To perform PROFORTION, or, as it is usually called, the RULE OF THREE, or the GOLDEN RULE, by Logarithms. RULE. In stating questions in this rule, the first and second terms are to be of the same name, and the third term is to be of the same name as that required. Let, Let, therefore, that term which is of the same name as the required one, be placed as the third term of the proportion. Then consider, from the nature of the question, whether the required term is to be greater, or less, than the third term. If it is to be greater, the least of the other two terms is to be placed as the first term of the proportion, and the other as the middle term: but if it ought to be less, the greater of the two terms is to be placed as the first, and the least as the middle term. Then, from the logarithms of the sum of the second and third terms, subtract the logarithm of the first term, and the remainder will be the logarithm of the answer. EXAMPLES. I. Required the value of 17 yards of cloth, when 52 yards cost 46%. ? First term log. 1,662758 140.06 If 9 yards of muslin cost 17. 11s. how many yards may be bought for 31. 12s.? If a ship sails 15 miles in 24 hours, how many miles will she run, at the same rate, in 19 hours? IV. A merchantman, distance 24 miles, going at the rate of 5 knots an hour, is pursued by a privateer, whose hourly rate of sailing is 7 knots: after two hours chase, the breeze freshened, and the merchantman's rate was increased to 6 knots, and the privateer's to 8.2. In what time will the privateer come up with the merchantman? As the privateer gained two miles an hour on the merchantman, at the end of the first two hours, the distance between them was 20 miles; during the remaining part of the chase, the hourly advance of the privateer was 2.2 knots. Therefore As the hourly rate Is to the distance So is 2.2 20. 1 hour To the time required 9.091 0.342423 1.301030 0.000000 0.958607 Hence from the time the breeze freshened, the privateer would come up with the merchantman in 9 hours 5 minutes; or 11 hours 5 minutes, from the commencement of the chase. PROBLEM VI. To perform INVOLUTION by Logarithms, that is, to find the Square, Cube, &c. of any given Number. RULE. Multiply the logarithm of the given number by the index of the power to which it is to be raised, and the product will be the logarithm of the power sought. EXAMPLES. To perform EVOLUTION by Logarithms; that is, to extract any proposed Root of a given Number. RULE. Divide the logarithm of the given number by the index of the power, and the quotient will be the logarithm of the root. If the given number is a decimal, and the arithmetical complement is used, prefix 1 to the index for the square root, 2 for the cube root, &c. To find the Tonnage of a Ship by Logarithms, according to the common Method. RULE. If the vessel is a ship of war, let fall a perpendicular from the foreside of the stern, at the height of the hause holes; but if a merchantman, the perpendicular is to be let fall from that part of the foreside of the stern which is at the same height above the keel as the wing transom also let fall another perpendicular from the back of the main post, at the height of the wing transom. Find the distance between these two perpendiculars, from which subtract three fifths of the extreme breadth; and also the product of the height of the wing transom above the upper edge of the keel by 2 inches, and the remander is the length of the keel for tonnage. To the logarithm of which, add the logarithm of the breadth, and that of the half-breadth and the constant logarithm 8 026872*, the sum, rejecting 10 from the index, will be the logarithm of the tonnage required. EXAMPLE. Let the length between the fore-part of the stem, and the after-part of the post, be 128 feet; the extreme breadth 32 feet; and the height of the wing transom 17 feet. Required the tonnage of the ship? The constant number 8.026872, is the arithmetical complement of the logarithm of 44, the common-divisor for finding the tonnage. This table, which is particularly adapted for the purpose of reducing the apparent to the true distance, by Method Second, vol. 1. page 153, contains the logarithmic sines to every tenth second, with the prop part to every single second, at the bottom of the page. Hence, the log. sine or co-sine of any given degree, minute, and second, and conversely, may be easily found by the following problems. PROBLEM I. To find the Log. Sine, or Co-sine, of any given Arch, expressed in Degrees, Minutes, and Seconds. RULE. Find the log sine, or co sine, answering to the given degree, minute, and next less tenth second; to which add the proportional part for the odd seconds, from the bottom of the table, if a sine is wanted ; but subtract the prop. part if a co-sine is required, and the sum or remainder will be the log sine or co-sine of the proposed angle. If the given arch exceeds 90°, the sine of its supplement is to be taken when the sine is wanted; or the sine of its complement when its co-sine is required. To find the Arch answering to a given Log. Sine or Co-sine. RULE. Find the arch answering to the next less sine, or next greater cosine, the difference between which, and that given, being found at |