Ex. 2. A ship from Cape Espicheli, in lat. 38° 25′ N., sails as follows: S.W. by W., 28 miles; W. by N., 55 miles; West, 47 miles; S.E. S., 25 miles; South, 101 miles; W. S., 72 miles: required the latitude in, also the course and distance made good.

We seek in the Traverse Table till the diff. of lat. 136-6, and dep. 180°5, are found opposite each other, in their respective columns; the nearest to these are 1805 and 136.0, which give the course (at the bottom of page, dep. being the most) S. 53° W., and distance 226′. This is an illustration of the remark in Rule XLI, page 114, that when the departure exceeds the difference of latitude, the course is more than 45°.

Lat. left

Diff. lat. 136.6

Lat. in (or arrived at)

38°25' N. = 2 17 S. 36 8 N.

The lat. in is found according to
Rule XXVI, 2o, page 80.

Ex. 3. A ship from lat. 37° 24' S., sails the following true courses:-S.W. by S., 20 miles; West, 16 miles; N. W. by W., 28 miles; S.S.E., 32 miles; E.N.E., 14 miles; S.W., 36 miles: required the lat. in, also the course and distance made good.

We seek in the several pages of the Traverse Table II, for the diff. lat. 50'7; and dep. 507; the nearest found to these are diff. lat. 50‘9, dep. 50′9, give course S. 45° W., distance 72 miles.

The diff. lat. and dep. being of equal amount, the course is 45°, or 4 points.

Lat. left
Diff. lat. 507 =
Lat. arrived at

37° 24' S. 51 S. 38 15 S.

The lat. sailel from being South, and the ship having sailed South, the ship has evidently increased her South lat., whence the sum of lat. from and diff. lat. is taken to obtain lat. in.-(See Rule XXVI, 1o, page 80.

Ex. 4. A ship from lat. 46° 20′ N., sails (all true courses) N. 72° E., 21 miles; N. 38° E., 17 miles; S. 26° W., 13 miles; S. 73° E., 19 miles; S. 1° W., 19 miles; S. 65° E., 48 miles; N. 76° E., 19 miles; N. 48° E., 48 miles: required the lat. in, also the course and distance made good.

Course due East, and dist. 1403, the same as the departure. (See Rule XLII, 6o, Note (b), pago 117).

The Traverse Table being filled up, the sums of the Northings and Southings are both 56-6, and being of contrary directions, show that the ship has returned to the same parallel of latitude which she sailed from. The sum of the Eastings is 146'3, and that of the Westings 6'0; their difference, 140'3, shows that the ship has gained so much to the Eastward, that being the greater. Consequently the Course is due East, and the Distance 1403, the same as the departure.

Ex. 5. A ship from a place in lat. 1o 5′ S., sails the following true courses:—N. 17° E., 13 miles; North, 38 miles; N. 27° E., 18 miles; N. 79° E., 25 miles; S. 83° W., 23 miles; S. 48° E., 25.2 miles; N. 48° W., 27'1 miles; N. 36° W., 21 miles: required the latitude also the course and distance made good.

The Traverse Table being completed, the sum of the Northings is 106.3, and the sum of the Southings is 197, the difference 86-6, and to that amount the ship has altered her latitude. The miles of departure in the East are 52-2, and those in the West column are also 52°2; but as the East and West departures destroy one another, there is no resulting departure, and therefore it is not necessary to refer to the Traverse Table. The ship is under the same meridian as she sailed from; consequently, the course is due North, and the distance sailed is equal to the diff. lat., viz., 866. This is according to Rule XLII, 6°, Note (a), page 117. Latitude left 1° 5' S. Diff. lat. 6,0)8,6.6

The ship being 1o 5', or 65 miles, S. of the equator, must evidently be in N. lat. after making 87 miles of Northing. Thus, in subtracting one of the quantities from the other, the difference takes the name of the greater. Rule XXVI, page 80.

The course is North, and dist. 86·6, the same as diff. lat.

I.

EXAMPLES FOR PRACTICE.

A ship from the Texel in lat. 52° 58' N., sails W. by N., 44 miles; S. by E., 45 miles; W. by S., 35 miles; S.S.E., 44 miles; W.S.W. } W., 42 miles: find diff. lat. and dep., the course and dist. made good, also the lat. arrived at.

2.

A ship from Heligoland, lat. 54° 12′ N., sails W.S.W., 12 miles; N.W., 24 miles; S. by W., 20 miles; N. W. by W., 32 miles; S. by E., 36 miles; W. by N. N., 42 miles; S.S.E.E., 16 miles; W. N., 45 miles: required diff. lat. and dep., course and dist. made good, also the lat. arrived at.

3. A ship sails from lat. 3° 50' N., sails S.S.W., 112 miles; S. by E., 86 miles; S.S.E., 112 miles; S. by W., 86 miles: find diff. lat. and dep., the course and dist. made good, also the lat. arrived at.

4. Yesterday we were in lat. 19° S., and since then have sailed S.E. S., 13 miles; S. by E., 19 miles; S.E. by E., 22 miles; E. by S. S., 32 miles; N.N.E., 20 miles; N. by W. W., 27 miles; N.E. by E. E., 24 miles; S.W. S., 10 miles.

5.

A ship from lat. 1° N., sails East, 8 miles; E. & N., 20 miles; S.E. by E., 33 miles; S. W., 31 miles; N.E. N., 43 miles; South, 28 miles; S. E., 21 miles; S. by W. W., 12 miles required diff. lat. and dep., course and dist. made good, and also the lat. in.

6. A ship from lat. 1° 10′ N. sails N. 40° W., 20 miles; S. 56° W., 51 miles; S. 19° W., 19 miles; S. 48° W., 16 miles; N. 85° E., 28 miles; S. 44° E., 15 miles; N. 22° W., 25 miles; S. 9° E., 54 miles: find diff. lat. and dep., course and dist. made good, also the lat. in.

7. A ship from lat. 47° 12′ N. sails S. 31° W., 16 miles; N. 72° E., 13′′1; S. 52° W., 15'; S. 44° E., 151; N. 44° W., 19'7; N. 77° E, 11'4; S. 40° W., 16'; S. 14° E., 6': required the course and dist. made, the lat. arrived at, and the dep. made.

8. Since leaving lat. 34° 11' N. we have sailed the following courses:-N. 36° W., 27′; N. 24° E., 30'; S. 75° W., 47'; S. 80° W., 29'; N. 72° W., 42'; N. 78° W., 34'; S. 12° E., 28': required the course and dist. made, the lat. arrived at, and the dep. made.

9. Since leaving lat. 36° 35′ S. the ship has sailed N. 84′ W., 18′; N. 89o W., 30′′4; N. 67° W., 29''9; N. 39° W., 33'9; N. 8° W., 25'9; N. 73° W., 34''9; N. 86° W., 44′′7 ; S. 65° E., 56': required the lat. arrived at, and the course and dist. made good,

PARALLEL SAILING.

126. When two places on the same parallel of latitude, or due East or West of each other, the distance between them, estimated along a parallel, or E. and W. (which is all departure), is converted into difference of longitude; or, on the other hand, the difference of longitude is converted into distance by Parallel Sailing.

Since the meridians are all parallel to the equator and meet at the poles, the distance between any two meridians, measured East and West, is less as the latter is greater-that is, the absolute number of miles, or of feet, in a degree of longitude is less as the latitude in which they are measured is greater. Hence, also, a given number of miles between two meridians correspond to a greater difference of longitude, as the latitude in which they are measured is greater. For example, two places in lat. 10° and distance 60 miles East and West from each other, have 60'9 diff. long. In lat. 60° N. or S., two places similarly situated have 2o o' diff. long., while at 73° the diff. long. is 3° 25'. Questions of this kind are solved by Parallel Sailing.

In this sailing then a ship's track is supposed to be on a parallel of latitude, as from A to B (see Fig. 31, page 122). The latitude of A or of B is represented by AD or BL.

The diff. long. between A and B is the arc of the equator, DL.

The arc A B is the departure (or meridian distance) between A and B.

CASE A.

127. Given the departure made good on a given parallel of latitude, to find the difference of longitude corresponding thereto.

1o. Take out of the Tables (XXIV, NORIE, or 67, RAPER) the log. of the departure, and the log. of secant of the latitude (Table XXV, NORIE, or Table 68, RAPER).

2o. Add these logs. together and find the natural number corresponding thereto (XXIV, NORIE, or 67, RAPER): the result is the diff. of longitude required. Ex. 1. A ship in the parallel of 50' N. latitude makes 240 miles of departure: find the corresponding difference of longitude.

With the latitude of the parallel as a Course, and the dep. as diff. lat., enter the Traverse Table, and the corresponding distance will be the diff. long. required.

EXAMPLE (above).—Entering the Traverse Table with 50° as Course, and one-half the departure 120 (the whole dep. 240 being too large for the Tables) as diff. lat., the corresponding dist. 187 multiplied by 2 = 374 will be the diff. long. required.