -18 2' | | Centauri | 1,2 | 13 49 23 | 4° 11' | 59 21' | +17 8' | | Arcturus | 1 | 14 6 13 | 2° 72' | 20 15' | -19 1' | | Centauri | 1 | 14 26 1 | 4° 45' | 59 58' | +16 1' | | Alphacca | 2 | 15 25 55 | 2° 53' | 27 25' | -12 5' | | Scorpius | 2 | 15 53 26 | 3° 47' | 10 13' | +10 3' | | Antares | 1 | 16 16 44 | 3° 64' | 25 57' | +8 7' | | Ras. Algethi | 2 | 17 5 13 | 2° 73' | 14 38' | -4 8' | | Ras. Alhague | 2 | 17 25 20 | 2° 77' | 12 43' | -3 0' | | Vega | 1 | 18 29 56 | 2° 02' | 38 35' | +2 6' | | Altair | 1,2 | 19 40 41 | 2° 92' | 8 19' | +8 5' | | Deneb | 2 | 20 34 22 | 2° 03' | 44 32' | +12 5' | | Gruis | 2 | 21 55 6 | 3° 85' | 47 57' | -8 8' | | Fomalhaut | 1 | 22 46 10 | 3° 33' | 30 42' | -19 0' | | Scheat | 2 | 22 53 45 | 2° 87' | 26 57' | +19 2' | | Markab | 2 | 22 54 27 | 2° 96' | 14 52' | +19 2' | | Andromedæ | 2 | 23 47 43 | 3° 07' | 27 56' | +20 0' | Inland Navigation.

Navigation of the ancients. See Phoenicia and Trade.

Inland Navigation, the method of conveying commodities from one part of a country to another by means of rivers, lakes, canals, or arms of the sea, penetrating far into the internal parts. See the article Canal.

The advantages of this mode of conveyance, in an extensive and populous country, are sufficiently obvious; whether we take into account the superior cheapness, facility, or quickness with which great quantities of goods can thus be carried from one place to another, or the advantages which may accrue to agriculture and other arts, by thus conveying manures, the produce of the ground, or heavy manufactured goods, to and from distant quarters; which would be altogether impracticable by land-carriage, without incurring a much greater expense than the commodities could bear. The good effects of inland navigation are particularly evident in the vast empire of China, and in the states of Holland. In both these countries, the multitude of canals undoubtedly contributes to the opulence of the inhabitants, both by the more free scope they give to trade, and the advantages derived from them to agriculture; not to mention, that by means of the canals themselves the ground is often meliorated and made capable of producing both corn and pasture, where otherwise it would perhaps yield neither, or at least very imperfectly. These countries, however, particularly Holland, are very flat, and thus very much fitted for this kind of navigation: Great Britain and Ireland are less so, on account of the greater inequality of their surface; though in them also the making of canals is now become very common, notwithstanding the immense expense with which such undertakings are attended.

In a late treatise upon this subject by Mr Edmund Leach surveyor, the author considers the advantages which might accrue to the kingdoms of Britain and Ireland, were their inland navigation improved as much as it might be; and he considers both countries as exceedingly proper for improvements of this kind on account of the number of fine streams of water they contain. "Every county (says he) in each kingdom is furnished with rivers and streams of water; very few, if any, of which but may be made navigable to within a mile or two of their sources" by the method he proposes.

The method of making canals hitherto practised, is so fully described under the article Canal, that nothing needs be said upon it in this place. Mr Leach observes, that all the artificial inland navigations made in this, as well as in other kingdoms, have been done by beginning them at the foot or mouth of a river, or at the utmost extent of the ebb or flow of the tide; proceeding thence upwards, either in the ordinary course of the river by sinking, widening, and cleansing it; then with the assistance of a lock and dam to raise the boats or other vessels to a higher level. This is next to be cut, widened, and cleansed as before, till we come to a third level, when the vessels are to be raised as in the former; proceeding thus, as it were step by step, till we arrive at the extent of the intended navigation. Otherwise the navigation is carried on by an artificial side-cut or canal near the course of the river, raised up by locks and dams as before. These methods, however, though probably of very great service to the country, are attended with many and considerable inconveniences.

1. Every river or other stream is the natural receptacle of all the springs, rains, and floods, which discharge themselves into it; by which means it sometimes discharges such torrents of water as sweep everything away before them. This not only interrupts the navigation for a time, but is often attended with the most intolerable expense.

2. The original making of the canal with a number of locks and dams, such as is described under the article Canal, must also be very expensive.

3. Some rivers run very rapidly in particular places; and if the descent of the bed be very great, it will require a number of the locks and dams already mentioned to make such a river navigable, though the navigation should extend but a few miles from the mouth.

4. The passing of the locks is always attended with a considerable difficulty, besides the loss of time and expense.

To obviate all these inconveniences, Mr Leach proposes to begin and carry on his canal in a manner precisely the reverse of that just now described. In his method, the canal is to be begun as near the source of the river as possible; that is, as near as there is a probability of having water in the driest seasons sufficient to fill the canal; of which there can be but little required, as a stream sufficient to turn an ordinary mill must be sufficient to answer all the purposes of such a canal. Thus the navigation may be carried up much higher than in the ordinary canals with dams and locks. If a canal be made properly, and the sides and bottom plastered with a cement that may be made of the same earth and clay used in making the trunk, very little water will be requisite to keep it full. Our author proposes that a stream sufficient to turn an ordinary mill will be sufficient to keep it full, even if the canal should be 20 miles long.

To construct a canal of this kind, we must make choice of a level as near the head of the river as possible; this level must be continued for a considerable way, so that the water may not have the least current, but may absolutely stagnate. Thus the original bed of the river will soon be left at a distance, and the stagnant canal will become higher and higher with respect to it, in proportion to the descent of the ground or rapidity of the river. Having proceeded this way as far as convenient, the vessels must be let down by a machine (to be afterwards described) into the bed of the original river, or from one elevated canal to another much lower; perhaps 10, 15, or 20 fathoms.

By this perfect stagnation of the water in the canals, there will be no danger of washing off the plastering from the sides and bottom, so that the only waste of water will be by evaporation, and what is required for working the machine above mentioned; which, as it is only required at particular times when vessels are to be raised or lowered, must be very trifling, and cannot bear any proportion to the constant supply at the head of the canal.

In order to keep the first level easily, and for a considerable way, it will be necessary to avoid the many valleys which are naturally to be met with near the original river, by the descent of other streams into it. Hence canals constructed on the plan recommended by our author must run out in serpentine windings a considerable way into the country, which will augment its length greatly beyond that of the original river; and thus a much greater number of people, and larger extent of country, will reap the benefit of it than if it was continued in a straight line. "One inconvenience (says our author) attending the old inland navigations that are carried on in or nigh the original rivers, is their being confined within too small a space of land, and where, for the most part, there is the least occasion either for water to water the land, or for manure to enrich the soil; whereas, by beginning a canal near the head of a river, and by continuing it on one common level, if the river have any thing of a descent, and runs pretty rapid, the original river and its valley is soon left, and the canal is removed up into the side of the hill, where there is generally most occasion for manure and water both; and the further on it is contained, and the more rapid the river runs, the further the navigation is removed from the valley and the original course of the river, till it is brought to the place designed for the purpose of transferring the vessels down at once by a machine into the original river, the sea, or into another canal."

Notwithstanding the advantages which attend this new mode of construction, our author acknowledges that there are some cases in which the old method only can be put in practice.

With regard to the size, form, and expense of canals of this kind, Mr Leach gives a computation from one which was intended to be made in the county of Cornwall, and which was called the Tamar canal. An act of parliament was obtained for this in 1774. A navigable canal was to be made from Bude Haven, in the county of Cornwall, on the Bristol Channel, to the navigable part of the river Tamar; the design being to open a communication between the English and Bristol Channels through the counties of Devon and Cornwall. By the act, it was determined that the canal should not exceed 63 feet in breadth, nor should the ground be cut more than 39 inches deep below the surface, excepting in places where docks or basins were to be made, or where cranes or other engines were to be erected for particular purposes. The breadth allowed (63 feet) was to comprehend the water itself, the two towing paths, one on each side, and the fences beyond them. The breadth of the canal itself was to be 21 feet at the surface and 12 at bottom; the depth on the under side, 39 inches, as already said; the towing paths 12 feet broad each; and beyond these an hedge or rail at the distance of six feet on each side from the paths. There were to be drains as usual in canals; one on the under side for carrying off the superfluous water from the canal; the other on the upper side for carrying off the water which might accidentally fall into it in rains, or by springs in the higher grounds. Thus the whole would not exceed the breadth proposed by act of parliament; while the canal itself would be sufficiently large for every necessary purpose. Mr Leach computes, that on this canal two boats, carrying 10 tons each, might very easily pass; and he recommends this as the proper size of canals in other parts of the kingdom.

Canals of this kind, our author observes, will, for the most part, be made in the side of an hill or rising ground; whence the greatest part of the earth dug out of it will be thrown upon the under side. Here it will elevate the ground sufficiently to form a towing path, together with an elevated space whereon to plant the hedge or fence. On the higher side, 12 feet of the ground must be cut down to within 18 inches of the surface of the water; and the earth and stones dug out in making this path will make the fence on the outside for keeping off the cattle. As by the act of parliament, however, there is a greater breadth allowed in some places for making docks or basins, our author calculates the whole breadth of the canal at 66 feet; and according to this breadth he makes his computations.

1. The first and principal expense is the digging of the trunk of the canal; but this must vary so much according to the nature of the ground, that no certain estimate of the expense can be made. Sometimes the ground may be soft, and easily cut, sometimes hard and rocky, sometimes marshy and boggy, &c. There must also be a very considerable difference in the expense of cutting the canal, according as the ground on one side of it is more or less elevated. Mr Leach, after making the proper calculations, supposes that in every perch of a canal of the kind under consideration there must be removed 40½ cubic yards, or 108 cubical feet, of earth. Agreements with the workmen are commonly made by the cubic yard. It is rare that a cubic yard can be removed for less than 2d. "nor (says our author) will any sort of ground require more than 9d. except in passing through a rock, which will not often happen, unless where, by the situation of the land through which the canal may pass, it should require to be cut more than four feet under the surface, as it may sometimes happen in cutting through a hill or neck of land, whereby the course of the canal may be shortened. According to this estimate, Mr Leach has formed a table of the expense of making canals from 2d. the cubic yard to 9d. The smallest expense of these per mile is L. 107 : 11 : 1½; the greatest L. 483 : 19 : 9. Sometimes, however, the expense may be even greater; so that the mile may cost L. 600, or near it; but this must be accounted an extraordinary expense, and not often to be expected.

In digging the trunk of the canal, care must be taken to preserve all the sand, stones, and gravel, for the purpose of making drains or gutters and the towing paths; also for the making of a wall, the outside face of which is to be 21 feet distant from the edge of the canal. The design of this is to prevent the earth and rubbish from being thrown too great a distance from the side of the canal; and likewise for the purpose of raising a bank on the lower side about 16 or 18 inches above the surface of the water in the canal, when it is full up to the drains. On the top of this bank the towing path is to be made 12 feet wide on that side of the canal; and if this wall should be raised four feet above the surface of the towing path, it would be a fence to keep the cattle off from that side. In making the towing path on the higher side, the surface of the land must be sunk to within 16 or 18 inches of the surface of the water in the canal when full; and the turf, earth, Inland Na- stones, &c., taken from thence in making the towing path on that side, will make the other fence of the canal.

2. Drains or gutters must be made under the towing paths, on both sides of the canal. On the upper side they must be made through the hedge or fence at all convenient places, for admitting springs and rivulets, as well as rain-water into the canal. On the higher side, these should be about 132 feet distant from one another, exclusive of those for the admission of springs. On the lower side, it is necessary to have drains at the distance of 66 feet from each other; which will be 80 drains per mile. These may be constructed for 2s. 6d. each, which amounts to L. 10 per mile. On the higher side they will cost about four shillings per drain; and, at the distance of 132 feet betwixt each drain, the expense will be L. 8 per mile.

In making the drains on the lower side, great care must be taken to have them truly level, and parallel to the horizon; the bottom part of the drain being exactly 39 inches perpendicular above the bottom of the canal; as on the true placing of these drains, and the true level of the bottom of the canal, entirely depends the regularity of the depths of the water.

3. The value of the ground through which the canal is to pass must also be considered. From the observations and memorandums which our author took in the year 1774, concerning the ground through which the Tamar canal was to pass, he concluded that one eighth part of the land through which the tract passed was worth 40s. per acre; another eighth, 20s.; a third eighth part, 10s.; and all the rest not worth more than five; and a great deal of it not more than two; the average of the whole being 1s. an acre. "Then (says he), if the good, indifferent, and bad land be worth on an average, 15s. an acre, and as 30 years value is a capital price for lands, it will, at that rate, amount to 22l. 10s. per acre. The stipulated breadth of the canal with the towing paths and hedges being, as already said, 66 feet, which is exactly a gunter's chain; then, 10 such chains in length and one in breadth form exactly a statute acre of land; and ten chains in length is a furlong; so that every mile of canal will thus take up eight acres of land, which, at 22l. 10s. per acre, amounts to 180l. per mile.

4. Bridges are likewise a considerable article of expense in the making of canals. Mr Leach supposes that there may be one common road bridge and two swivel bridges required per mile; the former may be erected for 60l. and the latter for 30l. each; so that the expense of all together will amount to 120l. per mile.

5. As in some particular places, on account of short turns, and where docks and basins and landing places are to be made, a greater wideness will be required, our author allows half an acre more land for every furlong on this account; which brings an additional expense of 90l. per mile.

6. It will always be necessary to have sluices for emptying the canal; each of which will cost 20l. Mr Leach allows a sluice and stop-gate for every mile; and as these cost 20l. each, we have thus a farther expense of 40l. per mile.

The wall on the lower side against the bank of earth and towing path may be built for three shillings per perch; and as every mile contains 320 perches, the expense of the wall will be 48l. per mile.

8. Our author calculates the towing path on the lower side at the same price; so that it also makes an addition of 48l. per mile; but as he estimates the higher towing path and hedge at 8s. per mile, this will amount to 128l. per mile.

Thus, according to our author's calculation, the whole expense of making a canal with its necessary appendages will amount to L. 944 : 1s. 4½ per mile; and by the calculation of another engineer, the particulars of which he also enumerates, it would not exceed L. 1032 : 13s. 4½. This our author thinks a moderate expense; but an objection naturally occurs from the great length to which such canals must necessarily run; a remarkable instance of which he gives in the proposed Tamar canal; where, though the distance betwixt Bude Haven and the navigable part of the river Tamar is no more than 28 miles in a straight line, the length of the canal would not have been less than 80 miles. This length, however, according to our author, "ought not to be an object of discouragement, but on the contrary an inducement and an encouragement to promote the extending of it as much as possible into the country, as it will bring a market to every man's door that it draws nigh in its passage, and be a means to improve a greater quantity of uncultivated lands; and the navigation will be thereby much more extensive, and a much larger number of inhabitants will reap the advantages thereof, than if it had been carried upon the original river."

We have already taken notice, that in the method of constructing canals just now mentioned, there will at certain intervals be places where the vessels must be raised and lowered by means of mechanical powers, instead of the dams and locks made use of in other canals. These machines are compounded of an inclined plane and wheel in axis. The inclined plane is a parallelogram whose length reaches from the end of one canal to the beginning of another, or to the sea or navigable river, to which the vessel is next to convey; the breadth ought to be 22½ feet. It may be made of good oak or deal plank, and sufficiently strong to bear the weight to be laid upon it; and it must be very strongly supported by beams of oak or other wood. It ought to be divided in the middle by a ledge or rib of 12 inches square, the side ribs being nine by 12 inches. The elevation must depend upon particular circumstances. Fig. 1 shows the inclined part of the machine; AB being the wooden part just described, placed between the side of the hill W and the navigable river F. According to the dimensions already given, the two paths A and B on which the vessels move are exactly ten feet wide. G represents the canal, brought perhaps from the distance of several miles to the top of the precipice WW. At the end of the canal, and quite across from R to R, must be built a very strong wall; in which are two sluices with flood-gates at K and L, to let out the water occasionally. Between the head of the plane AB, and the end of the canal G, is a horizontal platform divided into two parts, as is represented in the figure by the letters HI. At the end of the canal are six rollers. rollers M and N, of use in carrying the boats and lighters in and out of the canal. Near the end of the canal, at S and T, are two other sluices, with their flood-gates, for letting out a quantity of fluid to drive the other part of the machine. O and P represent the two ends of the towing paths, one on each side of the canal.

Fig. 2 shows the vehicle by which the lighters are conveyed up and down the inclined plane, by the two paths A and B, fig. 1. AA (fig. 2) represents part of the inclined plane, B the vehicle in the position in which it rolls up and down the two paths. C is the body of the vehicle, which is made hollow, to contain a quantity of water occasionally used as a counterbalance for its corresponding vehicle. DDD are three rollers between the bottom of the vehicle and the plane, for the purpose of rolling the boats up and down. HHH are six rollers; four on the horizontal part of the vehicle on which the boat E is to rest in its passage up and down the plane; the other two rollers are in a moveable part, which is fastened to the body of the vehicle with a pair of very strong hinges; and in the passage of the vehicle up and down the plane it turns up between the head of the boat and the plane, preventing the former from rubbing against the plane. When the vehicle gets up to the top, this moveable part falls down on the platform marked HI, becoming parallel with the horizontal part of the vehicle; after which it serves as a launch and passage to place the boat upon the rollers MN (fig. 1) at the end of the canal. This passage part of the vehicle, together with the three rollers at the end of the canal, is likewise of great use in towing a boat out of the canal, in order to place it on the horizontal part. At the bottom of the cavity of the vehicle is a large hole F, with a valve opening inwardly. Through this hole the water enters when the vehicle sinks into the navigable river F, for the purpose of receiving a boat on the top or horizontal part of the vehicle, till it is quite full; and will then sink entirely under water, while the boat is towed in on the horizontal part. A small rope K is fastened to the valve, on purpose to lift it up, and to keep it so while the vehicle and boat are ascending up the plane out of the canal, that so the water may discharge itself till as much as is necessary be got out; or till it becomes an equal balance for the corresponding vehicle and its contents, which are descending by the other path. Hence we see, that every machine must have two of these vehicles furnished with rollers as already described, and so constructed that one may be as nearly as possible a counterbalance to the other. As it is necessary that the vehicles should be watertight, the insides of them must be caulked very tight; and they should be spacious enough to hold as much water as will balance the largest boat with its contents. Here it may be observed, that every vessel will be balanced by as many cubic feet of water as it displaces by being put into the water when loaded. The quantity may easily be known, by observing how far the boat sinks in the water, and calculating the bulk of the part immersed.

The machine which puts the vehicles in motion, may either be constructed with an under-shot or breast-water-wheel; by an over-shot water-wheel; or by two walking-wheels, for men to walk in as in cranes, &c.

Fig. 3 shows a front view of the under-shot water-wheel movement; where A is the end of the axis or Inland Navigation cylinder of the cog or spur-wheel; the diameter of vibration, which axis is four feet, and its length not less than 22 feet, as it must be extended quite across the canal from one side to the other, and placed on the top of very strong supports on each side of the canal, about seven feet above the surface of the water, as the loaded boat is to pass backwards and forwards under the cylinder, and at a convenient distance from the wall RR (fig. 1) and placed between the two sluices S and T; on the end of which cylinder is the cog-wheel B (fig. 3). The wheel B is supposed to be 20 feet in diameter, having on its edge 120 cogs; and underneath the cog-wheel is the breast-water cone C, 24 feet in diameter from the tip of one aller-board to the tip of its opposite. On the end of the axis of the water-wheel D is a trundle two feet and an half in diameter, with 15 rounds or flaves contained therein. This must be placed between the two sluices S and T, to let the water out of the canal; which, falling on the float-boards, will turn the wheel round from the right hand towards the left, when the sluice on the left hand of the wheel is opened; but the contrary way when that on the right is opened.—The water falling upon the boards passes along with the wheel in the circular cavity EGF, and is discharged at G, whatever way the wheel may turn.

To the axis or cylinder of this machine, which must always be horizontal, are fixed two pair of strong ropes; the ends of each pair fastened to the upper part of the cylinder; it being necessary that they should act in contrary directions. Each must extend the whole length of the plane, and their strength must be proportioned to the weight necessary to be sustained. The two vehicles already mentioned are fastened to the other ends of the ropes; so that one pair of the ropes are wound up by the cylinder turning one way, and the other by its turning the contrary way. Thus, when one of the vehicles is at the upper part of the path A, ready to discharge its boat and cargo into the upper canal, the other boat will be at the foot of the path B, all under water in the lower canal, and ready for the reception of a boat to be towed in on the horizontal part of it; so that as one vehicle rolls up on one side of the plane, the other will roll down on the other side, and vice versa.

Fig. 4 shows the movement by means of an over-shot water-wheel. It consists of a water-wheel C, and two spur or cog-wheels A and B. The water-wheel is 18 feet in diameter, and has two rows of buckets placed contrariwise to one another, that it may turn round in contrary directions, according as the one or the other sluice, S or T, is opened. On its axis F is a trundle of three feet diameter, having 18 rounds or flaves which fall into the cogs of the second spur-wheel B, causing it to turn round in a direction contrary to that of the water-wheel. This second wheel is likewise 18 feet in diameter, with a trundle of three feet, having 18 rounds or flaves.—The diameter of the upper spur-wheel A is also 18 feet, but the diameter of its axis is six feet. On the edge of the wheel are 108 cogs. These fall in between the flaves of the axis of the other spur-wheel; and thus the third wheel turns round the same way with the water-wheel C. The cylinder of this upper spur-wheel... wheel must be placed across the canal betwixt the two sluices, on very strong supports, as explained in the former movement, and the two pair of ropes in the same manner.

The movement of the walking-wheel is shown (fig. 5.) A1 and A2 are two wheels for men to walk in, each of them 24 feet in diameter. B1 and B2 are the axes or cylinders of the two wheels, of equal lengths; viz. 11 feet each, and four in diameter.—At one end of each of the two cylinders C1 and C2, is a wheel of the same diameter with the cylinder. On the edges of these wheels are teeth of an equal number in each wheel; and as the teeth of the wheels mutually fall into each other, the revolutions of both must be performed in the same time. By this contrivance also the cylinders will turn different ways; and the ropes on the two different cylinders will constantly one pair be wound up, and the other wound down, by the natural moving of the machine. DDD is the frame that supports the whole, which must be made very firm and secure.

Let us now suppose, that there is a boat in the upper canal to be brought down, but none to go up for a balance. In this case, as one of the vehicles must be at the top to receive the boat, the other will be at the bottom to take in water. Let then any of the movements just described be set to work, and it is plain, that as the upper vehicle with its boat descends, the under vehicle will ascend with the water; the valve being in the mean time lifted up till a sufficient quantity of water has flowed out, to make the one nearly a counterbalance to the other; so that the vessel may slide down gently and without any violence.

If it happens that a boat is to go up while none is to come down, one of the vehicles being at the foot of the plane under water, and in readiness to have the boat towed upon its horizontal part, one of the sluices at K or L is to be opened, and a quantity of water let into the cistern of the upper vehicle sufficient to counterbalance the boat with its contents which is to ascend. This being done, the machine is set to work, the valve of the under vehicle kept open till the water is all discharged; and then the boat will roll up to the top of the plane.

From this description of the canal and machinery for raising and lowering the vessels, the reader can be at no loss to understand the principles on which it depends. It would be superfluous to adduce examples, or follow our author through his calculations relative to particular cases. We shall only observe, that the difference of time in which vessels may be raised or lowered by the machinery just described, in comparison with what can be done in the common way by dams and locks, must give a very favourable idea of the new method. According to Mr Leach's computations, a boat with its cargo weighing 10 ton might be raised by the walking-machine in 12 or 14 minutes, by the undershot-wheel in 15 minutes, and by the overshot-wheel in 30 minutes; and that through a space of no less than 30 fathoms measured on the inclined plane, or 114 feet perpendicular.