Scholium. There are various methods of solving the Problem of the Course. 10. Plane Sailing, where the distance run being small, the sea is regarded as a plane, and the common forms of plane trigonometry are employed. 20. Traverse Sailing, where several short courses are run in succession, and forms (421), (422), are applicable. 3°. Parallel Sailing, where the ship sails on a parallel of latitude, and the difference of longitude will be found by (610). 4°. Middle Latitude Sailing, where the distance run is moderate and nearly east and west; in this method the middle latitude is assumed to be equal to the half sum of the two extreme latitudes when both are of the same name. 5°. Mercator's Sailing, as explained in (609). EXERCISES. 10. A ship sails from 3° S. lat. and 9° W. lon., N. 40° W. 538 miles. Required her latitude and longitude. For the difference of latitude we have (606), log. cos 40° = 1.884254 log. 5382 730782 2o. In what direction must a ship sail from New York to arrive at London, and what distance will she make? 3. A ship sails from 20° N. L. and 70° W. L. for the Azores in 38° N. L. and 30° W. L. After performing a thousand miles of her course, she tacks for New York; in what direction must PROBLEM OF THE PLACE. 295 she steer, and what will be her distance to the last-mentioned port? 4°. If a ship could sail from the equator N. 45° E. till she should arrive at the pole, what would be her distance run, and what her progress in longitude? The distance run will be found = 7636'68 miles, a finite number, but the difference of longitude will be infinite; that is, the ship will make an infinite number of turns about the pole, spinning there at length, like a top, with an infinite velocity of rotation. 5°. Calculate, the following table of meridional parts, and employ it in solving the exercises just given, and in constructing a chart extending from the equator 60° each way. 10 60 Table of Meridional Parts. 10° 20° 30° 40° 50° 60° 70° 80° 0 603 1225 1888 2623 3474 4527 5966 8375 SECTION SECOND. Problem of the Place. Owing to the impossibility of sailing absolutely in the required course and measuring accurately the distance run, it becomes necessary to resort to astronomical observations for the purpose of determining from time to time the ship's position on the ocean. To this end the daily places of the most conspicuous of the heavenly bodies are computed and laid down in the Nautical Almanac* for Greenwich mean time, and for that of Paris in the Connaissances des Tems. These places are such as they would appear to an observer at the earth's centre, and are estimated in right ascension and declination. The declination of a heavenly body is its angular distance from the Equinoctial [Celestial Equator], plus if north, minus when south, and the distance measured, commonly in hours, from the Vernal Equinox, to the east upon the Equinoctial, is the right ascension of the same body. PROPOSITION I. Given the geocentric meridian altitude of a heavenly body and its declination, to find the latitude. S Let O be the centre of the earth, P the pole elevated above the horizon, H, and 90° distant from P the celestial equator, E; then denoting the zenith by Z and the position of any heavenly body in the same meridian by S, S, S, we have H Fig. 134. The latitude of a place, or the altitude of the elevated (611) pole, is equal to a heavenly body's meridian distance from the point of the horizon under the pole, increased or diminished by the polar distance of the same body, according as it is situated below or above the pole. Scholium I. The polar distance of any heavenly body is (612) equal to 90° its declination. Scholium II. The latitude may be found on shore, or when (613) the vessel is stationary, by observing that the elevation of the pole * Blunt's reprint is sufficient for all purposes except those of a fixed and regularly furnished observatory, and is much less expensive. The student should be possessed of such astronomical knowledge and acquaintance with the stars as may be acquired from elementary works like "Kendall's Uranography." + To make use of either ephemeris it will evidently be necessary to reduce the time of any observation to that of the meridian for which the ephemeris is calculated, by allowing 15 to the hour, and observing the equation of time. = the half sum of the upper and lower meridian altitudes of any circumpolar star; since give Ph=(Sh) - Sp and Pr= (S)+ Sp P = [(S)+(Sn)ı]• h Example. On the 20th of August, 1847, at noon, the meridian altitude of the sun's centre was found to be 62° 49′ 50′′; at the same instant the chronometer, regulated to Greenwich time indicated 2h. 45m. Required the latitude of the place. Interpolating (279) for the declination from the Nautical Almanac, we have the following x D1 = == -2 16'3 '6 PROPOSITION II Given the polar and zenith distances of a heavenly body, and the zenith distance of the pole [= colatitude], to find the time. The hour, found approximately by the longitude of the vessel and the chronometer, regulated to Greenwich time, is to be corrected by observation. Let A be the pole, B the position of the heavenly body, and C the zenith of the place; then the hour angle A will be found by (540) or by (541), the first being preferable when A is small. Fig. 135. (614) Example. In the preceding example, after sailing 3 hours due west, the zenith distance of the sun's centre was found to be 47° 43′ 51.3". Required the hour angle, A, or the time referred to the meridian of the place arrived at. |