To find the Time requisite to reach the Point B: As distance 5 miles, Log. ar. comp. = 9.301030 Is to 1 hour, or 60 minutes, Log. = So is AG 1.778151 To the time=3523'.34-35".389 Log.=1.548868 To find the Direction in which he should pull or steer : From the angle bAe 47:17:55", take away the angle b AE = 33:45, and the remaining angle EA e 13:32:55 is the direct course which he should steer; viz., E. 13:33: S., or E. b. S. & S. nearly. Hence it is evident, that if the waterman pulls in the direction of E. 13:33. S. or E. b. S. ‡ S. nearly, he will reach the intended point in the space of about 35 minutes and 23 seconds. SOLUTION OF PROBLEMS RELATIVE TO THE ERRORS OF THE LOG-LINE AND THE HALF-MINUTE GLASS, BY LOGARITHMS. The instruments generally employed at sea, for finding the distance run by a ship in a given time, are the log-line and the half-minute glass. Now, since a ship's reckoning is kept in nautical miles, of which 60 make a degree, the distance between any two adjacent knots on the log-line should bear the same proportion to a nautical mile that half a minute does to an hour; viz., the one hundred and twentieth part. And, since a nautical mile contains 6080 feet, the true length of a knot is equal to 6080 divided by 120; that is, 50 feet and 8 inches: but, because it is advisable at all times to have the reckoning a-head of the ship, so that the mariner may be looking out for the land in sufficient time, instead of his making it unexpectedly, or in an unprepared moment, 48 feet, therefore, is the customary measure allowed to a knot. And, to make up for any time that may be unavoidably lost, in turning the half-minute glass, its absolute measure should not exceed twenty-nine seconds and a half. The method of finding the hourly rate of sailing, or distance run in a given time by the log-line and the half-minute glass, is subject to many errors thus, a new log-line, though divided with the utmost care and attention, is generally found to contract after being first used; and, after some wear, it stretches so very considerably as to be out of due proportion to the measure of the half-minute glass. Nor is the halfminute glass itself free from error: for this instrument is so very liable to be affected by various changes of weather, from moist to dry, and conversely, that notwithstanding its being perfectly correct when first taken on board, yet it alters so sensibly at sea, that at one time it will run out in the short space of 26 or 27 seconds, and at another not till it has passed the half-minute by several seconds. Hence it becomes indispensably necessary to examine those instruments frequently; and, if found erroneous, to correct the ship's run accordingly. This may be done by means of the following rules, which are adapted to a log-line of 48 feet to a knot, and to a glass measuring 30 seconds. PROBLEM I. Given the Distance sailed by the Log, and the Number of Seconds run by the Glass; to find the true Distance, the Line being truly divided. RULE. To the arithmetical complement of the logarithm of the number of seconds run by the glass, add the logarithm of the distance given by the log, and the constant logarithm 1. 477121*; the sum of these three logarithms, abating 10 in the index, will be the logarithm of the true distance sailed. Example 1. Let the hourly rate of sailing be 11 knots, and the time measured by the glass 33 seconds; required the true rate of sailing? = 8.481486 Seconds run by the glass = 33, Log. ar. comp. True rate of sailing 10 knots, Log. = Example 2. 1.477121 1.000000 If a ship sails 198 miles by the log, and the glass is found, on examination, to run out in 26 seconds, required the true distance sailed? * This is the logarithm of 30 seconds, the true measure of the half-minute glass. T PROBLEM II. Given the Distance sailed by the Log, and the measured Length of a Knot; to find the true Distance, the Glass being correct. RULE. To the logarithm of the distance given by the log, add the logarithm of the measured length of a knot, and the constant logarithm 8.318759* ; the sum of these three logarithms, rejecting 10 in the index, will be the logarithm of the true distance sailed. Example 1. Let the hourly rate of sailing be 9 knots, by a log-line which measures 53 feet to a knot; required the true rate of sailing? Let the distance sailed be 240 miles, by a log-line which measures 43 feet to a knot; required the true distance sailed? Distance sailed by log = 240 miles, Log. =2.380211 = 1.633469 Given the measured Length of a Knot, the Number of Seconds run by the Glass, and the Distance sailed by the Log; to find the true Distance sailed. RULE. To the arithmetical complement of the logarithm of the number of seconds run by the glass, add the logarithm of the measured length of a knot, the logarithm of the distance sailed by the log, and the constant This is the arithmetical complement of the logarithm of 48, the generally-approved length of a knot. logarithm 9.795880*; the sum of these four logarithms, rejecting 20 from the index, will be the logarithm of the true distance sailed. Example 1. Let the hourly rate of sailing be 12 knots, the measured length of a knot 44 feet, and the time noted by the glass 25 seconds; required the true rate of sailing? 1.643453 Seconds run by the glass = 25, Log. ar. comp. 8. 602060 = 1.079181 9.795880 Let the distance sailed by the log be 354 miles, the measured length of a knot 52 feet, and the interval run by the glass 34 seconds; required the true distance sailed? Given the Number of Seconds run by any Glass whatever, to find the corresponding Length of a Knot, which shall be truly proportional to the Measure of that Glass. RULE. To the logarithm of 10 times the number of seconds run by the glass, add the constant logarithm 9. 204120, and the sum, abating 10 in the index, will be the logarithm of the proportional length of a knot, in feet, corresponding to the given glass. This is the sum of the two preceding constant logarithms; thus 1.477121 + 8.318759 9.795880. Example 1. Required the length of a knot corresponding to a glass that runs 27 seconds? Number of seconds 27 x 10 = 270 Log. = 2.431364 9.204120 True length of a knot, in feet, = 43.2 Log. = 1.635484 Example 2. Required the length of a knot corresponding to a glass that runs 34 seconds? Number of seconds 34 x 10 = 340 Log. = 2.531479 True length of a knot, in feet, = 54.4 Log. 9.204120 = 1.735599 SOLUTION OF A PROBLEM IN GREAT CIRCLE SAILING, Very useful to Ships going to Van Diemen's Land, or to New South Wales, by the way of the Cape of Good Hope. Great Circle Sailing is the method of finding the successive latitudes and longitudes which a ship must make; with the courses that she must steer, and the distances to be run upon such courses, so that her track may be nearly in the arc of a great circle, passing through the place sailed from and that to which she is bound. The angle of position is an angle which a great circle, passing through two places on the sphere, makes with the meridian of one of them; and shows the true position of each place, in relation to the intercepted arc of the great circle and the respective meridians of those places. The polar angle is an arc of the equator intercepted between the meridians, or circles of longitude, of two given places on the sphere. On the sphere, the shortest distance between two places is expressed by the arc of a great circle intercepted between those places: consequently the spiral, or rhumb line, passing through two places on the sphere, can never represent the shortest distance between those places, unless such rhumb line coincides with the arc of a great circle; and this can never happen but when the places are situate under the equator, or under a |