ENCYCLOPÆDIA BRITANNICA.

NAVIGATION.

History. NAVIGATION is the art of conducting a ship from one port or place to another.

HISTORY.

The profane poets refer the invention of the art of navigation to their heathen deities, though historians ascribe it to the Æginetes, the Phœnicians, the Tyrians, and the ancient inhabitants of Britain. Scripture refers the origin of so useful an invention to God himself, who gave the first specimen of navigation in the ark built by Noah under his direction.

The earliest record of the practice of the art of navigation is that of the Egyptians, who at a very remote period are said to have established commercial relations with India. This traffic was carried on between the Arabian Gulf and the western coast of India, across the Indian Ocean. It would appear, however, that this intercourse was of no long duration, and that the Egyptians soon confined themselves to overland traffic with their neighbours, even excluding from all access to their country those foreigners who would have traded with them by the Mediterranean Sea.

The Phœnicians were the most distinguished of the early navigators; their commercial relations with other nations were the most widely spread; and their capital, Tyre, was for ages the centre of ancient commerce and the "mart of nations." The narrowness and poverty of the little slip of ground they possessed along the coast, and the convenience of two or three good ports, naturally drove an enterprising and industrious people, stimulated by a genius for commerce, to seek by sea those riches which were denied them by land. Accordingly, Lebanon and the other neighbouring mountains furnishing them with excellent wood for ship-building, they in a short time became masters of a numerous fleet; and constantly hazarding new navigations, and settling new trades, they soon arrived at an incredible pitch of opulence and populousness, insomuch as to be in a condition to send out colonies. The principal of these was Carthage, which, keeping up the Phœnician spirit of commerce, in time not only equalled Tyre itself, but vastly

surpassed it, sending its merchant fleets through the Straits of Gibraltar, along the western coasts of Africa and Europe, and even, if we may believe some authors, to America itself.

At an early period of their history, although long subsequently to the rise of the Phœnician navigators, the Greeks learned and practised the art of navigating the adjacent seas, although they seem to have trusted almost entirely to the air as the instrument of propulsion. The celebrated voyage of the Argonauts belongs to a very early period; and in later times the Corinthians and Corcyreans disputed with Athens the empire of the Greek seas. At length Tyre, whose immense riches and power are represented in such lofty terms both by sacred and profane authors, was destroyed by Alexander the Great, upon which its navigation and commerce were transferred by the conqueror to Alexandria, a new city, admirably situated for these purposes, and intended to form the capital of the empire of Asia, of which Alexander then meditated the conquest. And thus arose the great navigation of the Egyptians, which was afterwards so much cultivated by the Ptolemies, that Tyre and Carthage were quite forgotten.

Egypt being reduced to 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 merchandise from the commercial capital of Egypt.

At length Alexandria itself underwent the fate of Tyre and Carthage, being surprised by the Saracens, who, in spite of the Emperor Heraclius, overspread the northern coast of Africa. By them the merchants were expelled, and Alexandria was, until lately, in a languishing state; though it always had a considerable share of the commerce of the Christian merchants trading to the Levant. A fresh impulse, however, has been given of late years to the trade of Alexandria by its having become an important post in the overland route to India.

The fall of Rome and its empire drew along with it not only the overthrow of learning and the polite arts, but also 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 braver amongst 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,—than they began to learn the advantages of navigation and commerce, and the methods of managing them, from the people they had 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 that we usually ascribe the invention and use of banks, book-keeping, exchanges, recharges, &c.

It does not appear which of the European people, after the settlement of their new masters, first betook themselves to navigation and commerce. Some think it began with the French, although the Italians seem to have the fairest title to this distinction, and are accordingly regarded as the restorers of navigation, 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 merit of this restoration; and it is to their advantageous situation for navigation that 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 these well screened, and almost inaccessible, the residence of some fishermen, who here supported themselves by a little trade in fish and salt, which they found in some of these islands. Thither the Veneti, a people inhabiting that part of Italy which stretches along the coasts of the gulf, retired, when Alaric, King of the Goths, and afterwards Attila, King of the Huns, ravaged Italy.

These new islanders, little imagining that this was to be their fixed residence, did not think of composing any body politic; but each of the seventy-two islands of this little archipelago continued a long time under its separate master, and each formed a distinct commonwealth. When their commerce had become considerable enough to occasion jealousy to their neighbours, they began to think of uniting into a body; and it was this union, first begun in the sixth century, but not completed till the eighth, that laid the sure foundation of the future grandeur of the state of Venice. From the time of this union, their fleets of merchantmen were sent to all the ports of the Mediterranean; and at last to those of Egypt, particularly Cairo, a new city built by the Saracen princes, on the eastern bank of the Nile, where they traded for the spices and other products of the Indies. Thus they flourished and increased their commerce, their navigation, and their conquests, till the league of Cambrai in 1508, when a number of jealous princes conspired to bring about their ruin. This 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. But jealousy soon broke out; and the two republics coming to an open rupture, there was almost continual war for three centuries. Towards the end of the fourteenth century, the battle of Chioza ended the strife. The Genoese, who till then had usually the advantage, now lost all; and the Venetians, reduced almost to despair, secured to themselves, by one happy and unexpected blow, the foremost place in commerce and the sole empire of the sea.

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 of which it was capable till the discovery of the East and West Indies, but also formed a new scheme of laws for its regulation, which still obtain under the name of Uses and Customs of the Sea.

This society is that famous league of the Hanse Towns, commonly supposed to have been instituted about the year 1164. (See HANSEATIC LEAGUE, and COMMERCE. For the present state of navigation in the various countries of the world, see under the name of each.)

We shall only add, that in examining the causes of commerce passing successively from the Venetians, Genoese, and Hanse Towns, to the Portuguese and Spaniards, and from these again to the English and Dutch, it may be established as a maxim, that the relation between commerce and navigation, or 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 decline together. Hence so many laws, ordinances, statutes, and edicts for its regulation; and hence particularly that celebrated act of navigation, which an eminent foreign author calls the palladium or tutelary deity of the commerce of England, which was long considered as the standing rule, not only of the British amongst themselves, but also as that of other nations with whom they trafficked.

The progress of political and commercial science, which gradually opened men's eyes to the great principle that all trade is most healthy and prosperous when subjected to the fewest possible, and those only the most necessary, restrictions, resulted in the total repeal of these famous navigation laws in 1846, since which period trade with England has been free and open to all the world; but so far from British shipping and commerce being injured or diminished, they have been more prosperous since that repeal than they were at any former period, when most carefully fostered by protective laws.

The art of navigation has been exceedingly improved in modern times, both with regard to the form of the vessels themselves, and also with respect to the methods of working them. The use of rowers is now entirely superseded by the improvements made in the formation of the sails, rigging, &c.; by which means ships can not only sail much faster than formerly, but can tack in any direction with the greatest facility; and of late years the extensive and still growing employment of steam as the propelling power, whether applied to paddle-wheels or screws, has still further placed the mariner beyond the adverse retarding influence of calm and contrary winds, and has introduced an element of certainty and punctuality in commercial intercourse unknown at any previous period, and invaluable as it affects the interests of commerce. It is also very certain 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 possess over the ancients consists in the mariner's compass, by which they are enabled to find their way with more facility in the midst of an immeasurable ocean, than 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 imagine to have been 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 Judæa were ignorant of this instrument; and

History. it is very improbable that such a useful invention, if it had once been commonly known to any nation, would have been forgotten, or perfectly concealed from such a prudent people as the Romans, who were so deeply interested in the discovery of it.

Amongst those who admit that the mariner's compass is a modern invention, it has been much disputed who was the inventor. Some attribute the honour of the discovery to Flavio Gioia of Amalfi in Campania, who lived about the beginning of the fourteenth century; whilst others contend that it came from the East, and was earlier known in Europe. But at whatever time it was invented, it is certain that the mariner's compass was not commonly used in navigation before the year 1420. In that year the science was considerably improved under the auspices of Henry, Duke of Visco, brother to the King of Portugal. In the year 1485, Roderick and Joseph, physicians to John II., King of Portugal, together with one Martin de Bohemia, a Portuguese native of the island of Fayal, and scholar of 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 both by Columbus and by Sebastian Cabot. The former certainly did observe the variation, without having heard of it from any other person, on the 14th of September 1492, and it is very probable that Cabot might have done 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 has since been observed that the variation alters in time. The use of the cross staff now began to be introduced amongst sailors. This ancient instrument is described by John Werner of Nuremberg, in his annotations on the first book of Ptolemy's Geography, printed in the year 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 two Spanish treatises were published in the year 1545,—one by Pedro de Medina, and the other by Martin Cortes,—which contained a complete system of the art, as far as it was then known. These seem to have been the oldest writers who fully handled the art; for Medina, in his dedication to Philip, Prince of Spain, laments that multitudes of ships daily perished at sea, because there were neither teachers of the art, nor books by which it might be learned; and Cortes, in his dedication, boasts to the emperor that he was the first who had reduced navigation into a compendium, valuing himself much on what he had performed. Medina defended the plane chart; but he was opposed by Cortes, who showed its errors, and endeavoured to account for the variation of the compass by supposing the needle to be influenced by a magnetic pole (which he called the point attractive), different from that of the world, which notion has been farther prosecuted by others, and is now generally accepted as true in the scientific world. Medina's book was soon translated into Italian, French, and Flemish, and for a long time served as a guide to foreign navigators. Cortes, however, was the favourite author of the English nation, and was translated in the year 1561; whilst 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 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, epact, moon's age, and tides; a description of the compass, an account of its variation, for the discovery of which Cortes said that 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, in order to discover where the ship was, they made use of a small table, which 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. Some other instruments were also 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; together with a complex machine to discover the hour and latitude at once.

About the same time proposals were made for finding the longitude by observations of the moon. In 1530 Gemma Frisius advised the keeping of time by means of small clocks or watches, which were then, as he says, newly invented. He also contrived a new sort of cross staff, and an instrument called the nautical quadrant, which last was much praised by William Cunningham in his Astronomical Glass, printed in the year 1559.

In the year 1537, Pedro Nunez, or Nonius, published a book in the Portuguese language, to explain a difficulty in navigation proposed to him by the commander Don Martin Alphonso de Susa. In this he exposed the errors of the plane chart, and likewise gave the solution of several curious astronomical problems, amongst which was that of determining the latitude from two observations of the sun's altitude and the intermediate azimuth. He observed, that although the rhumbs are spiral lines, yet the direct course of a ship will always be in the arc of a great circle, whereby the angle with the meridians will continually change; and hence 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 in reality the ship will thus describe a course without the rhumb line intended; and therefore his calculations for assigning the latitude, where any rhumb line crosses the several meridians, will be in some measure erroneous. He invented a method of dividing a quadrant by means of concentric circles, which, after having been much improved by Dr Halley, is used at present, and is called a nonus.

In the year 1577, William Bourne published a treatise in which, by considering the irregularities in the moon's motion, he showed the error 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 advised, in sailing towards high latitudes, to keep the reckoning by the globe, as there the plane chart was most erroneous. He despaired 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 doubted; but as he had shown how to find the variation at all times, he recommended to keep an account of the observations, as useful for finding the place of the ship; and this 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 Edward Wright,

History. entitled the Haven-finding Art. In this ancient tract is also described the method 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 which floats in the water, whilst the time is reckoned during which the line that is fastened to it is veering out. The inventor of this contrivance is not known; but it was first described in an account of an East India voyage published by Purchas in 1607, from which time it became famous, and was much taken notice of by almost all writers on navigation in every country. It still continues to be used as at first, although 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 the year 1581 Michael Coignet, a native of Antwerp, published a treatise, in which he animadverted on Medina. In this he showed, that as the rhumbs are spirals, making endless revolutions about the poles, numerous errors must arise from their being represented by straight lines on the sea charts; but although 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, which he admitted were then in common use amongst the sailors. He likewise described 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 sixtieth degree of latitude; and at the end of the book is delivered a method of sailing upon a parallel of latitude by means of a ring dial and a twenty-four hour glass. The same year the discovery of the dipping-needle was made by Robert Norman. In his publication on that subject 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; and he also made considerable improvements on the construction of compasses themselves, showing especially the danger of not fixing, on account of the variation, the wire directly under the fleur de lis, as compasses made in different countries have it placed differently. To this performance of Norman's is prefixed a discourse on the variation of the magnetic needle, by William Burrough, in which he shows how to determine the variation in many different ways, and also points out many errors in the practice of navigation at that time, speaking in very severe terms concerning those who had published upon it.

During this time the Spaniards continued to publish treatises on the art. In 1585 an excellent Compendium was published by Roderico Zamorano, and 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; nevertheless, the plane chart continued still to be followed, though its errors were frequently complained of. Methods of removing these errors had indeed been sought after; and Gerard Mercator seems to have been the first who found the true method of effecting this, so as to answer the purposes of seamen. He represented the degrees both of latitude and longitude by parallel straight lines, but gradually augmented the space between 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 formed an angle with the meridians, expressing the rhumb leading from the one to the other. But although in 1569 Mercator published a universal map constructed in this manner, it does not appear that he was

acquainted with the principles upon 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 Edward Wright, an Englishman.

Wright supposed, 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 alluded to relate, indeed, to the proportion between the distances of the parallels and meridians; but instead of proposing any gradual widening between 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; with this precaution, however, that the degrees of longitude in each should bear some proportion to those of latitude, and this proportion was to be deduced from that which the magnitude of the respective parallels bore to a great circle of the sphere. He added, that, in particular maps, if this proportion be observed with regard to the middle parallel, the inconvenience will not be great, although the meridians should be straight lines parallel to each other. But here he is understood 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, although 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 in 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 afterwards 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 thereof to navigators. In 1610 a second edition of Wright's book was published, with improvements. An excellent method was proposed of determining the magnitude of the earth; and 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 barleycorn. Amongst his other improvements may be mentioned the Table of Latitudes for Dividing the Meridian computed to Minutes, whereas it had been only divided to every tenth minute. He also published a description of an instrument which he calls the sea rings, by which the variation of the compass, the altitude of the sun, and the time of the day, may at once readily be determined in any place, provided the latitude is known. He also showed how to correct the errors arising from the eccentricity of the eye in observing by the cross staff. In the years 1594, 1595, 1596, and 1597, he amended the tables of the declinations and places of the sun and stars from his own observations made with a six-feet quadrant, a sea quadrant to take altitudes by a forward or backward observation, and likewise with a contrivance for the ready finding of 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

History. mistakes in the original, adding a large table of the variation of the compass observed in different parts of the world, in order to show that it was not occasioned by any magnetical pole.

These improvements soon became known abroad. In 1608 a treatise, entitled Hypomnemata Mathematica, was published by Simon Stevin for the use of Prince Maurice. In the portion of the work relating to navigation, the author treated of sailing on a great circle, and showed how to draw the rhumbs on a globe mechanically; he also set down Wright's two tables of latitudes and of rhumbs, in order to describe these lines more accurately; and even pretended to have discovered an error in Wright's table. But Stevin's objections were fully answered by the author himself, who showed that they arose from the rude method of calculating made use of by the former.

In 1624 the learned Willebrordus 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 has been long ago refuted; and Wright's title to the honour of those discoveries remains unchallenged.

Having shown how to find the place of the ship upon his chart, Wright observed that the same might be performed more accurately by calculation; but considering, as he says, that the latitudes, and especially the courses at sea, could not be determined so precisely, he forbore setting down particular examples; as the mariner may be allowed to save himself this trouble, and only to mark out upon his chart the ship's way, after the manner then usually practised. However, in 1614, Raphe Handson, amongst the nautical questions which he subjoined to a translation of Pitiscus's Trigonometry, solved very distinctly every case of navigation, by applying arithmetical calculations to Wright's Tables of Latitudes, or of Meridional Parts, as it has since been called. Although 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 methods of approximation without the assistance of Wright's division of the meridian line. The first was computed by the arithmetical mean between the cosines of both latitudes; and the other by the same mean between the secants, as an alternative when Wright's book was not at hand; although this latter is wider of the truth than the former. By the same calculations also he showed how much each of these compends deviates from the truth, and also how widely the computations on the erroneous principles of the plane chart differ from them all. The method generally used by our sailors, however, is commonly called the middle latitude, which, although it errs more than that by the arithmetical mean between the two cosines, 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 cosines, equivalent to the geometrical mean between the cosines themselves—a method since proposed by John Bassat. The computation by the middle latitude will always fall short of the true change of longitude, that by the geometrical mean will always exceed; but that by the arithmetical mean falls short in latitudes of about 45°, 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 period logarithms were invented by John Napier, Baron of Merchiston in Scotland, and proved of the utmost service to the art of navigation. From these 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,1 on which are described lines of logarithms, of logarithmic sines and tangents, of meridional parts, &c.; and he greatly improved the sector for the same purposes. He also showed how to take a back observation by the cross staff, by which the error arising from the eccentricity of the eye is avoided. He likewise described another instrument, of his own invention, called the cross bow, for taking altitudes of the sun or stars, with some contrivances for more readily finding the latitude from the observation. The discoveries concerning logarithms were carried into France in 1624 by 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, that of the tables of artificial sines and tangents, as modelled according to Napier's last form, erroneously attributed by Wingate to Briggs.

Gunter's scale was projected into a circular arch by the Reverend William Oughtred in 1633; and its uses were fully shown in a pamphlet entitled the Circles of Proportion, where, in an appendix, several important points in navigation are well treated. 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 Thomas Addison, in his treatise entitled Arithmetical Navigation, printed in the year 1625. He also gave two traverse tables, with their uses; the one to quarter points of the compass, and the other to degrees. Henry Gellibrand published his discovery of the changes of the variation of the compass, in a small quarto pamphlet, entitled A Discourse Mathematical on the Variation of the Magnetical Needle, printed in 1635. This extraordinary phenomenon he found out by comparing the observations which had been made at different times near the same place by Burrough, Gunter, and himself, all persons of great skill and experience in these matters. This discovery was likewise soon known abroad; for Athanasius Kircher, in his treatise entitled Magnes, first printed at Rome in the year 1641, informs us that he had been told of it by John Greaves, and then gives a letter of the famous Marinus Mersennus, containing a very distinct account of the same.

As altitudes of the sun are taken on shipboard by observing his elevation above the visible horizon, to obtain from these the sun's true altitude with correctness, Wright observed 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 he complains that the most effectual was out of his power to execute, and therefore he contented himself with a rude attempt, in some measure sufficient for his purpose. The dimensions of the earth deduced by him corresponded very well with the usual divisions of the log-line; nevertheless, as he did not write an express treatise on navigation, but only for correcting such errors as prevailed in general practice, the log-line did not fall under his notice. Richard Norwood, however, put in execution the method recommended by 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 which measurement, 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

1 See GUNTER'S SCALE.

History. fathoms or toises; which is very exact, as appears from many measurements that have been made since that time. Of all this Norwood gave a full account in his treatise called the Seaman's Practice, published in 1657. He there showed the reason why Snellius had failed in his attempt; and he also pointed out 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, however, has at length made its way into practice; and few navigators of reputation now make use of the old measure of forty-two feet to a knot. In this treatise Norwood also describes his own excellent method of setting down and perfecting a sea reckoning, by using a traverse table, which method he had followed and taught for many years. He likewise shows how to rectify the course, by taking into consideration the variation of the compass; 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 abridged as a manual for sailors, in a very small work called an Epitome; which useful performance has gone through a great number of editions. No alterations were ever made in the Seaman's Practice till the twelfth 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, No. 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, Bond published, in Norwood's Epitome, a very great improvement of Wright's method, from a property in his meridian line, whereby the divisions are more scientifically assigned than the author himself was able to effect. It resulted from this theorem, that these divisions are analogous to the excesses of the logarithmic tangents of half the respective latitudes augmented by 45^\circ above the logarithm of the radius. This he afterwards explained more fully in the third edition of Gunter's works, printed in 1653, where he observed that the logarithmic tangents from 45^\circ upwards increase in the same manner as the secants do added together, 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 to be 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^\circ; divide the difference of those numbers by the logarithmic tangent of 45^\circ 30', the radius being likewise rejected, and the quotient will be the meridional parts required, expressed in degrees." This rule is the immediate consequence of the general theorem, that the degrees of latitude bear to one degree (or sixty minutes, which in Wright's table stand for the meridional parts of one degree) the same proportion as the logarithmic tangent of half any latitude augmented by 45^\circ, and the radius neglected, to the like tangent of half a degree augmented by 45^\circ, with the radius likewise rejected. But here there was still wanting the demonstration of this general theorem, which was at length supplied by James Gregory of Aberdeen, in his Exercitationes Geometricæ, printed at London in 1668; and afterwards more concisely demonstrated, together with a scientific determination of the divisor, by Dr Halley, in the Philosophical

Transactions for 1695 (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 Roger Cotes, in his Logometria, first published in the Philosophical Transactions for 1714 (No. 388). 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^\circ 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^\circ 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 a book called the Seaman's Kalender, 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 forty-nine years to come. Thus he acquired such reputation, that his treatise entitled The Longitude Found, was, in the year 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 the year 1678 another treatise, entitled The Longitude not Found, made its appearance; and as Bond's hypothesis did not answer its author's sanguine expectations, the solution of the difficulty 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. (See MAGNETISM.) In 1700, however, Dr Halley published a general map, with curve lines expressing the paths where the magnetic needle had the same variation; which was received with universal applause. But as the positions of these curves vary from time to time, they should frequently be corrected by skilful persons, as was done in 1644 and 1756, by Mountain and Dodson. 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, 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, not a perfect sphere, but a spheroid, with the shortest diameter passing through the poles, a tract was published in 1741 by the Reverend Dr Patrick Murdoch, wherein he accommodated Wright's sailing to such a figure; and the same year Colin Maclaurin, 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, and in Delambre's Astronomia (t. iii., ch. xxxvi.).

Amongst 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

Practice of purpose were found to answer so well that he obtained the parliamentary reward. These have been improved by Arnold, Earnshaw, and many others, so as now to be almost in common use.

The works which have latterly appeared on navigation are those on the longitude and navigation by Mackay, Inman, Riddle, Norie, Jeans, and others; and these contain every necessary requisite to form the practical navigator.

PRACTICE OF NAVIGATION.

BOOK I.

CONTAINING THE VARIOUS METHODS OF SAILING.

The art of navigation depends upon mathematical and astronomical principles. The problems in the various modes of sailing are resolved either by trigonometrical calculations, or by tables or rules formed by the assistance of plane and spherical trigonometry. By mathematics the necessary tables are constructed and rules investigated for performing the more difficult parts of navigation.

The places of the sun, moon, and planets, 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 investigation of the rules required for this purpose belongs properly to the science of ASTRONOMY, to which the reader is referred. 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 this subject, it seems unnecessary to insert them in this place. The subject naturally divides itself under two heads:—First, The methods of conducting a ship from one port to another by help of rules, in which the log-line and compass are alone required, which is Navigation properly so called. Second, The method of ascertaining the ship's latitude and longitude, and variation of compass, by means of observations on the heavenly bodies; and the rules for that purpose deduced from astronomy, in order to correct the ship's place, and the courses derived from the former method, to which the name of Nautical Astronomy is generally applied. Although the reader is referred to the respective articles on the sciences on which navigation is founded in this work for complete information, we shall, nevertheless, endeavour to make our explanation of the several rules as complete as possible, even at the risk of repeating somewhat the substance of portions of our other articles.

CHAP. I.—PRELIMINARY PRINCIPLES.

SECT. I.—ON LATITUDE AND LONGITUDE; DEFINITION OF TERMS USED IN NAVIGATION; AND GENERAL EXPLANATIONS.
1. Latitude and Longitude.

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

The equator is a great circle on the earth's surface, every point of which is equally distant from the two poles or the extremities of the imaginary axis about which the earth makes her diurnal rotation. It therefore divides the earth into two equal parts, called the Northern and the Southern

Hemispheres, according as the North or the South Pole lies within them.

The latitude of a place is its distance from the equator, reckoned on a meridian in degrees, minutes, and seconds, and decimal parts of seconds (if necessary), being either north or south, according as it is the Northern or Southern Hemisphere. Hence it appears that the latitudes of all places are comprised within the limits 0^{\circ} and 90^{\circ} N., and 0^{\circ} and 90^{\circ} S.

The first meridian, which is a great circle passing through the poles, also divides the earth into two equal portions, called the Eastern and Western Hemispheres, according as they lie to the right or left of the first meridian; the spectator being supposed to be looking towards the north.

The longitude of a place is the arc of the equator intercepted between the first meridian and the meridian of the given place reckoned in degrees, minutes, and seconds; and is either east or west as the place lies in the Eastern or Western Hemisphere respectively to the first meridian. The longitude of all places on the earth's surface is comprised within the limits of 0^{\circ} and 180^{\circ} E., and 0^{\circ} and 180^{\circ} W.

On the supposition that the earth is a sphere, the length of all arcs of great circles upon it subtending an angle of 1^{\circ} at the centre are equal; hence 1^{\circ} of latitude or longitude is equal to one geographical or nautical mile, of which a degree contains 60. Hence intervals of latitude and longitude, reduced to minutes and parts of minutes, also represent the same number of nautical miles and parts of a nautical mile.

In the practice of navigation, the latitude and longitude of the place which a ship leaves, are called the latitude and longitude from; and the latitude and longitude of the place at which it has arrived, are called the latitude and longitude in.

A geometric diagram illustrating the definition of terms in navigation. It shows a spherical projection with a pole P at the top. A great circle representing the equator is shown as an arc V-U-T-S-R-Q. Several meridians are shown as arcs passing through points A, B, C, D, E, F. A path is traced from A to F. Perpendiculars BH, CI, DK, EL, &c. are drawn from points B, C, D, E, &c. to the equator. Other points labeled include M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z, &c.

Fig. 1.

2. Definitions of Terms used in Navigation, and Explanations.

Let QR...V be a portion of the equator, P the pole, and PAQ, PBR, PCS.....PFV be meridians supposed very near to one another, passing through points A, B, C, D, E, F, the line AF being the path traced out by a vessel in passing from A to F, such that it makes equal angles with every meridian over which it passes. From B, C, D, &c., let BH, CI, DK, EL, &c., be drawn perpendicular to the

two meridians between which they respectively lie; or, in other words, the arcs of small circles or parallels of latitude through the points B, C, &c. These are consequently all parallel to one another, and to FG the whole arc of the parallel at F included between the extreme meridians PAQ and PFV.

The constant angle at which the line AF is inclined to the successive meridians, viz., BAP, CBP, DCP, &c., is called the course. Also, if the small circles or parallels at B, C, &c., be continued to the meridian PAQ, the portion of this meridian intercepted between any two consecutive parallels, as cd, will be equal to CK, the distance between the parallels through C and D; and so on for all. Hence the sum of these distances, AH + BI + CK + DL + EM = AG; which is called the true difference of latitude, or true diff. lat. from A to F.

The corresponding arcs of parallels at different latitudes intercepted between the same meridians are not equal, but gradually decrease from the equator to the poles. Hence the sum of the arcs BH + CI + DK + EL + FM is less than QV, the intercepted arc of the equator, but greater than FG, the arc of the highest parallel intercepted between PAQ and PFV.

BH + CI + DK + EL + FM is called the departure; the arc QV of the equator is the difference of longitude; and AF, the curve described by the vessel in passing from A to F, is called the distance.

In navigation, each of the triangles ABH, BCI, &c., is considered as a plane triangle; and as each of them is right-angled, and contains, besides, one constant angle, viz., the course, the other angle in each must also be constant; and all the triangles will be equiangular and similar. Hence we have

\begin{aligned} AH : BH : AB &:: AH : BH : AB \\ BI : CI : BC &:: AH : BH : AB \\ CK : DK : CD &:: AH : BH : AB \\ &\&c. &\&c. &\&c. \\ EM : FM : EF &:: AH : BH : AB. \end{aligned}

And since, when any number of quantities are in continued proportion, as the first consequent is to its antecedent, so are all the consequents to all the antecedents; we have

\begin{aligned} AH + BI + CK + \&c. : BH + CI + DK + \&c. : AB + BC + CD + \&c. \\ :: AH : BH : AB. \end{aligned}
\begin{aligned} \text{But } AH + BI + CK + \&c. &= AG \text{ the true diff. lat.} \\ BH + CI + DK + \&c. &= \text{departure.} \\ AB + BC + CD + \&c. &= AF \text{ or the distance.} \end{aligned}

Hence the true difference of latitude, departure, and distance, may be considered as the sides of a right-angled triangle, similar to each of the small triangles; the angle of which, therefore, between the true difference of latitude and distance, is the course.

Take AB (fig. 2) = the true diff. lat. Draw BC at right angles to it = departure. Join AC. Then AC is the distance, and BAC is the course.

From this it appears, that when any two of the four quantities, true difference of latitude, departure, distance, and course are given, the remaining two can be found by solving the right-angled triangle ABC.

1. Given course (BAC) and distance (AC), to find true difference of latitude (AB), and departure (BC).

\begin{aligned} AB &= AC \cos BAC \\ BC &= AC \sin BAC, \end{aligned}
\begin{aligned} \text{i.e., true diff. lat. (in miles)} &= \text{dist.} \times \cos \text{ course} \quad (1.) \\ \text{and departure} &= \text{dist.} \times \sin \text{ course} \quad (2.) \end{aligned}
Figure 2: A right-angled triangle ABC. The right angle is at B. Side AB is vertical, side BC is horizontal, and side AC is the hypotenuse. A small circle is drawn around vertex A.
Fig. 2.
Or in logarithms,
\begin{aligned} \log. \text{ true diff. lat.} &= \log. \text{ dist.} + L. \cos \text{ course} - 10, \\ \text{where } L. & \text{ means tabular logarithm, i.e., logarithm increased by 10; and } \log. \text{ dep.} = \log. \text{ dist.} + L. \sin \text{ course} - 10. \end{aligned}

2. Given course (BAC) and true difference of latitude (AB), to find distance and departure.

\begin{aligned} AC &= AB \times \sec BAC \\ BC &= AB \times \tan BAC; \\ \text{or dist.} &= \text{true diff. lat.} \times \sec \text{ course} \quad (III.) \\ \text{dep.} &= \text{true diff. lat.} \times \tan \text{ course} \quad (IV.) \end{aligned}

3. Given course and departure, to find distance and true difference of latitude.

\begin{aligned} AC &= BC \times \csc BAC \\ AB &= BC \times \cot BAC; \\ \text{or dist.} &= \text{dep.} \times \csc \text{ course} \quad (V.) \\ \text{and true diff. lat.} &= \text{dep.} \times \cot \text{ course} \quad (VI.) \end{aligned}

4. Given distance and true difference of latitude, to find course and departure.

\begin{aligned} \cos BAC &= \frac{AB}{AC}; \\ \text{or cosine course} &= \text{true diff. lat.} \div \text{dist.} \quad (VII.) \end{aligned}

And the course having been found, we get

\text{dep.} = \text{dist.} \times \sin \text{ course by (II.)}

5. Given the distance and departure, to find the course and true difference of latitude.

\begin{aligned} \sin BAC &= \frac{BC}{AC}; \\ \text{or sin course} &= \text{dep.} \div \text{distance} \quad (VIII.) \end{aligned}

And then we have

\text{true diff. lat.} = \text{dist.} \times \cos \text{ course.}

6. Given the true difference of latitude and departure, to find the course and distance.

\begin{aligned} \tan BAC &= \frac{BC}{AB}; \\ \text{or tan course} &= \text{departure} \div \text{true diff. lat.} \quad (IX.) \end{aligned}

And having found the course, we have

\begin{aligned} \text{dist.} &= \text{true diff. lat.} \times \sec \text{ course by (III.)} \\ \text{or dist.} &= \text{dep.} \times \csc \text{ course by (V.)} \end{aligned}

Length of Arc of 1° of Parallels of Latitude.

We have already stated that the lengths of the parallels of latitude diminish as the latitude increases. In fact, it decreases in the ratio of cosine latitude to unity.

Let EQ be the equator, PCP' the polar diameter of the earth passing through C, and LML' any parallel of latitude. Let the angle ECL, or the latitude, be l, and let LM and EF be the arcs of the parallel and of the equator intercepted between the meridians PEP' and PFP'. Then angle ECF = angle LOM, because they both measure the angle between the planes of the two meridians. Hence

\begin{aligned} \text{arc LM} : \text{arc EF} &:: OL : CE :: OL : CL, \text{ because } CE = CL; \\ \text{or arc LM} &= \text{arc EF} \times \frac{OL}{CL} = \text{arc EF} \times \sin OCL \\ &= \text{arc EF} \times \cos LCE = \text{arc EF} \times \cos l. \end{aligned}

Hence if FE be the length of an arc 1° of longitude at the equator, or 60 miles, LM the length of an arc 1° of longitude in latitude l = 60 \times \cos l.

Figure 3: A diagram of a sphere showing the equator EQ, the polar diameter PCP', and a parallel of latitude LML'. Points P, Q, C, P', E, F, L, M, O, and C' are marked. The diagram illustrates the geometry of latitude and longitude on a sphere.
Fig. 3.
Middle Latitude.

The departure is less than VQ (fig. 1), the intercepted arc of the equator, or than the intercepted arc of the parallel through A; but it is greater than FG. But since the arc of the parallel gradually decreases from A to F, there is some point intermediate in position between A and F (y), the intercepted arc of the parallel of which will be exactly equal to the departure. The exact determination of this point is not very easy. Various methods have been proposed to determine this latitude nearly, with as little trouble as possible: first, by taking the arithmetical mean of the two latitudes for that of the mean latitude; secondly, by using the arithmetical mean of the cosines of the latitudes; thirdly, by using the geometrical mean of these cosines; and, lastly, by employing the latitude deduced from the mean of the meridional parts of the two latitudes. The first of these methods is the one usually employed. It has the merit of great simplicity; and as all the rules in navigation are approximate only, it may perhaps be depended on as much as any of these. Hence y may be considered as the middle point between A and F; and \text{dep.} = xyz.

But xyz = VQ \cos \text{lat. of } y;
or \text{dep.} = \text{diff. long.} \times \cos y, where y is the arithmetical mean of the latitudes of A and F = \frac{1}{2}(l+l'), if
l = \text{latitude of A,}
l' = \text{latitude of F.}
This is commonly called the middle latitude. Hence we have \text{dep.} = \text{diff. long.} \times \cos \text{middle latitude.}
By equation (1.) we have
true diff. lat. = dist. \times \cos \text{course.}

Whence it appears that if the middle latitude be considered as a course, and the departure as a true difference of latitude, the corresponding distance will be the difference of longitude. And conversely, treating the middle latitude as a course, and the difference of longitude as a distance, the corresponding true difference of latitude will be the departure.

Mercator's Chart—Meridional Parts.

The chart used at sea for tracing the ship's track exhibits the surface of the earth on a plane, in which the meridians are parallel, and consequently the distance between them throughout their length equal to the equatorial distance, instead of gradually decreasing as the latitude increases. In other words, FG is increased so as to become equal to VQ. Now, in order that on this chart all points may occupy the same relative position with respect to each other that the points corresponding to them do on the surface of the globe, the distance AF and the distance AG must be increased in the same proportion that BH + CI + DK + &c., i.e., the departure has been increased.

The distance AG so increased is called the meridional difference of latitude, or mer. diff. lat.; and the chart constructed on this principle is called Mercator's Chart. If for any latitude the meridional difference of latitude between this point and the equator be expressed in miles or minutes, the number of miles so expressed is called the meridional parts for that latitude. A table of meridional parts for every minute of latitude from 0^\circ up to 90^\circ, is given in every collection of nautical tables.

Construction of Table of Meridional Parts.

This table may be constructed approximately, by dividing the whole meridian from 0^\circ to 90^\circ into intervals of 1', and supposing the increase of the arc of the parallel of latitude, and consequently that of the arc of latitude, to take place at the end of the successive minutes.

Now we have seen that an arc of any parallel = corresponding arc of equator \times \cos \text{lat.}; and since the arcs of the successive parallels have all become equal to the correspond-

ing arc of the equator, they have all been increased in the ratio of \sec \text{lat.} to 1.

Hence, if the length of 1' of the meridian be 1, and a, b, c, d, the corresponding increased lengths between 0 and 1', between 1' and 2', 2' and 3', &c.,

a = 1 \times \sec 1'
b = 1 \times \sec 2'
c = 1 \times \sec 3'
d = 1 \times \sec 4'
&c. &c.

And a+b+c, \&c. = \sec 1' + \sec 2' + \sec 3' + \&c.; or the meridional parts in any arc of the meridian is equal to the sum of the secants of all the successive angles, differing by 1', from 1' up to the given latitude.

The true investigation is as follows:—Let m = the circular measure of the angle subtended at the centre by the meridional parts in the arc between the latitudes 0 and l; and let l become l+\delta l, and let \delta m be the corresponding increase of m. Then \delta m is proportional to the secant of l+\delta l;

\text{or } \frac{\delta m}{\delta l} = \sec(l+\delta l).
And ultimately taking the limit
\frac{dm}{dl} = \sec l;
\therefore m = \int_0^l \sec l dl = \int_0^l \frac{1}{\cos l} dl
= \int_0^l \frac{\cos^2 \frac{l}{2} - \sin^2 \frac{l}{2}}{\cos^2 \frac{l}{2} + \sin^2 \frac{l}{2}} dl = \int_0^l \frac{(1 + \tan^2 \frac{l}{2}) dl}{1 + \tan^2 \frac{l}{2}}
= 2 \int_0^l \frac{d \cdot \tan \frac{l}{2}}{1 + \tan^2 \frac{l}{2}} = \log_e \frac{1 + \tan \frac{l}{2}}{1 - \tan \frac{l}{2}}
= \log_e \left( \frac{\cos(45 - \frac{l}{2})}{\cos(45 + \frac{l}{2})} \right).
\text{Let } 45 - \frac{l}{2} = \frac{l_1}{2}, \text{ or } 90 - l = l_1;
45 + \frac{l}{2} = 90 - \frac{l_1}{2};
\text{and } m = \log_e \frac{\cos \frac{l_1}{2}}{\sin \frac{l_1}{2}} - \log_e \cot \frac{l_1}{2}
= 2 \cdot 3025851 \log_{10} \cot \frac{1}{2} l_1.
\text{Now } m = \frac{\text{arc}}{\text{radius}} = \frac{\text{arc (in minutes)}}{\text{radius (in miles)}} \\ = \frac{\text{meridional parts for lat. } l}{57 \cdot 29577 \times 60}
\therefore meridional parts for lat. l
= 57 \cdot 29577 \times 60 \times 2 \cdot 3025851 \times \log_{10} \cot \frac{1}{2} l_1.
In logarithms—
\log \text{ mer. parts for lat. } l = 3 \cdot 8934895 + \log_{10} (L \cot \frac{1}{2} \text{ colat.} - 10).
Whence we deduce the following
Rule for Finding the Meridional Parts.

Diminish the tabular logarithm of the cotangent of one-half the colatitude by 10. Find the logarithm of the remainder, and add to it the constant logarithm 3.8934895. The result is the logarithm of the meridional parts for the given latitude.

Preliminary principles. Ex.—Required the meridional parts for latitude 65^{\circ} and 65^{\circ} 20'.

\begin{array}{rcl} & (1.) & \\ L \cot 12^{\circ} 30' - 10 & = & 0.6542448 \\ \text{Log. } 0.6542448 & = & \overline{1}.8157403 \\ \text{Constant log.} & = & 3.8984895 \\ & & 3.7142298 \\ \therefore \text{Meridional parts,} & = & 5178.80 \\ & (2.) & \\ L \cot 12^{\circ} 20' - 10 & = & 0.6602609 \\ \text{Log. } 0.6602609 & = & \overline{1}.8197155 \\ \text{Constant log.} & = & 3.8984895 \\ & & 3.7182050 \\ \therefore \text{Meridional parts,} & = & 5226.43 \end{array}

Relations between Meridional Difference of Latitude and Difference of Longitude, Course, &c.

It appears that on Mercator's chart the difference of longitude, meridional difference of latitude, and increased distance, form the sides of a right-angled triangle, and are proportional to the departure, true difference of latitude, and distance, in the triangle ABC. The two triangles are therefore similar.

If, then, in AB produced (fig. 4) AB' be taken equal to mer. difference of latitude, and B'C' be drawn parallel to BC, meeting AC produced in C'; the sides of the triangle ABC' will be the meridional difference of latitude, difference of longitude, and increased distance.

Hence, in equations (1.), (2.), (3.), (4.), (5.), (6.), (7.), and (8.), we may substitute meridional difference of latitude for true difference of latitude, difference of longitude for departure, and increased distance for distance; and the equations will still hold.

The relation principally required is that which connects the meridional difference of latitude, difference of longitude, and course;

\text{or, diff. long.} = \text{mer. diff. lat.} \times \tan \text{ course.}

Parallel Sailing.

When the course is 90^{\circ}, or the ship sails in a parallel of latitude, the equations (1.), &c., give no result when applied to find the distance.

In this case we must apply the formula, arc of parallel = corresponding arc of equator \times \cos \text{ lat.} Here the arc of the parallel corresponds to the distance, and the arc of equator is difference of longitude, and we have for parallel sailing

\text{dist.} = \text{diff. long.} \times \cos \text{ lat.},

from which, any two of the quantities being given, the third may be found. Also, if d and d' be distances corresponding to the same difference of longitude in parallels l and l',

\begin{aligned} \text{we have } d &= \text{diff. long.} \times \cos l \\ d' &= \text{diff. long.} \times \cos l' \\ \text{or } d &= d' \cdot \frac{\cos l}{\cos l'} \end{aligned}

which enables us to find the distance on one parallel corresponding to a given distance on another, the difference of longitude being the same.

Middle-Latitude Sailing.

Since departure = diff. of longitude \times \cos \text{ mid. latitude}, if in equations (2.), (4.), (6.), (8.), we substitute this value for departure, we shall obtain the following equations:—

\begin{array}{l} \text{diff. long.} \times \cos \text{ mid. lat.} = \text{dist.} \times \sin \text{ course} \quad (XI.) \\ \text{diff. long.} \times \cos \text{ mid. lat.} = \text{true diff. lat.} \times \tan \text{ course} \quad (XII.) \\ \text{true diff. lat.} = \text{diff. long.} \times \cos \text{ mid. lat.} \times \cot \text{ course} \quad (XIII.) \\ \sin \text{ course} = \text{diff. long.} \times \cos \text{ mid. lat.} \div \text{dist.} \quad (XIV.) \end{array}

from which all the rules for middle-latitude sailing may be derived.

Traverse Tables.

A table in which the true difference of latitude and departure, corresponding to certain distances for every course, expressed in points and degrees, are laid down, is called a traverse table. It is very useful to enable the seaman to solve the several problems which occur in navigation by simple inspection. It must of course be calculated on some of the principles laid down in this chapter.

It is evident that as this table contains the relations of the sides and angles of a right-angled triangle, the solution of any right-angled triangle, whose sides represent any other quantities, will be given by it, by making the requisite changes.

Thus, the course remaining the same, if, for difference of latitude we look out in the table the meridional difference of latitude, the corresponding departure will be the difference of longitude; also, the difference of longitude can be found from the departure by these tables, by looking out the middle latitude as a course, and the departure as a true difference of latitude; then the corresponding distance is the difference of longitude.

SECT. II.—LONGITUDE AND LATITUDE.

PROB. I.—Given latitude from, and latitude in; to find the true difference of latitude.

Rule.—Under latitude in, with its proper name, i.e., N. or S., place latitude from. Subtract the greater from the less, if of the same name, and reduce to minutes. The result is the true difference of latitude required, and is of the same or different name with latitude in, according as latitude in is greater or less than latitude from. If they are of different names, add and affect with the name of latitude in.

Ex. 1.—A vessel sails from the Lizard, Lat. 49^{\circ} 58' N., to Cape St Vincent, Lat. 37^{\circ} 3' N.; what is the true diff. of latitude?

Latitude in..... 37^{\circ} 3' N.
Latitude from..... 49^{\circ} 58' N.
True diff. latitude..... 12^{\circ} 55' S. = 775 \text{ miles S.}

Ex. 2.—A vessel sails from New York, Lat. 40^{\circ} 42' N., to Liverpool, Lat. 53^{\circ} 25' N.; find the true diff. of latitude.

Latitude in..... 53^{\circ} 25' N.
Latitude from..... 40^{\circ} 42' N.
True diff. latitude..... 12^{\circ} 43' N. = 763 \text{ miles N.}

Ex. 3.—A ship sailed from Funchal, Lat. 32^{\circ} 38' N., to the Cape of Good Hope, Lat. 34^{\circ} 29' S.; what is the true diff. of latitude?

Latitude in..... 34^{\circ} 29' S.
Latitude from..... 32^{\circ} 38' N.
True diff. latitude..... 67^{\circ} 7' S. = 4027 \text{ miles S.}

PROB. 2.—Given the latitude from, and true difference of latitude; to find latitude in.

Rule.—If the latitude from and true difference be of the same name, add them (the true difference of latitude being turned, if necessary, into degrees and minutes); the sum is the latitude in of the same name. If of unlike names, under latitude from place the true difference of latitude, subtract the less from the greater; the result, with the name of the greater, is the latitude in.

Ex. 1.—A ship sailed from the Lizard, 49^{\circ} 58' N., and made good, in a northerly direction, 207 miles; what is the latitude in?

Latitude from..... 49^{\circ} 58' N.
True diff. latitude..... 3^{\circ} 27' N.
Latitude in..... 53^{\circ} 25' N.

Ex. 2.—A ship in Lat. 57^{\circ} 18' N. sailed due S. 3789 miles; what is the latitude in?

Latitude from..... 57^{\circ} 18' N.
True diff. latitude..... 63^{\circ} 9' S.
Latitude in..... 5^{\circ} 51' S.

PROB. 3.—To find the meridional difference of latitude; having given latitude from, and latitude in. Take the meridional parts for the two latitudes from the table for meridional parts; subtract if the names be alike, and add if the names be unlike. The result is the meridional difference of latitude, and is N. or S. according as the latitude in is N. or S. of latitude from.

Ex. 1.—Required the meridional diff. of latitude in sailing from Cape Finisterre, Lat. 42° 52' N., to Port Praya, in the island of Santiago, Lat. 14° 54' N.
Latitude in..... 14° 54' N. Mer. parts..... 904 N.
Latitude from..... 42 52 N. Mer. parts..... 2852 N.
Mer. diff. lat.... 1948 S.

Ex. 2.—To find the meridional diff. of latitude in sailing from Lat. 5° 35' N. to 8° 17' S.
Latitude in..... 8° 17' S. Mer. parts..... 499 S.
Latitude from..... 5 35 N. Mer. parts..... 335 N.
Mer. diff. lat.... 834 S.

PROB. 4.—To find the difference of longitude; having given longitude in, and longitude from.

Rule.—Under longitude in place longitude from; subtract, if of like name; reduce the result to minutes, and call it E. or W., according as longitude in is E. or W. of longitude from. Add, if of unlike names, and attach the name E. or W., according as the longitude in is E. or W. of longitude from. If the longitude found by this rule exceed 180° it must be subtracted from 360°, and affected with the contrary name.

Ex. 1.—A ship sails from Liverpool, Long. 2° 59' W., to New York, Long. 73° 59' W.; required the diff. of longitude.

Longitude in..... 73° 59' W.
Longitude from..... 2 59 W.
Diff. Longitude..... 71 0 W. = 4260 miles W.

Ex. 2.—A ship sails from Maskelyne's Isles, in Long. 167° 59' E., to Olinda, in Long. 35° 54' W.; find the diff. of longitude.

Longitude in..... 35° 54' W.
Longitude from..... 167 59 E.
203 53 W.
360 0
Diff. longitude..... 155 7 E. = 9367 miles E.

PROB. 5.—To find the longitude in; having given longitude from, and difference of longitude.

Rule.—Under longitude from place difference of longitude. If of like names, add, and the result is the longitude in of the same name as the longitude from; if of unlike names, subtract the less from the greater. The result, with the name of the greater, is the longitude in.

Ex. 1.—A ship from Long. 9° 54' E. sailed westerly till the difference of longitude was 1398 miles; what is the longitude in?

Longitude from..... 9° 54' E.
Diff. longitude..... 23 18 W.
Longitude in..... 13 24 W.

Ex. 2.—The longitude sailed from is 25° 9' W., and diff. of longitude 112° 6' W.; find longitude in.

Longitude from..... 25° 9' W.
Diff. longitude..... 18 46 W.
Longitude in..... 43 55 W.

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.)

The interval between two consecutive knots on the line—also technically called a knot—is supposed to be the same fraction of a nautical mile (6080 feet) that half a minute is

of an hour. Hence the proper length of a knot is \frac{6080}{120} = 51 feet nearly. 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 on the glass, it will be necessary to examine them from time to time; and the distance given by them must be corrected accordingly. The distance sailed, therefore, may 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. 1.—The distance sailed by the log, and the seconds run by the glass, being given; to find the true distance run, the line being supposed right.

Let the number of seconds in which the glass runs out be n, and let d and d' be the true distance and distance by log respectively. Then evidently the longer the time the glass is running, the less is the distance by log compared with the true distance, and conversely. Hence we have the proportion—

\begin{aligned} d : d' &:: 30' : n' \\ \text{or true dist.} &: \text{dist. by log} :: 30' : \text{number of seconds the glass is running;} \\ \text{or } d &= \frac{d' \times 30}{n}. \end{aligned}

Rule.—Multiply the distance given by log by 30, and divide by the number of seconds the glass is running; the result is the true distance run.

Ex. 1.—The hourly rate of sailing by the log is 9 knots, and the glass is found to run out in 35"; required the true rate of sailing.

\begin{array}{r} 9 \\ 30 \\ \hline 35)270(7.7 = \text{true rate of sailing.} \end{array}

Ex. 2.—The distance sailed by the log is 73 miles, and the glass runs out in 26"; required the true distance.

\begin{array}{r} 73 \\ 30 \\ \hline 26)2190(84.2 = \text{true distance.} \end{array}

PROB. 2.—Given the distance sailed by the log, and the measured interval between two adjacent knots on the line; to find the true distance, the glass running exactly 30".

Here evidently the true distance is greater or less than the distance by log, as the measured interval is greater or less than 51 feet.

Let a be the measured interval between the knots in feet, and d and d' the true, and measured distances as before.

Then 51 : a :: d' : d,
or 51 feet : measured distance in feet :: distance by log : true distance

\therefore \text{True distance} = \text{distance by log} + \frac{\text{measured distance}}{51}

Rule.—Multiply the distance given by log by the measured length of a knot, and divide by 51; the quotient is the true distance.

Ex. 1.—The hourly rate of sailing by the log is 5 knots, and the interval between knot and knot measures 53 feet; required the true rate of sailing.

\begin{array}{r} 53 \\ 5 \\ \hline 51)265(5.19 = \text{true rate of sailing.} \end{array}

Ex. 2.—The distance sailed is 85 miles, by a log line which measures 42 feet to a knot; required the true distance run.

\begin{array}{r} 85 \\ 42 \\ \hline 170 \\ 340 \\ \hline 51)3570(70 = \text{true distance run.} \\ 3570 \end{array}

PROB. 3.—Given the length of a knot, the number of seconds run by the glass in half a minute, and the distance

Preliminary principles. sailed by the log. To find the true distance we must evidently compound the ratios given in problems 1 and 2 and we have

n \times 51 : 30 \times a :: d : d; \text{or } d = d' \times \frac{30a}{51n} = \frac{d' \cdot 10a}{17n}.

Rule.—Multiply the distance given by the log by 10 times the measured distance between the knots, and divide by 17 times the number of seconds the glass is running.

Ex.—The distance sailed by the log is 159 miles, the measured length of a knot is 42 feet, and the glass runs out in 33"; required the true distance.

Distance by the log..... 159
10 times length of a knot..... 420
3180
17 times number of seconds..... 636
run by the glass..... 501
60780
(119.037 = true distance)

SECT. IV.—ON COURSES AND CORRECTIONS OF COURSES.

Mariner's Compass.

A ship is enabled to keep her course at sea by means of an instrument called the mariner's compass. It consists of a magnetic steel bar attached to the under side of a card, divided into points and quarter points, and supported by a fine pin, on which it turns freely within a box covered with glass. By reason of the directive property of the magnet, the north point, which is commonly denoted by a fleur de lis, is readily known. The circumference of the card is generally divided into thirty-two points, which, in the best compasses, are again subdivided into half points and quarters. These are reckoned sufficient for nautical purposes. On the inside of the box is drawn a dark vertical line called lubber's point. This point, or rather line, and the pin on which the card turns, are in the same line or plane with the keel of the ship; and hence the point on the circumference of the card opposite to lubber's point shows the angle which the ship's course makes with the magnetic meridian, called the course of the ship.

The annexed diagram (fig. 5) gives a general view of

Diagram of a mariner's compass showing a card with a magnetic needle, a box with a glass cover, and a lubber's point. The card is divided into points and quarter points, with North (N) and South (S) marked. The box is labeled with letters P, Q, R, S, T, U, V, W, X, Y, Z around its perimeter.

Fig. 5.

the compass. (For a full explanation of its magnetic properties, see MAGNETISM.) The names of the points, and the angles which they form with the meridian, are given in fig. 6, and as thus represented the instrument is called the steering compass.

The azimuth compass is the same instrument more nicely made. The circumference of the card is divided into degrees and parts by a vernier, and is fitted up with sight-vanes to take amplitudes and azimuths, for the purpose of determining the variation of the compass by observation. The variation is then applied to the magnetic course shown by the steering compass, whence the true course, with respect to the meridian, becomes known.

Besides the variation, the needle is also affected by the dip, which is likewise fully explained in the article MAGNETISM, as well as Mr Barlow's method of correcting the

Compass rose diagram showing the cardinal directions (N, S, E, W) and intermediate directions (NNE, NE, ENE, E, SSE, S, WSW, W, NNW, NW, N, SSW, S, WNW, W, NNE, NE, ENE, E, SSE, S, WSW, W, NNW, NW, N). The rose has 32 points, with the North point marked with a fleur-de-lis.

Fig. 6.

effects of local attraction, arising from the effects of the iron, guns, &c., in the vessel itself.

The compass course generally differs from the true course on account of three causes:—1. The variation of the compass; 2. The deviation of the compass; 3. The leeway. We shall now explain how these errors are to be applied.

1. The Variation of the Compass.—This is fully explained under the article MAGNETISM, which see. The mode of ascertaining its amount will be given hereafter.

PROB. I.—To find the true course, having given compass course.

Rule 1.—Allow easterly variation to the right, and allow westerly variation to the left.

Ex. 1.—The compass course is W.N.W., and variation 3½ pts. W.; find the true course.

Pts. qts.
Compass course..... 6 0 left of N.
Variation..... 3 1 left of N.
True course..... 9 1 left of N., or W. by S. S.

Ex. 2.—The compass course is S.W.¼W., and variation 2½ E.; find the course.

Pts. qts.
Compass course..... 4 3 right of S.
Variation..... 2 2 right of S.
True course..... 7 1 right of S., or W. by S. ¼W.

Ex. 3.—The compass course is N.W., the variation is 3½ E.; required the true course.

Pts. qts.
Compass course..... 4 0 left of N.
Variation..... 3 1 right of N.
True course..... 0 3 left of N., or N. ¼W.

PROB. II.—Given the true course, to find the compass course.

Rule 2.—Allow easterly variation to the left, and westerly to the right.

Ex. 1.—The true course is N.N.E.¼E., and variation 1½ W.; find the compass course.

Pts. qts.
True course..... 2 2 right of N.
Variation..... 1 2 right of N.
Compass course... 4 0 right of N., or N.E.

Ex. 2.—The true course is N.¼E., the variation 3½ E.; required the compass course.

Pts. qrs.
True course ..... 0 3 right of N.
Variation..... 3 1 left of N.
Compass course .... 2 2 left of N., or N.N.W. ¼ W.

2. The Deviation of the Compass.—This error arises from the effects of local attraction, and varies with every different position of the ship's head. Several methods are employed in order to ascertain its amount. That most commonly adopted is to place a compass on shore, out of reach of the ship's attraction, and to take the bearing of the ship's compass, or some other object in the same direction with it; while at the same time the bearing of the compass on shore is taken on board. If now 180° be added to the bearing of the shore compass, so as to bring it round to the opposite point, the difference between this augmented bearing and the bearing at the ship's compass will be the amount of deviation for that position of the ship's head.

Suppose the ship's head is N., and that the reading off at the shore compass is S. 17° 15' W., and that the reading off at the ship's compass is N. 20° E. Adding 180° to the bearing of the shore compass, we get S. 197° 15' W., or N. 17° 15' E.; and subtracting this from the bearing of the ship's compass, N. 20° E., we get the deviation equal to 2° 45' E., when the ship's head is N. The ship is now turned round, so that the head points successively to every point of the compass, and the deviation for each position found as before.

A table is then made, showing the deviation for every point of the compass. The deviation so found is treated exactly as the variation,—i.e., in correcting the compass course to find the true course, easterly deviation is allowed to the right, and westerly deviation to the left; and conversely, to find the compass course from the true course, easterly deviation is allowed to the left, and westerly deviation to the right. Hence it appears, that when both variation and deviation are given, we may consider the latter as a correction of the former—to be added to it if of the same name, and to be subtracted from it if of the opposite name.

Ex. 1.—The compass course is S.W. ¼ W., variation 1¼ E., and deviation ¼ W.; what is the true course?

Pts. qrs.
Compass course ..... 4 3 right of S.
Pts. qrs.
Variation ..... 1 3 right
Deviation ..... 0 2 left
1 1 right.
True course ..... 6 0 right of S., or W.S.W.

Ex. 2.—The compass course is W. ¼ N., the variation 2½ W., and deviation ¼ W.; what is the true course?

Pts. qrs.
Compass course ..... 7 2 left of N.
Pts. qrs.
Variation..... 2 2 left
Deviation..... 0 3 left
3 1 left.
True course ..... 10 3 left of N., or S.W. by W. ¼ W.

Ex. 3.—The true course is N.N.W. ¼ W., variation 1¼ E., and deviation ¼ W.; required the compass course.

Pts. qrs.
True course ..... 2 3 left of N.
Pts. qrs.
Variation..... 1 3 left
Deviation..... 0 1 right.
1 2 left.
Compass course ..... 4 1 left of N., or N.W. ¼ W.

The following table is taken from the monthly examination-papers at the Royal Naval College, Portsmouth, and will serve as a specimen of the tables which ought to be made for all ships:—

Deviation of the Compass of H.M.S. Vesuvius for different Preliminary positions of the Ship's Head.

Direction of Ship's Head. Deviation of Compass. Direction of Ship's Head. Deviation of Compass.
N. 2° 45' E. S. 3° 0' W.
N. by E. 4 57 S. by W. 4 20
N.N.E. 7 30 S.S.W. 5 0
N.E. by N. 9 0 S.W. by S. 6 7
N.E. 10 0 S.W. 7 0
N.E. by E. 10 53 S.W. by W. 7 27
E.N.E. 10 40 W.S.W. 7 50
E. by N. 9 53 W. by S. 8 20
E. 8 50 W. 8 50
E. by S. 7 15 W. by N. 8 10
E.S.E. 5 35 W.N.W. 6 50
S.E. by E. 3 40 N.W. by W. 5 40
S.E. 1 50 N.W. 4 50
S.E. by S. 0 20 E. N.W. by N. 3 20
S.S.E. 0 50 W. N.N.W. 1 40 W.
S. by E. 2 20 N. by W. 1 10 E.

3. Leeway.—The effect of the action of the wind upon the sails and hull of a ship is sometimes to produce a motion of the ship in a direction at right angles to that of the head or apparent course, as well as in this latter direction. The true course, therefore, is not that given by the compass, but that which is due to the composition of the two velocities of the ship,—viz., that in the direction of its head, and that at right angles to this direction. To obtain the true course from the compass course, therefore, we must add or subtract the angle of leeway, which is the angle between the compass course and the true course.

If the wind be on the right of a person on board ship who is looking towards the head, the real course is then evidently to the left of the direction of the ship's head,—i.e., of the apparent course. If the wind be on his left hand, the true course is to the right. In the former case the ship is said to be on the starboard tack, and in the latter on the port tack. Whence is derived the rule for obtaining the true course from the compass course.

Rule.—If the ship is on the starboard tack, allow leeway to the left; if on the port tack, allow leeway to the right. Conversely, to obtain the compass course from the true course on the starboard tack, allow leeway to the right; if on the port tack, allow it to the left.

There are many circumstances which prevent laying down accurate rules for the allowance of leeway. The construction of different vessels, their trim with regard to the nature and quantity of cargo, the position and area of sail set, the velocity of the ship and the swell of the sea, are all susceptible of great variation, and very much affect the leeway.

The following rules are usually given for the purpose:—

  1. 1. When a ship is close-hauled under all sail, the water smooth, and with a light breeze, allow no leeway.
  2. 2. When the top-gallant sails are hauled, allow one point.
  3. 3. Under close-reefed topsails allow two points.
  4. 4. When one topsail is hauled, allow two points and a half.
  5. 5. When both topsails are hauled, allow three points.
  6. 6. When the fore-course is hauled, allow four points.
  7. 7. When under mainsail only, allow five points.
  8. 8. Under balanced mizen, allow six points.
  9. 9. Under bare poles, allow seven points.

These rules, however, are not much to be depended upon. 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 taffarel, with its diameter at right angles to the ship's length, and dividing its circumference into points and quarters. The angle contained between the semi-diameter which points right aft,

Plane Sailing. and that which points in the direction of the wake, is the leeway. But the best and most rational way of finding the leeway is to have a compass or semicircle on the taffel, 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.

Ex. 1.—A ship's apparent course is S.S.W. \frac{1}{2} W., leeway \frac{1}{2} points, the wind being S.E. \frac{1}{2} E.; required the true course. In this case the wind is on the left of the vessel, or it is on the port tack, and leeway must be allowed to the right.

Pts. qrs.
Apparent course..... 2 3 right of S.
Leeway ..... 2 2 right of S.
True course..... 5 1 right of S., or S.W. by W. \frac{1}{2} W.

Ex. 2.—The apparent course is N.N.W., leeway \frac{1}{2} points, and wind E.N.E. Here the vessel is on the starboard tack.

Pts. qrs.
Apparent course..... 2 0 left of N.
Leeway ..... 1 2 left of N.
True course..... 3 2 left of N., or N.W. by N. \frac{1}{2} W.

Ex. 3.—The true course of a ship is S.E. \frac{1}{2} S., leeway \frac{1}{2} points, and wind N.N.E.; required the compass course. Here the ship is on the port tack.

Pts. qrs.
True course ..... 3 2 left of S.
Leeway ..... 2 2 right of S.
Compass course..... 1 0 left of S., or S. by E.

CHAP. II.—ON 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 considered parallel 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; and consequently the departure and difference of longitude are equal. In fact, in the right-angled triangle ABC (fig. 7), where AB is the true difference of latitude, BAC the course, AC the distance, and BC the departure, which is assumed equal to the difference of longitude; all the problems in sailing are solved by the relations of the sides and angles of the single right-angled triangle ABC. Except, however, for a small portion of the earth's surface near the equator, the departure cannot be assumed equal to the difference of longitude without very considerable error; and the longitude in cannot be at all depended on when found by this method. If, however, the departure be an element, this method is correct.

In fig. 7, A is the place from which the ship sails; AB the meridian, and equal to the true difference of latitude; BC perpendicular to the meridian, and equal to the departure. It is always possible and easy to construct a right-angled triangle when two parts, of which one is a side, beside the right angle, are given. Consequently problems in navigation may always be solved by construction, with the aid of the rule and compasses. In making constructions for this purpose, it is only necessary to attend to the following convention:—Let the upper part of the paper or plan on which the drawing is to be made represent the north; then the lower part will be south, the right-hand side east, and the

Figure 7: A right-angled triangle ABC. Vertex A is at the bottom left, B is at the top left, and C is at the top right. The right angle is at B. Side AB is vertical, side BC is horizontal, and side AC is the hypotenuse.
Fig. 7.

left-hand side west. This convention we have already tacitly assumed in treating of the corrections of the courses.

To make a Construction.

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 end of this line is to be marked as the position of the place, according as the course is southerly or northerly. From this point as centre, with the chord of 60^\circ (on the rule), an arc of a circle 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 on this arc, beginning from the meridian. A straight line drawn through this point and the point sailed from is the direction of the distance, which, if given, must be laid down on this line, beginning at the point sailed from. A straight line is to be drawn from the extremity of the distance perpendicular to the meridian; and hence the true difference of latitude and the departure will be found.

If the true difference of latitude be given, it is to be laid down on the meridian, beginning at the point from which the ship sailed; and a straight line drawn through the extremity of the difference of latitude, perpendicular to the meridian, to meet the distance produced, will limit the figure, and enable us to find the parts required.

If the departure be given, it is to be laid off on a parallel, and the line drawn through its extremity will limit the distance.

If the distance and true difference of latitude be given, through the extremity of the true difference of latitude draw a straight line perpendicular to the meridian; extend a pair of compasses to the given distance, place one of its points at the place from which the ship sailed, and let the other point be in the perpendicular line first drawn; join this point with the point from, and the triangle is determined, and the course and departure found.

If the departure and distance are given, with the point from as centre and the distance as radius, describe an arc of a circle. Let the departure be laid off on a parallel, so that one point is in the meridian and the other in the circle just described. Join this latter point with the point from, and the triangle is formed. The general mode of solving problems in plane sailing has already been given in chap. i., sect. 2. The following examples will show how the formulae are to be applied:—

Obs. It is to be distinctly understood that the above method cannot be applied to obtain the difference of longitude without very sensible error.

Ex. 1.—A ship from St Helena, in Lat. 15^\circ 55' S. sailed S.W. by S. 158 miles; required the latitude in, and departure.

By Construction.—Draw the meridian AB (fig. 8), and with the chord of 60^\circ describe the arc mn, and make it equal to the rhumb of three points, and through n draw AC equal to 158 miles; from C draw CB perpendicular to AB; then AB applied to the scale from which AC was taken will be found to measure 131.4, and BC 87.8.

Figure 8: A diagram for construction. It shows a vertical line AB representing the meridian. A horizontal line BC is drawn from B to the right. A line AC is drawn from A to C, making an angle with AB. A dashed line mn is drawn from the top of AB to the right, parallel to BC. The angle between AB and AC is labeled with a rhumb symbol.
Fig. 8.

By Calculation.—To find the true difference of latitude.

Log. cosine of course..... 3 pts. = 9.91985
Log. distance ..... 158 m. = 2.19866
-10
Log. true diff. lat..... = 2.11851
Hence, true diff. lat..... = 131.4 S.

To find departure.

Log. sin course ..... 3 pts. = 9.74474
Log. distance ..... 158 m. = 2.19866
-10
Log. 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 sine rhumbs (marked S. R.) will reach from the distance 158 to 131.4, the difference of latitude on the line of numbers; and the extent from 8 points to 3 points on sine rhumbs will reach from 158 to 87.8, the departure on numbers.1

Latitude St Helena..... 15° 55' S.
Diff. latitude ..... 2 11 S.
Latitude in..... 18 6 S.

Ex. 2.—A ship from St George's, in Lat. 35° 45' N., sailed S.E. 1/4 S., and the latitude by observation was 35° 7' N.; required the distance run, and departure.

Latitude St George's..... 35° 45' N.
Latitude in..... 35 7 N.
Diff. latitude..... 3 38=218 miles S.

By Construction.—Draw the portion of the meridian AB (fig. 9) equal to 218 miles; from the centre A, with the chord of 60°, describe the arc mn, which make equal to the rhumb of 3 1/4 points; through A 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 miles, and the departure BC 179 miles.

Figure 9: A geometric construction diagram for finding distance and departure. It shows a right-angled triangle ABC with the right angle at B. A vertical line segment AB represents the difference of latitude (218 miles). An arc mn is drawn from point A with a radius equal to the chord of 60 degrees. A line segment AC is drawn from A through the arc. A perpendicular line segment BC is drawn from B to AC. The distance AC is the hypotenuse, and BC is the departure.

Fig. 9.

By Calculation.—To find the distance.

L sec of the course..... 3 1/4 pts. = 10.11181
Log. true diff. latitude ..... 218 m. = 2.33846
-10
Log. distance ..... = 2.45027
Or distance ..... = 282

To find departure.

L tan of the course ..... 3 1/4 pts. = 9.91417
Log. true diff. latitude ..... 218 m. = 2.33846
-10
Log. departure ..... = 2.25263
Or departure ..... = 178.9

By Inspection.—Find the given difference of latitude 218 miles in latitude column, under the course of 3 1/4 points; opposite to which, in distance column, is 282 miles; in departure column 178.9 m.; the distance and departure required.

By Gunter's Scale.—Extend the compass from 4 1/4 points, the complement of the course, to 8 points on sine rhumbs; that extent will reach from the difference of latitude 218 miles to the distance 282 miles on numbers; and the extent from 4 points to the course 3 1/4 points on the line of tangent rhumbs (marked T. R.) will reach from 218 miles to 178.9, the departure on numbers.

Ex. 3.—A ship from Palma, in Lat. 28° 37' N., sailed N.W. by W., and made 192 miles of departure; required the distance run, and the latitude come to.

By Construction.—Make the departure BC (fig. 10) equal to 192 miles, draw BA perpendicular to BC, and from the centre C, with the chord of 60°, describe the arc mn, which make equal to the rhumb of 3 points, the complement of the course; draw a line through Cn, which produce till it meet BA in A. Then the distance AC being measured, will be equal to 231 m., and the difference of latitude AB will be 128.3 miles.

Figure 10: A geometric construction diagram for finding distance and latitude. It shows a right-angled triangle ABC with the right angle at B. A horizontal line segment BC represents the departure (192 miles). A vertical line segment BA represents the difference of latitude. An arc mn is drawn from point C with a radius equal to the chord of 60 degrees. A line segment AC is drawn from C through the arc. A perpendicular line segment BA is drawn from B to AC. The distance AC is the hypotenuse, and BA is the difference of latitude.

Fig. 10.

By Calculation.—To find the distance.

L cos of course ..... 5 pts. = 10.08015
Log. departure ..... 192 m. = 2.28330
-10
Log. distance..... = 2.36345

To find the true difference of latitude.

L cot of course..... 5 pts = 9.82489
Log. departure..... 192 m. = 2.28330
-10
Log. true diff. latitude ..... = 2.10819

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

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

Latitude of Palma ..... 28° 37' N.
Diff. latitude..... 2 8 N.
Latitude in ..... 30 45 N.

Ex. 4.—A ship from a place in Lat. 43° 13' N., sails between the north and east 285 miles, and is then by observation found to be in Lat. 46° 31' N.; required the course and departure.

Latitude sailed from ..... 43° 13' N.
Latitude by observation..... 46 31 N.
Diff. of latitude..... 3 18=198 miles.

By Construction.—Draw the portion of the meridian AB (fig. 11) equal to 198 miles; from B draw BC perpendicular to AB; then take the distance 285 miles from the scale, and with one foot of the compass in A describe an arc intersecting BC in C, and join AC. With the chord of 60° describe the arc mn, the portion of which contained between the distance and difference of latitude, applied to the line of chords, will measure 46°, the course; and the departure BC being measured on the line of equal parts, will be found equal to 205 miles.

Figure 11: A geometric construction diagram for finding course and departure. It shows a right-angled triangle ABC with the right angle at B. A vertical line segment AB represents the difference of latitude (198 miles). A horizontal line segment BC represents the departure. An arc mn is drawn from point A with a radius equal to the chord of 60 degrees. A line segment AC is drawn from A through the arc. The distance AC is the hypotenuse, and BC is the departure.

Fig. 11.

By Calculation.—To find the course.

Log. true diff. latitude ..... (198) + 10 = 12.29660
Log. distance..... 285 - 2.45484
L cos course ..... 46° = 9.84176
Or course is..... N. 46° 0' E.

To find the departure.

L sin course ..... 46° = 9.85693
Log. distance..... 285 = 2.45484
-10
Log. departure ..... = 2.31177

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

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

Ex. 5.—A ship from Fort-Royal, in the island of Grenada, in Lat. 12° 9' N., sailed 260 miles between the south and west, and made 190 miles of departure; required the course and latitude come to.

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

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

Figure 12: A geometric construction for plane sailing. It shows a right-angled triangle ABC with the right angle at C. A line segment BC is drawn perpendicular to AB. An arc of radius 260 miles is drawn from C to intersect AB at point A. A line segment AC is drawn. An arc of radius AC is drawn from A to intersect the line of chords (extension of AB) at point m. The angle between AC and AB is labeled as the course.
Fig. 12.

By Calculation.—To find the course.

Log. departure..... 190 + 10 = 12.27875
Log. distance..... 200 = 2.41497
L sin course ..... 46° 57' = 0.86378
Or course is..... S. 46° 57' W.

To find the true difference of latitude.

L cos course..... 46° 57' = 0.83410
Log. distance..... 200 = 2.41497
-10
Log. true diff. latitude ..... 177.3 = 2.24916

By Inspection.—Seek in the traverse table until the nearest to the given departure is found in the same line with the given distance 200. This is found to be in the page marked 47° at the bottom, which is the course; and the corresponding difference of latitude is 177.3.

By Gunter's Scale.—The extent of the compass, from the distance 200 to the departure 190 on the line of numbers, will reach from 90° to 47°, the course on the line of sines; and the extent from 90° to 43°, the complement of the course on sines, will reach from the distance 200 to the difference of latitude 177½ on the line of numbers.

Latitude Port-Royal ..... 12° 9' N.
Difference of latitude ..... 177 = 2 57 S.
Latitude in ..... 9 12 N.

Ex. 6.—A ship from a port in Lat. 7° 56' S. sailed between the south and east till her departure was 132 miles, and was then, by observation, found to be in Lat. 12° 3' S.; required the course and distance.

Latitude sailed from ..... 7° 56' S.
Latitude in, by observation 12 3 S.
Difference of latitude... 4 7 = 247

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

Figure 13: A geometric construction for plane sailing. It shows a right-angled triangle ABC with the right angle at B. A line segment BC is drawn perpendicular to AB. An arc of radius 247 miles is drawn from B to intersect AB at point A. A line segment AC is drawn. An arc of radius AC is drawn from A to intersect the line of chords (extension of AB) at point m. The angle between AC and AB is labeled as the course.
Fig. 13.

By Calculation.—To find the course.

Log. departure ..... 132 + 10 = 12.12057
Log. true diff. latitude ..... 247 = 2.39270
L tan course ..... 28° 7' = 0.52787
Or course is..... S. 28° 7' E.

To find the distance.

L sec course ..... 28° 7' = 10.05454
Log. true diff. latitude ..... 247 = 2.39270
-10
Log. 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°, 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 the sines, will reach from the difference of latitude 247 to the distance 280 on numbers.

The six problems whose solutions are illustrated above

are all that occur in the solution of the right-angled triangle Mercator's whose sides represent the true difference of latitude, de- Sailing. parture, and distance, and one of whose angles is the course.

CHAP. III.—ON MERCATOR'S SAILING.

We have already explained the principle of Mercator's Chart, and have shown that in every problem of navigation there is a second right-angled triangle similar to that whose solutions formed the subject of investigation in chapter ii.; and the sides of which, corresponding to the true difference of latitude and departure, are the meridional difference of latitude and difference of longitude.

We have already given a rule for finding the meridional parts for any given latitude, on the supposition that the earth is a perfect sphere. If the earth's oblateness be taken into account, and the compression, i.e., the ratio of the difference of the equatorial and polar semi-diameters to the equatorial semi-diameter, be taken as \frac{1}{345}, which it is very nearly, the meridional parts for latitude l will be given by the formula—

\text{Mer. parts} = 7915.705 \log. \tan. \left( 45^\circ + \frac{l}{2} \right) - 22.88 \sin l - 0.0508 \sin 3l - \&c.

Let AD (fig. 14) be the meridian, BAC the course, AB the true difference of latitude, AD the meridional difference of latitude; then

DE is the diff. long., and
ED = DA \times \tan BAC; or,
diff. long. = mer. diff. lat. \times \tan course;
or log. diff. long. = log. mer. diff. lat. +
L \tan course = 10.

Also, by similar triangles, ADE and ABC, we have

\begin{aligned} DE : BC &:: DA : BA; \text{ or,} \\ DE &= BC \times DA \\ &= BA \end{aligned}

\therefore diff. long. = dep. \times mer. diff. lat.; or,
true diff. lat.
log. diff. long. = log. dep. + log. mer. diff. lat. - log. true diff. lat.
Whence, from the departure and true and meridional differences of latitude, the difference of longitude may be found. The following examples will illustrate the mode of using these formulae:—

Ex. 1. A ship sails from Cape Finisterre, Lat. 42° 52' N., Long. 9° 17' W., to Port Praya, in the island of Santiago, Lat. 14° 54' N., and Long. 23° 29' W.; required the course and distance.

Lat. from..... 42° 52' N. Mer. parts ... 2852
Lat. in..... 14 54 N. Mer. parts ... 904
Diff. of lat..... 27 58 S. Mer. diff. lat. 1948 S.
1678 S.
Long. from ..... 9° 17' W.
Long. in..... 23 29 W.
Diff. long. .... 14 12 = 852 W.

By Construction.—Draw the straight line AD (fig. 15), to represent the meridian of Cape Finisterre, upon which lay off AB, AD, equal to 1678 and 1948, the true 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'.

Figure 15: A geometric construction for Mercator's sailing. It shows a right-angled triangle ADE with the right angle at D. A line segment DE is drawn perpendicular to AD. A line segment BC is drawn parallel to DE. A line segment AB is drawn from A to B on AD. A line segment AC is drawn from A to C on DE. The angle between AC and AD is labeled as the course.
Fig. 15.

By Calculation.—To find the course.

Log. diff. long..... 852 + 10 = 12.93044
Log. mer. diff. lat..... 1948 = 3.28959
L tan course ..... 23° 37' = 0.54085
Or course is..... S. 23° 27' W.

To find the distance.

L sec. course..... 23° 27' = 10.03798
Log. true diff. lat..... 1678 miles = 3.22479
-10
Log. distance..... 1831 = 3.26277

Ex. 2.—A ship from Cape Henlopen in Virginia, in Lat. 38° 47' N., Long. 75° 4' W., sailed 267 miles N.E. by N.; required the ship's present place.

By Construction.—With the course, and distance sailed, construct the triangle ABC (fig. 16), and the difference of latitude AB being measured, is 222 miles; hence the latitude in is 42° 29' N., and the meridional difference of latitude 293. Make AD equal to 293, and draw DE perpendicular to AD, and meeting AC produced in E; then the difference of longitude DE being applied to the scale of equal parts, will measure 196; longitude in is therefore 71° 48' W.

Figure 16: A geometric diagram showing a triangle ABC. Point A is at the bottom left, B is on the left side, and C is on the right side. A horizontal line segment BC is drawn. A vertical line segment AD is drawn from A upwards, passing through B. A line segment AC is drawn from A to C. A line segment DE is drawn from D perpendicular to AC, meeting it at point E.

Fig. 16.

By Calculation.—To find the true difference of latitude.

L cos course ..... 3 points = 9.91985
Log. distance..... 267 miles = 2.42651
-10
Log. true diff. latitude ..... 222 N. = 2.34636
Lat. from ..... 38° 47' N. Mer. parts...2528
True diff. lat. .... 3 42 N.
Lat. in ..... 42 29 N. Mer. parts...2821
Mer. diff. lat. 293

To find the difference of longitude.

L tan course..... 3 points = 9.82459
Log. mer. diff. lat..... 293 miles = 2.46687
-10
Log. diff. long. .... 195.8 E. = 2.29176
Long. from..... 75° 4' W.
Diff. long. .... 3 16 E.
Long. in ..... 71 48 W.

Ex. 3.—A ship from Port Canso in Nova Scotia, in Lat. 45° 20' N., Long. 60° 55' W., sailed S.E. 1/2 S., and, by observation, was found to be in Lat. 41° 14' N.; required the distance sailed, and longitude come to.

Lat. Port Canso ..... 45° 20' N. Mer. parts...3058
Lat. in, by observation 41 14 N. Mer. parts...2720
Diff. lat. .... 4 6 = 246 Mer. diff. lat. 338

By Construction.—Make AB (fig. 17) equal to 246, and AD equal to 338; draw AE, making an angle with AD equal to 3 1/2 points, and draw BC, DE perpendicular to AD. Now AC being applied to the scale, will measure 332, and DE, 306.

Figure 17: A geometric diagram showing a triangle ADE. Point A is at the top left, D is at the bottom left, and E is at the bottom right. A vertical line segment AD is drawn. A line segment AE is drawn from A to E. A horizontal line segment BC is drawn from B on AD to C on AE. A vertical line segment DE is drawn from D to E.

Fig. 17.

By Calculation.—To find the distance.

L sec. course..... 3 1/2 points = 10.13021
Log. true diff. lat. .... 246 miles = 2.39093
-10
Log. distance ..... 332 = 2.52114

To find the difference of longitude.

L tan course..... 3 1/2 points = 9.95729
Log. mer. diff. lat..... 338 miles = 2.52892
-10
Log. diff. long. .... 306.3 E. = 2.48621
Long. Port Canso from ..... 60° 55' W.
Diff. long. .... 5 6 E.
Long. in..... 55 49 W.

Ex. 4.—A ship sailed from Saltee, in Lat. 33° 58' N., Long. 6° 20' W., the corrected course was N.W. by W., and departure 420 miles; required the distance run, and the latitude and longitude in.

By Construction.—With the course and departure construct the triangle ABC (fig. 18); now AC and AB being measured, will be found to be equal to 476 and 224 respectively; hence the latitude in 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 is therefore 14° 56' W.

Figure 18: A geometric diagram showing a triangle ABC. Point A is at the bottom left, B is on the right side, and C is on the left side. A horizontal line segment BC is drawn. A vertical line segment AD is drawn from A upwards, passing through B. A line segment AC is drawn from A to C. A line segment DE is drawn from D perpendicular to AC, meeting it at point E.

Fig. 18.

By Calculation.—To find the distance.

L cos course ..... 5 1/2 points = 10.05457
Log. dep. .... 420 miles = 2.62325
-10
Log. dist..... 476.2 = 2.67782

To find true difference of latitude.

L cot course ..... 5 1/2 points = 9.72796
Log. dep..... 420 miles = 2.62325
-10
Log. diff. latitude..... 224.5 = 2.35121
Latitude Saltee from.... 33° 58 N. Mer. parts.... 2169
Diff. latitude ..... 3 44 N.
Latitude in ..... 37 42 N. Mer. parts.... 2445
Mer. diff. lat... 276

To find the difference of longitude.

L tan course ..... 5 1/2 points = 10.27204
Log. mer. diff. latitude ..... 276 miles = 2.44091
-10
Log. diff. longitude ..... 516.3 = 2.71295

Or—

Log. dep. .... 420 = 2.62325
Log. mer. diff. latitude ..... 276 = 2.44091
5.06416
Log. true diff. latitude..... = 2.35121
Log. diff. longitude ..... 516.3 = 2.71295

Ex. 5.—A ship from St Mary's, in Lat. 36° 57' N., Long. 25° 9' W., sailed on a direct course between the north and east 1162 miles, and was then, by observation, in Lat. 49° 57' N.; required the course steered, and longitude come to.

Latitude of St Mary's.... 36° 57' N. Mer. parts.... 2389
Latitude in ..... 49 57 N. Mer. parts.... 3470
Diff. latitude ..... 13 0 N. Mer. diff. lat... 1081 N.
780 N.

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

Figure 19: A geometric diagram showing a triangle ADE. Point A is at the bottom left, D is at the top left, and E is at the top right. A vertical line segment AD is drawn. A line segment AE is drawn from A to E. A horizontal line segment BC is drawn from B on AD to C on AE. A vertical line segment DE is drawn from D to E.

Fig. 19.

By Calculation.—To find the course.

Log. true diff. latitude..... 780 + 10 = 12.89209
Log. dist..... 1162 = 3.06521
L cos course ..... 47° 50' = 9.82688

To find difference of longitude.

L tan course ..... 47° 50' = 10.04302
Log. mer. diff. latitude..... 1081 miles = 3.03383
-10
Log. diff. longitude..... 1194 = 3.07685

Longitude from .....

Diff. longitude .....

Longitude in .....

Ex. 6.—From Aberdeen, in Lat. 57° 9' N., Long. 2° 8' W., a ship sailed between the south and east till her departure was 146 miles, and Lat. in 53° 32' N.; required the course and distance run, and longitude in.

Mercator's Sailing. Latitude Aberdeen..... 57° 9' N. Mer. parts..... 4199
Latitude in ..... 53 32 N. Mer. parts..... 3817
Diff. latitude..... 3 37 = 217 S. Mer. diff. lat.... 382

By Construction.—With the difference of latitude 217 miles, and departure 146 miles, construct the triangle ABC; 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.

Log. dep. .... 146 + 10 = 12.16435
Log. true diff. lat. ... 217 = 2.33646
I tan course... 33° 56' 9.82789

To find the distance.

I sec course ..... 33° 56' 10.08109
Log. true diff. latitude ..... 217 miles 2.33646
-10
Log. dist..... 261.5 2.41755

To find the difference of longitude.

Log. mer. diff. latitude ..... 382 = 2.68206
Log. dep..... 146 = 2.16435
4.74641
Log. true diff. latitude ..... 217 = 2.33646
Log. diff. longitude ..... 257 = 2.40995
Longitude from ..... 2° 8' W.
Diff. longitude ..... 4 17 E.
Longitude in ..... 2 9 E.

Ex. 7.—A ship from Naples, in Lat. 40° 51' N., Long. 14° 14' E., sailed 252 miles on a direct course between the south and west, and made 173 miles of westing; required the course made good, and the latitude and longitude in.

By Construction.—With the distance and departure make the triangle ABC 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 in is 37° 48' N., and meridional difference of latitude 237. Make AD equal to 237, and complete the figure, and the difference of longitude DE will measure 224; hence the longitude in is 10° 39' E.

Figure 21: A geometric diagram showing a right-angled triangle ABC with the right angle at C. A line segment AD is drawn from A perpendicular to BC, and a line segment DE is drawn from D parallel to BC. Points A, B, C, D, E are labeled.

Fig. 21.

By Calculation.—To find the course.

Log. dep. .... 173 + 10 = 12.23805
Log. dist. .... 252 = 2.40140
I sin course..... 43° 21' 9.83665

To find the true difference of latitude.

I cos course..... 43° 21' = 9.86164
Log. dist..... 252 miles = 2.40140
-10
Log. true diff. latitude ..... 183.2 = 2.26304
Latitude from (Naples) .... 40° 51' N. Mer. parts... 2690
True diff. latitude ..... 3 3 S.
Latitude in ..... 37 48 N. Mer. parts... 2453
Mer. diff. lat. 237

To find the difference of longitude.

I tan course ..... 43° 21' = 9.97497
Log. mer. diff. latitude..... 237 miles = 2.37475
-10
Log. diff. longitude ..... 223.7 = 2.34972
Longitude from..... 14° 14' E.
Diff. longitude..... 3 44 W.
Longitude in..... 10 39 E.

Ex. 8.—A ship from Terceira, in Lat. 38° 45' N., Long. 27° 6' W., sailed on a direct course, which, when corrected, was N. 32° E.,

and is found, by observation, to be in Long. 18° 24' W.; required the latitude come to, and distance sailed.

Longitude of Terceira..... 27° 6' W.
Longitude in ..... 18 24 W.
Diff. longitude..... 8 42 = 522

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

Figure 22: A geometric diagram showing a right-angled triangle ADE with the right angle at E. A line segment AB is drawn from A perpendicular to DE, and a line segment BC is drawn from B parallel to DE. Points A, B, C, D, E are labeled.

Fig. 22.

By Calculation.—To find meridional difference of latitude.

I cot course ..... 32° 0' = 10.20421
Log. diff. longitude..... 522 miles = 2.71767
-10
Log. mer. diff. latitude..... 835.2 N. = 2.92188
Latitude from Terceira .... 38° 45' N. Mer. parts... 2526
Mer. diff. lat. 835
Latitude in ..... 48 46 N. 3361
True diff. latitude .... 10 1 N. = 601 miles N.

To find the distance.

I sec course ..... 32° 0' = 10.07158
Log. true diff. latitude..... 601 miles = 2.77887
-10
Log. dist. .... 707.1 = 2.85045

CHAP. IV.—ON TRAVERSE SAILING, OR COMPOUND COURSES.

It is the first business of the navigator, when he is about to conduct a ship from one port to another, to calculate beforehand the course on which the vessel is to be steered, and the distance she must run on that course. If the sea is perfectly free from obstruction between the two ports, one course and one distance will suffice for this purpose. It very seldom happens, however, that the sea is free from obstruction; but rocks or shoals, islands or some part of a mainland, intervenes, and a change of course is thus rendered necessary. In this case, the course and distance of the vessel, supposing the navigation unobstructed, having been taken from the chart, the mariner will determine how many changes of course are necessary, and will proceed to calculate the several courses and distances which shall be equivalent to the one course and distance on which the vessel would sail if unobstructed. This calculation, it must be remarked, is very different from that of the course and distance actually made good on a given day, when, by reason of variation of winds and other causes, the course requires to be altered; although naturally the modes of making these calculations are similar. In the former case, however, the distances to be dealt with are very much greater, and the changes of course less frequent, than in the latter.

The investigations of this chapter are intended to guide the navigator in making his preliminary calculation; the mode of correcting the course and calculating the distance run in each day will form the subject of a subsequent investigation.

If a ship sail on 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 distances run into a single course and distance.

RULE 1.—Make a table sufficiently large to contain the several courses, &c. Divide this table into six columns; the courses are to be put in the first, and the corresponding distances in the second column; the third and fourth columns are to contain the differences of latitude, and the two last the departures.

The several courses and their corresponding distances being properly arranged in the table, find the true difference of latitude and departure answering to each in the traverse table, remembering that the true difference of latitude is to be put into a N. or S. column according as the course is in a northern or southern direction, and that the departure is to be put in E. or W. column according as the course is easterly or westerly. Add together these several quantities in each of the columns, and set the sum down at the bottom. The difference between the sums in the N. and S. columns will be the true difference of latitude made good, of the same name with the greater; and the difference between the sums of the E. and W. columns is the departure made good, of the same name with the greater sum.

Look in the traverse table for a true difference of latitude and departure agreeing 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 true difference of latitude is greater or less than the departure, since in the former case the course is less than 45° or 4 points, and in the latter case greater.

Having found the latitude, find also the meridional difference of latitude; and 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 from, will give the longitude in.

It is also easy to resolve a traverse by construction; and we now show how this may be done, although it is scarcely ever practised at sea.

Describe a circle with the chord of 60° as radius, and in it draw two diameters at right angles to each other, at whose extremities are to be marked the initials of the cardinal points, N. 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 the proper length; and so on 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.

To construct for the difference of longitude we must find by the table the meridional difference of latitude, and lay it off on the meridian, and then complete the triangle similar to that whose sides represent the true difference of latitude; distance and departure as usual.

Ex. 1.—A ship from Fayal, in Lat. 35° 32' N., and Long. 28° 35' W., sailed as follows:—E.S.E., 163 miles; S.W. 1/2 W., 110 miles; S.E. 1/2 E., 180 miles; and N. by E. 68 miles: required the latitude and longitude in, the course, and distance made good.

Course. Dist. Diff. of Lat. Departure.
N. S. E. W.
E.S.E. .... 163 ... 62.4 150.6 ...
S.W. 1/2 W. .... 110 ... 69.8 ... 85.0
S.E. 1/2 E. .... 180 ... 144.5 107.2 ...
N. by E. .... 68 66.7 ... 13.3 ...
66.7 176.7 271.1 85.0
66.7 66.7 85.0
S. 41 1/2° E. .... 281 210.0 186.1

Latitude from ..... 35° 32' N. Mer. parts..... 2569
True diff. latitude.... 3 30 S.
Latitude in ..... 35 2 N. Mer. parts..... 2247
Mer. diff. lat... 262

Now to course 41 1/2°, and opposite 131, half the meridional difference of latitude in latitude column, stands 115 in a departure column, which doubled gives 230 for difference of longitude.

Longitude from..... 28° 36' W.
Diff. longitude ..... 3 50 E.
Longitude in..... 24 46 W.

By Construction.—With chord of 60° describe the circle NESW (fig. 23), the centre of which represents the place the ship sailed from. Draw two diameters NS, EW, at right angles to each other,

the one representing the meridian, and the other the parallel of latitude of the place sailed from. Take each course from the line of rhumbs, lay it off on the circumference from its proper meridian, and number it in order, 1, 2, 3, 4. Upon the first rhumb C1, lay off the first distance 163 miles from C to A; through it draw the second distance AB parallel to C2, and equal to 110 miles; 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 arc Sn, which represents the course, being measured on the line of chords, will be found equal to 41 1/2°. From E draw EF perpendicular to CS produced; then CF will be the difference of latitude, and FE the departure made good, which, applied to the scale, will be found to measure 210 and 186 miles respectively. On CF produced lay off to the scale CG equal to 262, the meridional difference of latitude; and through G draw GH parallel to FE, meeting CE produced in H. Then GH is the difference of longitude; and, when applied to the scale, will be found to measure 230 miles.

A geometric diagram showing a circle with center C and diameter NS. A vertical line NS represents the meridian, and a horizontal line EW represents the parallel of latitude. Points A, B, D, E, F, G, H are marked on the circle and its extensions. Lines connect these points to form a series of parallel and perpendicular segments representing the ship's course and the resulting distance made good, difference of latitude, and difference of longitude.

Fig. 23.

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

RULE 2.—Complete the traverse table as before, to which annex five columns. Now, with the latitude from, 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 are to be put; and in the third, the meridional differences of latitude.

Then to each course, and corresponding meridional difference of latitude, find the difference of longitude by Ex. 4, chap. iii., 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.

Ex. 2.—A ship from Lat. 78° 15' N., Long. 28° 14' E., sailed the following courses and distances, viz.:—W.N.W. 154 miles, S.W. 90, N.W. 1/2 W. 89, N. by E. 110, N.W. 1/2 N. 56, S. by E. 78. The latitude in is required, and the longitude, by both methods; the bearing and distance of Huelvit's headland, in Lat. 79° 55' N., Long. 11° 55' E., is also required.

TRAVERSE TABLE. LONGITUDE TABLE.
Courses. Dist. Diff. of Latitude. Departure. Successive Latitudes. Merid. Parts. Meridional Diff. of Lat. Diff. of Longitude.
N. S. E. W. E. W.
W.N.W. 154 58.9 ... ... 142.3 78° 15' 7817 ... ... ...
S.W. 96 ... 67.9 ... 67.9 79 14 8120 303 ... 731.7
N.W. ¼ W. 89 55.4 ... ... 68.8 78 6 7774 346 ... 346.0
N. by E. 110 107.9 ... 21.5 ... 79 2 8056 282 ... 343.6
N.W. ¼ N. 56 45.0 ... ... 33.4 80 50 8576 620 123.6 ...
S. by E. ¼ E. 78 ... 73.4 26.3 ... 81 35 8970 294 ... 218.0
268.2 141.3 47.8 312.4 80 22 8504 466 166.7 ...
141.3 ... ... 47.8 290.3 1639.3
126.9 ... ... 264.6 ... 290.3
1349.0
By Rule I.
Lat. from 78° 15' N. Mer. parts. 7817 Long. from 28° 14' E.
Diff. lat. 2 7 N. Diff. long. 22 29 W.
Lat. in 80 22 N. Mer. parts. 8504 Long. in 5 45 E.
Mer. diff. lat. ... 687 To find the bearing and distance of Hacluit's headland—
Log. mer. diff. lat. 687 = 2.83896 Lat. H. H. = 79° 55' N. M. P. 8347 Long. 11° 55' E.
Log. dep. 264.6 = 2.42256 Lat. ship = 80 22 N. M. P. 8504 Long. 5 45 E.
Log. true diff. lat. 126.2 5.25952 Diff. lat. = 0 27 S. M. D. L. 157 Diff. long. 6 10 E.
Log. diff. long. 143.2 = 2.10346
23° 52' W.
Long. from 23 14 E.
Long. in 4 22 E.
The error of this method, in above example, is therefore 1° 23'.
CHAP. V.—OF PARALLEL SAILING.

When the course is 8 points or 90° from the meridian, —i.e., due E. or W.,—the true difference of latitude becomes = 0, and the rules we have investigated in chaps. iii. and iv. fail to give any result. In this case the ship sails on a parallel of latitude. We have already proved in chap. i, that, neglecting the earth's oblateness, the arc of a parallel, in any given latitude, intercepted between two meridians, is equal to the corresponding arc of the equator, in other words, the difference of longitude, multiplied by the cosine of the latitude.

Whence we derive these three formulae for parallel sailing:—

\begin{aligned} \text{Distance} &= \text{diff. longitude} \times \cos \text{ latitude}; \\ \text{Cos latitude} &= \text{distance} \div \text{diff. longitude}; \\ \text{Diff. longitude} &= \text{distance} \times \sec \text{ latitude}. \end{aligned}

Problems in parallel sailing may be solved by construction; for it is evident that we have only to construct a right-angled triangle whose hypotenuse is the difference of longitude, one of the sides the distance, and the angle between this side and the hypotenuse the latitude. Also it is evident, that in a traverse table, if we consider the latitude a course, and the difference of longitude a distance, the distance will be a true difference of latitude.

Ex. 1.—Required the number of miles contained in a degree of longitude in latitude 55° 58'.

By Construction.—Draw the indefinite right line AB (fig. 24); make the angle BAC equal to the given latitude 55° 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.

Figure 24: A geometric construction for parallel sailing. It shows a right-angled triangle ABC. The hypotenuse is AC, representing the difference of longitude at the equator (60 miles). The angle BAC is the latitude (55° 58'). A perpendicular line CB is drawn from C to the hypotenuse AB. The length of AB is the distance in a degree of longitude at the given latitude, which is 33.5 miles.
Fig. 24.

By calculation—

\begin{aligned} L \cos \text{ lat} &= 55^\circ 58' = 9.7479360 \\ \text{Log. miles in a degree} &= 60 = 1.7781513 \\ &= 10 \end{aligned}
\text{Log. miles in a deg. in lat. } 55^\circ 58' = 1.5260873

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

By Gunter's Scale.—The extent from 90° to 34°, the complement of the given latitude on the line of sines, will reach from 60 to 33.6 on the line of numbers.

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

Ex. 2.—Required the compass course and distance from A to B.

\begin{aligned} \text{Given lat. A} &= 17^\circ 30' \text{ S.} & \text{Long. A} &= 9^\circ 12' \text{ E.} \\ \text{lat. B} &= 17^\circ 30' \text{ S.} & \text{Long. B} &= 10^\circ 42' \text{ E.} \end{aligned}

Variation 1½ E., and deviation as in the table on p. 13. The true course is due E.

\begin{aligned} \text{Also, long. A} &= 19^\circ 12' \text{ E.} \\ \text{long. B} &= 10^\circ 42' \text{ E.} \\ \text{Diff. long.} &= 1^\circ 30' = 90 \text{ E.} \end{aligned}
\begin{aligned} \text{Log. diff. long.} &= 90 = 1.954243 \\ L \cos \text{ lat.} &= 17^\circ 30' = 9.979419 \\ &= 10 \\ \text{Log. dist.} &= 85.8 \text{ m.} = 1.933662 \end{aligned}

To find compass course.

Pts. qrs.
True course 8 0 right of N.
Variation 1 3 left of N.
Deviation by table 55° 58' E., or 0 6 1 right of N., or E.N.E. ¼ E.
0 3 left of N.
Compass course 5 2 r. of N., or N.E. by E. ¼ E.

Ex. 3.—A ship sails from Treguer in France, Long. 3° 14' W., to Gaspey Bay, Long. 64° 27' W., the common Lat. being 48° 47' N.; required the distance run.

Longitude from..... 3° 14' W.
Longitude in..... 64 27 W.
61 13=3673 W.
L cos latitude..... 48° 47' = 9.8188250
Log. diff. longitude..... 3673 = 3.5650209
-10
Log. distance run..... 2420 = 3.3838459

Ex. 4.—A ship from Cape Finisterre, Lat. 42° 52' N., Long. 9° 17' W., sailed due W. 342 miles; required the longitude in.

By Construction.—Draw the straight line AB (fig. 25), equal to the given distance 342 miles, and make the angle BAC equal to 42° 52', the given latitude; from B draw BC perpendicular to AB, meeting AC in C; then AC applied to the scale will measure 466½, the difference of longitude required.

L cosec lat..... 42° 52' = 10.13493
Log. distance..... 342 = 2.53403
-10
Log. diff. long. 466.6 = 2.66896
Long. Cape Finisterre..... 9° 17' W.
Diff. longitude..... 7 47 W.
Longitude in..... 17 4 W.
Figure 25: A right-angled triangle ABC with the right angle at B. Side AB is horizontal, BC is vertical, and AC is the hypotenuse. Point C is on the extension of AB.

Fig. 25.

Ex. 5.—A ship sailed due E. 358 miles, and was found by observation to have differed her longitude 8° 42'; required the parallel of latitude.

By Construction.—Make the line AB (fig. 26) equal to the given distance; to which let BC be drawn perpendicular, with an extent equal to 52½, the difference of longitude; describe an arc 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 48½, the required latitude.

Log. dist. .... 358 + 10 = 12.55383
Log. diff. long. 512 = 2.71767
L cos lat.... 48° 42' = 9.83621
Figure 26: A right-angled triangle ABC with the right angle at B. Side AB is horizontal, BC is vertical, and AC is the hypotenuse. Point C is on the extension of AB.

Fig. 26.

Ex. 6.—From two ports in Lat. 33° 58' N., distance 348 miles, two ships sail directly N. till they are in Lat. 48° 23' N.; required their distance.

By Construction.—Draw the lines CB, CE (fig. 27), 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 arc intersecting CE in E; then EF drawn from the point E, perpendicular to CP, will represent the distance required; which being applied to the scale, will measure 278½ miles.

By Calculation, as under:—

Log. given distance ..... 348 miles = 2.54168
L cos lat. in..... 48° 23' = 9.82226
12.36384
L cos lat. from ..... 33° 58' = 9.91874
Log. distance required..... 278.6 miles = 2.44510
Figure 27: A diagram showing a horizontal line CP. Points B and E are above CP. A vertical line BD is drawn from B to CP. A line CE is drawn from C through E to CP. An arc from C passes through E. A vertical line EF is drawn from E to CP.

Fig. 27.

Ex. 7.—Two ships, in Lat. 56° 0' N., distant 180 miles, sail due S.; and having come to the same parallel, are now 232 miles distant. The latitude of that parallel is required.

By Construction.—Make DB (fig. 28) 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 arc BE; and through M draw ME parallel to CD, intersecting the arc BE in E. Join EC, and draw EF perpendicular to CD; then the angle FEC will be the latitude required; which, being measured, will be found equal to 43° 53'.

Figure 28: A diagram showing a horizontal line CP. Points B and E are above CP. A vertical line DB is drawn from B to CP. A line CD is drawn from C through D to CP. An arc from C passes through B and E. A line ME is drawn from M through E to CD. A vertical line EF is drawn from E to CP.

Fig. 28.

By Calculation, as under:—

L cos lat. from..... 56° 0' = 9.74756
Log. distance on required parallel 232 miles = 2.36549
12.11305
Log. distance on known parallel... 180 = 2.25527
L cos latitude in ..... 43° 53' = 9.85778

CHAP. VI.—OF MIDDLE-LATITUDE SAILING.

It has been already explained in chap. ii. that the departure is greater than the intercepted arc of the parallel of the higher latitude, and less than that of the parallel of the lower of two places between which a ship sails; but that there is an intermediate parallel, the arc of which is exactly equal to the departure. This parallel is supposed to pass through the middle point between the extreme latitudes; and hence the latitude of this point is called the middle latitude. The relations between course, distance, departure, and true difference of latitude are to be found as in chap. iii.; and the relation between the departure and difference of longitude is given by the above considerations, viz.—

\text{Departure} = \text{diff. long.} \times \cos \text{ mid. latitude.}
\text{But departure} = \text{true diff. lat.} \times \tan \text{ course.}

Hence we get

\text{True diff. lat.} \times \tan \text{ course} = \text{diff. long.} \times \cos \text{ mid. lat.} \quad (\text{A.}) \text{also departure} = \text{distance} \times \sin \text{ course.}

Whence also

\text{Distance} \times \sin \text{ course} = \text{diff. long.} \times \cos \text{ mid. lat.} \quad (\text{B.})

If ABC (fig. 29) be the triangle for plane sailing, where AB is the true difference of latitude, AC the distance, BAC the course, and BC the departure; at C make BCD equal to the middle latitude, and produce CD to meet AB produced in D; then CD is evidently the true difference of longitude, and all the problems may be resolved and constructed by the two triangles which have a common side, viz., the departure BC.

Figure 29: A triangle ABC with the right angle at C. Side AB is vertical, AC is horizontal, and BC is the hypotenuse. A line CD is drawn from C perpendicular to AB, meeting it at D.

Fig. 29.

Also problems in middle-latitude sailing may be solved by the traverse table; for the relations between middle latitude, difference of longitude, and departure, are the same as those between course, distance, and true difference of latitude, and may therefore be found at once by inspection from the table.

Ex. 1.—Required the compass course and distance from the Island of May, in Lat. 56° 12' N. and Long. 2° 37' W., to the Naze of Norway, in Lat. 57° 50' N., and Long. 7° 27' E.; variation 2½ W.

Latitude Isle of May ..... 56° 12' N. 56° 12'
Latitude Naze of Norway ..... 57 50 N. 57 50
Difference of latitude..... 1 38=96° N. 114 2
Middle latitude ..... .....57 1
Longitude Isle of May..... 2° 37' W.
Longitude Naze of Norway..... 7 27 E.
Difference of longitude..... 10 4=604° E.

By Construction.—Draw the right line AD (fig. 30) to represent the meridian of the May; with the chord of 60° describe the arc 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 AD; make BA, equal to 96°, 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.

Figure 30: A diagram showing a vertical line AD. Points B and C are to the right of AD. A line CB is drawn from C perpendicular to AD. A line BA is drawn from B to AD. An arc from C passes through B and n. A line DC is drawn from D through n to C. A line mn is drawn from m to n on the arc.

Fig. 30.

By Calculation.—To find the course.
L \cos \text{ mid. latitude} 57^\circ 1' = 9.73591
\text{Log. diff. longitude} 694 miles = 2.78104
12.51695
\text{Log. true diff. latitude} 98 = 1.99123
L \tan \text{ course} 73^\circ 24' = 10.52572
Or course = N. 73^\circ 24' \text{ E.}
To find the distance.
L \sec \text{ course} 73^\circ 24' = 10.54411
\text{Log. diff. latitude} 98 miles = 1.99123
-10
\text{Log. distance} 343 = 2.53534
The true course is N. 73^\circ 24' \text{ E.}, or E.N.E. \frac{1}{2} E. nearly.
Pts. qrs.
True course 6 3 right of N.
Variation 2 2 right of N.
Or 9 1 right of N., or E.S.E. \frac{1}{2} E.
Or 6 3 left of S.
Deviation 0 2 left of S.
Compass course 7 1 or E. \frac{1}{2} S.

Ex. 2.—A ship from Brest, in Lat. 48^\circ 23' \text{ N.}, and Long. 4^\circ 30' \text{ W.}, sailed S.W. \frac{1}{2} W. 238 miles; required the latitude and longitude in.

By Construction.—With the course and distance construct the triangle ABC (fig. 31), and the difference of latitude AB being measured will be found equal to 142 miles; hence the latitude in is 46^\circ 1' \text{ N.}, and the middle latitude 47^\circ 12'. Now make the angle DCB equal to 47^\circ 12'; and DC being measured will be 281, the difference of longitude; hence the longitude in is 5^\circ 11' \text{ W.}

Figure 31: A geometric construction for finding latitude and longitude. It shows a triangle ABC where AB is the vertical side (difference of latitude), BC is the horizontal side (difference of longitude), and AC is the hypotenuse (distance). A line CD is drawn from C perpendicular to AB, and an angle DCB is constructed equal to the middle latitude.
Fig. 31.
By Calculation.—To find the difference of latitude.
L \cos \text{ course} 4 \frac{1}{2} pts. = 9.77503
\text{Log. distance} 238 miles = 2.37658
-10
\text{Log. true diff. latitude} 141.8 = 2.15161
Lat. Brest 48^\circ 23' \text{ N.} 45^\circ 23' \text{ N.}
Diff. latitude 2 22 S. Half.....1 11 S.
Latitude in 46 1 N. Mid. lat. 47 12 N.
To find the difference of longitude.
\text{Log. distance} 238 = 2.37658
L \sin \text{ course} 4 \frac{1}{2} pts. = 9.90483
12.28141
\text{Log. cos mid. latitude} 47^\circ 12' = 9.83215
\text{Log. diff. longitude} 281.3 = 2.44926
Long. Brest 4^\circ 30' \text{ W.}
Diff. longitude 4 41 W.
Longitude in 9 11 W.

Ex. 3.—A ship from St Antonio, in Lat. 17^\circ 0' \text{ N.} and Long. 24^\circ 25' \text{ W.}, sailed N.W. \frac{1}{2} N., till, by observation, her latitude was found to be 28^\circ 34' \text{ N.}; required the distance sailed, and longitude come to.

Latitude St Antonio 17^\circ 0' \text{ N.} 17^\circ 0' \text{ N.}
Latitude by observation 28 34 N. 28 34 N.
Diff. of latitude 11 34 = 694' 45 34 N.
Middle latitude 22 47 N.

By Construction.—Construct the triangle ABC (fig. 32), 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.

Figure 32: A geometric construction for finding distance and longitude. It shows a triangle ABC where AB is the vertical side (difference of latitude), BC is the horizontal side (difference of longitude), and AC is the hypotenuse (distance). A line CD is drawn from C perpendicular to AB, and an angle BCD is constructed equal to the middle latitude.
Fig. 32.
By Calculation.—To find the distance.
L \sec \text{ course} 31 pts. = 10.09517
\text{Log. diff. latitude} 694 miles = 2.84136
-10
\text{Log. distance} 864 = 2.93653
To find the difference of longitude.
L \tan \text{ course} 31 pts. = 9.87020
\text{Log. diff. latitude} 694 miles = 2.84136
12.71156
L \cos \text{ mid. latitude} 22^\circ 47' = 9.98472
\text{Log. diff. longitude} 558.3 = 2.74684
Long. St Antonio 24^\circ 25' \text{ W.}
Diff. longitude 9 18 W.
Longitude in 33 43 W.

Ex. 4.—A ship from Lat. 26^\circ 30' \text{ N.}, and Long. 45^\circ 30' \text{ W.}, sailed N.E. \frac{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. 33), and the distance and difference of latitude being measured will be found equal to 340 and 263 respectively. Hence the latitude in is 30^\circ 53', and middle latitude 28^\circ 42'. Now make the angle BCD equal to the middle latitude, and the difference of longitude DC applied to the scale will measure 246'.

Figure 33: A geometric construction for finding distance and longitude. It shows a triangle ABC where AB is the vertical side (difference of latitude), BC is the horizontal side (distance), and AC is the hypotenuse (distance). A line CD is drawn from C perpendicular to AB, and an angle BCD is constructed equal to the middle latitude.
Fig. 33.
By Calculation.—To find the distance.
L \sec \text{ course} 31 pts. = 10.19764
\text{Log. departure} 216 miles = 2.33445
-10
\text{Log. distance} 340.5 = 2.53109
To find the true difference of latitude.
L \cot \text{ course} 31 pts. = 10.08583
\text{Log. departure} 216 miles = 2.33445
-10
\text{Log. true diff. latitude} 263.2 = 2.42028
Latitude from 26^\circ 30' \text{ N.} 26^\circ 30'
Diff. latitude 4 23 N. Half.....2 12 N.
Latitude in 30 55 N. Mid. lat. 28 42 N.
To find the difference of longitude.
L \sec \text{ mid. latitude} 28^\circ 42' = 10.05693
\text{Log. departure} 216 miles = 2.33445
-10
\text{Log. diff. longitude} 246.2 = 2.39138
Longitude from 45^\circ 30' \text{ W.}
Diff. longitude 4 6 E.
Longitude in 41 24 W.

Ex. 5.—From Cape Sable, in Lat. 43^\circ 24' \text{ N.}, and Long. 65^\circ 39' \text{ W.}, a ship sailed 246 miles on a direct course between the S. and E., and was then by observation in Lat. 40^\circ 48' \text{ N.}; required the course, and longitude in.

Latitude Cape Sable 43^\circ 24' \text{ N.} 43^\circ 24' \text{ N.}
Latitude by observation 40 48 N. 40 48 N.
Diff. of latitude 2 36 = 156 S. Sum 84 12 N.
Middle latitude 42 6 N.

By Construction.—Make AB (fig. 34) 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^\circ 6', the middle latitude. Now DC will be found to measure 246, and the course or angle A will measure 50^\circ 39'.

Figure 34: A geometric construction for finding course and longitude. It shows a triangle ABC where AB is the vertical side (difference of latitude), BC is the horizontal side (distance), and AC is the hypotenuse (distance). A line CD is drawn from C perpendicular to AB, and an angle BCD is constructed equal to the middle latitude.
Fig. 34.
By Calculation.—To find the course.
\text{Log. diff. latitude} 156 + 10 = 12.19312
\text{Log. dist.} 246 = 2.39093
L \cos \text{ course} 50^\circ 39' = 9.88219
To find the difference of longitude.
\text{Log. dist.} 246 = 2.39093
L \sin \text{ course} 50^\circ 39' = 9.88834
12.27927
L \cos \text{ mid. latitude} 42^\circ 6' = 9.87039
\text{Log. diff. longitude} 256.4 = 2.4088

Longitude from ..... 65° 39' W.
Diff. longitude ..... 4 16 E.
Longitude in ..... 61 23 W.

Ex. 6.—A ship from Cape St Vincent, in Lat. 37° 2' N., Long. 9° 2' W., sails between the S. and W.; the latitude in is 18° 16' N., and departure 838 miles; required the course and distance run, and longitude in.

Latitude Cape St Vincent 37° 2' N. 37° 2' N.
Latitude in ..... 18 16 N. 18 16 N.
Difference of latitude 18 46 = 1126 S. Sum 55 18 N.
Middle latitude... 27 39 N.

By Construction.—Make AB (fig. 35) 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 364°, and the distance and difference of longitude, measured on the line of equal parts, will be found to be 1403 and 946 respectively.

Figure 35: A geometric diagram for middle latitude sailing. It shows a vertical line segment AB representing the difference of latitude. A horizontal line segment BC represents the departure. A line segment AC is drawn from A to C. A line segment CD is drawn from C to D, making an angle with CB equal to the middle latitude. The diagram is used to find the course and distance.

Fig. 35.

By Calculation.—To find the course.
Log. departure ..... 838 + 10 = 12.92324
Log. diff. latitude 1126 = 3.05154
L tan course ..... 36° 39' = 9.87170

To find the distance.
L sec course ..... 36° 39' = 10.09566
Log. diff. latitude ..... 1126 = 3.05154
-10
Log. dist. .... 1403 = 3.14720

To find the difference of longitude.
Log. departure ..... 838 = 2.92324
L sec mid. latitude ..... 27° 39' = 10.05266
-10
Log. diff. longitude ..... 946 = 2.97590
Longitude from ..... 9° 2' W.
Diff. longitude ..... 15 46 W.
Longitude in ..... 24 48 W.

Ex. 7.—A ship from Bordeaux, in Lat. 44° 50' N., and Long. 6° 35' W., sailed between the N. and W. 374 miles, and made 210 miles of westing; required the course, and the latitude and longitude in.

By Construction.—With the given distance and departure make the triangle ABC (fig. 36). Now the course, being measured on the line of chords, is about 341°, and the difference of latitude on the line of numbers is 309 miles; hence the latitude in is 49° 59' N., and middle latitude 47° 25'. Then make the angle BCD equal to 47° 25', and DC being measured, will be 310 miles, the difference of longitude.

Figure 36: A geometric diagram for middle latitude sailing. It shows a vertical line segment AB representing the difference of latitude. A horizontal line segment BC represents the departure. A line segment AC is drawn from A to C. A line segment CD is drawn from C to D, making an angle with CB equal to the middle latitude. The diagram is used to find the course and distance.

Fig. 36.

By Calculation.—To find the course.
Log. departure ..... 210 + 10 = 12.32222
Log. distance ..... 374 = 2.57287
L sin course ..... 34° 10' = 9.74935
To find the true difference of latitude.
L cos course ..... 34° 10' = 9.91772
Log. distance ..... 374 = 2.57287
-10
Log. diff. latitude ..... 309.4 = 2.49059
Latitude from ..... 44° 50' N. 44° 50' N.
True diff. latitude ..... 5 9 N. Half ..... 2 35 N.
Latitude in ..... 49 59 N. Mid. lat. 47 25 N.

To find the difference of longitude.
L sec mid. latitude ..... 47° 25' = 10.16963
Log. departure ..... 210 = 2.32222
-10
Log. diff. longitude ..... 310.3 = 2.49185

Longitude from ..... 6° 35' W.
Diff. longitude ..... 5 10 W.
Longitude in ..... 5 45 W.

Ex. 8.—A ship from Lat. 54° 55' N., Long. 1° 10' W., sailed between the N. and E. till, by observation, she was found to be in Long. 5° 28' E., and has made 220 miles of easting; required the latitude in, course, and distance run.

Longitude from ..... 1° 10' W.
Longitude in ..... 5 28 E.
Difference of longitude 6 38 = 396 E.

By Construction.—Make BC (fig. 37) equal to the departure 220, and CD equal to the difference of longitude 396; then the middle latitude BCD being measured, will be found equal to 56° 15'; hence the latitude come to is 57° 34', and difference of latitude 158'. Now make AB equal to 158, and join AC, which, applied to the scale, will measure 271 miles. Also the course BAC, being measured on chords, will be found equal to 54°.

Figure 37: A geometric diagram for middle latitude sailing. It shows a vertical line segment AB representing the difference of latitude. A horizontal line segment BC represents the departure. A line segment AC is drawn from A to C. A line segment CD is drawn from C to D, making an angle with CB equal to the middle latitude. The diagram is used to find the course and distance.

Fig. 37.

By Calculation.—To find the middle latitude.
Log. diff. of longitude ..... 396 + 10 = 12.59769
Log. departure ..... 220 = 2.34242
L sec mid. latitude ..... 56° 15' = 10.25527
Double middle latitude ..... 112° 30' N.
Latitude from ..... 54 55 N.
Latitude in ..... 57 34 N.
True diff. latitude ..... 2 38 = 158 miles N.

To find the course.
Log. departure ..... 220 + 10 = 12.34242
Log. diff. latitude ..... 158 = 2.19866
L tan course ..... 54° 19' = 10.14376

To find distance.
L sec course ..... 54° 19' = 10.23410
Log. diff. latitude ..... 158 = 2.19866
-10
Log. distance ..... 270.9 = 2.43276

Ex. 9.—A ship from a port in N. Lat., sailed S.E. 18. 438 miles, and differed her Long. 7° 28'; required the latitudes from and in.

By Construction.—With the course and distance construct the triangle ABC (fig. 38), and make DC equal to 448, the given difference of longitude. Now the middle latitude BCD will measure 48° 58', and the difference of latitude AB 324 miles; hence the latitude from is 51° 40', and latitude in 46° 16'.

Figure 38: A geometric diagram for middle latitude sailing. It shows a vertical line segment AB representing the difference of latitude. A horizontal line segment BC represents the departure. A line segment AC is drawn from A to C. A line segment CD is drawn from C to D, making an angle with CB equal to the middle latitude. The diagram is used to find the course and distance.

Fig. 38.

By Calculation.—To find the true difference of latitude.
L cos course ..... 31 pts. = 9.86979
Log. distance ..... 438 miles = 2.64147
-10
Log. true diff. latitude ..... 324.5 = 2.51125

To find middle latitude.
Log. distance ..... 438 ms. = 2.64147
L sin course ..... 31 pts. = 9.82708
Log. diff. longitude ..... 448 = 2.65128
L cos mid. latitude ..... 48° 58' = 9.81727
Mid. latitude ..... 48° 58' N.
Half diff. latitude ..... 2 42 S.
Latitude from ..... 51 40 N.
Latitude in ..... 46 16 N.

CHAP. VII.—OF OBLIQUE SAILING.

Oblique sailing is the application of oblique-angled plane triangles to the solution of problems at sea. This sailing

Oblique
Sailing.

will be found particularly useful in going along shore, and in surveying coasts and harbours.

Ex. 1.—At 11 a.m. the Girdle Ness bore W.N.W., and at 2 p.m. it bore N.W. by N.; the course during the interval S. by W. five knots an hour; required the distance of the ship from the Ness at each station.

By Construction.—Describe the circle NESW (fig. 39), and draw the diameters NS, EW at right angles to each other. From the centre C, which represents the first station, draw the W.N.W. line CF; and from the same point draw CH, S. by W., and equal to 15 miles, the distance sailed. From H draw HF in a N.W. by N. direction, and the point F will represent the Girdle Ness. Then the distances CF, HF will measure 19.1 and 26.5 miles respectively.

Figure 39: A circle with center C and points N, E, S, W on the circumference. A vertical diameter NS and a horizontal diameter EW are drawn. A line CF is drawn from C to the circle in the W.N.W. direction. A line CH is drawn from C to the circle in the S. by W. direction. A line HF is drawn from H to the circle in the N.W. by N. direction.
Fig. 39.

By Calculation.—In the triangle FCH are given the distance CH 15 miles, the angle FCH equal to 9 points, the interval between the S. by W. and W.N.W. points, and the angle CHF equal to 4 points, being the supplement of the angle contained between the S. by W. and N.W. by N. points. Hence CFH is 3 points; to find the distances CF, HF.

To find the distance CF.
Log. CH..... 15 m. = 1.17609
L sin CHF..... 4 pts. = 9.84948
4.02557
L sin CFH..... 3 pts. = 9.74474
Log. CF..... 19.07 m. = 1.29083
To find the distance FH.
Log. CH..... 15 m. = 1.17609
L sin FCH..... 9 pts. = 9.99157
11.16766
L sin CFH..... 3 pts. = 9.74474
Log. FH..... 25.48 m. = 1.42292

Ex. 2.—Running up Channel E. by S. per compass at the rate of 5 knots an hour. At 11 a.m. the Eddystone Lighthouse bore N. by E., and the Start Point N.E. by E.; and at 4 p.m. the Eddystone bore N.W. by N., and the Start N.W. by 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. 40) represent the first station, from which draw the N. by E. line CA, the N.E. by E. line CB, and the E. by S. line CD, which make equal to 25 miles, the distance run in the elapsed time. Then from D draw the N.W. by N. line DA, intersecting CA in A, which represents the Eddystone; and from the same point draw the N.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 73½°.

Figure 40: A geometric construction showing a point C with three lines CA, CB, and CD drawn at 45-degree angles. A line DA is drawn from D to intersect CA at A. A line DB is drawn from D to intersect CB at B. A line AB is drawn between A and B.
Fig. 40.
By Calculation—
The angle ACD=ACE+ECD=NCE-NCA+ECD = 8-1½+1 pt.=7½ pts.
BCD=NCE-NCB+ECD=8-5½+1 pt.=3½ pts.
ACB=ACD-BCD..... =4 pts.
ADC=N'DC-N'DA..... =7-3 pts.=4 pts.
N'DB..... =½ pts.
and CDB=N'DC+N'DB..... =7½ pts.
Also, CAD=16 pts.-ACD-ADC =16-7½-4 pts.=4½ pts.
and CBD=16-BCD-CDB..... =16-3½-7½=4½ pts.
To find AC.
Log. CD..... 25 m. = 1.39794
L sin ADC..... 4 pts. = 9.84948
11.24742
L sin CAD..... 4½ pts. = 9.86879
Log. AC..... 23.86 m. = 1.37763
To find BC.
Log. CD..... 25 m. = 1.39794
L sin BDC..... 7½ pts. = 9.99948
11.39742
L sin CBD..... 4½ pts. = 9.86818
Log. BC..... 32.30 m. = 1.50924
To find BAC.
BC-AC..... = 8.44
BC+AC..... = 56.16
Log. (BC-AC)..... 8.44 m. = 0.92634
L cot ½ACB..... 2 pts. = 10.38278
11.30912
Log. (BC+AC)..... 56.16 = 1.74943
L tan ½(BAC-ABC)..... 19° 56' = 1.55969
½(BAC+ABC)..... 67 30
∴ BAC..... 87 26
ABC..... 47 34

Hence BAF = BAC-CAF
= 87° 26' - 1½ pts.
= 87° 26' - 14' 4"
= 73° 22' = 73½° nearly.

To find AB, or the distance.
Log. BC..... 32.30 m. = 1.50924
L sin ACB..... 4 pts. = 9.84948
11.35972
L sin CAB..... 87° 26' = 9.99956
Log. AB..... 22.9 m. = 1.36016

Many other examples might be given. These and all other cases which can occur in practice are to be resolved by plane trigonometry, from calculating the triangles which the data of the given case afford.

CHAP. VIII.—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.

Ex.—The wind at N.W., 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 of each tack.

By Construction.—Draw the N.W. line CA (fig. 41) equal to 64 miles; from C draw CB W. by S., and from A draw AD parallel thereto and in an opposite direction. Bisect AC in E, and draw BED parallel to the N. by 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.

Figure 41: A geometric construction showing a point C with a line CA drawn at a 45-degree angle. A line CB is drawn from C in the opposite direction of CA. A line AD is drawn from A parallel to CB. A line BED is drawn from B through E to D, where E is the midpoint of AC. Points B and D are on CB and AD respectively.
Fig. 41.

CHAP. IX.—OF CURRENT SAILING.

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

In a calm, it is evident a ship will be carried in the direction and with the velocity of the current. Hence if a ship sails in the direction of the current, her rate will be augmented by the rate of the current; but if sailing directly against it, the distance made good will be equal to the difference between the ship's rate as given by the log and that of the current. And the absolute motion of the ship will be ahead 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.

It is evident from the above observations that we may consider the direction of the current in the light of a separate course; and by multiplying the rate of the current per hour by the number of hours it has been running, and treating this as a distance, we may estimate the ship's real place by any of the rules for compound courses.

Ex. 1.—A ship sailed N.N.E. at the rate of 8 knots an hour during 18 hours, in a current setting N.W. by W. 2½ miles an hour; required the course and distance made good.

By Construction.—Draw the N.N.E., line CA (fig. 42) equal to 18 × 8 = 144 miles; and from A draw AB parallel to the N.W. by 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'.

Figure 42: A geometric diagram showing a ship's course and drift. Point C is the origin. A line CA is drawn at an angle of 45 degrees (N.N.E.). Point A is at the end of CA. A line AB is drawn parallel to the N.W. by W. rhumb. Point B is at the end of AB. A line BC is drawn to find the distance made good. A dashed line extends from C to the right, and a dashed line extends from A to the right, forming a horizontal reference line.

Fig. 42.

  • The angle CAB..... = 9 pts.
  • CA..... = 144 m.
  • AB..... = 45 m.
  • CA + AB..... = 189 m.
  • CA - AB..... = 99 m.
  • Log. (CA - AB)..... 99 m. = 1.995635
  • L cot ¼ CAB..... 4½ pts. = 9.914173
  • 11.909808
  • Log. CA + CB..... 185 m. = 2.276462
  • L tan ¼ (ABC - ACB)..... 23° 15' = 9.633346
  • ¼ (ABC + ACB)..... = 39 22
  • ∴ ACB = 16° 7' and ABC..... = 62 37
  • NCA..... = 22 30
  • ∴ NCB the course..... = 6 23
  • Log. AB..... 45 m. = 1.653212
  • L sin CAB..... 9 pts. = 9.991574
  • 11.644786
  • L sin ACB..... 16° 7' = 9.443410
  • Log. BC..... 159 m. = 2.201376

For first course we have course N.N.E., or 2 points.
And distance..... 144 Whence, from traverse table,
True diff. latitude..... 133.0 N. Departure..... 55.1 E.

For second course we have course N.W. by W., or 5 points.
And distance..... 45 And from traverse table,
True diff. latitude..... 25 N. Departure..... 37.4 W.

  • True diff. latitude..... 158 N.
  • Departure..... 17.7 W.
  • Log. departure..... 17.7 + 10 = 11.24797
  • Log. true diff. latitude..... 158 = 2.19866
  • L tan course..... N. 6° 23' E. = 9.04931
  • L sec course..... 6° 23' - 10 = .00271
  • Log true diff. latitude..... 158 m. = 2.19866
  • Log. distance..... 159 = 2.20137

Ex. 2.—A ship from Lat. 35° 20' N. sailed 24 hours in a

current setting N.W. by N., and by account is in latitude 35° 42' N., having made 44 miles of easting; but the latitude by observation is 35° 58' N.; required the course and distance made good, and the drift of the current.

Figure 43: A geometric diagram showing a ship's course and drift. Point C is the origin. A line CA is drawn. Point A is at the end of CA. A line AB is drawn parallel to the N.W. by N. rhumb. Point B is at the end of AB. A line CB is drawn to find the distance made good. A dashed line extends from C to the right, and a dashed line extends from A to the right, forming a horizontal reference line.

Fig. 43.

By Construction.—Make CE (fig. 43) equal to 22 miles, the difference of latitude by dead reckoning, 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 N.W. 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 41½°.

By Calculation—

  • ABF..... = 3 pts.
  • BF = CD - CE..... = 16 miles.
  • To find AB.
  • Log. BF..... 16 m. = 1.20412
  • L sec ABF..... 3 pts. - 10 = 0.08015
  • Log. AB..... 19.2 m. = 1.28427
  • Or drift of current..... 19.2 miles.
  • To find AF.
  • Log. BF..... 16 m. = 1.20412
  • L tan ABF..... 3 pts. - 10 = 9.82489
  • Log. AF..... 10.7 m. = 1.02901
  • Hence BD = AE - AF = 44 - 10.7 = 33.4.
  • To find the course.
  • Log. BD..... 33.4 + 10 = 11.52244
  • Log. CD..... 38 = 1.57978
  • L tan course..... N. 41° 14' E. = 9.94266
  • To find the distance.
  • Log. sec course..... 41° 14' E. - 10 = 0.12376
  • Log. CD..... 38 = 1.57978
  • Log. distance..... 50.5 m. = 7.0354

By Traverse Table.—Taking the current course first, true difference of latitude 16, and course N.W. by N., we find in the traverse table the corresponding distance 19.3, and departure 10.7.

Again, for second course, we have true difference of latitude 38, and departure 44 - 10.7 = 33.4 E.

Points. Course Distance. Diff. of Latitude. Departure.
N. S. E. W.
3 N.W. by N. 19.3 16 ... ... 10.7
N. 41° E. ... 51 38 ... 33.3 ...

Whence the course and distance are found as above.

Or, from the traverse table to nearest degree and minute, we find in the columns of distance and angle, opposite to difference of latitude 38.5, and departure 35.5—distance 51, and angle 41°.

CHAP. X.—OF THE DAY'S WORK AND SHIP'S JOURNAL.

The most usual application of the principles laid down in the preceding chapters, is to ascertain from the several courses and distances run by a ship in the interval between the noons of two successive days, the ship's place at the noon of the latter day, i.e., its latitude and longitude; its latitude and longitude being given for the noon of the preceding day. This constitutes a day's work; and the ship's place deduced therefrom is called her place by account or dead reckoning. The day aboard ship, like the astronomical day, commences at noon; and the ship's position is always calculated at every noon. In the Royal Navy, the log is hove once in every hour; but in most trading-vessels only once in every two hours. A record of the knots, and tenths of knots, run every hour or every two hours, the course, the direction of the wind, the leeway, and everything which affects the ship's place, is kept in the journal, which, for this purpose, is usually divided into six or seven columns. The first column on the left hand contains the hours

from noon to noon; the second and third, the knots and tenths of knots sailed every hour, or every two hours; the fourth contains the courses steered; the fifth, the direction of the wind; and the sixth, when there are seven columns, contains the leeway; and the last contains general remarks, including phenomena, variation, &c., &c.

The mode of forming a table showing the deviation of the compass for the several positions of the ship's head, has already been given.

The courses steered, as entered in the log-book, must be corrected for variation, deviation, and leeway. The setting and drift of current, and the heave of the sea, are to be marked down. These are to be corrected for variation only. In the day's work, it is usual to treat a current as an independent course and distance. If the ship does not sail from a place whose latitude and longitude are known (which rarely happens), the bearing of some known place is to be observed, and its distance found, which is usually done by estimation. The ship is then supposed to have taken her departure from this place, in a course exactly opposite to the observed bearing, and to have run the estimated distance on it. If there be any reason to suspect the correctness of the estimated distance, it will be easy to obtain the true distance as follows:—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 again to be observed. We shall then have a triangle, all of whose angles are known from the observed bearings, and one of its sides, viz., the distance the ship has sailed. The other two sides, viz., the distance of the ship from the place of departure at each of the observations, can be immediately found, as in problem 1 on "Oblique Sailing." The distances for each course may be obtained by adding together the hourly distances. The courses being thus corrected, and the distances found, the latitude and longitude may be found by any of the methods explained in chap. iv. As the differences of latitude are not usually great, the traverse table may generally be made use of for finding the latitude in; and having found the middle latitude, the longitude may be obtained by the middle latitude method.

The following example will enable the reader to apply the directions we have just given:—

Ex.—September 12, 1857, at noon, a point of land in Lat. 64^{\circ} 20' S., and Long. 59^{\circ} 40' E., bore by compass S.E., distant 15 miles (ship's head being E.), afterwards sailed as by the following log-account; find the latitude and longitude in, on September 13, at noon.

H. K. \frac{1}{2}ths. Course. Wind. Leeway. Remarks.
1 4 7 W. by N. N.E. 2 P.M.
2 6 4 ... ... ...
3 5 8 ... ... ...
4 4 ... ... ... ... Variation of the compass,
5 7 ... ... ... ... 1\frac{1}{2} W.
6 6 5 ... ... ...
7 7 4 ... ... ...
8 4 4 ... ... ... (For deviation see the table, page 13.)
9 5 6 N.N.E. N.W. 1\frac{1}{2}
10 6 ... ... ... ...
11 5 5 ... ... ...
12 6 ... ... ... ...
1 7 4 ... ... ... A.M.
2 4 2 ... ... ...
3 6 ... S.E. E.N.E. 2\frac{1}{2} During the last 7 hours a current set the ship N.W. at the rate of two knots an hour.
4 5 5 ... ... ...
5 4 ... ... ... ...
6 6 ... ... ... ...
7 4 4 ... ... ...
8 5 5 ... ... ...
9 3 8 S. by W. W. 2\frac{1}{2}
10 4 5 ... ... ...
11 7 ... ... ... ...
12 5 5 ... ... ...
Departure course, N.W. being the opposite to S.E.
Compass course..... 4 pts. 0 qrs. left of N.
Variation..... 1 2 "
Deviation..... 0 3 right, or
True course..... 4 3 N.W. \frac{1}{2} W.
Dist..... 15.
First course, W. by N.
Compass course..... 7 pts. 0 qrs. left of N.
Variation..... 1 2 "
Deviation..... 0 3 "
Leeway (wind N.E. on starboard tack)..... 2 0 "
Or true course..... 11 1 left of N.
Dist..... 4 3 right of S., or S.W. \frac{1}{2} W.
Second course, N.N.E.
Compass course..... 2 pts. 0 qrs. right of N.
Variation..... 1 2 left.
Deviation..... 0 3 right.
Leeway (wind N.W. on port tack)..... 1 3 right.
True course..... 3 0 right of N., or N.E. by N.
Dist..... 34.7.
Third course, S.E.
Compass course..... 4 pts. 0 qrs. left of S.
Variation..... 1 2 left.
Deviation..... 0 1 right.
Leeway (wind E.N.E. on port tack)..... 2 1 "
True course..... 3 0 left of S. or S.E. by S.
Dist..... 31.4.
Fourth course, S. by W.
Compass course..... 1 pt. 0 qrs. right of S.
Variation..... 1 2 left.
Deviation..... 0 2 "
Leeway (wind W. starboard tack)..... 2 2 left.
True course..... 3 2 left of S., or S.E. by S. \frac{1}{2} E.
Dist..... 20.8.
Current N.W.
Compass course..... 4 pts. 0 qrs. left of N.
Variation..... 1 2 "
Deviation..... 0 0 "
Dist..... 5 2 left of N. or N.W. by W. \frac{1}{2} W.

Enter these in a table as under:—

Points. Course. Distance. Diff. Lat. Departure.
N. S. E. W.
4\frac{1}{2} N.W. \frac{1}{2} W. 15 8.9 ... ... 12.0
4\frac{1}{2} S.W. \frac{1}{2} W. 46.2 ... 27.4 ... 26.9
3 N.E. 6 N. 34.7 29.1 ... 19.4 ...
3 S.E. 6 S. 31.4 ... 25.8 17.2 ...
3\frac{1}{2} S.E. by S. \frac{1}{2} E. 20.8 ... 16.2 13.3 ...
5\frac{1}{2} N.W. by W. \frac{1}{2} N. 14 6.6 ... ... 12.3
44.6 68.4 49.9 61.2
44.6 ... 49.9
True diff. lat. S. 23.8 Dep. W. 11.3
Lat. from..... 64^{\circ} 20' S. Lat. from..... 64^{\circ} 20' S.
T. D. Lat..... 0 23 S. Half..... 0 12 S.
Lat. in..... 64^{\circ} 43' S. Mid. Lat..... 64^{\circ} 32' S.
Log. departure..... 11.3 = 1.05308
L. sec mid. lat..... 64^{\circ} 32' - 10 = 0.36654
Log. diff. long..... 26.2 = 1.41962
Long. from..... 59^{\circ} 40' E.
Diff. long..... 0 26 W.
Long. in..... 59^{\circ} 14' E.

In this example the true differences of latitude and departures are taken by inspection from the traverse table.

When a ship is bound for a distant port, the bearing and distance of the port must be found. This may be done by calculation or by a chart. If islands, capes, or headlands intervene, it will be necessary to find the several courses and distances between each successively. The true course between the places must be reduced to the compass course

Sea Charts. by making the requisite allowances for variation and deviation, as already explained.

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 to which the vessel would be otherwise subjected, to the endangering of her masts and straining her timbers, &c. When a ship is brought to, the tiller or wheel is put down over to the leeward, which brings her head round to the wind. The wind having then little power over the sails, the ship loses her way through the water; and the action of the water on the rudder ceasing, 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 oscillating motion, coming up to the wind and falling off from it again alternately. The middle point between those upon which she comes up and falls off is taken for her apparent course; and the leeway, variation, and deviation are to be allowed from this to find the true course.

It is generally found that the latitude by account does not agree with that by observation. On considering the imperfections of the common log-line, and the uncertainty with regard to variation, an exact agreement of latitudes cannot be expected. When the difference of longitude is to be found by dead reckoning, and 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 is 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 to be in error. This last correction will, according to different authors, place the ship upon opposite sides of her meridian by account. As these corrections, therefore, are no better than guessing, they should be absolutely rejected.

If the latitudes do not agree, the navigator should 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 in 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 may be corrected in the same way.

It will be proper for the navigator to determine the longitude of the ship by 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. The modes of finding the latitude and longitude of a ship by observation, and the variation of the compass, will be given in the next book.

CHAP. XI.—OF SEA CHARTS.

The charts usually employed in the practice of navigation are the Plane and Mercator's charts. The former of these is adapted to represent a portion of the earth's surface near the equator, where the change in the lengths of corresponding arcs of the parallel is very small; and the other for all portions of the earth's surface. (For a particular description of these, see the articles CHART and GEOGRAPHY.) We shall here only describe their use.

Use of the Plane Chart.

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

Rule.—Take the least distance of the given place from the nearest parallel of latitude; this distance applied to the graduated meridian from the extremity of the parallel will give the latitude of the place. In the same way the longitude is found by taking the least distance from the nearest meridian, and applying it to the graduated parallel.

Thus the distance between Bonavista and the parallel of 15^{\circ} being laid from that parallel on the graduated meridian, will reach to 16^{\circ} 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; if a parallel ruler be used, keeping the edge of one ruler passing through the places fixed, move the other until it passes through the centre of one of the compasses on the chart; the point of the compass through which this edge passes will show the course.

Or, generally, let a line on the edge of another ruler be placed so as to be parallel to the first ruler, and to pass through the centre of a compass; it will cut the circumference in a point which will determine the course.

The interval between the places being applied to the scale will give the distance.

Thus the course from Palmas to St Vincent will be found to be about S.S.W. \frac{1}{2} W., and the distance 13\frac{1}{2}^{\circ} or 795 miles.

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

Rule.—Lay a ruler over the given place parallel to another ruler laid over one of the compasses, with one edge passing through the centre, and the other the point on the circumference which shows the course, and lay off on it the distance taken from the scale; it will give the point representing the ship's present place.

Thus, supposing a ship has sailed S.W. by W. 160 miles from Cape Palmas; then by proceeding as above, it will be found that she is in Lat. 2^{\circ} 57' N.

The reader will have no difficulty in solving various other problems by means of this chart, being, in fact, only the construction of the various problems in plane sailing on this chart.

Use of Mercator's Chart.

The method of finding the latitude and longitude of a place, and the course or bearing between two given places, is the same as in the plane chart, which see.

PROB. I.—To find the distance between two given places on the chart.

CASE 1.—When the given places are under the same meridian.

Rule.—The difference or sum of their latitudes, according as they are on the same or on opposite sides of the equator, will be the distance required.

CASE 2.—When the given places are under the same parallel.

Rule.—If that parallel be the equator, the difference or sum of their longitudes, according as they are on the same or on opposite sides of the first meridian, is the distance; otherwise take the distance between the places, lay it off upwards and downwards from the given parallel, and the intercepted degrees will be the distance between the places.

Or take an equal extent of a few degrees on the meridian on each side of the parallel; and the number of extents and parts of an extent contained between the places, multiplied by the length of an extent, will give the required distance.

CASE 3.—When the given places differ both in latitude and longitude.

Rule.—Find the difference of latitude between the given places, and take it from the equator or graduated parallel; then lay a ruler over the places, and move one point of the compass opened to the difference of latitude just found along the edge of the ruler till the other just touches a parallel; then the distance from the point of the compass on the ruler to the point of intersection of the ruler and the parallel, applied to the equator, will give the distance between the places in degrees and parts of a degree, which, multiplied by 60, will give it in miles.

PROB. II.—Given the latitude and longitude in; to find the ship's place by the chart.

Rule.—Lay a ruler over the given latitude, and lay off the given longitude from the first meridian by the edge of the ruler, and the ship's present place will be obtained.

PROB. III.—Given the course sailed from the given place, and the latitude in; to find the ship's present place on the chart.

Rule.—Lay a ruler over the place sailed from, in the direction of the given course; its intersection with the parallel of latitude in, will give the ship's present place.

PROB. IV.—Given the latitude and longitude of the place left, and the course and distance sailed; to find the ship's present place on the chart.

Rule.—Lay a ruler over the given place, in the direction of the given course, take the distance sailed from the equator, and put one point of the compass opened to this distance at the intersection of the ruler with any parallel, and the other point will reach to a certain place by the edge of the ruler. This point being kept fixed, 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 is known; and the intersection of the edge of the ruler with the parallel of this latitude will give the ship's present place.

The above problems sufficiently illustrate the use of Mercator's Chart. The reader will have no difficulty in solving other problems by means of it.

BOOK II.

CONTAINING THE METHODS OF FINDING THE LATITUDE AND LONGITUDE OF THE SHIP AT SEA, THE VARIATION OF THE COMPASS, AND TIME OF HIGH WATER.

CHAP. I.—DESCRIPTION AND USE OF INSTRUMENTS USED IN OBSERVATIONS.

SECT. I.—OF HADLEY'S SEXTANT AND QUADRANT.

The principal difference between these instruments is in the extent of the angle which can be observed by them; and in the more elaborate and careful workmanship of the latter of the two. Indeed the quadrant is only available for taking observations which determine the latitude. The distances of the moon from the sun or other heavenly body, which are frequently used for the determination of the longitude, can only be observed by the help of the sextant.

Allowing for these differences, the principle on which the quadrant and sextant are constructed is the same. In the Royal Navy sextants are almost exclusively in use, although quadrants are still employed for the observation of altitudes in many trading vessels. The sextant, therefore, will first be described, and afterwards those points in which the quadrant differs from the sextant will be explained.

The reader is supposed to be aware of the ordinary laws with regard to the propagation and reflection of light, viz.,—

that in the same medium, light is propagated in straight lines, the smallest conceivable quantity of which that can be stopped or propagated alone is called a ray; and that when a ray of light is incident on a plane reflecting surface, it is bent or reflected after incidence in such manner, that the incident and reflected rays and the straight line perpendicular to the mirror at the point of incidence (called the normal to the surface) lie all in one plane; and that the incident and reflected rays make equal angles with the normal or the surface.

Diagram of a sextant showing the optical principle. A horizontal line represents the plane of the instrument. A ray from point S strikes a mirror at point A, reflecting to point B. Another ray from point P strikes a mirror at point C, reflecting to point D. The mirrors are perpendicular to the plane. Points M, I, and O are on an arc below the plane, representing the arc of the instrument. The diagram illustrates how the sextant measures the angle between two objects S and P.
Fig. 44.

Let MO (fig. 44) be an arc of a circle, CO and CM two radii, and CI be a moveable radius carrying a plane mirror, silvered through its whole extent, firmly fixed to it; EFG another mirror, the lower part of which FG only is silvered, while the upper part EF is unsilvered, so that a ray reflected from the lower portion FG in direction FH, and a direct ray PFH through the unsilvered part EF, may be seen together by an eye at K. This mirror is fixed to the radius CM in such a manner that when the moveable radius occupies the position ACO, the two mirrors ACB and EFG are both perpendicular to the plane of the instrument, and parallel to one another.

Let now S and P be two distant objects whose angular distance is required to be found. Let the instrument be placed so that its plane passes through S and P, and that a ray from P, passing through the unsilvered glass EF, may be seen directly by an eye at K; and while in this position let the bar be moved round C, CI carrying the mirror with it until a ray from S, falling on ACB, is reflected in the direction CF, and again reflected by FG in the direction FH; so that to the eye at K the images of the two objects S and P are seen together, or coincide.

Produce SA to meet PFH in H; then SHP is the angle through which the ray SA has been deflected, and is also the angular distance between S and P. Let A'B' be the new position of the mirror AB; then ACA' is the angle through which the mirror has turned, and consequently also the angle through which CI has moved.

Now angle of deflection

\text{SHP} = \text{SCF} - \text{CFH} \\ = 180^\circ - 2\text{FCB}' - (180^\circ - 2\text{EFC})

because by law of reflection,

\text{SCA} = \text{FCB}'; \text{ and therefore}
\text{SCF} = 180^\circ - \text{SCA}' - \text{FCB}' \\ = 180^\circ - 2\text{FCB}',

and \text{EFC} = \text{GFH}; and therefore

\text{CFH} = 180^\circ - \text{EFC} - \text{GFH} \\ = 180^\circ - 2\text{EFC};
\therefore \text{SHP} = 2\text{EFC} - 2\text{FCB}';

But \text{EFC} = \text{FCB}, because EFG is parallel to ACB;

\text{or SHP} = 2\text{FCB} - 2\text{FCB}' = 2\text{ACA}'

= twice the angle through which CI has moved.

Hence if the arc OM be divided into degrees, and each degree marked as two degrees, the reading off of the arc OI will be the angle between the distant objects S and P.

Observation Instruments. An instrument constructed on this principle, whose circular arc or limb is a sixth part of a circle, and therefore capable of measuring angles up to 120^\circ, is called a sextant; if the limb contain only an eighth part of a circle, it is a quadrant, and can only measure angles up to 90^\circ.

The Sextant.
A detailed technical drawing of a sextant, labeled Fig. 45. The instrument features a large graduated arc (limb) labeled AA, which is part of a larger frame labeled PLM. A central index, labeled N, is mounted on a pivot and can be moved along the arc. The index is equipped with a vernier scale, labeled OQ, for precise reading. A horizontal mirror, labeled I, is attached to the index. Below the mirror is a horizon glass, labeled F. The instrument is supported by a sturdy frame with various adjustment screws, including a finger-screw labeled C and an adjusting-screw labeled B. The entire instrument is mounted on a base with a curved limb labeled L and M.
Fig. 45.
  1. (1.) PLM (fig. 45) is the frame of the sextant.
  2. (2.) AA the graduated arc or limb.
  3. (3.) N the index, carrying the vernier OQ.
  4. (4.) I the index-glass.
  5. (5.) F the horizon-glass.
  6. (6.) D the coloured or dark glasses between the index-glass and horizon-glass.
  7. (7.) E the coloured glasses behind the horizon-glass.
  8. (8.) K the tube or collar in which the telescope is inserted.

The frame of the sextant. The frame of the sextant consists of an arc AA, firmly attached to the two radii LP, MP, which are bound together by braces, as shown in the figure, to prevent warping and liability to bend.

The index. The index N is a flat bar of brass, and turns on the centre of the sextant; at the lower end of the index there is an oblong opening; to one side of this opening the vernier scale is attached to subdivide the divisions of the arc; at the end of the index there is a piece of brass which bends under the arc, carrying a spring to make the vernier scale lie close to the divisions. It is furnished with a finger-screw C, by which the index is fixed in any position to the limb of the instrument. There is also an adjusting-screw B attached to the index, capable of moving it with greater accuracy than the hand; this screw does not act until the index is fixed by the finger-screw C. Care must be taken not to force the adjusting-screw when it arrives at either extremity of its adjustment. When any considerable movement is required to be given to the index, the screw C at the back of the sextant must be set free; but where the index is brought nearly to the divisions required, this back screw should be tightened, and then the index gradually moved by the adjusting-screw.

The index-glass. Upon the index, and near its axis of motion, is fixed a plane speculum or mirror of glass I, quicksilvered. It is set in a brass frame, which is firmly fixed by a strong cock to the centre plate of the index, with its face 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. As has already been observed, this glass

is to receive the rays from the sun or other object, and reflect them upon the horizon-glass. It is furnished with screws at its back, the object of which is to replace it in a perpendicular position, if by any accident it has been deranged.

To the radius PL is attached a small speculum F, whose surface is parallel to the index-glass when zero on the index coincides with zero on the limb. The under part only of this speculum is silvered, the upper half being left transparent, and the back part of the frame cut away, that nothing may impede the sight through the unsilvered 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. It is set in a brass frame, which turns on axes and 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 easy, 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; this screw G is in some instruments under the glass, in others behind it, and in others at the side.

There are four coloured glasses at D, tinged red and green, each of which is set on a separate frame that the brightness of the solar image and the glare of the moon, and may be used separately or together as occasion may require. There are three more such glasses placed behind the horizon-glass at E, to weaken the rays of the sun or moon when viewed directly through the horizon-glass. The paler glass is sometimes used in observing altitudes at sea to take off the strong glare of the horizon.

The sextant is furnished with a plane tube K; and in the order to render objects distinct, it has two telescopes—one a Galileo's telescope, representing the objects erect in their natural position; the longer one, an astronomical telescope, shows them inverted. It has a large field of view; and has parallel wires placed in the principal focus, where a true image of the object viewed by it is seen; thus rendering the position of the image more exact and more easy to be read off, and is that which should be used in taking observations at sea when great accuracy is required. A little use will soon accustom the observer to the inverted position, and to manage the instrument with ease. By a telescope the contact of the images is more perfectly distinguished; and by the place of the images in the field of view, it is easy to perceive whether the sextant is held in the proper position 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 to it by two screws; by turning one 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 which slides in a socket, and, by means of a screw at the back of the sextant, may be raised or lowered so as to move the centre of the telescope to that part of the horizon-glass which shall be deemed most fit for observation. Tinged glasses are provided to screw on the eye-end of either of the telescopes or the plane tube.

The limb of the sextant is divided from right to left into 120 primary divisions, which are to be considered as degrees; the degree is subdivided in some cases into two equal parts, each of which is 30'; in others into three equal parts, each of which is 20'; and in others again into six equal parts, each of which is 10'. If the zero of the index stand exactly at one of the divisions of the limb, the

Observation Instruments. reading off in that case is immediately known. If, however, the zero of the index do not stand exactly at one of the divisions, but distant from it by a small space, the value of this space is known by means of the divisions of the vernier-plate to the left of 0.

Reading off. The vernier contains a space equal to nineteen divisions on the limb, and is divided into twenty equal parts; hence the difference between a division on the vernier and a division on the limb is one-twentieth of a division of the limb, or 1', if the interval between divisions on the limb is equal to 20'. Or supposing the limb divided into intervals of 10', and that fifty-nine divisions of the limb correspond to sixty divisions of the vernier; it is then evident that the difference between a division of the instrument and of the vernier is \frac{1}{60}th part of 10', i.e., 10". This is the most usual kind of division.

To find the actual reading off in any particular case, we must observe which division of the vernier coincides with a division of the limb; the number denoting this, multiplied by the value of the difference between a division of the limb and of the vernier, will give the additional reading. Suppose, for instance, the nearest division of the limb to the zero of the vernier to be 25° 30', and the eighth division of the vernier to be coincident with a division of the limb, the additional angle will be 80" or 1' 20", and the reading off will be 25° 31' 20".

Adjustments of the sextant. The adjustments of the sextant are to set the mirrors perpendicular to the plane of the instrument, and parallel to one another when the index is at zero; and to set the axis of the telescope parallel to the plane of the instrument.

Adjustment 1.—To set the index-glass perpendicular to the plane of the sextant.

Set the index towards the middle of the limb, and hold the sextant so that its plane is 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 the limb seen directly, the speculum is perpendicular to the plane of the instrument. If they do not appear in the same plane, i.e., if the image be seen above or below the arc itself, its position must be gradually and carefully changed by means of the screws at its back until the error is rectified.

Adjustment 2.—To set the horizon-glass perpendicular to the plane of the instrument.

Place the instrument horizontal, and direct the sight to a distant well-defined object, as the sun, so as to view it directly; then move the index until the image of the object seen by reflection is on the field of view, and move the index backwards and forwards so as to make the image pass over the object. If it pass exactly over the object, the fixed mirror is perpendicular to the plane of the instrument; if not, move the screw G until their exact coincidence takes place.

Adjustment 3.—To set the horizon-glass parallel to the index-glass when the zero of the index or vernier-plate coincides with the zero of the graduations of the limb.

Set 0 on the index exactly to 0 on the limb, and fix it in that position by the screw on the under side of it; hold the sextant with its plane vertical, and direct the sight to a well-defined part of the horizon; then if the horizon seen on the silvered part coincides with that seen through the transparent part, the horizon-glass is adjusted; but if the horizons do not coincide, the position of the glass must be altered by moving a screw placed near the fixed reflector, which gives it a motion about an axis perpendicular to the plane of the instrument.

This adjustment is seldom made, as turning the adjusting-screw too often renders this part of the instrument very apt to get out of order. It is usual, therefore, to determine the error in the reading called the Index Error.

To do this, direct the sight to the horizon, and move the index until the reflected horizon coincides with that seen

by direct vision; then the difference between 0 on the limb and 0 on the vernier-plate will be the index error, which is to be added when 0 of the vernier is to the right of 0 on the limb; otherwise subtracted.

A more accurate method than the above is to measure the sun's apparent diameter twice with the index placed alternately on the right and on the left of the zero point of the graduated limb. 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 0 is less or greater than the diameter measured with the index to the right of the beginning of the divisions. Care must be taken to measure the sun's horizontal diameter, as the vertical diameter is often affected with refraction. This must be done by keeping the plane of the instrument at right-angles to the vertical diameter of the sun.

For example, on January 2, 1857, the sun's diameter, measured with the index first to the right and secondly to the left of the zero point of division, was 33' and 32' 20" respectively, and the index error obtained by taking the semidifference is 20".

Adjustment 4.—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, or two stars of the first magnitude, be selected, whose distance is not less than 90° or 100°; make the contact of these as perfect as possible at the wire nearest the plane of the instrument; fix the index in this position; move the sextant until 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. If, however, the objects are apparently separated, or overlap one another, 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. In practice, this adjustment is usually made by means of the sun and moon. The mode of bringing the limbs of the sun and moon into contact will be explained when the use of the sextant is treated of. 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 difference of readings off in these two positions will be the angular distance between the wires. Other and better methods will readily occur to the observer on land.

The Quadrant.

It has been already observed, that this instrument differs from a sextant in the extent of the divided limb and in its rougher manufacture. It is only calculated for observing altitudes. Fig. 46 represents a quadrant of the common construction.

The frame, index, index-glass, and F the fore horizon-glass, are much the same as in the sextant. There is, besides, another horizon-glass G, called the back horizon-glass attached to the same radius as F. Instead of a tube or telescope, the quadrant is furnished with vanes or sights H and I. There are but three coloured glasses, two of which are red and the other green. They are fixed at K, as shown in the figure, when the fore horizon-glass is used.

Observation Instruments. If the back horizon-glass be used, they are transferred to N. The back horizon-glass is silvered at both ends, but has a

A detailed technical drawing of a sextant, labeled Fig. 66. The instrument is a large, triangular frame with a curved base. It features a central mirror (M) and a graduated arc (limb) at the bottom. Various adjustment screws and sighting vanes (H, I) are visible. The drawing is labeled with letters A through N to identify specific parts of the instrument.
Fig. 66.

transparent slit in the middle through which the horizon may be seen. Each of the horizon-glasses is set in a brass frame, to which there is an axis passing through the wood-work, 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 or perpendicular, according as it is the fore or back horizon-glass, 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 the other 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.

The sight-vanes H and I are perforated pieces of brass, designed to direct the sight parallel to the plane of the quadrant. 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.

The limb is divided into ninety primary divisions, which are considered as degrees, and each degree subdivided into three equal parts, which are therefore of 20' each. The vernier-plate is generally so divided as to enable the observer to read off accurately to minutes.

These consist in setting the mirrors perpendicular to the plane of the instrument, and the fore horizon-glass parallel, and the back horizon-glass perpendicular to the index-glass, when the zero of the index or vernier-plate coincides with zero of the graduations on the limb.

The adjustments for the index-glass and fore horizon-glass are performed nearly in the same way as for the sextant. The index error, however, must be ascertained by bringing the horizon by reflection into the same line with the horizon seen directly. The method by taking the distance of two stars of the first magnitude, or the sun and moon, is inapplicable here.

The back horizon-glass is so seldom used, that for its adjustments and the mode of taking observations with it, the reader is referred to Norie's Navigation, and other works in which this subject is treated.

The altitude of an object may be determined by either instrument, and is the reading off on the limb, with the proper index error applied, when by reflection that object appears to be in contact with the horizon. The distance between the sun and moon, or other heavenly bodies, may

be observed by the sextant when the limbs of the bodies whose distance is required appear to be in contact. If the quadrant be used for taking the altitude of the sun, when it is 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 be bisected 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 on the silvered part of the glass adjacent to the line of separation of the two parts.

With the quadrant two different methods of taking observations may be employed. In the first, the observer faces the sun, and looks to that part of the horizon which is immediately under the sun, and the observation is therefore called the fore observation. In the other method, the observer's back is towards the sun, and he looks to the part of the horizon opposite to that which is under the sun; and this is consequently called the back observation. It is not to be employed except when the horizon under the sun is obscured, or rendered indistinct by fog or other impediment.

In all cases of taking altitudes, it must be considered that it is necessary to be quite sure that the distance of the sun or other body from the horizon is the least possible, otherwise it would not be the altitude that is observed. Consequently, after the instrument has been placed as nearly as possible in a vertical position, and a contact made, a motion about the line of sight of the sun must be communicated to the instrument, so as to keep the image always in the same part of the silvered mirror, the plane of the instrument being inclined. In this way we keep the angular distance of the sun from the line through the eye by which it is viewed the same, and the sun's image describes a small circle, whose angular radius is this distance. The horizon being fixed and viewed directly, will always occupy the same position. If, then, on giving this vibratory motion to the instrument, the arc described by the sun touches the horizon, the angular distance observed is the altitude. If it should cut the horizon, so that a portion of the sun's image goes below it, the index must be moved back until this arc simply touches the horizon. In the back observation with the quadrant, and in observing with the sextant furnished with the inverting telescope, the images are inverted, and the arc described by the sun's image lies below the horizon, to which line it is convex.

The motion must be given round the axis passing through the observer's eye and the sun. To do this, a motion about the axis of vision must be given to the instrument, and at the same time the observer must turn himself about upon his heel; for the motion about the line of sight of the sun may be resolved into these two motions; and the observer has no means of giving the requisite motion directly by one movement. When the sun is near the horizon, the line from the eye to the sun will not be far removed from the axis of vision, and the principal motion of the instrument will be performed on this axis; while that part of the motion made about the vertical axis will be small. On the contrary, if the sun be near the zenith, the line from the eye to the sun is nearly vertical and perpendicular to the axis of vision; hence the motion about the vertical axis is the greatest, and that about the axis of vision very trifling. In intermediate positions of the sun the motions of the instrument about these two axes will be more equally divided. When the distance between the moon and sun, a planet or a star, is to be observed, the sextant must be so held that its plane may pass through the eye of the observer and both objects; and the reflected image of the brighter of the two is to be brought into contact with the other seen directly. To effect this, therefore, it is evident that when the brighter object is to the right of the other, the face of the sextant

must be held upwards, and 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 the left hand, and the motion of the index regulated with the right hand. Sometimes a sitting posture will be found convenient for the observer, particularly when the reflected object is to the right of the direct one. In this case the instrument is supported by the right hand; the elbow may rest on the right knee; the right leg at the same time resting on the left knee. If the sextant be provided with a ball and socket, and a staff, one of whose ends is attached thereto, and the other rests in a belt fastened round the observer's body, the greater part of the weight of the instrument will be supported by his body. In all cases where the sextant is used, when the contact is nearly made, the index should be fixed by the under screw, and the remaining small motion given by the adjusting screw.

Error may arise from two kinds of causes: one inherent in the construction of the instrument—as defect of parallelism or perfect planeness in the fore and back surfaces of the mirrors, as also of the coloured glasses, and of the true circular form of the arc, and true centring, for which no remedy can be provided; and the other arising from the bending and elasticity of the index or moveable radius.

The parallelism of the two surfaces of the mirrors may be tested by viewing through them obliquely a distant distinct object. If the image is perfect and well defined, the surfaces are parallel; otherwise not.

To ascertain whether the surfaces of the mirrors are plane, observe the angle between two distant objects which are nearly of the same altitude, the image of the left-hand object being brought into contact with the right-hand object viewed directly; then move the instrument in its own plane so as to bring the image of the right-hand object into contact with the left-hand object viewed directly. If they continue in contact, the surfaces are plane; otherwise not.

To test the form of the dark glasses, measure the sun's diameter to the right and to the left of the zero point with different combinations of the glasses. If the sum of the diameters so measured be nearly equal to four times the semidiameter given in the Nautical Almanac, the form of the glasses is satisfactory. For the true centering of the arc, and its truly circular form and correct graduation, the navigator must trust entirely to the skill of the maker.

By reason of the bending and elasticity of the index, and the resistance it meets with in turning round the centre, its extremity, on being pushed round the arc, will sensibly advance before the index-glass begins to move, and may be seen to recoil when the force acting on it is removed. Mr Hadley, in order to remedy this defect, which he seems to have apprehended, gave special directions that the index be made broad at the end next the centre, and the centre or axis itself have as easy a motion as is consistent with steadiness; that is, an entire freedom from looseness or shake, as the workmen term it. By strictly complying with these directions, the error in question may indeed be greatly diminished, so as to be nearly insensible, when the index is made strong, and the proper medium between the two extremes of a shake at the centre on the one hand, and too much stiffness there on the other, is nicely hit; but it cannot be entirely corrected, for to more or less of bending the index will always be subject, and some degree of resistance will remain at the centre, unless the friction there could be totally removed, which is impossible.

Of the reality of the error to which he is liable from this cause, the observer, if he is provided with an instrument furnished with an adjusting screw for the index, may thus satisfy himself:—After finishing the observation, lay the instrument on a table, and note the angle; then cautiously

loosen the screw which fastens the index, and it will immediately, if the instrument is not remarkably well constructed, be seen to start from its former situation, more or less according to the perfection of the joint and strength of the index. This starting, which is due to the index recoiling after being released from the confined state it was in during the observation, will sometimes amount to several minutes; and its direction will be opposite to that in which the index was moved by the screw at the time of finishing the observation. But how far it affects the truth of the observation depends on the manner in which the index was moved in setting it to 0, for adjusting the instrument, or in finishing the observations necessary for finding the index error.

The easiest and best rule to avoid these errors seems to be this:—In all observations made by Hadley's quadrant or sextant, let the observer take notice constantly to finish his observations by moving the index in the same direction which was used in setting it to 0 for adjusting, or in the observations necessary for finding the index error. If this rule is observed, the error arising from the spring of the index will be obviated. For as the index was bent the same way, and in the same degree, in adjusting as in observing, the truth of the observations will not be affected by this bending.

To Observe the Sun's Altitude at Sea.

Turn down one of the dark glasses before the horizon-glass (if the instrument be the quadrant, the fore horizon-glass is to be used) according to the sun's brightness; 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 instrument a slow vibratory motion about the axis of vision, as already described; move the index until the upper or lower limb of the sun is nearly in contact with the horizon at the lowest or highest part of the arc (according as the image is seen erect or inverted) described by this motion; and complete the contact by the tangent-screw, if the sextant be used—if not, by moving the index. The reading off of the limb will be the altitude of the sun.

To Observe the Moon's Altitude at Sea.

Turn down the green glass, and observe the moon in the silvered part of the horizon-glass, the eye being directed towards the horizon; move the index gradually, and proceed as already described in the case of the sun, until the enlightened limb is in contact with the horizon at the lowest or highest point of the arc described by the vibratory motion. The reading off will be the altitude of the moon's observed limb. If the lower limb be observed, the moon's semidiameter must be added; and if the upper limb be observed, it must be subtracted from the observed altitude, in order to obtain the altitude of the moon's centre. If the observation is made in the day-time, the coloured glass is not to be used.

To Observe the Altitude of a Star or Planet.

Put the index to zero; then direct the sight to the star so as to see it through the unsilvered part of the horizon-glass; turn the instrument a 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 will be seen to descend; continue to move the index gradually, until the star is in contact with the horizon at the lowest point of the arc described by the vibratory motion; in the case of the sextant, clamping the index when the contact is nearly made, and completing it with the adjusting or tangent screw.

Observation Instruments. To find the Altitude of the Sun on Shore with an Artificial Horizon.

A flat dish, containing a small quantity of mercury, is generally used for this purpose. The surface of the mercury is horizontal, and is a good reflector of the sun's rays. Let the observer so stand that he may receive on his eye rays from the sun which have been reflected from the surface of the mercury. He will then, according to the principles of optics, see the reflected image of the sun as much below the surface of the mercury as the sun is above it. If, then, he looks at the image through the unsilvered part of the horizon-glass, instead of at the horizon, and brings the image of the sun reflected at the index-glass and horizon-glass into contact with this, it is evident that the angle observed will be double of the sun's altitude. The details of this process are the same as have been already explained. Having read off the angle when the contact has been made, it must be corrected for the index error; and the result, divided by 2, will be the apparent altitude of that limb of the sun which has been observed.

To Observe the Distance between the Moon and any Celestial Object with the Sextant.

1. Between the Sun and Moon.—Put the telescope in its place, and the wires parallel to the plane of the instrument; and if the sun is very bright, raise the plate before the silvered part of the speculum; direct the telescope to the transparent part of the horizon-glass, or to the line of separation of the silvered and transparent parts, according to the brightness of the sun; and turn down one of the coloured glasses. Then hold the sextant so that its plane produced may pass through the sun and moon, having its face upwards or downwards, according as the sun is to the right or left of the moon; direct the sight through the telescope to the moon, and move the index till the limb of the sun is nearly in contact with the illumined limb of the moon; now clamp the index, and, by a gentle motion of the instrument, make the image of the sun move alternately to one side and the other of the moon; and when in that position, where the limbs are nearest each other, make the contact of the limbs perfect by the tangent-screw; this being effected, read off the degrees and parts of a degree shown by the index on the limb, using the magnifying-glass; and thus the angular distance between the nearest limbs of the sun and moon is obtained.

2. Distance between the Moon and a Planet or Star.—Direct the middle of the field of the telescope to the line of separation of the silvered and transparent parts of the horizon-glass; if the moon is very bright, turn down the lightest-coloured glass, and hold the sextant so that its plane may be parallel to that passing through the eye of the observer and both objects; its face being upwards if the moon is to the right of the star, but downwards if it be to the left. Now direct the sight through the telescope to the star, and move the index till the moon appears by the reflection to be nearly in contact with the star; clamp the index, and turn the adjusting or tangent screw till the coincidence of the star and the enlightened limb of the moon is perfect; and the reading off of the limb at the index will be the observed distance between the moon's enlightened limb and the star.

The contact of the limbs must always be observed in the middle, between the parallel wires.

It is sometimes difficult for those not much accustomed to observations of this kind, to find the reflected image in the horizon-glass; it will perhaps, in this case, be found more convenient to look directly to the object, and, by moving the index, to make its image coincide with that seen directly.

SECT. II.—OF THE CORRECTIONS TO BE APPLIED TO OBSERVED ALTITUDES AND DISTANCES.

1. Parallax.

In order that the place of a heavenly body may be fixed in space, it is necessary to suppose that all observations are taken from one point. This point is the centre of the earth. Consequently the places of the sun, moon, and planets, whose distances from the earth are measurable, as observed, must be reduced to what they would be if seen from the centre. The correction for this object is called parallax. The fixed stars are at so great a distance from the earth that they have no sensible parallax, and this correction is not to be applied to them.

Let C (fig. 47) be the centre of the earth, P the place of the observer on its surface, and Z his zenith; and let S be a heavenly body whose position is observed. Then ZPS is the observed zenith distance, or complement of the observed altitude, and ZCS the true zenith distance, —i.e., the zenith distance as observed at the earth's centre.

Then clearly the angle ZPS is greater than the angle ZCS by the angle PSC, which is also in the plane of the vertical circle through S. It is evident from the figure that a heavenly body is depressed by parallax; and the observed altitude is less than the true altitude by a certain amount depending on the altitude, which is called the correction of parallax. The correction of parallax, therefore, must always be applied with the positive sign. Its amount may be easily found; for, let \angle ZPS = z', and ZCS = z, and PSC the parallax = p; then, in triangle PSC,

\sin PSC = \frac{PC}{SC} \sin SPC = \frac{PC}{SC} \sin SPZ.

Let PC = r, the radius of the earth, and SC = R, the distance of the heavenly body from C; also PSC is always very small, and \sin PSC = PSC = p very nearly.

\text{Hence } p = \frac{r}{R} \sin z'.

r and R being invariable, p is greatest when z' = 90^\circ, or when S is in the horizon. Let P be the value of p in this position; then

P = \frac{r}{R},
\text{and } p = P \sin z' = P \cos a',

if a' be the observed altitude. This will illustrate the principle on which tables of correction for parallax for different heavenly bodies, as sun, moon, &c., for different angles of altitude, are calculated.

2. Refraction.

Refraction is a correction to be applied in consequence of the rays from every heavenly body being bent or refracted as they pass through the successive layers of the earth's atmosphere, in consequence of which they describe curvilinear paths, having their convexity turned towards the zenith of the observer. The tangent to this curve at the eye of the observer is the direction in which he sees the object, and is evidently, from what has been said, bent towards the zenith. Hence the effect of refraction is to raise the heavenly bodies in the heavens above their true

Diagram illustrating parallax. A circle represents the Earth with center C. A vertical line ZC represents the zenith line. Point P is on the surface of the Earth. A horizontal line represents the horizon. Point S is a heavenly body. The angle ZPS is the observed zenith distance. The angle ZCS is the true zenith distance. The angle PSC is the parallax. The diagram shows that the observed zenith distance ZPS is greater than the true zenith distance ZCS by the parallax angle PSC.
Fig. 47.

places; and the correction must therefore be applied to the observed altitudes of all bodies with a negative sign.

The law which this correction follows is very complex; it is very great when the body is near the horizon, and vanishes when it is in the zenith. The best position of heavenly bodies for observation, so far as this correction is concerned, is near the zenith.

A table of this correction for every 5' of altitude is calculated, and is to be found in all nautical tables.

3. Correction for Semidiameter.

When the sun or moon, or a planet, is observed, the altitude of one of the limbs is the observed altitude, the altitude of the centre must be obtained by adding or subtracting the semidiameter of the observed body according as the lower or upper limb is observed. The semidiameters of the sun and moon are continually changing; and tables are given in the Nautical Almanac of their values at noon of every day at Greenwich, and in the case of the moon at midnight also. Their values at any intermediate hour may be calculated from these, as will be more fully shown hereafter.

This correction is not to be applied when a fixed star is the object observed.

4. Correction for Dip.

The altitude is supposed to be observed from the true horizon—i.e., from a horizontal plane through the place of the observer. This, however, is never the case, for the observer's eye must then be on the earth's surface. It is, in fact, always elevated above it, and the apparent horizon is depressed in consequence below the true horizon.

The nature of this correction will be easily seen from

fig. 48. B the place of observation at a distance AB=d feet above the surface, BH' touching the surface at H'; then BH' is the plane of the sensible horizon; S a heavenly body to be observed. Draw Ba parallel to AH, the true horizon of A, and let BH' and AH intersect in J. Then, since AB is very small compared with BS and AS, the \angle ABS may be considered as equal to \angle HAS. Also let R = earth's radius, CA or CH'. Hence observed altitude of S

Diagram for Figure 48 illustrating the correction for dip. It shows a circle representing the Earth with center C. Point A is on the surface, and point B is above it at distance d. A horizontal line AH represents the true horizon. A line BH' is tangent to the surface at H', representing the sensible horizon. A line BS represents the line of sight to a heavenly body S. A line BA is drawn parallel to AH. The angle between BS and BH' is the observed altitude, and the angle between BS and AH is the true altitude. The angle between BA and AH is the dip angle ACH'.

Fig. 48.

\begin{aligned} &= H'BS = \angle ABS + \angle BH' \\ &= \angle ABS + \angle ABH = \angle ABS + \angle ACH'. \end{aligned}

Or true altitude = observed altitude - ACH'.

\text{Now } \tan ACH' = \frac{BH'}{CH'} = \frac{\sqrt{(2R+d)}d}{R}
= \sqrt{\frac{2d}{R}} \text{ nearly, because } \frac{d^2}{R^2} \text{ may be neglected in comparison of } \frac{d}{R}
\text{Also } ACH' = \tan ACH' \text{ nearly;}
\therefore ACH' = \sqrt{\frac{2d}{R}}

Now R in feet = 6120 \times 3958.

Hence ACH' in seconds

= 57.29577 \times 60 \times 60 \times \sqrt{\frac{2d}{6120 \times 3958}}

Hence, putting d=1, 2, 3, \&c., feet respectively, we can find the dip when the eye is 1, 2, 3, &c., feet respectively above the surface.

Whence tables for the dip at different elevations may be calculated.

5. Index Error.

This error has already been explained.

It will be observed that all observed altitudes are affected with refraction and dip, of which the first is always subtractive, and the second is subtractive in all observations, except when the back horizon-glass of the quadrant is made use of, when it is additive.

Parallax and semidiameter affect only those bodies whose distance from the earth is not very great, and which have a sensible diameter, as the sun and moon; but no fixed stars. Parallax is always additive; and the semidiameter is to be added or subtracted according as the lower or upper limb of the heavenly body is observed.

CHAP. II.—PRELIMINARY PROBLEMS IN NAUTICAL ASTRONOMY, AND USE OF NAUTICAL ALMANAC AND TABLES.

SECT. I.—OF TIME.

A day is the interval between two successive transits of a heavenly body over the meridian, and derives its name from the body whose motion is observed; and of whatever denomination it be, it is divided into 24 hours, each hour into 60 minutes, and each minute into 60 seconds. The interval between two successive transits of the sun is a solar day; of the moon, a lunar day; of a fixed star, a sidereal day. The earth's revolution about her axis is performed always in the same time; hence if all the heavenly bodies retained the same position with regard to one another, all days of whatever name would be of the same length. The sun (or, more strictly speaking, the earth), the moon, and planets are always varying their position with respect to one another and the fixed stars; and they move with velocities not only different from each other, but variable in different parts of their own orbits. The length of the day, therefore, determined by these bodies is variable. In order to obtain a definite and uniform measure of time, the mean solar day, which is the average of all the apparent solar days in the year, is employed. An imaginary body, called the mean sun, is supposed to describe the equator uniformly with the true sun's average or mean daily motion, and the interval between two successive transits of this imaginary sun is a mean solar day.

Clocks and chronometers are adapted to mean solar time; so that a complete revolution through twenty-four hours of the hour-hand of one of these instruments would exactly correspond with the revolution of the earth about her axis with regard to the mean sun. Time reckoned in mean solar days and parts of mean solar days is called mean solar time.

The true sun sometimes passes the meridian before and sometimes after the mean sun; the difference in time of the transits of the true and mean suns is called the equation of time, and is sometimes to be added and sometimes to be subtracted from the one of these times to obtain the other, as is pointed out in the Nautical Almanac.

As the earth revolves about her axis from E. to W., different meridians successively come under the mean sun; and since, after the lapse of twenty-four mean solar hours,

Nautical Astronomy the same meridian again comes under the sun; it follows that for every 15° of longitude between the places, there is a difference of 1h of mean time; and that mean noon is 1h earlier for every 15° of E. longitude, and 1h later for every 15° of W. longitude.

A sidereal day is the interval between two successive transits over the same meridian of the vernal equinox, or the first point of Aries. This is not strictly a uniform measure because of the change of the position of this point of Aries in consequence of precession and nutation. Time, therefore, so reckoned, ought strictly to be called apparent sidereal time; and mean sidereal time to be reckoned from the transit not of the true but of the mean equinoctial point. The smallness of the fluctuations to which a clock regulated to apparent sidereal time, compared with one regulated to mean sidereal time, is subject, amounting at the utmost to 2m 3s in 19 years, has prevented the practical inconvenience of this being felt; no clock being sufficiently perfect to go for so long a period without requiring frequently to be re-adjusted.

The sun's apparent revolution in his orbit round the earth takes place in 365.242218 mean solar days; the mean sun, therefore, describes an angle of 59° 8' 33" in a mean solar day; hence, in the interval between two successive transits of the mean sun over the same meridian, the earth revolves through 360° 59' 8' 33"; but in a sidereal day the earth revolves through 360°. Whence we obtain the proportion—

A sidereal day : a mean solar day :: 360° : 360° 59' 8" 33".

Whence also it is evident that, if the same interval of absolute time be expressed in mean solar and also in sidereal time, it will be greater in the latter denomination than in the former.

Tables are calculated for the acceleration of sidereal on mean solar time, and, conversely, for the retardation of mean solar on sidereal time, which are useful for converting intervals of sidereal into mean solar time, and conversely. These are to be found at page 516 of the Nautical Almanac.

Astronomical mean time always begins at noon, and goes on to the succeeding noon through 24 hours; so that a clock which shows astronomical mean time, set for any place (as Greenwich), shows 0h 0m 0s when the mean sun is on the meridian, and shows 24 hours, or 0h 0m 0s, when the mean sun is again on the meridian.

A clock set to show sidereal time shows 0h 0m 0s when the mean equinoctial point is on the meridian, and shows 24 hours, or 0h 0m 0s, again when this point is next on the meridian.

Sidereal Time, in the Nautical Almanac, is the sidereal time at mean noon for the meridian of Greenwich; or, in other words, the hour angle of the equinoctial point when the mean sun is on this meridian; and is very useful in all cases where mean time is to be deduced from observations of heavenly bodies.

The civil day is reckoned from midnight to midnight, and is divided into 12 hours, from midnight to noon, called A.M.; and into 12 hours, from noon to midnight, called P.M. Evidently time P.M., and astronomical mean time up to midnight, are the same. But for time A.M., add 12 hours, and date the day one back.

Thus, June 3, 7h 5m 27s P.M. is, June 3, 7h 5m 27s astronomical time; but, June 3, 5h 12m 37s A.M. is, June 2, 17h 12m 37s A.M.

Conversely, if the astronomical time is given, and the corresponding civil time is required, if under 12 hours, the given time is the same time P.M. for the same date; if above 12 hours, subtract 12 hours from it, and date the day one forward, and the result is civil time A.M.

Thus, September 18, 7h 10m 15s is, September 18,

7h 10m 15s P.M.; but September 18, 22h 13m 45s astronomical is, September 19, 10h 13m 45s A.M.

PROB. 1.—To convert degrees, or parts of the equator, into time.

Rule.—Divide the degrees by 15; the result gives hours. Multiply the remaining degrees, if any, by 4; the result is minutes. Divide the number of minutes by 15 for seconds, and then multiply the remaining minutes by 4 for seconds. Divide the seconds by 15 for seconds, and decimals of seconds, if necessary.

Ex. 1.—Reduce 25° 45' to time. Divide 25 by 15, the quotient is 1, with remainder 11. Hence, by the rule—

\begin{array}{r} 25^\circ = 1\text{ h } 44\text{ m} \\ 45' = 0\text{ } 3 \\ \hline \text{or } 25^\circ 45' = 1\text{ h } 47\text{ m.} \end{array}

Ex. 2.—Reduce 135° 48' 18" to time.

\begin{array}{r} 135^\circ = 9\text{ h } 16\text{ m } 0\text{ s} \\ 48' = 0\text{ } 3\text{ 12} \\ 18'' = 0\text{ } 0\text{ 1.2} \\ \hline \text{or } 135^\circ 48' 18'' = 9\text{ h } 19\text{ 13.2.} \end{array}

PROB. 2.—To convert time into degrees, minutes, and seconds.

Rule.—Multiply the given time by 10, and add to the result half of the product. The result will be the corresponding degrees, minutes, and seconds.

Ex.—Convert 3h 4m 28s into degrees, &c.

\begin{array}{r} 3\text{ h } 4\text{ m } 28\text{ s} \\ 10 \\ \hline 30\text{ 44 } 40 \\ \text{One half} = 15\text{ 22 } 20 \\ \hline 45\text{ 7 } 0 \\ \text{Answer} \dots 45^\circ 7' \end{array}

PROB. 3.—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 turn it into astronomical time; convert the longitude of the known meridian into time, which add to the given time if the place be W. of Greenwich, and subtract from the given time if it be E.; and, if necessary, increase the time by 24 hours, reckoning the day one back. If the resulting time exceed 24 hours, put the date one day forward, and subtract 24 hours.

Ex. 1.—It is 6h 15m P.M., June 23, in a ship whose longitude is 76° 45' W.

\begin{array}{r} 76^\circ 45' = 5\text{ h } 7\text{ m.} \\ \text{Given time, June 23} \dots 6\text{ h } 15\text{ m.} \\ \text{Longitude} \dots 6\text{ 7 W.} \\ \hline \text{Time at Greenwich} \dots 11\text{ 22 June 23.} \end{array}

Ex. 2.—The time at a ship in longitude 139° 48' 15" E. is, July 24, 5h 25m 20s P.M.; required the Greenwich time.

\begin{array}{r} 139^\circ = 9\text{ h } 16\text{ m } 0\text{ s} \\ 48' = 0\text{ } 3\text{ 12} \\ 15'' = 0\text{ } 0\text{ 1} \\ \hline 9\text{ 19 } 13 \end{array}

This being greater than the given time, we next add 24h to the latter, and we have—

\begin{array}{r} 23\text{ h } 25\text{ m } 20\text{ s} \\ 9\text{ 19 } 13 \\ \hline \text{Time at Greenwich} \dots 20\text{ 6 } 7 \text{ July 23.} \\ \text{Or civil time} \dots 8\text{ 6 } 7 \text{ A.M. July 24.} \end{array}

Ex. 3.—The time at a ship in longitude 175° 45' W. is, June 28, 4h 18m 12s A.M.; what is the Greenwich time?

\begin{array}{r} 175^\circ = 11\text{ h } 40\text{ m} \\ 45' = 0\text{ } 3 \\ \hline \text{Longitude} = 11\text{ 43 W.} \end{array}
\begin{array}{r} \text{Time at ship is, June 27} \dots 16\text{ h } 18\text{ m } 12\text{ s} \\ \text{Longitude} \dots 11\text{ 43 } 0\text{ W.} \\ \hline \text{Time} \dots 23\text{ 1 } 12 \text{ June 27.} \\ \text{Or} \dots 4\text{ 1 } 12 \text{ June 28.} \end{array}

Ex. 4.—The time at a ship in longitude 178° E. is, October 14, 3h 15m A.M.; what is the time at Greenwich?
3h 15m A.M. on Oct. 14... = Oct. 13, 15h 15m 0s
Longitude 178° ..... = 11 52 0 E.
Time at Greenwich. = Oct. 13, 3 23 0

PROB. 4.—To find the Greenwich time by a chronometer whose error on Greenwich mean time is known.

Rule.—Add or subtract the error of the chronometer to the time shown by it, according as it is slow or fast; the result is Greenwich time. Sometimes 12 hours must be added to this result, and the day reckoned one back. The best way to obviate this error, is to obtain Greenwich time approximately, by Problem 3, by help of the ship's mean time and longitude by account; if the difference between the two (Greenwich) dates, so found in these two ways, is nearly 12 hours, then the date by chronometer must be increased by 12 hours, and the day reckoned one back if necessary, so as to make the two dates agree in the day and hour nearly.

Ex. 1.—June 15, 1857, at 7h 45m P.M. mean time nearly, in longitude 45° W., a chronometer showed 10h 53m 12s, being 5h 15m fast; required the Greenwich time.

By Chronometer. By Rule 3.
June 15, Chronometer 10h 53m 12s Ship, June 15... 7h 45m
Error on Greenwich } 0 3 15 Longitude..... 3 0 W.
mean time (fast).....
Greenwich time ..... 10 49 57 Greenwich time, 10 45

Ex. 2.—October 12, 1857, at 11h 45m P.M., in longitude 110° 35' W., a chronometer showed 7h 0m 40s, being 9h 5m slow; required the Greenwich date.

By Chronometer. By Rule 3.
Oct. 12..... 7h 0m 40s Ship, Oct. 12..... 11h 45m 0s
Error (slow)..... 0 9 5 Longitude ..... 7 22 20
7 9 45 Greenwich, Oct. 12... 19 7 20
12 0 0
Greenwich, Oct. 12, 19 9 45

PROB. 5.—Given the Greenwich date, to find the date under a given meridian.

Rule.—The Greenwich date being reduced to the preceding noon, reduce the longitude to time, and subtract it from Greenwich date, increased by 24 hours, and put one day back if necessary, if the longitude be W.; and add if the longitude be E.

Ex. 1.—An eclipse of the moon commenced at Greenwich, December 25, 1852, at 22h 14m 36s; at what time did it begin at a place in longitude 175° E.?

Greenwich time, Dec. 25..... 22h 14m 36s
Longitude..... 11 40 E.
33 54
(i.e.) December 26..... 9 54

Ex. 2.—On the 6th of December 1857, at 9h at Greenwich, the distance of the moon from the sun was 106° 44' 31"; at what time will the distance be the same in longitude 168° 15' 43" W.?
Long. = 11h 13m 2s 86 W. Greenwich, Dec. 5..... 33h 0m 0s
Add 24 to Greenwich time. Longitude..... 11 13 2 86 W.
Date required, Dec. 5..... 21 46 57 14.

SECT. II.—OF THE SUN'S DECLINATION, RIGHT ASCENSION, EQUATION OF TIME, AND SEMIDIAMETER AND SIDEREAL TIME.

All of these quantities are given in the Nautical Almanac for the noon of every day. They all vary, and consequently their value at any time intermediate between two successive noons is different from that at either noon. For ordinary purposes, it is sufficient to suppose that the change of these quantities is proportional to the time from the preceding noon. Hence, if the difference of the value between the preceding and succeeding noon is taken from the Nautical Almanac, a simple proportion will give the quan-

tity to be applied to the value at the preceding noon, which must be added if the quantity is increasing, and subtracted if it be decreasing. As the values of the quantities are given for Greenwich noon, it is necessary to reduce the ship's time to Greenwich time, in order to obtain their proper value.

This process may be materially shortened by the use of proportional logarithms, which are given in nautical tables, and to which the reader is referred for an explanation. For example, to take out the sun's declination at a given time in a given place, get a Greenwich date, as already explained; take out from the Nautical Almanac the declination at two successive noons between which the date lies; take their difference; take out from the tables the logarithm of the Greenwich date of the sun; add to it the proportional logarithm for the change in 24 hours;—the result is the proportional logarithm of the change of declination for the given time. This will be found from the table, and added or subtracted, according as the declination is increasing or decreasing. Sometimes the declination at two successive noons will be of different names; the difference must then be found by adding them.

The very same mode applies to finding the equation of time, and also the sun's apparent right ascension.

The sun's semidiameter may always be taken to be that corresponding to the nearest noon.

The right ascension of the mean sun, called in the Nautical Almanac the sidereal time, may be found for any other meridian than Greenwich in the same way. Since, however, the motion of the mean sun is uniform throughout the year, the change in any given number of hours, minutes, and seconds is the same for all days, and can be found at once from the "Table of Time Equivalents" given in the Nautical Almanac.

Opposite to each of these quantities in the Nautical Almanac is given the difference for one hour, which enables us to find the quantity for any other hour than noon, by multiplying this hourly difference by the number of hours in the date from noon, and adding to it the proportional parts for the additional minutes and seconds, and applying this difference to the value for the preceding noon, with a positive or negative sign, according as the quantity is increasing or decreasing.

The following examples will illustrate these directions:—

Ex. 1.—October 11, 1857, in Long. 36° 45' E., at 7h 45m A.M.; required the sun's declination, right ascension, equation of time, and sidereal time, i.e., the right ascension of the mean sun when crossing the meridian of observer.

October 11, 7h 45m A.M. civil is, October 10... 19h 45m
Longitude..... 2 27 E.
Greenwich date, Oct. 10..... 17 18

To find Declination.—(1.) By proportional logarithms, from Nautical Almanac.

Declination, Oct. 10..... 6° 44' 0" 9 S.
" " 11..... 7 6 43 5 S.
Diff. for 24h..... 0 22 42 5
Greenwich date, logarithm of (O)..... = 14217
Proportional logarithm for 22h 42m 6..... = 80925
Prop. log. for 16h 22m..... 1 04142
Declination at noon, October 10. 6° 44' 0" 9 S.
Increase in 17h 18m..... 0 16 22
Declination required..... 7 0 22 9 S.

(2.) By hourly differences.

Difference for 1h, October 10, 55h 77 from Nautical Almanac. 17h
39739
5677
96509
Nautical
Astronomy
965.09..... = 16° 5' 09 difference for 17h.
15m is \frac{1}{2}..... 14.19
3m is \frac{1}{2}..... 2.83
16 22.11
Declination, October 10.. 6° 44 0.9
Declination required.. 7 0 23.01

To find right ascension.

Nautical
Astronomy
Right ascension, Greenwich, October 10 ..... 13h 3m 8s 30
" " " 11 ..... 13 6 50.02
0 3 41.22
Greenwich date logarithm of \odot..... = -14217
Prop. log. for 3m 41s..... = 1.58903
Prop. log. for 2m 39s..... 1.83120
Sun's right ascension at noon, October 10 ..... 13h 3m 8s 30
Increase in 17h 18m ..... 0 2 39
Sun's right ascension required ..... 13 5 47.80

Or, from Nautical Almanac

Nautical
Astronomy
Difference for 1h..... 9.216
17
64512
9216
158.672
15m is \frac{1}{2}..... 2.304
3m is \frac{1}{2}..... .461
159.437 or 2m 39s nearly.
Sun's right ascension, October 10, noon ..... 13h 3m 8s 30
Increase for 17h 18m ..... 0 2 39
Right ascension required ..... 13 5 47

To find equation of time.

Nautical
Astronomy
Equation of time, October 10..... 12m 59s 81
" " 11..... 13 15.15
0 15.34
Greenwich date log. of \odot..... = -14217
Prop. log. for 15s 3..... = 2.84873
Prop. log. for 11s..... 2.92090
Equation of time, October 10..... 12m 59s 81
Increase in 17h 18m..... 0 11
13 10.81

Or, from Nautical Almanac

Nautical
Astronomy
Increase for 1h..... 0m 630
17
4473
639
10.863
15m is \frac{1}{2}..... 129
3m is \frac{1}{2}..... 25
11.017 or 11s nearly,

which, added to the equation of time at mean noon of October 10, gives 13m 10s 81 for equation of time required.

Nautical
Astronomy
Right ascension mean sun, October 10 ..... 13h 16m 8s 61
" " " 11 ..... 13 20 5.17
Difference ..... 0 3 56.56
Greenwich date of \odot..... = -14217
Prop. log. for 3m 56s..... = 1.66051
Prop. log. for 2m 50s..... 1.80268
October 10..... 13h 16m 8s 61
Increase..... 0 2 50
Mean sun's r. ascen. or sidereal time, 13 18 58.61

Or, by table of time equivalents,

Nautical
Astronomy
Mean sun's right ascension, October 10 ..... 13h 16m 8s 61
Correction for 17h..... 0 2 47
Correction for 18m..... 0 0 3
Mean sun's right ascension required.. 13 18 58.61

Ex. 2.—To find the sun's declination at a place in Long. 97° 45' W., at 3h 46m p.m., on 20th March 1857,—Greenwich date, March 20, 10h 17m.

Nautical
Astronomy
Sun's declination, March 20 ..... 6° 3' 35s S.
" " 21 ..... 0 20 6.1 N.
Difference ..... 0 23 41.6
Nautical
Astronomy
Greenwich date logarithm \odot..... = -36808
Prop. log. for 23h 41m..... = .88983
Prop. log. for 10m 9s..... 1.24891
Declination ..... 6° 33s 5 N.

Or, by hourly difference,

Nautical
Astronomy
Difference for 1h, March 20..... 59s 23
10
Nautical
Astronomy
Difference for 10h..... 592.30
Nautical
Astronomy
15m is \frac{1}{2}..... 14.80
Nautical
Astronomy
2m is \frac{1}{2}..... 1.97
Nautical
Astronomy
609.07
Nautical
Astronomy
Correction..... 10m 9s
Nautical
Astronomy
And the declination, subtracting from this ..... 3 35.5
Nautical
Astronomy
Gives, as before ..... 6° 33s 5 N.

SECT. III.—OF TAKING OUT MOON'S DECLINATION, RIGHT ASCENSION, SEMIDIAMETER, AND HORIZONTAL PARALLAX.

The moon's declination and right ascension are given for every hour in the Nautical Almanac; and the difference for 10m at the beginning of every hour is recorded. From these data, the moon's declination and right ascension at any hour can be readily obtained; for by multiplying the difference for 10m by the number of intervals of 10m in the given time since the last hour, and taking parts for the additional minutes and seconds, and adding the result if the declination be increasing, and subtracting if decreasing, to the declination at the last preceding hour, we shall find the declination. The difference must always be added to the right ascension. Or they may be found by a table of logistic logarithms: thus, add together the logistic logarithms of the minutes and seconds in the Greenwich date, and the proportional logarithms of the difference; the result will be the proportional logarithm of the correction to be applied.

The moon's semidiameter and horizontal parallax are given in the Nautical Almanac for every 12h,—viz., from mean noon to mean midnight, and mean midnight to mean noon, of every day at Greenwich. Hence we may proceed as for the sun's right ascension and declination, except that we must always take out of the table the declination or right ascension for the mean noon or mean midnight, according as the time lies between noon and midnight or midnight and noon; and if the Greenwich date exceed 12h, we must subtract 12h from it, and reckon the difference from the preceding midnight.

Ex. 1.—August 10, 1857, in Long. 84° 30' E., at 7h 44m 15s p.m.; required the moon's right ascension and declination.

Nautical
Astronomy
Ship, August 10 ..... 7h 44m 15s
Longitude ..... 5 38 0 N.
Greenwich date, August 10, 2 6 15
Nautical
Astronomy
Moon's r. ascen., Aug. 10, 2h, 1h 8m 43s 49 Decl., 9° 22' 48s 7 N
" " 3 1 10 50.91 " 9 38 55.1 N.
Difference..... 0 2 7.42 0 16 6.4
Nautical
Astronomy
Log. log... 6m 15s = .98227 Log. log... 6m 15s = .78227
P. log. for.. 2 7 = 1.92962 P. log. for.. 16 6 = 1.04845
Nautical
Astronomy
P. log. for.. 0 13 2.91189 P. log. for.. 1 41 2.03072
Nautical
Astronomy
Moon's r. ascen., Aug. 10, 2h, 1h 8m 43s 49 Decl., 12h, 9° 22' 48s 7
Correction ..... 0 0 13
Nautical
Astronomy
Moon's right ascen. required, 1 8 56.49 Decl. req., 9 24 29.7
Or for right ascension— Or for declination—
Nautical
Astronomy
Difference for 1h ... 2m 7s 42 Difference for 10m ... = 161.05
Nautical
Astronomy
6m is \frac{1}{2}..... 12.742
15s is \frac{1}{2}..... .530
Nautical
Astronomy
13.272
Or correction is 13s, as before. 100.62

Or correction is 1m 41s, as before.

Ex. 2.—September 12, 1857, in Long. 149° 18' W., at 5h 36m p.m.; required the moon's semidiameter and horizontal parallax.

Nautical
Astronomy
Ship, September 12..... 5h 36m 0s
Longitude..... 9 57 12 W.
Greenwich, September 12.. 15 33 12
Nautical Astronomy Moon's Semidiameter. Moon's Horizontal Parallax.
Sept. 12, midnight .... 15° 53' 2" Sept. 12, midnight .... 58° 10'
Sept. 13, noon .... 15 49 3 Sept. 13, noon .... 57 55 9
Difference ..... 0 3 7 Difference ..... 0 14 1
And both are decreasing.
Greenwich date log } = 52895 Greenwich date } = 52895
3h 35m ..... log } .....
Prop. log. for 3h 7 .... 3 46522 Prop. log. for 14h 1 = 2 88420
Prop. log. for 1h 1 .. 3 99117 Prop. log. for 4h 2 = 3 41315
Moon's semidiameter } = 15° 52' 1" Horiz. parallax } = 58° 5' 8"
at 15h 33m 12s .. at 15h 33m 12s ..
These problems may also be solved by simply taking the proportional parts. Thus, for the horizontal parallax—
Difference for 12h ..... 14° 1'
3h is \frac{1}{4} ..... 3 5
30m is \frac{1}{60} ..... 58
4 68
Or the correction is 4".
And for semidiameter—
Difference for 12h ..... 3° 7'
3h is \frac{1}{4} ..... 92
30m is \frac{1}{60} ..... 15
1 07
Or the correction is 1° 07'.
SECT. IV.—OF LUNAR DISTANCES.
PROB.—Having given a lunar distance, to find its Greenwich date.

Rule.—Lunar distances are found for every third hour of Greenwich mean time, and are recorded in the Nautical Almanac, as also the proportional logarithms of the differences of the distances at intervals of every three hours. To find, then, the Greenwich date corresponding to a given distance, find in the table the nearest distance preceding in order of time the given distance, and take the difference between it and the given distance; from the proportional logarithm of this difference subtract the proportional logarithm of the said nearest distance; the remainder will be the proportional logarithm of the correction to be applied to the Greenwich date answering to the nearest distance in order to obtain the approximate Greenwich time.

The distances are here supposed to increase uniformly, which is not the case. It is therefore necessary to apply a correction to the time above obtained, if great accuracy is required. This may be found by help of a "Table showing the Corrections required on account of Second Differences," in the Nautical Almanac, as follows:—(1.) Find the approximate interval by the preceding rule. (2.) Take the difference between the proportional logarithms which stand opposite to the distances which include the given distance. (3.) Look down the column in the table at the head of which is this difference of proportional logarithms, and along the column at the left hand side of which stands the approximate interval; the number so found is the correction, which is to be added to the approximate time if the proportional logarithms are decreasing, and subtracted if they are increasing.

Ex.—Required the time at Greenwich at which the reduced distance of the moon from Jupiter will be 42° 55' 18" on the 23d December 1857.

From the table, it appears this will be between noon and 3h. The nearest distance preceding therefore is,—

Distance at noon..... 43° 58' 19" Propor. log..... 2590
Reduced distance..... 42 55 18
Difference..... 1 3 1 Propor. log..... 4558
Approximate interval 1h 54m 24s " ..... 1968

The difference of proportional logarithms at noon and 3h is 12; and on looking in the table for second differences down the column headed (2), and along the column on the left of which is 2h, the

nearest to the approximate interval, we find 3h the correction, which is to be added, as the proportional logarithms are decreasing. Hence the Greenwich time is 1h 54m 27s.

SECT. V.—OF MEAN AND APPARENT SOLAR TIME, AND SIDEREAL TIME.

PROB. 1.—Given mean solar time at any given place, to find apparent solar time; and, conversely, given apparent, to find mean solar time.

Find the time at Greenwich, and correct the equation of time for this date. The equation of time so corrected applied to the given time, with the proper sign (as shown in the Nautical Almanac), will give the time required.

Ex. 1.—June 16, 1857, mean solar time 8h 15m P.M., in longitude 25° W.; to find the apparent time.

Ship, June 16..... 8h 15m
Longitude ..... 1 40 W.
Greenwich, June 16..... 9 55
Equation of time (p. 11, Nautical Almanac) subtractive—
June 16..... 19° 43'
" 17..... 32° 34'
Difference..... 12° 91'
Greenwich date, log. ⊙ ..... = 33385
Propor. log. for 12h 9 ..... = 2 92283
Propor. log. for 5h 3..... 3 30668
Equation of time..... − 0h 0m 24s 73
Ship, June 16 ..... 8 15 0
Apparent time, ship ..... 8 14 35 27

Ex. 2.—Oct. 15, 1857, at 6h 51m P.M., apparent time, in Long. 113° 45' E.; find mean time.

Ship, Oct. 15..... 6h 51m
Longitude ..... 7 35 E.
Greenwich, Oct. 14..... 23 16
Equation of time (p. 1, Nautical Almanac)—
Oct. 14..... − 13m 57s 92
" 15..... 14 11 12
Difference ..... 0 13 20
Difference for 1h..... 550
23
1650
1100
Difference for 23h..... 12 650
15m is \frac{1}{4}..... 137
1s is \frac{1}{60}..... 909
12 796
Or 12° 8 nearly.
Hence mean time at ship, 6h 50m 47s 2 P.M., Oct. 15.
PROB. 2.—Given solar time, to find sidereal time.

Rule.—If the given time be mean solar, find the Greenwich date; correct the right ascension of mean sun for this date, and add to the mean time at ship; the result is sidereal time.

If the given time is apparent solar, convert it into mean solar time, as in Problem 1, and proceed as before.

Ex.—Nov. 18, 1857, at 9h 35m 30s A.M., mean time, in Long. 18° 7' W.; required the sidereal time.

Ship, Nov. 17 ..... 21h 35m 30s
Longitude ..... 1 12 28 W.
Greenwich, Nov. 17th ..... 22 47 58
Right ascension of mean sun, Nov. 17..... 15 45 57 72
Correction for 22h ..... 0 3 38 84
" 47m ..... 0 0 7 72
" 58s ..... 0 0 0 16
Right ascension of mean sun..... 15 49 42 44
Ship mean time..... 21 35 30
Sidereal time ..... 13 25 12 44
Rejecting 24h in the result.

Nautical Astronomy By help of the last problem, we may find what bright stars in the catalogue of the Nautical Almanac will be on the meridian of a given place next after any time, mean or apparent.

For, having found the sidereal time at a ship, it is evident that the star whose right ascension is next greater, is the first that will pass the meridian in question.

It is sometimes required to find all the bright stars that will pass a given meridian between two specified hours. We must in that case find the sidereal time for both hours, and all the bright stars in the catalogue whose right ascensions lie between these limits will pass the given meridian between the specified hours.

Ex. 1.—To find what bright star will pass the meridian in Long. 15^{\circ} 7' W., on November 18, 1857; next after 9^{\text{h}} 35^{\text{m}} 30^{\text{s}} A.M. (See Example to Problem 2.)

Looking in the catalogue of bright stars in the Nautical Almanac, we find that the bright star whose right ascension is the next greater than 13^{\text{h}} 25^{\text{m}} 12^{\text{s}} 44^{\text{s}}, is \beta Corvi, which is the star required.

Ex. 2.—What bright stars in the Nautical Almanac passed the meridian of a place in Long. 64^{\circ} E., between the hours of 6 and 9, on March 10, 1857?

Ship, March 10..... 6^{\text{h}} 0^{\text{m}} 0^{\text{s}} 0^{\text{h}} 0^{\text{m}} 0^{\text{s}}
Longitude..... 4 16 0 E. 4 16 0 E.
Greenwich, March 10..... 1 44 0 4 44 0
(1.) Right ascension of mean sun—
Greenwich, March 10..... 23h 12m 25s 64
Correction for 1h 0m..... 0 0 9.8565
" 0 44..... 0 0 7.2281
23 12 42.7236
Ship, March 10..... 6 0 0
Sidereal time..... 5 12 42.7246 (L.)
(2.) Right ascension of mean sun—
Greenwich, March 10..... 23h 12m 25s 64
Correction for 4h 0m..... 0 0 39.4259
" 0 44..... 0 0 7.2281
23 13 12.2940
Ship..... 9
Sidereal time..... 8 13 12.2940 (M.)

On looking at the catalogue, we find that the stars whose right ascensions lie between these limits are from \beta Tauri to 15^{\circ} Argus, which are the stars required.

PROB. 3.—Given sidereal, to find mean solar time.

Rule.—Take out of the Nautical Almanac the right ascension of the mean sun for noon of the given day. Subtract this from the sidereal time, increased, if necessary, by 24 hours; the remainder is mean time nearly. By help of the table which gives the relation of mean solar and sidereal time already referred to, correct this approximate time, and the result will be the mean time required.

Ex.—July 21, 1857, when a sidereal clock showed 6^{\text{h}} 45^{\text{m}} 35^{\text{s}} find the mean time.

Sidereal time..... 6^{\text{h}} 45^{\text{m}} 35^{\text{s}}
Right ascension mean sun at mean noon..... 7 56 47.66
Mean time nearly..... 22 48 47.34
Correction for 22h..... 3^{\text{h}} 46^{\text{m}} 69^{\text{s}}
" 48m..... 0 7.88
" 47s..... 0 0.13
0 3 54.60
Mean time..... 22 44 52.74

By means of the above we can find the error of a clock or chronometer by comparing the time shown by it with that of a sidereal clock in an observatory. Thus, Greenwich, November 4, 1857, a sidereal clock showed 18^{\text{h}} 15^{\text{m}} 30^{\text{s}}, and a chronometer showed 3^{\text{h}} 28^{\text{m}} 15^{\text{s}}, the sidereal clock being 3^{\text{m}} 13^{\text{s}} 5 slow; required the error of the chronometer.

Sidereal clock..... 18^{\text{h}} 15^{\text{m}} 30^{\text{s}}
Error (slow)..... 0 3 13.5
Sidereal time..... 18 18 43.5
R. ascen. of mean sun at mean noon..... 14 54 42.49
Mean time nearly..... 3 24 1.01
Correction for 3h..... 29.57
" 24m..... 3.94
" 1s..... 0
0 0 33.51
Mean time..... 3 23 27.50
Chronometer..... 3 28 15.0
Error (fast)..... 0 4 48.5

PROB. 4.—To find at what time any heavenly body will pass the meridian of a given place on a given day.

Rule.—From the right ascension of the given star taken out of the Nautical Almanac, increased, if necessary, by 24 hours, subtract the right ascension of the mean sun at mean noon (or sidereal time, as it is called in the Nautical Almanac). This gives the mean time at the ship approximately. Apply the longitude to this, and thus get a Greenwich date, and by this date correct the right ascension of the mean sun, and subtract the quantity so found from the star's right ascension; this will be the time required.

Ex.—At what time will \beta Corvi pass the meridian of a place in Long. 62^{\circ} 20' E., on July 24, 1857?

Right ascension \beta Corvi, July 24..... 12h 26m 53.88
Right ascension of mean sun at mean noon (sidereal time, Naut. Almanac)..... 8 8 37.33
Sidereal time ship, approximate..... 4 18 16.55
Longitude..... 4 9 20.0 E.
Greenwich date, July 24..... 6h 8m 56.55
Right ascension of mean sun at mean noon 8 8 37.33
Correction for 8m..... 0 0 1.31
" 56s..... 0 0 0.15
Right ascension of mean sun..... 8 8 38.79
" \beta Corvi..... 12 26 53.88
Time of star's passing the meridian..... 4 18 16.09

SECT. VI.—OF THE MOON'S AUGMENTATION—CORRECTION OF SEMIDIAMETER ON ACCOUNT OF REFRACTION—THE MOON'S MERIDIAN PASSAGE, &c., &c.

1. Of the Moon's Augmentation.—The moon's semidiameter given in the Nautical Almanac is calculated as if seen from the earth's centre. Moreover, on account of the comparatively moderate distance of the moon from the earth, the moon's semidiameter subtends an angle at the earth's surface which sensibly varies with the altitude, being greatest when the moon is in the zenith, and least when in the horizon of the observer. The correction to be applied to the semidiameter from this cause is called the augmentation. It is computed for every degree of altitude, and is to be found in all nautical tables.

2. Contraction of Moon's Semidiameter on account of Refraction.—The correction for refraction varies very rapidly near the horizon, and consequently the moon's lower limb is sensibly more raised than the upper; and the moon's apparent form is that of an ellipse instead of a circle. Hence, in taking a lunar distance when the moon is near the horizon, the moon's semidiameter to be added to the observed distance, as taken from the tables, is greater than it ought to be; and a correction must be applied in consequence of the moon's diameter being an oblique diameter of the elliptic face, and not the horizontal or greatest diameter. This correction is always subtractive, and is given in nautical tables.

3. Of the Moon's Meridian Passage.—In west longitude the moon crosses the meridian later with regard to the sun than at Greenwich, and in east longitude earlier; because the moon's distance from the sun in right ascension is constantly changing. Hence to the Greenwich meridian passage of the moon must be applied a correction for every other place, depending on the longitude. All nautical tables contain this correction.

The corrections to be applied to the observed altitudes and distances of the heavenly bodies have been now sufficiently explained, and the reader will, with a little experience, easily understand how the several corrections are to be taken out of the tables. Index error, dip, and refraction are common to all observed altitudes, and must be applied in the order in which they are here set down. If the body observed be a star, this is all that is required.

If a planet, the parallax for the given altitude, taken from the proper table, must be added. If the observed body be the sun, after the dip apply the sun's semidiameter, and then correct for refraction and parallax. These two corrections are frequently given together, under the name of correction in altitude, which is always subtractive, because the refraction is always greater than the parallax. If the observed body be the moon, the semidiameter and horizontal parallax cannot be considered as invariable for 24 hours; they must therefore be corrected for the proper Greenwich date, as already pointed out. To the semidiameter thus corrected must be added the augmentation, from the proper table. The rest of the process is exactly the same as for the sun.

Ex. 1.—Feb. 1, 1857, at 2h 46m P.M., in Lat. 53° 20' N., and Long. 13° 20' E., the observed altitude of the sun's lower limb was 13° 35', the index error was +1° 20', and the height of the eye above the sea 18 feet; to find the sun's true altitude.

Observed altitude..... 13° 35' 0"
Index correction ..... + 0 1 20
Dip ..... 13 33 20
San's apparent diameter ..... 0 3 33
Apparent altitude..... 13 32 47
Correction in altitude ..... + 0 16 15
San's true altitude ..... 13 49 12
0 3 45
13 45 17

Ex. 2.—August 24, 1857, in Long. 18° 20' W., the observed meridian altitude of the moon's lower limb was 54° 30' 20", the index correction was +2° 20", and height of the eye above the sea 20 feet; required the moon's true altitude.

August 24. Time at ship ..... 0h 0m 0s
Longitude ..... 1 13 20 W.
Greenwich date, August 24 ..... 1 13 29
Moon's semidiameter, 24th, Noon..... 14° 54' 4"
Midnight..... 14 52 0
0 24
Greenwich date, log. moon ..... 99101
Prop. log. for 2h 4m..... 3.65321
4.64722
Part..... 0° 3
Hence moon's semidiameter ..... 14° 54' 1"
Moon's horizontal parallax ..... 54 33 8
Augmentation..... 0 11 3
Hence semidiameter ..... 15 5 4
Observed altitude..... 54° 30' 20"
Index correction ..... + 0 2 20
Dip..... 54 32 40
Semidiameter ..... 0 4 24
Apparent altitude ..... 54 28 16
Correction in altitude..... 0 15 54
54 43 21 4
0 30 30
0 0 19
True altitude ..... 55 14 10 4

CHAP. III.—ON FINDING THE LATITUDE.

SECT. I.—OF FINDING THE LATITUDE BY MERIDIAN ALTITUDES.

1. By a single Meridian Altitude above the Pole.

Subtract the true meridian altitude (corrected from observed meridian altitude) from 90°; the result is the zenith distance—which mark N. or S., according as the zenith is north or south of the observed heavenly body: find also the declination, which must be marked N. or S. If the zenith distance and the declination have the same name, their sum will be the latitude, with the name of either. If they be of different names, take their difference, which will be the latitude, with the name of the greater.

The process will differ in detail, according to the nature of the body observed.

(1.) If it be a fixed star, the altitude must be corrected only for index correction, dip, and refraction; and the declination is given at once by the tables.

(2.) If it be a planet, the declination must be corrected by getting a Greenwich date, applying the proportional increase or decrease for this date. If great accuracy is required, the semidiameter and horizontal parallax must also be found, and the altitude corrected for these.

(3.) If it be the sun, the Greenwich date must be got, and the declination corrected for this; the altitude must be corrected for index, dip, semidiameter, refraction, and parallax.

(4.) If it be the moon, a Greenwich date must be got. If it be only known that the observed altitude is a meridian altitude, the meridian passage at Greenwich must be corrected for the longitude, and the Greenwich date obtained by applying the longitude in time as already shown. The moon's semidiameter, horizontal parallax, and declination, must be corrected for the Greenwich date, and the true altitude and declination thus found. When the true altitude and declination are found, the remainder of the process is according to the rule given above, and is the same for all.

Ex. 1.—May 20, 1857, the observed meridian altitude of Castor was 49° 20' 30" (zenith N. of star), the index correction was -3° 40", and the height of the eye above the sea was 18 feet; required the latitude.

Observed altitude..... 49° 20' 30"
Index error..... - 0 3 40
Dip..... 49 16 50
Correction for refraction..... - 0 4 11
True altitude..... 49 12 39
Zenith distance..... 49 11 49
From Nautical Almanac, declin. of Castor
αo Geminorum on May 20 .....
90 0 0
Latitude..... 40 48 11 N.
32 12 1 7 N.
73 0 12 N.

Ex. 2.—October 10, 1857, the observed meridian altitude of α Ophiuchi was 54° 20' 30" (zenith, N. of the star), the index correction was -3° 20", and the height of the eye above the sea was 16 feet; required the latitude.

Observed altitude..... 54° 20' 30"
Index correction..... - 0 3 20
Dip..... 54 17 10
Correction in altitude ..... - 0 3 56
Zenith distance..... 54 13 14
Declination of α Ophiuchi..... 0 4 42
Latitude ..... 54 12 32
90 0 0
35 47 23 N.
12 40 2 4 N.
48 27 30 4 N.

Ex. 3.—January 6, in Long. 59° 30' E., the observed meridian altitude of the sun's lower limb was 60° 20' 30" (zenith N. of the sun), the index correction was +4° 10", and the height of the eye above the sea was 10 feet; required the latitude.

January 6, ship..... 0h 0m 0s
Longitude..... 3 58 0 E.
Greenwich, January 5..... 20 2 0
Sun's declination, Jan. 6..... 22° 35' 43" 6 S.
" " 6..... 22 28 35 9
Diff..... 0 7 7 7
-07816
140300
148116 Part..... 0 6 6
True declination..... 22 29 37 6 S.
Sun's semidiameter..... 16° 18' 2
Observed altitude ..... 60° 20' 30"
Index..... +0 4 10
Dip..... -0 3 7
Semidiameter..... 0 16 18.2
Correction in altitude ..... -0 0 28
Zenith distance ..... 29 22 35.8 N.
Declination..... 22 29 37.6 S.
Latitude..... 6 52 59.2 N.

Ex. 4.—April 4, in Long. 55° 20' E., the observed meridian altitude of the moon's lower limb was 58° 40' 10" (zenith S. of the moon), the index correction was +3' 20", and height of the eye above the sea was 9 feet; required the latitude.

Moon's meridian passage, April 4..... 9h 0m 5s
" " 3..... 8 13.4
Correction ..... 0 47.1
Meridian passage (ship)..... 8 53.5 = 8h 53m 30s
Longitude..... 3 41 20 E.
Ship, April 4 ..... 5 12 10
Moon's
Semidiameter.
Horizontal
Parallax.
Declination.
Ap. 4, Noon.....15° 14' 9" Noon..... 55° 49' 7" 5h 17m 25s 42h 6 N.
Mid.....15 10.4 Mid..... 55 33.4 6h 17 13 19.4
Diff..... 0 4.5 Diff..... 0 16.3 0 12 23.2
36318 36318 69298
338021 282124 116243
374339 318442 185541
Part..... 0 2 Part 0 7.1 Part... 0 0 2.31
Semi..... 15 12.9 H.P. 55 42.6 Decln. 17 23 11.6 N.
Augmentn.. 0 12.9
Semidiam... 15 25.8
Observed altitude..... 58° 40' 10"
Index..... + 0 3 20
Dip..... - 0 2 57
Semidiameter..... 0 15 25.8
Correction..... 0 27 49
Zenith distance ..... 29 23 58.8
Declination..... 17 23 11.6 N.
Latitude..... 13 12 49.6 S.

2. By a single Observed Altitude below the Pole.

Rule.—Correct the altitude, and to it add 90°, and from this sum subtract the declination; the remainder is the latitude.

Ex. 1.—June 4, 1857, the observed meridian altitude of \beta Chamaeleontis, under the S. pole, was 29° 56' 40", the index correction was -2° 40", and the height of the eye above the sea was 15 feet; required the latitude.

Observed altitude..... 29° 56' 40"
Index correction..... -0 2 40
Dip..... -0 3 49
Correction in altitude..... -0 1 41
True altitude..... 29 48 30
Declination of \beta Chamaeleontis..... 78 31 30 S.
Latitude..... 41 17 0.8.

Ex. 2.—Nov. 20, 1857, the observed meridian altitude of \alpha Draconis under the N. pole, was 31° 40' 20", the index correction was +1' 20" and the height of the eye above the sea 20 feet; required the latitude.

tion was +1' 20" and the height of the eye above the sea 20 feet; required the latitude.

Observed altitude ..... 31° 40' 20"
Index..... +0 1 20
Dip..... -0 4 24
Correction in altitude..... -0 1 34
True altitude..... 31 35 42
Declination \alpha Draconis, Nov. 10..... 61 50 9 N.
Latitude..... 39 45 33 N.

In the following example, the altitude is observed by means of an artificial horizon.

June 8, 1857, in Long. 1° 6' W., the observed altitude of the sun's lower limb (in quicksilver horizon) was 121° 9' 4", index correction 1' 20"; required the latitude.

Ship, June 8 ..... 0h 0m 0s
Longitude..... 0 4 24 W.
Greenwich, June 8..... 0 4 24
Sun's declination, June 8..... 22° 52' 43" N.
" " 9..... 22 57 50.8
255630 0 0 7.3
154487 0 0 0.9
409117 22 52 43.9 N.
Semidiameter..... 15° 47' 3"
Observed altitude..... 121° 9' 40"
Index..... -0 1 20
Dip..... 2 121 8 20
Semidiameter..... 0 15 47.3
Correction in altitude..... -0 0 28
True altitude..... 121 8 29.3
Zenith distance ..... 29 10 30.7 N.
Declination..... 22 52 43.9 N.
Latitude..... 52 3 14.6 N.

SECT. II.—BY OBSERVED ALTITUDES OUT OF THE MERIDIAN.

1. By Observations of the Altitude of the Pole-Star (a Polaris).

The tables for this purpose are given in the Nautical Almanac, pp. 513 to 515.

Rule.—From the observed altitude, corrected for the error of the instrument, refraction, and dip of the horizon, subtract 2°.

Reduce the mean time of observation at the place to the corresponding sidereal time.

With the sidereal time found, take out the first correction with its proper sign. If the sign be +, the correction must be added to the reduced altitude; but if it be -, it must be subtracted; in either case the result will give an approximate latitude.

With the altitude and sidereal time of observation, take out the second correction; and with the day of the month and the same sidereal time, take out the third correction. These two corrections, added to the approximate latitude, will give the latitude of the place.

Ex.—January 20, 1857, at 9h 40m P.M. (mean time nearly), in Long. 30° 20' E., the observed altitude of Polaris was 31° 40' 20", the index correction was -2° 20", and height of the eye above the sea, 10 feet; required the latitude.

Ship, Jan. 20 ..... 9h 40m 0s
Longitude ..... 2 1 20 E.
Greenwich, Jan. 20 ..... 7 38 40
Sidereal time at mean noon, Greenwich,
Jan. 20
19h 59m 14s.42
Acceleration for 7h 39m 0 1 15
Mean time at place 9 40 0
Sidereal time 5 40 29.42
Observed altitude 31o 40' 20"
Index 0 2 20
Dip 31 38 0
Correction in altitude 0 3 7
True altitude of Polaris 31 34 53
Reduced altitude 0 1 34
With argument, 5h 40m (first cor.) 31 33 19
Approximate latitude 0 2 0
Arguments, 30o 5h 40m (second cor.) 31 31 19
Arguments, Jan. 20, 1857, 5h 40m (third cor.) 0 32 13
Latitude 0 0 31
31 2 16
31 1 53 N.

3. By Observed Altitudes of a Heavenly Body very near the Meridian.

The sun is the body most usually observed for the purpose. The apparent time from noon or hour angle is supposed to be known; as also the latitude by account, which is supposed to differ from the true latitude only by a small quantity.

Rule.—Add together the tabular logs. of cosines of the latitude by account, and of the declination of the observed body. To these add log. rising, if a table of log. rising be used; or, 6.301030 + log. haversine hour angle, if a table of haversines be used; or twice the log. sine of half the hour angle with the same constant log., if a table of log. sines only be used.

In all cases cast out the tens from the result, and the remainder is the logarithm of a natural number, which, added to the natural sine of the true altitude, will give the natural cosine of the meridian zenith distance. The declination applied to this—as already pointed out in preceding problems—will give the latitude. The hour angle in this rule must be found by applying to the mean time at the ship the equation of time corrected for the Greenwich date.

If the latitude thus found differ much from the latitude by account, the work must be repeated, using the latitude found instead of latitude by account.

Ex.—October 11, 1857, in Long. 114° E., and Lat. by account 46° 10' N., the chronometer showed 4h 38m 10s, being slow on Greenwich 14m 10s. The altitude of the sun's lower limb was observed to be 35° 35' 20", the index error was -2° 20', and the height of the eye above the sea 16 feet; required the latitude.

Chronometer 4h 38m 10s Equation of time
Error (slow) 0 14 10 October 10 0h 12m 59.81
4 52 20 " 11 0 13 15.15
12 0 0 0 0 15.34
Greenw. Oct. 10 16 52 20 15318
284873
300191 0 0 10.8
Sun's declination Equa. of time 0 13 10.61
October 10 6° 44' 0".9 Chron. showed 4 38 10
" 11 6 43.5 Error (slow) 0 14 10
15318 0 21 42.6 Greenw. m. t. 12h 4 52 20
284873 Longitude 7 36 0 E.
107165 0 14 59 24 28 20
Declination S. 6 58 59.9 Equation of time 0 13 10.61
Semidiameter 0 16 4 Hour angle 0 41 30.61
Lat. by acct. 46 10 0 L cosine 9.840459
Declination 6 58 57.9 L cosine 9.996766
Log. 0 41 30.61 19.837225
Log. rising 3.213724
3.050949

Nat. number by table of log. rising, 1124.

Or 19.837225
Hour angle 41h 30m 61s, log. haversine 7.912484
Constant log 6.301030
Natural number, 1124 3.050739
Or 19.837225
Half-hour angle, 20h 45m 3s, 2 log. sin 17.912484
Constant log 6.301030
Natural number, 1124 3.050739
Observed altitude 36° 35' 20"
Index correction 0 2 20
Dip 36 33 0
Semidiameter 0 3 56
Correction in altitude 36 29 4
0 16 4
36 45 8
0 1 11
36 43 57
Mer. zen. dist. 53° 16' 11" N.
Declination 6 59 0 S.
True lat. 45 16 11 N.
Nat. sine 597080
Nat. cosine 598204

4. By two Observed Altitudes of the Sun, and the Time between; having given also the Latitude by Account (Douce's Method).

Rule.—To find the hour angle corresponding to the greater meridian altitude—i.e., the hour from apparent noon, at which the greater altitude is observed.

To the log. secant of the latitude by account, add the log. secant of the sun's declination (the mean between the declinations at the first and second observation); this, rejecting 20, is called the logarithm ratio. To this add the log. of the difference of the natural sines of the two altitudes, and the logarithm of the half-elapsed time from the proper column (taken from nautical tables, which are furnished with special tables for this purpose).

Find this sum in the column of middle times, and take out the time answering thereto; the difference between this and the half-elapsed time will be the time from noon, when the greater altitude was observed.

Or from any table:—

Find the logarithm ratio as before.

Subtract each of the true altitudes from 90°, to get the true zenith distances, and take half their sum and half their difference; and to the logarithm ratio add the sum of the log. of the sines of this semi-sum and semi-difference, and from this sum subtract the log. sine of the half-elapsed time; the remainder is the log. sine of middle time, which find from a table of log. sines, and the difference of the middle time and half-elapsed time is the time from noon, at which the greater altitude is observed.

We have now an observed altitude and the time from noon or hour angle corresponding; the remainder of the solution is therefore the same as in the last case, except that the log. ratio, being the sum of the logarithms of the secants, may be used instead of the sum of the logarithms of the cosines, with the negative sign. If the latitude so found differs much from the latitude by account, the work must be repeated, using the computed latitude for the latitude by account.

This rule is only approximate, and must be used under the following restrictions:—1. The observations must be taken between nine in the forenoon and three in the afternoon. 2. If both the observations be made 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. 3. If one observation be in the forenoon and the other in the afternoon, the interval must not exceed 4h; and in all instances the nearer to noon the greater altitude is observed the better. 4. If the sun's meridian zenith distance be less than the latitude, the limitations are still more con-

Finding
the Lat-
tude.
tracted. 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 3½h. The observations must be taken still nearer to noon if the latitude exceed the zenith distance in a greater proportion.1

As the ship is generally in motion between the two observations, it is necessary to apply to the first altitude a correction for its run in the interval, so as to obtain what the altitude would have been if observed at the same time as the second.

If \theta be the angle between the course of the ship and the bearing of the sun, the correction for run is y = \pm m \cos \theta, where m is the number of minutes in the angle subtended by the ship's run, i.e., number of miles run, and \theta the angle between the bearing of the sun and the course; and the sign + or - is to be used according as the ship is moving towards or from the heavenly body: in the former case \theta is less than 90°, and in the latter greater. It is evident, from the form of the expression, that this correction may be found in a traverse table by looking out the difference between the course and bearing, or what it wants of 8 points for a course, and the ship's run as a distance; the corresponding difference of latitude, multiplied by 60, will then be the number of seconds in the correction for run, which is to be added or subtracted according as the angle is less or greater than 8 points or 90°. Thus, the bearing of the sun was S.E., the run was S.S.E. 18 miles; required the correction for run. Here the angle between the bearing and the course is 2 points, and the distance 18 miles; on entering a traverse table with these, we find the true difference of latitude 16.6. Hence the correction for run is +16.12".

Ex.—April 20, 1857, in Lat. 44° 20' N., Long. 35° 50' E., the following double altitude of the sun was observed:—

Mean Time, nearly. Chronometer. Obs. Alt. Sun's L.L. True Bearing.
10h 40m A.M. 8h 30m 20s 53° 5' 45" S.E. by S.
2 40 P.M. 0 29 40 41 0 30 S.W. by W.

The run of the ship in the interval was S.S.E. 25 miles, the index correction was -4' 40", and the height of the eye above the sea 15 feet; required the true latitude at the last observation.

The angle between the true bearing at first observation and ship's course is 1 point; the true difference of latitude, corresponding to distance 25 and course 1 point in traverse table, is 24.5. Hence correction for run to be applied to the greater altitude is +24' 30".

First observation— Second observation—
Ship, April 19, 22h 40m 0s Ship, April 20, 2h 40m 0s
Longitude..... 2 23 20 E. Longitude..... 2 23 20 E.
Greenw., Ap. 19, 20 16 40 Greenw., Ap. 20, 0 16 40
Sun's Declin. Semidiam. Sun's Declin. Semidiam.
April 19..... 11° 15' 16".7 15° 56' 9" 11° 35' 53".8 15° 56' 6"
" 20..... 11 35 53.8 11 58 19.6 " 20..... 11 35 53.8 11 58 19.6
0 20 37.1 1 93668 0 20 15.8 1 93668
.07319 .94105 .07319 .94105
1 01424 0 17 25 2 88517 0 0 14.1
True decl., 11 32 41.9 11 35 7.9 True decl., 11 35 7.9 11 35 7.9
2 23 8 49.6 11 34 24.8 2 23 8 49.6 11 34 24.8
mean declination.
First obs. alt. 53° 5' 45" Second obs. alt. 41° 0' 30" Index..... -0 4 40 Index..... -0 4 40
53 1 5 40 55 50 Dip..... -0 3 49 Dip..... -0 3 49
52 57 16 40 52 1 Semidiameter.. +0 15 55.9 Semidiameter.. 0 15 56.6
53 13 12.9 41 7 57.6 Cor. in alt..... -0 0 38 Cor. in alt..... 0 1 0
53 12 34.9 True alt.... 41 6 57.6 Cor. for run... +0 24 30 True alt.... 41 6 57.6
True alt.... 53 37 4.0
Lat. by acct. N. 44° 20' 0" Log. secant..... -145520
Mean declin. S. 11 34 24 Log. secant..... .008947
Log. ratio ..... Log. ratio ..... -154467
Greater alt., 53° 37' 4".9 Nat. sine, 80508
Less alt. .... 41 6 57.6 " " 65758
Diff..... 14850 Log. .... 4.171726
Chronometer .... 12h 29m 40s Log. ratio, 0.154467
" .... 8 30 20
2 3 59 20 Elapsed time.
1 59 40 Half elapsed time ..... 0.30213
0 49 4 Middle time ..... 4.62832
1 10 35 Time from noon at which greater altitude is observed.
Log. rising..... 3.67277
Log. ratio ..... -154467
Nat. number..... 3298 3.518303
Nat. sine gr. alt. 80508
83806 Nat. cos. .... 33° 3' 48" N. mer. zen. dist.
11 32 41 decl. at greater alt.
Approx. lat. 44 35 9
Latitude..... 44° 35' 9" Log. secant ..... -147522
Declination .... 11 34 24.8 Log. secant ..... .008947
Log. ratio ..... Log. ratio ..... -156469
Diff. nat. sine... 14850 Log. .... 4.171726
Log. ratio ..... -156469
1h 59m 40s Half-elapsed time .... .30213
0 49 18 Middle time..... 4.63032
1 10 22 Time from noon at which greater altitude is observed.
Log. rising..... 3.67196
Log. ratio..... -15647
3.51449
Nat. number... 3269
N. sine gr. alt... 80508
83777 Nat. cosine... 33° 5' 40" N. mer. zen. dist.
11 32 41 sun's decl.
True lat., 44 38 21 N.
Or, if a table of logarithmic sines, &c., only be used, we have—
1st true corrected alt... 53° 37' 4".9 2d true alt... 41° 0' 57".6
90 0 0 90 0 0
Zenith dist. .... 36 22 55.1 48 53 2.4
48 53 2.4
2 85 15 57.5
Semi-sum zenith dist. .... 42 37 58.7 Log. sine ..... 9.830813
2) 12° 30' 7".3
Semi-difference ..... 6 15 3.6 Log. sine .....
9.036953
Log. ratio .....
-154467
19.022233
Half-elapsed time .... 1h 59m 40s Log. sine ..... 9.897874
0 48 44 9.324339
2) 1 10 56
Half-hour angle... 0 35 23 2 log. sine ..... 18.375800
Constant log.... 6.301090
24.676836
Log. ratio ..... -15647
Nat. number..... 3314 4.320366
Nat. sine gr. alt... 80508
83822 Nat. cosine... 33° 2' 48" mer. zen. dist.
Decl..... 11 32 41
Approx. lat. 44 35 29

Similarly, by taking this approximate latitude, we shall get the true latitude, 44° 38' nearly, as before.

1 See Mackay's Treatises on Longitude and Navigation, &c.; Regisbie Tables, 3d edition; Mendoza Ries's Tables; Norie's and Riddle's Treatises on Navigation, &c.

5. By any Two Altitudes of the same or different Heavenly Bodies, and the Polar Angle between them.

In this case the declinations of the heavenly body or bodies at the two observations are supposed to be known.

If the same body be observed at different times, the polar angle is the elapsed time measured sidereally, if necessary. If different bodies be observed at the same time, this angle is the difference of their right ascensions.

If different bodies be observed, but not at the same time, to the right ascension of the first observed body, add the elapsed time measured sidereally, and the difference between this sum and the right ascension of the second observed body is the polar angle required.

The polar angle being given, the following rule will give the latitude:—

(1.) If the sun be the body observed: from the declination find the polar distance by subtracting it from 90°, if of the same name with the latitude; and adding it, if of an opposite name.

Add together log. sine polar distance at the greater bearing, log. sine polar distance at lesser bearing, and log. haversine of polar angle; reject 10 from the index; the result is the haversine of an arc, which look out, and call arc (1).

(2.) If two stars be observed: find the polar distances; add together log. sine polar distance at greater bearing, log. sine polar distance at lesser bearing, and log. haversine of the polar angle; the result is the haversine of an arc, which find from the table, and add its versed sine to the versed sine of the difference of the polar distances; the result is the versed sine of an arc, which is arc (1), as before.

Find the difference of the polar distances, and take the difference and the sum of this quantity and of arc (1). Add together log. cosecants of the arc (1) and of polar distance at greater bearing, and the halves of the log. haversines of the two arcs just obtained; the sum, rejecting 10 in the index, is the log. haversine of an arc, which take from tables, and call arc (2).

Again, take the difference of arc (1) and the zenith distance at greater bearing, and take the sum and difference of this arc and the zenith distance at less bearing. Add together log. cosecants of arc (1) and of zenith distance at greater bearing, and the halves of the log. haversines of the arcs just found; and the result, rejecting 10 in the index, is the log. haversine of an arc, which call arc (3).

Arc (4) is the difference between arc (2) and arc (3); or the sum if the arcs joining the heavenly bodies at the two observations passes between the zenith and the pole.

Add together log. sine polar distance at greater bearing, log. sine zenith distance at greater bearing, and log. haversine arc (4); the sum, rejecting 10 in the index, is the log. haversine of an arc, which call arc (5).

Add together versed sine of the difference of the zenith distance and polar distance at greater bearing, and the versed sine of arc (5). The result is the versed sine of the colatitude. Find this from the tables, subtract it from 90°, and the result is the latitude.

This method of finding the latitude, which is at once the most general and accurate, is due to the Rev. Dr. Inman, to whose excellent work on Navigation the reader is referred for further information respecting it.

Ex. 1.—March 25, 1857, in Lat. 54° 47' N., and Long. 98° E., the following double altitude of the sun was observed:—

Mean time, nearly. Chronometer. Obs. Alt. L. L. Bearing.
9h 30m A.M. 10h 26m 26° 0' 40" S.E.
1 45 P.M. 2 39 84 6 20 S.W. by S.

Ship, March 24..... Ship, March 25.....
Longitude..... 6 32 E. Longitude..... 1h 45m
Greenwich, Mar. 24, 14 58 Greenwich, Mar. 24, 10 13
Sun's declination, Sun's declination,
March 24..... 1° 30' 57" 2 March 24..... 1° 30' 57" 2
" 25..... 1 54 32 9 " 25..... 1 54 32 9
Diff..... 0 23 35 7 Diff..... 0 23 35 7
20509 09053
88235 88235
1° 08' 45" Part. 0 14 43 9° 58' 59" Part. 0 18 54
T. dec. gr. bearing 1 45 40 7 N. Dec. less bear. 1 49 51 2
90 0 0 90 0 0
P. dis. gr. bearing 88 14 19 3 P. dis. less. bear. 88 10 8 8
Sun's semidiameter..... 16' 14"
Sun's altitude at gr. bearing. Sun's altitude at lesser bearing.
Obs. alt..... 26° 0' 40" Obs. altitude..... 34° 6' 20"
Index correction +0 2 10 Index correction +0 2 10
25 2 50 34 8 30
Dip..... -0 4 11 Dip..... -0 4 11
25 58 39 34 4 19
Semidiameter... 0 16 4 Semidiameter... 0 16 4
25 14 43 34 20 23
Cor. in alt. for refr. and parallax -0 1 50 Cor. in altitude -0 1 18
25 12 53 True altitude... 34 19 5
Cor. for run... +0 16 35 90 0 0
True altitude... 25 29 29 Zen. dis. less. bear. 55 40 55
90 0 0
Zen. dis. gr. bear. 63 30 31
To find arc (1).
P. dist. gr. bearing 88° 14' 19" 3 L. sin..... 9.999794
P. dist. less. bear. 88 10 8 8 L. sin..... 9.999778
Chronometer..... 10h 26m
14 39
4 1 3 Elapsed time or polar angle.
L. haversine..... 9.439255
L. haversine arc (1)... 9.438827
Arc (1)..... 63° 12' 52"
To find arc (2).
Arc (1)..... 63° 12' 52" Log. cosec..... 0.49296
P. dist. gr. bear. 88 14 19 3 Log. cosec..... 0.00206
Difference..... 25 1 26 7
P. dist. less. bear. 88 10 8 8
Sum..... 113 11 35 3 ½ log. haversine.... 4.921590
Difference..... 63 8 42 1 ½ log. haversine.... 4.718970
Log. haversine arc (2)... 9.690062
Arc (2)..... 85° 50' 12"
To find arc (3).
Arc (1)..... 63° 12' 52" Log. cosec..... 0.49296
Zen. dist. gr. bear. 63 30 31 Log. cosec..... 0.45176
Difference..... 0 17 39
Zen. dist. less. bear. 55 40 55
Sum..... 55 58 34 ½ haversine..... 4.671433
Difference..... 55 23 16 "..... 4.667217
Log. haversine arc (3)..... 9.436127
Arc (3)..... 62° 59' 48"
To find arc (4).
Arc (2)..... 85° 50' 12"
Arc (3)..... 62 59 48
Arc (4)..... 25 50 24
To find arc (5).
P. dist. gr. bearing 88° 14' 19" 3 L. sin..... 9.999794
Zen. dist. " 63 30 31 L. sin..... 9.951824
Difference..... 24 43 48 3 Log. haversine.... 8.698906
Log. haversine arc (5)..... 8.650524
Arc (5)..... 24° 9' 12"
Nat. ver. sine..... 24° 9' 19" = 0.069450
Nat. ver. sine..... 24 43 48 3 = 0.091710
Nat. ver. colatitude..... = 0.181100
Colatitude..... 35° 1' 16"
90 0 0
Latitude..... 54 58 44

Finding the longitude. Ex. 2.—Nov. 30, 1857, in Lat. by account 30° N., the following altitudes of stars were taken at the same time; required the true latitude.

True alt. \beta Orionis. 42° 45' 15". Bearing. S.W. True alt. \alpha Hydræ. 39° 35' 15". Bearing. S.S.E.
From Nautical Almanac we find—
\beta Orionis, R.A. 5h 7m 44s Declination, S..... 8° 21' 57" 90 0 0
N. P. D. at G. B. .... 98 21 57 8 2 30
\alpha Hydræ, R. A. 9 20 36 Declination S..... 90 0 0 98 2 30
Polar angle ... 4 12 52 N. P. D. at L. B. .... 98 2 30 98 2 30
P. D. at G. B. ... 98 21 57 I. sine ..... 9.995353 9.995353
" L. B. ... 98 2 30 I. sine ..... 9.995708 9.995708
" Diff. .... 0 19 17 Haversine polar angle .... 9.438871 9.438871
Haversine ..... 9.429952 9.429952
Arc ..... 62° 29' 58".
Nat. ver. sine ..... 62° 29' 58" = 0.537993
248
Nat. ver. sine ..... 0 19 17 = 0.000016
Nat. ver. sine arc (1) ..... 0.538255
Arc (1) ..... 62° 30' 1".
To find arc (2).
Arc (1) ..... 62° 30' 1" Log. cosec .... .052069
P. D. at G. B. .... 98 21 57 " ..... .004646
Diff. .... 35 51 55
P. D. at L. B. .... 98 2 30
Sum ..... 133 54 25 ½ haversine.... 4.953876
Diff. .... 62 10 34 " ..... 4.712946
Haversine arc (2) ..... 9.733537
Arc (2) ..... 94° 45' 8".
To find arc (3).
True alt. at G. B. .... 42° 45' 15"
90 0 0
Z. D. at G. B. .... 47 14 45
True alt. at L. B. .... 39 35 15
90 0 0
Z. D. at L. B. .... 50 24 45
Arc (1) ..... 62° 30' 1" Log. cosec .... .052069
Z. D. at G. B. .... 47 14 45 " ..... .134142
Diff. .... 15 15 16
Z. D. at L. B. .... 50 24 45
Sum ..... 65 40 1 ½ Haversine.... 4.734159
Diff. .... 35 9 29 " ..... 4.480537
Haversine arc (3) ..... 9.400407
Arc (3) ..... 60° 11' 18".
To find arc (4).
Arc 2 ..... 94° 45' 8"
Arc 3 ..... 60 11 18
Arc (4) ..... 34 33 50
To find arc (5).
P. D. at G. B. 68° 21' 57" I. sine ..... 9.995354
Z. D. at G. B. 47 14 45 " ..... 9.865858
Diff. .... 51 7 12 Haversine arc (4) .. 8.945729
Log. haversine arc (5) ..... 8.806941
Arc (5) ..... 29° 20' 4".
Nat. ver. sine ..... 51° 7' 12" = 0.372263
" ..... 29 20 4 = 0.128225
Nat. ver. sine, colatitude ..... 0.500534
Colat. .... 60° 2' 7"
90 0 0
Lat. N. .... 29 57 53

CHAP. IV.—ON FINDING THE LONGITUDE BY OBSERVATION.

SECT. I.—INTRODUCTION.

The observations necessary to determine the longitude by this method are the altitudes of the sun or other heavenly body, or the distance between the sun and moon, the moon and a planet, or the moon and a fixed star near the ecliptic,

together with the altitude of each. The planets used in the Nautical Almanac for this purpose are the following:—Venus, Mars, Jupiter, and Saturn. The stars are, \alpha Arietis, Aldebaran, Pollux, Regulus, Spica Virginis, Antares, \alpha Aquila, Pomalhaut, and \alpha Pegasi; and the distances of the moon's centre from the sun, and from one or more of these planets and stars, are contained in the xiii. to xviii. pages of the month, at the beginning of every third hour mean time by the meridian of Greenwich. The distance between the moon and one of these objects is observed with a sextant; and the altitudes of the objects are taken as usual with a sextant or a Hadley's quadrant.

In the practice of this method it will be found convenient to be provided with three assistants. Two of these are to take the altitudes of the sun and moon, or moon and star, at the same time that the principal observer is taking the distance between the objects; and the third assistant is to observe the time, and write down the observations. In order to obtain accuracy, it will be necessary to observe several distances, and the corresponding altitudes, the intervals of time between them being as short as possible; and the sum of each divided by the number will give the mean distance and mean altitudes; from which the time of observation at Greenwich is to be computed by the rules to be explained.

If the sun or star from which the moon's distance is observed, be at a proper distance from the meridian, the time at the ship may be inferred from the altitude observed at the same time with the distance. In this case the chronometer is not necessary; but if that object be near the meridian, the chronometer is absolutely necessary, in order to connect the observations for ascertaining the mean time at the ship and at Greenwich 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 time being known from which the altitudes of the bodies may be computed, the longitude may be determined.

A set of observations of the distance between the moon and a star or planet, 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 estimated 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 cusps, 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 transparent part of the horizon-glass.

SECT. II.—TO FIND MEAN TIME BY A SINGLE OBSERVATION OF THE SUN OR MOON, OR OTHER HEAVENLY BODY.

PROB. 1.—Given the latitude of a place, the altitude, and declination of the sun, to find the true time and the error of the chronometer.

Rule.—If the latitude and declination are of different signs, take their sum; otherwise take their difference. From the natural cosine of this sum or difference, take the natural sine of the altitude (corrected); find the logarithm of the remainder, and to it add the log. secants of the lati-

Finding the longitude and declination; the sum will be the log. rising of the horary distance of the sun from the meridian, and hence the apparent time will be known; by applying to it the equation of time, the mean time may be found, and the error of the chronometer.

Or, if a table of log. haversines be used, take the difference or sum of the latitude and declination as before. Under this difference place the zenith distance of the sun, and take their sum and difference.

Take the sum of half of the log. haversines of these quantities and of the log. secants of the latitude and declination; the result, rejecting the tens, is the log. haversine of the hour angle.

Or, if a table of sines only be used, add together one-half of the log. sines of half the arcs above found, of the log. secants of latitude and declination; the result is the log. sine of one-half the hour angle.

Ex.—March 25, 1857, at 3h 30m p.m. nearly, in Lat. 53° N., and Long. 35° 20' W., when a chronometer showed 5h 46m 20s, the observed altitude of the sun's lower limb was 23° 12' 10", and the index correction was -4' 20", and the height of the eye above the sea 20 feet; required the true time, and error of chronometer.

Ship, March 25..... 3h 30m 0s
Longitude..... 2 21 20
Greenwich, March 25..... 5 51 20
Sun's declination.
March 25..... 1° 54' 32" N. Sun's semidiam.... 16' 3" 7
" 26..... 2 18 6 Equation of time + 6m 3s 20
Diff..... 0 23 33.1 Diff. for 1h..... 0.767
5
81334 3.835
88328 Diff. for 30m is \frac{1}{2}..... .383
149562 0 5 44 " 20 is \frac{1}{2}..... .256
Sun's dec... 2 0 16.9 " 1 is \frac{1}{2}..... .013
" 20 is \frac{1}{2}..... .004
Equation, 5m 58.703 = 4.491
Sun's obs. altitude..... 23° 12' 10"
Index correction..... - 0 4 20
23 7 50
Dip..... - 0 4 24
23 3 26
Semidiameter..... + 0 16 3.7
23 19 29.7
Correction in altitude..... - 0 2 13
True altitude..... 23 17 16.7
90 0 0
Zenith distance..... 66 42 43.3
Lat. N..... 53° 0' 0" Log. sec..... .220537
Declin. N..... 2 0 16.9 " ..... .000266
Diff..... 50 59 43.1
Nat. cosine..... .0623350
Nat. sin. 23° 17' 16", .385349
.0234031
Log..... 4.369272
Log. rising, 4.390075
Hour angle..... 3h 29m 24s
Equation of time..... + 0 5 59
Ship's mean time..... 3 35 23
Longitude..... 2 21 20
Greenwich mean time..... 5 56 43
Chronometer showed..... 5 46 20
0 10 23
Or error of chronometer is 10m 23s slow on Greenwich.
Or—
Lat..... 53° 0' 0" N. Log. sec..... .220537
Declination... 2 0 16.9 N. Log. sec..... .000266
50 59 43.1
Zenith dist.... 66 42 43.2
Sum..... 117 42 26.3 \frac{1}{2} haversine... 4.932397
Diff..... 15 43 0.1 " 4.135845
Log. haversine hour angle..... 5.289045
Hour angle..... 3h 29m 24s, as before.

By taking another observation at the interval of a few days, the daily rate of the chronometer may be found.

PROB. 2.—Given the latitude of a place, and the altitude of a known fixed star, to find the mean time of observation and error of chronometer.

The right ascension of the mean sun or sidereal time must be corrected for the Greenwich date of observation. The star's hour angle must then be found as in the last problem. To the hour angle thus found add the star's right ascension, and from the sum (increased, if necessary, by twenty-four hours) subtract the right ascension of the mean sun; the remainder is the mean time at the place at the instant of observation.

Ex.—January 16, 1857, at 8h p.m. (mean time, nearly), in Lat. 49° 57' N., and Long. 32° 10' W., when a chronometer showed 10h 23m 30s, the observed altitude of Regulus E. of meridian was 8° 20' 30", the index correction was -5' 20", and the height of the eye above the sea, 20 feet; required the mean time, and error of chronometer on Greenwich mean time.

Ship, Jan. 16..... 8h 0m 0s
Longitude..... 2 8 40 W.
Greenwich, Jan 16..... 10 8 40
Observed altitude..... 8° 20' 30"
Index correction..... - 0 5 20
8 15 10
Dip..... - 0 4 24
8 10 46
Refraction..... - 0 6 27
True altitude..... 8 4 19
90 0 0
True zenith distance..... 81 55 41
Star's right ascension..... 10h 0m 46s
Star's declination..... 12° 39' 50"
Right ascension of mean sun..... 19h 43m 28s
Correction for 10h..... 0 1 38.6
" 8h..... 0 0 1.3
" 40h..... 0 0 0.1
True right ascension of mean sun, 19 45 8
Latitude..... 49° 57' 0" N. Log. sec..... .188493
Declination.... 12 39 50 N. Log. sec..... .010696
Diff..... 36 57 10
Zenith dist.... 81 55 41
Sum..... 118 52 51
Half..... 59 26 25.5 Log. sin..... 9.935054
Difference..... 44 58 31
Half..... 22 29 15 Log. sin..... 9.582011
2) 19.716854
Log. sin... 9.58427
Hence half hour angle..... 3h 4m 49s
2
6 9 38
24 0 0
Star's hour angle..... 17 50 22
" right ascension..... 10 0 46
27 51 8
Sun's right ascension..... 19 45 8
Ship's mean time..... 8 6 0
Longitude..... 2 8 40 W.
Greenwich mean time..... 10 14 40
Chronometer..... 10 23 30
Error (fast on Greenwich)..... 0 8 50

SECT. III.—TO FIND THE LONGITUDE BY MEANS OF A CHRONOMETER.

In order to find the longitude at sea by means of a chronometer, its daily rate in mean solar or sidereal time must be established by observations made at some particular place, and its error ascertained for the meridian of that or of any other known place.

An observatory is the most proper and convenient place for this purpose, as there the rate and error may be both determined with the utmost accuracy by equal altitudes, or transits over the meridian of the sun or stars. But if an observatory is not adjacent, the rate and error of the chronometer may be found by altitudes taken daily for several

NAVIGATION.

Finding days from the horizon of the sea, or by the method of re-
the Longi- flection from an artificial horizon.

tude. If by these observations the daily rate is found to be nearly the same,—that is, if the chronometer gains or loses nearly the same portion of absolute time daily,—it may be depended on for finding the longitude; but if its rate is unequal, it must be rejected, as the longitude inferred from it cannot be expected to be accurate.

It would be proper to have two chronometers, and that they should be wound up at different stated times of the day, so that if one should be found stopped, either through neglect in winding up or otherwise, it may be set by the other, observing to apply the former interval of time between them, and the change in their rates of going in that interval.

PROB.—To find the longitude of a ship at sea by a chronometer.

Let several altitudes of the sun or a fixed star or planet be observed, and find the true mean altitude, with which, and the ship's latitude, and declination of observed body, compute the mean time of observation as in sect. ii.

To the mean of the times of observation, as shown by the chronometer, apply the error and accumulated rate. Hence the mean time under the meridian of the place where the error and rate were established will be known; to which apply the difference of longitude in time between that place and Greenwich, and the mean time of observation under the meridian of Greenwich will be obtained. The difference between the time at the place of observation and that at Greenwich will be the longitude of the ship in time; and it is east or west according as the time is later or earlier than the Greenwich time.

Ex. 1.—May 30, 1857, at 3h p.m. (mean time nearly), in Lat. 30° 20' S., and Long. by account 155° 10' E., when a chronometer showed 4h 40m 50s, the observed altitude of the sun's L.I. was 22° 10', the index correction was -7° 10', and the height of the eye above the sea was 20 feet; required the longitude.

On May 20 the chronometer was fast on Greenwich mean time 4m 50s, and its daily rate was 2m 5 losing.

Ship, May 30 ..... 3h 0m 0s
Longitude ..... 10 20 40 E.
Greenwich, May 29 ..... 16 39 30
Chronometer daily rate ..... 2m 5 losing
9
22.5
12h is \frac{1}{2} ..... 1.2
4h is \frac{1}{2} ..... .4
30m is \frac{1}{2} ..... .05
9m 30s is \frac{1}{2} nearly... .0166
Accumulated rate ..... 24.1666
Chronometer showed ..... 4h 40m 50s
4 41 14.166
Original error (fast) ..... 0 4 50
4 36 24.17
12 0 0
Greenwich mean time ..... 16 36 24.17
Sun's declination, Equation of time, subtractive—
May 29, N.... 21° 39' 43".9 3m 1s 73
30..... 21 48 46.1 2 54.19
0 9 2.2 0 7.54
0.15987 0.15967
1.25383 3.15835
1.41350 0 6 57 3.31803 0 5.2
Sun's decl. at obs. 21 46 40.9 Equation of time, 2 56.53
Sun's semidiameter, 15° 48".5.
Observed altitude ..... 22° 10' 0"
Index correction ..... 0 7 10—
22 2 50
Dip ..... 0 4 24
21 58 26
Semidiameter ..... 0 15 48.5
22 14 14.5
Correction in altitude ..... 0 2 14—
22 12 0.5
Latitude.... 30° 20' 0" S. Log. secant..... .063933
Declin.... 21 45 18.2 N. " ..... .029140
52 5 18.2
Nat. cosine 52 5 18.2 = 61434.7
Nat. sine 22 12 0.5 = 37784.1
23650.6 Log..... 4.373831
Log. rising..... 4.466909
Hour angle..... 3h 0m 3s
Equation of time..... 0 2 56
2 67 7
Ship's mean time..... 24 0 0
25 57 7
Chronometer..... 16 36 24.17
10 20 42.83
Longitude..... 155° 10' 42".45 E.

Ex. 2.—August 20, 1857, at 0h 30m A.M. (mean time nearly), in Lat. 50° 20' N., and Long. 142° 15' E., when a chronometer showed 2h 41m 13s, the observed altitude of \alpha Aquilæ, west of meridian, was 36° 59' 50", the index correction was +6' 20", and height of the eye above the sea 20 feet; required the longitude.

On August 4, at noon, the chronometer was slow on Greenwich mean time 17m 50s, and its daily rate was 4m 5 losing.

Ship, August 19 ..... 12h 30m 0s
Longitude ..... 9 29 0 E.
Greenwich August 19..... 3 1 0
Interval..... 15h 3m Daily rate..... 4.5
15
225
45
67.5
3h is \frac{1}{2} ..... .5
Accumulated rate ..... 1 8
Chronometer showed..... 2 41 13
2 42 21
Original error..... 0 17 50
Greenwich mean time..... 3 0 11
Observed altitude..... 36° 59' 50
Index ..... 0 6 20+
37 6 10
Dip..... 0 4 24—
37 1 46
Refraction..... 0 1 17—
True altitude..... 37 0 29
90 0 0
True zenith distance..... 52 59 31
Right ascension of mean sun (sidereal time)—
Aug. 19 ..... 9h 51m 7.80
Part for 3h..... 0 0 29.57
11h..... 0 0 0.03
9 51 37.4
Star's right ascension..... 19h 43m 51.33
Star's declination..... 8° 29' 43".N.
Latitude..... 50° 20' 0" N. Log. secant.... .194962
Declination.... 8 29 43 N. Log. secant.... .004791
41 50 17
Zenith dist... 52 59 31
Sum ..... 94 49 48
Half ..... 47 24 54 Log. sine..... 9.867039
Difference.... 11 9 14
Half ..... 5 34 37 Log. sine..... 8.987594
2)19.054386
Log. sine half-hour angle..... 9.527193
Half-hour angle..... 1h 18m 41".6
2
Hour angle ..... 2 37 23.2
Star's right ascension ..... 19 43 51.33
22 21 14.53
Right ascension of mean sun ..... 9 51 37.4
Ship, mean time ..... 12 29 37.13
Greenwich ..... 3 0 11
9 29 26.13
Longitude..... 142° 21' 31".E.

SECT. IV.—TO FIND THE LONGITUDE BY MEANS OF A LUNAR DISTANCE.

PRON. 1.—Given the apparent distance between the moon and sun, or a fixed star or planet, and the apparent and true altitudes of these bodies, to find the true distance, or, as it is called, to clear the distance.

Rule 1. (Borda's method).—Add together the logarithmic cosines of the true altitudes, the logarithmic secants of the apparent altitudes, the logarithmic cosines of one-half the sum of the apparent altitudes and apparent distance, and of this last arc, less the apparent distance. From one-half of this sum subtract the logarithmic cosine of one-half of the true altitudes; the result, rejecting 10 in the index, is the logarithmic sine of an arc. Find the logarithmic cosine of this arc, and add to it the logarithmic cosines of one-half of the true altitudes; the result, rejecting 10 in the index, is the logarithmic sine of one-half of the true distance required. Or,—

Rule 2.—Find the auxiliary angle A (which is given in the nautical tables of Inman, Norie, Riddle, and others).

Take the versed sine of the difference of the true altitudes. To it add the versed sines of the sum and difference of the auxiliary angle A and the apparent distance; under the difference of the apparent altitudes place the auxiliary angle A, and take their sum and difference; add together the versed sines of these two arcs, and subtract from the former quantity; the result is the versed sine of the true distance.

Ex.—Required the true distance of the moon from the sun, having given—

App. alt. sun ... 34° 21' 32" True alt. sun ... 34° 20' 14"
App. alt. moon ... 57 11 25 True alt. moon ... 57 40 11
App. distance of centres ... 35° 47' 24"
By Rule 1—
True alt. moon ... 57° 40' 11" L. cos. .... 9.728227
True alt. sun ... 34 20 14 L. cos. .... 9.916839
App. alt. moon ... 57 11 25 L. sec. .... 2.66131
" sun ... 34 21 32 L. sec. .... 2.083267
91 32 57
App. distance ... 35 47 24
2) 127 20 21
63 40 10 L. cos. .... 9.646941
35 47 24
27 52 46 L. cos. .... 9.946353
2) 39.587758
19.793879
Sum of true alt. ... 92° 0' 25"
Half ... 45 0 12 L. cos. .... 9.841746
L. sin. .... 9.952133
Arc ... 63° 35' 10"
L. cos 63° 35' 10" 9.648216
L. cos 45 0 12 9.841746
L. sin half true dist. .... 9.489962
Half true distance ... 17° 59' 37"
2
True distance ... 35 59 14

Rule 2.—Auxiliary angle A (taken from the tables) being 60° 25' 16".

Moon's true alt. ... 57° 40' 11"
Sun's true alt. ... 34 20 14
Diff. .... 23 19 57 Vers. .... 0.081777
A .... 60 25 16
App. distance ... 35 47 24
Sum .... 96 12 40 Vers. .... 11.08192
Diff. .... 24 37 52 Vers. .... 0.090990
App. alt. sun ... 34 21 32
" moon ... 57 11 25 12.80069
Diff. .... 22 49 53
A .... 60 25 16
Sum .... 83 15 9 Vers. 0.882506
Diff. .... 37 35 23 Vers. 0.207601
True dist. 35° 59' 18" 10.90107
Nat. vers. .... 0.190862

PRON. 2.—To find the longitude by a lunar observation. Find a Greenwich date, and to this date take out the moon's horizontal parallax and semidiameter. Increase the semidiameter for the augmentation corresponding to the moon's altitude.

Find the apparent and true altitudes of the centre of the moon and of the sun or star which is observed, and the apparent central distance. From these elements compute the true distance as in the last problem, and find the mean time at Greenwich corresponding to it, as shown in chap. ii., sect. iv., p. 38.

If the sun or star be not too near the meridian at the time of observation, compute the mean time at the ship from its altitude; if it be too near, compute the mean time from the moon's altitude. The difference between the mean times of observation at the ship and Greenwich will be the longitude of the ship in time, which is E. or W. according as the time at the ship is later or earlier than the Greenwich mean time.

Ex. 1.—January 3, 1857, at 2h 30m P.M., in Lat. 49° 20' N., and Long. 15° 40' E., the following lunar was taken:—

Obs. alt. Sun's L. L. Obs. alt. Moon's L. L. Obs. dist. N. L.
24° 40' 10" 21° 36' 40" 90° 28' 10"
Index error... +0 1 20 -0 1 10 0 1 20

Height of eye above the sea, 18 feet; required the longitude.

Ship, Jan. 3..... 2h 30m 0s Sun's Semidiameter.
Longitude..... 1 2 40 E. 10' 18" 2
Greenwich, Jan. 3... 1 27 20
Sun's declination. Equation of time.
Jan. 3, S. 22° 48' 36" 6 +4m 54s 03
" 4, S. 22 42 24 6 5 21.38
1.21885 0 6 14 1.21885
1.46055 2.59726
2.67940 0 0 22 6 3.81611
22 48 16
1.7
4 55.73
Moon's semi., 3, noon 16' 9" 2 Moon hor. par. 69' 8" 9
" mid. 16 10 7 " 59 14 2
0 1 5 0 5 3
91782 91782
3.85733 3.30915
4.77515 4.22697
0 0 1 0 0 7
16 9 3 59 9 6
Aug. 0 6
16 16 3
Sun's Alt. Moon's Alt. Lunar Dist.
Obs. alt. 24° 40' 10" 21° 36' 40" 90° 28' 10"
Ind. cor. +0 1 20 -0 1 10 +0 1 20
24 41 30 21 35 30 90 29 30
Dip. .... 0 4 11— -0 4 11 Sun's semi... 0 16 18
24 37 19 21 31 19 Moon's semi... 0 16 15
Semi .... 0 16 18 0 16 15 App. dis... 91 2 3
24 53 37 21 47 34
Cor. in alt. 0 1 57— +0 52 23
8
True alt. 24 51 40 Tr. alt. 22 40 5
90 0 0
65 8 20 true zenith distance.
Aux. angle A..... 60° 11' 46"
0 0 2
0 0 0
60 11 48

To find ship's mean time.

Sun's decl. .... 22° 48' 10" S. Sec. .... 0.035348
Latitude .... 49 20 0 N. Sec. .... 0.185981
72 8 16
Sun's zen. dist. .... 65 8 20
Sum .... 137 16 36
Half .... 68 38 18 Sin. .... 9.959099
Diff. .... 6 59 56
Half .... 3 29 58 Sin. .... 8.785604
2) 18.67022
9.489011
Half-hour angle..... 1h 11m 39s.6
2
Hour angle..... 2 23 19
Equation of time ..... 0 4 53.7+
Ship mean time..... 2 28 12.7

To find Greenwich mean time.

Sun's true alt. .... 24h 51m 40s
Moon's true alt. .... 22 40 5
Diff. .... 2 11 35
Vers. .... 0000726
Apparent dist. .... 91 2 3
6
Auxiliary angle ..... 60 11 48
Sum ..... 151 13 51
Vers. .... 1876447
120
Difference ..... 30 50 15
Vers. .... 0141338
Sun's appar. alt. .... 24 53 37
Moon's do. .... 21 47 34
2018674
Difference ..... 3 6 3
Auxiliary angle..... 60 11 48
Sum ..... 63 17 51
Vers. .... 0550421
220
Difference..... 57 5 45
Vers. .... 0456581
183
1007405
(1.) True distance ..... 90h 38m 44s 1011269
(2.) Distance at noon..... 89 52 28 054
(3.) Distance at 3 P.M. .... 91 30 14 215
Diff. (1.) and (2.), 6h 46m 16s Prop. log. .... 59000
" (2.) and (3.), 1h 37m 46s " .... 26508
Interval..... 1h 25m 11s 32492
∴ Greenwich mean time ..... 1 25 11
Ship mean time ..... 2 28 13
Longitude..... 1 3 2
Or..... 15h 45m 30s E.

Ex. 2.—February 28, 1857, at 7h 40m P.M., in Lat. 55°, and Long. 57° 20' W., the following star lunar was taken:—

Obs. Alt. Pollex Obs. Alt. Moon Obs. Dist. Further
E. of Mer. Lower Limb. Limb.
60h 10m 0s 46h 10m 0s 70h 40m 20s
Index error -0 2 10 +0 1 20 -0 3 10

The height of the eye above the sea was 20 feet; required the longitude.

Ship, Feb. 28..... 7h 40m 0s
Longitude ..... 3 49 20 W.
Greenwich, Feb. 28..... 11 29 20

Right ascension of mean sun,

Feb. 28..... 22h 33m 04.09 Star's R. A. .... 7h 36m 35s
Part for 11h ... 0 1 48.42 Declination.... 23 22 10 N.
" 29h ... 0 0 4.76
" 20h ... 0 0 .05
22 34 53.32
Moon's semi. 28, noon... 16h 23m 2 Hor. par. noon... 59h 59m 8
" mid... 16 20.0 " mid... 59 48.3
0 3.2 0 11.5
-01911 -01911
3.52827 2.97273
3.54738 Part 0 3.1 2.99184
16 20.1 59 48.3
Aug. 0 12.5
18 32.6
Star's Altitude. Moon's Altitude. Apparent Distance.
Obs. alt. 60h 10m 0s Obs. alt. 46h 10m 0s Obs. .... 70h 40m 20s
Ind. cor. -0 2 10 +0 1 20 -0 3 10
46 11 20 70 37 10
Dip .... -0 4 24 -0 4 24 Semi..... -0 16 32.6
App. alt. 60 3 26 46 6 55 App. dis. 70 20 37.4
Refr. .... -0 0 34 0 16 32.6
True alt. 60 2 52 App. alt. 46 23 28.6
50 0 0 Cor. alt. 0 39 48
33
T. Z. dis. 29 57 8 True alt. 47 3 49.6
Auxiliary angle A..... 60h 23m 38s
21
0
60 23 59

To find ship's mean time.

Star's dec..... 25h 22m 10 N. Sec ..... 0.055265
Latitude ..... 55 0 0 N. Sec ..... 0.241469
Diff. .... 28 37 50
Star's zen. dis. .... 29 57 8
Sum..... 58 34 58 Half havers.... 4.675736
Diff. .... 3 19 18 Half havers.... 3.462142
Hav. hour angle 8.434852
Hour angle (E. M.) ..... 22h 44m 1s
Star's R. A..... 7 36 35
30 20 36
R. A. mean sun..... 22 34 53
Ship mean time..... 7 45 43

To find Greenwich mean time.

Star's true altitude .... 60h 2m 52s
Moon's " ..... 47 3 49.6
Diff. .... 12 59 2.4
Vers. .... 0025564
App. distance ..... 70 20 37
A ..... 60 23 59
Sum..... 130 44 36
Vers. .... 1652539
133
Diff. .... 9 56 38
Vers. .... 0014991
30
Star's app. altitude..... 60 3 25
Moon's " ..... 46 23 22.6
13 39 57.4
A ..... 60 23 59
Sum..... 74 3 50.4
Vers. 0725202
261
Diff. .... 46 44 1.6
Vers. 0314605
7
1040075
Vers. true distance ..... 0653184
True distance ..... 60h 42m 13s
Distance at 9h ..... 71 16 1
" midnight ..... 69 27 42
First diff. .... 1h 33m 48s Prop. log..... .28307
Second ..... 1 48 10 Prop. log..... .22058
.05149
Time from 9h ..... 2h 36m 14s
9 0 0
Greenwich mean time ..... 11 36 14
Ship mean time ..... 7 45 43
3 50 31
Longitude...57° 37' 45" W.

CHAP. V.—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 article MAGNETISM, where the method of communicating magnetism to compass-needles is also fully described.)

Besides the variation, there is also the deviation of the compass arising from the local attraction of the iron on board ship, of which we have already given an account in the former part of this article. This deviation is always taken into account in ships of the Royal Navy, but not

always in ships belonging to the mercantile navy. In the latter case the variation is the whole difference between the observed bearing of the sun and the compass bearing; in the former, allowance must be made for deviation. We shall take deviation into account, as it is easy to omit it when it is not required.

To correct the variation for deviation, it will be sufficient to place under the variation, when determined by observation with its proper name, the deviation with a name opposite to its true name. Add these when the names are alike, and subtract the less from the greater if the names are different; and the remainder, with the name of the greater, is the true variation.

Prin. 1.—Given the latitude of a place, and the sun's declination, and the sun's magnetic amplitude, to find the deviation. [Obs.—The amplitude is the distance from the east point at which it rises, or from the west point at which it sets.]

Rule.—To the log. secant of latitude add log. sine of sun's declination; the sum, rejecting 10 from the index, will be the log. sine amplitude, which is east if the body is rising, and west if it be setting. The variation is the difference between the true and magnetic amplitudes if these be of the same name, and their sum if of different names. Also the variation is east if the true bearing is to the right of the compass bearing, west if the true bearing is to the left of the compass.

Ex.—May 18, 1857, about 5h 25m A.M., in Lat. 51° 5' N., Long. 143° W., the sun rose by compass E. 6° 40' S., the ship's head being E.; required the variation.

Ship, May 17..... 17h 25m 0s
Longitude..... 9 32 0 W.
26 57 0
24 0 0
Greenwich, May 18..... 2 57 0
Sun's declin., 18... 19° 36' 18" N.
" 19... 19 49 36" 1
0.91039 0 12 58.1
1.14244 0 0 0
2.05283 0 1 38
19 37 44 Sin..... 9.526303
Latitude..... 51 5 0 N. Sec..... 0.201909
Sin amplitude. 9.728212
True bearing..... E. 32° 20' 0" N.
Compass..... E. 6 40 0 S.
39 0 0 W.
Deviation..... 8 50 0 W.
Variation..... 47 50 0 W.

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 fifteen minutes and the dip of the horizon. Thus, if an observer be elevated 18 feet above the level of the sea, the amplitude should be taken at the instant the altitude of the sun's lower limb is 19' 11".

Prin. 2.—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.—Find the polar distance by adding 90° to the declination if the altitude and declination have unlike names, and subtracting the declination from 90° if they have the same name.

Take the difference of the latitude and altitude, and obtain the sum and difference of this quantity and the polar distance.

To the log. secants of the altitude and declination add the halves of the log. haversines of the last two arcs; the result, rejecting 10 from the index, is the haversine of the true bearing, which take from the table. Or if a table not contain-

ing haversines is used, to the log. secants as before add the log. sines of half the arcs obtained as before; one-half of the result is the log. sine of half the true bearing. Double the arc taken out of the table is the true bearing.

Mark the true bearing N. or S., according as the latitude is N. or S., and E. or W., according as the observed body is E. or W. of the meridian. The variation then can be found as before.

Ex.—On May 5, 1857, about 8h 10m 0s A.M., mean time, in Lat. 51° 10' N., Long. 140° W., the sun bore by compass S. 65° 25' E., and the observed altitude of the sun's lower limb at the same time was 25° 30' 10", the index correction was +1' 20", and the height of the eye above the sea 15 feet, the ship's head being N.W.; required the variation of the compass.

Ship, May 4..... 20h 10m 0s
Longitude..... 9 20 0
Greenwich, May 5..... 5 39 0
Sun's semidiameter..... 0° 15' 53"
Sun's declination, May 5..... 16° 19' 13" 2 N.
" " 6..... 16 35 8
0.63985 0 16 55
1.02696 0 3 53
1.66681 16 23 6
90 0 0
N.P.D..... 73 36 54
Sun's altitude—
Obs. altitude..... 28° 30' 10"
Index correction..... + 0 1 20
28 31 30
Dip..... 0 3 49
28 27 41
Semidiameter..... 0 15 53
28 43 34
Correction in altitude..... 0 1 38
True altitude..... 28 41 56
Latitude..... 51° 10' 0" Sec..... 202693
Altitude..... 28 41 56 Sec..... 056923
Diff..... 22 28 4
Polar dist..... 73 36 54
Sum..... 96 4 58 Half havers..... 4.871355
Diff..... 51 8 50 Half havers..... 4.635130
9.766101
True bearing..... N. 99° 37' 22" E.
Magnetic bearing..... N. 114 35 0 E.
14 57 38 E.
Deviation..... 4 50 0 E.
Variation..... 19 47 38 E.

CHAP. VI.—OF THE TIDES.

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

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

NAVIGATION.

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

Moon's Transit. Moon's Horizontal Parallax. Moon's Transit.
60'59'58'57'56'55'54' Moon's Transit. Moon's Transit. 60'59'58'57'56'55'54'
h.m. m.m.m.m.m.m.m. h.m. h.m.m.m.m.m.m. h.m.
00- 4- 3- 2- 0+ 2+ 4+ 6 120 60- 56- 58- 60- 62- 64 180
10065431+ 1+ 2 100 1005254565860 100
20087654- 3- 1 200 2004951535557 200
300101098765 300 3004648505153 300
400121211101098 400 4004344454749 400
50015141413121211 500 5003839404143 500
1017171616161515 130 703233333435 190
10020201919191918 100 1002727282829 100
20022222222222222 200 2002222222222 200
30024242525252525 300 3001818171616 300
40027272828282929 400 400111110108 400
50029303131313233 500 50066542 500
2031323333343536 140 801+ 1+ 2+ 3+ 5 200
10034353636373839 100 100+ 24579 100
20036373839404243 200 2005791114 200
30038394041424446 300 300810121518 300
40040414344464850 400 4001113161821 400
50042434546485052 500 5001316182023 500
3044454749515355 150 901417192124 210
10046474951545658 100 1001518202326 100
20048495153565861 200 2001719222528 200
30050525456586164 300 3001618212427 300
40052545658616467 400 4001618212427 400
50053555760636669 500 5001618212327 500
4055575962656972 160 1001517202327 220
10056586163667073 100 1001417202225 100
20057606365687275 200 2001316182023 200
30058616466697376 300 3001215171922 300
40059626567707478 400 4001113161821 400
50060626567707579 500 500911141619 500
5060636568717579 170 11079121417 230
10060636568727680 100 10068101215 100
20060636568717580 200 200467911 200
30059626567707478 300 30024679 300
40058616365687276 400 40002457 400
50057606265687174 500 500- 2- 1124 500
60- 56- 58- 60- 62- 65- 69- 72 180 120- 1- 1+ 0+ 0+ 2 240

TABLE II.—For finding the Height of the Tide.

Time of Transit. (PART I.) (PART II.)
Moon's Hor. Par. 60' Moon's Hor. Par. 57' Moon's Hor. Par. 54' Time from H. W. Multi. Time from H. W. Multi.
Multipliers. Multipliers. Multipliers. h.m. h.m.
00 0.995a + 0.1498 0.883a + 0.1175 0.795a + 0.0825 00 1.0003 10
040 1.104a + 0.0382 0.970a + 0.0303 0.874a + 0.0215 010 0.9983 20
120 1.138a + 0.0005 1.000a + 0.0005 0.901a + 0.0005 020 0.9933 30
20 1.104a + 0.0382 0.970a + 0.0303 0.874a + 0.0215 030 0.9853 40
240 0.995a + 0.1498 0.883a + 0.1175 0.795a + 0.0825 040 0.9743 50
320 0.853a + 0.3193 0.750a + 0.2503 0.676a + 0.1765 050 0.9594 0
40 0.668a + 0.5275 0.587a + 0.4135 0.529a + 0.2905 10 0.9414 10
440 0.460a + 0.7493 0.413a + 0.5875 0.372a + 0.4125 110 0.9214 20
520 0.284a + 0.9583 0.250a + 0.7505 0.225a + 0.5275 120 0.8974 30
60 0.133a + 1.1275 0.117a + 0.8835 0.105a + 0.6215 130 0.8714 40
640 0.034a + 1.2385 0.030a + 0.9705 0.027a + 0.6825 140 0.8434 50
720 0.000a + 1.2775 0.000a + 1.0005 0.000a + 0.7035 150 0.8125 0
80 0.034a + 1.2385 0.030a + 0.9705 0.027a + 0.6825 20 0.7795 10
840 0.133a + 1.1275 0.117a + 0.8835 0.105a + 0.6215 210 0.7745 20
920 0.284a + 0.9583 0.250a + 0.7505 0.225a + 0.5275 220 0.7685 30
100 0.460a + 0.7493 0.413a + 0.5875 0.372a + 0.4125 230 0.6705 40
1040 0.668a + 0.5275 0.587a + 0.4135 0.529a + 0.2905 240 0.6315 50
1120 0.853a + 0.3193 0.750a + 0.2505 0.676a + 0.1765 250 0.5916 0
120 0.995a + 0.1498 0.883a + 0.1175 0.795a + 0.0825 30 0.5516 20
TO FIND THE TIME OF HIGH-WATER.

Rule.—Let the approximate time of high-water be found, by taking the corrections for the moon's horizontal parallax for the nearest noon or midnight from Table I. Again, to this time and the given longitude take from the Nautical Almanac the moon's horizontal parallax. Also to the time of the moon's transit over the meridian of Greenwich apply the variation answering to the longitude and daily variation between the given and preceding day if the longitude is E. Subtract this from the transit over the meridian of Greenwich, and the remainder will be the time of transit over the meridian of the given place. But if the longitude be W., the correction answering to the longitude and daily variation of transit between the given and following day must be added to the time of transit over the meridian of Greenwich, to obtain the time of transit over the meridian of the given place. To the time of high-water, if new and full moon at the given place, add the reduced time of transit over the meridian of the same place, and to the sum apply the equation from the table answering to the time of transit and horizontal parallax formerly found; the result will be the true mean time of high-water required. The apparent time may be found by applying the equation of time, with its proper sign.

Ex. 1.—Required the time of high-water at Leith on Wednesday the 10th of May 1837, in Long. 3^{\circ} 11' W.

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

Apparent time of transit of upper meridian... 4^{\text{h}} 54^{\text{m}}
Equation from the table to horizontal parallax
54^{\circ} 16', and transit 4^{\text{h}} 54^{\text{m}}, subtract..... -1 18
Remainder ..... 3 36
Time of high-water at new and full moon ..... +2 20
Apparent time of high-water..... 5 56
Equation of time..... -0 4
Mean time of high-water..... 5 52

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

Ex. 2.—Required the time of high-water at Aberdeen on the 21st of June 1837, in Long. 2^{\circ} 6' W.

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

Here the horizontal parallax of the moon will be 60^{\circ} 30', and the mean time of transit on the given day 15^{\text{h}} 32^{\text{m}}. But as this transit exceeds 12^{\text{h}}, it will be necessary to take the time of transit over the under meridian, or, what comes to the same thing, half the

sum of the transits on the given and preceding days, or, \frac{1}{2}(14^{\text{h}} 32^{\text{m}} + 15^{\text{h}} 32^{\text{m}})..... =15^{\text{h}} 32^{\text{m}}
Correction from the table..... -0 44
Remainder..... 14 18
High-water at new and full moon ..... +1 10
Sum exceeding 12^{\text{h}}..... 15 28
By rule, subtract ..... 12 25
Apparent time of high water ..... 3 3
Equation of time..... +0 1
Mean time..... 3 4

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

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

Ex. 4.—Required the height of the tide at 3^{\text{h}} 15^{\text{m}} after high-water.

By part 2, 20.8 \times 0.5..... = 10.4 feet.

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

In the preceding pages we have not considered it our province to supply the reader with the tables which have been made use of in the solution of the several problems—e.g., traverse tables, tables of meridional parts, corrections for dip, parallax, refraction, and log. haversines, log. rising, middle time, half-elapsed time, &c. For these, and for further information, we refer him to the works on Navigation already mentioned. Some of the rules we have given will be found to differ in some respects from those given in these works. These discrepancies, however, arise entirely from the slightly different form in which the formulae on which the rules are based are made to appear, but not at all on any difference of principle. We have generally retained the forms of the formulae which are best known to astronomical students. For example, we have retained Borda's method of clearing the distance, in lieu of the slight deviations from it which are frequently employed, not merely as being equally correct with the latter, but as possessing some historical interest. In clearing the distance by natural versed sines, it is not difficult to put the formulae in such a form that the sum only of the versed sines may be taken. We have retained the formula in which two of the versed sines appear with a negative sign, as being that with which the readers of Hymer's Astronomy are already acquainted. Whenever the reader is familiar with astronomical formulae, we recommend him to study the rules given in these pages with the appropriate formulae before him; these he will find in any work on astronomy treated mathematically. In all cases, however, it is believed that, by a careful study of the rules and directions contained in this article, the reader will find himself possessed of all the information which will enable him to navigate a ship.