To find the Diff. of Longitude. To find the Longitude in. Diff. of Lat. — 4 oo Sum of Latitude 78 oo Is to Radius 90° 10.ooooo Is to Tang. Course 36.52 9.87561 So is Diff. of Lat. 240 2.38oz. So is Diff of Lat. 240 2.386.21 12.38021 12.255.2 2.477 I 2 9.890.50 To the Co-fine Cou. 36° 52' 9.90309 To Diff of Long. 231.6 2.36472 C - A S E VI. One Latitude, Course, and Boparture given, to find the Distance, Difference of Latitude and Difference of Longitude. A Ship sails E. S. E. from a certain Port in Latitude 50° 18' S. and Longitude 10° 16' E. until her Departure from the Meridian be , #. Miles; I demand her Distance sailed, and the Latitude and ongitude she is in : To One Latitude, Distance sailed, and Departure from the Meridian given, to find the Course, Difference of Latitude, and Difference of Longitude. A Ship in Latitude 49° 30' N, and Longitude 14° 40'W., sails South Eastward 645 Miles, until her Departure from the Meridian be 500 Miles; I demand the Course steered, and the Latitude and Longitude the Ship is in To find the Course it will be, To find the Diff of Lat, it will be, M. . . . Lat. Lat. left is 49° 30'N. Lat. left 49 30 Diff. Lat. 407, or 6 47 S. Lat. in 42 43 Lat, in 4. 43 N. Sum is - 92 13 * * * - Middle Latitude 46 6 M E R C A To R's SA I L ING. LANE SAILING, as has been before observed, supposes the Earth and Sea to be in the Form of a Bowling-Green, on which the Meridians are parallel, and the Degrees of Latitude and Longitude equal in all Places; but the Earth and Sea include a round Body or Globe, on which the Degrees of Latitude are equal in all Places, and the Degrees of Longitude decrease from the Equator in Proportion to the Sine Complement of the Latitude. Though the Meridians all meet at the Poles, and the Parallels to the Equator continually decrease, and that in Proportion to the Co-Sines of their Latitudes ; yet in old Sea Charts the Meridians were drawn parallel to each other, and consequently the Parallels of Latitude made equal to the Equator, and so a Degree of Longitude on any Parallel, as large as a Degree on the Equator: Also in these Charts the Degrees of Latitude were still represented (as they are in themselves, equal to each other, and to those of the Equator ; by these Means the Degrees of Longitude being increased beyond their just Proportion, and the more so the nearer they approach the Pole, the Degrees of Latitude at the same Time remaining the same, it is evident, Places must be very erroneously marked down upon those Charts, with respect to their Latitude and Longitude, and consequently their Bearing from one another must be very false. To remedy this Inconvenience, so as still to keep the Meridians parallel, it is plain we must protract or lengthen the Degrees of Latitude in the same Proportion as those of Longitude are, that so the Proportion in Easting or Westing may be the same with that of Northing or Southing; and consequently the Bearing of Places from each other to be the same upon the Chart as upon the Globe * OW Now to discover how the Meridians are expanded from the Equator in Proportion to the Degrees of Longitude decreasing towards the Poles; Let A B D in the annexed Scheme, represent the Quarter of the Meridian ; F B, the Radius of a Parallel of Latitude; now C G, which is equal to F B, is to CD, as a Degree on the Parallel is to a Degree on the Meridian, or any great Circle; and as C G is to CD, so is CB to CE, the Secant of the Latitude of the Parallel. Therefore in a Projećtion of the Globe where the Meridians are kept parallel, it is evident, a Degree on the Meridian at any Parallel must be - - i equal to the Secant of the Latitude of that Pa- C G D rallel; and the Distance of any Point, upon the Meridian, from the Equator, is equal to the Sum of the Secants contained between it and the Equator. Hence it is evident, that by a continual Addition of Secants, beginning at the Equator, a Table of Meridional Parts may be composed for every Degree and Minute in the Quadrant. Therefore the Meridional Difference of Latitude between any two Places may be easily found, by finding the Meridional Parts answering to both Latitudes, and either adding or subtracting, according as the Case requires; that is, if both North or both South, subtracted; but if one §. and the other South, added gives the Meridional Difference of Latitude between them : but the Meridional Difference of Latitude between any Place and the Equator is found, by taking the Meridional Parts belonging to the Latitude of that Place. To find the Meridional Parts belonging to any Number of Degrees and Minutes of Latitude required. In the Table of Meridional Parts, seek the Degrees on the upper Part of the Table, and in the left or Right Hand Column the Minutes marked on the Top with M. opposite to which, and under the Degrees, are the Meridional Parts required. Suppose the Meridional Parts belonging to 57° 18' were required 2 Look in the Table under 57°, and opposite to 18 stands 4216, the Meridional Parts of 57° 18'. The same may be observed of any Degrees and Minutes required. The Solution of the following Problems, as F jo-. well as all other Trigonometrical Operations in Navigation, depend upon the fourth Proposition of the sixth Book of Euclid; where it is demonforated, that Triangles which are fimilar or alike their like Sides are Proportional : Therefore in the annexed Triangles A B C, and A F G, the Radius, Sine, and Sine Complement, or the Radius Tangent, and Secant, form a right-angled Triangle ; and the Sine Tangent and Secant of M 2 any any Arch in one Circle, is in Proportion to the Sine Tangent and Secant of the same Arch in another Circle, as the Radius of the one is to the Radius of the other. Let A B represent the proper Difference of Latitude; B C the Departure; A C the Distance; and the Angle B A C the Ship's Course. Produce A B to F, to represent the Meridional or enlarged Difference of Latitude; and parallel to B C draw F G, to represent the Difference of Longitude. It is plain that A B is in proportion to B C, as A F is to FG ; and that the Sine Tangent and Secant of the Triangle F AG, is as the Radius AC is to d.o.o. AG : Wherefore, as the Sine Complement of the Course is to the Meridional Difference of Latitude, so is the Sine of the Course to the Difference of Longitude, and the contrary. . And as the proper Difference of Latitude is to the Departure, so is the Meridional Difference of Latitude to the Difference of Longitude. Hence it will be easy to reduce Departure into Difference of Longitude, and Difference of Longitude into Departure. Therefore, all Cases in Mercator's Sailing are worked by Geometry, Trigonometry, Gunter’s Scale, Inspection, and the Tables exactly the same as in Plane Sailing, by only considering the Meridional Difference of Latitude as proper Difference of Latitude; and the Difference of Longitude as Departure: For it is no more than enlarging the Difference of Latitude that the Difference of Lonitude may be in Proportion to the Departure, as the Meridional #. of Latitude is to the proper Difference of Latitude, the Course continuing the same ; and the Sine Complement, or Tangent Complement of the Course bears the same Proportion to the Meridional Difference of Latitude that the Sine or Tangent of the Course does to the Difference of Longitude, and therefore is found |