Hence the ship's distance run is 487.8 miles, and her departure from the meridian is 405.6 easterly. BY GUNTER. Extend from radius or 4 points to the course 5 points on the line marked TR, that extent will reach from the difference of latitude 271 to the departure 405.6 on the line of numbers. 2dly. Extend from the complement of the course 3 points to the radius 8 points on the line SR, that extent will reach from the difference of latitude 271 to the distance 488 on the line of numbers. BY INSPECTION. Find the course among the points or degrees, and the difference of latitude in its column, against which will stand the distance and departure in their columns. Now, as the difference of latitude 271 is too great to be found in the tables, I divide it by 2, and that gives 135.5 which I find over S. E. by E. or 5 points in the latitude eolumn; against that stand 244, for the distance and 202.9 for the departure, which multiplied by 2 give the distance 488, and the departure 405.8. CASE III Course and departure from the meridian given, to find the distance, and difference of latitude. If a ship sails N. E. by E. E. from a port in 3° 15' south latitude, until she depart from her first meridian 406 miles, I demand the distance sailed, and the latitude she is in? BY PROJECTION. Draw the meridian AB, upon which erect B the perpendicular BC, and set off thereon the departure 406 easterly from B to C: with the chord of 60°, on C as a centre,, describe an arch, and set off thereon the complement of the course DE; through D and C draw the line CDA, cutting the meridian in the point A; then AC measured' on the same scale before used, gives the distance 449, and AB 192, the difference of latitude. From the latitude left 3° 15 S. Subtract the difference of latitude 192 miles, or 3 12 N. 0 03 S. The remainder being 3, shows the ship is in the latitude of BY GUNTER. Extend from radius or 4 points to the co-course 2 points on the line marked TR; that extent will reach from the departure 406 to the difference of latitude 192 on the line of numbers. 2dly. Extend from the course 5 points to radius on the sines, that extent will reach from the departure 406 to the distance 449. miles on the line of numbers. BY INSPECTION. Find the course either among the points or degrees, and the departure in its column, against which will stand the distance and difference of latitude in their respective columns. Thus with the course 5 points, and half the departure 203* I find 224.5 or the distance, and 96.0 for the difference of latitude, which, being doubled, give the distance 449, and the difference of latitude 192.0, as before. CASE IV. Distance and difference of latitude given, to find the course and departure. Suppose a ship sails 488 miles, between the south and the east, from a port in 2° 52′ south latitude, and then by observation is in 7° 23′ south latitude; what course has she steered, and what departure has she made? From the latitude by observation 7° 23′ take 2° 52′ the latitude left, the remainder 4° 31'=271 miles, is the difference of latitude. Hence the course is S. E. b. E. and the departure 405.8. BY GUNTER.. The extent from the distance 488, to the difference of latitude 271 on the line of numbers, will reach from radius, or 90° to 33° 44′ the co-course on the line of sines. And the extent from radius, to 56° 16' on the line of sines, will reach from the distance 488 to the departure 405,8 on the line of numbers. BY INSPECTION. Seek in the tables till against the distance, taken in its column is found the given difference of latitude in one of the following columns, adjoining to it will stand the departure; which, if less than the difference of latitude, the course is to be found at the top ; but if greater, the course is to be found at the bottom. Thus half the distance 244, and half the difference of latitude 135.5, are found to correspond to a course of 5 points or S. E. b. E. and to the departure 202.9, which being doubled, gives 405.8, as before. CASE V. Distance and departure given, to find the course and difference of latitude. Suppose a ship sails 483 miles between the north and west, from the latitude of 32° 25′ north, until her departure is 405 miles, what course has she steered, and what latitude is she in? *The nearest numbers in the table are 202 5 and 203.4, and as the number 203 is nearly a mean of these two values, I take the mean 224.5 of the corresponding distances 224, 225, and the mean 96 of the corresponding departures 95.8 and 96.2; these doubled give the true distance 449, and departure 192. It may also be known whether the course be marked at the top or bottom of the table, by observing whether the difference of latitude and departure correspond with the marks at the top or bottom. Thus the half distance 244, and half difference of latitude 135.5 correspond to the course 5 points, because the column :n which 135.5 is found, is marked latitude at the bottom; the same may BY PROJECTION. Draw the line AB equal to the departure 405, and perpendicular thereto the line BC to repre- A sent the meridian, then take the distance 488 in your compasses, and fixing one foot in A as a centre, describe an arch cutting BC in C, join AC and it is done; for the angle ACB will be the course, and BC the difference of latitude. BY LOGARITHMS. To find the course. As the distance 488 Is to radius So is the departure 405 B To the sine of course 56° 6' Hence the course is N. 56° 6' W. or N. W. by W. nearly. To the latitude sailed from 32° 25′ add the difference of latitude 272, or 4° 32', the sum 36° 57′ is the latitude the ship is in. BY GUNTER. Extend from the distance 488 to the departure 405 on the line of numbers, that extent will reach from radius to the course 56° 6' on the line of sines. 2dly. Extend from radius to the complement of the course 33° 54′ on the line of sines, that extent will reach from the distance 488 to the difference of latitude 272 on the line of numbers. BY INSPECTION. Seek in the tables till against the distance, taken in its column, is found the given departure in one of the following columns; adjoining to it will stand the difference of latitude; which, if greater than the departure, the course is to be found at the top; but if less, the course is to be found at the bottom. Thus half the distance 244, and half the departure 202,5, agree nearly to a course of 5 points or N. W. by W. and a difference of latitude, 135,5, which being doubled, is 271, the difference of latitude, nearly as before. CASE VI. Difference of Latitude and Departure given to find the Course and Distance. A ship sails between the north and west till her difference of latitude is 271 miles, and her departure is 406 miles; I demand her course and distance? Hence her course is N. 56° 17′ W. or N. W. by W. and the distance sailed is 488.2 miles. 2.43297 As radius 10.00000 2.43297 10.25564 10.17556 To the distance 488.2 2.68861 BY GUNTER. Extend from the difference of latitude 271 to the departure 406 on the line of numbers, that extent will reach from radius to 56° 17′ the course on the line of tangents. 2dly. For the distance we must consider it as radius, (unless there is a line of secants on the scale) and extend from the course 56° 17′ to the radius or 90° on the line of sines, that extent will reach from the departure 406, to the distance 488 on the line of numbers. BY INSPECTION. Seek in the tables till the given difference of latitude and departure are found together in their respective columns, then against them will be the distance in its column, and the course will be found at the top of that table if the departure be less than the difference of latitude, otherwise at the bottom. Thus with half the difference of latitude 135.5, and half the departure 208, enter the tables, and these numbers will be found to correspond nearly to 5 points or N.W. by W. course, and a distance equal to 244 miles, which being doubled gives the sought distance, 488, Questions to exercise the learner in the foregoing Rules. Question I. A ship in 2° 10' south latitude, sails N. by E. 89 leagues; what latitude is she in, and what is her departure? Answer. Latitude in 2° 12′ N. and departure 17,36 leagues. Question II. A ship sails S. S. W. from a port in 41° 30' north latitude, and then by observation is in 36° 57′ north latitude, I' demand the distance run and departure? Answer. Distance run 98,5 leagues, departure 37,7 leagues. Question III. A ship sails S. S. W. W. from a port in 2° 30' south latitude, until her departure be 59 leagues-I demand her distance run and latitude in? Answer. Distance run 125,2 leagues, latitude in 8° 1' south. Question IV. If a ship sail 360 miles south westward from 21° 59′ south latitude, until by observation she be in 24° 49′ south latitude, what is her course and departure? Answer. The course is S. W. by W. half W. or S. 61° 49′ W. and her departure from the meridian is 317,3 miles. Question V. Suppose a ship sails 354 miles north eastward from 2° 9′ south latitude, until her departure be 150 miles, what is her course and latitude in? Answer. Her course is N. 25° 4' E. or N. N. E. E. nearly, and she is in latitude 3° 12′ N. Question VI. Sailing between the north and the west, from a port in 1° 59' south latitude, and then arriving at another port in 4° 8' north latitude, which is 209 miles to the westward of the first port-I demand the course and distance from the first port to the second? Answer. The course is N. 29° 40′ W. or N. N. W. W. nearly, and the distance of the ports is 422,4 miles, or 140,8 leagues. Question VII. Four days ago we were in lat. 3° 25′ S. and have since that time sailed in a direct course N. W. by N. at the rate of 8 miles an hour -required our present latitude and departure? Answer. Latitude in 7° 14′ N. Departure 426,7 miles. Question VIII. A ship in the latitude 3° 52′ S. is bound to a port bearing N. W. by W. W. in the latitude of 4° 50' N. how far does that port lie to the westward, and what is the ship's distance from it? Answer. The port lies 939,2 miles to the westward, and the direct distance 1065 miles. Question IX. A ship from the latitude of 48° 17′ N. sails S. W. by S. until she has depressed the north pole 2 degrees, what direct distance has she sailed, and how many miles has she got to the westward? TRAVERSE SAILING. 1000 A TRAVERSE is an irregular track which a ship makes by sailing on se veral different courses; these are reduced to a single course by means of two or more cases of Plane Sailing, either by geometrical construction, or by arithmetical calculation.* The geometrical construction is performed as follows: Describe a circle with the chord of 60°, to represent the compass, and lay off on its circumference the various courses sailed. From the centre, upon the first course set off the first distance, and mark its extremity; through this extremity, and parallel to the second course, draw the second distance of its proper length; through the extremity of the second distance, and parallel to the third course, draw the third distance of its proper length; and thus proceed till all the distances are drawn. A line, drawn from the extremity of the last dis tance to the centre of the circle, will represent the distance made good; a line, drawn from the same point, perpendicular to the meridian, will represent the departure; and the part of the meridian intercepted between this and the centre, will represent the difference of latitude. The arithmetical calculation to work a traverse is as follows: Make a traverse table consisting of six columns; title them, Course, Dist. N. S. E. W. begin at the left side, and write the given courses and distances in their respective columns. Find the difference of latitude and departure for each of these courses, by Gunter's Scale, or by Tables I. or II. (as in Case I. Plane Sailing) and write them in their proper columns; that is, when the course is southerly, the difference of latitude must be set in the column S. when northerly in the column N. The departure, when westerly, in the column W. and when easterly in the column E. Add up the columns of northing, southing, easting, and westing; take the difference between the northing and southing, and also between the easting and westing; the former difference will be the difference of latitude, which will be of the same name as the greater; and the latter will be the departure, which will be also of the same name as the greater. With this difference of latitude and departure, the course and distance made good are to be found as in Case VI. Plane Sailing. EXAMPLE I. Suppose a ship takes her departure from Block Island, in the latitude of 41° 10' N. the middle of it bearing N. N. W. distance by estimation 5 leagues, and sails S. E. 84, W. by S. 16, W. N. W. 39, and S. by E. 40 miles; required the latitude she is in, and 'her bearing and distance from Block Island? * This method of reducing compound courses to a single one is perfectly accurate in sailing on a plane, and is nearly so in sailing a short distance on the spherical surface of the earth; and though in this case it is liable to a small error in high latitudes, yet in general the rule is sufficiently accurate for reducing the several courses and distances sailed in one day to a single course and distance.> |