factor on A, if it happen that the other factor on B fall beyond the division, on either A or b, divide it by 10, or 100, &c, till the quotient found on В fall under some division on the line A, and multiply this faid division by the same 10, or 100, &c, for the product required. EXAMPLE 11. So when 250 is to be multiplied by 56 : Having set i on B to 250 on A, although 56 be found on B, it is beyond the end of A; therefore dividing it by 10, I find that opposite to the quotient 5-6 on B, is the division 1400 on A; which being multiplied by 10, we obtain 14000 for the product required. EXAMPLE III. But if 250 were to be multiplied by 1120: Having set i to 250 as before, 1120 is beyond the end of B, but being divided by 100, opposite to the quotient 11.2 on B I find 2800 on A, which being multiplied by 100, we have 280000 for the product required. PROBLEM II. To find the Quotient of Two Numbers. Rule. Set i on B to the divisor on A, then against the dividend on A, is the quotient on B. EXAMPLE 1. To divide 300 by 25. Having set i on b to 25 on A, opposite 300 on a I find 12 on B, the quotient required. Note. When the dividend falls beyond the erd of the line A, let it be divided by 10, 100, or some other power of 10 till it fall within the line, and use the quotient instead of it, multiplying the result by the same power of 10 as before. EXAMPLE II. So if 14000 inust be divided by 56. Having set i to 56, the dividend cannot be found on A till it be divided by 100, the quotient being 140, opposite to which I find 2.5 on B, which being QO4 multi multiplied by 100 we obtain 250 for the quotient required. PROBLEM III. To work the Rule-of-Three on the Sliding Rule : or baving Three Numbers given, to find a Fourth, which mall be to the Third as the Second is to the First. RULE. Set the first term on B, to either the second or third on A; then against the remaining term on B, stands the fourth term required on A. EXAMPLE. If 8 yards of cloth cost 24 shillings, what will 96 yards cost at the same rate ? Having set 8 on B to 24 on A, opposite 96 on B, I find, on A, 288 shillings, or 141 8s, which is the answer, PROBLEM IV. To Extreet the Square Root by the Sliding Rule. RULE. The first i on c ftanding against the first on D, on the stock, opposite the given number on c is its root on D. EXAMPLE. To find tlie side of a square, which shall be equal to a triangle, or circle, &c, whose area is 225; or, to extract the root of 225. Here opposite 225 on c stands 15 on D, which is the answer required. To Extract the Cube Root by the Sliding Rule. Rule. The line d upon the slide being set straight with E; find the given number on É, and opposite to it will be its cube root on D. EXAMPLE. To find the side of a cube equal to any other solid whose content is 3375; or to find the cube root of 3375 Here opposite 3375 on E, stands 15 on d, which is the answer required. Note. It is evident that the same lines, as are used in these two last problems, will serve to find the square or the cube of any given number, by taking the given number on the contrary lines. , To find a Mean Proportional between Two Given Numbers. RULE. Set one of the given numbers on c to the like or fame number on D, then against the other given number on c, is the number required on D. EXAMPLE. To find the side of a square whose - area shall be equal to that of a parallelogram whose length is 9, and its breadth 4 feet; or, to find a mean proportional between 4 and 9. Having set 4 on c to 4 on D, against 9 on c stands 6 on D, which is the number fought. PROBLEM VII. To find a Number which mall be to a Given Number, in a given Duplicate Proportion ; or baving given Three Numbers, to find a Fourth, which shall be to the Third, as the Square of the Second is to the Square of the First. RULE. Set the third number on c to the first on D, then against the second on D, will be found, on c, the fourth required, EXAMPLE. If the area of a parallelogram, or any other figure, be 120; it is required to find the area of a similar figure, their like dimensions or sides being as 2 to 3 Similar figures being as the squares of their like dimensions, by setting 120 on c to 2 on D, against 3 on D, stands 270 on c, for the number fought. PROBLEM VIII. To find a Number which shall be to a "Given Number, in a given Şubduplicate Proportion ; or having given Three Numbers, to find a Fourth, which Mall be to the Third, as the Root of the Second is to the Root of the First. Rule. Set the first number on c to the third on d, then against the second on c, will be found the fourth on D. EXAMPLE. The side of a regular figure is 2, and its area 120; it is required to find the side of a similar figure whose area is 270. . The roots of the areas of similar figures being as their lides, we must find a number which shall be to 2, as the root of 270, is to the root of 120. Therefore, having set 120 on c to 2 on D, against 270 on c, will be found 3 on D, which is the number fought. To find a Number in a given Triplicate Proportion to a is to the Cube of the Firsi. RULE. Set the first number on the slide D, to the third number on E, then against the second on D, is inches, contain 100 gallons, what will be the content of a similar cask whose length is 36 inches? Similar solids being as the cubes of their like sides, the content required must be to 100 gallons, as 363 is to 40°. Therefore setting 40 on D to 100 an E, against 36 on D, will be found 72-9 gallons on E, which is the content required. To find a Number in a given Subtriplicate Proportion to a Given Number; or, baving Three Numbers given, to find a Fourth, which shall be to the Third, as. the Cube Root of the Second, is to the Cube Root of the First. Rule. Set the third number on D, to the first on E, then against the second on E, will stand the fourth on D. EXAMPLE. What is the length of a cask whose content is 72-9 gallons, supposing the length of a similar calk to be 40 inches, and its content 100 gallons ? Since the dimensions of similar solids are as the cube roots of their contents, we must find a number which shall be to 46, as the cube root of 72'9 is to the cube root of 100. Therefore, having set 40 on D to 100 on E, against 72.9 on E, will be found 36 on D, which is the length required. PRO |