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fore the angle ABC is equal to the angle BAD.-Euclid, Book I., Prop. 29. And because the straight line CD falls upon the two parallel straight lines CB and AD, it makes the angle ADB equal to the angle BCD, by the aforesaid proposition. And because the two triangles ADF and BCF have, thus, two angles of the one equal to two angles of the other, viz., the angle FAD to the angle FB C, and the angle ADF to the angle BCF; and the side A F of the one equal to the side B F of the other: therefore the remaining sides AD and DF of the one are equal to the remaining sides BC and CF of the other, each to each; and the third angle AFD of the one equal to the third angle B FC of the other.-Euclid, Book I., Prop. 26. Now, since the two triangles A FD and B FC are, thus, evidently equal to one another, we have only to compute the unknown sides of one, viz., of the triangle AFD, where the three angles are given, and the side AF, to find the sides AD and FD; thus, the difference between N.E. b. N. and E.N.E., is 3 points. the angle FA D, measured by the arc ab; the difference between E.N.E. and E.S.E. (the opposite point to W.N.W.), is 4 points the angle ADF, measured by the arc bg; and the difference between W.N.W. and N.E. b. N., is 9 points the angle AF D, measured by the arc ad: hence, by oblique angled trigonometry, Problem I., page 177,

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Hence it is evident that the ship must first run 62. 42 miles on the larboard tack; then 70. 70 miles on the starboard tack; and, again, 62.42 miles on the larboard tack, before she can reach her intended port.

Example 3.

A ship that can lie within 6 points of the wind is bound to a port bear

ing N.N.W., distance 120 miles, which it is intended she shall make on four tacks, with the wind at N. b. W. The coast, which is to the eastward, trends in a direction nearly parallel to the bearing of the port, so that the ship must go about as soon as she reaches the straight line joining the two ports; required the course and distance to be run upon each tack, on the supposition that the ship's progress is not affected by either leeway or currents?

Solution.-Since the wind is N. b. W., and the land trends in a N.N.W. direction, the first board, therefore, must be on the starboard tack; and, as the ship can lie within 6 points of the wind, the course on the starboard tacks will be W. b. N., and that on the larboard tacks N.E. b. E.

In the annexed diagram, let the N.N.W. line AB, 120 miles, represent the bearing and distance between the ship and the port to which she is bound; let the W.b.N. line A D represent the first board

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on the starboard tack, and FC, parallel to A D, the second board on that tack; let the N.E. b. E. line D F represent the first board on the larboard tack, and, parallel thereto, the line C B = the second board on this tack. And, since the ship is to make her port in four tacks, without going to the eastward of the line AB, therefore, at the end of the second tack, she must reach the point F, which bisects or divides the distance A B into two equal parts, of 60 miles each; thus making A F = to AB.

Now, because the straight line A B falls upon the two parallel straight lines AD and FC, it makes the angle BFC equal to the interior and opposite angle FAD: and, because the straight line AB falls upon the two parallel straight lines FD and CB, it makes the angle AFD equal to the interior and opposite angle C B F,-Euclid, Book I., Prop. 29. And, since the two triangles AFD and FBC have, thus, two angles of the one equal to two angles of the other, viz., the angle AFD to the angle FBC, and the angle FAD to the angle BF C, and the side A F of the one equal to the side FB of the other, therefore the remaining sides 'A D and D F of the one, are equal to the remaining sides FC and CB of the other, each to each; and the third angle ADF of the one equal to the third angle FCB of the other.-Euclid, Book I., Prop. 26.

The two triangles ADF and FC B, being, thus, clearly equal to one

another in every respect, we have only to compute the unknown sides of one, viz., of the triangle A FD, where the three angles are given, and the side A F 60 miles, to find the sides AD and DF; thus the difference between N.N.W. and W. b. N., is 5 points the angle FAD, measured by the arc ab; the difference between W. b. N. and S.W. b. W. (the opposite point to N.E. b. E.), is 4 points the angle AD F, measured by the arc ae; and the difference between N.N.W. and N.E. b. E., is 7 points = the angle AFD, measured by the arc bd.

Hence, by oblique angled trigonometry, Problem I., page 177,

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As the angle D = 4 points, Log. co-secant = 10.150515
Is to the side A F= 60 miles, Log.
So is the angle A = 5 points, Log. sine =

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To the side DF 70. 55 miles, Log.

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1.778151

9.919846

1.848512

From this it is manifest, that the ship must first run 83. 22 miles upon the starboard tack; then 70.55 miles upon the larboard tack; then 83. 22 miles again upon the starboard tack; and 70.55 miles upon the larboard tack, before she can reach the port to which she is bound.

SOLUTION OF CASES IN CURRENT SAILING.

Current Sailing is the method of determining the true course and distance made good by a ship, when her own motion is affected or combined with that of the current in which she sails.

A current is a progressive motion of the water, causing all floating bodies thereon to move in the direction to which its stream is impelled. The setting of a current is that point of the compass towards which the water runs ; and the drift of a current is the rate at which it runs per hour.

When a ship sails in the direction of a current, her velocity will be equal

to the sum of her own proper motion and the current's drift; but when she sails directly against a current, her velocity will be expressed by the difference between her own proper motion and the drift of the current: in this case, the absolute motion of a ship will be a-head, if her proper velocity exceeds the drift of the current; but if it be less, she will make sternway. When a ship's course is oblique to the direction of a current, her true course and distance will be compounded of the course and distance. given by the log, and of the observed setting and drift of the current.

When a ship's course and distance by the log, and the setting and drift of the current in which she sails are given, the true course and distance made good may be found by a trigonometrical solution of the triangles forming the figure; but the easiest and most expeditious method of finding the course and distance made good, particularly when a ship sails upon different courses, is by resolving a traverse, in which the setting and drift of the current are to be esteemed as an additional course and distance to those exhibited by the log.

Example 1.

If a ship sails S.W. b. W., at the rate of 4 knots an hour, in a current setting S.S.E. E., at the rate of 14 miles an hour; required the course and distance made good in 24 hours?

Solution.-4 x 24 96 miles, the distance sailed, by log, in 24

hours;

And 12" × 24 = 42 miles, the observed drift of the current in 24 hours.

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difference between S.S.E. E. and N.E. b. E. (the opposite point to S.W. b. W.), measured by the arc ab, to find the angles A and C, and the true distance A C. Hence, by oblique angled trigonometry, Problem III., page 179,

To find the Angles A and C :

As AB+BC = 138 miles,

BC= 54 miles,

Log. ar. comp. = 7.860121
Log. =

-

Is to AB
1.732394
So is sum of the angles=43:35:37" Log. tangent=9.978673

To diff. of the angles = 20.25.59 Log. tangent=9.571188

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To the true distance AC 106.7 miles, Log. = 2.027989

To find the Course made good:

From the angle SAB S.W. b. W., or 56:15, subtract the angle CAB 23:9:38", and the remainder, 33°5:21" the angle SAC, is the course made good.

Hence the course made good is S. 33:5 W., or S.W. b. S. nearly, and the distance 106 miles nearly.

To find the Course and Distance made good by the Traverse Table :

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