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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.

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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.

From Table 3 take the meridional parts for the latitudes to the nearest mile and find their difference by adding, which is the meridional difference of latitude and takes the same name as the proper difference of latitude. '

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 point is east of the longitude of the ship.

Enter Table 2 with the meridional difference of latitude 361'.8 S. and the difference of longitude 90'.5 E. and where they are found to agree nearly, the meridional difference in the latitude column and the difference of longitude in the departure column will give the course S. 14° E.

Enter Table 2 with the course S. 14° E. and find the proper difference of latitude 363'.5 in the latitude column. The distance 374'.7 is found opposite in the distance column.

When the places under consideration lie so near the equator it is not necessary to use Mercator 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.

MERCATOR SAILING BY LOGARITHMS The course, distance, etc., found by inspection should not be used unless the distance is small and an approximate result will answer. Its most frequent use is in finding the course and distance made good from noon to noon, which should be exact. However, in any case, the elements wanted should be found by inspection first, which not only serves as a check, but saves time in selecting logarithms from Tables 42 and 44.

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 Mercator sailing:

Ex.

Given.

To Find.

Methods.

Co. Both lats.

Dist. and longs. Dep.

Tang. co. = dif. long. • mer. dif. lat.
Dist.

- sec. co. X prop. dif. lat.
Dist.

= prop. dif. lat. - cos. co.

- prop. dif. lat. X tang. co. Dep. = (prop. dif. lat. X dif. long.) = mer. dif. lat.

Dep.

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Tang. co. = dep. ; prop. dif. lat.
Dist.

= prop. dif. lat. X sec. co.
Dist.

Dist.

| = dep. : sin. co. Dif. long.

Dif. long. = mer. dif. lat. X tang. co.

Dif. long. = (dep. X mer. dif. lat.) = prop. dif. lat.
Dep.
Dep.

dist. X sin. co.
Prop. dif. lat. Prop. dif. lat. = dist. X cos co.
Dif. long Dif. long. = mer. dif. lat. X tang. co.
Dist.
Dist.

= prop. dif. lat. = cos. co.
Dep.
Dep.

= prop. dif. lat. X tang. co.
Dif. long. Dif. long. = mer. dif. lat. X tang. co.
Co.
Cos. co.

= prop. dif. lat. • dist.
Dep.

dist. X sin. co.
Dif. long. Dif. long. = mer. dif. lat. X tang. co.
Prop. dif. lat. Dist.
Prop. dif. lat. = dep. • tang. co.

dep. : sin.co.
Dist.

Dif. long. = mer. dif. lat. X tang. co.
Dif. long.

Dif. long. = (dep. X mer. dif. lat.) ; prop. dif. lat.

co.

Both 5 lats. and

dist.

One lat. co. and dep.

One lat. 7 dist. and

dep.

Co.

Sin. co. = dep. • dist.
Prop. dif. lat. Prop. dif. lat. = dist X cos. co.
Dif. long.

Dif. long. = mer. dif. lat. X tang. co.
Dif. long. = (dep. X mer. dif. lat.) ; prop. dif. lat.

EXAMPLE NUMBER I

Both latitudes and longitudes to find the course and distance : A ship is in latitude 36° 51'00" N. and longitude 70° 55' 45" W. Find the course and distance to Barnegat light in latitude 39° 45' 52" N. and longitude 74° 06' 24" W.

Lat. ship — 36° 51'00" N.
Lat. Barnegat 39 45 52 N.
Dif. 174'.9= 2 54 52 N.

Mer. parts
Mer. parts
Dif.

— 2367':3

2589.5
222 .2 N.

Long. ship - 70° 55' 45" W.
Long. Barnegat 74 06 24 W.
Dif. 190'.6= 3 10 39 W.
Dif. long.
Mer. dif.
Co. N. 40° 37' W.
Co. N. 40° 37' W.
Prop. dif. lat.
Dist.

190'.6 W. 2.28012 222.2 N. -- 2.34674 tang.

9.93338

sec. 174.9 230'.4

.11971 + 2.24270 = 2.36250

EXPLANATION

Find the difference of latitude by subtracting the latitude of the ship from that of Barnegat light, convert it into miles and name it north, because the point is north of the latitude of the ship.

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

From Table 3 take the meridional parts for the latitudes to the nearest mile and find their difference, which is the meridional difference of latitude and takes the same name as the proper difference of latitude.

From the logarithm of the difference of longitude 190'.6 W. subtract the logarithm of the meridional difference of latitude 222'.2 N., which gives the logarithm tangent of the course N. 40° 37' W.

To the logarithm secant of the course N. 40° 37' W. add the logarithm of the proper difference of latitude 174'.9, which gives the logarithm of the distance 230'.4.

EXAMPLE NUMBER 2

Both latitudes, one longitude and departure to find the course and distance, and longitude in: A ship left a point in latitude 37° 55' 25" N. and longitude 74° 56' 22" W. and arrived in latitude 35° 57' 25" N., the departure being 216'.5 E. Find the course and distance and longitude of the ship...

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