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current, finds that at the end of 24 hours she has made a change of latitude of 3 degrees, and a departure of 25 miles to the west; required the setting and drift of the current.

(349.) A ship has made by the reckoning S. E. 20 miles; but by observation it is found that, owing to a current, she has actually gone SSW. 28 miles; required the setting and drift of the current in the time the ship has been running.

(350.) A ship's course to her port is NW. by W., and she is running by the log 9 miles an hour, but meeting with a current setting W. S. 3 miles an hour; what course must she steer in the current that her true course may be NW. by W., and at what rate will she approach the port?

(351.) In a tide running WNW. 3 miles an hour, I wish to weather a cape bearing NE. by N. 21 miles; what course must I steer so as to clear the point, the ship going 8 miles an hour by the log, and in what time will I reach the cape? (352.) If a ship sail due W. 9 miles an hour by the log in a current setting NW. by W. 3 miles an hour; what is her true

course and hourly rate of sailing?

(353.) A ship at sea in the night has sight of Scilly light bearing NW. by N. distant 15 miles, it being then flood-tide setting ENE. 2 miles an hour, and the ship running at the rate of 5 miles an hour; required what course and distance she must sail to reach the Lizard, which bears from Scilly E. S. distant 51 miles.

(354) In a river two miles broad, in which the ebb tide is running E. by S. at the rate of 3 miles an hour, a waterman who can row at the rate of 5 miles an hour, intends to cross the river to a point on the opposite side, SSW. from his present position. In what direction must he pull that he may reach the intended place in the shortest time possible, and in what time will he reach it?

(355.) The conditions being all the same as in the last question, except that the point on the opposite side bears direct south; required the course and the time required to cross.

CHAPTER IX.

ON PLYING TO WINDWARD.

260. Windward Sailing is the method of gaining an intended port by the shortest and most direct method possible, when the wind is in a direction unfavourable to the course the ship ought to steer, so as to reach the port directly.

261. In order to attain this important point, it is evident the ship must sail on different tacks; the object, therefore, of this sailing is to find the proper courses to be steered, and the distance to be sailed on each board, that the vessel may arrive at the intended port the soonest possible. The following examples will shew the rules to be observed in this sailing.

Example I.

A ship is bound to a port 34 miles direct to windward, the wind being NNE.; she can lie within 6 points of the wind on each tack; required how far she must sail on each tack so as to reach the port in two boards.

Draw AB, NNE.

=

A

B

Solution.

34 miles, and through A draw AC and AE, each making an angle of 6 points with AB, the direction of the port; through B draw BC parallel to AE, and produce AC and BC till they meet in C; draw CD perpendicular to AB, and AD will evidently be the half of AB, and the angles CAB and CBA

being equal, the side AC = BC (Euc. I. 6). Again, in the triangle ADC, right-angled at D (Art. 126),

=

AC - AD × sec. DAC = 17 × sec. 6 points. .. AC

=

17 x 2.6131259

= 44·4231403 miles, the distance to be sailed on each board; hence,

262. RULE I.-When the ship has to sail directly against the wind. Multiply half the distance into the secant of the angle within which the ship can sail of the wind, and the product will be the distance to be sailed on each tack.

263. RULE II.-By inspection; with the angle within which the ship can sail of the wind as a course, and half the distance of the port as a difference of latitude, enter the Traverse Table, and opposite to half the distance of the port, in the Dif. Lat. column, will be found, in the Distance column, the distance to be sailed on each tack.

Example II.

The wind being SSW., a ship is bound to a port bearing S. by E., distant 54 miles, which it is proposed to reach in two boards, each being within 5 points of the wind, the first being in the SE. quarter; required the distance to be run on each board.

Solution.

Draw AS, a meridian, and A, SSW., to represent the direction

of the wind, and AB, S. by E., to represent the direction of the port; and make AB = 54 miles, the distance of the port; draw AC and AE, making angles of 5 points with the direction of the wind; and through B draw BC parallel to AE,

and meeting AC in C;

MSS

B

then AC = the distance to be sailed on the starboard tack, and CB= the distance to be sailed on the port tack. Angle CAB= 5-32 points, and angle ABC = EAB = 51+3=81 points; hence also, ACB = 5 points.

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264. If BD be drawn through B parallel to AC, to meet AE in D, it is plain that the ship may reach the port either by first sailing on the starboard tack from A to C, and then on the port tack from C to B; or, by first sailing on the port tack from A to D, and then on the starboard tack from D to B.

265. If it were required to reach the port B in three boards, there are two ways in which this might be done, either by first sailing any portion of the line AC, and then sailing across parallel to AD or CB till the ship reach the line DB, and

then sailing the remaining part of the starboard tack on DB; or, by first sailing any portion of AD, and then along on the starboard tack parallel to AC or DB, till she reach the line CB, and then sailing the remaining part of the port tack along CB. In all the four cases above mentioned, the distance or distances sailed on the starboard tack is equal to AC = 64.632 miles, and the distance or distances sailed on the port tack is equal to AD or CB=30.615 miles.

Example III.

The wind being at N. E., a ship is bound to a port bearing NNE., distant 68 miles, which she intends to make at four boards. The coast which is to westward also tends NNE., so that the ship must go about as soon as she reaches the straight line joining the two ports; required the course and distance on each board, the ship making her way good within 6 points of the wind.

Solution.

266. Draw AB, NNE. = 68 miles; then draw AF, making an angle of 6 points with the direction of the wind; and hence

NE

B

E

an angle of 4 points with AB; draw also BF, making an angle of 6 points with the direction of the wind, and therefore an angle of 7 points with AB; produce these lines to meet in F; then the angle at F = 16 points(7+4)= 4 points. Take any point D in AB, and through it draw DE parallel to AF, and DC parallel to BF; then the four boards are evidently AC, CD, DE, and EB; but they are together equal to AF + FB, the two distances on the port tack-namely, AC and DE-being together equal to AE, and the

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