CHAPTER HI. TRIGONOMETRIC SERIES. A. TRIGONOMETRIC FUNCTIONS. 66. For the engineer, and especially the electrical engineer, a perfect familiarity with the trigonometric functions and trigonometric formulas is almost as essential as familiarity with the multiplication table. To use trigonometric methods efficiently, it is not sufficient to understand trigonometric formulas enough to be able to look them up when required, but they must be learned by heart, and in both directions; that is, an expression similar to the left side of a trigonometric for- mula must immediately recall the right side, and an expression similar to the right side must immediately recall the left side of the formula. Trigonometric functions are defined on the circle, and on the right triangle. Let in the circle, Fig. 28, the direction to the right and upward be considered as positive, to the left and downward as negative, and the angle a be counted from the positive hori- zontal OA, counterclockwise as positive, clockwise as negative. The "projector s of the angle a, divided by the radius, is called sin a; the projection c of the angle a, divided by the radius, is called cos a. The intercept t on the vertical tangent at the origin A, divided by the radius, is called tan a; the intercept ct on the horizontal tangent at B, or 90 deg., behmd A, divided by the radius, is called cot a. (1) &i Thus, in Fig. 28, .sma=-; c cos «=-; tana = -; ct cot a = - r TRIGONOMETRIC SERIES. 95 In the right triangle, Fig. 29, with the angles a and /?, opposite respectively to the cathetes a and h, and with the hypotenuse c, the trigonometric functions are : sinQ: = cos^ = -; cosa:=sin^=- c c tan a = cot cot Q:=tan /? = (2) By the right triangle, only functions of angles up to 90 deg., or -, can be defined, while by the circle the trigonometric functions of any angle are given. Both representations thus must be so familiar to the engineer that he can see the trigo- B 1- y^ s y t 1 0 C 1 A ^ < Fig. 28. Circular Trigonometric Functions. Fig. 29. Triangular Trigono- metric Functions. nometric functions and their variations with a change of the angle, and in most cases their numerical values, from the mental picture of the diagram. 67. Signs of Functions. In the first quadrant, Fig. 28, all trigonometric functions are positive. In the second quadrant, Fig. 30, the sin a. is still positive, as s is in the upward direction, but cos a is negative, since c is toward the left, and tan a and cot a also are negative, as t is downward, and d toward the left. In the third quadrant. Fig. 31, sin a and cos a are both 96. ENGINEERING MATHEMATICS. negative: s being downward, c toward the left; but tan a and cot a are again positive, as seen from t and ct in Fig. 31. \ ct B / f^ '^^ ^ 1 v c 0 N t Fig. 30. Second Quadrant. Fig. 31. Third Quadrant. In the fourth quadrant, Fig. 32, sin a is negative, as s is downward, but cos a is again positive, as c is toward the right; tan a and cot a are both negative, as seen from t and ct in Fig. 32. In the fifth quadrant all the trigonometric functions again have the same values as in the first quadrant. Fig. 28, that is, 360 deg., or 2?:, or a multiple thereof, can be added to, or subtracted from the angle «, without changing the trigonometric functions, but these functions repeat after every 360 deg., or 2;:; \ ct h B C^ A . I v ^ Xj — + t Fig. 32. Fourth Quadrant. that is, have 2;: or 360 deg. as their period. SIGNS OF FUNCTIONS Function. Positive. Negative. sin a COS a tan (X cot a 1st and 2d 1st and 4th 1st and 3d 1st and 3d 3d and 4th quadrant 2d and 3d 2d and 4th 2d and 4th (3) TRIGONOMETRIC SERIES. 97 68. Relations between sin a and cos a. Between sin a and cos a the relation, sin^ a+cos^ a = l, exists; hence, sin a Vl cos^ a ; (4) (4a) COS a = Vl — sin^ a. Equation (4) is one of those which is frequently used in both directions. For instance, 1 may be substituted for the sum of the squares of sin a and cos a, while in other cases sin^ a +cos2 a may be substituted for 1. For instance, 1 sin^a + cos^cK /aiTi^\2 /sin a\' Vcos a/ cos^ a cos- a Relations between Sines and Tangents. + I=tan2a: + 1. hence tan a = cot a = cot a = tan a = sm a cos a ' I COS a sm a (5) tan a ■ 1 cot a' (5a) As tan a and co^ a are far less convenient for trigonometric calculations than sin a and cos a, and therefore are less fre- quently applied in trigonometric calculations, it is not neces- sary to memorize the trigonometric formulas pertaining to tan a and cot a, but where these functions occur, sin a and and cos a are substituted for them by equations (5), and the calculations carried out with the latter functions, and tan a or cot a resubstituted in the final result, if the latter contains sin a . . - , or its reciprocal. cos a In electrical engineering tan a or cot a frequently appears as the starting-point of calculation of the phase of alternating currents. For instance, if a is the phase angle of a vector 98 ENGINEERING MATHEMATICS. quantity, tan a is given as the ratio of the vertical component over the horizontal component, or of the reactive component over the power component. In this case, if m . ,. . tan ex = a sin a = a and cos « = Va^ + h^ cot a = c "d' sin a = d and COS a = or, if Vc^+d^' Vc^+d^' (5c) The secant functions, and versed sine functions are so little used in engineering, that they are of interest only as curiosities. They are defined by the following equations: sec a = cos a' 1 cosec a = - , sm a ski vers a = l — sin a, cos vers a = 1 — cos a. 69. Negative Angles. From the circle diagram of the trigonometric functions follows, as shown in Fig. 33, that when changing from a positive angle, that is, counterclockwise rotation, to a negative angle, that is, clockwise rotation, s, t, and ct reverse their direction, but c remains the same; that is, sin (— «) = —sin a, cos ( — a)= +COS a, tan ( — a) = —tan a, cot { — a) = —cot a. (6) cos a thus is an " even function," while the three others are ^' odd functions." TRIGONOMETRIC SERIES. 99 Supplementary Angles. From the circle diagram of the trigonometric functions follows, as shown in Fig. 34, that by changing from an angle to its supplementary angle, s remains in the same direction, but c, t, and ct reverse their direction, and all four quantities retain the same numerical values, thus, sin {7z—a)= +sin a, cos (tt— a) = — cos a, tan (n—a) == —tan a, cot (tt— a) = — cot a. (7) Fig. 33. Functions of Negative ^ Fig. 34. Functions of Supplementary Angles. Angles. Complementary Angles. Changing from an angle a to its complementary angle 90 — a, or ^ — «, as seen from Fig. 35, the signs remain the same, but s and c, and also i and d exchange their numerical values, thus. sm I 9 ~ « ) = COS a-, lit \ . cosi ^ — a I ==sm a, tani 7y—a\= cot a, cotf|-aj tan a. (8) 100 ENGINEERING MATHEMATICS. 70. Angle {a±7i). Adding, or subtracting 7: to an angle a, gives the same numerical values of the trigonometric functions Fig. 35. Functions of Complemen- Fig. 36. Functions of Angles Plus tary Angles. or Minus tz. as a, as seen in Fig. 36, but the direction of s and c is reversed while t and ct remain in the same direction, thus, sin (a ±71:) = — sin a, cos (a ±7[) = —cos a, tan (a ±7r) = +tan a, cot {a±7L)= +cot a. J (9) ■1 \ ct B ct ^ 2 \ t 0 m \^\s ) \ A * t FiG. 37. Functions of Angles+ ^. Fig. 38. Functions of Angles Minus |-. Angle(a±|). Adding ^, or 90 deg. to an angle a, inter- changes the functions, s and c, and t and ct, and also reverses TRIGONOMETRIC SERIES. 101 the direction of the cosiije, tangent, and cotangent, but leaves the sine in the same direction, since the sine is positive in the second quadrant, as seen in Fig. 37. Subtracting ^, or 90 deg. from angle a, interchanges the functions, s and c, and t and ct, and also reverses the direction, except that of the cosine, which remains in the same direction; that is, of the same sign, as the cosine is positive in the first and fourth quadrant, as seen in Fig. 38. Therefore, sm cos tan cot sm cos tan cot («+!)= + COS a, / '^\ ("+27 = — sin a, ("+1)=- — cot a, («+!)= —tan a, / '^\ ("-2)= — cos a, Hh +sin a, Hh — cot a, Hh — tan a. (10) (11) Numerical Values. From the circle diagram. Fig. 28, etc., follows the numerical values : sin 0° = 0 sin 30° = i sin 45° = ^\/2 sin 60° = iV3 sin 90° =1 sinl20° = i\/3 etc. cos 0° = = 1 cos 30° = = +\/3 COS 45° = = i\/2 COS 60° = = i COS 90° = = 0 COS 120° = etc. = -i tan 0° = = 0 tan 45° = = 1 tan 90° = = 00 tan 135° = = -1 etc. cot 0° = oo cot 45° =1 cot 90° = 0 cot 135°= -1 etc. (12) 102 ENGINEERING MATHEMATICS. 71. Relations between Two Angles. The following relations are developed in text-books of trigonometry : sin (a +/?) =sin a cos /? + cos a sin /?, sin (a — /3)=sin a cos /9— cos ct sin /?, cos (a +/?) =cos a cos /?— sin a sin ft cos (a — /5) =cos a cos /?+sin a sin /?, . Herefrom follows, by combining these equations (13) in (13) pairs : cos a cos/? = i{cos (a+ft +cos (a — /?)!, sin a sin /? = J{cos (a — /?) — cos (aH-^}), sin a cos/? = J{sin (a+/?)+sin (a — /?)}, cos a sin/5 = J{sin (0:+^) — sin (a — /?)|. (14) By substituting ai for (a+/3), and /?i for {a — ^8) in these equations (14), gives the equations, sin cti+sin ft sin «! — sin ft . ai+ft «i-ft, 2 sm — p: cos — ~ — 2 sin cosai+cosft= 2 cos cos ai — cos/9i 2 «i-ft 2 «i+ft 2 «i+ft «l+ft; COS— 2- COS 2 n'l-ft. (15) These three sets of equations are the most important trigo- nometric formulas. Their memorizing can be facilitated by noting that cosine functions lead to products of equal func- tions, sine functions to products of unequal functions, and inversely, products of equal functions resolve into cosine, products of unequal functions into sine functions. Also cosine functions show a reversal of the sign, thus: the cosine of a sum is given by a difference of products, the cosine of a differ- ence by a sum, for the reason that with increasing angle the cosine function decreases, and the cosine of a sum of angles thus would be less than the cosine of the single angle. TRIGONOMETRIC SERIES. 103 Double Angles. From (13) follows, by substituting a for ^: sin 2a: = 2 sin a cos a, .... (16) COS 2a = cos^ a — sin^ «, = 2cos^ a-1, = l-2sin2a. fron 1 follow sin^ 1- cos^ a and cos^ a- 1 + cos^ a (16a) 72. Differentiation. -T- (sin x)= +C0S X, -T- (cos x) = —sin a:. dx ^ ' (17) The sign of the latter differential is negative, as with an increase of angle a, the cos a decreases. Integration. 1 sin a:da:=— cos a, I cosadcK= +sina. Herefrom follow the definite integrals : sin (a + a)da: = 0; Xc+2jt cos (a:+a)da=0; sin (a + a)da = 2 cos (c + a) ; Xc + jr COS (a +a)da = - 2 sin (c +a) ; (18) (18a) (186) 104 ENGINEERING MATHEMATICS, I Oin rv/lrv =f\' I sin ada = 0 ; ! r cos ada=0; smada= +1; cos ada = +1. (18c) (ISd) 73. Binomial. One of the most frequent trigonometric operations in electrical engineering is the transformation of the binomial, a cos a + 6 sin a, into a single trigonometric function, by the substitution, a = c cos p and 6 = c sin p; hence, a cos a + 6 sin a =c cos (a — 7?), . . . (19) where h . . (20) 'c = Va^ + h^ and tanp = -; . . u or, by the transformation, a = c sin q and h = c cos q, a cos a:+6 sin a==c sin («+g), . . where c = \^+b^ and ism q = T-' 74. Polyphase Relations. (21) (22) X^ / 2mi7z\ ^ 2^(^os[a+a±—^j=0, ^ . / • 2mi7t\ ^ 2^ sm[a+a±—^j=0, (23 where m and n are integer numbers. Proof. The points on the circle which defines the trigo- nometric function, by Fig. 28, of the angles (a-\-a± 1, TRIGONOMETRIC SERIES. 105 are corners of a regular polygon, inscribed in the circle and therefore having the center of the circle as center of gravity. For instance, for n = 7, m = 2, they are shown as Pi, P2, . . . P7, in Fig. 39. The cosines of these angles are the projections on the vertical, the sines, the projections on the horizontal diameter, and as the sum of the projections of the corners of any polygon, Fig. 39. Polyphase Relations. FiG. 40. Triangle. on any line going through its center of gravity, is zero, both sums of equation (23) are zero. 2^ cos [a+a±—^j cos {^a+h±—^j=- cos {a-h), (24) 2^ sm(^a+a±— ^jsm^a + 6±^^j=-cos {a-h), 2^ sm (a+ai-^— jcos / a: + 6±— -— 1 =- sm {a-h). These equations are proven by substituting for the products the single functions by equations (14), and substituting them in equations (23). 75. Triangle. If in a triangle a, ./?, and y are the angles, opposite respectively to the sides a, 6, c. Fig. 40, then, sin a-^-sin /?4-sin y = a^h^c^ (25) %%%%%% i 1 1 ^ 106 ENGINEERING MATHEMATICS. COS )- = 2ah or \ c^ = a--\-h'^—2ahQosr. ah sin y (26) Area = 2 c^ sin a sin ^ sin 7- (27) B. TRIGONOMETRIC SERIES. 76. Engineering phenomena usually are either constant, transient, or periodic. Constant, for instance, is the terminal voltage of a storage-battery and the current taken from it through a constant resistance. Transient phenomena occur during a change in the condition of an electric circuit, as a change of load; or, disturbances entering the circuit from the outside or originating in it, etc. Periodic phenomena are the alternating currents and voltages, pulsating currents as those produced by rectifiers, the distribution of the magnetic flux in the air-gap of a machine, or the distribution of voltage around the commutator of the direct-current machine, the motion of the piston in the steam-engine cylinder, the variation of the. mean daily temperature with the seasons of the year, etc. The characteristic of a periodic function, y=f{x), is, that at constant intervals of the independent variable x, called cycles or periods, the same values of the dependent variable y occur. Most periodic, functions of engineering are functions of time^ or of space, and as such have the characteristic of univalence; that is, to any value of the independent variable x can corre- spond only one value of the dependent variable y. In other words, at any given time and given point of space, any physical phenomenon can have one numerical value only, and obviously, must be represented by a univalent function of time and space. Any univalent periodic function, ' • ; ■ y-m, (1) TRIGONOMETRIC SERIES. 107 can be expressed by an infinite trigonometric series, or Fourier series, of the form,\ y = ao-\-ai cos cx+a2Cos 2cx-\-a3 cos 3cx + . . . . + &1 sin cx+fe2sin2cx+63 sin 3cx + . . . ; .... (2) or, substituting for convenience, ex = 0, this gives { 2/ = o^o+ai cos ^+a2 cos 2<9 + a3 cos 3^ + . . . +&1 sin (9+62 sin 2/9 + 63 sin 3^ + . .. ; (3) or, combining the sine and cosine functions by the binomial (par. 73), y = ao+cicos (<9-/?i) +C2 cos (2(9-/?2) +C3 cos (3/9-^3) +• ■ = oo+cisin(/9 + ri)+C2sin(2^ + r2)+C3sin (Sd i- ys) + • . • ' where or Cn-- = v^ '+hr?] tan ^n- hn an' tan Tn-- On hn (5) The proof hereof is given by showing that the coefficients a^ and 5 „ of the series (3) can be determined from the numerical values of the periodic functions (1), thus, y=m=Me). (6) Since, however, the trigonometric function, and therefore also the series of trigonometric functions (3) is univalent,, it follows that the periodic function (6), y=fo{d)f must be uni- valent, to be represented by a trigonometric series. 77. The ; most ; important periodic functions in electrical engineering are the alternating currents and e.m.fs. Usually they are, in first approximation, represented by a single trigo- nometric function, as : i = io cos {O—ix))] or, e = eo sin (d—d); that is, they are assumed as sine waves. 108 ENGINEERING MATHEMATICS. f ■ . Theoretically, obviously this condition can never be perfectly attained, and frequently the deviation from sine shape is suffi- cient to require practical consideration/ especially in those cases, where the electric circuit contains electrostatic capacity, as is for instance, the case with long-distance transmission lines, underground cable systems, high potential transformers, etc. (However, no matter how much the alternating or other periodic wave differs from simple sine shape — ^that is, however much the wave is '' distorted," it can always be represented by the trigonometric seriesj(3). As illustration the following applications of the trigo- nometric series to engineering problems may be considered: {A) The determination of the equa^ori^of_the_,periodic function; that is, the evolution of~tRe constants a^ and b^ of the trigonometric series, if the numerical values of the periodic function are given. Thus, for instance, the wave of an alternator may be taken by oscillograph or wave-meter, and by measuring from the oscillograph, the numerical values of the periodic function are derived for every 10 degrees, or every 5 degrees, or every degree, depending on the accuracy required. The problem then is, from the numerical values of the wave, to determine its equation. While the oscillograph shows the shape of the wave, it obviously is not possible therefrom to calculate other quantities, as from the voltage the current under given circuit conditions, if the wave shape is not first represented by a mathematical expression. . It therefore is of importance in engineering to translate thejicite or the table "^ of numerical values of a periodic function into a mathematical expression thereof. • ' , (B) If one of the engineering quantities, as the e.m.f. of an alternator or the magnetic flux in the air-gap of an electric machine, is given as a general periodic function in the form of a trigonometric series, to determine therefrom other engineer- ing quantities, as the current, the generated e.m.f., etc. A. Evaluation of the Constants of the Trigonometric Series from , the Instantaneous Values of the Periodic Function. 78. Assuming that the numerical values of a univalent periodic function y=fo{d) are given; that is, for every value of 6, the corresponding value of y is known, either by graphical representation, Fig. 41; or, in tabulated form. Table I, but TRIGONOMETRIC SERIES. 109 the equation of the periodic function is not known. It can be represented in the form, 2/ = ao+ai cos ^+a2Cos 2/9 + a3 cos3<9 + . . .+anCos nd + . . . + &1 sin /9 + 62sin2^ + 6j^sin 3^ + . . .+6„sin nd + . . . , (7) and /the problem now as, to determine the coefficients ao, ai, a2 . L. 6i, 62 . • . . Fig. 41. Periodic Functions. TABLE I. 1 e 1 y e y e y e y 0 -60 90 + 90 180 + 122 270 + S5 10 -49 100 + 61 190 + 124 280 + 65 20 -38 110 + 71 200 + 126 290 + 35 30 -26 120 + 81 210 + 125 300 + 17 40 -12 130 + 90 220 + 123 310 0 50 0 140 + 99 230 + 120 320 -13 60 + 11 150 + 107 240 + 116 330 -26 70 + 27 160 + 114 250 + 110 340 -38 80 + 39 170 + 119 260 + 100 350 -49 90 + 50 180 + 122 270 + 85 360 -60 Integrate the equation (7) between the hmits 0 and 27r : Jr2n ~ r2iz r'^fT' r^iz I ydQ^a^X d<9 + ai I cos <9d<9 + a2 ( cos 2/9^/9 + . . . X2z r2n cosn^c?(9 + . . .+61 j ^mddO + sin 2ddd + . . . +6n j sin nddd + ..A 2n -"^ , /sin 2^ /2'^ —qlq I d I +ai/sin^/ 0 / /o / ^ /o /sinn/9 /2'^ , , / . /2 ■^an / + . . .-61 /cos d ^cos2/9 /2'^ ^ /cos 2Q /^^^ /cos nd /^^ / 2/0 * ' * V n /o 110 ENGINEERING MATHEMATICS. All the integrals containing trigonometric functions vanish, as the trigonometric function has the same value at the upper limit 2;r as at the lower limit 0, that is, /cos nd /2'^ 1 /^^r~/o =^(cos2n;r-cos0)=0; /sin nO /^' 1 , . ^ . „. ■ and the result is ydO = ao/ ^/ = 2;rao ; / /o hence £ 2 Jo ' ^o = 2^\ ydO (8) ydO is an element of the area of the curve y, Fig. 41, and Jydd thus is the area of the periodic function y, for one 0 ^ period; that is, «o = 2;:^, (9) where >4 = area of the periodic function y=fo{0), for one period; that is, from d = Oto 0 = 2?:. A 2r is the horizontal width of this area A, and pr- thus is the area divided bj^ the width of it; that is, it is the average height of the area ^ of the periodic function y; or, in other words, it is the average value of y. Therefore, ao = avg. (y)o^" (10) The first coefficient, ao, thus, is the average value of the instantaneous values of the periodic function y, between ^=^0 and 6 = 271. Therefore, averaging the values of y in Table I, gives the first constant ao. 79. To detei-mine the coefficient an, multiply equation (7) by cos nd, and then integrate from.O to 27r, for the purpose of making the trigonometric functions vanish. This gives TRIGONOMETRIC SERIES. Ill ( y cos nOdO = ao ( cos nOdO+ai ( cos n6 cos Odd + Jo J Jo Jo +a2 I cos n^ cos 2/9(i/9 + . . .+an j cos^ nddd + .,. + 61 I cosn^sm(9(i/9+?>2 I cosn^sin2i9d(9 + . . . + ?>n I COS n^ sin n^d^ + . . . Hence, by the trigonometric equations of the preceding section : r2^ r2r. r2n I 2/cosn(?d^=ao I cosn(9d^+ai I i[cos(n+l)/9+cos(n-r^]c?^ Jo Jo Jo + a2 I ^"i[cos (n + 2)^+cos (n-2)^]d/9 + . . . X27Z i(l+cos2n^)d^ + .. . + hi\'^\[mi{n + \)d-^m{n-l)d]dd +&2 I i[sin(n+2)^-sin(n-2)/9]d/9 + . . . Jo X27Z hsm2nddd+: . . All these integrals of trigonometric functions give trigo- nometric functions, and therefore vanish between the Umits 0 and 2?:, and there only remains the first term of the integral multiplied with a„, which does not contain a trigonometric function, and thus remains finite : an\ 7^dO = anLjj ^arjt, and therefore, p. I y cos n0dd-=an7:] Jo hence 1 n^ (^n = - \ y COS nOdd (11) ^ Jo 112 ENGINEERING MATHEMATICS. If the instantaneous values of y are multiplied with cos ndj and the product 2/n = 2/ cos nd plotted as a curve, y cos nddd is an element of the area of this curve, shown f or n = 3 in Fig. 42, r and thus I y cos nddd is the area of this curve ; that is. 71 (12) Fig. 42. Curve of y cos ZQ. where An is the area of the curve 2/ cos nd^ between ^ = 0 and (9^27r. ] 7 A \ As 2?: is the width of this area A^ -r^ is the average height of this area; that is, is the average value of y cos n/9, and -An 71 thus is twice the average value of y cos nd\ that is. "vi^ an = 2 avg. (^ cos n<9)o 2;r . . (13) Fig. 43. Curve of y sin 3^. (The coefficient an of cos nQ is derived by multiplying all the instantaneous values of y by cos nd, and taking twice the average of the instantaneous values of this product y cos nQ.\ TRIGONOMETRIC SERIES. 113 80. hn is determined inJhG_amlQ^u^jnaiij}£r by multiply- ing y by sin nO and integrating from 0 to 27r; by the area of the curve y sin nd, shown in Fig. 43, for n = 3, /^2;r r2K r2n I y sin nOdd = ao ( sin nOdd +ai | sin n^ cos Qdd Jo Jo Jo +a2 I sin n^ cos 2^c?<9 + . . .+an ( sin n<9 cos n^d^ + . . . Jo Jo r^Tt r2iz + 61 I ^mnddd-\-h2\ sinni., &2 • • • , are calculated as the averages : 2;r tto^avg. {y)^-^^; ai=2 avg. 0/ cos 6)^^""; a2 = 2 avg. {y cos 26) ^ "; a„ = 2avg. {ycosn6)Q ": 6i = 2avg. (?/sin i^)/"; 62 = 2 avg. (y sin 2^)0""; 6n = 2avg. (?/sin 72^)0^"; (18) Hereby any individual harmonic can be calculated, without calculating the preceding harmonics. For instance, let the generator e.m.f. wave, Fig. 44, Table II, column 2, be impressed upon an underground cable system Fig. 44. Generator e.m.f. wave. of such constants (capacity and inductance), that the natural frequency of the system is 670 cycles per second, while the generator frequency is 60 cycles. The natural frequency of the TRIGONOMETRIC SERIES. 115 circuit is then close to that of the 11th harmonic of the generator wave, 660 cycles, and if the generator voltage contains an appreciable 11th harmonic, trouble may result from a resonance rise of voltage of this frequency; therefore, the 11th harmonic of the generator wave is to be determined, that is, an and 6ii calculated, but the other harmonics are of less importance. Table II d y cos 11« sin no y cos 110 J/ sin lie 0 10 20 5 4 20- + 1.000 -0.342 -0.766 0 + 0.940 -0.643 + 5.0 -1.4 -15.3 0 + 3.8 -12.9 30 40 50 22 19. 25 + 0.866 + 0.174 -0.985 -0.500 + 0.985 -0.174 + 19.1 + 3.3 -24.6 -11.0 + 18.7 - 4.3 60 70 80 29 29 30. + 0.500 + 0.643 -0.940 -0.866 + 0.766 + 0.342 + 14.5 + 18.6 -28.2 -25.1 + 22.2 + 10.3 90 100 110 38 46 3& 0 + 0.940 -0.643 - 1 . 000 + 0.342 + 0.766 0 + 43.3 -24.4 -38.0 + 15.7 + 29.2 120 130 140 41 50 32. -0.500 + 0.985 -0.174 -0.866 -0.174 + 0.985 -20.5 + 49.2 -5.6 — 35 . 5 - 8.7 f 31.5 150 160 170 30 33 7 -0.866 + 0.766 + 0.342 -0.500 -0.643 + 0.940 -26.0 + 25.3 + 2.2 -15.0 -21.3 180 -5. + 6.6 Total Divided by 9 + 34.5 + 3.83 = ai, -29.8 -3.31=b„ In the third column of Table II thus are given the values of cos 11^, in the fourth column sin 116, in the fifth column y cos 11^, and in the sixth column y sin 11^. The former gives as average +1.915, hence an--= +3.83, and the latter gives as average —1.655, hence 6ii=— 3.31, and the 11th harmonic of the generator wave is an cos 11^ +6n sin 11<9 = 3.83 cos 11^-3.31 sin 11^ = 5.07 cos (11^+410), 116 ENGINEERING MATHEMATICS. hence, its effective value is 5.07 V2 3.58, while the effective value of the total generator wave, that is, the. square root of the mean squares of the instanta- neous values y, is e = 30.5, thus the 11th harmonic is 11.8 per cent of the total voltage, and whether such a harmonic is safe or not, can now be deter- mined from the circuit constants, more particularly its resist- ance. 82. In general, the successive harmonics decrease; that is, with increasing n, the values of an and bn become smaller, and when calculating a^ and hn by equation (18), for higher values of n they are derived as the small averages of a number of large quantities, and the calculation then becomes incon- venient and less correct. Where the entire series of coefficients an and hn is to be calculated, it thus is preferable not to use the complete periodic function y, but only the residual left after subtracting the harmonics which have already been calculated; that is, after ao has been calculated, it is subtracted from y, and the differ- ence, yi=y—cio, is used for the calculation of ai and bi. Then ai cos ^ + ?>i sin ^ is subtracted from yi, and the difference, 2/2 = ?/i — (tti cos 6+bi sin 6) = ?/— (tto + ai cos ^ + 61 sin /9), , is used for the calculation of a2 and 62. Then a2 cos 2^ +&2 sin 20 is subtracted from 1/2, and the rest, ys, used for the calculation of a.^ and 63, etc. In this manner a higher accuracy is derived, and the calcu- lation simplified by having the instantaneous values of the function of the same magnitude as the coefficients a„ and b„. As illustration, is given in Table III the calculation of the first three harmonics of the pulsating current, Fig. 41, Table I: TRIGONOMETRIC SERIES. 117 83. In electrical engineering, the most important periodic functions are the alternating currents and voltages. Due to, the constructive features of alternating-current generators-, alternating voltages and currents are almost always symmet- rical waves; that is, the periodic function consists of alternate half -waves, which are the same in shape, but opposite in direc- tion, or in other words, the instantaneous values from 180 deg. to 360 deg. are the same numerically, but opposite in sign, from the instantaneous values between 0 to 180 deg., and each cycle or period thus consists of two equal but opposite half cycles, as shown in Fig. 44. In the earlier days of electrical engineering, the frequency has for this reason frequently been expressed by the number of half -waves or alternations. In a symmetrical wave, those harmonics which produce a difference in the shape of the positive and the negative half- wave, cannot exist; that is, their coefficients a and h must be zero. . Only those harmonics can exist in which an increase of the angle d by 180 deg., or tz, reverses the sign of the function. This is the case with cos nO and sin nd, if n is an odd number. If, however, n is an even number, an increase of ^ by tt increases the angle nd by 2n or a multiple thereof, thus leaves cos nd and sin nO with the same sign. The same applies to Gq. There- fore, symmetrical alternating waves comprise only the odd harmonics, but do not contain even harmonics or a constant term, and thus are represented by i/ = ai cos O+as cos S^ + as cos 5^ + . . . + 61 sin e + hssm 3(9 +65 sin 5^ + . . . (19) When calculating the coefficients a„ and hn of a symmetrical wave by the expression (18), it is sufficient to average from 0 to 7:; that is, over one half-wave only. In the second half-wave, cos nO and sin nd have the opposite sign as in the first half-wave, if n is an odd number, and since y also has the opposite sign in the second half -wave, y cos nd and y sin nd in the second half-wave traverses again the same values, with the same sign, as in the first half-wave, and their average thus is given by averaging over one half-wave only. Therefore, a symmetrical univalent periodic function, as an 118 ENGINEERING MATHEMATICS. Table e V 2/i = 2/-Oo J/, cos d 2/i sin d ci = alecs'? + 6i sin e 2/i=2/,-c> 0 10 20 -60 -49 -38 -111 -100 -89 -111 -98 -84 0 -17 -30 -84 -85 -83 -27 -15 -6 30 40 50 -26 -12 0 -77 -63 -51 -67 -48 -33 -38 -40 -39 -79 -72 ' -63 + 2 9 12 60 70 80 + 11 27 39 -40 -24 -12 -20 -8 -2 -35 -23 -12 -52 -40 -26 12 16 14 90 100 110 50 61 71 -1 + 10 20 0 -2 -7 -1 + 10 + 19 -11 + 4 18 10 6 + 2 120 130 140 81 90 99 30 39 48 -15 -25 -37 + 26 + 30 + 31 32 45 58 -2 -6 -10 150 160 170 107 114 119 56 63 68 -49 -59 -67 + 28 + 22 • + 12 67 75 81 -11 -12 -13 180 190 200 122 124 126 71 73 75 -71 -72 -71 0 -13 -26 84 85 83 -13 -12 -8 210 220 230 125 123 120 74 72 69 -64 -55 -44 -37 -47 -53 79 72 63 — 5 0 + 6 240 250 260 116 110 100 65 59 49 -32 -9 -28 -56 -48 52 40 26 13 19 23 270 280 290 85 65 35 34 + 14 -16 0 + 2 -5 -34 -14 + 15 11 -4 -IS 23 18 + 2 300 310 320 + 17 0 -13 -34 -51 -64 -17 -33 -49 + 30 + 39 + 41 -.32 -45 -58 -2 -6 -6 330 340 350 -26 -38 -49 -75 -89 -100 -65 -84 -99 + 37 + 30 + 17 -67 -75 -81 -S -14 -19 Total . .- Divided by 36 ... f 1826 + 50.7 = ao Total Divided by 18 .-1520 -84.4 = ai -204 -11.3 = 6i Total Divided by 18 . . . TRIGONOMETRIC SERIES. 119 III. i'2 C03 20 1/2 sin 20 C2 = 02 cos 2(5 + bism20 V3 = y)^C2 J/3 COS SO 2/3 sin 3(? d -27 0 -15 -12 -12 0 0 -14 — 5 -12 -3 -3 -1 10 — 5 -4 -7 + 1 0 +1 20 + 1 + 2 -1 + 3 0 + 3 30 + 2 + 9 + 4 + 5 _2 + 4 40 -2 + 12 11 + 1 -1 0 50 -6 + 10 13 -1 + 1 0 60 -12 + 10 15 + 1 -1 0 70 -13 + 5 16 _2 + 1 + 2 80 -10 0 15 — 5 0 + 5 90 -6 — 2 12 -6 -3 + 5 100 -2 -1 7 — 5 -4 + 2 110 + 1 + 2 + 1 -3 -3 0 120 + 1 + 6 -4 _9 -2 -1 130 -2 + 10 -11 + !• 0 + 1 140 — 5 + 10 -13 + 2 0 + 2 150 -9 + 8 -15 + 3 -1 + 3 160 -12 -4 -16 + 3 -3 + 1 170 -13 0 -15 + 2 _2 0 ISO -11 -4 -12 0 0 0 190 -6 -6 -7 -1 0 -1 200 -2 -4 -1 -4 0 -4 210 0 0 + 4 -4 -2 -4 220 -1 + 6 11 -5 -4 -2 230 -6 + 11 13 0 0 0 240 -15 + 12 15 + 4 + 4 + 2 250 -22 + 8 16 + 7 + 3 + 6 260 -23 0 15 + 8 0 + 8 270 -17 -6 12 + 6 -3 + 5 280 — 2 -1 7 .— 5 + 4 _2 290 + 1 + 2 + 1 -3 + 3 0 300 + 1 + 6 -4 -2 + 2 + 1 310 -1 + 6 -11 + 5 _2 -4 320 -4 + 7 -13 + 5 0 — 5 330 -11 + 9 -15 + 1 0 -1 340 -18 + 6 -16 -3 -3 + 1 350 -270 + 120 Total -33 + 27 -15.0 = 0, + 6.7 = 62 Divided by 18 -1.8 = 03 + 1.5 = 63 1 120 ENGINEERING MATHEMATICS. alternating voltage and current usually is, can be represented by the expression, y = ai cos O+a^i cos 3 ^+05 cos 5 d + aj cos 76+. . , + bi sin 0+h3 sin 3 ^+65 sin 5 6/ + 67 sin 7 6 +...; (20) where, ai =,2 avg. (y cos 6)q''] ?)i = 2 avg. (?/ sin /9)o'^; as =2 avg. (?/ cos ^6)0^; 63 =2 avg. (y sin 3/9)''; as = 2 avg. {y cos 5^)o''; ?>5 = 2 avg. (y sin 56>y; a7 = 2 avg. (y cos 7^)0"; 67 = 2 avg. (y sin 7/9)^'^. (2.1) 84. From 180 deg. to 360 deg., the even harmonics have the same, but the odd harmonics the opposite sign as from 0 to 180 deg. Therefore adding the numerical values in the range from 180 deg. to 360 deg. to those in the range from 0 to 180 deg., the odd harmonics cancel, and only the even har- monics remain. Inversely, by subtracting, the even harmonics cancel, and the odd ones remain. Hereby the odd and the even harmonics can be separated. If y = y{6) are the numerical values of a periodic function from 0 to 180 deg., and y' = y{6-\-n) the numerical values of the same function from 180 deg. to 360 deg., y2{6) = h\y{d)+y{6+n)], .... (22) is a periodic function containing only the even harmonics, and yi{0) = h\y{d)-y{6+n)\ (23) is a periodic function containing only the odd harmonics ; that is : yi{6)=ai cos 6+ as cos 3^+ 0.5 cos 5/9 + . . . +6isin ^ + bo sin 3 ^ + 65sin 5^+ ...; . . (24) i/2(/9)=ao+a2Cos 2(9 + a4 cos 4/9 + . . . + b2sin2^+d4sin4/9 + ..., (25) and the complete function is y{6)=y,{6)+y2(6) (26) TRIGONOMETRIC SERIES. 121 By this method it is convenient to determine whether even harmonics are present, and if they are present, to separate them from the odd harmonics. Before separating the even harmonics and the odd har- monics, it is usually convenient to separate the constant term ao from the periodic function y, by averaging the instantaneous values of y from 0 to 360 deg. The average then gives ao, and subtracted from the instantaneous values of y, gives yo(d)=y{e)-ao (27) as the instantaneous values of the alternating component of the periodic function; that is, the component yo contains only the trigonometric functions, but not the constant term. 2/0 is then resolved into the odd series yi, and the even series 2/2. 85. The alternating wave 2/0 consists of the cosine components : u{6)=ai cos d + a2 cos 2d + a3 cos Sd + a^ cos 4^ + . . ., (28) and the sine components : v{d)=hi sin I9 + &2 sin 2^4- 63 sin 3/9+ 64 sin 4d + . ..; (29) that is, yo{d)=u{d)+v(d). (30) The cosine functions retain the same sign for negative angles ( — ^), as for positive angles ( + ^), while the sine functions reverse their sign; that is, u{-d)=+u(d) and v{-d) = -iid). . . . (31) Therefore, if the values of 2/0 for positive and for negative angles d are averaged, the sine functions cancel, and only the cosine functions remain, while by subtracting the values of 2/0 for positive and for negative angles, only the sine functions remain; that is, yo{d)+yo{-d)=2u(d),] (32) yo(d)-yo{-d)=2v{d);\ hence, the cosine terms and the sine terms can be separated from each other by combining the instantaneous values of 2/0 for positive angle d and for negative angle (—6), thus: (33) iie)=h\y„{e)-yoi-e)].\ 122 ENGINEERING MATHEMATICS. Usually, before separating the cosine and the sine terms, u and v. first the constant term n^ is separated, as discussed above; that is, the alternating function yo = y—aQ used. If the general periodic function y is used in equation (33), the constant term Gq of this periodic function appears in the cosine term ii, thus: u{0) = h\ y(0) +{-d)}^ao+ai cos ^ +a2 cos 20 -{-as cos 3^ + . . ., while v{0) remains the same as when using ?/q. 86. Before separating the alternating function yg into the cosine function u and the sine function v, it usually is more convenient to resolve the alternating function y^ into the odd series i/i, and the even series y2, as discussed in the preceding paragraph, and then to separate 2/1 and ?/2 each into the cosine and the sine terms : Ui{d)--^^\yi{d)+yi{—d)\=aiCOsd+a3Cos3d+a5Cos56-\-. . Vi{0) = i\yi(0)-yi(~0)\-bisin0+h3sm30+b5smod+.. U2(6) = i{y2(d)+y2{-0)\=a2cos20+a4cosid + . . : 1 V2(d)=-h{y2(d)-y2{-d)]=h2sm2d+h4sm40 + . . . J (34) (35) In the odd functions ui and Vi, a change from the negative angle (—0) to the supplementary angle {n— 0) changes the angle of the trigonometric function by an odd multiple of z or 180 deg., that is, by a multiple of 2;r or 360 deg., plus 180 deg., which signifies a reversal of the function, thus : Ux(d)=\\y,{0)-yi{^-0)\A .... (36) v,{0) = hAym+yi{^-0)\.\ However, in the even functions U2 and V2 a change from the negative angle (— ^) to the supplementary angle (tc—O), changes the angles of the trigonometric function by an even multiple of tt; that is, by a multiple of 27r or 360 deg.; hence leaves the sign of the trigonometric function unchanged, thus: U2{0) = i\y2{d)+y2{7z-d)\, ] . ... (37) V2(d) = i\y2(0)-y2{7:-d)\. J TRIGONOMETRIC SERIES. 123 To avoid the possibility of a mistake, it is preferable to use the relations (34) and (35\ which are the same for the odd and for the even series. 87. Obviously, in the calculation of the constants an and hn, instead of averaging from 0 to 180 deg., the average can be made from —90 deg. to +90 deg. In the cosine function u{d), however, the same numerical values repeated with the same signs, from 0 to —90 deg., as from 0 to +90 deg., and the multiphers cos nO also have the same signs and the same numerical values from 0 to —90 deg., as from 0 to +90 deg. In the sine function, the same numerical values repeat from 0 to —90 deg., as from 0 to +90 deg., but with reversed signs, and the multipliers sin nd also have the same numerical values, but with reversed sign, from 0 to —90 deg., as from 0 to +90 deg. The products u cos nd and v sin nd thus traverse the same numerical values with the same signs, between 0 and — 90 deg., as between 0 and +90 deg., and for deriving the averages, it thus is sufficient to average only from 0 to -, or 90 deg.: that is, over one quandrant. Therefore, by resolving the periodic function y into the cosine components u and the sine components v, the calculation of the constants a„ and bn is greatly simplified; that is, instead of averaging over one entire period, or 360 deg., it is necessary to average over only 90 deg., thus: ai = 2 avg. (wi cos d)o^: hi =2 avg. (vi sin ^)o^ ; a2 = 2 avg. (u2 cos 2d)o^] 62 = 2 avg. {v2 sin 2^)^i^ a3=2 avg. {us cos 3/9)o2"; 63 = 2 avg. {vs sin Sd)o^ ; ^ (33) ai = 2 avg. (u4 cos 4^)0^ ; 64 = 2 avg. {V4. sin 4:d)o^ as = 2 avg. {us cos 5d)o^ ; 65 = 2 avg. (vs sin 5^)o^ etc. etc. where ui is the cosine term of the odd function 2/1; U2 the cosine term of the even function 1/2; u^ is the cosine term of the odd function, after subtracting the term with cos 0; that is, n3 = ui — ai cos 0, 124 ENGINEERING MATHEMATICS. analogously, W4 is the cosine term of the even function, after subtracting the term cos 2d; U4 = U2—a2 cos 26, and in the same manner, Ue = U4^—a4 cos id, and so forth; vi, V2, V5, V4, etc., are the corresponding sine terms. When calculating the coefficients an and bn by averaging over 90 deg., or over 180 deg. or 360 deg., it must be kept in mind that the terminal values of y respectively of u or v, that is, the values for ^ = 0 and 6 = 90 deg. (or 6 = ISO deg. or 360 deg. respectively) are to be taken as one-half only, since they are the ends of the measured area of the curves an cos n6 and bn sin n6, which area gives as twice its average height the values Un and bn, as discussed in the preceding. In resolving an empirical periodic function into a trigono- metric series, just as in most engineering calculations, the niost important part is to arrange the work so as to derive the results expeditiously and rapidly, and at the same time accurately. By proceeding, for instance, immediately by the general method, equations (17) and (18), the work becomes so extensive as to be a serious waste of time, while by the system- atic resolution into simpler functions the work can be greatly reduced. 88. In resolving a general periodic function y(6) into a trigonometric series, the most convenient arrangement is: 1- To separate the constant term aQ, by averaging all the instantaneous values of y(d) from 0 to 360 deg. (counting the end values at ^ = 0 and at ^ = 360 deg. one half, as discussed above) : «o = avg. {2/(^)|o^ (10) and then subtracting Aq from y(6), gives the alternating func- tion, TRIGONOMETRIC SERIES. 125 2. To resolve the general alternating function Void) into the odd function yi{d), and the even function y2(d), yi{d)-i{yo(e)-y,{d+7r)\; .... (23) 2/2W=il2/oW+2/o(^+^)! (22) 3. To resolve yi(0) gnd y2(.d)) into the cosine terms u and the sine terms v, um-h[yx{Q)+yi{-0)\ vi{d) = \{y,{d)-yi{-o)\ . U2{0)^\\y2id)^y2{-d)\ V2{d)-i{y2{0)-y2(-d)] 4. To calculate the constants ai, a2, as. . .; &i, 62, ^3- • . by the averages, a.„ = 2avg. (ucosn^)o2; . ^ ... (38) hn = 2 avg. (v sin n<9)o^. If the periodic function is known to contain no even har- monics, that is, is a symmetrical alternating wave, steps 1 and 2 are omitted. (34) (35) Fig. 45. Mean Daily Temperature at Schenectady. 89. As illustration of the resolution of a general periodic wave may be shown the resolution of the observed mean daily temperatures of Schenectady throughout the year, as shown in Fig. 45, up to the 7th harmonics.* * The numerical values of temperature cannot claim any great absolute accuracy, as they are averaged over a relatively small number of years only, and observed by instruments of only moderate accuracy. For the purpose of illustrating the resolution of the empirical curve into a trigonometric series, this is not essential, however. 126 ENGINEERING MATHEMATICS. Table IV (1) e (2) y (3) y — ao = yo yi (5) ?/2 Jan. 0 10 20 - 4.2 - 4.7 - 5.2 -12.95 -13.45 -13.95 -13.10 -13.55 -13.65 + 0.15 + 0.10 -0.30 Feb. 30 40 50 - 5.4 - 3.8 - 2.6 -14.15 - 12 . 55 -11.35 -13.55 -12.35 -11.20 -0.60 -0.20 -0.15 Mar. 60 70 80 - 1.6 + 0.2 + 1.8 -10.35 - 8.55 - 6.95 - 9.75 - 7.65 - 6.05 -0.60 -0.90 -0.90 Apr. 90 100 110 + 5.1 + 9.1 + 11.5 - 3.65 + 0.35 + 2.75 - 3.35 - 0.35 + 1.75 -0.30 + 0.70 + 1.00 May 120 130 140 + 13.3 + 15.2 + 17.7 + 4.55 + 6.45 + 8.95 + 3.90 + 5.85 + 8.15 + 0.65 + 0.60 + 0.80 June 150 160 170 + 19.2 + 19.5 + 20.6 + 10.45 + 10.75 + 11.85 + 10.10 + 10.80 + 12.15 + 0.35 -0.05 -0.30 July 180 190 200 + 22.0 + 22.4 + 22.1 + 13.25 + 13.65 + 13.35 Aug. 210 220 230 + 21.7 + 20.9 + 19.8 + 12.95 + 12.15 + 11.05 Sept. 240 250 260 + 17.9 + 15.5 + 13.8 + 9.15 + 6.75 + 5.15 Oct. 270 280 290 + 11.8 + 9.8 + 8.0 + 3.05 + 1.05 - 0.75 Nov. 300 310 320 + 5.5 + 3.5 + 1.4 - 3.25 - 5.25 - 7.35 Dec. 330 340 350 - 1.0 - 2.1 - 3.7 - 9.75 -10.85 -12.45 Total. Divided 315.1^ 8 . 75 = Oq fby36: TRIGONOMETRIC SERIES. Table V. 127 (1) 0 (2) (3) (4) Vl (5) 2/2 (6) W2 (7) ■ -90 -80 -70 + 3.35 + 0.35 - 1.75 -0.30 + 0.70 + 1.00 -60 -50 — 40 - 3.90 - 5.85 - 8.15 + 0.65 + 0.60 + 0 80 -3.0 -20 -10 0 + 10 + 20 -10.10 -10.80 -12.15 -13.10 -13.55 -13.65 + 0.35 -0.05 -0.30 + 0.15 + 0.10 -0.30 + 0.15 -0.10 -0.17 0 + 0.20 -0.12 -13.10 -12.85 -12.23 0 -0.70 -1.42 + 30 + 40 + 50 - 13 . 55 -12.35 -11.20 -11.82 -10.25 - 8.53 -1.73 -2.10 -2.67 -0.60 -0.20 -0.15 -0.12 + 0.30 + 0.22 -0.47 -0.50 -0.37 + 60 + 70 + 80 - 9.75 - 7.65 - 6.05 - 6.82 - 4.70 - 2.85 -2.93 -2.95 -3.20 -0.60 -0.90 -0.90 + 0.02 + 0.05 -0.10 -0.62 -0.95 -0.80 + 90 - 3.35 0 -3.35 -0.30 -0.30 0 128 ENGINEERING MATHEMATICS. r/} , W l-H t— I > « H p^ j/j hp pq W < ;«^ H rn C CJ ^^^ t~ Hfi c <& t^ X 0 rH CO !^ 0 CO rH r-l CM ii >^ (M CO CO ,^ on So^o 0 rH C^ ^00 (N CD q "^9 9 U5 3 odd 1 1 1 d d d 1 1 + (6 G <6 + 1 + S CD d) 1 1 1 ^^^ 0 rHiCI c cs X 00 II , v_^ "^ 10 t^ tH rt< to CO CO o § iC 10 1-1 l:^ CO GO l^ (M 0 --5 q rH cvi =^ CO 0 rH w W T-H 0 0 (>1 0 0 000 s d d d (6 <6 d> <6 <6 <6 6 6 6 1 1 1 + 1 + + + 1 + + + 10 10 LC 10 to to ic oc 00 (N CO rf^ to (M 00 ,_^ T-H 0 0 CO 0 (M >-H T-( CO £3 . . . . . . 0 000 odd 000 1 1 + 1 + 1 + + 1 «Ci •0 10 iC to to to CO CO CO CO CO 00 CO 00 CO •-^ 02 CO (N oe 10 CO 22 ;^ ^ 0 00 -^CO ^0 tH 1—1 T-H 0 0 rJH ^ S CO i d d d d d d d d r-i r^ d + + + + + + + + + + + X 0 vo (M 0 CO 00 0 to T-l (N (M CO i-H 10 rH i-t to s§ d d d + 1 + d d d 1 1 1 6 6 6'^ 1 1 1 <5i 00 10 CX) 0 to 0 'tl ^ 0 m (N 0 rfi 10 rH 0 CO to CO ^0 S" CO CO (N ^* d GO 0 ^ CM 0 e T-l T-H 1-i 1 1 1 1 1 ' 1 1 1 ^^^ r-♦^^ <& X II r« V ^ ^0 0 UO 0 10 CO CO ^ s s ^ to 00 " rH CO »0 (N 00 ^ t^ -* CM . (O • 3 CO (N i-H 0 t^ 10 CO rH 0 0 • CO 7 7 'i' 7 ' ' 1 1 1 to CO rH 1 1 1 <& ic 0 CO CO 0 CM Tt< 00 -^ CO CO rfH rf t-- So C5 0 00 !>. CO 10 CO tH 0 ^ 6 <6 (6 d> d> (6 d> d CM 0 ^ CO (M to CO CM 0 to T-H 00 (M 00 CM to 00 t- 00 ■ ^^ 0 >> £i 3 CO 1 + + + II d d d 1 1 1 <*> 2| o Ci CO i§§ X » o o 1 1 o o o + 1 + o o o o + 111 o o o 1 1 1 t^ lO i-O O 1-H O ^ § ;^ S S e^ o o O 1 1 O O o + 1 o o o o 1 + 1 1 CO B2 rt^ (N t^ ^ ^ o t^ (M Tt< O 1-H 1-H ^4 o o o 1 1 o o o 1 1 1 ^666 + + + CO o s^ ^BS X II lO t^ Tti CD O 1-H ^ o o 1 1 o o o 1 + ! o o o 1 1 + o o o 1 1 1 (M GO CO (M r^ CO (M O O tH § 2S § § ss o O O 1 1 o o o 1 + 1 o o o o 1 + + 1 ^ 00 -* CO .-H CD CO lO CD rH lO CO CO 00 CO 00 T-( (N CO S-. o O 1-1 i-H (N (N 5- 562. 5a5: -1.1 =277065- 78a5- 56.2565; 32.5 =2770tt7 + 109.267-1 102. 5a7: 3.5 =277067-109.207 -1102. 567. (12) solved, these equations give ai = 13.12 61 = 0.07 03 = - 5.03 63 = - 0.30 ^5 = -18.72 65 = - 1.15 a7 = 19.30 67 = 3.37 hence, ^ = 13. 12 cos <9-5. 03 cos 3^- 18. 72 cos 5^ + 19. 30 cos 7^ +0.07 sin^-0.30 sin 3^-1.15 sin 5/9+3.37 sin 7^ = 13.12 cos ((9-0.3°)-5.04 cos (3^-3.3°) -18.76 cos (5^-3.6°) +19.59 cos (7(9-9.9°). (13) ■ . (14) TRIGONOMETRIC SERIES. 143 96. The effective value of this current is given as the square root of the sum of squares of the effective values of the indi- vidual harmonics, thus : W2?+2? = 21.6an>p. As the voltage between Hne and neutral is 25,400 effective, this gives Q = 25,400X21. 6 = 540, 000 volt-amperes, or 540 kv.- amp. per line, thus a total of 3Q = 1620 kv.-amp. charging cur- rent of the transmission line, when using the e.m.f. wave of these old generators. It thus would require a minimum of 3 of the 750-kw. generators to keep the voltage on the line, even if no power whatever is delivered from the line. If the supply voltage of the transmission line were a perfect sine wave, it would, at 44,000 volts between the Unes, be given by ei = 36sin d, (15) which is the fundamental, or first harmonic, of equation (9). Then the current i would also be a sine wave, and- would be given by ii = ai cos ^+61 sin 0, = 13.12 cos ^+0.07 sin 0, = 13.12 cos (<9-0.3)°, (16) and its effective value would be r 13.12 ^^ 1 1 = — 7^ = 9.3 amp. V2 (17) This would correspond to a kv.-amp. input to the line 30i = 3 X 25.4 X 9.3 = 710 kv.-amp. The distortion of the voltage wave, as given by equation (1), thus increases the charging volt-amperes of the line from 710 144 ENGINEERING MATHEMATICS, kv.-amp. to 1620 kv.-amp. or 2.28 times, and while with a sine wave of voltage, one of the 750- kw. generators would easily be able to supply the charging current of the line, due to the Fig. 47. wave shape distortion, more than two generators are required. It would, therefore, not be economical to use these generators on the transmission line, if they can be used for any other purposes, as short-distance distribution. Fig. 48. In Figs. 47 and 48 are plotted the voltage wave and the current wave, from equations (9) and (14) repsectively, and TRIGONOMETRIC SERIES. 145 the numerical values, from 10 deg. to 10 deg., recorded in Table XII. In Figs. 47 and 48 the fundamental sine wave of voltage and current are also shown. As seen, the distortion of current is enormous, and the higher harmonics predominate over the fundamental. Such waves are occasionally observed as charg- ing currents of transmission lines or cable systems. 97. Assuming now that a reactive coil is inserted in series with the transmission line, between the step-up transformers and the line, what will be the voltage at the terminals of this reactive coil, with the distorted wave of charging current traversing the reactive coil, and how does it compare with the voltage existing with a sine wave of charging current? Let L= inductance, thus x = 27r/L = reactance of the coil, and neglecting its resistance, the voltage at the terminals of the reactive coil is given by ^—4 (i«) Substituting herein the equation of current, (11), gives e^ = x{ai sin d+Sas sin Sd+5a5 sin 5d+7a7 sin 76 1 -6icos/9-363Cos3/?-565Cos5<9-767Cos7<91; J hence, substituting the numerical values (13), e' = x{ 13.12 sin (9-15.09 sin 3<9-93,6 sin 5^ + 135.1 sin 7d 1 -0.07 cos ^+0.90 cos 3(9 +5.75 cos 5^-23.6 cos 7^i = 0:113.12 sin (^-0.3°) -15.12 sin (3^-3.3°) -93.8 sin (5^-3.6°) +139.1 sin (7^-9.9°)}. This voltage gives the effective value (19) K20) E' = a:VJi 13.122 + 15.122 +93.82 + 139.121=119.40:, while the effective value with a sine wave would be from (17), ^/ = a:/i = 9.3a:; hence, the voltage across the reactance x has been increased 12.8 times by the wave distortion. 146 ENGINEERING MATHEMATICS. The instantaneous values of the voltage e' are given in the last column of Table. XII, and plotted in Fig. 49, for x = l. As seen from Fig. 49, the fundamental wave has practically Fig. 49. vanished, and the voltage wave is the seventh harmonic, modi- fied by the fifth harmonic. Table XII e e i e' d c i *•' 0 10 20 -0.10 + 2.23 3.74 + 8.67 + 5.30 - 0.86 - 17 + 46 + 3 90 100 110 27.41 31.77 40.57 - 4.15 + 26.19 + 24.99 -200 -106 + 119 ■ 30 40 50 7.47 17.35 31.70 + 7.39 + 30.39 + 38.58 + 131 -116 + 36 120 130 140 42.70 33.14 18.03 - 8-10 -38.79 -36.65 + 182 + 93 - 96 60 70 80 42.06 40.33 32.87 + 15.66 -19.01 -29.13 + 167 + 159 - 54 150 160 170 6.99 2.88 1.97 -13.41 + 2.43 - 1.00 -138 - 31 + 54 90 27.41 - 4.15 -200 180 + 0.10 - 8.67 + 17 CHAjfTER IV. MAXIMA AND MINIMA. 98. In engineering investigations the problem of determin- ing the maxima and the minima, that is, the extrema of a function, frequently occurs. For instance, the output of an electric machine is to be found, at which its efficiency is a max- imum, or, it is desired to determine that load on an induction motor which gives the highest power-factor; or, that voltage Y /^ R V / 1 V. p A ^ X p ^f Q N \ ^ ^+- ^ x \ -^ / \ p y y / ^4 P. / / ^ 0 / X Fig. 50. Graphic Solution of Maxima and Minima. which makes the cost of a transmission line a minimum; or, that speed of a steam turbine which gives the lowest specific steam consumption, etc. The maxima and minima of a function, y=f{x), can be found by plotting the function as a curve and taking from the curve the values x, y, which give a maximum or a minimum. For instance, in the curve Fig. 50, maxima are at Pi and P2, minima at P3 and P4. This method of determining the extrema of functions is necessary, if the mathematical expression between 147 148 ENGINEERING MATHEMATICS. X and y, that is, the function y=f{x), is unknown, or if the function y=f{x) is so complicated, as to make the mathematical calculation of the extrema impracticable. As examples of this method the following may be chosen: 16 -1^ ^^ . ■ ■ — io- — 8- / / o c L/ ■ i2 / / o / X ^ } i \ 1 0 1 2 1 t 1 5 1 i 2 »- i 1 i 28 3Q Fig. 51. Magnetization Curve. Example i. Determine that magnetic density (B, at which tlie permeability /it of a sample of iron is a maximum. The relation between magnetic field intensity 5C, magnetic density (35 and permeability jk cannot be expressed in a mathematical equation, and is therefore usually given in the form of an 1400 1200 ^ , -■ b. ■V -"ml ^ N \ -800- -600- ^ X ■ s \, / / \ \, / \ £B ■ I \ \ i 1 > ( 5 i i • ) 1 D 1 I 'Vt ^ 4 1 5 Fig. 52. Permeability Curve. empirical curve, relating (B and JC, as shown in Fig. 51. From this curve, corresponding values of (B and 3C are taken, and their ratio, that is, the permeability /^ =— , plotted against (B as abscissa. This is done in Fig. 52. Fig. 52 then shows that a maximum MAXIMA AND MINIMA. 149 occurs at point n^^^, for (B = 10.2 kilolines, /i = 1340, and minima at the starting-point P2, for (B = 0, ju = 370, and also for (B = oo , where by extrapolation /x = l. Example 2. Find that output of an induction motor which gives the highest power-factor. While theoretically an equation can be found relating output and power-factor of an induction motor, the equation is too compUcated for use. The most convenient way of calculating induction motors is to calculate in tabular form for different values of slip s, the torque, output, current, power and volt -ampere input, efficiency, power-factor, etc., as is explained in " Theoretical Elements of Electrical Engineering," third edition, p. 363. From this P, 3 0 P^l / 7=^ ^^K \ s. / \ \, / / \ \ 2C 00 30 00 40 P on 50 00 60 [KJ V /atts Cos el 0.90 0.86 0.84 Fig. 53. Power-factor Maximum of Induction Motor. table corresponding values of power output P and power- factor cos 6 are taken and plotted in a curve. Fig. 53, and the maximum derived from this curve is P = 4120, cos ^ = 0.904. For the purpose of determining the maximum, obviously not the entire curve needs to be calculated, but only a short range near the maximum. This is located by trial. Thus in the present instance, P and cos 6 are calculated for s = 0.1 and 8 = 0.2. As the latter gives lower power-factor, the maximum power-factor is below s = 0.2. Then s = 0.05 is calculated and gives a higher value of cos 6 than s = 0.1; that is, the maximum is below s = 0.1. Then s = 0.02 is calculated, and gives a lower value of cos d than s = 0.05. The maximum value of cos 6 thus lies between s = 0.02 and s = 0.1, and only the part of the curve between s = 0.02 and s = 0.1 needs to be calculated for the determination of the maximum of cos d, as is done in Fig. 53. 99. When determining an extremum of a function y=f(x), by plotting it as a curve, the value of x, at which the extreme 150 ENGINEERING MATHEMATICS. occurs, is more or less inaccurate, since at the extreme the curve is horizontal. For instance, in Fig. 53, the maximum of the curve is so flat that the value of power P, for which cos d became a maximum, may be anywhere between P = 4000 and P=4300, within the accuracy of the curve. In such a case, a higher accuracy can frequently be reached by not attempting to locate the exact extreme, but two points of the same ordinate, on each side of the extreme. Thus in Fig. 58 the power Po, at which the maximum powder factor cos = 0.904 is reached, is somewhat uncertain. The value of power-factor, somewhat below the maximum, cos ^ = 0.90, is reached before the maximum, at Pi = 3400, and after the maximum, at P2 = 4840. The maximum then may be calculated as half-way between Pi and P2, that is, at Po = 4{Pi+P2! = 4120 watts. This method gives usually more accurate results, but is based on the assumption that the curve is symmetrical on both sides of the extreme, that is, falls off from the extreme value at the same rate for lower as for higher values of the abscissas. Where this is not the case, this method of inter- polation does not give the exact maximum. Example 3. The efficiency of a steam turbine nozzle, that is, the ratio of the kinetic energy of the steam jet to the energy of the steam available between the two pressures between which the nozzle operates, is given in Fig. 54, as determined by experiment. As abscissas are used the nozzle mouth opening, that is, the widest part of the nozzle at the exhaust end, as fraction of that corresponding to the exhaust pressure, while the nozzle throat, that is, the narrowest part of the nozzle, is assumed as constant. As ordinates are plotted the efficiencies. This curve is not symmetrical, 'but falls off from the maximum, on the sides of larger nozzle mouth, far more rapidly than on the side of smaller nozzle mouth. The reason is that wdth too large a nozzle mouth the expansion in the nozzle is carried below the exhaust pressure p2, and steam eddies are produced by this overexpansion. The maximum efficiency of 94.6 per cent is found at the point Po, at which the nozzle mouth corresponds to the exhaust pressure. If, however, the maximum is determined as mid- way between two points Pi and P2, on each side of the maximum, MAXIMA AND MINIMA, 151 at which the efficiency is the same, 93 per cent, a point Po' is obtained, which lies on one side of the maximum. With unsymmetrical curves, the method of interpolation thus does not give the exact extreme. For most engineering purposes this is rather an advantage. The purpose of deter- mining the extreme usually is to select the most favorable operating conditions. Since, however, in practice the operating conditions never remain perfectly constant, but vary to some extent, the most favorable operating condition in Fig. 54 is not that where the average value gives the maximum efficiency Hi IL ^ ^^ > ^ "^ k 90^- u Q. 8fi^ ^ ^ \ / ^ \ \ / \ \ \ y \ \ \ 0 6 0 -^ 0 No 8 zzle 0 Oper 9 ing 0 1 1 1 2 ... 1.3 Fig. 54. Steam Turbine Nozzle Efficiency; Determination of Maximum. (point Po), but the most favorable operating condition is that, where the average efficiency during the range of pressure, occurr- ing in operation, is a maximum. If the steam pressure, and thereby the required expansion ratio, that is, the theoretically correct size of nozzle mouth, should vary during operation by 25 per cent from the average, when choosing the maximum efficienc}^ point Po as average, the efficiency during operation varies on the part of the curve betw^een Pi (91.4 per cent) and P2 (85.2 per cent), thus averaging lowTr than by choosing the point Po'(6.25 per cent below Po) as average. In the latter case, the efficiency varies on the part of the curve from the Pi'(90.1 per cent) to P2'(90.1 per cent). (Fig. 55.) 152 ENGINEERING MATHEMATICS. Thus in apparatus design, when determining extrema of a function y=f{x), to select them as operating condition, consideration must be given to the shape of the curve, and where the curve is unsymmetrical, the most efficient operating point hes not at the extreme, but on that side of it at which the curve falls off slower, the more so the greater the range of variation is, which may occur during operation. This is not always realized. 100. If the function y=f(x) is plotted as a curve. Fig. 50, at the extremes of the function, the points Pi, P2, P3, Pi of curve Fig. 50, the tangent on the curve is horizontal, since 96— ^ ^ -^ "X N 5 ^ r^ \ V 0 88-^- Q ^ r^' \ \ ^ \ > C H \ t' \ 82-^ \ \ 78 ■ 0 6 0 ,7 0 No 8 -\ Dper 9 ing 1 ,0 1 1 1 .2 J Fig. 55. Steam Turbine Nozzle Efficiency; Determination of Maximum. at the extreme the function changes from rising to decreasing (maximum. Pi and P2), or from decreasing to increasing (min- imum, P3 and P4), and therefore for a moment passes through the horizontal direction. In general, the tangent of a curve, as that in Fig. 50, is the line which connects two points P' and P" of the curve, which are infinitely close together, and, as seen in Fig. 50, the angle 6, which this tangent P'P'' makes with the horizontal or X-axis, thus is given by: tan 6 = P'VJy P'Q dx MAXIMA AND MINIMA. 153 At the extreme, the tangent on the curve is horizontal, that is, 4-<9 = 0, and, therefore, it follows that at an extreme of the function, y-m, (1) %'« <^) The reverse, however, is not necessarily the case; that is, dv if at a point x, y : -^ = 0, this point may not be an extreme ; that is, a maximum or minimum, but may be a horizontal inflection point, as points P5 and Pq are in Fig. 50. With increasing x, when passing a maximum (Pi and P2, Fig. 50), y rises, then stops rising, and then decreases again. When passing a minimum (P3 and P4) y decreases, then stops decreasing, and then increases again. When passing a horizontal inflection point, y rises, then stops rising, and then starts rising again, at P5, or y decreases, then stops decreasing, but then starts decreasing again (at Pq). The points of the function y=f{x), determined by the con- dii dition, ^- = 0, thus require further investigation, whether they represent a maximum, or a minimum, or merely a horizontal inflection point. This can be done mathematically: for increasing x, when passing a maximum, tan 6 changes from positive to negative; that is, decreases, or in other words, -7- (tan /9)<0. Since dy . . d^y tan ^ =-T-, it thus follows that at a maximum -71 < 0. Inversely, at a minimum tan 6 changes from negative to positive, hence d d^y increases, that is, -7- (tan d)>0; or, -^^> 0. When passing a horizontal inflection point tan 6 first decreases to zero at the inflection point, and then increases again; or, inversely, tan 6 first increases, and then decreases again, that is, tan 6 = dy J- has a maximum or a minimum at the inflection point, and d d^y therefore, -7- (tan ^) = t-2 = 0 at the inflection point. 154 ENGINEERING MATHEMATICS, In engineering problems the investigation, whether the dv solution of the condition of extremes, j^. = 0? represents a minimum, or a maximum, or an inflection point, is rarely required, but it is almost always obvious from the nature of the problem whether a maximum of a nnnimum occurs, or neither. For instance, if the problem is to determine the speed at w^hich the efficiency of a motor is a maximum, the solution speed = 0, obviously is not a maximum but a mimimum, as at zero speed the efficiency is zero. If the problem is, to find the current at which the output of an alternator is a maximum, the solution i = 0 obviously is a minimum, and of the other two solutions, i\ and io, the larger value, i2, again gives a minimum, zero output at short-circuit current, while the inter- mediate value i\ gives the maximum. 10 1. The extremes of a function, therefore, are determined by equating its diiTerential quotient to zero, as is illustrated by the following examples : Example 4. In an impulse turbine, the speed of the jet (steam jet or water jet) is Si. At what peripheral speed So is the output a maximum. The impulse force is proportional to the relative speed of the jet and the rotating impulse wheel; that is, to {S1-S2). The power is impulse force times speed S2\ hence, P=^kS2{S^-S2), (3) and is an extreme for the value of S2, given by -j^ =^0; hence, Si-2>S2 = 0 and S2 = ^] (4) that is, when the peripheral speed of the impulse wheel equals half the jet velocity. Example 5. In a transformer of constant impressed e.m.f. eo==2300 volts; the constant loss, that is, loss which is independent of the output (iron loss), is P, = 500 watts. The internal resistance (primary and secondary combined) is r = 20 MAXIMA AND MINIMA. 155 ohms. At what current i is the efficiency of the transformer a maximum; that is, the percentage loss, A, a mininuun? The loss is P==P,+n2 = 500 +20^2 (5) The power input is Pi =ei = 2300i; .... (6) hence, the percentage loss is, ^~Pi~ ci ' ^ -^ and this is an extreme for the value of current i, given by hence. or, di Pi — ri^-=0 and i = ^/-^ = 5 amperes, ... (8) and the output is Po = 6^ = 1 1,500 watts. The loss is, P = P^ + n2 = 2P^ = 1000 watts; that is, the t^r loss or variable loss, is equal to the constant loss P,-. The percentage loss is, P VT^ ^ = -fr= — —=0.087 = 8.7 per cent, ■it c and the maximum efficiency thus is, l-;-=0.913 = 9].3 per cent. 102. Usually, when the problem is given, to determine those values of x for which y is an extreme, y cannot be expressed directly as function of x, y=f{x), as was done in examples (4) and (5), but y is expressed as function of some other quan- ties, y=fiu, V . .), and then equations between w, v . . and x are found from the conditions of the problem, by which expres- sions of X are substituted for ii, ?; . ., as show^n in the following example : Example 6. There is a constant current io through a cir- cuit containing a resistor of resistance ro. This resistor ro 156 ENGINEERING MATHEMATICS. is shunted by a resistor of resistance r. What must be the resistance of this shunting resistor r, to make the power con- sumed in r, a maximum? (Fig. 56.) Let i be the current in the shunting resistor r. The power consumed in r then is, P = ri^ (9) The current in the resistor ro is io — i, and therefore the voltage consumed by ro is roiio — i), and the vohage consumed by r is ri, and as these two vohages must be equal, since both To Fig. 56. Shunted Resistor. resistors are in shunt with each other, thus receive the same voltage, n = ro(^o-^), and, herefrom, it follows that. Substituting this in equation (9) gives, ^=(FT^- (11) dP and this power is an extreme for ;t7^0; hence: (r + rpyroV - rroHo^2 (^ + ^o) _ ^ {r + ror ' hence, r = ro; (12) that is, the power consumed in r is a maximum, if the resistor r of the shunt equals the resistance ro. MAXIMA AND MINIMA. 157 The current in r then is, by equation (10), and the power is, p^roio^ 103. If, after the function y=-f{x) (the equation (11) in dv example (6)) has been derived, the differentiation ;7- = 0 is immediately carried out, the calculation is very frequently much more complicated than necessary. It is, therefore, advisable not to differentiate immediately, but first to simplify the function y=f{x). If y is an extreme, any expression differing thereform by a constant term, or a constant factor, etc., also is an extreme. So also is the reciprocal of y, or its square, or square root, etc. Thus, before differentiation, constant terms and constant factors can be dropped, fractions inverted, the expression raised to any power or any root thereof taken, etc. For instance, in the preceding example, in equation (11), p^rroV (r+ro)2' the value of r is to be found, which makes P a maximum. If P is an extreme, r ^'^(r+ro)2^ which differs irom P by the omission of the constant factor roV, also is an extreme. The reverse of 2/1, (r + ro)^ 2/2 = ^-7 — , is also an extreme. (2/2 is a minimum, where 2/1 is a maximum, and inversely.) Therefore, the equation (11) can be simplified to the form : 2/2= — =r+2ro+Yy 158 ENGINEERING MATHEMATICS. and, leaving out the constant term 2ro, gives the final form, 2/3 = r+'^' (13) This differentiated gives, dr r^ ' hence, 104. Example 7. From a source of constant alternating e.m.f. e, power is transmitted over a line of resistance tq and reactance xq into a non-inductive load. What must be the resistance r of this load to give maximum power? If 1 = current transmitted over the line, the power delivered at the load of resistance r is P = ri^ (14) The total resistance of the circuit is r+ro; the reactance is xo] hence the current is i = --=J=, (15) \\r + roY+XQ^ and, by substituting in equation (14), the power is ^=(7^7^?-.? ^''^ if P is an extreme, by omitting e- and inverting, 2/1= ^ (r + ro)2+Xo^ r+2ro+ , is also an extreme, and likewise, 2.'2 = r+- ^ , is an extreme. MAXIMA AND MINIMA. 159 Differentiating, gives: dr f^ ' r = \W+Xo^ (17) Wherefrom follows, by substituting in equation (16), p Vro^ + xo^e^ (ro + Vro2+Xo2)2 + xo2 e2 2(ro + Vro2+Xo2) (18) Very often the function y=f{x) can by such algebraic operations, which do not change an extreme, be simplified to such an extent that differentiation becomes entirely unnecessary, but the extreme is immediately seen; the following example will serve to illustrate : Example 8. In the same transmission circuit as in example (7), for what value of r is the current i a maximum? The current i is given, by equation (15), 1 = V(r+ro)2+Xo2' Dropping e and reversing, gives, 2/i = v(r+ro)2 + Xo2; Squaring, gives, 2/2 = (r+ro)2+xo2; dropping the constant term xq^ gives 2/3 = (r+ro)2; (19) taking the square root gives y4 = r+ro; 160 ENGINEERING MATHEMATICS. dropping the constant term tq gives 2/5 = r; (20) that is, the current i is an extreme, when ?/5=-r is an extreme, and this is the case for r=0 and r = qo : r = 0 gives, (21) as the maximum value of the current, and r = oo gives i = 0, as the minimum value of the current. With some practice, from the original equation (1), imme- diately, or in very few steps, the simplified final equation can be derived. 105. In the calculation of maxima and minima of engineer- ing quantities x, y, by differentiation of the function y=f(x)j it must be kept in mind that this method gives the values of X, for which the quantity y of the mathematical equation y =f(x) becomes an extreme, but whether this extreme has a physical meaning in engineering or not requires further investigation; that is, the range of numerical values of x and y is unUmited in the mathematical equation, but may be limited in its engineer- ing application. For instance, if a: is a resistance, and the differentiation of y=f{x) leads to negative vahies of x, these have no engineering meaning; or, if the differentiation leads to values of x, which, substituted in y=f{x), gives imaginary, or negative values of y, the result also may have no engineering appUcation. In still other cases, the mathematical result may give values, which are so far beyond the range of indus- trially practicable numerical values as to be inapplicable. For instance : Example 9. In example (8), to determine the resistance r, which gives maximum current transmitted over a trans- mission line, the equation (15), 1 = V(r + ro)2+xo2 MAXIMA AND MINIMA. 161 immediately differentiated, gives as condition of the extremes: di^ 2(r+ro) dr~ 2|(r + ro)2+xo2!V(r + ro)2+a:^~ ' hence, either r + ro = 0; (22) or, (r+ro)2+Xo2 = «) . .' (23) the latter equation gives r = oo ; hence i = Q, the minimum value of current. The former equation gives r=-ro, , (24) as tne value of the resistance, which gives maximum current, and the current would then be, by substituting (24) into (15), ^=£ • • . (25) The solution (24), however, has no engineering meaning, as the resistance r cannot be negative. Hence, mathemetically, there exists no maximum value of i in the range of r which can occur in engineering, that is, within the range, 0< r< oo. In such a case, where the extreme falls outside of the range of numerical values, to which the engineering quantity is limited, it follows that within the engineering range the quan- tity continuously increases toward one limit and continuously decreases toward the other limit, and that therefore the two limits of the engineering range of the quantity give extremes. Thus r = 0 gives the maximum, r = oo the minimum of current. io6. Example lo. An alternating-current generator, of generated e.m.f. e = 2500 volts, internal resistance ro = 0.25 ohms, and synchronous reactance a:o = 10 ohms, is loaded by a circuit comprising a resistor of constant resistance r = 20 ohms, and a reactor of reactance x in series with the resistor r. What value of reactance x gives maximum output? If i = current of the alternator, its power output is P-=n2 = 20i2; (26) 162 ENGINEERING MATHEMATICS. the total resistance is r + ro = 20.25 ohms; the total reactance is xH-a:o = 10+x ohms, and therefore the current is V(r + ro)2 + (x + .To)2' and the power output, by substituting (27) in (26), is (27) p_ re^ _ 20X2500^ (r + ro2)+(x + a:o)2 20.252 + (10 +x)2- * ' ^^^-^ Simplified, this gives 2/i = (r + ro)2 + (x+xo)2; (29) 2/2 = (^+:co)2; hence, g = 2(x+xo)=0; and a;=-.'Co=-10ohms: . .... (30) that is, a negative, or condensive reactance of 10 ohms. The power output would then be, by substituting (30) into (28), re' 20+25002 ^ = (y. + roY ^ 20.252 ^^^^^ ^ ^^^ ^^- • • ^^^^ If, however, a condensive reactance is excluded, that is, it is assumed that x >0, no mathematical extreme exists in the range of the variable x, which is permissible, and the extreme is at the end of the range, x = 0, and gives P=f /TL 2 = 245 kw (32) (r + ro)2+xo2 107. Example 11. In a 500-kw. alternator, at voltage 6 = 2500, the friction and windage loss is P«, = 6 kw., the iron loss Pi = 24: kw., the field excitation loss is Py=6 kw., and the armature resistance r=0.1 ohm. At what load is the eflSciency a maximum? MAXIMA AND MINIMA. 163 The sum of the losses is: P = P^+P,. +P/ + n;2^36,000+0.1i2. . . . (33) The output is Po = e^ = 2500^; (34) hence, the efficiency is Pq ei _ 2500?: '^~Po + P~ei+P^ + Pi + P/ + n2 36000 +25002 +0.H2' ^'^^^ or, simplified, P^+Pi + Pf^ . 2/1= ^ — +^^; hence, d^/i „ P^+Pi+Pf and. di i = ^/ — ^^^ ^^ — — - \/~rn"~ == 600 amperes, (36) and the output, at which the maximum efficiency occurs, by substituting (36) into (34), is P = ei = 1500kw., that is, at three times full load. Therefore, this value is of no engineering importance, but means that at full load and at all practical overloads the maximum efficiency is not yet reached, but the efficiency is still rising. io8. Frequently in engineering calculations extremes of engineering quantities are to be determined, which are func- tions or two or more independent variables. For instance, the maximum power is required which can be delivered over a transmission line into a circuit, in which the resistance as well as the reactance can be varied independently. In other words, if y-){u,v) (37) 164 ENGINEERING MATHEMATICS. is a function of two independent variables u and v, such a pair of values of u and of v is to be found, which makes y a maximum, or minimum. Choosing any value uo, of the independent variable u, then a value of v can be found, w^hich gives the maximum (or minimum) value of y, which can be reached for u=uo* This is done by differentiating y=f(uo,v), over v, thus: ^^=0. (38) From this equation (38), a value, v=Muo), (39) is derived, which gives the maximum value of y, for the given value of Uo, and by substituting (39) into (38), y=f2{uo), (40) is obtained as the equation, which relates the different extremes of y, that correspond to the different values of Uo, with uq. Herefrom, then, that value of uo is found w^hich gives the maximum of the maxima, by differentiation: ^ = 0 (41) duo ^ Geometrically, y=f{u,v) may be represented by a surface in space, with the coordinates y, u, v. y =f(uo,v), then, represents the curve of intersection of this surface with the plane uo = constant, and the differentation gives the maximum point of this intersection curve. y=f2(:Uo) then gives the curve in space, which connects all the maxima of the various inter- sections with the Uo planes, and the second differentiation gives the maximum of this maximum curve y=f2{uo), or the maximum of the maxima (or more correctly, the extreme of the extremes). Inversely, it is possible first to differentiate over u, thus, ^'0 c^) MAXIMA AND MINIMA. 165 and thereby get u=Mvo), (43) as the value of u, which makes y a maximum for the given value of v = vo, and substituting (43) into (42), y=f4{vo), (44) is obtained as the equation of the maxima, which differentiated over vo, thus, ^-"■- ■ <«) gives the maximum of the maxima. Geometrically, this represents the consideration of the intersection curves of the surface with the planes i; = constant. The working of this will be plain from the following example : 109. Example 12. The alternating voltage e = 30,000 is impressed upon a transmission line of resistance ro = 20 ohms and reactance .xo = 50 ohms. What should be the resistance r and the reactance x of the receiving circuit to deliver maximum power? Let i = current delivered into the receiving circuit. The total resistance is (r+ro); the total reactance is (x+Xo); hence, the current is i= , ^ - (46) V(r+ro)2 + (a: + Xo)2 The power output is P-ri^', (47) hence, substituting (46), gives ^^{r + ro)^ + {x + Xoy * ^^^^ (a) For any given value of r, the reactance x, which gives maximum power, is derived by j- = 0. P simplified, gives yi = {x + xoy; hence, . . . .-j^ = 2(x + xo)=0 and x=-xo; . . . (49) 166 ENGINEERING MATHEMATICS. that is, for any chosen resistance r, the power is a maximum, if the reactance of the receiving circuit is chosen equal to that of the line, but of opposite sign, that is, as condensive reactance. Substituting (49) into (48) gives the maximum power available for a chosen value of r, as : re^ " (r + ro)2' or, simplified, (50) hence, (r+ro)2 ro^ y2 = —^ and 2/3 = r+— ; -;^=l-72- and r = ro, .... (51) and by substituting (51) into (50), the maximum power is, (6) For any given value of x, the resistance r, which gives maximum power, is given by -r- = 0. P simplified gives, (r + ro)2 + (x+Xo)2 2/1= y2 = r + r ro^ + (x+Xoy r ' r=Vro2 + (x+Xo)2, (53) which is the value of r, that for any given value of x, gives maximum power, and this maximum power by substituting (53) into (48) is, n vro2 + (x + Xo)2e2 "0 = [ro + Vro2 + (a: + Xo) ^f + (x + Xo)^ .... (54) 21ro + Vro2 + (a:+:io)2!' MAXIMA AND MINIMA, 167 which is the maximum power that can be transmitted into a receiving circuit of reactance-^'ifr The value of x, w^hich makes this maximum power Pq the dP highest maximum, is given by "j7~=0- Pq simplified gives y3=-ro + V ro- + U + j:o)2: 2/6 = (x+Xo)^; y7 = {x+Xo); and this value is a maximum for (x+xo)=0', that is, for x=—Xo (55) Note. If x cannot be negative, that is, if only inductive reactance is considered, x = 0 gives the maximum power, and the latter then is -^ max — ~; / — ^ ; .... (56) the same value as found in problem (7), equation (18). Substituting (55) and (54) gives again equation (52), thus, P = — 1 10. Here again, it requires consideration, whether the solution is practicable within the limitation of engineering constants. With the numerical constants chosen, it would be ■'max — 4 OQ 11,250 kw.; 6 i=7y- = 750 amperes, 168 ENGINEERING MATHEMATICS. and the voltage at the receiving end of the line would be e' =i\/r'^ + x2 = 750 V2(F+ 50^ = 40,400 volts ; that is, the voltage at the receiving end would be far higher than at the generator end, the current excessive, and the efficiency of transmission only 50 per cent. This extreme case thus is hardly practicable, and the conclusion would be that by the use of negative reactance in the receiving circuit, an amount of power could be dehvered, at a sacrifice of efficiency, far greater than economical transmission would permit. In the case, where capacity was excluded from the receiv- ing circuit, the maximum power was given by equation (56) as Si? ^max = , -=-6100 kw. III. Extremes of engineering quantities x, y, are usually determined by differentiating the function. and from the equation^ y=m, (57) !"-» (»> deriving the values oi x, which make y an extreme. Occasionally, however, the equation (58) cannot be solved for X, but is either of higher order in x, or a transcendental equation. In this case, equation (58) may be solved by approx- imation, or preferably, the function, dy ax, (59) is plotted as a curve, the values of x taken, at which z = 0, that is, at which the curve intersects the X-axis. For instance : Example 13. The e.m.f. wave of a three-phase alternator, as determined by oscillograph, is represented by the equation, e = 36000jsin ^-0.12 sin (3^-2.3°)-23 sin (5^-1.5°) + 0.13 sin (7^-6.2°)! (60) MAXIMA AND MINIMA. • 169 This alternator, connected to a long-distance transmission line, gives the charging current to the line of i = 13.12cos (^-0.3°) -5.04 cos (Bd*- 3.3°) -18.76 cos (5i9-3.6°) + 19.59 cos (7(9-9.9°) .... (61) (see Chapter III, paragraph 95). What are the extreme values of this current, and at what phase angles 6 do they occur? The phase angle 6, at which the current i reaches an extreme value, is given by the equation ?ro (62) Fig. 57. Substituting (61) into (62) gives, di z = ^^=-13.12 sin (^-0.3°) +15.12 sin (3^-3.3°) +93.8 sin (5^-3.6°)-137.1 sin (7^-9.9°) = 0. . . . (63) This equation cannot be solved for 6. Therefore z is plotted as function of d by the curve. Fig. 57, and from this curve the values of d taken at which the curve intersects the zero line. They are: <9 = P; 20°; 47° 78°; 104°; 135°; 162°. 170 ENGINEERING MATHEMATICS. For these angles 6, the corresponding values of i are calculated by equation (61), and are: 1*0= +9; -1; +39; -30; +30; -42; +4 amperes. The current thus has during each period 14 maxima, of which the highest is 42 amperes. 112. In those cases, where the mathematical expression of the function y=f{x) is not known, and the extreme values therefore have to be determined graphically, frequently a greater accuracy can be reached by plotting as a curve the differential of y=f{x) and picking out the zero values instead of plotting y=f{x), and picking out the highest and the lowest points. If the mathematical expression of y=f{x) is not known, obvi- dv ously the equation of the differential curve z=— (64) is usually not known either. Approximately, however, it can fre- quently be plotted from the numerical values of y=f{x), as follows: If Xi, X2, X3 . . . are successive numerical values of x, and 2/1; V^, 2/3 •• • the corresponding numerical values of y^ dv approximate points of the differential curve 2 = -7- are given by the corresponding values: X2+X1 ^ X3+3'2 , X4 + X3 as auscissas : o^ o^ o***> ,. . y2-yi ys-y2 y^-y^ as ordmates : ; ; . . . a;2 — xi ' x^ — X2 X4^—Xz 113. Example 14, In the problem (1), the maximum permea- bility point of a sample of iron, of which the (B, 3C curve is given as Fig. 51, was determined by taking from Fig. 51 corresponding values of (B and 3C, and plotting [JL=-zp„ against (B in Fig. 52. A considerable inaccuracy exists in this method, in locating the value of (B, at which /i is a maximum, due to the flatness of the curve, Fig. 52. MAXIMA AND MINIMA, 171 The successive pairs of corresponding values of (B and 3C, as taken from Fig. 51 are given in columns 1 and 2 of Table I. Table T. (B Kilo Lines. 3C J;z 1 (B 0 1 2 0 1.76 2.74 370 570 730 + 200 160 0.5 1.5 3 4 5 3.47 4.06 4.59 865 985 1090 135 120 105 2.5 3.5 4.5 6 7 8 5.10 5.63 6.17 1175 1245 1295 85 70 50 5.5 6.5 7.5 9 10 11 6.77 7.47 8.33 1330 1340 1320 35 + 10 -20 8.5 9.5 10.5 12 13 14 9.60 11.60 15.10 1250 1120 930 70 130 190 11.5 12.5 13.5 15 20.7 725 205 14.5 In the third column of Table I is given the permeability, j«=— , and in the fourth column the increase of permeabilitV; per (B=l, Jfi; the last column then gives the value of (B, to which Jfi corresponds. In Fig. 58, values Jji are plotted as ordinates, with (B as abscissas. This curve passes through zero at (B = 9.95. The maximum permeability thus occurs at the approximate magnetic density (B = 9.95 kilolines per sq.cm., and not at (B = 10.2, as was given by the less accurate graphical determination of Fig. 52, and the maximum permeability is jjLo = lSiO. As seen, the sharpness of the intersection of the differential curve with the zero line permits a far gi'eater accuracy than feasible by the method used in instance (1). 114. As illustration of the method of determining extremes, some further examples are given below: 172 ENGINEERING MATHEMATICS. Example 15. A storage battery of n = 80 cells is to be connected so as to give maximum power in a constant resist- ance r = 0.1 ohm. Each battery cell has the e.m.f. eo = 2.1 volts and the internal resistance ro = 0.02 ohm. How must the cells be connected? Assuming the cells are connected with x in parallel, hence n . - m series. The internal resistance of the battery then is n - To X nro ,11, n — — = ~;:5- ohms, and the total resistance of the circuit is -^ro + r. XX x^ -\ \ 7 ! • ^^ "1 r K) ■^^ ^ ^- Kile -line s "1 '• 3 '. ( I J 1( 1 1 3 1 ' t ' 1 \ \ \ Fig. 58. First Differential Quotient of (B,/« Curve Th 71/ The e.m.f. actine; on the circuit is — eo, since - cells of e.m.f. ^ X ' X Cq are in series. Therefore, the current delivered by the battery is, n 1 = n ro+r and the power which this current produces in the resistance r,is, rn^eo^ MAXIMA AND MINIMA, 173 This is an extreme, if is an extreme, hence, and nro y= \-rx dx x^ -^-*: that is, ^=\/-::r = 4 cells are connected in multiple, and -=^/ — = 20 cells in series. X \ro 115. Example 16, In an alternating-current transformer the loss of power is limited to 900 watts by the permissible temper- ature rise. The internal resistance of the transformer winding (primary, plus secondary reduced to the primary) is 2 ohms, and the core loss at 2000 volts impressed, is 400 watts, and varies with the 1.6th power of the magnetic density and there- fore of the voltage. At what impressed voltage is the output of the transformer a maximum? If e is the impressed e.m.f. and i is the current input, the power input into the transformer (approximately, at non- inductive load) is P = ei. If the output is a maximum, at constant loss, the input P also is a maximum. The loss of power in the winding is n2 = 2i2. The loss of power in the iron at 2000 volts impressed is 400 watts, and at impressed voltage e it therefore is (2000) X400' and the total loss in the transformer, therefore, is 174 ENGINEERING MATHEMATICS, herefrom, it follows that, and, substituting, into P^ei: P = e^450-200(^)'' Simplified, this gives, 2000 g3.6 and, differentiating. and Hence, e ^^4 5e-^:^^ = 0 de 20001-6 ^' \2000/ ^•-'^• = 1.15 and e = 2300 volts. 2000 which, substituted, gives P = 2300V450-200Xl.25 = 32.52kw. . ii6. Example 17. Ina 5-kw. alternating-current transformer, at 1000 volts impressed, the core loss is 60 watts, the i^r loss 150 watts. How must the impressed voltage be changed, to give maximum efficiency, (a) At full load of 5-kw; (h) at half load? The core loss may be assumed as varying with the 1.6th power of the impressed voltage. If e is the impressed voltage, ^ = is the current at full load, and ii = is the Current at e ' e half load, then at 1000 volts impressed, the full-load current is ^^ = 5 amperes, and since the i-r loss is 150 watts, this ^ves MAXIMA AND MINIMA. 175 the internal resistance of the transformer as r = 6 ohms, and herefrom the i^r loss at impressed voltage e is respectively, ., 150X106 ^ .^ 37.5X106 ^^ j^2 = and rii^ = ^ watts. Since the core loss is 60 watts at 1000 volts, at the voltage e it is Pi=60x(^Y^^j watts. The total loss, at full load, thus is r. r. -o nr. ( ^ \ l'^ + 150 X 10^ and at half load it is 37.5X106 P,, = P,4-n.^ = 60x(j^)^" Simplified, this gives •6 2'=(ll)o)" +2.5XlO«Xe- ■ 2/. = (l^)"Vo.625Xl06Xe-2; hence, differentiated, e3-6^- 3.125 X 106 X 10001-6 = 3.125 XlOi«^ e3-6 = 0.78125Xl06xl000i-6 = 0.78125Xl0i6-8; hence, e = 1373 volts for maximum efficiency at full load. and e = 938 volts for maximum efficiency at half load. 117. Example 18. (a) Constant voltage 6 = 1000 is im- pressed upon a condenser of capacity C = 10 mf., through a reactor of inductance L = 100 mh., and a resistor of resist- ance r = 40 ohms. What is t.hp maximum value of the charg- ing: current? 176 ENGINEERING MATHEMATICS. (b) An additional resistor of resistance r' = 210 ohms is then inserted in series, making the total resistance of the con- denser charging circuit, r = 250 ohms. What is the maximum value of the charging current? The equation of the charging current of a condenser, through a circuit of low resistance, is {" Transient Electric Phenomena and Oscillations,'^ p. 61) : '-ii--'**i. where and the equation of the charging current of a condenser, through a circuit of high resistance, is {" Transient Electric Phenomena and Oscillations," p. 51), 5 I where =^F#■ Substituting the numerical values gives: (a) 1 = 10.2 £-200^ sin 980 i- {b) i = 6.667{ £-5oo«_ ,-20oo«j_ Simplified and diiferentiated, this gives : (o) 2=-^' = 4.9 cos 980^-sin 980^=0; hence tan 980^ = 4.9 980i = 68.5° =1.20 1.20 +n;r t- Q^g sec. (h) ^ = ^ = 4e-2000/_ ,-500^ = 0; ^ ' at MAXIMA AND MINIMA. 177 hence, e+i50o< = 4, 1500^=!-^— = 1,38, log £ ^ = 0.00092 sec, and, by substituting these values of t into the equations of the current, gives the maximum values: 1.20 + n;r (a) i = 10e 4:9— = 7.83£-o-6^" = 7.83X0.53" amperes; that is, an infinite number of maxima, of gradually decreasing values: +7.83; -4.15; +.2.20; -1.17 etc. (6) i=6.667(£-o-46- £-1-84) =3.16 amperes. M 118. Example 19. In an induction generator, the fric- tion losses are Pf= 100 kw.; the iron loss is 2000 kw. at the ter- minal voltage of e = 4 kv., and may be assumed as proportional to the 1.6th power of the voltage; the loss in the resistance of the conductors is 100 kw. at i = 3000 amperes output, and may be assumed as proportional to the square of the current, and the losses resulting from stray fields due to magnetic saturation are 100 kw. at e = 4 kv., and may in the range considered be assumed as approximately proportional to the 3.2th power of the voltage. Under what conditions of operation, regard- ing output, voltage and current, is the efficiency a maximum? The losses mav be summarized as follows: Friction loss. P/=100kw.; Iron loss. P^. + 200(J)''; i^r loss, ^-l^<3oVo)'' Saturation loss, ^-ioo(l)^'^ hence the total loss isPL = Pf-\-Pi+Pc+Ps -™i>«(l)'^(i)^0 3-2 178 ENGINEERING MATHEMATICS. The output is P = ei; hence, percentage of loss is 2 /e\3-2] M\(\ J 1 1 Ol 1 II ^ \ ™l-«-^,e3-2-=0; (e \i'^ 2 J 1 = -p^ and e = 5.50 kv. , and, by substitution the following values are obtained : i^ = 0.0323; efficiencv 96.77 per cent; current i = 8000 amperes; output P = 44,000 kw. 1 19. In all probability, this output is beyond the capacity of the generator, as hmited by heating. The foremost limita- tion probably will be the i^r heating of the conductors; that is. MAXIMA AND MINIMA. 179 the maximum permissible current will be restricted to, for instance, 2 = 5000 amperes. For any given value of current i, the maximum efficiency, that is, minimum loss, is found by differentiating, 3-2 100 ; = /+KJ)"'"(3o'oor^(^y over e, thus : SimpHfied, X gives 41. e*^ ' 43-2 y-e{^+{s-mJ\+^>^^-'-^^^^-'- hence, difTerentiated, it gives de e2j ^\3000/ ' 4i-6e0.4 ' 43.2 ~ '^^ ' \3000/ ' ~'~/tl-6-0.4 ' /13.2 ~^^y (4) Ml) =nr+(3ooo)t' /g\i.6 -3+>j64 4-55 3Q^ 11 For 1 = 5000, this gives: (|y"^ = 1.065 and e = 4.16kv.; hence, >l = 0.0338, Efficency 96.62 per cent. Power P=20,800 kw. Method of Least Squares. 120. An interesting and very important application of the theory of extremes is given by the method of least squares, which is used to calculate the most accurate values of the constants of functions from numerical observations which are more numer- ous than the constants. If y-fix), (1) 180 ENGINEERING MATHEMATICS. is a function having the constants a, h, c . . , and the form of the function (1) is known, for instance, t/ = a + 6x+cx2, (2) and the constants a h, c are not known, but the numerical values of a number of corresponding values of x and y are given, for instance, by experiment, Xi, X2, Xs, x^. . . and i/i, y-z, 2/3, 2/4 • • • , then from these corresponding numerical values Xn and yn the constants a, b, c . . . can be calculated, if the numerical values, that is, the observed points of the curve, are sufficiently numerous. If less points Xi t/i, X2, 2/2 •• • are observed, then the equa- tion (1) has constants, obviously these constants cannot be calculated, as not sufficient data are available therefor. If the number of observed points equals the number of con- stants, they are just sufficient to calculate the constants. For instance, in equation (2), if three corresponding values Xi, 2/1; X2, 2/2; X3, 2/3 are observed, by substituting these into equation (2), three equations are obtained: 2/i=a + ?)Xi+cxi2; y2 = a + hx2+cx2^; y3 = a3-\-hx+cx3^, (3) which are just sufficient for the calculation of the three constants a, h, c. Three observations would therefore be sufficient for deter- mining three constants, if the observations were absolutely correct. This, however, is not the case, but the observations always contain errors of observation, that is, unavoidable inac- curacies, and constants calculated by using only as many observations as there are constants, are not very accurate. Thus, in experimental work, always more observations are made than just necessary for the determination of the constants, for the purpose of getting a higher accuracy. Thus, for instance, in astronomy, for the calculation of the orbit of a comet, less than four observations are theoretically sufficient, but if possible hundreds are taken, to get a greater accuracy in the determination of the constants of the orbit. MAXIMA AND MINIMA, 181 If, then, for the determination of the constants a, h, c of equation (2), six pairs of corresponding values of x and y were determined, any three of these pairs would be sufficient to give a, h, c, as seen above, but using different sets of three observations, would not give the same values of a^ h, c (as it should, if the observations were absolutely accurate), but different values, and none of these values would have as high an accuracy as can be reached from the experimental data, since none of the values uses all observations. 121. If y=f{x), (1) is a function containing the constants a, b, c . . ., which are still unknown, and Xi, yi; X2, 2/2; x^, y^; etc., are corresponding experimental values, then, if these values were absolutely cor- rect, and the correct values of the constants a,h, c . . . chosen, ?/i =/(xi) would be true ; that is, /(xi)-2/i=0; ] (5) /(X2)- 2/2 = 0, etc. J Due to the errors of observation, this is not the case, but even if a, 6, c . . . are the correct values, 2/1^/(0:1) etc.; (6) that is, a small difference, or error, exists, thus /(xi)- 2/1 = ^1; ] (7) /(x2)-2/2 = ^2, etc.: j If instead of the correct values of the constants, a, h, c . . ., other values were chosen, different errors ^1, 82 . . . would obviously result. From probability calculation it follows, that, if the correct values of the constants a, h, c . . . are chosen, the sum of the squares of the errors, di^ + d2^-\-ds^ + . . (8) is less than for any other value of the constants a, 6, c . . . ; that is, it is a minimum. 182 ENGINEERING MATHEMATICS, 122. The problem of determining- the constants a, 6, c. . ., thus consists in finding a set of constants, which makes the sum of the square of the errors d a minimum; that is, 2= 202= minimum, (9) is the requirement, which gives the most accurate or most probable set of values of the constants a,h,c... Since by (7), d=f(x)-y, it follows from (9) as the condi- tion, which gives the mpst probable value of the constants a,h,c...; . z=li\f(x) — y]^ = minimum; .... (10) that is; the least sum of the squares of the errors gives the most probable value of the constants a, h, c . . . To find the values of a, 6, c . . ., which fulfill equation (10), the differential quotients of (10) are equated to zero, and give This gives as many equations as there are constants a,h,r. . . ., and therefore just suffices for their calculation, and the values so calculated are the most probable, that is, the most accurate values. Where extremely high accuracy is required, as for instance in astronomy when calculating .from observations extending over a few months only, the orbit of a comet which possibly lasts thousands of years, the method of least squares must b(^ used, and is frequently necessary also in engineering, to get from a limited number of observations the highest accuracy of the constants. 123. As instance, the method of least squares may be applied in separating from the observations of an induction motor, when running light, the component losses, as friction, hysteresis, etc. MAXIMA AND MINIMA. 183 In a 440-volt 50-h.p. induction motor, when running light, that is, without load, at various voltages, let the terminal voltage e, the current input i, and the power input p be observed as given in the first three colunms of Table I: Table I e i p ih- Vn PO calc. J 148 220 320 8 11 19 790 920 1500 13 24 72 780 900 1430 746 962 1382 + 32 - 62 + 48 410 440 473 23 26 29 1920 2220 2450 106 135 168 1810 2085 ^280 1875 2058 2280 - 35 + 2.7 0 580 640 700 43 56 75 3700 5000 8000 370 627 1125 3330 4370 6875 3080 3600 4150 + 250 + 770 ■■ + 2725 The power consumed by the motor while running light consists of: The friction loss, which can be assumed as con- stant, a; the hysteresis loss, which is proportional to the 1.6th power of the magnetic flux, and therefore of the voltage, he^-^\ the eddy current losses, which are proportional to the square of the magnetic flux, and therefore of the voltage, ce^; and the i^r loss in the windings. The total power is^ p = a-{-he\-^-\-ce^-\- ri?. (12) From the resistance of the motor windings, r = 0.2 ohm, and the observed values of current i, the i^r loss is calculated, and tabulated in the fourth column of Table I, and subtracted from p, leaving as the total mechanical and magnetic losses the values of po given in the fifth column of the table, which should be expressed by the equation : p = a-\-he^-^-\-ce'' (13) This leaves three constants, a, h, c, to be calculated. Plotting now in Fig. 59 with values of e as abscissas, the current i and the power po give curves, which show that within the voltage range of the test, a change occurs in the motor, 184 ENGINEERING MATHEMATICS. as indicated by the abrupt rise of current and of power beyond 473 volts. This obviously is due to beginning magnetic satura- tion of the iron structure. Since with beginning saturation a change of the magnetic distribution must be expected, that is, an increase of the magnetic stray field and thereby increase of eddy current losses, it is probable that at this point the con- I Po r / f- / , ''o / / 1 / y / / « / / Annr\ i/ r / / / ^ }" / -3000 .^ / y / 30 y / > y n ^^ y y ■^ _^ ^ y^ ^ ^ ^ -<■ e= Volt 5 ^1 1( L 2t ?o 3( 0 4( )0 5( ?0 6( 0 7( 0 Fig. 59. Excitation Power of Induction Motor. stants in equation (13) change, and no set of constants can be expected to represent the entire range of observation. For the calculation of the constants in (13), thus only the observa- tions below the range of magnetic saturation can safely be used, that is, up to 473 volts. From equation (13) follows as the error of an individual observation of e and po : a = a + 6ei-fi4-ce2-po; (14) MAXIMA AND MLMIMA 185 hence, thus: 2=S^2=.5-^ + 5ei.6+c62-po2}=minimum,. (15) da db 2ia+6ei-«+^e2-po!=0; i:\a+be^'^+ce^-po\e'-^' = 0; (K)) and, if n is the number of observations used (n = 6 in Ihis instance, from e = 14S to e=-473), this gives the following equations : (17) Substituting in (17) the numerical values from Table I gives. hence, a + 11.7 b 103 + 126 c 10^ = 1550; a + 14.6 b 103 + 163 c 103 = 1830: a + 15.1 b 103 + 170 c 103 = 1880; a = 540; 6 = 32.5x10-3; \ c = 5x10-3, (18) (19) and po = 540 +0.0325 £^i-« +0.005 e2 (20) The values of po, calculated from equation (20), are given in the sixth column of Table I, and their differences from the observed values in the last column. As seen, the errors are in both directions from the calculated values, except for the three highest voltages, in which the observed values rapidly increase beyond the calculated, due probably to the appearance of. a 186 ENGINEERING MATHEMATICS. loss which does not exist at lower voltages — the eddy current? caused by the magnetic stray field of saturation. This rapid divergency of the observed from the calculated values at high voltages shows that a calculation of the constants, based on all observations, would have led to wrong values, and demonstrates the necessity, first, to critically review series of observations, before using them for deriving constants, so as to exclude constant errors or unidirectional deviation. It must be realized that the method of least squares gives the most probable value, that is, the most accurate results derivable from a series of observations, only so far as the accidental errors of observations are concerned, that is, such errors which follow the general law of probability. The method of least squares, however, cannot eliminate constant errors, that is, deviation of the observations which have the tendency to be in one direction, as caused, for instance, by an instrument reading too high, or too low, or the appearance of a new phenomenon in a part of the observation, as an additional loss in above instance, etc. Against such constant errors only a critical review and study of the method and the means of observa- tion can guard, that is, judgment, and not mathematical formalism.