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(193.) The ship's rate of sailing is measured by a log-line. The log-line is a cord about 300 yards long, which is wound

round a reel, one end being attached to a piece of thin board called a log. This board is in the form of a sector of a circle, the arc of which is loaded with lead sufficient to give the board a vertical position when thrown upon the water. This is designed to prevent the log from being drawn along after the vessel while the line is running off the reel.

The time is measured by a sand-glass, through which the sand passes in half a minute, or the 120th part of an hour.

The log-line is divided into equal parts called knots, each of which is 50 feet, or the 120th part of a nautical mile. Now, since a knot has the same ratio to a nautical mile that half a minute has to an hour, it follows, that if the motion of a ship is uniform, she sails as many miles in an hour

as she does knots in half a minute. If, then, seven knots are observed to run off in half a minute, the ship is sailing at the rate of seven miles an hour.

PLANE SAILING.

(194.) Plane sailing is the method of calculating a ship's place at sea by means of the properties of a plane triangle. The particulars which are given or required are four, viz., the

distance, course, difference of latitude, and departure. Of these, any two being given, the others may be found.

Let the figure EPQ represent a portion of the earth's sur

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that the curvature of the earth may be neglected. Through the points of division draw the meridians Pb, Pc, &c., and the parallels eb, fc, &c. Then, since the course is every where the same, each of the angles eAb, fbc, &c., is equal to the course. The distances Ae, bf, &c., are the differences of latitude of A and b, b and c, &c. Also, eb, fc, &c., are the departures for the same distances. Hence the difference of latitude from A to B is equal to

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C __Departure

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Construct the triangle A'B'C' so that A'b'e' shall be equal to Abe, b'c'f' shall be equal to e bef, c'd'g' equal to cdg, and d'B'h' equal to dBh. Then A'B' represents the distance sailed, B'A'C' the course, A'C' the difference of latitude, and B'C the departure; that is, the distance, difference of latitude, and departure are correctly represented by the hypothenuse and sides of a right-angled triangle, of which the angle opposite to the departure is the Of these four quantities, any two being given, the others may be found. Plane sailing does not assume the earth's surface to be a plane, and does not involve any error even in great distances.

course.

Diff. Lat.

Course

Distance

A

EXAMPLES.

1. A ship sails from Vera Cruz N.E. by N. 74 miles. Required her departure and difference of latitude.

According to the principles of right-angled triangles, Art. 44. Radius distance :: sin. course: departure.

:: cos. course: diff. latitude.

The course is three points, or 33° 45'; hence we obtain

Departure =41.11 miles.

Diff. latitude=61.53 miles.

2. A ship sails from Sandy Hook, latitude 40° 28′ N., upon a course E.S.E., till she makes a departure of 500 miles. What distance has she sailed, and at what latitude has she arrived? By Trigonometry, Art. 44,

Sin. course departure:: radius: distance,

:: cos. course : diff. latitude

Ans. Distance =541.20 miles.

Diff. latitude=207.11 miles, or 3° 27'.

Hence the latitude at which she has arrived is 37° 1′ N.

3. The bearing of Sandy Hook from Bermuda is N. 42° 56' W., and the difference of latitude 486 miles. Required the distance and departure.

By Trigonometry, Art. 46,

Radius: diff. latitude :: tang. course: departure,

:: sec. course: distance.
Ans. Distance =663.8 miles.

Departure=452.1 miles.

4. A ship sails from Bermuda, latitude 32° 22′ N., a distance of 666 miles, upon a course between north and east, until she finds her departure 444 miles. What course has she sailed, and what is her latitude?

By Trigonometry, Art. 44,

Distance: radius :: departure : sin. course,
Radius: distance :: cos. course: diff. latitude.

Ans. Latitude- 40° 38′ N.
Course N. 41° 49′ E.

5. The distance from Vera Cruz, latitude 19° 12' N., to Pensacola, latitude 30° 19′ N., is 820 miles. Required the bearing and departure.

By Trigonometry, Art. 45,

Distance: radius :: diff. latitude: cos. course,
Radius: distance sin. course departure.

:

Ans. Bearing N. 35° 34' E.

Departure=476.95 miles.

6. A ship sails from Sandy Hook upon a course between south and east to the parallel of 35°, when her departure was 300 miles. Required her course and distance.

By Trigonometry, Art. 47,

Diff. latitude: radius :: departure: tang. course,
Radius diff. latitude :: sec. course: distance.

Ans. Course S. 42° 27' E.

Distance 444.5 miles.

TRAVERSE SAILING.

(195.) A traverse is the irregular path of a ship when sailing on different courses.

The object of traverse sailing is to reduce a traverse to a single course, when the distances sailed are so small that the curvature of the earth may be neglected. When a ship sails on different courses, the difference of latitude is equal to the difference between the sum of the northings and the sum of the southings; and, neglecting the earth's curvature, the departure is equal to the difference between the sum of the eastings and the sum of the westings. If, then, the difference of latitude and the departure for each course be taken from the traverse table, and arranged in appropriate columns, the difference of latitude for the whole time may be obtained exactly, and the departure nearly, by addition and subtraction; and the corresponding distance and course may be determined as in plane sailing.

EXAMPLES.

1. A ship sails on the following successive tracks:

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Find the course and distance for the whole traverse.

We form a table as below, entering the courses from the table of rhumbs, page 138, and then enter the latitudes and departures taken from the traverse table.

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Hence the course is found by plane sailing N. 48° 36′ E.,

and the distance =122.2 miles.

The proportions are

Diff. latitude radius :: departure tang. course,

:

Radius: diff. latitude :: sec. course: distance.

2. A ship leaving Sandy Hook makes the following courses and distances:

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Required her latitude, the distance made, and the direct

Course.

Ans. Latitude=38° 1' N.

Distance 193.7 miles.
Course S. 40° 47′ E.

3. A ship from Pensacola, latitude 30° 19', sails on the fol lowing successive courses:

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