II. From the above Table construct a Napier's curve, and give the courses you would steer by standard compass to make the following courses correct magnetic: 1. N.N.W. W. Answer:-1. N. 51° W. 2. E. by S. S. 3. S.W. W. 4. E. N. 3. S. 271 W. 4. S. 64° E. III. Suppose you have steered the following courses by standard compass; find the correct magnetic courses from the curve drawn :-- IV. You have taken the following bearings of distant objects by your standard compass as above; with the ship's head as given, find the correct magnetic bearings : I. From the following Table, find the correct magnetic bearing of the distant object, and thence the deviation: Answer:-Correct magnetic bearing, S. 40° E. Answer.-Deviations:-1. 4° E. 2. 10° E. 3. 18' E. 4. 19° E. 6. 24° W. 7. 19° W. 8. 6° W.、 5. 2° W. II. From the above table construct a Napier's curve, and give the courses you would steer by standard compass to make the following courses correct magnetic . 1. E. S. 2. Answer:-1. N. 82° E. 2. S. E. 3. W. by S. S. 4. N. W. III. Suppose you steer the following courses by standard compass, find the correct magnetic courses from the curve drawn: 1. S.W. by W. W. 2. N. 5° W. 3. S. 8° E. 4. N. 80 E. Answer:-1. S. 40° W. 2. N. W. 3. S. 5° E. 4. S. 83 E. IV. You have taken the following bearings of distant objects by your standard compass as above; with the ship's head as given, find the correct magnetic bearings. EXAMPLE X. I. From the following table find the correct magnetic bearings of the distant object, and thence the deviation. Answer:-Correct magnetic bearing S. 88° W. Answer.-Deviation:- 1. 2° W. 2. 10° E. 3. 11° E. 4. 6° E. 5. 2° E. II. Given the correct magnetic course, to find the courses to steer by standard compass :— 1. S. 87° E. 2. S. 83° W. 3. N. 8° W. 4. N. 57° E. N. 851° W. 3. Answer.-1. N. 811° E. 2. III. Given courses by standard compass, to find correct magnetic : N. 4° W. 4. N. 47° E. N. 86° W. 3. N. 63° E. 4. S. 9° W. IV. You have taken the following bearings of distant objects by standard compass with ship's head as given below, find the correct magnetic bearing: Correct magnetic bearings:-1. N. 77° E. 2. S. 8 E. 3. S. 81° E. 4. S. 88° W. EXAMPLE XI. I. From the following Table find the correct magnetic bearing of the distant object, and thence the deviation: II. From the above Table construct a Napier's curve and give the courses you would steer by standard compass to make the following courses correct magnetic:— III. Suppose you steer the following courses by standard magnetic courses from the curve drawn: compass, find the correct IV. You have taken the following bearings of distant objects by your standard compass as above; with the ship's head as given find the correct magnetic bearing:— I. From the following Table find the correct magnetic bearing of the distant object, and thence the deviation: Answer :—Deviations:-1. 1° W. 2. 3° E. 3. 11° E. 4. 14° E. 5. 3° E. 6. 11° W. 7. 13° W. 8. 6° W. II. Given the correct magnetic courses, to find the courses to steer by standard compass:1. N. 41° W. 2. N. 66° E. Answer:-1. N. 361° W. 2. N. 60° E. 3. S. 39° E. 4. S. 79° W. III. Suppose you steer the following courses by standard compass, find the correct IV. You have taken the following bearing of distant objects by standard compass; with the ship's head as given below, find the correct magnetic bearings: Answer:-Deviations:-1. 6° E.; 2. 14° W.; and correct magnetic bearings, 1. N. 68° E.; 2. N. 64° W. 243. 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 4 points or 45°; but at the bottom of the Table, when it exceeds 4 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 4 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 or Raper.) 244. USE OF THE TABLE. Given the course and distance, to find the difference of latitude and departure. RULE LXVI. 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. Obs. When the course is found at the bottom of the page, care must be taken to see that the diff. of lat. and the dep. are taken from the proper column above the words departure and diff. lat. It must be carefully remembered that when the course is less than 4 points or 45°, the diff. lat. exceeds the dep.; but when it is more than 4 points or 45° the dep, exceeds the diff. lat. EXAMPLES. Ex. I. 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. 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. Inasmuch as the sine of an angle is the cosine of its complement, 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 the top and at the bottom of the page, and using the top reading for courses from 0° 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. 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 in the adjacent column (marked at the bottom dep. and diff. lat.) dep. 86°4, and diff. lat. 462, 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. at top, 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 252.6, 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 32'1 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 767 over Lat., and 157.3 over Dep., which was required. (a) To find diff. lat. and dep. when there are tenths in the distance. 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. however, must be increased by 1, if the figure cut off is 5, or upwards. Ex 1. thereto. EXAMPLES. The tenths, Course 3 points, distance 2013; required the diff. lat. and dep. corresponding With course 3 points, and dist. 2013, taken as 203, we get the diff. of lat. 156'9, dep. 128.8; 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. 15'7, and dep. 12'9. It will be observed that the tenths are increased by 1, in each case, as the figures cut off in both cases exceeds 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, 15'4, 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. A ship sails E.N.E., distance 29'5; find diff. of lat. and dep. corresponding. 295 1129 diff. lat., 272'5 dep. .'. 29'5 = 11°3 diff. lat., 27''3 dep. After dropping the tenths, and removing the decimal point one place to the left, we have diff. lat. 113 and dep. 27.3. The reason of this rule is that the Traverse Table is entered with a distance ten times as great as the given distance, and the resulting diff. lat. and dep. is divided by ten. This is done by merely imagining the decimal point to be removed one place to the right before entering the table, and then one place to the left after taking out the results (see p. 26, (10). |