(Look out rather 7 points and 140 distance, the diff. lat. and dep. for which are 69 and 139.8; move the decimal points one place to the left,) and put down the result to the nearest tenth, which are 7 and 140. Insert them in the spaces left unmarked under N. and W. The second course being due W. 8.2', the departure will be 8.2 (the same as the distance). With third course 6 points and distance 9·9 (looking for 99, and making the proper change in decimal points) the diff. lat. is 3.8' and dep. 9.2'. In a similar manner find difference latitude and departure for the other courses. When the four columns are added up, it appears that the ship has sailed N. 42·7' and S. 7·8'; therefore upon the whole the true difference latitude is 34.9' N.; and her departure has been 14.5' E. and 84.6′ W.; hence the departure made good in the 24 hours is 701′ W. (5.) To find the latitude in, apply the true difference latitude to the latitude from, in the usual manner, to obtain the latitude in. (Rule e.) (6.) To find the longitude in.* With the latitude from and latitude in, find middle latitude. Add together log. secant mid. lat. and log. departure; the result (rejecting 10 in index) is the log. difference longitude, which, found in If, therefore, the traverse table is entered with complement of mid. lat. as a course, and with the given departure, the distance corresponding thereto will be the difference of longitude nearly. the tables, and applied to the longitude from, gives the longitude in. Thus: To find course and distance on Mercator's chart between two known places. To find course. Apply the edge of a ruler to the two places, and then ascertain at what degree a straight line (as the edge of the same or another ruler) placed parallel thereto, and passing through the centre of some adjacent compass, cuts the circumference. This will indicate the bearing or course required. To find distance. This may, in general, be found by applying the distance on the chart to the side of chart, so that the chart distance may be so placed that the middle point may coincide with the middle parallel between the two places; then the degrees of lat. covered by the chart distance will be the distance nearly. More concise methods for solving many of the preceding problems might have been given, by employing the traverse table, &c. But these will suggest themselves to the learner after he has had some practical experience in nautical matters. The following examples formed part of examination papers given at the Royal Naval College, at the monthly examination, in navigation, of lieutenants and assistant-masters in her Majesty's Royal Navy. 113. Required the course and distance from A to B. 115. June 20, 1852, at noon, a point of land in latitude 47° 12′ N. and longitude by account 3° 10′ W. bore by compass E.N.E., the ship's head being South by compass, distant 17 miles (variation of the compass 3 W.); afterwards sailed as by the following log account: required the lat. and long. in, on June 21, at noon. 116. Required the course and distance from A to B. Variation of compass 2 W. For local dev. see p. 16. 118. March 5, 1852, at noon, a point of land in latitude 57° 12′ N. and longitude by account 75° 34′ W. bore by compass E.N.E. (ship's head being N. by compass), distant 18 miles (variation of the compass 1 W.); afterwards sailed as by the following log account; required the lat. and long. in, on March 6, at noon. 119. Required the course and distance from A to B. 121. June 1, 1852, in longitude 18° 28′ E., and latitude 34° 28′ S., a point of land bore N.W. (ship's head N. by compass), distant 10 miles (variation of the compass 24 W.); afterwards sailed as per log: required the latitude and longitude in, on June 2. |