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From the logarithm of the departure 342'.O subtract the logarithm tangent of the course S. 67° 30' W., which gives the logarithm of the proper difference of latitude 141'.7.

From the latitude left subtract the proper difference of latitude 141'.7, which gives the latitude of the ship.

From Table 3 take the meridional parts for both 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.

To the logarithm of the meridional difference of latitude 183'.2 S. add the logarithm tangent of the course S. 67° 30' W. and the result will be the logarithm of the difference of longitude 442.3, which is added to the longitude left and gives the longitude of the ship.

To the logarithm secant of the course S. 67° 30' W. add the logarithm of the proper difference of latitude 141'.7, which gives the logarithm of the distance 370'.2..

EXAMPLE NUMBER 7

One latitude, one longitude, distance and departure to find the course, latitude and longitude in: A ship was sailed 428'.0 to the southward and made a departure 214'.o east from a point in latitude 20° 15'00" N. and longitude 74° 08' 01" W. Find the course, latitude and longitude of the ship.

Dep.
Dist.
Co. S. 30° oo' E.

214.0 E. 2.23041 428.0 S. — 2.63144 sin. = 9.69897

Dist.
Co. S. 30° oo' E.
Prop. dif. lat.

428'.0

cos. 370'.7

2.63144 +9.93753 = 2.56897

1233.0

Lat. left 20° 15'00" N. Mer. parts
Dif. 370'.7=- 6 10 42 S.
Lat. ship

14 04 18 N. Mer. parts

Dif. Mer. dif.

286'.I Co. S. 30° oo' E.

tang. Dif. long.

3° 42'54" E. = 222.9

- 846.9

386.1 S.

2.58670 +9.76144 =2.34814

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EXPLANATION From the logarithm of the departure 214'.0 E. subtract the logarithm of the distance 428'.0, which gives the logarithm sine of the course S. 30° E.

To the logarithm of the distance 428'.0 add the logarithm cosine of the course S. 30° E., which gives the logarithm of the proper difference of latitude 370'.7.

From the latitude left subtract the proper difference of latitude 370'.7 and the result will be the latitude of the ship.

From Table 3 take the meridional parts for both 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.

To the logarithm of the meridional difference of latitude 386'.I add the logarithm tangent of the course S. 30° E. and the result will be the logarithm of the difference of longitude 222'.9 E., which subtracted from the longitude left gives the longitude of the ship.

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. Find the course and distance.

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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.

From the logarithm of the difference of longitude 90'.5 subtract the logarithm of the meridional difference of latitude 361'.8, which gives the logarithm tangent of the course S. 14° 03' E.

To the logarithm secant of the course S. 14° 03' E. add the logarithm of the proper difference of latitude 363.5, which gives the logarithm of the distance 374.7.

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.

EXAMPLE NUMBER 9

Find the course and distance from Sandy Hook light in latitude 40° 27' 42" N., and longitude 74° 00' 09" W. to Cape of Good Hope light in latitude 34° 21' 12" S., and 18° 29' 26" E.

2644'.2 +2183 .7

4827.9 S.

Lat. Sandy Hk. 40° 27' 42" N. Mer. parts
Lat. Cape G. H. + 34 21 12 S. Mer. parts
Dif. 4488'.9=74 48 54 S. Dif.
Long. Sandy Hook 74° 0' 09" W.
Long. Cape Good Hope + 18 29 26 E.
Dif.

5549'.6=92 29 35 E.

Dif. long.
Mer. dif.
Co. S. 48° 59' E.

5549'.6 3.74429 4827.9–3.68377 tang. 0.00052

Co. S. 48° 59' E.
Prop. dif. lat.
Dist.

sec. 18291 4488.9+ 3.65215 6840.0= 3.83506

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