In the annexed diagram, let the triangle ABC be a figure in plane sailing, in which AC represents the distance, A B the difference of latitude, BC the departure, and the angle A the course. Again, let DBC be a figure in parallel sailing, in which DC represents the difference of longi. tude, BC the meridional distance, and the angle C the middle latitude. Hence, the parts concerned form two connected right angled triangles, in which the departure or meridional distance BC is a side common to both. Now, in one of these triangles, there will be always two terms given, and in the other one term, at least, to find the required terms. The required parts in that triangle which has two terms given, may be readily found by the analogies for right angled plane trigonometry, page 171 to 177; and, hence, the unknown terms in the other triangle. When the departure BC is not under consideration, the two connected triangles may be considered as one oblique angled triangle, and resolved as such. In this case, if the course, distance, middle latitude, and difference of longitude, are the terms in question, any three of them being given, the fourth may be found by one direct proportion. Thus, in the oblique angled triangle A CD, the side AC is the distance; the angle A, the course; the angle BCD, the middle latitude; and, consequently, the angle D its complement, and the side D C the difference of longitude. Now, if any three of these be known, the fourth may be found by one of the following analogies; viz., 1. As co-sine middle latitude = C: sine of course A :: distance = AC difference of longitude = DC. 2. As sine of course A co-sine middle latitude = C:: difference of longitude DC: distance = AC. = 3. As distance = AC: difference of longitude = DC: co-sine of middle latitude = C: sine of course = A. 4. As difference of longitude = DC: distance = AC:: sine of course A co-sine of middle latitude = C. Again, if the course, middle latitude, difference of latitude, and difference of longitude, be the terms under consideration, the resulting analogies will be, 5. As difference of latitude = AB: difference of longitude = DC :: co-sine of middle latitude = C: tangent of course = A. 6. As difference of longitude = DC: difference of latitude = AB :: tangent of course = A co-sine of middle latitude = C. 7. As co-sine of middle latitude = C: tangent of course = A :: difference of latitude = AB: difference of longitude = A C. 8. As tangent of course = A: co-sine of middle latitude = C:: difference of longitude = DC; difference of latitude = A B. == In these four analogies, it is evident that the course must be a tangent, because the difference of latitude A B is concerned. Note. Since the sine complement of the middle latitude=the angle D, is expressed directly by the co-sine of the angle BC D, therefore, with the view of abridging the preceding analogies, the co-sine of the middle latitude has been used instead of its sine complement; and, in the operations which follow, the same term will be invariably employed. Remark. The middle latitude between two places is found by taking half the sum of the two latitudes, when they are both of the same name, or half their difference if of contrary names. PROBLEM I. Given the Latitudes and Longitudes of two Places, to find the Course and Distance between them. Diff. of latitude = Sum of latitudes 8: 6-486 miles. Diff. of long. = 7:40=460 ms. 74:12 + 2 = 37:6′ = the middle latitude. Here, since the departure is not in question, the parts concerned come under the 5th analogy in page 222: hence, The course being thus found, the distance may be determined by trigonometry, Problem II., page 172: hence, Hence, the true course from Oporto to Porto Santo is S. 37:3: W., or S.W. S. nearly, and the distance 609 miles. To find the Course and Distance by Inspection in the general To middle latitude 37 as a course, and one-fourth the difference of longitude = 115, as a distance, the corresponding difference of latitude is 91.8 tude the meridional distance. Now, one-fourth the difference of lati121.5, and the meridional distance 91. 8 in a departure column, are found to agree nearest at 37°, under distance 152. Hence, the course is S. 37: W., and the distance 152 x 4 = 60S miles. PROBLEM II. Given the Latitude and Longitude of the Place sailed from, the Course, and Distance; to find the Latitude and Longitude of the Place To find the Difference of Latitude AB: = Here the course A, and the distance = A C, being given, the difference of latitude = AB may be found by trigonometry, Problem I., page To find the Difference of Longitude = CD: Here, since the departure is not concerned, the parts in question come under the 1st analogy in page 222: hence, To find the Difference of Latitude and Difference of Longitude by Inspection: Under or over one-fifth of the given distance = 118, and opposite to the course 4 points, is difference of latitude 74.9, and departure 91.2. Tabular difference of latitude 74.9 x 5 = 374.5, the whole difference of latitude; whence the latitude in, is 45:55? N., and the middle latitude 42:48. Now, to middle latitude 42°, and departure 91. 2, in a latitude column, the corresponding distance is 123 miles; and to middle latitude 43°, and departure 91. 2, the distance is 125 miles: hence, the difference of distance to 1 of latitude, is 2 miles; and 2 x 4860 1'.6, which, added to 123, gives 124. 6; this, being multiplied by 5 (the aliquot part), gives 623 miles the difference of longitude, or 10:23. E. Q = PROBLEM III. Given both Latitudes and the Course; to find the Distance and the With the course = To find the Distance AC: A, and the difference of latitude = A B, the distance is found by trigonometry, Problem II., page 172; as thus: As radius = 90:0: Log. co-secant = 10.000000 Is to the diff. of latitude = 256 miles Log. = To find the Difference of Longitude = CD: Here, since the departure is not in question, the parts concerned fall under the 7th analogy, page 222: hence, |