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C lay out the first course S W by S and the distance 24 miles according to an adopted scale; then from a lay off the second course N N W, 57 miles, thence SE by E } E, 84 miles, and, finally, the last course south, 35 miles. The point b is then the place at which the ship has arrived. Connect b with C, also draw db perpendicular to NS; Cd is then the difference of latitude, d b the departure. Cb the distance, and the angle

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dCb the course made good by the ship. By measuring the different parts of the triangle thus formed they will be found to agree exactly with the results obtained by the Traverse Tables.

EXAMPLE. - A steamer leaves Grand Marais Harbor, latitude 46° 41' N and longitude 85° 58' W, and runs the following courses and distances: NNE 48 miles, WNW 25 miles, S W 36 miles. Find the course and distance made good and the latitude and longitude of the place arrived at.

SOLUTION. – Arrange a traverse similar to the one in the preceding illustration, and take out from the tables the differences of latitude and departures corresponding to each course and distance. Thus,

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D. Lat. made good 28.4' N Dep. made good = 30.2 W Entering the Traverse Tables with the difference of latitude and departure thus found, the corresponding course made good is found to be N 47° W and the distance 41 miles. To find the latitude in we apply the difference of latitude to the latitude left; thus, Lat. left (Grand Marais Harbor) = 46° 41' N

D. Lat. 28.4' N

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For the longitude in we find first the middle latitude; thus,

Lat. left 46° 41' N
Lat. in 47° 9.4' N

Sum 93° 50.4

sum M. Lat. 46° 55.71 and thence the difference of longitude by entering the Traverse Tables with the middle latitude (47°) and the departure (30.2). The difference of longitude (44') thus found is then applied to the longitude left as usual. Thus, Long. left (Grand Marais Harbor) 85° 58' W

D. Long.

44' W whence,

Long. in

86° 42' W. Ans. .

65. In the preceding example, had it been required to find the course and distance from the place arrived at to some other place, for example, the eastern extremity of the Manitou Island, latitude 47° 25' N and longitude 87° 35' W, we would proceed as follows:

Lat. arrived at = 47° 9.4' N

Long. arrived at = 86° 42' W Lat. Manitou Island = 47° 25 N Long. Manitou Island 87° 35' W D. Lat. = 15.6' N

D. Long 53' W Sum of Lats. 94° 34.4 sum = M. Lat. 47° 17.2

Resulting Dep. 36.1' W The course and distance corresponding to a difference of latitude of 15.6' and a departure of 36.8' is 67° and 40 miles, respectively. Hence, the course from the place arrived at to Manitou Island is N 67° W and the distance 40 miles.

It must be borne in mind that the course thus found is true and must be corrected for deviation, variation, and leeway, if necessary.

66. It is evident that the course between the two places may have been found more quickly by taking it directly from the chart. This, of course, is true in cases where the distance is small and the scale of the chart is large, but when the distance is considerable and the scale of the chart is small, much inaccuracy may result, especially if the chart is constructed on any other projection than that of Mercator's. In such cases, therefore, the Traverse Tables should be used in connection with the middle-latitude method for finding the

courses.

This process of finding the true course, and thence, by the application of variation, deviation, and leeway, the course to steer, or compass course, is commonly known as shaping the course.

EXAMPLE. – A steamer leaves Buffalo, N. Y., for a western port. The weather is misty, reducing the distance of the visible horizon to about 2 miles, and weather indications point to a stiff breeze from the southwest with rain. As soon as the breakwater at Buffalo, latitude 42° 52.7' N and longitude 78° 54' W, is passed, the patent log is put out. From the breakwater, a true course S W by W is kept; when 35 miles on that course has been covered the steamer is forced, by some accident, to run off N W by W } W 13 miles. At this point the course is again changed to SSW I W, on which she runs a distance of 22 miles. The course is then changed to S W by WW. Find the latitude and longitude in after a distance of 23 miles has been covered on the last course.

SOLUTION. - When the courses and distances run, and the corresponding differences of latitude and departures are properly inserted in their respective columns of a traverse, the table will appear thus:

TRAVERSE TABLE

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Lat. in 42° 9.7' N. Ans.
Sum of Lats. 84° 62.4

M. Lat. 42° 31.2 In this case the middle latitude is nearly 42°; hence, we take out the difference of longitude for both 42 and 43°; for the former, with the nearest departure (70.6), it is 95'; and for the latter, with the nearest departure (70.9), it is 97'. Therefore, the difference of longi

95 + 97 tude corresponding to 421° must be

- 96'; this we apply to

2 the longitude left. Thus,

Long. left 78° 54' W

D. Long. (96') 1° 36' W whence,

Long. in

80° 30' W. Ans. 67. Courses Should be True. – Hitherto, in the examples given, the courses have been true; in actual practice, however, the courses are compass courses and must be corrected for leeway, deviation, and variation. In cases where the distances run are small and the variation the saine for all courses, the magnetic courses, corrected for leeway, may be used in entering the Traverse Tables and the result will be the magnetic course made good, but it must then be corrected for variation. To avoid mistakes, however, each course should be reduced to true before it is entered in the traverse. The student should make this an invariable rule.

68. Taking the Departure. – At the commencement of a voyage, before the ship loses sight of land, her position is always determined, according to some of the methods already described, by the bearing and distance of some object (headland, lighthouse, lightship, etc.), the latitude and longitude of which is known. This operation is usually known as taking the departure.

69. The Departure Course. – When working a traverse, this bearing is reversed and entered, together with the distance, in the traverse table and is treated like the rest of the courses, the deviation applied being that due to the direction of the ship's head at the moment of taking the departure. Thus, if the bearing of an object is N E by E and the distance is 10 miles, it is entered in the traverse as S W by W distance 10 miles. The bearing thus reversed and entered in the traverse is usually known as the departure course. In the following example the departure course will be introduced:

EXAMPLE 1.-A schooner leaves Michigan City, Ind., bound for Ludington, Mich., latitude 43° 57.5' N and longitude 86° 28' W. When the light at the east-side entrance to Michigan City Harbor, latitude 41° 43.5' N and longitude 86° 55' W, is about to disappear, its magnetic bearing is found to be SW, and its estimated distance 12 miles; the ship at the time of taking the departure is headed north. From the point where the bearing was taken the following compass courses and distances were sailed: N by W, 56 miles; N W N, 23 miles; N E E, 16 miles; N by W 1 W, 20 miles. The variation for all courses is į point E, and the deviation for the different positions of the ship's head is as follows:

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Find the latitude and longitude arrived at and the compass course and distance from that point to the place of destination, assuming the variation to be į point E and the deviation point W.

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