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2. A ship sails E. by N. 7.5 knots an hour, in a current setting S.W. 4 knots an hour. What will be her course and distance made good in 24 hours ?

Ans. The course will be S. 73° 12' E., or E.S.E. } E. nearly, and the distance 113.5 miles.

3. A ship sailing at the rate of 9 knots an hour, and wanting to double a cape bearing from her N.W. by W., finds she is in a current setting S.S.W. 3} miles an hour. What course must she steer to counteract the effect of tie current ?

Ans. Course to steer for the cape is N. 33° 50' W., or N.W. by N. nearly.

4. A ship sailing by her log 9 miles an hour is bound to a port which lies N.W. by N. from her, distant 56 miles, and finds she is in a current setting N.E. 1 N. 3 miles an hour. What course must she steer in the current, and distance make good ; and how long will it take her to arrive at her

port?

Ans. The course to be steered is N. 51° 21' W., or N.W. i W. nearly ; the distance to run is 53.61 miles; and the time it will take nearly 6 hours.

5. A ship bound from Bombay to England, being on the edge of the Bank of Agulhas on April 21st, at noon, was by observation in lat. 35° 3'S., and in long. by chronometer, 26° 52' E.; on the 22nd the latitude, by observation at noon, was 35° 13' S., and the longitude by chronometer, 25° 5' E., having sailed by her reckoning N. 81° W. 39 miles. Required the set and drift of the current.

Ans. The current set S. 71° 49' W., or W. 18° II' S., and its drift in 24 hours was 51:57 miles, being at the rate of 2.15 per hour.

6. A ship bound to a port bearing S.E. is in a current setting N. by W. Il miles per hour. Find the courses on which she must be laid to make good the course to the port, when her rate is respectively 6, 7, 8, and 9 knots per hour; and give the ship's hourly rate of approach to the port, with each rate of sailing.

Ans. (6) Course S. 37° 1' E. and 4.69 knots; (7) course S. 38° 10' E. and 5.7 knots; (8) course S. 39° 2' E. and 6.71 knots; (9) course S. 39° 41' E. and 7.72 knots.

NOTE.—Current sailing in practice is done on the chart in a correct and expeditious manner and the calculations given above are for the benefit of students who wish to learn the methods by calculation.

OBLIQUE SAILING AND TAKING THE

DEPARTURE.

In the navigation of the ship this is an important problem. In previous Editions of Norie's “ Epitome” the subject is discussed under the head of OBLIQUE SAILING, which is stated to be the application of obliqueangled plane triangles to various cases at sea, as in coasting along shore, approaching or leaving the land, surveying coasts or harbours, etc. ; but much may be done by inspection, as will be shown presently.

In this kind of sailing, to set an object means to observe what rhumb, or point of the compass, is directed to it; and the bearing of an object is the rhumb on which it is seen; also the bearing of one place from another is reckoned by the name of the rhumb passing through those two places : hence the bearing of two places from each other will be upon opposite points of the compass; thus, if one place bears E.N.E. from another, the latter will bear W.S.W. from the former, being in the same line, but in opposite directions.

A great variety of examples might be given in this sailing; hence, in this place, it is sufficient to select those only that appear to be useful in practice, and advantageous to the student who wishes to understand the problem on the basis of the solution of plane triangles.

Example 1.–Sailing down the English Channel, I observed the Eddystone to bear N.W. by N.; and after sailing W.S.W. 18 miles, I found it bore from me N. by E. Required the distance of the ship from the Eddystone at both stations.

BY CONSTRUCTION

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Describe the circle N.W.S.E., to represent the compass, and draw the diameters W.E. and N.S. at right angles to each other; draw the N.W. by N., W.S.W., and N. by E. rhumb lines, and on the W.S.W. line lay off 18 from A to B, taken from any scale of equal parts; through B draw B C parallel to the N. by E. line, meeting the N.W. by N. line A C in C; then will A represent the place of the ship at her first station, B her place at the second station, and C the place of the Eddystone; A C will le the ship's distance from the Eddystone at the first station, measuring 21 miles, and B C the distance at the second station, measuring 25 miles.

W

5 pts

W.S.W: 18 m

BY CALCULATION

In the plane triangle A B C are given the side A B 18 miles, the angle C A B equal to 7 points (being the angle contained between N.W. by N. and W.S.W.); the angle A B C equal to 5 points (being the angle contained between N. by E. and E.N.E. the opposite to the W.S.W. rhumb); and the angle B C A equal to 4 points (the angle between S. by W. and S.E. by S.); to find the sides A C and B C.

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Hence the distance of the Eddystone from the ship's first station is 21:17 miles, and from the second station 24:97, or 25 miles nearly.

See another form under Example II.

Example 11.Coasting along shore, I observed two capes: the first bore N. by E., and at the same time the second bore N.E. I E.; now, by the Chart, these capes bear from each other N.W. W., and S.E. | E., distant 21 miles. Required my distance from both places at that time.

NOTE. - The bearings must all be considered to be true.

BY CONSTRUCTION

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Having drawn the compass N.W.S.E., the centre of which is to represent the ship’s place, draw the N. by E. and N.E. E. rhumb lines A B and AC, being the bearings of the capes from the ship; draw likewise the N.W. I w. and S.E. I E. line, the bearing of the capes from each other, on which from A to D lay off 21 miles, the distance between the capes ; through D draw D C parallel to the N. by E. line, and through C draw C B parallel to the N.W. 1 W. and S.E. 1 E. line; then B will represent the place of the first cape, C the second cape, A B the distance of the first cape from the ship, measuring 31 miles, and A C the distance of the second cape, measuring

2nd Cape

N.EE

33 pes

27 miles.

BY CALCULATION

in the plane triangle A B C are given the angle B A C 38 points (the angle between N. by E. and N.E. E.); the angle A B C 54 points (the angle between S. by W. and S.E. | E.); and the angle A B C 7 points (the angle between N.W. 1 W. and S.W. W.); and the side B C 21 miles; to find the sides A B and A C.

с

C =

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a

a

To find the Side A B or c.

To find the Side A C or b.
sin. C
a sin. C
6 sin. B

a sin. B
sin. A

sin.
A

sin. A

sin. A log. c = log. a + L. sin. C

L sin. A log. b = log. a + L sin. B L sin. A a or side B C 21 m. Log. 1.322219 a or side B C 21 m. Log. 1322219 2 C 7 pts. Sin. 9.991574

ZB 51 pts.

Sin. 9.933350
II 313793

II.255569
L A 3 pts.
Sin. 9.827084 L A 3 pts.

Sin. 9.827084 b or side A C 30.67 m. Log. 1.486709 b or side A C 26-82m. Log. 1.428485

Hence the distance of the ship from the first cape is 30.67 miles, and from the second cape 26-82, or 27 miles nearly. See another form under Example I.

Example III.-Being close in with a point of land at A, I ran 27 miles on a direct course to the westward, and then found the point of land at B to bear N.N.W.; now the bearing of B from A (by Chart) is W. | N., and the distance 29 miles. Required the course steered, and the distance of the ship from B.

PY CONSTRUCTION

N

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Describe a circle, and divide it into 4 equal parts by the diameters N.S. and W.E., the extremities of which will represent the cardinal points of the compass; and the centre of the 8 W., N. 29.9.

E circle the place the ship sailed á from (A); draw the W. 1 N. line A B equal to 29 miles, then will B represent the place of the point B; through B draw BC parallel to the N.N.W. line, and with the distance run, 27 miles in the compasses, set one foot in A, and with the other describe an arc cutting B C in C, and draw the line A C; then will C represent the ship's place, B C the distance of the ship from the point B, measuring 19 miles, and the angle S A C the course steered from the south, measuring 533°.

N.N. W.

27 m.

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BY CALCULATION

In the plane triangle A B C are given the side A B, equal to 29 miles ; the side A C 27 miles; and the angle A B C 54 points (the angle between E. 1S. and S.S.E.); to find the angle B A C, and the side B C.

The angle being given opposite the less side, this is the ambiguous case as shown in figure.

a

a

168 32

a

.. a =

ölno

a' =

To find the Angle A
sin. A
sin. A

x sin. B
sin. B b

b L sin. A

log. a + L sin. B – log. b c or side A B 29 log. 1.462398 B (51 pts.) 64° 41' 64° 41' LB 52 points sin. 9956163

2 с

76 9

103 51 II'418561

Sum

140 50 b or side A C 27 log. 1.431364

180

180 ZC 76° 9'

sin. 9.987197 LA ...... 39 10 or A' 11 28 180 Or Z C' 103 51 To find the Side B C or a.

To find BC' or a'.
sin. A
b sin. A
a sin. A'

b sin. A'
b sin. B
sin. B

sin. B

sin. B log. a = log. b + L sin. A — L sin. B b or side A C 27 log. 1.431364

1.431364 LA 39° 10' sin. 9.800427 LA II° 28' sin. 9:298412 II.231791

10°729776 LB 51 pts. sin. 9956163

9'956163 a or side B C 18.86 log. 1.275628

a' or side B C' 5.938 log. 0-773613 Now the bearing of A B, which is W. | N. or W. 2° 49' N., subtracted from the angle B A C 39° 10', makes the bearing of A C to be W. 36° 21' S.; hence the course steered is S. 53° 39' W. or S.W. W. nearly, and the distance of the ship from the point B is 18.86, or 19 miles nearly.

Otherwise, if the lay of the land between A and B permitted, the course might be as follows: W. 2° 49' N. subtracted from B AC 11° 28' makes the bearing of A C to be W. 8° 39' S. ; hence the course steered would be S. 81° 21' W., and the distance from the point B 5.94, or 6 miles nearly.

Example IV.-Coasting along, at noon, a point C bore N. 29° E., and another point D bore N. 20° W.; running N.W., at the rate of 7 knots an hour, at 2 p.m. the point C bore N. 72° E., and the point D bore N. 9° E. Required the bearing and distance of the point D from C.

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BY CONSTRUCTION Draw the compass as before, and let the centre A represent the first station, from which draw the first bearing A C, N. 29° E., and the second bearing A D, N. 20° W.; also draw the B N.W. line A B equal to 14 miles, the distance run in 2 hours; then will B represent the second station; through B draw B C parallel to N. 72° E., and B D parallel to N. 9° E., meeting the lines A C and A D in C and D; join D C; then will the line C D be their distance, measuring 18 miles nearly, and the bearing of D from C will be N. 70° W., or W.N.W. I W.

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