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EXPLANATION Find the difference of latitude by subtracting the latitude of the point left from that of the ship, convert it into miles and name it south, because the ship has sailed south.

Enter Table 2 with the distance 392'.o and the difference of latitude 66'.8 S. and find them to agree in their respective columns, and the departure 386'.o E. will be found in the departure column and the course S. 80° E. at the bottom of the page.

Add together the latitude of the point left and that of the ship. Then divide by 2, which will give the middle latitude, the nearest degree of which is 561/2°.

With the middle latitudes 56° and 57° enter Table 2 and find half of the departure 193'.o E. in the latitude columns and the corresponding numbers in the distance columns added together will be the difference of longitude 699'.o E., which is subtracted from the longitude of the point left because the ship has sailed east and in doing so has decreased the longitude.

EXAMPLE NUMBER 6

One latitude, one longitude, course and departure to find the distance, latitude and longitude in: A ship was sailed S. 67° 30' W. from a point in latitude 40° 37' 02" N. and longitude 69° 37' 05° W., making 342'.o departure. Find the distance, latitude and longitude of the ship.

Co. S. 67° 30' W., _ S Dif. lat. 141'.6 S.
Dep. 342'. W.S | Dist 370.0
Lat. left 40° 37' 02" N. Long. left 69° 37' 05" W.
Dif. 141'.6 — 2 21 36 S. Dif. 443'.o = + 7 23 00 W.
Lat. ship

38 15 26 N. Long. ship 77 00 05 W. Lat. left 40° 37' 02" N. Lat. ship + 38 15 26 N.

2)78 52 28 Middle lat. = 39 26 14=39° 1/2 dep. 342'.o=dif.long. 443'.o.

EXPLANATION Enter Table 2 with the course S. 67° 30' W. and find the departure 342'. W. in the departure column and the difference of latitude 141'.6 S. and the distance 370'.o will be found in their respective columns.

The difference of latitude 141'.6 S. subtracted from the latitude left will give the latitude of the ship.

Add together the latitude of the point left and that of the ship. Then divide by 2, which will give the middle latitude, the nearest degree of which is 39/2°.

With the middle latitudes 39° and 40° enter Table 2 and find the departure 342'.o. in the latitude columns and the corresponding numbers in the distance columns added together and divided by 2 will be the difference of longitude 443'. W., which is added to the longitude of the point left because the ship has sailed west and in doing so has increased the longitude.

EXAMPLE NUMBER 7

One latitude, one longitude, distance and departure to find the course, latitude and longitude in: A ship was sailed 428'. to the southward and made departure 214'. E. from a point in latitude 20° 15'00" N. and longitude 74° 08' 01" W. Find the course, latitude and longitude of the ship.

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Lat. left. 20° 15'00" N. Long. left 74° 08' 01" W.
Dif. 370'.7 =- 6 10 42 S. Dif. 224.0=- 3 44 00 E.
Lat. ship 14 04 18 N. Long. ship 70 24 01 W.
Lat. left. 20° 15'00" N.
Lat. ship +14 04 18 S.

2)34 19 18 Middle = 17 09 39=17° dep. 214'.0 E.= dif. long. 224.0 E.

EXPLANATION Enter Table 2 with the distance 428'.o S. and departure 214'. E. and find them to agree in their respective columns and the difference of latitude 370'.7 S will be found in the latitude column and the course S. 30° E. at the top of the page.

The difference of latitude 370'.7 S. subtracted from the latitude left will give the latitude of the ship.

Add together the latitude left and that of the ship. Then divide by 2, which will give the middle latitude, the nearest degree of which is 17o.

With the middle latitude 17° enter Table 2 and find the departure 214'. E. in the latitude column and the corresponding number in the distance column will be the difference of longitude 224'.o E., which is subtracted from the point left because the ship has sailed east, and in doing so has decreased the longitude.

Points itu de 25° 35" dastance: A shipta) and bot

EXAMPLE NUMBER 8 Both latitudes (one north and the other south) and both longitudes to find the course and distance: A ship is in latitude 4° 32' 45" N. and longitude 25° 35' 45" W. Find the course and distance to a point in latitude 1° 30' 45" S. and longitude 24° 05' 15" W. Lat. ship

4° 32' 45" N. Long. ship. 25° 35' 45" W. Lat. point +1 30 45 S. Long. point – 24 05 15 W. Dif. 363.5 = 6 03 30 S. Dif. 90'.5 = I 30 30 E. Greater lat. 2)4° 32' 45" Middle lat. 2 16 22=2° dif. long. 90'.5 = dep. 90'.4.

{ = Co. S. 14° E. distance 375'.O. Dep. 90.4 )

This method will satisfy the ordinary demands of practice; but should not be used when the latitudes of the places under consideration lie on opposite sides of the equator. In such cases the course, distance, etc., should be found by Mercator sailing.

EXPLANATION

Find the difference of latitude by adding the latitude of the point to that of the ship, convert it into miles and name it south because the point is south of the latitude of the ship.

When one latitude is north and the other south, the middle latitude can best be found by taking half the greater latitude, which has been done in this case, and gives the middle latitude to the nearest degree, which is 2°.

Find the difference of longitude by subtracting the longitude of the point from that of the ship, convert it into miles, and name it east because the longitude of the point is east of the longitude of the ship.

With 2° enter Table 2 and find the difference of longitude 90'.5 in the distance column and the corresponding number in the latitude column will be the departure 90'.4.

Enter Table 2 with the difference of latitude 363'.5 and departure 90'.4 and they will be found to agree nearly, giving the course S. 14° E. distance 375'.O.

When the places under consideration lie so near the equator, it is not necessary to use middle latitude sailing unless requested to do so at an examination, as plain sailing, in which the degrees of latitude and longitude are supposed to be equal, will give a result practically the same.

MIDDLE LATITUDE SAILING BY LOGARITHMS

For ordinary purposes the course and distance found by inspection is quite satisfactory; but when the distance between the two points is too great or an exact result is wanted, it is necessary to use logarithms and is the method usually required at examinations.

A general description of the various operations is not given here; but in its place an individual explanation follows each example.

The following table contains the necessary rules for the solution of any problem by middle latitude sailing :

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