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514. Method II.-The second method of solution is by Mercator's sailing.

In this method the surface of the earth is considered to be plane, the meridians being parallel lines, and also the parallels of latitude, as in plane sailing; and since by this hypothesis the distance between the meridians is increased, the lengths of the arcs of the meridians are increased in the same proportion, so that the distances between the parallels of latitude for every successive minute are continually increasing with the latitude; and the relative bearings of places are thus preserved. This method is so accurate, that it may be used without sensible error for any distance on the earth's surface.

The lengths of the meridians from the equator to any latitude are thus increased, and the increase is greater the higher the latitude. For instance, the increased distance of the parallel of 10°, instead of being 600 miles, is found to be 603 miles; and that of the latitude of 50°, instead of 50 × 60 = 3000 miles, is 4527 miles. The increased lengths of the meridians, from the equator to any latitude, are called the meridional parts, from the manner in which they are computed; and their numerical values are contained in tables.

515. The difference of the meridional parts for any two latitudes, is called the meridional difference of latitude; and the true difference of latitude is sometimes, for distinction, called the proper difference of latitude.

The analogies peculiar to this method are:

516. The difference of latitude is to the departure, as the meridional difference of latitude to the difference of longitude.'

Dif. lat. departure = mer. dif. lat. : dif. long.

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517. The meridional difference of latitude is to the difference of longitude, as radius to the tangent of the course.' Mer. dif. lat. : dif. long. = radius: tan course.

These two proportions may be varied so as to have any term last; and the first three being known, it could be found. Some of these proportions would, however, be practically

useless. These proportions can be obtained from two right

angled triangles, ABC, ADE; in
which AB and AD are the pro-
per and the meridional differences
of latitude, BC the departure, DE
the difference of longitude, AC S
the distance and A the course.

From the triangle ABC, other A
proportions can be obtained; but
as they are the same as in plane
sailing, they are not to be de-
pended on for any distance exceed-

D Dif. Long. E

Dif. Lat. }

A.

Dep.

Distance.}<

ing a few miles, nor in high latitudes, even for small distances. 518. PROBLEM I.-Given the place left and that bound for, to find the course and distance.

1. By Middle Latitude Sailing.

:

To find the course.

Dif. lat. dif. long = cos mid. lat. : tan course.
To find the distance.

Cosine course: radius = dif. lat. : distance.
2. By Mercator's Sailing.

To find the course.

Mer. dif. lat. : dif. long.

-radius: tan course.

distance.

To find the

Cosine course: radius dif. lat. : distance.

The second proportion in the two methods is the same. EXAMPLE.-Required the bearing and distance of New

York from Liverpool.

By Mid. Lat. Sailing.

Liverpool, lat. = 53° 25′ N.
N. York, lat. = 40 42 N.

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Cos course 75° 16′ 9-40538 Cos course 75° 10′ 9.40825

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1. What is the bearing and distance of a place in latitude 71° 10′ N., longitude 26° 3′ E., from another place in latitude 60° 9' N., and longitude 0° 58′ W.?

Ans. Course N. 45° 19′ E., distance 940 miles, by mid. lat. sailing.

Course N. 44° 48′ E., distance 938.5 miles, by Mercator's sailing.

2. A ship having arrived at a place in latitude 49° 57′ N., longitude 5° 14′ W., and is bound for another place in latitude 37° N., longitude 25° 6' W.; required the bearing and distance of the latter place from the former.

Ans. Course S. 48° 4′ W., distance 1163 miles, by mid. lat. sailing.

Course S. 47° 54′ W., distance 1159 miles, by Mercator's sailing.

519. PROBLEM II.-Given the place sailed from, the course and distance, to find the place arrived at.

By Middle Latitude Sailing.

To find the difference of latitude.
Radius: cos course distance : dif. lat.

To find the difference of longitude.

Cos mid. lat. : sin course

distance: dif. long.

By Mercator's Sailing.

To find the difference of latitude.

The analogy is the same as in the preceding method.
Radius: cos course distance: dif. lat.

Or,

To find the difference of longitude.

Radius: tan course mer. dif. lat. : dif. long.

By the first proportion in these two methods, the difference of latitude is found, and by the second, the difference of longitude; and hence the latitude and longitude of the place in are known.

EXAMPLE.-A ship from a place in latitude 25° 40′ S., and longitude 35° 12′ W., sails SW. b S. 246 miles; required the latitude and longitude in.

To find the difference of latitude by both methods.

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within less than a mile of each other. The place arrived at is in latitude 29° 5' S., and longitude 37° 46′ W.

EXERCISES.

1. A ship from a place in latitude 50° 30′ N., and longitude 145° 20′ W., sails 450 miles SSW.; what place is it in? Ans. Latitude 43° 34′ N., longitude 149° 33′ W., by mid. lat. sailing, and 149° 33′ W., by Mercator's sailing.

2. A ship from latitude 51° 15′ N., and longitude 9° 50′ W., sails SW. S. till the distance run is 1022 miles; what are the latitude and longitude in?

Ans. Latitude 37° 5′ N., longitude 23° 2′ W., by mid. lat. sailing, and 23° 8′ W. by Mercator's sailing.

520. PROBLEM III.-Given the latitude left, the difference of latitude and departure, to find the difference of longitude.

By Middle Latitude Sailing.

Cos mid. lat. : radius

dep. : dif. longitude.

By Mercator's Sailing.

Dif. lat. : dep. = mer. dif. lat. : dif. long.

EXAMPLE.-A ship on a course between the south and west from latitude 54° 24′ N., and longitude 36° 45′ W., has made 346 miles of difference of latitude, and 243 miles of departure; what is the latitude and longitude in?

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