6. N.W. by W. W.5 L. of N. 61° 41' L. of N. N.W. by W. Deviation = 10 2 L. W.N.W N. 71 43 W. Change for 1 pt. = 9° 18′ W. 10 16 W. 58 4)58 2)3 32 Dev. on pt. 1 46 W. Correct magnetic course N. 7 24 W. Half the sum of the deviations for N. and N. by W. is taken for deviation on N. W.; both deviations being of the same name. EXAMPLES FOR PRACTICE. Correct the following courses steered, as given in the Table, page 89. 1. N.E. by N. 2. North 3. N. by W. W. 10. 6. S.S.W. W. 7. N.N.E. E. 17. W. S. 14. N.W. W. 21. S. by E. 23. N.W. 27. East The following Examples are designed to show the method of correcting courses for leeway, variation, and deviation. Ex. 1. Course steered E.N.E., wind S.E.; leeway 24 points; variation 1 W.; and the deviation 2 points E.: required the true course. Here the course by compass is E.N.E. or ...... 6 points right of North. their difference, viz. (2 E.-1 W.= = 1 E.) ...... +1 points right of North. Ex. 2. Course by compass N.N.W., wind N.E.; leeway 2 points; variation 4 points W.; deviation 1 point E.: find true course. Here the ship's course is N.N.W. or 2 points left of North. - 1 E.) +2 left. Therefore the sum is course corrected for variation and deviation The ship being on the starboard tack the leeway is applied to the left, and hence is 4 points left of North. 2 left. Sum 7 points left of North. True Course W. by N. Ex. 3. Course steered S.W. by S., wind W. by N.; leeway 2 points; variation 3 points E.; deviation 1 point E.: required the true course. Ex. 4. Course by compass W. by S.; variation 3 E.; deviation 13° 50′ W.; wind S. by W.; leeway 1 points: required the true course. Since no course can exceed 90° from either N. or S., when the course as in this example exceeds 90°, the result must be taken from 180°, and its name changed. (See Rule XXXIX, page 86.) Subjoined are a few Examples for Exercise in the above problem. From the following Compass Courses find the True Courses : 22 21 2 3233-22 E. E. I 11. S.W. by W. S. by E. 2112 2 E. W. 28° 17° E. 7° E. 16. E. S. N.N.E.E. 2 pts. 21° W. 4° W. 3 pts. E. 5° E. 14° W. 10° E. 12° W. 2 pts. THE TRAVERSE TABLE. In all collections of tables for the use of navigators there is inserted a table containing the true difference of latitude and departure, corresponding to certain distances (at intervals of one mile) up to 300 nautical miles, for every course, at intervals of a quarter point, and also of degrees, from o° to a right angle (90°). Tables I and II (Raper or Norie). In these Tables the course is found at the top of the Table, when under four points or 45°; but at the bottom of the Table, when it exceeds four points or 45°. The first column contains the distance to 60 miles, the second column contains the difference of latitude, expressed in minutes and tenths, and the third column, similarly expressed, contains the departure; but if the course exceeds four points or 45°, the second column contains the departure, and the third column the difference of latitude. The other columns are a continuation of the former, exactly upon the same principle, and extending to 300 miles of distance.* (See Tables I and II, Norie and Raper.) * This table is constructed by solving a right-angled triangle, of which one angle represents the course, and the hypothenuse the distance; by giving these different and successive values, the corresponding values of the other two sides are found, which sides represent the true difference of latitude and departure. It is evident that the difference of latitude and departure for any course, are the departure and difference of latitude for the complement of that course, and hence the table is compactly arranged by interchanging the headings of the columns containing these elements at USE OF THE TABLE. Given the course and distance, to find the difference of latitude and departure. RULE XLII. With the course open the Tables, and under or above the proper number of points (or degrees) and opposite the distance, will be found the difference of latitude and departure. When the course is found at the bottom of the page, care must be taken to see that the diff, of lat. and dep. are taken from the proper column above the words departure and diff. lat. EXAMPLES. Ex. J. A ship sails N.W. N. a distance of 78 miles: required the difference of latitude and departure by inspection. The given course is 3 points; and referring to Table I, we find the page assigned to this course to be page 14, Norie, or page 436, Raper's Navigation, in which against 78, in column headed Dist., stands 60·3 under the head Lat., and 49'5 under the head Dep. We conclude, therefore, that for the given course and distance, the difference of latitude is 60 3 miles, and the departure 49'5 miles. Ex. 2. Suppose the course to be 5 points, and the distance 98 miles. Then, since the course here exceeds four points, we look for it at the foot of the page (page 10, Norie, or 432, Raper), and against 98 in the distance column we find 864 in the adjacent departure column, and 46'2 in the difference of latitude column, so that the difference of latitude made is 46°2, and the departure 86.4. Ex. 3. Course N.E. by N., distance 129 miles: find diff. lat. and dep. Enter Table I, and find 3 points at the top, and in one of the columns marked Dist. find the distance 129, then in the columns opposite to this, marked lat. and dep., stands the difference of latitude 107 3, and departure 71*7. Ex. 4. Course E. by N. N., distance 264 miles: find diff. of lat. and dep. Open Table I at 6 points, found at the bottom, and opposite the distance 264 stands departure 2526, and difference of latitude 76·6. Ex. 5. A ship sails N. 40° E., 50 miles: required the diff. of lat. and departure. The course being less than 45°, is found at the top, and the distance being under 60 miles, is found in the left hand column; therefore, on the page (56 Norie) is 40° at the top, and opposite to 50 in the distance column (marked Dist.), is 38·3 under Lat., and 321 under Dep., the difference of latitude and departure required. Ex. 6. A ship sails N. 64' W., 175 miles; required the diff. of lat, and departure. The course being more than 45°, is found at the bottom, in page 42, and opposite to the distance 175 miles, is 76·7 over Lat., and 157.3 over Dep., which was required. the top and at the bottom of the page, and using the top reading for courses from 。° to 45°; and the bottom reading for courses from 45° to 90°. This table may be used for a great number of problems, depending for their solution on the relation of the several parts of a right-angled triangle, and, since all the relations between any two quantities may be expressed as functions of some angle, in terms of the sine, cosine, or tangent, it may be used, in fact, as a general proportional table. (a.) When there are tenths in the distance, in order to find diff. lat. and dep. in such case, take the distance as an entire number of miles, i.e., as a whole number, and find the corresponding diff. lat. and dep., from each of which cut off the right hand figure, or tenths, and remove the decimal point one place to the left hand, which will give the required diff. lat. and dep. in miles, and tenths of a mile. The tenths, however, must be increased by 1, if the figure cut off is 5, or upwards. Course 3 points, distance 2013; required the diff. lat. and dep, corresponding thereto. Ex. 1. With course 3 points, and dist. 2013, taken as 203, we get the diff. of lat. 156·9, dep. 1288; now cut off the right hand figure of each (the 9 and 8), and shifting the decimal point one place to the left, we have diff. lat. 157, and dep. 129. It will be observed that the tenths are increased by 1, in each case, as the figures cut off in one case exceeds 5, and in the other amounts to 5. Ex. 2. Required the diff. lat. and dep, corresponding to course 4 points, and dist. 24'3 miles. With course 4 points, and dist. 243 (as 243 miles), we find diff. lat. 154°2, and dep. 1878 hence we obtain, after dropping the tenths, and removing the decimal point in each one place to the left, 154, and 18.8, for the required quantities. The tenths in the dep., it will be observed, are increased by 1, since the figure dropped exceeds 5. Ex. 3. N. 3 pts. W., and dist. 20.6 miles, give diff. lat. 17.1 N., and dep. 114 W. Ex. 4. N. 65° E., and dist. 21.5 (as 215), give diff. lat. 90'9, and dep. 194'9, which is diff. lat. 9.1 N., and 19.5 E. It will be observed that the tenths are increased by 1, in each case, as the figure dropped exceeds 5. (b.) If the distance exceeds the limits of the Traverse Table, take the half, the third, &c., so as to bring it within the limits, taking care to multiply the corresponding quantities by 2, 3, &c. Ex. 5. Let the course be 3 points, and distance 435; required the corresponding diff. lat. and dep. 435 divided by 3 gives 145. Course 3 points, and dist. 145 give diff. lat. 116.5 and dep. 86.4 X 3 Diff. lat. 349'5 X 3 Dep. 259'2 If the distance had been 43'5, the diff. lat. would have been 350, and the dep. 25'9. (c.) But when the distance is between 300 and 600, we may take out diff. of lat. and dep. for 300, and for the excess of 300, take the sum of the quantities thus found, cut off the last figure and remove the decimal point as before. Ex. 6. Course 5 points, and distance 526: required the corresponding diff. of lat. and departure. Course 5 points, and dist. 300, give diff. lat. 128.3, and dep. 271.2 226 96.6 224'9 204'3 475'5 |