6. A ship from the Lizard, in lat. 49° 58′ N., sails to the westward on a direct course, till she arrives in lat. 48° 11′ N., and finds she has made 87 miles of westing: required her true course, and distance run.

Ans. Course S. 39° 7′ W., or S.W. S. nearly, and distance 137.9 miles.

7. A ship from Ascension Island, in lat. 7° 56′ S., sails N.N.W. W. (true) 244 miles required her latitude in, and departure made good.

Ans. Lat. in 4° 27′ S., and departure 125 4 miles.

8. A ship from Cape St. Vincent, in lat. 37° 3′ N., sails between the south and west, till her difference of latitude is 69 miles, and her departure 215 miles required her true course, distance and latitude in.

Ans. Course S. 72° 12′ W., or W.S.W. W. nearly, distance 225.7 miles, and lat. in 35° 54' N.

9. A ship from the Lizard, in lat. 49° 58′ N., sails 456 miles to the westward, and then finds she is 360 miles to the southward of the Lizard : required her true course, departure, and latitude in.

Ans. Course S. 37° 52′ W., or S. W. S. nearly, departure 279.9 miles, and lat. in 43° 58′ N.

10. A ship from Cape Hatteras, in lat. 35° 15′ N., sails in the north-east quarter 226 miles, and then finds that she is 198 miles to the eastward of the Cape required her true course, and latitude in.

Ans. Course N. 61° 11′ E., or N.E. by E. E., and lat. in 37° 4′ N.

II. If a ship take her departure at 6h. p.m. from Cape Verd, in lat. 14° 45′ N., and sails W.S.W. W. (true) at the rate of 7 miles an hour until the next day at noon: what will be her distance run, departure, and latitude in?

Ans. Distance 126 miles, departure 120-6, and lat. in 14° 8' N.

12. A ship from lat. 55° 30′ N. sails S. 33° 45′ W. (true) during 20 hours, and then finds by observation she is in lat. 53° 17′ N.: required her rate of sailing per hour, and the departure she has made.

Ans. Departure 88.87 miles, distance run in 20 hours 160 miles, and therefore her rate per hour 8 miles.

13. Find the true course, supposing a ship to sail in the north-east quarter, until her departure is twice her difference of latitude.

Ans. Course N. 63° 26' 6" E.

TRAVERSE SAILING

When a ship, either from contrary winds or other causes, is obliged to sail on different courses, the irregular or zigzag track she makes is called a Traverse, or Compound Course; and the method of reducing these courses and distances into a resultant course and distance is called resolving a traverse.

To resolve a Traverse, make a Table (as in Example I.), and divide it into six columns; in the first of these set down the various courses, and opposite to them, in the second column, the corresponding distances: the third and fourth columns are to be marked, one N. and the other S. at the top, and are to contain the differences of latitude; the fifth and sixth are to be marked E. and W., to contain the departures.

The difference of latitude and departure corresponding to each course and distance can be found by plane sailing, as already indicated, but this is needless.

The most common method of resolving the traverse is by inspection, using the Traverse Table in Norie's Tables, to which you must now turn; set the difference of latitude and departure down opposite the distance in their proper columns, observing that the difference of latitude must be placed in the N. column if the course be northerly, and in the S. column if the course be southerly; and that the departure must be placed in the E. column if the course be easterly, and in the W. column if it be westerly. When the course is due north, south, east, or west, set down the whole distance in the column answering to its proper heading. This done, add up the columns of northing, southing, easting, and westing, and set down the sum of each at the bottom; then the difference between the sums of the north and south columns will be the difference of latitude made good, of the same name as the greater; and the difference between the sums of the east and west columns will be the departure made good, also of the same name as the greater.

With this difference of latitude and departure made good, find the direct course and distance made good, as in plane sailing; or again turn to the Traverse Tables as shown in the sequel.

Example I.-Suppose a ship from Start Point, in lat. 50° 13′ N.. sail the following true courses and distances :-W.S.W. 51 miles, W. by N. 35 miles, S. by E. 45 miles, S.W. by W. 55 miles, and S.S.E. 41 miles required her direct true course and distance sailed, and her latitude in.

BY CONSTRUCTION

With the chord of 60° describe the circle NES W, to represent the compass; draw the diameters N S and E W at right angles, the one representing the meridian, and the other the parallel the ship sailed from: take the courses from the line of rhumbs, and lay them off from the meridian in their respective quarters, and number them in order, 1, 2, 3, 4, etc.; thus, from S. to I lay off 6 points for the first course W.S.W.; from N. to 2 lay off 7 points for the second course W. by N.; from S. to 3 lay off one point

for the third course S. by E.; from S. to 4 lay off 5 points for the fourth course S.W. by W.; from S. to 5 lay off 2 points for the fifth course S.S.E.; and from the centre of the circle draw rhumb lines to each of these points, which may be produced to any length that is necessary. Upon the first rhumb C 1, lay off the first distance 51 miles, from C to A; then will A represent the ship's place at the end of the first course; through A draw A B

parallel to the second course C 2 (Prob. IV. Geom.), and make it equal to the second distance 35 miles; through B draw B C parallel to C 3, and equal to 45 miles; through C draw C D parallel to C 4, and equal to 55 miles; and through D draw D E parallel to C 5, and equal to 41 miles. Through E draw the line E F parallel to the east and west line W E, meeting N S produced to F, and join C E. Then will C F be the difference of latitude made good, measuring 125'; É F is the departure 102', C E the distance 162 miles, and the angle E C F the course 391° or 3 points.

Obs. For practice you can verify, by plane sailing, the results taken from Traverse Table, and set down in the above columns of difference of latitude and departure.

The Course and Distance made good may be calculated as already shown in plane sailing, and note that the course takes its name from the greater difference of latitude and departure columns.

by which the direct course made good is S. 39° 19′ W. or S.W. S.; and the distance 162 miles.

But the usual method is to determine the Course and Distance by Inspection.

By Inspection.-Seek in the Traverse Table till the difference of latitude 125.3, and departure 102.6, are found side by side, each in its proper column; the nearest to these will be 125.1 and 101.3, which give the course S. 39° W. at the top; and the distance 161 in distance column.

Ans. Course S. 39° 19′ W., distance 161 miles, lat. in 48° 8′ N.

Example II-A ship from the North Foreland, in lat. 51° 23' N., and bound to the Texel, which lies in lat. 53° 2′ N., and 115 miles to the eastward, sails the following true courses and distances :-N.E. 35 miles, E. by S. 25 miles, N.E. by E. E. 40 miles, North 21 miles, and N.W. by W. 30 miles; required her true course and distance made good, the latitude she is in, and the departure made; also the direct true course and distance from the ship to the Texel.

BY CONSTRUCTION

With the chord of 60° describe a circle, through the centre of which draw the north and south, or meridian line, N S, and at right angles to it the east and west line W E; lay off from the meridian, upon the circumference of the circle, the courses in their proper quarters, number them in order, and draw a rhumb line from the centre to each of the points; then, on the first rhumb line, which is N.E. or 4 points, lay off the distance 35 miles from the centre, and through the end of this first distance draw a line parallel to the second rhumb, E. by S. 7 points, on which lay off the second distance 25 miles; proceed thus till all the courses and distances are laid off, and through the end of the last distance draw the line A B parallel to the line W E, meeting S N produced to B; then will A represent the ship's place; CB will measure the difference of latitude made good 76-4; BA the departure 59'6; C A the distance run 97 miles; and the angle B C A the course steered, N. 38° E.

To find what course the ship must steer, and what distance she must run before she can arrive at the Texel, lay off from C to D 99 miles, the difference of latitude between the Foreland and the Texel, and through D draw D F parallel to W E, and equal to the departure 115; then will F

represent the situation of the Texel: through A draw A E parallel to C D, and join A F; then A E will be the northing, or difference of latitude 23', and E F the easting, or departure 55'4, the ship has to make good; A F the distance the ship has to sail 60m. ; and the angle E A F, measuring 671°, the course from the ship to the Texel.