BY INSPECTION. With the mid lat 44° as a course, and diff long 1293 miles, as a distance, the dep is found in the lat column to be 930 miles. Then, with diff lat 744 miles, and dep 930 miles, in their proper columns, the course is N 51° E, and dist 1195 miles. BY GUNTER'S SCALE. Extend from rad to 46°, the comp of mid lat, on the line of sines, that extent will reach from the diff long 1293 miles to the dep 930 miles on the line of numbers. Extend from this dep to 744 miles, the diff lat on the line of numbers, and that extent will reach from rad, or 45°, to the course, about 514o, on the line of tangents. Extend from rad to the comp of the course, on the line of sines, that extent will reach from the diff lat to the dist 1195 on the line of numbers. D B E Draw the vertical line A D, on which lay off A D 1041 miles, the mer diff lat, at D erect a perpendicular, and on it take DE 1293 miles, the diff long. From A on A D lay off A B 744 miles, the proper diff lat, draw BC parallel to D E, and join A E, cutting B C in C, then A C will be the distance, ▲ A the course, and B C the departure. A C measured on a line of equal parts will be found 1190 miles, and A measured by the scale of chords will be found about 514°. Hence the dist is 1190 miles, and the course N 514° E. BY INSPECTION. In Table 3. with the mer diff lat 1041 miles in the lat column, and the diff long 1293 miles in the dep column, the course is found at the bottom of the page to be between 51° and 52°, or nearly 51+o, with this course, and the proper diff lat in the lat column, the dist îs found to be about 1185 miles. BY GUNTER'S SCALE. Extend from the mer diff lat 1041 miles to the diff long 1293 on the line of numbers, that extent will reach from rad to the course 51° on the line of tangents. Extend from rad to the comp of the course on the line of sines, and that extent will reach from the proper diff lat 744 miles (towards the right) to the dist 1192 miles on the line of numbers. 2. A ship sails in the N W quarter 248 miles, till her departure is 135 miles, and her diff long 310 miles, required her course, the latitude left, and the latitude come to? BY CONSTRUCTION. Draw the meridian line A D, and at C in that line erect a perpendicular, on which take B C equal to the departure 135 miles. From B, with the diff long 310 miles, cut A Din D; and from the same point B, with the distance 248, cut A D in A, and join B D, A D. Then BD will repre- B D ent parts, we find A C 208 miles, ▲ A 33°, and 4 D B C 64° 11'. Hence the course is N 33° W, and as half the diff lat added to the middle lat gives the greater, and subtracted leaves the less latitude, the lat left is 62° 27′, and the lat come to 65° 55'. BY INSPECTION. In Table 3. with the dist 248 miles, and the dep 135 miles in their own columns, we find the course at the top of the page to be 33°, and the diff lat 208 miles; and with the diff long 310 miles in the distance column, and the dep 135 miles in the lat column, the mid lat is found at the bottom of the table, as a course nearly 64°. Hence the latitudes are found as above. BY GUNTER'S SCALE. Extend from the dist 248 miles to the dep 135 miles, on the line of numbers that extent will reach from rad to the course 33° on the line of sines. Extend from the diff long 310 miles to the dep 135 miles, on the line of numbers, that extent will reach from rad to the comp of 64° 11', the mid lat on the line of sines. Extend from rad to the comp of the course on the line of sines, that extent will reach from the distance to the diff lat, 208 miles, on the line of numbers. Note. The preceding example cannot be solved by Mercator's sailing. 3. If a ship sail from Cape Finisterre S WW till her diff of longitude is 134 miles, required her distance and latitude in? BY CONSTRUCTION. Draw the meridian line A D, and at D erect the perpendicular, D E, and make it equal to 134 miles, the diff long. Make the angle AED equal to 3 points the complement of the course, and let E A meet AD in A, then A D measured on the scale of equal parts from which DE was taken, will be found A about 121 miles, the mer diff lat. Now A D subtracted from the mer parts of the latitude of Cape Finisterre will leave 2737, the mer parts of 41° 27', B the latitude arrived at. Hence the proper diff lat is 89 miles. Make A B equal to 89 miles, and draw B C D parallel to D E, then A C will be the distance, and equal to about 132 miles. BY INSPECTION. Enter Table 3. with the course 44 points at the bottom of the table, and the diff long 134 miles, in the dep column; and corresponding to it in the lat column will be found 121.5, the mer diff lat. Find the lat in and diff lat as above, and enter the table with the course, and corresponding to 89 in the lat column will be found the dist, about 132.5 miles in its own column. BY GUNTER'S SCALE. Extend from rad to the complement of the course 3 points on the line of tan rum, that extent will reach from the diff long 134 miles on the line of numbers towards the left, to the mer diff lat 121 miles. Extend from rad to the complement of the course on the line of sines, and that extent will reach from 89 miles, the diff lat, to 131.5 miles, the distance. Latitude Cape Finisterre 42° 56′ N mer parts....2858 Note. The preceding example cannot be solved by middle latitude sailing. EXAMPLES FOR EXERCISE. 1. If a ship sail from Lisbon W SW + W 168 miles, required her latitude and longitude? Answer, lat 37° 57′ N, long by Mercator's sailing 12° 59′ W, by mid lat 12° 56′ W. 2. If a ship sail from Cape Race S SEE 216 miles, required her latitude and longitude? Answer, lat 43° 35′ N, and long 50° 26′ W. 3. Required the course and distance from lat 41° 46′ S, long 10° 34′ E, to Rio Janeiro ? Answer, by Mercator's sailing, course N 67° 13′ W, dist 2923 miles, by mid lat course N 67° 23′ W, and dist 2943 miles. 4. A ship from the lat 40° 12′ N, long 18° 3′ W, is bound for England, after sailing N E b N N 248 miles, required her course and distance to the Lizard? Answer, by Mercator's sailing, course N 48° 47′ E, dist 557 miles, by mid lat course N 48° 49′ E, and dist 557 miles. 5. If a ship sail from lat 40° 5′ N, long 28° 14' W, ENE E till her difference of longitude is 320 miles, required her distance, latitude, and departure? Answer, lat 41° 18′ N, dist 251.5 miles, and dep 240.6 miles. 6. A ship from Cape Horn sails in the southwest quarter, till she arrives in lat 60° 10′, having made 218 miles difference of longitude, required her course and distance? Answer, by Mercator's sailing, course S 24° 34′ W, dist 277 miles, by mid lat course S 24° 35′ W, and dist 277 miles. 7. A ship from the Cape of Good Hope sails N Wb W till she arrives in lat 30° 4′ S, required her course and distance to St Helena ? Answer, course N 46° 32′ W, and dist 1234 miles. 8. Required the course and distance lat 53° 18′ N long 0° 55′ E to the Naze? Answer, course N 36° 31′ E, and dist 348 miles. 9. If a ship leave Cape Clear bearing N EN 16 miles, and sail SWS 150 miles, required her course and distance to St. Mary's, Azores? Answer, course S 38° 11′ W, and dist 935 miles. 10. A ship bound for Cork was met at sea by another, which had sailed from Cape Ortegal, N W b N 186 miles, required the ship's course and distance to Cork harbour. Answer, course N 13° 9' E, and dist 341 miles. 11. A ship from lat 40° 41′ N, long 16° 37′ W, sails in the NE quarter till she arrives in lat 43° 57′ N, and has made 248 miles departure, required her course, distance, and longitude in? Answer, course N 51° 41' E, dist 316 miles, and long in 11° W. 12. How far must a ship sail N EE from lat 44° 12′ N, long 23° W, to reach the parallel of 47° N, and what from that point will be the bearing and distance of Ushant? Answer, she must sail 265 miles, and her course and distance to Ushant will then be N 80° 32′ E, and dist 535 miles. 13. A ship from the Cape of Good Hope steers ES 446 miles, required her place, and her course and distance to Kerguelen's Land? |