globe, by placing it as for a right sphere, and bring ing a or, or any of the intermediate points, to the zenith; when it will be evident, that if we are to go from one of these points, a, to the other, %, or to any point on the equator, we must continue our course due east, to arrive at a, or vice versa. From hence we may deduce this consequence, that if a ship, under the equator, sails due east or west, she will continue under the equator.

In the two foregoing cases, the course being an arc of a great circle (the meridian or equator) is the shortest and most convenient way it can sail.

If two places lie under the same parallel, the course from one to the other is due east or west: this may be seen upon the globe, by the following method: bring any point of a parallel to the zenith, and stretch a thread over it, perpendicular to the meridian; the thread will then be a tangent to the parallel, and stand east and west from the point of contact. Hence, if a ship sails in any parallel, due east or west, she will continue in the same parallel. In this case the most convenient course, though not the shortest, from one to the other, is to sail due east or west.

If two places lie neither under the equator, nor on the same meridian, nor in the same parallel, the most convenient, though not the shortest course, from one to the other, is a rhumb.

For if we would, in this case, attempt to go the shortest way, in a great circle drawn through the twa places, we must be perpetually changing our course. Thus, in fig. 8, whatever is the bearing of K from A,

the bearings of all the intermediate points, as B, C, D, &c. will be different from it, as well as different from each other, as may be easily seen upon the globe, by bringing the first point, A, to the zenith, and observing the bearing of K from each of them. Thus, suppose, when the globe is rectified to the horizon of A, the bearing of K from A be north-east, and the angle of position of K, with regard to A, be 45 degrees; if we bring B to the zenith, we shall have a different horizon, and the bearing and angle of position from K to B will be different from the former; and so on of the other points B, C, D, &c. they will each of them have a different horizon, and K will have a different bearing and angle of position.

From hence we may draw this corollary, that when two places lie one from the other, towards a point not cardinal, if we sail from one place towards the point of the other's bearing, we shall never arrive at the other place. Thus, if K lies north-east from A, if we sail from A towards the north-east, we shall never arrive at K.

A rhumb upon the globe is a line drawn from a given place A, so as to cut all the meridians it passes through at equal angles; the rhumbs are denominated from the points of the compass, in a different manner from the winds. Thus, at sea, the north-east wind is that which blows from the north-east point of the horizon towards the ship in which we are; but we are said to sail upon the N.E. rhumb, when we go towards the north-east,

The rhumb ABCDK, plate 13, fig. 8, passing through the meridians LM, NO, PQ, makes the angle LAB, NBC, PCD, equal; from whence it follows, that the direction of a rhumb is in every part of it towards the same point of the compass: thus, from every point of a north-east rhumb upon the globe, the direction is towards the north-east, and that rhumb makes an angle of 45 degrees, with every meridian it is drawn through.

Another property of the rhumbs is, that equal parts of the same rhumb are contained between parallels of equal distance of latitude; so that a ship continuing in the same rhumb, will run the same number of miles in sailing from the parallel of 10 to the parallel of 30, as she does in sailing from the parallel of 30 to that of 50.

The fourth thing mentioned to be required in navigation was, to know, at any time, what point of the globe the ship is upon. This depends upon four things: 1. The longitude; 2. The latitude; 3. The course the ship has run; 4. The distance, that is, the way she has made, or the number of leagues or miles she has run in that course, from the place of the last observation. Now, any two of these being known, the rest may be easily found.

Having thus given some general idea of navigation, we now proceed to the problems, by which the cases of sailing are solved on the globe.

Given the difference of latitude

PROBLEM XLIII.

and difference of longitude, to find the course and distance sailed.*

Example. Admit a ship sails from a port A, in latitude 38 deg. to another port B, in latitude 5 deg. and finds her difference of longitude 43 deg.

Let the port A be brought to the meridian, and elevate the globe to the given latitude of that port 38 deg. and fixing the quadrant of altitude precisely over it on the meridian, move the quadrant to lie over the second port B, (found by the given difference of latitude and longitude) then will it cut in the horizon 50 deg. 45 min. for the angle of the ship's course to be steered from the port A. Also, count the degrees in the quadrant between the two ports, which you will find 51 deg. this number multiplied by 60, the nautical miles in a degree, will give 3060 for the distance run.

PROBLEM XLIV. Given the difference of latitude and course, to find the difference of longitude, and distance sailed.

Example. Admit a ship sails from a port A, in 25 deg. north latitude, to another port B, in 30 deg. south latitude, upon a course of 43 deg.

Bring the port A to the meridian, and rectify the

* See Martin, on the Globes.

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 intersects 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