the latitude column will be found the departure. In this case the nearest number to 24.7 is 25; the corresponding departure, as found in the latitude column, is therefore 18.6. We now have both the difference of latitude and the departure between the two places, and may, consequently, find the course and distance by seeking in the Traverse Tables for a place where the difference of latitude (45.3) and the departure (18.6) stand beside each other in their respective columns. This is found on the page for 22°, the nearest coinciding numbers being 45.4 and 18.4. The course from Port Stanley to Ashtabula Harbor is therefore S 22° E, because the difference of latitude was south and the difference of longitude east; the required distance is found in the distance column opposite the two numbers; in this case it is 49 miles. Ans. EXAMPLE 2.- Required the course from Grand Marais Harbor to Battle Island, the former being situated in latitude 46° 41' N and longitude 85° 58' W, the latter in latitude 48° 45.5' N and longitude 87° 33' W. Find, also, the distance between the two places. SOLUTION. - Proceed precisely as in the solution of the preceding example. Find first the difference of latitude and longitude and the middle latitude. Thus, Lat. Grand Marais Harbor 46° 41' N Long. = 85° 58' W Lat. Battle Island 18° 45.5' N Long. = 87° 33' W Or = 95/ W 1 sum = M. Lat. 47° 43.2 Then enter the Traverse Tables with the degree nearest the middle latitude (48°) and the difference of longitude (95') in the distance column. In this case, as the middle latitude exceeds 45°, the table is entered from the bottom; the corresponding departure, as found in the latitude column, is 63.6. Then seek for the place in the tables where the difference of latitude (124.5) and the departure (63.6) stand beside each other in their respective columns. The nearest numbers found are 124.7 and 63.6 on the 27° page, and the distance corresponding to it is 140. Hence, the course from Grand Marais Harbor to Battle Island is N 27° W, and the distance 140 mi. Ans. NOTE. – The correctness of the results obtained in the preceding examples may be verified on any Hydrographic Office chart or charts of the Great Lakes that the student may possess. It is recommended that the student obtain charts embracing Lakes Erie, Huron, Michigan, and Superior, and verify all problems given in connection with the sailings. By practising in this manner a high degree of efficiency in the use of charts will be attained. The charts referred to can be purchased at low cost from the Hydrographic Office, Navy Department, Washington, D. C., or from any of the Branch Offices in lake ports. WORKING A TRAVERSE 60. In the examples of sailing hitherto introduced, the student will notice that only a single course has been considered. But since a vessel may change her course several times during a trip, or in the interval of a few hours, it becomes necessary to have a method, or process, by which the several courses and distances run for a certain interval of time may be reduced to the single course and distance that the vessel would have made had she gone directly from the place left to the place arrived at. This method, or process, called working a traverse, is explained in the articles that follow. 61. Explanation. -Suppose a ship to have sailed from A to E, Fig. 24, according to the courses and distances shown in the figure, that is, from A to B, from B to C, from C to D, and from D to E; N S and W E representing, respectively, the direction of the meridians and parallels. Then, by laying out the difference of latitude and the departure for each course, a number of small right triangles are formed, and by dropping a perpendicular from A to the parallel KE, a larger right triangle A K E is formed, in which the side A Krepresents the total difference of latitude and KE the total departure corresponding to the total distance A E and the course K A E made good by the ship. Now, by examining the figure it will be seen that the total difference of latitude A K and the total departure K’ E are, respectively, the algebraic sum of the several differences of latitude and departures made on each of the separate courses run by the ship. To prove this, let easterly departures and northerly differences of latitude be denoted by +, and westerly departures and southerly differences of latitude be denoted by -. We then have: - A a + C'C- n D + D m = and ta B – B C' + C n + m E = AK. Now, by substituting the numerical values of the different KE, departures and differences of latitudes, according to the scale attached, and taking their algebraic sum, we get the whole departure, KE= -16+ 41 – 172 + 251 = 322; and the whole difference of latitude, A K = +27. – 159 + 24 + 20 = 561 miles. The accuracy of this may be verified by actual measurements on the figure. 164 276 244 564 m 25 20 N K 324 16 FIG. 24 62. Constructing a Traverse. – From the foregoing it is evident that in order to find the total departure and difference of latitude made good by a ship that has sailed on several courses, it is necessary to find the departure and difference of latitude corresponding to each separate course and distance. This is conveniently found by the Traverse Tables, and the several departures and differences of latitude thus found are then entered into a table called a traverse, or traverse table, consisting of four columns headed N, S, E, and W, respectively, similar to the one shown below: The figures in each column are then added together, when the difference of the sums of the N and Scolumns will be the total difference of latitude made good, and is given the same name as the greater sum; likewise will the difference between the sum of the E and W columns be the total departure made good, to be named the same as the greater sum. From this total departure and difference of latitude is then found the course and distance made good according to plane sailing. 57 63. Illustration. – A ship has steamed the following courses and distances: S W by S, 24 miles; NNW, miles; S E by E) E, 84 miles; and south, 35 miles. To find the course and distance made good, proceed as follows: Construct a traverse similar to the one shown; the first column to contain the courses, the second the distances corresponding to each course; the third and fourth the differences of latitude; the fifth and sixth the departures. The table having been constructed, enter the Traverse Tables and take out the difference of latitude and departure for each course and distance run, and insert them in their proper columns. Thus, when the course is southerly the difference of latitude must be entered in the column headed S; when northerly, in the column headed N; the departure, when easterly, must be entered in the column headed E; and when westerly, in the column headed W. For the beginner, a good idea is to make a horizontal dash in all places where no figures are to be entered, as shown in the table. All the data having been properly inserted in its respective columns, the table will appear thus: TRAVERSE TABLE D. Lat. nade good = 41.9 S 39.0 E = Dep. made good. Then add the difference of latitude in each column and subtract the lesser sum from the greater, the remainder will be the total difference of latitude, with the same name as the greater sum. Proceed similarly with the departures. The result by the traverse in this example shows that the whole difference of latitude made good is 41.9 miles south and the departure 39 miles east. From these data the course and distance made good are found by seeking in the Traverse Tables for the place where they stand beside each other in their respective column. This occurs on the page of 43° where 41.7 and 38.9 are the nearest. Hence, the course made good is S 43° E, or nearly SES, and the distance made good is 57 miles. 64. Solution by Construction. The problem may also be solved by construction in the following manner: On a sheet of paper draw two lines NS and WE, Fig. 25, to represent the direction of the cardinal points. From the center |