BY GUNTER. 1st. The extent from the course 6 points, to the radius 4 points, on the line marked TR, will reach from the departure 160, to the difference of latitude 66.3, on the line of numbers. 2dly. The extent from 6 points, to the radius, or 8 points, on the line marked SR, will reach from the departure 160, to the distance 173.2, on the line of numbers. 3dly. The extent from the complement of the middle latitude 39° 17', to the radius 90°, on the sines, will reach from the departure 160, to the difference of longitude 252.7, on the line of numbers. BY INSPECTION. Find the course among the points or degrees, Table I. or Table II. (as in Case III, Plane Sailing), and the departure in its column, corresponding to which, in the columns of distance and difference of latitude, will be found the distance and difference of latitude respectively; then with the middle latitude as a course, seek the departure in the column of latitude, corresponding to which, in the distance column, will stand the difference of longitude. Thus, I enter Table I., above E. S. E., or 6 points, and seek for the departure 160, the nearest to which is 159.8; the corresponding numbers give the distance 173, and the difference of latitude 66.2 miles. Enter Table II. with the middle latitude 50° 43', or (51° nearly) as a course, and seek for the departure 160, in the latitude column, opposite to which, in the distance column, will be found the difference of longitude 254 miles, nearly. CASE VII. One latitude, distance sailed, and departure from the meridian given, to find the course, difference of latitude, and difference of longitude. A ship in the latitude of 49° 30′ N., and longitude of 25° 0′ W., sails south-easterly 215 miles, until her departure from the meridian be 167 miles; required the course steered, and the latitude and longitude the ship is in. BY PROJECTION. Draw the line BD equal to the departure 167 miles, and perpendicular thereto draw the meridian line ABC; take an extent equal to the distance 215, in your compasses, and with one foot in D, as a centre, describe an arc cutting AB in A; join AD; then will AB be the difference of latitude 135.4 miles, and BAD the course, S. 50° 58′ E. Hence we have the latitude in, and middle latitude; make the angle BDC equal to the middle latitude, and draw DC cutting ABC in C; then DC will be the difference of longitude 251.5 miles, BY LOGARITHMS. * The correction of this latitude in the table is 1', making the corrected middle latitude 48° 24/. BY GUNTER. 1st. The extent from the distance 215, to the departure 167, on the line of numbers, will reach from the radius 90°, to the course 50° 58′ on the line of sines. 2dly. The extent from radius 90°, to the complement of the course 39° 02', on the line of sines, will reach from the distance 215, to the difference of latitude 135.4, on the line of numbers, 3dly. The extent from the complement of the middle latitude 41° 37', to the radius 90°, on the line of sines, will reach from the departure 167, to the difference of longitude 251.5, on the line of numbers. BY INSPECTION. As in Case V. Plane Sailing, find the course by seeking in Table II. till against the distance, in its column, is found the given departure in one of the following columns, adjoining to which, in the other column, will be the difference of latitude, which if greater than the departure, the course will be at the top, but if less the course will be found at the bottom. Then take the middle latitude as a course, and find the departure in the column of difference of latitude, against which, in the distance column, will be found the difference of longitude. Thus the distance 215, and the departure 167, are found nearly to correspond to a course of 51 degrees, and a difference of latitude of 135.3 ; then with the middle latitude 48°, as a course, I enter the table, and seek for the departure 167, in the latitude column; the distance corresponding 250 is the difference of longitude nearly. In all the preceding examples, we have used the middle latitude, without any correction, in computing the difference of longitude; but when absolute accuracy is required, this latitude must be corrected. We have given in the following table the value of this correction in the most common cases. It requires no particular explanation: one example will serve to show its use. Suppose, therefore, the two latitudes to be 40° and 60°. Here the middle latitude is 50°, and the difference of latitude 20°; the tabular correction corresponding to these numbers is 57'; adding this to 50°, we get the corrected middle latitude 50° 57', which is to be used instead of 50°, when great accuracy is required. We have inserted in the notes at the bottom of the pages, in the preceding examples, the values of this correction, but have not introduced it into the calculations, because it is generally unnecessary on account of its smallness. TABLE, This Table contains the correction, in minutes, to be added to the Middle Latitude to obtain the corrected Middle Latitude. MID. LAT. DIFFERENCE OF LATITUDE. 1° 2° 3° 4° 5° 6° 7° 8° 9° 10° 12° 14° 16° 18° 20° MID. LAT. 555 554 765 696 444 8 8 11 9 11 778 7 9 10 11 14 499 6 6 7 499 9 12 15 11 14 8 10 12 15 19 13 16 20 22 8 12 15 19 24 25 32 26 34 36 46 57 32 43 57 72 90 66 59 72 47 50 555 40 33 43 14 18 26 35 46 58 37 49 40 88 110 70 58 76 98124 72 This Table is to be entered at the top with the difference of the two latitudes, and at the side with the middle latitude; under the former, and opposite to the latter, is the correction, in minutes, to be added to the middle latitude, to obtain the corrected QUESTIONS FOR EXERCISE. Question I. Required the bearing and distance between two places, one in the latitude 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′ Ñ., longitude 51° 52′ W. Quest. VII. Suppose a ship sailing from a place in the latitude of 49° 57′ N., and longitude of 30° W., makes a course good of S. 39° W., and then, by observation, is in 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′ E., latitude 42° 42′ N., longitude 12° 59′ W. MERCATOR'S SAILING. THE calculations by Middle Latitude Sailing are sufficiently exact for a short run, or a day's work, and are to be preferred in all cases where the difference of latitude is stnall 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 same 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 III., 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. * In 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 arc, 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. ..... Meridional difference of latitude Meridional parts 2786 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 calculating 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 10 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 merid. diff. of lat. :: tangent course: diff. of long. Radius departure:: cotangent course: proper diff. of lat. Distance: radius :: departure: sine course. Hence we obtain the other latitude and merid. difference of latitude. Radius merid. diff. of lat. :: tangent course: diff. of long. Prop. diff. of lat. : departure :: mer. diff. of lat. : diff. long. CASE I 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-house, 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. |