The course and distance in the last part of this problem may be found by either Mercator or middle latitude sailing, by inspection, but the former would be preferable, as the course is more than 45° and by doing so save the trouble of close interpolation to convert the difference of longitude into departure.

DAY'S WORK Example 3.-Farralon light bore E. by N. distant 9.7 miles. Ship’s head N. W. by W. Deviations as per table on page 16. Throughout the day a current set S. E. (magnetic) 78 of a mile per hour. Variation 19° 30' E. The ship was then sailed as follows:

Find the course and distance made good and the ship's position. From that point find the course and distance to Cape Flattery light

FARRALON LIGHT
Lat. 37° 41' 51" N.
Long. 123° 00'07" W.

CAPE FLATTERY LIGHT
Lat. 48° 23' 30" N,
Long. 124° 44' 6" W.

Dif. lat. 182'.5 N. ? = Co. N. 29° W. dist. 200'.O.
Dep. 100.4 W. S -

Lat. F. L. 37° 41' 51" N.
Lat. ship + 40 44 21 N.

2)78 26 12
Middle lat. 39 13 06
Middle lat. 39° dep. 100'.4 = dif. long. 129'.o=2° 0'00" W.

The traverse table is not used in the practice of navigation, but its use must be learned that more practical means may be used to keep account of a ship’s track.

Usually the following method is not wanted at an examination, but should be used in practice. The former is quite satisfactory so far as the final result is concerned, which is not known until the end of the day. By this the ship's position may be known at any moment. In other words, by one the ship takes the man and by the other the man takes the ship. +

+

+

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Lat. Far. L.
Dif.
Lat. noon
Dif.
Lat. 4 P.M.
Dif.
Lat. 8 P.M.
Dif.
Lat. Mdt.
Dif.
Lat. 4 A.M.
Dif.
Lat. 8 A.M.
Dif.
Lat. noon
Dif. (current)
Lat. noon

37° 41'.8 N.

Long. Far. L.
+ 2.8 N. = 38° dep. 9'.3= dif.
37 44.6 N.

Long. noon
39.0 N. = 38 dep. 23.0= dif.
38 23.6 N.

Long. 4 P.M.
35.5 N. = 39 dep. 35.5= dif.
38 59.1 N.

Long. 8 P.M.
22.1 N. = 39 dep. 32.8= dif.
39 21.2 N.

Long. Mdt.
50.7 N. = 40 dep. 8.5= dif.
40 11.9 N.

Long. 4 A.M.
30.3 N. = 40 dep. 16.8= dif.
40 42.2 N.

Long. 8 A.M.
+ 21.0 N. = 41 dep. .5= dif.
41 03.2 N.

Long. noon
- 18.9 S. = 41 dep. 9.0= dif. (current)
40 44.3 N.

Long. noon

123° 00'.1 W.
+_11.8W.
123 11.9 W.

29.2 W.
123 41.1 W.
- 45.7 W.
124 26.8 W.

_42.2 W.
125 09 o W.

_11.1 E.

124 57.9 W.
+ 21.9 W.

125 19.8W.
+ .7W.

125 20.5 W. - 11.9 E.

125 08.6 W.

+

The result is two noon positions, the first without considering the current, which in the other is taken into account.

The current could be considered on each course, but such a degree of precision is seldom necessary.

The ship's position differs slightly from that found by the preceding method, but is quite near enough to meet the demands of practice.

Lat: ship

Lat. Far. L. — 37° 41'.8 N. Long. Far. L. — 123° 00'. i W.

40 44 .3 N. Long. ship 125 08.6 W. Dif. 182'.5= 3 02.5 N. Dif. 128.5 = 2 08.5 W. Mid. lat. 39o dif. long. 128'.5= dep. 99'.9 Dif. lat. 182'.5 N. 1 Dep. 99.9 W. S FCO. N. 29 w. dist. 209 Lat. ship — 40° 44'.3 N. Mer. parts — 2665'.2 Lat. Cape F. 48 23.5 N. Mer. parts

3308.5 Dif. 459'.2= 7 39.2 N. Dif.

643 -3 N. Long. ship 125° 08'.6 W. Long. Cape F. — 124 44.1 W. Dif. 24'.5= 24.5 E. Dif. long.

24'.5 1.38917 Mer. dif. lat.

643-3—2.80841 Co. N. 2° 11' E.

tang. 8.58076 Co. N. 2° 11' E.

sec. .00032 Prop. dif. lat.

459'.2 + 2.66200 Dist.

459.5 = 2.66232

The last course and distance has been found by Mercator sailing, using logarithms; but while by middle latitude sailing a result nearly the same is obtained, the former method should be favored when the distance is great or near the limits of Table 2.