1. Lat. left 34° 4' S., long. left 12° 5′ E., dif. of lat. 145 miles S., dif. of long. 365 miles W. Ans. Lat. in 36° 29′ S., long. in 6° 0′ E. 2. Lat. left 20° 40′ N., long. left 178° 14′ W., dif. of lat. 216 miles S., dif. of long. 420 miles W. Ans. Lat. in 17° 4' N., long. in 174° 46' E. Navigation is divided into different branches, according to the methods of calculation that are employed. PLANE SAILING. 500. In plane sailing the surface of the earth is considered to be a plane, the meridians being equidistant lines, and the parallels of latitude also equidistant, cutting the meridians perpendicularly. This supposition, though incorrect, will lead to no error, so far as the nautical distance, difference of latitude, and departure, are concerned; for, as appears from the explanation following the example given below, these elements will be the same, whether they are lines drawn on a plane, or equal lines similarly related drawn on a sphere. As the north is on the upper side of the figure of the mariner's compass, and the upper side of maps, the top of a page is considered to be directed towards the north; and hence the upper parts of diagrams in navigation are considered to be the northern parts of the figure. Hence, a vertical line, BC, will denote the difference of latitude; a horizontal line, AB, the departure; and the oblique line or hypotenuse, AC, the nautical distance; and angle C the course, and A the CO-course. Hence 501. If any two of the four parts, namely, the nautical distance, departure, difference of latitude, and course, are Distance. Diff. of Latitude given, the other two can be found by the A Departure B rules of right-angled trigonometry." There will therefore be six cases, of which the first, however, is the most important. These cases may also be calculated very easily, and with sufficient accuracy, by means of the table of the difference of latitude and departure, or, as it is sometimes called, a traverse table; this method of solution is called inspection. They can also be solved by construction, as in the problems from article 145 to 149; or by means of logarithmic lines, as explained from article 180 to 183. 502. PROBLEM I.-Of the course, distance, difference of latitude, and departure, any two being given, to find the other two. EXAMPLE.-A ship, from a place in latitude 56° 14′ N., sails SW. W. 425 miles; required the latitude in and the departure. The proportions are the same as in the first case of rightangled trigonometry, only the nautical terms are used for the angles and sides of the triangle. Construction. Let BC be the meridian, and make angle C 4 points = 50° 37', and CA = 425, and draw AB perpendicular to BC. Then measure AB and BC. By Gunter's Logarithmic Lines. When the course is given in points, use sine rhumbs or tangent rhumbs instead of the lines of sines and tangents. To find the departure. The distance from radius, or 90° on the line S. Rhumb to 4 pts., will extend on the line of numbers from 425 to 328, the departure. = To find the difference of latitude.-The distance from 90° to co-course 31 pts. (as sine 31 pts. is cosine 4 pts.) on the line S. Rhumb, will extend on the line of numbers from 425 to 270, the difference of latitude. By Inspection. In the traverse table in the page containing the course 4 points, and opposite to the distance 42.5, is the departure 32.8 and the difference of latitude 26-9; and these being multiplied by 10, as the distance is 425, give 328 and 269. The distance 42.5 is taken, because the distance in the table does not exceed 300. Let AC, BD, be the parallels of the latitude left and reached, BC, DA, their meridians, and AGB their nautical distance, which therefore is at every point equally inclined to the meridian. Let the distance AB be divided into a great number of minute equal parts AG, GH,...and let Gg, Hh,...be portions of parallels of latitude, and Ag, Gh,... portions of meridians passing through the points A, G, H. Then, since these parts differ insensibly from straight lines, and the angles GAg, HGh,...are equal, therefore the parts AG, portional to Ag, Gh, and hence AG: GH+ : Ag + Gh + = ... ... rad: cos course; hence, or as AB: AD. AB: AD rad: cosine course. F E h H G A GH,...are pro- = It is similarly shown that AG: Gg AG + GH + ... Hh + ... distance: departure; and hence, : Gg rad: sin course distance: departure. 503. The distance, difference of latitude, departure, and course, are therefore related as the sides and angles of a plane right-angled triangle, and their various relations are therefore determinable in the same manner as those of the sides and angles of the triangle. The following exercises, which illustrate the six cases, are to be performed by construction, calculation, and logarithmic lines, and by inspection. EXERCISES. 1. A ship, from a place in latitude 49° 57′ N., sails SW. W. 244 miles; required the departure and latitude in ? Ans. Departure 203, latitude 47° 42′ N. 2. A ship sails SE. 6 E. from a place in 1° 45′ north latitude, and is then found by observation to be in 2° 46′ south latitude; required the departure and distance. Ans. Departure 405-6, distance 487·8. 4 3. A ship sails NE. 6 E. E. from a port in latitude 3° 15' S., till its departure is 406 miles; what is the distance sailed and the latitude in? Ans. Distance 449, latitude 0° 3′ S. 4. A ship sails between the south and east 488 miles from a port in latitude 2° 52′ S., and then by observation it is found to be in latitude 7° 23' S.; what course has she steered, and what departure has she made? Ans. The course 56° 16' or SE. 6 E., departure 405.8. 5. A ship has sailed between the north and west from the island of Bermudas, in latitude 32° 25′ N., till her distance is 488 miles and departure 405 miles; what has been her course, and what is the latitude in ? Ans. The course N. 56° 6'′ W., latitude in 36° 57′ N. 6. A ship sails between the north and west till its difference of latitude is 271 miles, and departure 406 miles ; what is the course and distance sailed? Ans. Course N. 56° 17′ W. or NW. ¿ W., and distance 488. 504. PROBLEM II.-Given several successive courses and distances sailed by a ship between two places, to find the single course and distance, that is, the equivalent course and distance, between the place left and that arrived at. 'Find the difference of latitude and departure for each rhumb and distance, and then the whole difference of latitude and departure, and the course and distance corresponding to these two elements." The difference of latitude and departure for each course and distance are to be found by the last problem, the method by the traverse tables being the most expeditious; then these are arranged in a table called a traverse table, the courses being in the first column, the distances in the second, the north and south differences of latitude, marked N. and S., in the third and fourth, and the east and west departure, marked E. and W., in the fifth and sixth columns. The difference between the sums of columns N. and S., or of the northings and southings, will be the whole difference of latitude of the same name as the greater; and the difference between the sums of the columns E. and W., or of the eastings and westings, will be the whole departure of the same name as the greater." This method is usually called traverse sailing, and is used to compute the result of a whole day's sailing, called a day's work; for, from contrary winds and other causes, a ship is generally obliged to sail on several tacks in the course of a day. Having found the whole difference of latitude and departure for the day's work, the course and distance corresponding to them, that is, the equivalent course and distance, are then to be computed. The equivalent course and distance are also called the course and distance made good. EXAMPLE. A ship from Cape Clear, lat. 51° 25′ N., sails S.6 W. 20 miles, SE. 12 miles, SW.6S. 18 miles, WNW.N. 14 miles, and SSW. 24 miles; required the equivalent course and distance, and the latitude in. |