QUESTIONS FOR EXERCISE. Question I. Required the bearing and distance between two places, one in the atitude of 37° 55′ N., and longitude of 54° 23′ W.; the other in the latitude of 32° 38′ N., and longitude of 17° 5′ W. Answer. S. 80° 9′ E., and N. 80° 9′ W., distance 1854 miles. Quest. II. Required the direct course and distance, from a place in the latitude of 36° 55′ S., and longitude of 20° 0' E., to another place in the latitude of 32° 38′ S., and longitude of 8° 54′ W. Ans. N. 79° 46′ W., distance 1447 miles. Quest. III. A ship from the latitude of 37° 30′ S., and longitude of 60° E., sails N. 79° 56′ W. 202 miles; required the latitude and longitude in. Ans. Latitude 36° 55′ S., longitude 55° 50′ E. Quest. IV. A ship from the latitude of 34° 35' N., and longitude of 45° 16′ W., sails S. 83° 36′ E., 101 miles; required her latitude and longitude. Ans. Latitude 34° 24' N., longitude 43° 14′ W. Quest. V. A ship in the latitude of 49° 57' N., and longitude of 15° 16′ W., sails south-westerly till her departure is 789 miles, and latitude in 39° 20′ N.; required the course, distance, and longitude in. Ans. Course S. 51° 05′ W., distance 1014 miles, longitude in 33° 45′ W. Quest. VI. A ship in the latitude of 42° 30′ N., and longitude 58° 51′ W., sails S. E. by S. 591 miles; required the latitude and longitude in. Ans. Latitude 34° 19′ N., longitude 51° 52′ W. Quest. VII. Suppose a ship sailing from a place in the latitude of 49° 57′ N., and ongitude of 30° W., makes a course good of S. 39° W., and then, by observation, is ir the latitude of 45° 31′ N.; required the distance run, and longitude in. Ans. Distance 342.3, longitude 35° 20′ W. Quest. VIII. A ship in the latitude of 50° 10′ S., and longitude of 30° 00′ E., sails E. S. E. until her departure is 957 miles; required her distance sailed, and latitude and longitude in. Ans. Distance 1036 miles, latitude 56° 46′ S., longitude 56° 48′ E. Quest. IX. A ship in the latitude of 49° 30′ N., and longitude of 25° 00′ W., sails south-easterly 645 miles, until her departure from the meridian be 500 miles; required the course steered, and the latitude and longitude the ship is in. Ans. Course S. 50° 49′ F... latitude 42° 42′ N. longitude 12° 59 W MERCATOR'S SAILING. THE calculations by Middle Latitude Sailing are sufficiently exact for a snort run ɔr a day's work, and are to be preferred in all cases where the difference of latitude is small in comparison with the difference of longitude; but this method is liable to great errors in calculating the situations of places differing greatly in latitude and longitude, particularly in high latitudes. To remedy this inconvenience, a chart was invented and published in the year 1566, by GERARD MERCATOR, a Flemish geographer, in which all the meridians are parallel to each other, but proportionally lengthened so as to conform to the spherical figure of the earth. The principles on which this chart is constructed were first explained in the year 1599, by Edward Wright, an Englishman, and are as follows: By Theorem II. of Parallel Sailing, the distance of two meridians corresponding to a degree or mile of longitude, in any latitude, is to the length of a corresponding degree or mile of the meridian, as the cosine of the latitude is to the radius, that is (by Art. 56, Geometry), as radius is to the secant of the latitude. Hence, if the meridians are supposed to be parallel to each other, or the distance of the meridians to remain the sane in every latitude, the degree or mile of latitude must be increased in proportion to the secant of the latitude. Therefore, if the radius be supposed to be equal to one mile, the length of the first mile of latitude from the equator will be represented by the secant of 1'; the second mile, by the secant of 2; the third mile, by the secant of 3', &c. Therefore the length of the expanded arc of the meridian may be found by a continual addition of secants, to every degree and minute of the quadrant, as in Table II., by means of which the chart (called Mercator's Chart) may be constructed, and all the cases of Mercator's Sailing may be projected and calculated. * lu using this table, the degrees are to be found at the top or the bottom, and the miles at the side; in the angle of meeting will be the length of the corresponding expanded are, usually called the meridional parts. If you wish to find the arc of the expanded meridian intercepted between any two parallels, or, as it is usually called, the meridional difference of latitude, you must, when both places are on the same side of the equator, subtract the meridional parts of the least latitude from the meridional parts of the greatest; the remainder will be the meridional difference of latitude: but if they are on different sides of the equator, the sum of the meridional parts of both latitudes will be the meridional difference of latitude required. EXAMPLE I. Required the meridional parts corresponding to the latitude of 42° 34′. Look in the bottom or top of the table for 42°, and in the right or left hand column, marked (M), for 34'; under the former and opposite the latter stand 2828, the meridional parts corresponding to 42° 34'. EXAMPLE II. Required the meridional difference of latitude between Cape Cod, in the latitude of 42° 03' N., and the island of St. Mary, in the latitude of 36° 59′ N. Cape Cod's latitude.... 42° 03′ N. 36° 59′ N. Meridional parts 2786 Meridional difference of latitude 395 *The manner of constructing this chart will be particularly explained hereafter. It may be observed, that the smaller the subdivisions of the arc of the meridian are, the greater will be the accuracy of the calculated length of the expanded arc of the meridian. To be perfectly accurate, the arc ought to be subdivided into the smallest quantities possible. Attention was paid to this circumstance in calculaung EXAMPLE III. Required the meridional difference of latitude between a place in the latitude of 35° 12′ N., and the Cape of Good Hope, in the latitude of 34° 22′ S. From these principles it follows, that in sailing upon any course, the true or proper difference of latitude is to the departure as the meridional difference of latitude is to the difference of longitude. Hence if MI (in the figure of Case I. following) be the proper difference of latitude, IO the departure, MO the distance, the angle IMO the course, and we take MT equal to the meridional difference of latitude, and draw TH parallel to IO to cut MO continued in H, the line TH will represent the difference of longitude for (by Art. 53, Geometry) MI: IO :: MT : TH. Now, in the triangle MTH, by making MT radius, we have MT: radius :: TH: tangent TMH; that is, the meridional difference of latitude is to radius, as the difference of longitude is to the tangent of the course. By making MH or TH radius, we shall have other analogies, which, being combined with those in Plane Sailing, furnish the solutions of the various cases of Mercator's Sailing contained in the following table. SOLUTIONS. Having both lats. the mer. diff. lat. is found by Table III. Radius: dist. :: cosine course: prop. diff. of lat. Hence we Distance: radius :: proper diff, of latitude : cosine course. The latitudes and longitudes of two places given, to find the direct course and distance between them. Required the bearing and distance from Cape Cod light-ho se, in the latitude of 42° 03′ N., and longitude 70° 04' W., to the island of St. Mary one of the Western Islands, in the latitude of 36° 59' N., and longitude of 25° 10′ W. Draw the meridian MT equal to the meridional difference of latitude 395 miles; set off also upon it MI equal to the proper difference of latitude 304 miles; perpendicular to MT draw TH and 10; make TH equal to the difference of longitude 2694 miles draw MH cutting IO in Ó; then will the angle TMH be the course S. 81° 40′ E., and OM the distance 2098 miles. 1st. Extend from the meridional difference of latitude 395, to the difference of longitude 2694, on the line of numbers; that extent will reach from the radius or 45°, to the course 81° 40', on the line of tangents. 2dly. Extend from the complement of the course 8° 20′, to radius 90°, on the line of sines; that extent will reach from the proper difference of latitude 304, to the distance 2098, on the line of numbers. BY INSPECTION. With the meridional difference of latitude and difference of longitude used as difference of latitude and departure, find the course, by inspecting the tables until those numbers are found to correspond; with this course and the proper difference of latitude, find the corresponding distance. Thus one tenth of the meridional difference of latitude and difference of longitude are found to agree nearly to a course of 74 points; this course and one tenth of the proper difference of latitude 30.4, is found to correspond to the distance 207; this multiplied by 10 gives the distance 2070, differing a little from the result by logarithms, owing to the neglect of a few minutes in the course. CASE II. Both latitudes and the departure given, to find the course, distance, and difference of longitude. A ship in the latitude of 49° 57′ N., and longitude of 15° 16′ W., sails south-westerly until her departure is 197 miles, and then, by observation, is in the latitude of 47° 18′ N.; required her course, distance, and longitude in. Latitude left.... 49° 57' N. 47 18 N. Meridional parts.... 3470 .... BY PROJECTION. Distance A With the proper difference of latitude and departure, project as in Case VI. Plane Sailing, by drawing the meridian AEB, on which take AE equal to the proper difference of latitude 159 miles; erect ED perpendicular to AE, and make it equal to the departure 197 miles; join AD, and continue it towards C; make AB equal to the meridional difference of latitude 241 miles, and draw BC perpendicular to AB, to cut AC in C, and it is done. For AD will be the distance 253.2 miles, BC the difference of longitude 298.6 miles, and the angle BAC will be the course-S. 51° 06′ W. D Departure Diff Long. B To find the distance. 10.00000 As radius... To the distance 253.2..... Longitude left...... Longitude in ........... 2.40347 15° 16' W. 4 59 W 20 15 W The difference of longitude may als be found by saying, As proper difference of latitude departure:: meridional differ ence of latitude : difference of longitude. BY GUNTER. 1st. The extent from the difference of latitude 159, to the departure 197, on the line of numbers, will reach from radius 45°, to the course 51° 06', on the line of tangents. 2dly. The extent from the course 51°06', to radius 90°, on the sines, will reach from the departure 197, to the distance 253.2, on the line of numbers. 3dly. The extent from the radius 45°, to the course 51° 06', on the line of tangents, will reach from the meridional difference of latitude 241, to the difference of longitude 298.6, on the line of numbers. BY INSPECTION. Find the course by Plane Sailing, Case VI., by seeking in the tables with the proper difference of latitude and departure till they are found to agree in their respective columns, corresponding to which will be the distance in its column, and the course will be found at the top of that column if the departure is less than the proper difference of latitude, otherwise at the bottom; with the same course find the meridional difference of latitude in the latitude column, corresponding to which, in the departure column, will be the true difference of longitude. Thus with the true difference of latitude and departure 159, and 197, I find the course 51°, and the distance 253; in the same table, opposite to half of the meridional difference of latitude 120.5, I find the departure 148.8, which, being multiplied by 2, gives the difference of longitude 298 miles, nearly. CASE III. One latitude, course, and distance given, to find the difference of latitude and difference of longitude. A ship in the latitude of 42° 30′ N., and longitude of 58° 51′ W., sails S. W. by S. 300 miles; required the latitude and longitude in. BY PROJECTION. Draw the meridian ABC and ADE, making an angle with it equal to the course 3 points; make AD equal to the distance sailed 300 miles, and from D let fall upon AB the perpendicular BD; then will BD be the departure, and AB the difference of latitude 249.4 miles. Hence we have both latitudes, and the meridional difference of latitude, to which make AC equal, and draw CE parallel to BD, meeting ADE in E; then will CF be the difference of longitude 218.5 miles. *This logarithm, by the preceding operation, was found equal to 10.09307, differing a little from the log. tang, of 51'06 |