SOLUTION OF CASES IN WINDWARD SAILING. Windward Sailing is the method of reaching the port or place bound to by the shortest route, when the wind is in a direction contrary to the direct course between the ship and the place to which she is bound. When the wind is opposed to the course which a ship should steer from any one port to another, she is obliged to sail upon different tacks, closehauled to the wind, in order to reach the port bound to. The object, therefore, of this method of sailing, is to find the proper course to be conned on each tack, so that the ship may arrive at the place to which she is bound, in the shortest time possible. Example 1. A ship that can lie within 6 points of the wind is bound to a port 50 miles directly to windward, which it is intended she shall reach on two tacks; the first being on the starboard tack, and the wind steady at N. b. E.; required the course and distance to be run upon each tack? be made good on the larboard tack. Then, in the triangle ABC, the three angles are given to find the side AB or BC, which sides are mutually equal to each other, because the triangle is isosceles, and its vertex at B the angle comprehended between those sides. Thus, the difference between N. b. E., and N.W. b. W., is 6 points the angle BAC, measured by the arc ab; the difference between E. b. N., and S.E. b. E. (the opposite point to N.W. b. W.), is 4 points, measured by the arc de; and the difference between N. b. E., and E. b. N., is 6 points, measured by the arc bd and since the distance AC is given 50 miles, the side A B, or As the angle B = 4 points, Log. co-secant = To the distance AB 65.33 miles, Log. = 10. 150515 1.698970 9.965615 1.815100 Hence, it is evident that the ship must run 65.33 miles on the starboard tack, and 65.33 miles on the larboard tack, before she can reach her intended port, Example 2. A ship that can lie within 6 points of the wind is bound to a port bearing N.E. b. N., distance 90 miles, which it is intended she shall reach on three tacks, with the wind steady at north; required the course and distance to be run upon each tack, the first course being on the larboard tack? the line BC= the second board on that tack. And, since the ship is to make her port in three tacks, it is evident that the board on the starboard tack, represented by the W.N.W. line CD (parallel to dg), must bisect the line A B in the point F; and that, therefore, A F and FB are equal to one another, each being equal to 45 miles = half the line, or distance AB. Now, since the straight line A B falls upon the two parallel straight lines CB and AD, it makes the alternate angles equal to one another; there Since the angles A and C are equal to one another, the sides which subtend, or are opposite to those angles (viz., BC and A B), are also equal to one another.-Euclid, Book I., Prop. 6. fore the angle A B C 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 AFD and BFC 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 a b; 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, 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 pórt 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 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 A B. Now, because the straight line AB 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 F BC, and the angle FAD to the angle B F C, and the side A F of the one equal to the side FB of the other, therefore the remaining sides 'A D and DF of the one, are equal to the remaining sides FC and C B 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 FCB, 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 AF 60 miles, to find the sides AD and D F; 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, 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 |