But since m is infinitely small, m2, m3, &c. are infinitely smaller, and the error of rejecting them in (259) is less than any assignable quantity; whence and by (254), M = 1+2 m 1 e = (1 + 2 m ) 2 m whence e has the same value as in (163 and 164). II. The value e (164) gives by (13) (260) log. (2.71828) 0.43429 so that we have by (257) Mer. parts of L = 7915.7 log. cotan. (45° — L), (262) which agrees with the explanation of Table III. given in the Preface to the Navigator. Correction for middle latitude. 45. EXAMPLES. 1. Calculate the meridional parts of latitude 45° 48'. 2. Calculate the meridional parts of latitude 28° 14'. Ans. 1767. 3. Calculate the meridional parts of latitude 83° 59′. Ans. 10127. 46. Problem. To calculate the correction for middle latitude sailing. Solution. If the angle DBC (fig. 19) were exactly what it should be in order that the hypothenuse BD should be the difference of longitude, and the leg BC the departure, it would be the corrected middle latitude, and we should have diff. long. sec. cor. mid. lat. X departure sec. cor. mid. lat. X diff. lat. Xtang. course, (263) which, compared with (258) gives, by dividing by tang. course, mer. diff. lat. sec. cor. mid. lat. X diff. lat. (264) Correction for middle latitude. whence sec. cor. mid. lat. = mer. diff. lat. (265) If, from the corrected middle latitude, calculated by this formula, the actual middle latitude is subtracted, the correction of the middle latitude is obtained, as in the table on p. 76 of the Navigator. The meridional difference of latitude should be obtained for these calculations, not from the tables of meridional parts, but directly from the tables of logarithmic sines, &c., by means of (257); and when the difference of latitude is less than 14°, tables should be used in which the logarithms are given to seven places of decimals. 47. Corollary. A formula, adapted to calculation by logarithms of five places, can be obtained by the following process. Let L。 the middle latitude = ≥ (L + L') Ο By changing, in (248), the small letters to large ones, we obtain log. 1 + sin. 1⁄2 l。 sec. L。 −2 log. e [sin. 1⁄2 l。 sec. 1。 Ο 1 — sin. 1⁄2 l。 sec. L。 +(sin. 27 sec. L)+(sin. 2 l, sec. Lo)5+&c.] (268) which gives, by substitution in (267 and 266), 1. Find the correction for middle latitude sailing, when the middle latitude is 35°, and the difference of latitude 14°. sec. 10.08879 corrected mid. lat. — 35° 24' correction = 35° 24′ 35° 24'. - 2. Find the correction for middle latitude sailing, when the middle latitude is 66°, and the difference of latitude 10°. (sin. 5° sec. 66°)3 0.00984 7.99297 (0.00984)=0.00328 (sin. 5° sec. 66°) 5 0.00045 6.65495 (0.00045)=0.00009 (sin. 5° sec. 66°)7=0.00002 5.31693 (0.00002) = 0.00000 3. Find the correction for middle latitude sailing, when the middle latitude is 30°, and the difference of latitude 4°. |
