1 ft. Draw the Meridian of the Lizard A B, which make equal to the Meridional Difference of Latitude 267 1 Miles, on B erect the Perpendicular B C, upon which set off the Diff. of Longitude 3263 Miles. Draw the Line C A ; then will the Angie B A C be the Course 50°42'. 2dly. Set off the proper Diff. . . of Latitude 2205, upon the Me-C ridian, from A to D, and E D parallel to B C, to cut A C in E ; then will EA be the Distance 3481 Miles. To find the same First. “The Extent from the Meridional Diff. of Lat. 2671 to the Diff of Long. 3263 on the Line of Numbers, will reach from Radius, or 45° to 53°42' on the Line of Tangents.’ Secondly. “The Extent from Co-Course 39° 18' to Radius or 90° on the Line of Sines, will reach from the Proper Diff. of Lat. 22.05 to 3481, the Distance, on the Line of Numbers.” RULE. Look for the Meridional Difference of Latitude, and Difference of Longitude, as if they were really Difference of Latitude and Departure in Plane Sailing, and the Course will be found among the Points or Degrees at the Top or Bottom ; then, instead of the Meridional Difference of Łatitude, look for the proper Difference of Latitude, in the Column marked Lat, and in the Column marked Dist. will be the Distance required. Thus, with the 20th Part of the Mer. Diff. of Lat. 2671, and Diff. of Long. 3263, viz. 133 and 163. I the nearest Numbers are which, C -Z/2Zong, E. By P R O J E C T I O N. With the Proper Difference of Lat, and Departure, projeć, the same as in Case VI. in Plane Sailing; extend the Meridian A E to B, and make A B equal to the Meridional Diff. of Lat. and draw a Line parallel to the Departure D. E.; produce the Distance A D to cut this Parallel; and CD will be the Difference of Longitude. Hence the Course will be found to be S. 51° 5' W. Distance 1014, and Diff. of Long. I I 14 Miles. To find the same r By P R O J E C T I O N. Draw a Meridian, A B the lower End A will represent the Ship's Place in her first Latitude. Take the Proper Difference of Latitude 266 in your Compasses, and with one Foot in A, the Ship's Place, lay the other upon the Meridian Meridian ; then from A to E, take the Merid. Diff. of Latitude 396 in your Compasses, and with one Foot in A, the Ship's Place, as before, lay the other upon the Meridian at B ; and upon these two Points raise the Perpendiculars DE and C B ; a Line drawn from the Ship's Place, making an Angle with the Meridian equal to 39° the Ship's Course, will cut the two Perpendiculars at D and C; the first will be the Departure, which terminates the Distance A D 342, and the other will be the Difference of Longitude C B = 32. I Miles. . . . From what has been said, it is plain, that any Case in Mercator's Sailing may be projected as a Right-angled Triangle, by only confidering the Difference of Longitude, or 19eparture, as the Base; the Meridional, or Proper Difference of Latitude, as the Perpendicular; the Hypothenue cut by the Departure as Distance; and the Angle which that makes with the Perpendicular, the Course; for in all Cases in Mercator's Sailing, the Meridional Difference of Latitude bears the same Proportion to the Difference of Longitude, that the Proper Difference of Latitude does to the Departure. ; These Instructions being well understood, will be sufficient to inform the Learner how to construct any of the following Cases: To the Distance 343.3 2,53438. To the Diff of Lon. 320,7 2,50607 * * Ilongitude left 5° 14' W. Difference of Long, 32.1; or 5 21 W. Longitude in - io 35 W. . By G U N T E R. 1st. ‘The Extent from Co-Sine Course 51°, to Radius on the Line of Sines, will reach from the Proper Difference of Latitude 266, to the Distance 342.3 on the Line of Numbers.” 2dly. “The Extent from Co-Sine Course 5.1°, to Sine Course 39° on the Line of Sines, will reach from the Meridian Difference of Latitude 396, to the Difference of Longitude 321, on the Line of Numbers.” tinder the same Degrees, and in the Lat. Column, look for half the Merid. Diff. of Lat. 198, against that, in the Dep. Column, stands ióo.5, doubled is 321, the Diff of Longitude nearly as before. - - - - C A S E IV. * One Latitude, Cours, and Distance given, to find the Difference of Latitude, and Difference of Longitude. A Ship in Latitude 42° 30' N. and Longitude 18° 31' W. fails S. W. by S. 591 Miles. I demand the Latitude and Longitude the Ship is in - - To find the Diff. of Latitude it will be, By G ty. N T E R. ist. ‘The Extent from Radius to 5 Points, the Com. of the Course on the Line marked S R, will reach from the Distance 591, to the Difference of Latitude 491.4, on the Line of Numbers.” 2dly: ‘The Extent from Co-Course 5 Points, to the Course 3 Pts. on the Line marked S R, will reach from the Merid. Difference of Latitude 628, to the Difference of Longitude 419.6, on the Line of Numbers.” Under the Course 3 Points, and opposite a tenth of the Distance 59 in the Latitude Column stands 49, I, which multiplied by 19, is 491, the Difference of Latitude; then find # of the Meridian Dif- . ference of Latitude 157, in the Latitude Column, against which Ítands 105 in the Departure Column, which multiplied by 4, gives 420, the Difference of Longitude. N C A S E |