CHAP. II.-ON PLANE SAILING. B C 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. Fig. 7. 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 This convention we have already Plane left-hand side west. tacitly assumed in treating of the corrections of the courses. Sailing. 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° (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 formula 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. 10.08015 - 2.28330 -10 230.9 = 2.36345 772 A B C 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. n A .43° 13′ N. .46 31 N. 3 18 198 miles. 9.82489 2.28330 m .10 .28° 37' N. 2 8 N. .30 45 N. 285 (198) + 10 = 12.29660 .285 2.45484 L cos course 46° 9.84176 Or course is........... N. 46° 0′ E. To find the departure. ******... ..... Fig. 11. 46° = = = 9.85693 2.45484 -10 Log. departure ..... ..............205 = 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 te 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. Log. distance...... 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. Plane Sailing. Plane By Construction.-Draw BC perpendicular to AB, and equal to are all that occur in the solution of the right-angled triangle, Mercator' Sailing. the given departure 190 miles; then from the centre C, with the distance C Latitude sailed from 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 260. This is found to be in the page marked 47° at the bottom, which is the course; and the corresponding difference of latitude is 177.3. By Gunter's Scale.-The extent of the compass, from the distance 260 to the departure 190 on the line of numbers, will reach from 90° to 47°, the course on the line of sines; and the extent from 90° to 43°, the complement of the course on sines, will reach from the distance 260 to the difference of latitude 177 on the line of numbers. By Calculation.-To find the course. Log. departure L tan course ..12° 9' N. 177 2 57 S. 9 12 N. Latitude in .... 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. 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. Fig. 12. L sec course = To find the distance. .132+ 10 9.83419 B = n = 12-12057 2.39270 9.72787 m Fig. 13. B 28° 7' Log. distance. 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. 10.05454 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 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. Fig. 16. By Calculation.-To find the true difference of latitude. L cos course 3 points = 9.91985 Log. distance.... = 2.42651 -10 = 2.34636 Mer. parts...2528 Log. diff. long. 1831 23° 27' = 1678 miles = -10 38° 47′ N. 3 42 N. 42 29 N.. To find the difference of longitude. 3 points 293 miles 267 miles 222 N. Lat. Port Canso 45° 20' N. Lat. in, by observation 41 14 N. Diff. lat. 4 6=246 B Log. distance 332 *****........ To find the difference of longitude. L tan course........ Log. mer. diff. lat....................... D Ex. 3.-A ship from Port Canso in Nova Scotia, in Lat. 45° 20' N., Long. 60° 55′ W., sailed S E.S., and, by observation, was found to be in Lat. 41° 14′ N.; required the distance sailed, and longitude come to. = -10 .195.8 E. = 2.29176 75° 4′ W. 3 16 E. 71 48 W. B D 10-03798 3.22479 Mer. parts...2821 Mer. diff. lat. 293 C Mer. parts...3058 Mer. parts...2720 Mer. diff. lat. 338 Fig. 17. 3 points 10-13021 246 miles = 2.39093 -10 = 9.82489 2.46687 = Log. diff. long............... 306.3 E. 5 E 3 points= 338 miles = - 10 2.52114 9.95729 2.52892 2.48621 E Ex. 4.-A ship sailed from Sallee, in Lat. 33° 58′ N., Long. 6° 20′ W., the corrected course was N.W. by W.W., and departure 420 miles; required the distance run, and the latitude and longitude in. VOL. XVI. 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. longitude in is therefore 14° 56′ W. Fig. 18. The By Calculation.-To find the distance. By Calculation.-To find the course. Log. dist..... 780+10 L cos course 47° 50' L tan course Log. diff. longitude............ 1194 -10 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 5.06416 2.35121 2.71295 D Fig. 19. 12.89209 = 3.06521 B C E Mercator's Sailing, ........ 217 Log. diff. longitude ............. 257 L sin course.. ....... By Calculation.-To find the course. Log. dep. I cos course......... Log. dist......... 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° 30' E. Fig. 20. 33° 56' = 10-08109 2.33646 217 miles = -10 261.5 = 382 146 B To find the true difference of latitude. = == 2.58206 2.16435 4.74641 2.33646 = = 2.40995 2° 8' W. 4 17 E. 2 9 E. Ex. 7.-A ship from Naples, in Lat. 40° 51' N., Long. 14° 14' E., CHAP. IV.-ON TRAVERSE SAILING, OR COMPOUND E 2.41755 C 183.2 = 2.26304 Log. true diff. latitude Latitude from (Naples).... 40° 51' N. Mer. parts... 2690 True diff. latitude 3 3 S. 37 48 N. Latitude in .......... To find the difference of longitude. 173+10= 12.23805 252 = 2.40140 43° 21' = 9.83665 Fig. 21. 43° 21' = 9.86164 2.40140 252 miles = <-10 A Mer. parts... 2453 Mer. diff. lat. 237 D 43° 21' = 9.97497 2.37475 237 miles = -10 223.7 = 2 34972 14° 14' E. 3 44 W. 10 30 E. E Log. diff. longitude Longitude from.. Diff. longitude...... Longitude in...... 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., B 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 colunin; the third and fourth columns are to contain the differences of latitude, and the two last the departures. |