SOLUTION. - First reduce each compass course to true by applying corrections for variation and deviation. First Course Compass course = Or Thus, Second Course N by W Deviation = Magnetic course = Npoint W Variation = point E True course = North Deviation Variation = The true courses, together with the reversed bearing, are then entered in a traverse; thus, plotted the courses By actual measure On the chart of Lake Michigan, Fig. 26, are and distances recorded in the preceding example. ment on that chart, according to instructions given in Lake Navigation, Part 2, the student will find that the result thus obtained will agree exactly with that obtained by calculation. EXAMPLE 2.-A steamer coming from St. Mary's River, is bound for Eagle Harbor, Mich., latitude 47° 27.6' N and longitude 88° 9′ W. After passing White Fish Point, latitude 46° 46.3′ N and longitude 84° 57' W, the light situated at that point is watched for a departure, and just before it goes out of sight its bearing by compass is found to be S 56° E. The range of visibility of the light is 16 statute, or nearly 14 nautical, miles. The deviation of the point on which the ship was headed when the bearing was taken is 12° E, and the variation is 1° W. When the light disappeared from sight, the steamer is put on a compass course N 75° W; the deviation on that point is 15° E and a strong northwesterly wind produces a leeway estimated at point. After a distance of 53 miles has been run on that course, an accident to the machinery compels the ship to be hove-to on the starboard tack and headed S 66° W by compass. On this course a distance of 12 miles is covered, the leeway being 14 points, variation 1o E, and the deviation 8° W. After repairs have been effected, the steamer is headed N 54° W by compass, the deviation on that point being 10° E, and the variation 2o E. Find the position of the steamer after a distance of 43 miles has been run on the last course, and find the true course and distance from that point to the port of destination. SOLUTION. As before, we reduce the bearing and compass courses to true; thus, Bearing First Course N 75° W Compass bearing S 56° E Deviation = 12° E = = S 44° E 1° W Compass course = = 8° = N 83° W = Variation True course = S 45° W The data in this case, together with the traverse, may conveniently be arranged in the following tabular form: For the true course and distance to the port of destination, we proceed as follows: The nearest degree corresponding to the given D. Lat. and Dep. is 70°. Hence, true course is S 70° W, and distance is 37 mi. Ans. EXAMPLES FOR PRACTICE 1. A steamer, after passing the breakwater at Buffalo, N. Y., latitude 42° 52.7' N and longitude 78° 54′ W, steers a true course S W by W1⁄2 W. Find the latitude and longitude in after a distance of 48 miles has been covered Ans. = Lat. in 42° 30.1' N. 2. Find the true course and distance from Grand Haven, Mich., latitude 43° 3.2 N and longitude 86° 15' W, to Sheboygan, Wis., Lat. 43° 45′ N, Long. 87° 41.2′ W. Ans. True course = N 56° W, or N W by W. 3. The true bearing of the lighthouse on the north end of Devil's Island, Wis., latitude 47° 4.8' N and longitude 90° 43.2' W, is S W, distance 10 miles. From the point where the bearing was taken, the following true courses and distances were run: East 49 miles, and N 60° E 18 miles. Find (a) the ship's position and (b) the true bearing and distance from that position to the lighthouse at Fourteen Miles Point, latitude 46° 59′ N and longitude 89° 7.2′ W. 4. What compass course should be steered from Marquette breakwater, latitude 46° 32′ N and longitude 87° 22.7 W, to the lighthouse on Caribou Island, latitude 47° 20′ N and longitude 85° 49′ W, assuming the mean value of the variation to be 2° E and the deviation 16° W? How long would it take a steamer making a speed of 10 knots per hour to cover the distance between the two places? |