the first partial product. To locate it in this case, note that if the 2'3'0.'2'5'8'5093 15.70796 2302.59 1151 29 161 18 1 3616.89 multiplier were 1.57+, the first partial product would be 230.259; but the multiplier is 10 times 1.57+, and the first partial product is therefore 10 × 230.259 = 2302.59. After multiplying by 9 and obtaining 21 for the fifth product, cut off one more figure of the multiplicand, and since 6 X 2 = 12, write 1 for the sixth partial product. This method of multi plying is slightly more accurate than that at (c) in the preceding example, when both factors were reduced to 5+1 = 6 figures. The product correct to 5 significant figures is 3616.9. Ans. 35. Division may be performed in a somewhat similar manner. 15.7+. The quotient correct to 5 significant figures is 15.708. As another example, find the quotient of 13643.765 ÷ 3.141593 to 5 significant figures. Limiting the divisor and dividend to 6 figures each, the dividend will not contain the divisor; hence, instead of cutting off another figure from the divisor, add another figure to the dividend, and then proceed with the division as shown. The quotient evidently contains 4 integral places, since the quotient of 13643 3 will contain 4 integral places. The quotient correct to 5 significant figures is 4342.9. Ans. 36. Constants and Variables.-A large majority of the formulas used in practice contain one or more quantities that remain the same no matter what the conditions are that govern the problem. For instance, the formula in Art. 1 contains the quantity .7854. Now, no matter what the values of D, d, l, and w are, the quantity .7854 remains the same; for this reason, it is called a constant. The other quantities in this formula are called variables, because their values vary or change for different cylinders. For all castiron cylinders, w=.2604, and, hence, for all cast-iron cylinders, w is also a constant, its value being .2604. It has been shown that for multiplication and division, the numbers used may be limited to one more figure than the number of significant figures required in the result without loss of accuracy. What is true in this respect of multiplication and division is also true of addition, subtraction, powers, and roots. Consequently, in all practical applications of formulas, the number of significant figures used in all the quantities contained in the formulas may be limited to one more than is contained in the constant having the smallest number of significant figures, and the result should be limited to the same number of significant figures as this constant 22.5G contains. For example, the formula A = P+ 8.62 two constants, both having 3 significant figures; hence, the values of G and P may be limited to 3 + 1 = 4 significant figures, and the result when found should be expressed to 3 significant figures. The formula d 1.54aD+2.6 contains two constants, one of which contains but two significant figures; hence, the values of a and D may be limited to 3 significant figures and the value of d when found should be limited to 2 significant figures. = contains 37. In the examples that follow, unless for some reason a special exception is made, all applications of formulas will be made under the assumption that all the quantities, both constants and variables, have been limited to one more significant figure than the number of significant figures contained in the constant having the least number of significant figures; if there are no constants, then all quantities should ordinarily be limited to 5 significant figures, and the results will be expressed to 4 significant figures. = EXAMPLE.-In the formula s = 48EI' suppose that W = 1200 pounds, 108 inches, E = 1,500,000 pounds, and I 8 inches; find the value of s. and d = bd3 - 12 in which b = 3 inches SOLUTION. In this formula, the constants 48 and 12 are exact, and are therefore correct to any number of significant figures; the number 1,500,000 is correct to only 2 significant figures, and it is doubtful even if the second figure is correct. Consequently, it is useless to employ more than 3 significant figures in the calculation. First calculate the value of I, obtaining I = 3 X 83 = 128. Substituting this value of I and the other values given in the above formula, The formula just given gives the value of s in inches, s being the deflection of a beam having a certain shape and under certain conditions of loading, etc. Therefore, the deflection in this case would be taken as .16 inch, and this value is as close as can be obtained, although it might be a trifle more or a trifle less. Note that instead of cubing the number 108, the cube was expressed as the product of three factors, in order to employ cancelation. APPROXIMATE METHOD FOR FINDING ROOTS 38. Cube and Fifth Roots of Numbers.-In certain formulas used in engineering, it is sometimes necessary to extract the cube root or, less frequently, the fifth root of a number. These roots may be found by an extension of the method explained in Arithmetic for finding the square root; but it entails a great amount of labor, and for practical purposes an approximate method answers the purpose equally well and is much easier to apply. Of the many approximate methods that have been recommended, the following is, perhaps, the simplest and most accurate. It was discovered by Charles Hutton, a famous English mathematician. Let n = the number whose root is to be found; let r = index of the root, and let a be a number a little greater or a little less than the exact value of the root, so that a' is a little greater or a little less than n. Then, [ Gr n = Dar] ̄(r + 1)n + (r − 1)a′] (r For cube root, r = a, nearly. 3, and formula (1) reduces to (1) the The more nearly a approaches in value to n the more accurate will be the value obtained for the root. In order to save labor in finding a, the following table has been calculated; it gives the cubes and fifth powers of 1.1, 1.2, 1.3, etc. up to 9.9, and is used as described below. Suppose it is desired to find the cube root of 34,586. The first step is to point off the number into periods of 3 figures each, 3 being the index, beginning with the decimal point and going to the right and left. If the number contains an integral part, point off that part only; but if it is a pure decimal, point it off to the right. Thus, the above number, when pointed off, becomes 34'586. If it had been .034586, it would become .034'586 when pointed off. The next step is to move the decimal point so that it will follow the first period that contains a digit, and the given number then becomes 34.586. The given number, after shifting the decimal point will be called the altered number. Of course, if the integral part of the given number contains not more than 3 figures, it is not necessary to shift the decimal point. = = Now referring to the table, and looking in the column headed n3, the number 34.586 is found to lie between 32.768 3.23 and 35.937 3.33; the cube root of 34.586 therefore lies between 3.2 and 3.3, and one of these two numbers is to be selected for a in applying formula (2). To decide which one, find which of the two cubes just mentioned is nearest in value to the altered number. The easiest way to do this is to = |
