Traverse The several courses and their corresponding distances Sailing, or being properly arranged in the table, find the true difference Compound of latitude and departure answering to each in the traverse Courses. table, remembering that the true difference of latitude is to be put into a N. or S. column according as the course is 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. Latitude from......... 38° 32′ N. Mer. parts...... 2509 Mer. parts...... 2247 Mer. diff. lat... 262 Now to course 4140, 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 Fig. 23. 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. 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 equal to 262, the meridional difference of latitude; and through G 186 miles respectively. On CF produced lay off to the scale CG 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. B E H 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. 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. 96, N.W. W. 89, N. by E. 110, N.W. N. 56, S. by E.gE. 78. The latitude in is required, and the longitude, by both methods; the bearing and distance of Hacluit's headland, in Lat. 79° 55 N., Long. 11° 55' E., is also required. Traverse Sailing, or Compound Courses. 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 formulæ for parallel sailing: Distance = diff. longitude x cos latitude; Cos latitude = distance +diff. longitude; Diff. longitude = distance x sec latitude. Problems in parallel sailing may be solved by construction; for it is evident that we have only to construct a rightangled triangle whose hypothenuse is the difference of longitude, one of the sides the distance, and the angle between is side and the hypothenuse 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. Log. diff. long. L cos lat.... Log. dist...... True course............. Variation... 55° 58' 60 Diff. of Longitude. To find the bearing and distance of Hacluit's headland- M. P. 8347 Also, long. A = 19° 12 E. long. B 10 42 E. Diff. long. E. ... ... ... Now, opposite to 78.5, half the meridional difference of latitude, and 185-0, half the difference of longitude, stands the course 67°; and opposite to the difference of latitude 27, the distance is 69 miles. Hence Hacluit's headland bears S. 67° E., distant 69 miles. To find compass course. Pts. qrs. 8 123.6 166-7 | 290-3❘ 1639-3 290.3 Variation 1 E., and deviation as in the table on p. 13. The true course is due E. Log. miles in a deg. in lat. 55° 58'...33.58 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. W. 731.7 346.0 343.6 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. Long. 11° 55' E. Diff. long. 6 10 E. 370 There are two lines on the other side of the scale, with respect to Gunter's line, adapted to this particular purpose, one of which is entitled chords, and contains the several degrees of latitude; the other, marked M. L., signifying miles of longitude, is the line of longitudes, and shows the number of miles in a degree of longitude in each parallel. The use of these lines is therefore obvious. 218.0 Ex. 2.-Required the compass course and distance from A to B. Long. A= 9° 12′ E. 28° 14' E. 22 29 W. 5 45 E. ... 1349.0 9.7479360 = 1.7781513 <-10 0 right of N. 1 3 left of N. = 1 30 90 E. 90 = 1.954243 17° 30′ 9.979419 - 10 85.8 m. 1.933662 right of N., or E.N.E.E. left of N. r. of N., or N.E by E.JE Parallel Log. diff. long. 466.6 48° 47' By Calculation L cosec lat........... 42° 52′ = 10-13493 = 2.53403 3° 14' W. ....64 27 W. Log. dist.......358+10-12-55388 L cos lat....46° 42' Longitude in........ ..17 4 W. 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.-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 278 miles. By Calculation, as under: Log. given distance L cos lat. in......... 61 13 3673 W. 3-5650209 <-10 2420 = 3.3838459 CHAP. VI. OF MIDDLE-LATITUDE SAILING. Log. distance run ....... Ex. 4.-A ship from Cape Finisterre, Lat. 42° 52' N., Long. 9° 17' W., sailed due W. 342 miles; required the longitude in. It has been already explained in chap. ii. that the the given distance 342 miles, and make the By Construction. Draw the straight line AB (fig. 25), equal to departure is greater than the intercepted arc of the 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 522', 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 463°, the required latitude. By Calculation Fig. 25. 9° 17' W. 7 47 W. = 271767 B =9-83621 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.-Make DB (fig 28) A 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'. DF P 348 miles = 2.54158 ..48° 23' = 9-82226 12.36384 =9-91874 L cos lat. from ...33° 58' Log. distance required..... 278-6 miles = 2.44510 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 Calculation, as under:- FDP Log. distance on known parallel... 180 If ABC (fig. 29) be the triangle for plane 9.74756 2.36549 12-11305 = 2-25527 9-85778 Departure diff. long. x cos mid. latitude. True diff. lat. x tan course = diff. long. x cos mid. lat. (A.) Distance x sin course = diff. long. x cos mid. lat. (B.) A = Latitude Isle of May Fig. 29. Also problems in middle-latitude sailing Ex. 1.-Required the compass course and distance from the Island 56° 12' 56° 12' N. 1 A Longitude Isle of May.. Difference of longitude........10 4=604' E. By Construction.-Draw the right line AD (fig. 30) to B n Fig. 30. MiddleLatitude Sailing. Longitude from Diff. longitude.............. Longitude in Ex. 8.-A ship from Lat. 54° 56′ N., Long. 1° 10′ W., sailed between the N. and E. till, by observation, she was found to be in Long. 5° 26' E., and has made 220 miles of easting; required the latitude in, course, and distance run. Longitude from Longitude in Difference of longitude 6 36=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', B 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°. .1° 10' W. .5 26 E. ...0° 35' W. .5 10 W. ..5 45 W. L cos course.......... Log. distance ....... To find distance. 54° 19' 10.23410 158 =2.19866 -10 ..............270·9 = 2.43276 Log. distance.. Ex. 9.-A ship from a port in N. Lat., sailed S.E.S. 438 miles, and differed her Long. 7° 28'; required A the latitudes from and in. ........... Log. diff. longitude.. By Construction.-With the course and dis. tance construct the triangle ABC (fig. 38), and make DC equal to 448, the given difference of longitude. Now the middle latitude BCD will B measure 48° 58', and the difference of latitude AB 324 miles; hence the latitude from is 51° 40', and latitude in 46° 16'. By Calculation.-To find the true difference of latitude. L cos mid. latitude .220+10= 12-34242 ...158 = 2.19866 ..54° 19' 10.14376 D Log. true diff. latitude........324.5 To find middle latitude. Mid. latitude 34 pts. 438 miles Fig. 87. Fig. 38. = 9.86979 =2·64147 -10 =2.51126 438 ms. = 2.64147 3 pts. 9-82708 12.46855 448 =265128 .48° 58' = 9.81727 .48° 58′ N. 2 42 S. .51 40 N. 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. |
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