Now, by Problem II., page 108,

The difference of latitude 89.3, and the departure 58. 2, are found to agree nearest abreast of 33:, under or over distance 107.

Hence the course made good is S. 33: W., or S.W. b. S., and the distance 107 miles; which nearly agrees with the above result.

Example 2.

Suppose a ship sails N.W. 65 miles, W.N.W. 70 miles, and N. b. E. 71 miles, in a current that sets S.E. b. S. 36 miles in the same time; required the true course and distance made good?

the

Solution. With the difference of latitude and departure, thus found, course and distance made good may be determined by Problem II., page 108; as thus:

The difference of latitude 112.5, and the departure 76. 8, are found to agree nearest abreast of 34: under or over 136.

Hence the direct course made good is N. 34: W., or N.W. b. N. nearly, and the distance 136 miles.

To find the Course and Distance made good by Calculation :

This may be done by means of the 5th analogy, page 237; as thus:

To find the true Course

So is the true course=34:19:12"Log. secant =, 10.083072

To the distance = 136.2 miles, Log. =

2.134224

Hence the course made good is N. 34:19. W., or N.W. b. N. nearly, and the distance 136 miles.

Example 3.

There is a harbour 2 miles broad, in which the tide is running N.W. b. N. at the rate of 3 miles an hour, Now, a waterman who can pull his boat at the rate of 5 miles an hour, wishes to cross the harbour to a point on the opposite side bearing E.N.E.; required the direction in which he should pull, so as to meet with the least possible resistance from the force of the tide in gaining the intended point, and the time that it will take him to reach that point?

With the chord of 60: describe the arch NESW; draw the north and south line NS, and, at right angles thereto, the east and west line WE; make the arc Na= 3 points, and draw the N.W. b. N. line A a D, which make equal to 3 miles (taken from any scale of equal parts), to represent the direction of the harbour; perpendicular thereto draw the N.E. b. E. line Ab C, which make equal to 2 miles, to represent the breadth of the harbour; and, from the point C, draw the line C G parallel to A D, which lines will represent the eastern and western shores of the harbour respectively.

Make N c equal to 6 points, and draw the E.N.E. line Ac F, cutting CG in B; then will B represent the point to which the waterman intends to cross. Take 5 miles in the compasses; place one foot on the point D; and where the other falls upon the E.N.E. line A F, there make a point, as at F, and draw the line DF; parallel to which, draw the line A G, and it will represent the distance and direction in which the waterman must pull to gain the point B: for in the time that he would reach the point G, by pulling at the rate of 5 miles an hour, the tide, running at the rate of 3 miles an hour, would carry him to the point B; because B G bears the same proportion to 3 miles an hour that AG does to 5. Now, AG, being applied to the same scale of equal parts from which the other sides were taken, will measure 2.95 miles, and the angle GA E, or eAE, being applied to the line of chords, will measure 13:33'; hence the direction in which he should pull, is E. 13:33: S, or E. b. S. S. nearly.

Now, in the triangle ADF, given the side AD = 3 miles, the side DF 5 miles, and the angle DAF 9 points (being the difference between E.N.E. and N.W. b. N., measured by the arc a c), to find the angle A F D. Hence, by oblique angled trigonometry, Problem I., page 177,

Now, because the straight line A F falls upon the two parallel straight lines DF and AG, it makes the alternate angles equal to one another; therefore the angle DFA is equal to the angle FAG,-Euclid, Book I., Prop. 29; but the angle DFA is known to be 36:2:55"; therefore the angle FAG, measured by the are ce, is also equal to 36:2.55"; and if to the angle FAG we add the angle BAC 11:15 (being the difference between N.E. b. E, and E.N.E., measured by the arc be), the sum = 47:17:55" is the angle CA G, measured by the arc be. Then,

In the right angled triangle A CG, given the angle CAG 47:17:55% and the side A C= 2 miles, the breadth of the harbour, to find the side AG equal to the distance which the waterman must pull before he can reach the point B. Hence, by right angled trigonometry, Problem II., page 172, making AC radius,

To find the Time requisite to reach the Point B:

To the time 3523'.34-35". 389 Log.=1.548868

To find the Direction in which he should pull or steer :

From the angle b Ae

47:17:55", take away the angle b AE = 33:45, and the remaining angle É Ae 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;

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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?

If a ship sails 198 miles by the log, and the glass is found, on exainination, 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.

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