placed that this line shall be parallel to the keel of the ship. Consequently, if b be placed towards the bow of the vessel, the point which it marks on the card will show the compass course, for the line NS is always on the magnetic meridian, and EW is east and west. The course is generally read to quarter points, and as a quadrant contains eight points, each point is equal to 90° 811° 15'; and a quarter point = 11° 15'÷ 4 = 2° 48′ 45′′. The table of Rhumbs, after the Traverse Table, shows the degrees in each course, to quarter points. 13. A ship's rate of sailing is determined by means of an instrument, called the log, and an attached line called the log line. The log is a piece of wood in the form of a sector of a circle, the rim of which is loaded with lead, so that when it is heaved into the sea it assumes a vertical position. The log line is so attached as to hold the log square against the water, that it may not be drawn along after the ship as the line unwinds from the reel, by the ship's forward motion. The time in which the log line unwinds from the reel, is noted by a sand-glass, through which the sand passes in half a minute; that is, in the one hundred and twentieth part of an hour. For convenience, the log line is divided into equal parts, marked by knots, and each part is equal to the one hundred and twentieth part of a nautical or geographical mile*. Now, since half a minute is the one hundred and twen tieth part of an hour, and each knot indicates the one hun dred and twentieth part of a mile, it follows that the number of knots reeled off while the half minute glass runs out, will indicate the rate of the ship's sailing per hour. * A geographical mile is one minute, or one-sixtieth of a degree, measured on the equator. Taking the diameter at 7916 English miles, the geographical mile will be about 6079 feet; that is, one-sixth greater thar. the English mile, which SECTION II. OF PLANE SAILING. 14. Let the diagram EPQ represent a portion of the earth's surface, P the pole, and EQ the equator. Let AB be any rhumb line, or track described by a ship in sailing from A to B. B Conceive the path of the ship to be divided into very small parts, and through the points of division draw meridians, and also the parallels of latitude b'b, c'c, d'd, e'e, and B'B: a series of triangles will thus be formed, but so small that each may be considered as a plane triangle. In these triangles, the sum of the bases Ab' + bc' + cd' + de' + ef = AB', which is equal to the difference of latitude between the points A and B. Also, b'b + c'c + d'd + e'e+ƒB= BB', which is equal to the distance that the ship has departed from the meridian AB'P, and is called the departure in sailing from A to B. Therefore, the distance sailed, the dif ference of latitude made, and the departure, may be represented by the hypothenuse, the base and perpendicular of a rightangled triangle, of which the angle opposite the departure is the course. B' When any of the four parts abovenamed are given, the other two can be determined. This method of determining A the place of a ship reduces all the elements to the parts EXAMPLES. 1. A ship from latitude 47° 30' N. has sailed S. W. by S. 98 miles. What latitude is she in, and what departure has she made? Let be the place sailed from, CB the meridian, and BCA the course, which we find from the table of rhumbs to be equal to 33° 45'; then AC will be the distance sailed, equal to 98 miles. Also, AB will be the departure, and CB the difference of latitude. Then by the formulas for the solution of right angled triangles, = Dif. lat. 81.48 miles 81.48 minutes = 1° 22' S. In latitude 46° 08'. Departure, 54.45 miles. 2. A ship sails 24 hours on a direct course, from lattude 38° 32' N. till she arrives at latitude 36° 56' N. The course is between S. and E. and the rate 5 miles an hour. Required the course, distance, and departure. Lat. left 38° 32′ N. 24 × 5132 miles distance. Diff. 1° 36′ 96 miles. As dist. 132 ar. c. 7.879426 | As radius ar. c. 0.000000 : diff. lat. 96 :: radius 132 2.120574 1.982271: dist. 10.000000:: sin course 43° 20′ 9.836477 Hence, the course is S. 43° 20′ E., and the departure 90.58 miles east. 3. A ship sails from latitude 3° 52' S. to latitude 4° 30′ N., the course being N. W. by W. W.: required the dist ance and departure. Ans. Dist. 1065 miles; dep. 939.2 miles W. 4. Two points are under the same meridian, one in latitude 52° 30' N., the other in latitude 47° 10' N. A ship from the southern place sails due east, at the rate of 9 miles an hour, and two days after meets a sloop that had sailed from the other: required the sloop's direct course, and distance run. Ans. Course S. 53° 28′ E.; dist. 537.6 miles. 5. If a ship from latitude 48° 27′ S., sail S. W. by W. 7 miles an hour, in what time will she reach the parallel of 50° south? Ans. 23.914 hours. SECTION III. OF TRAVERSE SAILING. 15. When a ship, in going from one place to another, sails on different courses, it is called Traverse Sailing. The determination of the distance and course, from the place of departure to the place of termination, is called compounding or working the traverse. This is done by the aid of the "Traverse Table," which has already been explained, and the method of working the traverse, is in all respects similar to that adopted in the Prob. of Art. 34, page 123. EXAMPLES. 1. A ship from Cape Clear, in lat. 51° 25′ N., sails, 1st, S. S. E. E. 16 miles; 2d, E. S. E. 23 miles; 3d. S. W. by W. W. 36 miles; 4th, W. N. 12 miles; 5th, S. E. by E. E. 41 miles required the distance run, the direct : We first form the table below, in which we enter the courses, from the table of rhumbs, omitting the seconds, and then enter the latitudes and departures, taken from the tra verse table, to the nearest quarter degree. Thus, in taking the latitude and departure for 25° 18′ we take for 2510. The dif ference of latitudes gives the line AG, and the dif ference of departures the line GF Difference of latitude 59.66 miles = 1° 00′ S. |