The Course and Distance made good, if desirable to calculate it by plane sailing, will be, course N. 37° 58′ E. and distance 96.9 miles.

By Inspection. The difference of latitude 76-4, and departure 59.6, being sought in Traverse Table in their respective columns, give the course N. 38° E., and distance 97 miles.

To find the direct Course and Distance from the Ship to the Texel

Latitude of N. Foreland 51° 23′ N.
Diff. lat. made good, 76m.=1 16 N.

Latitude in
Latitude of Texel

Whole departure

115m. E.

....

52 39 N. 53 2 N. 0 23 N.

Departure made good
Departure to make good

59.6 E.

55°4 E.

Diff. lat. to make good

By calculation, plane sailing, with difference of latitude 23 N. and departure 55'4 E. the direct course from the ship's position to the Texel is N. 67° 27′ E. and distance 59.98 miles.

By Inspection. The difference of latitude 23', and departure 55' 4, being sought in the Traverse Table, the nearest numbers answering to them will be over the course N. 6 pts. E., and opposite the distance 60 miles.

Ans. True course N. 37° 58′ E.; distance 96-9 miles; lat. in 52° 39′ N.; departure 59.6 miles. True course to Texel N. 67° 27′ E.; distance 59.98 miles.

Examples for Practice

NOTE. The courses in these Examples are all supposed to be true courses, except in Example 8.

1. A ship from the Lizard, in lat. 49° 58′ N., sails as follows:-S. by W. 42 miles, W.S.W. 36 miles, West 18 miles, E.S.E. 22 miles, South 34 miles, and N.E. 21 miles. Required the latitude arrived at, and the course and distance made good.

Ans. Lat. in 48° 35′ N. ; the course made good S. 16° 27′ W. or S. by W. W. nearly; and the distance 86.13 miles.

2. A ship from lat. 9° 26′ N. sails N.E. 20 miles, North 33 miles, N.N.W. 15 miles, East 25 miles, N.E. by N. 42 miles, and S. W. W. 28 miles. Required her course and distance made good, and the latitude in.

Ans. Course N. 24° 12′ E., or N.N.E. E. nearly; distance 85.63 miles; and lat. in 10° 44′ N.

3. A ship from the Cape of Good Hope, in lat. 34° 22′ S., sails S.W. S. 27 miles, S.E. by E. 45 miles, S. W. by S. 48 miles, West 23 miles, and S.S.W. W. 18 miles. Required her course and distance made good, and her latitude in.

Ans. Course S. 24° 45′ W., or S.S.W. W. nearly; distance 112 miles; and lat. in 36° 4′ S.

4. A ship from lat. 1° 12' S. sails E. by N. N. 56 miles, N. E. 80 miles, S. by E. E. 96 miles, N. E. 68 miles, E.S.E. 40 miles, N.N.W. W. 86 miles, and E. by S. 65 miles. Required her course, distance, and latitude in.

Ans. Course N. 51° 47′ E., or N.E. E. nearly; distance 193.8 miles, and lat. in o° 48′ N.

5. A ship from lat. 46° 18′ N. sails N. 25° W. 50 miles, N. 74° E. 64 miles, S. 52° W. 36 miles, N. 35° E. 40 miles, N. 69° W. 75 miles, and S. 47° E. 48 miles. Required her course, distance, and latitude in.

Ans. Course North; distance 67.7 miles; and lat. in 47° 26' N.

6. A ship from lat. 51° 30′ N., running at the rate of 8 knots an hour, sails W.S.W. 3 hours, N.W. 2 hours, West 4 hours, S.W. by S. 24 hours, and N.W. W. 2 hours. Required her course, distance, and latitude in.

Ans. Course West; distance 90.7 miles; and lat. in 51° 30′ N.

7. A ship from a port in lat. 38° 42′ N., bound to another port situated in lat. 36° 32′ N., and 137 miles to the eastward, sails the following courses: S. by W. W. 55 miles, S.W. S. 37 miles, South 60 miles, E.S.E. 40 miles, S.E. S. 32 miles, and N.E. by E. E. 58 miles. Required her course and distance made good, and the latitude in; also the course and distance to her intended port.

Ans. The course made good is S. 23° 38′ E., and the distance 169 miles; the lat. in 36° 7′ N.; the course to the intended port, N. 70° 8′ E., and the distance 73.56 miles.

8. The course (correct magnetic) from Beachy Head to Selsea Bill is N. 67° W. and distance 40 miles; from Selsea Bill to St. Catherine's Point N. 86° W. 21 miles; from St. Catherine's Point to Portland Lights N. 69° W. 44 miles; from Portland Lights to the Start N. 85° W. 49 miles. Required the correct magnetic course and distance from Beachy Head to the Start.

Ans. The course is N. 75° 51′ W. or W.N.W. & W., and distance 152.2 miles.

9. Suppose a ship to sail upon the following courses and distances :—S.E. by S. 29 miles; N.N.E. 10 miles; E.S.E. 50 miles; E.N.E. 50 miles; S.S.E. 10 miles; N.E. by N. 29 miles; West 25 miles; S.S.E. 10 miles; W.S.W. W. 42 miles; North 110 miles; E. N. 62 miles; North 7 miles; West 62 miles; North 10 miles; West 8 miles; South 10 miles; West 62 miles; South 7 miles; E. S. 62 miles; South 110 miles; W.N.W. W. 42 miles; N.N.E. 10 miles; and West 25 miles. Required her course and distance made good.

Ans. The ship has returned to the place she sailed from.

NOTE.

This example is taken from "Robertson's Elements of Navigation," the figure forming the outline of an anchor

10. A ship in lat. 1° o' S. is bound to a port in lat. 1° 10' N., distant 220 miles in the north-west quarter, but meeting with contrary winds, runs the following courses:-N.E. by N. 63 miles, N.W. W. 85 miles, North 96 miles, and N.N.W. 87 miles. Required the ship's place; and the course and distance to her port.

Ans. The lat. in is 3° 46′ N.; the course to her port S. 36° 42′ W. or S.W. S., and distance 194.4 miles.

II. Two ships (A) and (B) part company in lat. 1° 44′ S., and meet again at the end of two days, having run as follows:

(A) N.N.E. 96 miles, W.S.W. 96 miles, E.S.E. 96 miles, and N.N.W. 96 miles.

(B) N.N.W. 96 miles, E.S.E. 96 miles, W.S.W. 96 miles, and N.N.E. 96 miles. Required the latitude arrived at, with the direct course and distance of each ship.

Ans. There being no departure made good, and the difference of latitude being N., therefore the course made good is North, and the distance is 104 miles; also, the ships meet on the equator.

PARALLEL SAILING

In Plane Sailing it was observed that the meridians are considered as being all parallel to each other, and the length of a degree on the meridian and parallel everywhere equal, on the supposition of the earth, or at least any small part of it, being a plane: but as the earth is a sphere, or globe, and the meridians meet at the poles, it is evident that the distance between any two meridians must vary in every latitude, their greatest distance apart being at the equator, on which the difference of longitude is measured.

PARALLEL SAILING is the method of finding the distance between two places in the same latitude, when their difference of longitude is known; or, otherwise, of finding the difference of longitude corresponding to the meridian distance or departure made good, when a ship sails due east or west.

Difference oF LONGITUDE between any two places is the arc of the equator contained between the meridians passing through the two places.

The MERIDIAN DISTANCE between two places is the arc of their parallel of latitude contained between the meridians of the places. It is also the departure made, or the distance run, on the parallel.

The Equator is the great circle on which longitude is reckoned; it is also the origin whence latitude is measured, polarwise, through successive parallels of latitude. When a ship sails 60 nautical miles on the equator she changes her position by 60' (or 1°) of longitude; but since the parallels of latitude are small circles which successively decrease in circumference as they are nearer to the poles, it follows that the farther the parallel is from the equator, the smaller must be the number of nautical miles measuring the circumference, while, at the same time, that circumference ranges through 360° of longitude, as does the equator. As a specific instance in point, 360° of longitude are represented on the equator by 21,600 nautical miles; but the same number of degrees of longitude, on the parallel of 60° N. or S., are represented by only 10,800 nautical miles; thus the circumference of the earth on the parallel of 60° is only half what it is on the equator, and consequently half a nautical mile on that parallel is the equivalent of 1' of longitude. It is by the methods of parallel sailing that the relation between the meridian distance and the difference of longitude between two places on the same parallel is determined.

A

R

Parallel sailing is particularly useful in making small or low islands, in which case it is usual to run into the latitude, and then steer due east or west. The principles upon which parallel sailing depend may be thus illustrated. Let PAEC represent a section of one-fourth part of the earth, C being the centre, and P the pole; then P A E will be part of a meridian; PC the polar, and E C the equatorial radius; also let P B O represent part of another meridian, A and B two places in same parallel, being equally distant from the equator E OQ; then will A B be the meridian distance between A and B, and E O their difference of longitude; join A C; then the arcs E A or O B or the angle E C A will be the latitude of A or B ; and A P or B P their co-latitude. Now EOQ C is the plane of the equator, and A BSR is a plane parallel to the equator; there

fore A R is parallel to E C, and B R parallel to O C, and the angles E CO and AR B are equal; and as circles and similar arcs of circles are in direct ratio to their radii, therefore—

But arc A B is the departure or meridian distance; arc E O is the difference of longitude; and angle A CE is the latitude; therefore

by one or other of which formulæ all cases in parallel sailing may be computed.

By Construction.—If a triangle, as A B C (see Fig. in Example), be so constructed that the hypotenuse A C may represent the difference of longitude in miles, the base A B the meridian distance, and the angle C A B (between the hypotenuse and base) equal to the latitude; then any two of these parts being given, the other may be found by the ordinary solution of a right-angled triangle.

Given the Difference of Longitude between two Places, both on the same Parallel of Latitude, to find their Distance apart

Example. A ship in lat. 36° 58′ N., and long. 20° 25′ W., is bound to St. Mary's, one of the Western Islands, in the same latitude, and in long. 25° 13′ W.; what distance must she run to arrive at the island?