the W. N. W. line, and DE equal to 40 miles, parallel to the S. by E. line; then will E represent the place of the ship after sailing her several courses. Join EL, and draw EM perpendicular to LM; then will LE be the distance of Block Island, 66.8 miles; and the angle ELM=12° 16′, will be the course made good; LM the difference of latitude, and EM the departure. Though this method of finding the difference of latitude and departure by logarithms is accurate, yet the calculations may be more easily made by the tables of difference of latitude and departure, as in Case I. Plane Sailing. ing from it, and the remainder, 65.3, shows how far the ship is to the southward of A ship from Mount-Desert rock, in the latitude of 43° 50′ N., sails for Cape Cod, in the latitude of 42° 3′ N., its departure from the meridian of Mount-Desert rock being supposed to be 84 miles west; but by reason of contrary winds, she is obliged to sail on the following courses, viz. south 10 miles, W. S. W. 25 miles, S. W. 30 miles, and W. 20 miles. Required the bearing and distance of the two places, the course and distance sailed by the ship, and the bearing and distance of her intended port. Let C represent Mount-Desert rock; draw the meridian CF, which make equal to 107 miles, the difference of latitude between the two places, and perpendicular thereto, the line FE, equal to the departure, 84 miles; then is E the place of Cape Cod. With the chord of 60° sweep about the centre, C, a circle, S. W., to represent the compass, and upon it note the various courses sailed. The first course being south, the distance, 10 miles, is set off from C towards F upon the meridian, and this point represents the place of the ship after sailing her first course; continue setting off the various courses and distances as in the last example, viz. W. S. W. 25 miles, S. W. 30 miles, and west 20 miles. to the point A; then will A represent the place of the ship after sailing these courses. Join CE, AC, AE; draw AB perpendicular to the meridian CF, and AD parallel thereto; then will AC=76.2 miles be the distance made good; AE 69.1 miles, the distance of Cape Cod from the ship; CE the distance of the two places: 136 miles; ACB=57° 36′, the course made good; EAD=16° 34', the course to Cape Cod; and ECF the course from Mount-Desert rock to Cape Cod 38° 8', &c. BY LOGARITHMS. To find the bearing and distance of the two places by Case VI. Plane Sailing. To find the bearing. Whence the course from Mount-Desert rock to Cape Cod is S. 38° 8 W., distance 136 miles. The same may be found by the scale, or by inspection. Subtract the difference of latitude made good by the ship, 40.8 miles, from the whole difference of latitude, 107 miles, and there remain 66.2 miles, which is the difference of latitude between the ship and Cape Cod. In the same manner, by subtracting the ship's departure, 64.3 miles, from the whole departure, 84 miles, there remain 19.7 miles for the departure between the ship and Cape Cod. With this difference of latitude, 66.2, and departure, 19.7, the bearing of Cape Cod is found, by Case VI. Plane Sailing, S. 16° 34' W., and its distance, 69.1 miles. All the preceding calculations may be made by logarithms, by the scale, or by inspection. But we shall leave them to exercise the learner, and for the same purpose shall add the following example. EXAMPLE HII. A ship in the latitude of 37° 10′ N., is bound to a port in the latitude of 33° 0′ N, which lies 180 miles west of the meridian of the ship; but by reason of contrary winds, she sails the following courses, viz. S. W. by W. 27 miles, W. S. W. W. 30 miles, W. by S. 25 miles, W. by N. 18 miles, S. S. E. 32 miles, S. S. E. E. 27 miles,. S. by E. 25 miles, S. 31 miles, and S. S. E. 39 miles. Required the latitude the ship is in, and her departure from the meridian, with the course and distance to her intended port. The difference of latitude and departure made on each course, are given in the adjoined traverse table; hence it appears that the difference of latitude made good is 169.4 miles; the departure, 47.4 miles; and by Case VI. Plane Sailing, the course S. 15° 38′ W., and distance, 175.9 miles; and the course to the intended port, S. 58° 42′ W., distance 155,2 miles; the latitude being in 34° 21' N. TRAVERSE TABLE. Diff. of Lat. Departure. Instead of putting the course S. S. E. 32 miles, and S. S. E. 39 miles, you might make one entry PARALLEL SAILING. P IN Plane Sailing, the earth is considered as an extended plane; but this supposition is very erroneous, because the earth is nearly of a spherical figure, in which the meridians all meet at the poles; consequently the distance of any two meridians measured on a parallel of latitude (which distance is called the meridian distance) decreases in proceeding from the equator to the poles. To illustrate this, let PB represent the semi-axis of the earth, B the centre, P the pole, PCA a quadrant of the meridian, AB the radius of the equator, and CD (parallel thereto) the radius of a parallel of latitude. Then it is evident that CD will be the cosine of AC, or the cosine of the latitude of the point C, to the radius AB; now, if the quadrantal arc PCA be supposed to revolve round the axis PB, the point A will describe the circumference of the equator, and C the circumference of a parallel of latitude; and the former circumference will be to the latter as AB to CD (as may easily be deduced from Art. 55, Geometry), that is, as radius to the cosine of the latitude, or the point C; hence it follows, that the length of any arc of the equator intercepted between two meridians, is to the length of a corresponding arc of any parallel intercepted between the same meridians, as radius is to the cosine of the latitude of that parallel. Hence we obtain the following theorems. THEOREM I. A B The circumference of the equator is to the circumference of any other parallel of latitude, as radius is to the cosine of that latitude. THEOREM II. As the length of a degree of the equator is to the meridian distance corresponding to a degree on any other parallel of latitude, so is radius to the cosine of that parallel of latitude. THEOREM III. As radius is to the cosine of any latitude, so are the miles of difference of longitude between two meridians (or their distance in miles upon the equator) to the distance of these two meridians on that parallel of latitude in miles. As the cosine of any THEOREM IV. latitude is to radius, so is the length of any arc on that parallel of latitude (intercepted between two meridians) in miles to the length of a similar arc on the equator, or miles of difference of longitude. THEOREM V. As the cosine of any latitude is to the cosine of any other latitude, so is the length of any are on the first parallel of latitude in miles, to the length of the same arc on the other in miles. By means of Theorem III. the following table was calculated, which shows the meridian distance corresponding to a degree of longitude in every latitude; and may be made to answer for any degree or minute by taking proportional parts. The following Table shows for every degree of latitude how many miles distant the two meridians are, whose difference of longitude is one degree. LAT. MILES. LAT. MILES. LAT. MILES. LAT. MILES. LAT. MILES. When a ship sails east or west on the surface of the earth supposed to be spherical, she describes a parallel of latitude, and this is called Parallel Sailing. In this case, the distance sailed (or departure) is equal to the distance between the meridians sailed from and arrived at in that parallel; and it is easy, by Theorem IV. (preceding) to find the difference of longitude from the distance, or the distance from the difference of longitude, as will appear plain by the following examples. CASE I. The difference of longitude between two places in the same parallel of latitude being given, to find the distance between them. Suppose a ship in the latitude of 49° 30', north or south, sails directly east or west, until her difference of longitude be 3° 30′; required the distance sailed. BY PROJECTION. Take the sine of 90° from the plane scale, and, with one foot of the compasses on P (fig. 1) as a centre, describe the arc EQ with the difference of longitude, 210 miles, in the compasses, and one foot in E, as a centre, describe an arc cutting EQ in Q; join PE, PQ. Take the sine of the complement of the latitude 40° 30′ in your compasses, and with one foot in P, as a centre, describe the arc FG, cutting PE, PQ, in F, G; then the length of the chord FG being measured on the same scale of equal parts, will be the departure 136.4 miles. Or this projection may be made in the following manner. Draw AD (fig. 2) of an indefinite length; make the angle DAC equal to the latitude 49° 30′, and AC equal to the difference E P of longitude 210 miles; draw CD perpendicular to AD; then will the line AD be the distance or departure required. |