direction NE as at the beginning. Let us now suppose a ship leaving the first meridian at the point where the great circle crosses the equator to the northward and steering NE, her course may for a short space be considered as coinciding with the great circle; but when she arrives at a meridian 1° to the eastward of the first, not parallel to it but pointing to the N pole, their common point of meeting, the direction of her course must be a little inclined to the northward, so as still to form an angle of 45° with the second meridian. Proceeding NE until she come to the third meridian, 1° to the eastward of the second, her course must again be a little deflected to the northward in order to form with this third meridian an equal angle of 45°; by which process her course will not only depart to the northward of the circumference of the great circle on which she set out, but if we suppose her to sail quite round the globe in a NE direction, continually bisecting the angle formed by the meridian running northerly, and the parallels of latitude running easterly, her course will describe a winding or spiral curve line on the surface of the globe, continually approaching the north pole but never falling into it, inasmuch as the NE course always lying midway between N and E, it never can deviate into due N, which last course alone can in correct geometrical language bring the ship to the N pole. It is this constant tendency of the ship's course to approach the pole and deviation from the direction of the preceding portions of her course which occasion the inaccuracy in the ordinary computations by the middle latitude sailing, giving results constantly less than the truth. In solving problems by middle latitude sailing the following rules are to be observed. CASE 1st.-Given the latitudes and longitudes of two places; required the course or bearing and the distance between them. Required the bearing or course and distance between Cape VOL, II. 2 a Clear Clear, the southern extremity of Ireland, in N. lat. 51° 19′, W. long. 9° 23' and isle of Ushant on the coast of France, in N. lat. 48° 29', W. long. 5° 5. With these data to find the course we have the following Again to find the distance between the two given places, From Cape Clear, therefore, to Ushant is a distance of 238 nautical miles, and the course is S 44° 21' E, or SE 39 S. CASE CASE 2d. Given the latitude and longitude sailed from, the course and distance, required the latitude and longitude come to. A ship from Cape St. Vincent in N lat. 37° 3', W long. 8° 59', sails SW by W 560 miles; required the latitude and longitude come to. As radius To co-sine of course So distance To diff. of lat. Lat. sailed from 37.03 Hence we find that the ship will have arrived at a point in N. lat. 32° 52', and in W. long, 18° 27', as was required to be discovered. CASE 3d.-Given both latitudes and the course, to find the distance and difference of longitude. A ship from the Lizard in N. lat. 49° 58', W. long: 5° 11' sails SWW until she come into N. lat. 43° 20', required the distance run, and the longitude come to. To secant of course, 4 points = 50° 37′ = 10.19706 CASE 4th.-Given both latitudes and distance, to find the course and difference of longitude. A ship from the Spurn in N. lat. 53° 41′, E. long. 0° 17', sails 220 miles in the NE quarter, and then finds by observation |