TRAVERSE SAILING.] 



NAVIGATION. 



1055 



she is found by observation to be in latitude 18 49' S. : 

 required her distance run, and departure. 

 Ans. Distance, 274 milea. Departure, 212 miles E. 



5. A ship from latitude 34 23' S. sails between the 

 south and west till she reaches latitude 36 34' S. and 

 finds that she has made75 miles of departure. Required 

 her course and distance run. 



Ans. Course S. 29 47' W. Distance, 151 miles. 



6. A ship from latitude 3 16'N. sails S.W. by W.JW., 

 till she has made 35(5 miles of departure. Required the 

 distance sailed and latitude in. 



Ans. Distance, 415 miles. Lat. i, 17' S. 



7. A ship in latitude 3 52' S., is bound to a port 

 tearing N.W. by W. W., in latitude 4 30' N. How- 

 far does that port lie to the westward, and what is the 

 ship's distance from it 1 in other words, what departure 

 and distance must the ship make to reach it ? 



Ans. Departure, 939 miles W. Distance, 1065 miles. 



8. A ship, from latitude 42 18' N., sails S. 25 W. a 

 distance of 150 miles. Required her departure and lati- 

 tude in. 



Ans. Departure 634 miles W. Lat. in 40 % N. 



9. If a ship sail from latitude 48 27' S. on a S.W. by 

 W. course at the rate of 7 knots an hour, in how many 

 hours will she arrive at latitude 50 S. ? 



Ans. In 23j5 hours. 



10. A ship, from latitude 55 W N., sails S.W. by S. 

 for 20 hours, and then finds by observation that she is 

 in latitude 53 17' N. Required her hourly rate of sail- 

 ing, and the departure she has made. 



Ans. Rate, 8 miles an hour. Departure, 88 '87 miles W. 



11. A ship sails for 18 hours on a single course be- 

 tween the S. and \V. from latitude 38 32' N. to latitude 

 36 66' N. , at the rate of 7J miles an hour. Required the 

 course, distance, and departure. 



Ans. Course, S. 43 20' W. Distance, 132 miles. 

 Departure, 90-58 W. 



12. A ship sails for 53J hours on a S. E. f E. course 

 from latitude 52 W N. to latitude 47 10' N. Re- 

 quired the average rate of sailing per hour, and the 

 departure made. 



Ans. Rate, 10 miles an hour. Departure, 432 miles K 



COMPOUND COURSES, OR TRAVERSE SAILING. When, 

 from contrary winds or other causes, a ship's track 

 from one place to another is made up of several single 

 courses, the zig-zag path it takes is called a Traverse, or 

 a Compound Course ; and the determination of the single 

 course and distance, from the pb>ce left to the place 

 arrived at, is called working or resolving the traverse. 



To work a traverse, it is only necessary to find the 

 difference of latitude and departure for each distinct 

 course, as in the foregoing article ; to take the aggregate 

 of these for the whole difference of latitude and depar- 

 ture, and thence to find the corresponding single course 

 and distance. 



The most orderly way of proceeding is to form a little 

 traverse table, consisting of six columns, to receive the 

 proper entries for course, distance, diff. lat. N. and S., 

 and dep. E. and W., as in the specimen in Example 1 

 following. When the entries are completed, the two diff. 

 lat. columns are added up separately, and the difference 

 of the results taken : this difference is the whole diff. 

 lat., which is N. or S. according as the N. or S. column 

 gives the greater result. In like manner, the results of 

 the two departure columns being found, their difference 

 is the resultant du[>arture, to be used with the whole 

 difference of latitude, to determine the direct course and 

 distance. 



1. A ship from latitude 51 25' N. has sailed the fol- 

 lowing courses, namely 



1st, S.S.K. i E., 16 miles. 



2nd,E.S.E, 23 miles. 



3rd, S.W. I, W. W., 36 miles. 



4th, W. J N., 12 miles. 



5th, S.E. (>E iE., 41 miles. 



Required the latitude in, and the direct course and 

 distance to ruxch it. 



TRAVERSE TABLE. 



The first two columns of this table are occupied with 

 the given courses and distances ; in the other four are 

 inserted the diff. lat. and dep. corresponding to each 

 course and distance, taken by inspection from the Tra- 

 verse Table. The results of these latter columns show 

 that the difference of latitude made is 59 6 miles S., and 

 the departure 19 -6 miles E. And from these the lat. in, 

 and the direct course and distance from the lat. left to 

 the place reached, is found by computation as follows : 



Latitude left. . . .51 25' N. 



Diff. lat. 59 -6 miles . . 1 0' S. 



Latitude in . . . .60 25' N. 

 1. To find the direct course. 



By logarithms. 



As diff. lat. = 69-6 . . . -1-7752 

 : departure = 19 6 . . 1 -2923 

 : : radius 10 



tan. course, 18 12 7 . . 9 5171 



Without logarithms, 

 tan. course = dep. -Hdiff. lat. 

 5.9 -6) 19 -6( -3288 = tan. 18 12' 

 1788 



172 

 1192 



528 

 477 



51 

 48 



Hence the direct course is S. 18 12' E. 

 It thus appears that if the ship had left her port on 

 the course S. 18 12' E., and had kept this course un- 

 altered for a run of 63 miles (see next page), she would 

 have arrived at the place reached by the above traverses. 

 This conclusion, however, is not rigorously true, though 

 near enough for practice. (See the remarks subjoined to 

 Example 8). 



2. To find the distance. 



By logarithms. 

 As sin. course, 18 12 7 . . -9 '4946 



: radius . . . . .10 

 :: departure = 19 -6 . . . 1'2923 



: distance = 62-75 17977 



Without logarithms, 

 dist. = dep. -j- sin. course 

 sin. 18 12^ -3,1,2,3)19-6 (62 '75 

 18738 



Hence the nautical distance 

 is 62-75 miles. 



862 

 625 



237 

 219 



13 

 16 



