so included are to be multiplied by another quantity. For example, 3 xb+c, b+cxa, vb + c Xa, etc., are always written as here printed. Before proceeding further, we will explain one other device that is used by formula makers, and which is apt to puzzle one who encounters it for the first time, It is the use of what mathematicians call primes and subs., and what printers call superior and inferior characters. As a rule, formula makers designate quantities by the initial letters of the names of the quantities. For example, they represent volume by v, pressure by p, height by h, etc. This practice is to be commended, as the letter itself serves in many cases to identify the quantity that it represents. Some authors carry the practice a little further and represent all quantities of the same nature by the same letter throughout the book, always having the same letter represent the same thing. Now, this practice necessitates the use of the primes and subs. above mentioned when two quantities have the same name, but represent different things. Thus, consider the word pressure as applied to steam at different stages between the boiler and the condenser. First, there is absolute pressure, which is equal to the gauge pressure in pounds per square inch plus the pressure indicated by the barometer reading (usually assumed in practice to be 14.7 pounds per square inch, when a barometer is not at hand). If this be represented by p, how shall we represent the gauge pressure? Since the absolute pressure is always greater than the gauge pressure, suppose we decide to represent it by a capital letter, and the gauge pressure by a small (lower-case) letter. Doing so, P represents absolute pressure, and p gauge pressure. Further, there is usually a "drop" in pressure between the boiler and the engine, so that the initial pressure, or pressure at the beginning of the stroke, is less than the pressure at the boiler. How shall we represent the initial pressure? We may do this in one of three ways, and still retain the letter p or P to represent the word pressure: First, by the use of the prime mark; thus, p' or P' (read p prime and p major prime) may be considered to represent the initial gauge pressure or the initial absolute pressure. Second, by the use of sub. figures; thus, p1 or P1 (read p sub. one and p major sub. one). Third, by the use of sub. letters: thus, p; or P¡ (read p sub. i and P major sub. i). Likewise, p' (read p second), p2, or p, might be used to represent the gauge pressure at release, etc. Sub. letters have the advantage of still further identifying the quantity represented; in many instances, however, it is not convenient to use them, in which case primes and subs. are used instead. The prime notation may be continued as follows: p''', piv, pv, etc.; it is inadvisable to use superior figures, for example, p1, p2, p3, pa, etc., as they are liable to be mistaken for exponents. The main thing to be remembered by the reader is that when a formula is given in which the same letters occur several times, all like letters having the same primes or subs. represent the same quantities, while those that differ in any respect represent different quantities. Thus, in the formula w1, w2, and 3 represent the weights of three different bodies; 81, 82, and 83 their specific heats; and t1, t2, and to their temperatures; while t represents the final temperature, after the bodies have been mixed together. It is very easy to apply the above formula when the values of the quantities represented by the different letters are known. All that is required is to substitute the numerical values of the letters, and then perform the indicated operations. Thus, suppose that the values of w1, 81, and t1 are, respectively, 2 pounds, .0951, and 80°; of w2, 82, and t2, 7.8 pounds, 1, and 80°, and of wз, 83, and tg, 34 pounds, .1138, and 780°; then, the final temperature t is, substituting these values for their respective letters in the formula, 2.0951 X 80 +7.8 × 1 × 80 + 34 × .1138 × 780 2 X .0951 +7.8 X 1 + 3 X .1138 15.216+624 + 288.483 .1902+7.8 + .36985 927.699 = 110.97°. In substituting the numerical values, the signs of multiplication are, of course, written in their proper places; all the multiplications are performed before adding, according to the rule previously given. The reader should now be able to apply any formula involving only algebraic expressions that he may meet with, not requiring the use of logarithms for their solution. We will, however, call his attention to one or two other facts which he may have forgotten. Expressions similar to 160 660 25 sometimes occur, the heavy line indicating that 160 is to be divided by the quotient obtained by dividing 660 by 25. If both lines were light it would be impossible to tell whether 160 was to be divided by 160 660 25 or whether was to be divided by 25. If this latter result 660 160 were desired, the expression would be written 660 In every case the heavy line indicates that all above it is to be divided by all below it. In an expression like the following, 160 the heavy line 660' 7+ 25 is not necessary, since it is impossible to mistake the opera 25 line becomes necessary in order to make the resulting expression clear. Fractional exponents are sometimes used instead of the radical sign. That is, instead of indicating the square, cube, fourth root, etc. of some quantity, as 37 by 37, 37, † 37, etc. these roots are indicated by 37a, 37, 37a, etc. Should the numerator of the fractional exponent be some quantity other than 1, this quantity, whatever it may be, indicates that the quantity affected by the exponent is to be raised to the power indicated by the numerator; the denominator is always the index of the root. Hence, instead of expressing the cube root of the square of 37 as 372, it may be expressed 373, the denominator being the index of the root; in other words, 372 = 37. Likewise, (1+a2b)3 may also be written (1 + a2b), a much simpler expression. We will now give several examples showing how to apply some of the more difficult formulas that the reader may encounter. The area of any segment of a circle that is less than (or equal to) a semicircle is expressed by the formula. 3.1416; angle obtained by drawing lines from the center to the extremities of arc of segment; c = chord of segment; h = height of segment. EXAMPLE.-What is the area of a segment whose chord is 10 in. long, angle subtended by chord is 83.46°, radius is 7.5 in., and height of segment is 1.91 in.? SOLUTION.-Applying the formula just given, =40.968-27.95 = 13.018 sq. in., nearly. The area of any triangle may be found by means of the following formula, in which A represent the lengths of the sides: = the area, and a, b, and c EXAMPLE. What is the area of a triangle whose sides are 21 ft., 46 ft., and 50 ft. long? SOLUTION.-In order to apply the formula, suppose we let a represent the side that is 21 ft. long; b, the side that is 50 ft. long; and c, the side that is 46 ft. long. Then, substituting in the formula, The above operations have been extended much further than was necessary; this was done in order to show the reader every step of the process. The Rankine-Gordon formula for determining the least load in pounds that will cause a long column to break is in which P = load (pressure) in lb.; S = ultimate strength (in lb. per sq. in.) of material composing column; A = area of cross-section of column in sq. in.; q = a factor (multiplier) whose value depends on the shape of the ends of the column and on the material composing the column; / length of the column in in.; G = least radius of gyration of cross-section of column. = The values of S, 9, and G2 are all given in tables to be found in books on strength of materials. EXAMPLE.-What is the least load that will break a hollow wrought-iron column whose outside diameter is 14 in., inside diameter 11 in., length 20 ft., and whose ends are flat? = 150,000, and q = 1 25,000 for flat SOLUTION. For steel, S ended steel columns; A, the area of the cross-section, .7854(d12 — d22), d1 and d2 being the outside and inside diam. = eters, respectively; l 20 X 12 240 in.; and G2 = Substituting these values in the formula, SA 150,000 .7854(142-112) = P = 150,000 X 58.905 8,835.750 7,915,211 lb. 1+.1163 1.1163 |