NWN 100 miles, required the equivalent course and distance, and the latitude in? Answer, lat in 47° 20′ N, course N 42° 58′ W, and dist 317.2 miles. 9. If a ship sail from lat 55° 4′ N, SE b E 24, SE b S 46, SES 31, E b N 14, and E & S 18 miles, required her course, distance, and latitude in? Answer, lat in 53° 50′ N, course S 51° 7′ E, and dist 118 miles. 10. If a ship sail from Halifax ES 23, S EbE 30, E b N 45, and NEN 25 miles, required her latitude in, and course, and distance made good? Answer, lat in 44° 54′ N, course N 84° 42′ E, and dist 107 miles. 11. Yesterday, on leaving the Lizard, the land bore from us NE 18 miles, since that time we have sailed WSW 14, S Wb W 26, SWS 37, SSW W 29, and W 15 miles, required our present latitude, and the course and distance which we have made? Answer, lat in 48° 31′ N, course S 48° 20 W, and distance 131 miles. 12. On leaving the Cape of Good Hope for St. Helena, we took our departure from Cape Town bearing S Eb S 12 miles; after running N W 36, and N W b W 140 miles, required our latitude in, and the course and distance which we have made? Answer, lat in 32° 3′ S, course N 52° 41′ W, and distance 187 miles. ON PLYING TO WINDWARD. WHEN the wind blows so near the point towards which a ship is bound that she is obliged to steer on different tacks, it becomes a question how far she ought to steer on each tack to reach her port, or to avoid any danger that may lie in her way. Questions of this kind however are very simple; for, with a given wind, the course of a ship, close hauled, on either tack, is easily found; and she must steer on one of the tacks till the bearing of the place which she wishes to reach, is the same as the course on the other tack, if she reach the place in two boards. When the bearing of the port is given, the angle which the course of the ship on either tack makes with that bearing is known; and hence when the distance of the port is also given, the distance on each tack will be obtained by the resolution of a plane triangle, of which one side (the given distance) and all the angles, are given, to find the other two sides. This computation may be conveniently made by the following rule, which obviously results from what has here been said. As the sine of the angle included between the two courses is to the given distance, so is the sine of the angle included between the bearing of the port and the course on either tack to the distance to be sailed on the other tack. Thus if a port bear W N W 20 miles, and the wind blows from W b S, and the ship can lie within 6 points of the wind, then the course which she can make close hauled on the larboard tack will be N W b N, and on the starboard tack S b W, and she may be steered on either tack that is most convenient till the bearing of the port be the same as the course on the other tack. The first of these courses makes an angle of 3 points with the bearing of the port, the second makes an angle of 9 points, and the angle included between the two courses is 12 points. Hence sin 12 points: 20 miles :: sin 3 points : 15.71 miles, the distance to be run on the starboard tack. And sin 12 points: 20 miles :: sin 9 points: 27.74 miles, the distance to be run on the larboard tack. It may be observed that whatever number of boards it may be found expedient that a ship should make, the sum of the distances on each tack will be the same as if the place had been reached on two boards only. EXAMPLES FOR EXERCISE. 1. A ship which can lie within 5 points of the wind is bound to a port bearing SE 18 miles, the wind being at S E, required her course and distance on each tack to reach her port, close hauled, in two boards? Answer, on the larboard tack S b W 19:39 miles, and on the starboard tack E b N 5'656 miles. 2. Wanting in two boards to reach the mouth of a river, which bore NW + W 10 miles, I found my ship could lie within 5+ points of the wind (then at N N W) on either tack, but that on the larboard tack she made point leeway, while on the starboard tack she made 14 points, required the course and distance on each tack? Answer, on the starboard tack W b S 17.91 miles, and on the larboard N E 13.91 miles. 3. Wishing to go round a point, which bore N N W 15 miles, but the wind being at W b N, I was obliged to ply to windward; I found my ship would make way within 6 points of the wind, required the course and distance on each tack? Answer, on the larboard tack N b W 17·65, and on the starboard SW b S 4.138 miles. 4. If a ship can lie within 6 points of the wind on the larboard, but within 5 points on the starboard tack, required her course and distance on each tack to reach a port lying S b E 22 miles, the wind being at S W? Answer, on the starboard tack SbEE, 23.66, and on the larboard W NW 2.79 miles. 5. From a ship, A, the mouth of a river bore N N E, and from another, B, it bore N b W, distant from each 18 miles. If each ship can lie within 5 points of the wind, and sail with the given wind, close hauled, 4 miles an hour, which will reach the harbour sooner, and how much, the wind being at N? Answer, A will reach the river about 27 minutes 43 seconds before B. 6. The wind is at W S W, and a ship sailing within 5 points of it 24 miles per hour, makes on each tack 24 points leeway, in what time will she advance 30 miles directly to windward ? Answer, in 5 days 2 hours and 30 minutes. OF SAILING IN CURRENTS. D с Ir a ship at B, sailing in the direction B A, were in a current which would carry her from B to C in the same time that in still water she would sail from B to A, then, by the joint action of the current and the wind, she would in the same time describe the diagonal B D of the parallelogram A B C D. For her being carried by the current in a direction parallel to B C would neither alter the force of the wind, nor the position of the ship, or the sails, with respect to it; the wind would therefore continue to propel the ship in a direction parallel to A B, in the manner as if the current had no existence. Hence, as she would be swept to the line CD by the independent action of the current, in the same time that she reached the line A D by the independent action of the wind on her sails, she would be found at D, the point of intersection of the lines AD and CD, having moved along the diagonal B D. Now the log heaved from the ship in the ordinary way can give no intimation of a current; for the line withdrawn from the reel is only the measure of what the ship sails from the log; and, consequently, as the log itself as well as the ship will move with the current, the distance shewn by the log in a current is merely what it would have been if the ship had been in still water. The velocity and direction of a current at sea, or its drift and setting, are generally determined by heaving the log from a boat; having first, by means of a line, sunk a pot loaded with iron, or some heavy piece of metal, to a considerable depth, by which means the boat is kept nearly stationary. Then the line withdrawn by the log in a given time is evidently the measure of the current's drift in the same time; and the direction in which the current carries the log, is the direction or setting of the current. If the ship sail in the direction of the current the whole effect of the current will be to increase the distance; but if she sail against the current, the difference between the rate of sailing given by the log and the drift of the current, will be the distance which the ship actually goes; and she will move forward if her rate of sailing be greater than the drift of the current, but otherwise her motion will be retrograde, or she will be carried backwards in the direction of the current. Problems relating to the oblique action of a current upon a ship may be resolved by the solution of an oblique angled plane triangle, such as A B D in the above figure, where if A B represent the distance which a ship would sail in still water, and A D the drift of the current in the same time, B D will be the actual distance sailed, and A B D the change in the course produced by the current. A great variety of problems might be proposed relative to currents, but the chief ones of any practical importance are the following. 1. To determine a ship's actual course and distance in a current, when her course and distance by the compass and the log, and the setting and drift of the current are given. 2. To find the course to be steered through a known current, the required course in still water, and the ship's rate of sailing being known. 3. To find the setting and drift of a current, from a ship's actual place, compared with that deduced from the compass and the log. The first of these cases may be conveniently resolved, by considering the ship as having performed a traverse, the setting and drift of the current being taken as a separate course and distance. EXAMPLE. If a ship sail W 28 miles in a current, which in the same time carries her N N W S miles, required her true course and distance? First solution by direct computation from the oblique angled triangle. Let B A, in the above figure, represent the distance run by the log, and A D the drift of the current, then the angle BAD will be 10 points, and consequently half the sum of the angles A B D and ADB will be 3 points. Hence A B+ A D (36) : A B − A D (20) : : tan 3 points: tan 20° 22′, half the difference of the angles A D B and A B D, whence A B D is 13° 23′, which allowed from the west towards the north gives the course N 76° 37′ E. And as sin A B D (20°22′): A D (8) :: sin B AD (10 points): BD 31·93, the true distance. N As 739 rad: : 31.06: tan 76° 37′, the course, rad: cosec 76° 37′ :: 31·06: 31.93, the distance. EXAMPLES FOR EXERCISE. 1. If a ship sail E 7 miles an hour by the log, in a current setting ENE 2.5 miles per hour, required her true course, and hourly rate of sailing? Answer, course N 84° 8' E, and rate 9.358 per hour. 2. A ship has made by the reckoning NW 20 miles, but by observation it is found that, owing to a current, she has actually gone N NE 28 miles, required the setting and drift of the current in the time which the ship had been running? Answer, setting N 64° 48′ E, and drift 14.1 miles. 3. A ship's course to her port is W N W, and she is running by the log 8 miles an hour, but meeting with a current setting WS 4 miles an hour, what course must she steer in the current that her true course may be W NW? Answer, course N 44° 39′ W. 4. In a tide running N W b W 3 miles an hour, I wished to weather a point of land which bore NE 14 miles, what course must I steer so as to clear the point, the ship going 7 miles an hour by the log, and what time shall I be in reaching the point? Answer, course N 69° 51′ E, and time 2 hours 25 minutes. 5. From a ship in a current steering W SW 6 miles an hour by the log, a rock was seen at 6 in the evening, bearing S WS 20 miles. The ship was lost on the rock at 11 P. M. required the setting and drift of the current? Answer, setting S 75° 10′ E, and drift 3·11 miles per hour. 6. If a ship sail due W 85 per hour by the log, in a current setting S W b W, 4 miles per hour, required her true course, and hourly rate of sailing? Answer, course S 79° 21' W, and rate 12.04 miles per hour. |