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To find the Side A C or b.
c sin. B
.log. 1•262226 In the plane triangle A B D are given the side A B 14 miles; the angle A D B equal to 29° (the angle between S. 20° E. and S. 9° W.); and the angle A B D equal to 126° (the angle between N. 9° E. or E. 81° N. and S. 45° E.); to find the side A D.
To find the Side A D or b'.
c sin. B
log. c + L sin. B — L sin. D Side c 14 m.
..log. I•146128 < B 126°
11.054086 < D 29°
...sin. 9•585571 Side A D 23:36
..log. 1•368515 In the plane triangle A C D are given the side A C 18.29; the side A D 23:36; and the included angle C A D equal to 49° (the angle between N. 29° E. and N. 20° W.); to find the angle A CD, and the side C D. Side A D or c 23:36
180° Side A Cord 18.29
sum of _* 65°30' = } (C+D) For the Angle ACD.
For Side C D osv. (log. (c-d) + L tan.
c sin. A L tan. 3 (C-D) =
.: CD = T/C + D) — log. (c + a)
log. CD= log. c + L sin. A-L sin. C C-d 5:07
Log. 0.705008 Side C 23:36 Log. 16368473 } (C+D)65° 30' Tan. 10°341296 Ľ A 49°
Sin. 9•877780 11.046304
II.240253 c+ d 41.65 Log. 1.619615
LC 80° 27' Sin. 9.993939 } (C—D) 14° 57' Tan. 9-426689
Side C D 17.88 Log. I'252314 1 (C+D) 65 30 Sum = LACD 80 27
Sum because _ A C D is opposite the greater side.
Now the angle A C D 80° 27' added to 29°, the bearing of A C from the south, gives the bearing of C D = S. 109° 27' W., which, subtracted from 180°, leaves the bearing N. 70° 33' W. or W.N.W. 1 W. nearly : hence the bearing of D from C is W.N.W. 1 W., and the distance 18 miles nearly.
TAKING A DEPARTURE BY A SINGLE BEARING AND DISTANCE
In this case the point of land or other object is set by compass, and its distance is estimated by the eye.
This is the general method of taking a departure, and it is sufficiently accurate when the ship is leaving the land, bound on an oversea voyage. But since distances are almost always over-estimated, probably to the extent of a fifth or more of the whole, this method should never be relied upon when coasting.
TAKING A DEPARTURE (AND DETERMINING THE DISTANCE) BY TWO
BEARINGS OF THE SAME OBJECT In this case the ship's course, supposed to continue unaltered between the two times of observation, lies more or less across the line of direction of the object.
Take the bearing of the object, and note the number of points contained between it and the ship's head (or course) : after the bearing of the object has altered not less than two or three points, again take the bearing of the object, and note the number of points contained between it and the ship’s head. Each of these is a “ difference” between the course and bearing at the instant of observation.
(A) To find the distance when the second bearing was observed Enter Table* (p. 322) with the first difference at the side, and the second difference at the top, and take out the number corresponding thereto, as a multiplier : reserence to the Table sufficiently explains the mode of entry. Multiply the number got from the Table by the number of miles run in the interval of time between taking the two observations, and the product is the distance (in miles) of the ship from the object at the time the second bearing was taken.
Norie's Tables now contain a new table," Distance off by two Bearings and Distance run between them.” It is entered with the angle between the ship's course and the first bearing in degrees at the top, and the angle between the first and second bearings in degrees at the side. Under the former, and opposite the latter, take out the number corresponding to any two arguments, and this number multiplied by the distance the ship has sailed between the first and second bearings will be the distance off at the time the second bearing was taken.
* CONSTRUCTION OF THE TAPLE: To the L sine of the difference between the course and first bearing, add the L co-secant of the difference between the first and second bearings. Their sum is the logarithm of a natural number, which is the multiplier in the Table.
TABLE FOR FINDING THE DISTANCE OF AN OBJECT BY TWO BEARINGS AND THE
1.56 1.39 1.26 1.16 1.07 1.00 0.94 0.89 0.84 0.80 0.77 0.740.720.69 0.68 0.66 0.64 0.63 0.62 0.61 0.61 0.600.600.600.600.600.600.60 0.61 0.61 0.62'0-63
2.60 2.33 2.111.93 1.79 1.67 1.571.48 1.411.34|1.29 1.241.20 1.16 1.13
12-612-34 2.121.941.80 1.681.571.49 1.41|1:35 1.29 1.241.20 1.16
12.61 2.34 2.121.94|1.80 1.68 1.58 1.49|1.411.351.29 1.24 1.20 8
2.61 2.34 2.121.941.80 1.68 1:57|1.49 1.411.35 1.29 1.24
2.60 2:33 2.111.941.79 1.67|1.571.481.411.34 1.29
2:36 2 111.921.76
DISTANCE RUN BETWEEN THEM.
Difference between the Course and Second Bearing in Points of the Compass.
41 5 +
61 17 3 1 3 8
9 1 है 1 10
111 f 112 11
Difference between the course and First Bouring in Points of the Compas.
Example.- The Eddystone bore N.W. by N.; after running W.S.W. 18 miles, it bore N. by E. Find the distance of the ship from the Eddystone at the time when the last bearing was taken.
Difference between N.W. by N. and W.S.W. is 7 points ; difference between W.S.W. and N. by E. is II points. Opposite 7 at the side, and under 11 at the top, stands 1:39, which multiplied by 18 (miles) gives 25 miles, the distance of the ship from the Eddystone when the last bearing was taken (see Example 1, p. 316).
(B) To find the distance when the FIRST bearing was observed Take the supplement of each difference, that is subtract each from 16 points. Then, enter Table p. 322 with the supplement of the second difference at the side, and the supplement of the first difference at the top; take out the number corresponding thereto, and multiply it by the distance run ; the product is the distance (in miles) of the ship from the object at the time the first bearing was taken.
Example. Find the distance of the ship from the Eddystone at the time the first bearing was taken, the elements being as in the previous Example.
First difference being 7 points, the supplement is 9 points; the second difference being 11 points, the supplement is 5 points. The distance run is as before, 18 miles. Then, 9 at the top, and 5 at the side, gives 1.18, which multiplied by 18 gives 21.24 miles as the distance of ship from the Eddystone at the time when the first bearing was taken (see Example 1, p. 316).
N.B.- In iron ships and steamers the bearings are affected to the extent of the deviation due to the direction of the ship's head at the time of observing the bearings; and thus the deviation must be applied as well as the variation of the compass, for true readings; only the deviation for magnetic readings.
General Rule for computing the distance of the ship from the object at the tinie of taking the second bearing.—Note the number of points (or degrees) contained between the first bearing of the object and the course of the ship; when the bearing of the object has altered not less than 2 or 3 points, note the distance run. Then,
Add together the logarithm of the distance run, the L sine of the angle between the first bearing and the course, and the L co-secant of the angle between the two bearings; the sum (rejecting tens in the index) will be the logarithm of the distance from the object when the last bearing was taken.
E ample 1.-Cabrera Island Light bore N.N.W. at 2 a.m.; and at 4 a.m. it bore N.E. by N.; in the interval of the observations the course was west, and the distance run 12 miles. Find the distance of the Lighthouse when the second bearing was taken.
Angle between the bearings (N.N.W. and N.E. by N.) = 5 points.
Then, dist. run 12 m.
5 points Dist. of Cabrera Is. Light 13'3 m.
Example 2.-At noon, Ascension Island bore S. 80° E.; and at 5 p.m. it bore S. 46° E.; in the interval of the observations the course was N.N.W., and the distance run 19 miles; find the distance of the island when the second bearing was taken.
Angle between the bearings (S. 80° E., and S. 46° E.) 34
Angle between first bearing and course (S. 80° E., and N. 22° 30' W.) 122° 30'.
RULE by Inspection and the Traverse Table.—When the angle between the first bearing and the course is more or less than 8 points, the method by Inspection from the Traverse Table is as follows :
Enter Traverse Table with the angle between the first bearing and the course, as a course; and the distance run (in distance column); take out the departure.
N.B.-If the angle is more than 8 points (or 90°) take its supplement; i.e. subtract it from 16 points (or 180°), and enter the Traverse Table with this difference as a course.
Then, enter Traverse Table again, with the difference of bearings as a course, and the departure (just found) in departure column ; the distance of the object at the time of taking the second bearing will be found in the distance columnn.
Angle between ist bearing in Trav. Dep. II'I in Example 1 (p. 323).- and course 6 points.
Dist. 12 m., in dist. col. 5 Tab.give] dep. col. Diff. of bearings, 5 points. in Trav. Dist. 13.5 m.
Dep. II'I in dep. col. Tab. give) in dist. col.
By Table p. 322: Difference between course and first bearing is 6 points; difference between course and second bearing is ii poirs; hence I'II X 2 =
Exainple 2 (p. 324). —
Angle between ist bearing)
and course, 1221o. (Take in Trav. Dep. 16 in 57.1°, the supplement.) Tab. givel dep. col. Dist. 19 m., in dist. col.
Diff. of bearings, 34° in Trav. Dist. 28.5 m.
If the angle between the first bearing and the course be exactly 8 points (or 90°) the calculation is shorter than by the preceding Rule p. 323 ; thus