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without any progressive motion. Draw AC=22 miles S. and C will represent the ship's real place ; join BC. It follows, therefore, that while the ship has been describing the line AB, with respect to the water, the water itself has moved through a space equal to the distance BC in the direction of that line. Draw AD parallel to BC, and on the same side of AB as BC is drawn.

The line BC will be the space passed through by the current in four hours, and the angle CAD will be its direction or course.

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Computation.

In the triangle ABC are given the sides AB, AC, and the included angic A=33° 45'. Therefore,

Sum of sides AB+ AC=42 ar. co. 8:37675 : Difference of sides 9C-AB-2

030103 :: Tangsum not included=73° 7' 10'51800

Zs

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: Tang. their difference

8 55

919578

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sum to difference -- 82 2 == greater ZB, because opposite to the greater side.

Sine B-82° 2' ar. co.
: Side AC=22
:: Sine LA=33° 45'.

0:00.420
1'34242,
974474

: Side BC=123

1'09136

The bearing of the line BC may be gathered from its position with respect to BA, which lies N. 33° 45' W. from B, and BC is 82° 2' to the left of BA ; that is, BC is 115° 47' from the north, or 64° 12' from the south. Now, the distance BC=12 miles, passed by the current in 4 hours, gives 3 milcs to one hour. Consequently the current sets S. 64° 12' W. or nearly WSW S. at the rate of 3 miles an hour.

From the consideration of this example it is shewn, that a ship under way in a current will, at the end of any given time, be found at the same place, as if she had been in still water, after describing like courses and distances, together with another course and distance answering to the direction and velocity of the current. For the ship, under the action of the current, arrives at the same place C, as if she had been in still water, by sailing through AB and BC; and it is evidently of no importance, in this case,

provided

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provided the time continues unaltered, whicdier AB be made good on a single course, or on a number of courses.

7. A sloop is bound from the main land of Africa to an island bearing WbN. distant 22 leagues, a current setting NNW. 2 miles an hour. What is the course to arrive at the island in the shortest time, supposing the sloop to sail at the rate of 6 knots an hour ; and likewise what time will she take ?

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Construct the horizon with its meridian and paraliel of latitude. Draw AC=22 leagues =66 miles WbN. and draw AG NNW. Upon AG set off AK=25 miles, or any multiple of 2 miles ; and from K, with KD=60 miles, or a like multiple of 6 miles, describe an arc cutting AC, prolonged if necessary, in D. Join KD. From C draw CB parallel to KA, and from A draw AB parallel to KD and meeting CB in B.

The

The angle SAB is the course, and AB the distance of the run corresponding with the time. But if the velocity of the vessel be not greater than that of the current, that is, if KD be not greater than KA, the arc will not intersect AC, and it will be impossible for her to reach the island.

Computation. The sloop will arrive in the shortest time at the island by sailing in a right line. And, by the remark on the sixth example, her situation at the end of the time will be the same, as if she had described her apparent course and distance in still water, together with another course and distance corresponding with the motion of the current. Now, it is evident, that the apparent distance, or distance by log, AB, must be to the drift of the current BC, as the velocity of the vessel is to the velocity of the current, that is, as 6 to 2 In the triangle DKA, the sides DK, KA, are respectively like multiples of 6 and 2, and are, therefore, in that ratio. And the triangles DKA, ABC, are similar, because the alternate angles KDA, CAB, are equal. Therefore KD : AK :: 6 : 2 :: AB : BC. Consequently AB is the line, which, if described apparently by the vessel, would bring her to C by reason of the drift BC. Now to find the direction and length of this line AB.

In the triangle AKD are given AK and KD, together with the ZA=ZNAC-ZNAC=78° 45'--22° 30'= 56° 15'. Therefore, Side DK 66

8:18046 : Sine ZA = 56° 15' 991985 :: Side AK = 25

1'39794

ar. CO.

: Sine ZKDA= 180 21

949825 In the similar triangles AKD, CBA, the KAD LACB=56° 15', and the KDA=<CAB=18° 21'.

Therefore, Therefore, in the triangle ABC is given the side AC=66 miles, and two angles. Consequently the B is likewise known, being =105° 27'. Whence,

Şine B =1050 24' ar. co. 0'0 1588
: Side AC= 66

1'81954
:: Sine C=56° 15'

991985

: Side AB=569

1'75527

The angle NAB is equal to NAC+2 CAB=78° 45' +18° 21'=97° 6', and the angle SAB, or the apparent course, is equal to the supplement of ZNAB, which is 82° 54'. And the apparent distance, or distance by log, =56'9, or 57 miles, at the rate of 6 knots an hour, gives the time 9 hours. That is to say, the sloop will reach the island directly in 9 hours, by steering S. 82° 54' W. or nearly WS.

TRAVERSE SAILING.

The application of plane trigonometry to the solution of nautical questions has been exemplified. But the solution by logarithms is thought too operose for daily use, and it is seldom that mariners avail themselves of the logarithmic lines on Gunter's Scale. It is now a long time since tables of difference of latitude and departure, commonly called Traverse Tables, have been calculated; by the help of which any person, though totally ignorant of the trigonometrical proportions, 'may readily find the parts of any right-angled triangle, provided two of them exclusive of the right angle be given. For, since the distance sailed on any course is the hypotenuse of a right-angled triangle, of which the difference of latitude and departure are the

legs,

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