13. 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.

Of Plane Sailing.

Conceive the path of the ship to be divided

B

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+cc+d'd+ee+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 difference of latitude made, and the departure, are correctly represented by the hypothenuse and sides of a right angled triangle, of which the angle opposite the departure is the course.

When any two of the four things above named are given, the other two can be determined. This method of determining the place of a ship reduces all the elements to the parts of a plane triangle, and hence is called plane sailing.

B'

EXAMPLES.

1. A ship from latitude 47° 30′ N. has sailed S. W. by S. 98 miles.

Let C 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,

2. A ship sails 24 hours on a direct course, from latitude 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.

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 distance and departure. Ans. Dist. 1065 miles; dep. 938.9 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 sloops 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°

Of Traverse Sailing.

14. 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 is in all respects similar to that adopted in the Prob. of Art. 147, p. 115.

1. A ship from Cape Clear, in lat. 51° 25′ N., sails, 1st, S. S. E. E. 16 miles; 2nd, E. S. E. 23

miles; 3rd, S. W. by W. W. 36 miles; 4th, W. N. 12 miles; 5th, S. E. by E.

EXAMPLES.

E. 41 miles: required the distance run, the direct course, and the latitude.

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 traverse table, to the nearest quarter degree. Thus, in taking the latitude and departure for 25° 18' we take for 251°. The difference of latitudes gives the line AG, and the difference of departures the line GF.

E

D

Traverse Table.

Latitude left

51° 25' N.

=

Difference of latitude 59.66 miles 1° 00' S.

Then, by formulas for the solution of right angled triangles, we have,

19.64 1.293141

As AG, diff. lat. 59.66 1.775683 | As sin. course 18° 13' 9.495005 : radius, 10.000000 : departure 19.64 1.293141 :: radius

:: departure

: tangt. course 18° 13' 9.517458

distance

10.000000

62.83 1.798136

Therefore the direct course is S. 18° 13′ E., and the distance 62.83 miles.

Of Plotting.

15. There is yet another method of finding the direct course and distance, much practiced by seamen, although it does not afford a high degree of accuracy. It is a method by plotting, which requires the use of a mariner's scale and a pair of dividers.

One of the scales marked on the mariner's scale, is a scale of chords, commonly called a scale of rhumbs, being divided to every quarter point of the compass; and there is also a second scale of chords divided to degrees. Both of these scales are constructed in reference to the same common radius, so that the chords on the scale of rhumbs correspond to those on the scale of marked chords. The manner of using the scales will appear in plotting the last example.

To construct this traverse, describe a circle with a radius equal to the chord of 60° and draw the meridian NS. Then take from the line of rhumbs the chord of the first course 21 points, and apply it from S to 1, to the right of NS, since the course is southeasterly, and draw S1; take, in like manner, the chord of the second course, 6 points, from S to 2, and lay it off also to the right of the meridian line. Apply the chord of the third course, 5 points, from S to 3, to the left of the meridian; the fourth course, 7 points from N to 4, to the left of NS, this course being northwesterly; and, lastly, apply the chord of the fifth course, 5 points, from S to 5, to the right of NS, and join all the lines as in the figure.

In the direction A1, lay off the distance AH=16 miles from a scale of equal parts, and through the extremity H, draw HC parallel to 42, and lay off HC=23 miles. Draw CD parallel to 43, and lay off CD=36 miles; then draw DE parallel to A4, and lay off 12 miles; and lastly draw EF parallel to 45, and lay off 41 miles, and F will be the place of the ship. Hence, we conclude that AF will be the distance made good, and GAF

Applying, then, the distance AF to the scale of equal parts, we find it equal to 623 miles; and applying the chord Sa to the scale of chords we find the course GAF=181°.

2. A ship sails from a place in latitude 24° 32′ N., and runs the following courses and distances, viz. 1st, S. W. by W. dist. 45 miles; 2nd, E. S. E. dist. 50 miles; 3rd, S. W. dist. 30 miles; 4th, S. E. by E. dist. 60 miles; 5th, S. W. by S. W. dist. 63 miles required her latitude, and the direct course and distance from the place left to the place arrived at, and the construction of the traverse.

3. A ship from lat. 28° 32′ N. has run the following courses, viz. 1st, N. W. by N. 20 miles; 2nd, S. W. 40 miles; 3rd, N. E. by E. 60 miles; 4th, S. E. 55 miles; 5th, W. by S. 41 miles; 6th, E. N. E. 66 miles: required her latitude, the distance made good, and the direct course, also the construction of the traverse. Ans. Dist. 70.2 miles, course E.

4. A ship from lat. 41° 12′ N. sails S. W. by W. 21 miles; S. W. S. 31 miles; W. S. W. S. 16 miles; S. 3 E. 18 miles; S. W. W. 14 miles; then W. N. 30 miles: required the latitude, the direct course, and the distance.

Ans.

Lat. 40° 05', course S. 52° 49′ W.
Dist. 111.7 miles.

5. A ship runs the following courses, viz.

1st, S. E. 40 miles; 2d, N. E. 28 miles; 3d, S. W. by W. 52 miles; 4th, N. W. by W. 30 miles; 5th, S. S. E. 36 miles; 6th, S. E. by E. 58 miles: required the direct course, and distance made good.

Ans. {

4

Direct course S. 25° 59′ E., or S. S. E. E., nearly.
Distance 95.87 miles.

6. A ship sails, 1st, N. W. by W. W. 40 miles; 2nd, N. W. by N., 41 miles; 3rd, N. by E. 16.1 miles; and 4th, N. E. † E. 32.5 miles: required the distance made, and the direct course.

Ans. Course 21° 54' West of North. Dist. 94.6 miles. These examples will, perhaps, suffice to illustrate the principles of plane sailing.

The longitude, made on any course, cannot be determined by these methods, for this being the arc of the equator intercepted between two meridians, cannot be found under the supposition that the meridians are parallel.

The most simple case of finding the difference of longitude is when the ship sails due east or due west: this is called Parallel