globe to the latitude thereof, 25 deg. where fix the quadrant of altitude, and place it so as to make an angle with the meridian of 43 deg. in the horizon, and observe where the edge of the quadrant interseets the parallel of 30 deg. south latitude, for that is the place of the port B. Then count the number of degrees on the edge of the quadrant intersected between the two ports, and there will be found 73 deg. which, multiplied by 60, gives 4380 miles for the distance sailed. As the two ports are now known, let each be brought to the meridian, and observe the difference of longitude in the equator respectively, which will be found 50 deg. N. B. Had this problem been solved by loxodromics, or sailing on a rhumb, the difference of longitude would then have been 52 deg. 30 min. between the two ports, PROBLEM XLV. Given the difference of latitude and distance run, to find the difference of longitude and angle of the course. Example. Admit a ship sails from a port A, in latitude 50 deg. to another port B, in latitude 17 deg. 30 min. and her distance run be 2220 miles. Rectify the globe to the latitude of the place A, then the distance run, reduced to degrees, will make 37 deg. which are to be reckoned from the end of the quadrant lying over the port A, under the meridian ; then is the quadrant to be moved, till the 37 degrees coincide with the parallel of 17 deg. 30 min. north latitude ; then will the angle of the course appear in the arch of the horizon, intercepted between the quadrant and the meridian, which will be 32 deg. 40 min.; and by making a mark on the globe for the port B, and bringing the same to the meridian, you will observe what number of degrees pass under the meridian, which will be twenty, the difference of longitude required. PROBLEM XLVI. Given the difference of longitude and course, to find the difference of latitude and distance sailed. Example. Suppose a ship sails from A, in the latitude 51 deg. on a course making an angle with the meridian of 40 deg. till the difference of longitude be found just 20 degrees ; then rectifying the globe to the latitude of the port A, place the quadrant of altitude so as to make an angle of 40 deg. with the meridian; then observe at what point it intersects the meridian passing through the given longitude of the port B, and there make a mark to represent the said port; then the number of degrees intercepted between that and the port A, will be 28, which will give 1680 miles for the distance run: and the said mark for the port B, being brought to the meridian, will have its latitude there shewn to be 27 deg. 40 min. PROBLEM XLVII. Given the course and distance sailed, to find the difference of longitude, and difference of latitude. Example. Suppose a ship sails 1800 miles from a port A, 51 deg. 15 min. south-west, on an angle of 45 deg. to another port B. Having rectified the globe to the port A, fix the quadrant of altitude over it in the zenith, and place it to the south-west point in the horizon ; then upon the edge of the quadrant under 30 deg. (equal to 1800 miles from the port A) is the port B; which bring to the meridian, and you will there see the latitude; and, at the same time, its longitude on the equator, in the point cut by the meridian. In all these cases, the ship is supposed to be kept upon the arch of a great circle, which is not difficult to be done, very nearly, by means of the globe, by frequently observing the latitude, measuring the distance sailed, and (when you can) finding the difference of longitude ; for one of these being given, the place and course of the ship is known at the same time; and therefore the preceding course may be altered, and rectified without any trouble, through the whole voyage, as often as such observations can be obtained, or it is found necessary. Now if any of these data are but of the quantity of four or five degrees, it will suffice for correcting the ship’s course by the globe, and carrying her directly to the intended fort, according to the following problem. PROBLEM XLVIII. To steer a ship upon the arch of a great circle by the given difference of latitude, or difference of longitude, or distance sailed in a given time. Admit a ship sails from a port A, to a very distant port Z, whose latitude and longitude are given, as well as its geographical bearing from A; then, First, having rectified the globe to the port A, lay the quadrant of altitude over the port Z, and draw thereby the arch of the great circle through A and Z; this will design the intended path or track of the ship. Secondly, having kept the ship upon the first given course for some time, suppose by an observation you find the latitude of the present place of the ship, this added to, or subducted from the latitude of the port A, will give the present latitude in the meridian ; to which bring the path of the ship, and the part therein, which lies under the new latitude, is the true place B of the ship in the great arch. To the latitude of B rectify the globe, and lay the quadrant over Z, and it will shew in the horizon the new course to be steered. Thirdly, suppose the ship to be steered upon this course, till her distance run be found 300 miles, or 5 degrees ; then, the globe being rectified to the place B in the zenith, laying the quadrant from thence over the great arch, make a mark at the 5th degree. from B, and that will be the present place of the ship, which call C; which being brought to the meridian, this new its latitude and longitude will be known. Then rec. tify the globe to the place C, and laying the quadrant from thence to Z, the new course to be steered will appear in the horizon. Fourthly, having steered some time upon course, suppose, by some means or other, you come to know the difference of longitude of the present place of the ship, and of any of the preceding places, C, B, A ; as B, for instance; then bring B to the meridian, and turn the globe about, till so many degrees of the equator pass under the meridian as are equal to the discovered difference of longitude; then the point of the great arch cut by the meridian is the present place D of the ship, to which the new course is to be found as before, And thus, by repeating these observations at proper will find future places, E,F,G, &c. in the great arch; and by rectifying the course at each, your ship will be conducted on the great circle, or the nearest way from the port A to Z, by the use of the globe only intervals, you |