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tures are usually filled up from a table already computed to every quarter point of the compass, and to all distances from one mile up to a hundred or 120; so that, by entering this table with any given course and distance, the proper difference of latitude and departure is found by inspection. Most books on navigation contain also a second and more enlarged traverse table, being computed to every course from a quarter of a degree up to forty-five degrees. This latter table we have not thought it necessary to insert in our collection, but the former we have given (Table Iv.), and its use is fully explained in the introduction prefixed.

But there is another mode of finding the direct course and distance, much practised by seamen, viz. by construction. To facilitate this construction the mariner's scale is employed, which is a two-foot flat rule exhibiting several scales on each side, by help of which and a pair of compasses the usual problems in sailing may be all solved. One of these scales is a scale of chords, commonly called a scale of rhumbs, being confined to every quarter point of the compass; and another is a more enlarged scale of chords, being to every single degree. Both these scales are constructed in reference to the same common radius, so that the chords on the scale of rhumbs belong to that circle whose radius equals the chord of 60° on the scale of chords; and the method of laying down a traverse from these scales, and one of equal parts, and of thence measuring the equivalent single course, and distance made good, will be at once understood from the following examples.

EXAMPLES.

1. A ship sails from a place in lat. 24° 32′ N., and has run the following courses and distances, viz.

1st, S.W. by W., distance 45 miles; 2d, E.S.E., distance 50 miles; 3d, S.W., distance 30 miles; 4th, S.E. by E., distance 60 miles; 5th, S.W. by S. 4 W., distance 63 miles: required her present latitude, with the direct course and distance from the place left to the place arrived at.

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It

appears from the results of this table that the difference of latitude

made by the ship during the traverse is 149.2 S.

Lat. left

Diff. lat.

Lat. in

=

2° 29′ S.

24°32′ N.

2 29 S.

22 3 N.

It appears also that the departures east are equal to the departures west, so that the ship has returned to the meridian she sailed from, consequently the direct course from the place left to that come to is due south, and the distance is equal to the difference of latitude, which is 149.2 miles.

The construction of this traverse is as follows.

A

With the chord of 60°, taken from the line of chords on the mariner's scale, describe the horizon circle, and draw the north and south line N.S. From the line of rhumbs take the chords of the several courses, and as these are all southerly they must be laid off from the south point S, those which are westerly to the left, and those which are easterly to the right, their extremities being marked 1, 2, 3, &c. in the order of the courses. This done, lay off from any convenient scale of equal parts, and in the direction A1 the distance AB sailed on the first course; then in the

D

E

direction parallel to A2, the distance BC sailed on the second course; in the direction parallel to A3, the distance CD on the third course; in the direction parallel to A4, the distance DE on the fourth course; and, lastly, in the direction parallel to A5, the distance EF on the third course; then F will represent the plane of the ship at the end of the traverse; FA, being applied to the scale of equal parts, will show the distance made good, and the chord of the arc included between this distance, and the meridian, being applied to the line of rhumbs, will show the direct course. In the present case the intercepted arc will be O, showing that F is on the meridian of A.

2. A ship from Cape clear, in lat. 51° 25′ N., sails 1st, S.S.E. E., 16 miles; 2d, E.S.E., 23 miles; 3d, S.W. by W. W., 36 miles; 4th, W. & N., 12 miles; 5th, S.E. by E. † E., 41 miles: required the distance made good, the direct course, and the latitude in?

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therefore, as the difference of latitude is south, and the departure east, the direct course is S. 18° 12′ E., and the distance made good 62-74 miles.

To construct this traverse, describe, as before, the horizon circle, with a radius equal to the chord of 60°, and taking from the line of rhumbs the chord of the first course, 24 points, apply it from S. to 1, to the right of S.N., as this course is south-easterly; apply, in like manner, the chord of the second course, 6 points from S. to 2, 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 chord of the fourth course, 74 from N. to 4, to the left of N.S., this course being north-westerly, and, lastly, apply the chord of the fifth course, 54 points, from S. to 5, to the right of S.N. In the direction A1, lay off the distance AB = 16 miles from a scale of equal parts; in the direction parallel to AQ, lay off the distance BC = 23 miles; in the direction parallel to A3, lay off CD 36; in the direction parallel to A4, lay off DE = 12 miles; and, lastly, in the direction parallel to A5, lay off EF = 41, then F will be the place of the ship at the end of the traverse; consequently, AF will be the distance made good, and the angle FAS the direct course; applying, therefore, the distance AF to the scale of equal parts, we shall find it reach from 0 to 62; and applying the distance Sa to the line of chords, we shall find it reach from 0 to 18°.

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3. A ship from lat. 28° 32′ N., has run the following courses, viz. 1st, N.W. by N., 20 miles; 2d, S.W., 40 miles; 3d, 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 present latitude, the distance made good, and the direct course from the place left to that come to.

The direct course is due east, and distance 70.2 miles, the ship being in the same latitude at the end as at the beginning of the traverse.

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. & E., 18 miles; S.W. W., 14 1 miles; and W. N., 30 miles: required the latitude of the place arrived at, and the direct course and distance to it.

Lat. 40° 5' N.; course S. 52° 49′ W.; distance 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. Direct course S. 25° 59′ E., or S.S.E. E. nearly; distance 95.87 miles.

These examples will, perhaps, suffice to illustrate the principles of plane sailing, in which, course, distance, difference of latitude, and departure, are the only things concerned. The determination of the difference of longitude made on any course cannot be effected by these principles, for this element is not the same as if the meridians were all parallel to each other, as is the case with the other elements. The finding of the difference of longitude is the easiest when the ship sails due east or due west, that is, upon a parallel of latitude; this is called parallel sailing.

Parallel Sailing.

(66.) The theory of parallel sailing is comprehended in the following proposition, viz.

The cosine of the latitude of the parallel is to the distance run as the radius to the difference of longitude. This may be demonstrated as follows.

In the figure, at page 66, let IQH represent the equator, and BDA any parallel of latitude; CI will be the radius of the equator, and cB the radius of the parallel. Let BD be the distance sailed, then the difference of longitude will be measured by the arc IQ of the equator, and since (Geom., prop. 12, Cor. 2, B. 7) similar arcs are to each other as the radii of the circles to which they belong, we have

CB: CI: dist. BD : diff. long. IQ.

M

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