MIDDLE LATITUDE SAILING.

MIDDLE LATITUDE SAILING relates to the conversion of the departure into difference of longitude, and the difference of longitude into departure, when the ship's course lies obliquely across the meridian, that is, when besides departure, she makes difference of latitude.

Suppose a ship, in going on the same course, from latitude 40° to latitude 44°, makes 100 miles departure: this departure, if made good altogether in latitude 40°, would give 130'5 difference of longitude by Rule XXXIII, page 84; and again, if made good in latitude 44°, it would give 139 difference of longitude. Now, since the ship has sailed between these two parallels, and not on either of them exclusively, her real difference of longitude must be between 130'5 and 139, and therefore we may conclude it to be not far from that which would result from a departure made good altogether in the middle parallel; hence the name Middle Latitude Sailing. Middle latitude sailing then, is founded on the consideration that the arc of the parallel of middle latitude of two places intercepted between their meridians, is nearly equal to the departure. If we conceive the ship to sail along this middle parallel, we may apply the principle of parallel sailing to the cases in point. In parallel sailing the departure (or distance) and difference of longitude are connected by the relation, dep. diff. of long. X cos. lat. When the ship's course lies obliquely across the meridian, making good a difference of latitude, a modification of this formula gives the formula for middle latitude sailing, dep. (nearly)= diff, of long, X cos. mid, lat. ; or, in logarithms, log, dep, = log, diff. of long. + log. cos. mid. lat. - IO. Middle latitude sailing has thus the same two cases as parallel sailing, and accordingly the rules for inspection, computation already given, Rule XXXIII, page 84, apply equally to this sailing, observing merely to read middle latitude for latitude.

To find the latitude and longitude in, the course and distance from a known place being given by Traverse Table and Middle Latitude.

1o.

RULE XXXV.

With the given course and distance enter the Traverse Table, and take out true difference of latitude and departure (see page 75).

2o.

With difference of latitude and latitude from, find latitude in (see Rule XXII, page 53).

3°. Get the middle latitude, as directed, Rule XXIII, page 54. 4°. With the middle latitude as course, look in the difference of latitude column for the departure, the corresponding distance at the top is the difference of longitude.

With difference of longitude and longitude from get longitude in, as in Rule XXV, page 56.

5°.

NOTE. When the departure to be looked for as difference of latitude at the middle latitude, is beyond the limits of the Table, one-half, one-third, &c., must be used, and the resulting difference of longitude multiplied by the divisor, in order to get the whole difference of longitude.

EXAMPLES.

Ex, I.

A ship from lat. 52° 6′ N., long. 35° 6′ W., sailed S.W. by W., 256 miles; required her latitude and longitude in.

Explanation.-The difference of latitude and departure are found as described in page 75. The latitude in is found by Rule XXII, page 53; and thence the middle latitude, by adding the latitude from and latitude in together, and dividing by 2 (see Rule XXIII, page 54). The departure exceeding the limits of the Tables, the half is taken. Then with middle latitude as a course, and half the departure, in difference of latitude column, half the difference of longitude is found in the distance column. This being doubled (as half the departure was taken) and divided by 60, gives the difference of longitude expressed in degrees and minutes. The ship is in West longitude, sailing West, add difference of longitude to longitude left to obtain longitude in (Rule XXV, page 56).

This is the usual case at sea of working the Day's Work.

Ex. 2.

A ship from lat. 48° 27'4 N., and long. 29° 12' W., sails N.E. by N., 22'5 miles required the latitude in, also the longitude in.

Course N.E. by N.3 pts.; then 3 pts, and dist. 22.5 give diff. lat. 18.7, and dep. 125 (see page 75).

Ex. 3. Lat, from 58° 13′ N., long. from 3° 33′ E., course S.S.E. 1 E., distance 213 miles required latitude and longitude in.

Course 2 points, and dist. 213 miles, give diff. lat. 192'6, and dep. 91*1. 6,0)19,2'6

3 12.6
or 3° 13' S.
Lat. from 58 13 N.

Lat. in 55 o N.

2)113 13

Mid. lat. 56 36

The mid. lat. is here 5610.

Mid. lat. 56° as course, and dep. 91'1 as diff, lat., give in dist. column 163; and mid. lat, 57° as course, and dep. 91'o, give in dist. column 167. Then 163+167= = 330, which divided by 2, gives diff. long. 165 miles. 6,0)16,5

2 45 or 2° 45′ E,

Long. from 3 33 E.

Long, in 6 18 E.

Ex. 4. A ship from the Lizard, in lat. 49° 57' N., sails W.S.W,, 163 miles, variation

2 points W.: required the latitude come to, and difference of longitude.

W.S.W. by compass is (allowing 23 points westerly variation) S.W. S. true, which in Table II, and dist. 163, gives diff. lat. 126, and dep. 103'4.

Ex. 5. Sailed from A, in lat. 50° 48′ N., long. 1° 10′ W., S. 41° E., 275 miles. Entering Traverse Table II with dist. 275 miles, and course 41°, the true diff. lat. is 207′ 5, or 3° 27'5 S.; applying this to lat, from, the lat. in is 47° 20' 5 N. The corresponding dep. is taken out at the same opening, which is 180'4. The mid. lat., or half sum of lat. from and lat, in, is 49° to the nearest degree. The dist. corresponding to 49° as a course, and 180'4 in diff. lat. column, is found to be 275', in degrees 4° 35' E., which is the diff. long. Applying this to the long. from 1° 10' W., we have the long. in 3° 25′ E.

EXAMPLES FOR PRACTICE.

In each of the examples following, the latitude and longitude arrived at are required to be found, having given the latitude and longitude from, with the course and distance sailed.

MERCATOR'S SAILING.

MERCATOR'S Sailing, like middle latitude sailing, relates to finding the difference of longitude a ship makes when sailing on any oblique rhumb, and is a perfectly general and rigorously true method, which the other is not.

Mercator's sailing is characterised by the use of the Table of meridional parts, and the chart constructed by means of it called Mercator's chart. With the assistance of this Table, the rules of plane trigonometry suffice for the solution of all the problems.

In the triangle ACB let A be the course, AB the distance, AC the true difference of latitude, CB the departure; then corresponding to AC the table of meridional parts gives AC', the meridional difference of latitude, and completing the right-angled triangle AC'B', C'B' will be the difference of longitude. In addition then to the three canons of plane sailing which can be deduced from the triangle ACB, the triangle AC'B' gives the characteristic canon of Mercator's sailing (since C'B' AC' tang. A) diff, long. mer. diff. lat. + tang, course.

C'

B

B'

Given the latitudes and longitudes of two places, to find the course and distance between them.

RULE XXXVI.

1o. Find the true difference of latitude, according to Rule XX, page 50. Find the meridional difference of latitude, Rule XXI, page 52.

2o.

3°. Next find the difference of longitude, Rule XXIV, page 54.

4°. To find the course.-From the log. of diff. of longitude (increasing

its index by 10), subtract the log. of meridional diff. of lat.: the remainder is

the tangent of course, which take out of the tables, and place before it the letter of diff. of lat., and after it the letter of diff. of long.

5°. To find the distance.-To the secant of course (rejecting 10 from the index), add the log. of diff. of lat.: the sum is the log. of distance, the natural number corresponding to which find in the tables.

EXAMPLES.

Ex. 1. Required the course and distance from Tynemouth light to the Naze of Norway.

The lats. having like names their difference is taken and reduced to miles, and since the lat. to is to the north of lat. from, the ship must sail to the northward, whence the diff. lat. is marked N. The difference of the meridional parts is taken as the lats. are both of the same name. The longs. being of different names the sum is the diff. of long. which is reduced to miles, and since the ship has to pass from West long. to The diff. of long. is therefore marked E.

East long., she must steer Eastward to do so.

To find the course we look for the tabulated logarithmic tangent nearest, and not exceeding, the given tangent, which we find to be 10:198325, the corresponding angle to which is 57° 39′; we next take the difference of the tabulated tangent thus found and the given tangent, annex two cyphers, and divide by the difference (466) found in corresponding column of difference, the quotient (38) is the additional seconds for the tangent, whence the course is N. 57° 39′ 38′′ E. (see Rule, pages 40 and 41). The secant of course is next taken out (see Rule, page 40).

Ex. 2. Required the course and distance from A to B.

Both latitudes being of the same name (north), take the difference and reduce to miles. Take out the meridional parts for each lat. (Table IV, Norie), and take also the difference of these, since the lats. are of the same name. The longs. being of like names the diff. of long. is found by taking the less long. from the greater, which also is reduced to miles,