ENGINEERING LIBRARY 
 
A GRAPHICAL TREATMENT 
 
 OF THE 
 
 INDUCTION MOTOR 
 
 ""0 
 
 Translated from the Second Edition 
 G. H. ROWE and R. E. HELLMUND 
 
 NEW YORK 
 
 McGraw Publishing Company 
 
 1906 
 
&NGINEERIMG LIBRARY 
 
 Copyrighted, 1906, 
 
 oy tne 
 
 McGRAW PUBLISHING COMPANY 
 New York 
 
CONTENTS. 
 
 PAGE 
 
 Introduction 1 
 
 General Theory 1 
 
 Diagrams of the Induction Motor. 5 
 
 Current Diagram 6 
 
 Field Diagram 7 
 
 Circle Diagram 8 
 
 Determination of Input, etc., from the Diagram 10 
 
 Friction and Iron Losses 10 
 
 Copper Losses 10 
 
 Electrical Input 13 
 
 Torque 13 
 
 Output 14 
 
 Slip 14 
 
 Electrical Efficiency 15 
 
 Practical Application of the Diagram 17 
 
 Note on Correction to be Applied 22 
 
 Examples 23 
 
 2 H.P. Motor 24 
 
 7 H.P. Motor 29 
 
 12 H.P. Motor. 33 
 
 Induction Motor as Generator 34 
 
 Single Phase Motor. 36 
 
 838991 
 
A GRAPHICAL TREATMENT 
 OF THE INDUCTION MOTOR. 
 
 The object of the method described in the following 
 pages is the experimental determination of the charac- 
 teristic properties of induction motors. It consists es- 
 sentially in the practical application of the circle dia- 
 gram, first described by the writer in 1894.* It is 
 based on two simple and quickly performed experimen- 
 tal tests on the finished motor. 
 
 The method shows at a glance the main properties 
 of a motor and its commercial excellence. The writer 
 has used the circle diagram several years in the testing 
 room for comparing calculated values with results ob- 
 tained from tests, and it has well served its purpose. 
 
 Before describing the method and its applications, 
 perhaps I may be allowed to present briefly the theory 
 of the induction motor, and the derivation cf the 
 diagram. 
 
 GENERAL THEORY OF THE INDUCTION MOTOR. 
 
 The induction motor is in principle a transformer. 
 The exciting member A (Fig. 1) represents the inducing 
 or primary circuit ; the short-circuited member B repre- 
 
 *Elektrotechnische Zeitschrift, Oct. 11, 1894, p. 561. 
 
 1 
 
2 THE INDUCTION MOTOR. 
 
 sents the induced or secondary circuit. The alter- 
 nating fields produced by the current in the exciting 
 coils combine in the well known manner to form a 
 rotating field. The motor is, therefore, a transformer 
 with a rotating field, and the load of the system is deter- 
 mined at any time by the difference between the con- 
 stant speed of rotation of the field produced in the 
 stationary member .4, and that of the rotating short- 
 circuited member B. The turning of the rotor is due 
 
 FIG. 1. 
 
 to the torque existing between the rotating field and 
 the currents produced in the short-circuited member 
 by rotation in the field. These currents, and therefore 
 also the torque, vary directly with the product of the 
 slip and the field interlinked with the secondary con- 
 ductors. 
 
 The slip is defined as the difference between the 
 speed of rotation of the exciting field, and that of the 
 short-circuited secondary. The variation of the slip in 
 
THE INDUCTION MOTOR. 3 
 
 the induction motor has the same significance as the 
 variation of load of the ordinary transformer, due to 
 a change of its external resistance. 
 
 With constant impressed electromotive force, the 
 field produced by the current in the primary winding 
 is constant for all loads, as in the transformer. This, 
 however, is not true in the short-circuited member, 
 which is the real work transmitting element. Herein 
 lies the essential difference between the induction motor 
 and the transformer without leakage, in which, as is 
 well known, the total primary field passes through the 
 secondary. Since there must, of necessity, be an air- 
 gap between the short circuited windings and the pri- 
 mary windings, not all of the magnetic lines < induced 
 in the primary member pass through into the secondary. 
 But a part <f> s passes directly through the space be- 
 tween the two windings back into the primary mem- 
 ber, and only the difference < A = <fi <p s passes into 
 the short-circuited member. This field </) A produces, in 
 the latter, currents which lag, according to the law of 
 induction, a quarter period behind the field, i.e., they 
 are in quadrature with the inducing field, and there- 
 fore the torque produced is proportional to the pro- 
 duct of induced currents and primary field. 
 
 The lines <p s are called leakage lines or leakage field, 
 and since they are caused by the primary current 7 t 
 
 *In fact there is not only a primary leakage, but also a 
 secondary leakage that is a flux which is interlinked with the 
 secondary conductors only. It is however admirable, and 
 simplifies further considerations, to assume that the primary 
 and secondary leakage fluxes are combined to form one single 
 leakage flux i. e. the primary leakage flux S . The inexact- 
 ness introduced thereby is of no practical importance in the 
 results obtained from the following derivation. ROWE HELL- 
 MUND. 
 
4 THE INDUCTION MOTOR. 
 
 only, they must be proportional to it. If we designate 
 the reluctance of the leakage path by p s , then we have 
 
 The main field, as in the transformer, is of constant 
 amplitude = <. Therefore, the armature field ^> A is 
 given by the difference between the main field and 
 the leakage field < A = < - </>* (Fig. 2), and is pro- 
 
 FIG. 2. 
 
 duced by the difference between the ampere turns of 
 the primary and secondary circuits. If we call the 
 resulting magnetizing current, I, and the reluctance 
 of the path of the secondary field p, then 
 
 / 
 < A x- 
 
 This magnetizing current is thus not constant, as 
 in the transformer, but decreases with increasing load 
 since with increasing load the current 7 t and therefore 
 c6c increases. </> A will therefore decrease. 
 
THE INDUCTION MOTOR. j 
 
 If no current is produced in the secondary mem- 
 ber, the primary winding carries only the magnetizing 
 current, and therefore the primarv current equals the 
 magnetizing current, or 
 
 /--/ 
 
 This occurs when the motor is running unloaded, and 
 the slip is zero. 
 
 As the load increases, we have the following phe- 
 nomena: The slip increases, and with it the electro- 
 motive force induced in the secondary member, and 
 the current 7 2 resulting therefrom increases. This 
 secondary current 7 2 exerts a demagnetizing effect on 
 the field, and therefore causes, as in the transformer, 
 a corresponding increase of the current in the primary 
 member. The demagnetizing influence of the current 
 7 2 acts only on the armature field. The strength of the 
 leakage field increases, however, in direct ratio with 
 the increase of the primary current. Therefore, the 
 increase of the load, and the increase of the secondary 
 current, have the following consequences: 
 
 (1) The secondary field is decreased by the demag- 
 netizing action of 7 2 . 
 
 (2) Indirectly, the leakage field is increased, due to 
 an increase of 7 X . 
 
 Thus, the leakage field is directly proportional to 
 I v The armature field equals the difference between 
 the main and leakage fields, and therefore the sum of 
 the armature field and leakage field must be equal 
 to the main field and constant. 
 
 DIAGRAM OF THE INDUCTION MOTOR. 
 
 As in the well known transformer diagram, the pri- 
 mary current, and the secondary current in the short- 
 circuited secondary, can be represented by two sides 
 
6 THE INDUCTION MOTOR. 
 
 of a triangle (Fig. 3), of which the resulting third side 
 is the magnetizing current 7. 
 
 This magnetizing current produces the secondary 
 field, which interlinks with both the exciting and short- 
 circuited windings. The secondary field produces the 
 currents in the short-circuited secondary; it being in 
 fact the only field which passes into the secondary. 
 The secondary currents, therefore, lag exactly a quarter 
 period behind the secondary field. As shown before, 
 the main difference between the transformer without 
 leakage and the induction motor, lies in the fact that 
 
 Impressed E.M.F. 
 FIG. 3. 
 
 the magnitude of the magnetizing current changes, and 
 that the current triangle changes its position. 
 
 We know (Fig. 3) that the constant main field, which 
 induces a counter electromotive force corresponding to 
 the impressed electromotive force is at right angles to 
 the impressed electromotive force in a circuit without 
 resistance, and without iron losses. Under these con- 
 ditions, in a transformer without leakage, the magnet- 
 izing current is therefore at right angles to the im- 
 pressed electromotive force, and is constant. 
 
 Further, the current triangle always has the same 
 relative position to the electromotiv? force. 
 
THE INDUCTION MOTOR 7 
 
 In the induction motor, nowever, the main field must 
 be resolved into two parts. One part, the leakage field, 
 is in phase with the primary current, and proportional 
 to it. The other part, the secondary field, must there- 
 fore be variable, and cannot be in phase with the 
 main field. The magnetizing current which produces 
 this secondary field, must therefore vary in amplitude 
 and direction. It is possible to follow these changes 
 by the graphical combination of the fields (Fig. 4). 
 
 The diagram proposed by the writer is based on 
 
 Impressed E.M.F. 
 
 FIG. 4. 
 
 these assumptions, and is represented in Fig. 5. 
 
 The leakage field is directly proportional to the pri- 
 mary current, and in phase with it, or 
 
 <f> s oc^- = AC' 
 
 Ps 
 
 Thus, we can represent the leakage field according to 
 a definite scale by the vector AC' = I r The main 
 field may be represented by a line A D, constant in 
 
8 
 
 THE INDUCTION MOTOR. 
 
 magnitude and in quadrature with the electromotive 
 force. 
 
 The third side C' D, must therefore give the second- 
 ary field, or 
 
 < A = = C'L> 
 and C' D must be in quadrature with c' C' = / 2 , and 
 
 D 
 
 Impressed E.M.F. 
 
 FIG. 5. 
 
 parallel with A c' = I. Further, there must be a con- 
 stant relation between C' D and A c' according to an 
 equation previously given, or 
 
 C'D_<f> I 
 
 ~A T ~ T~ & 
 
 Ac' I o 
 
 In order to satisfy this equation when the load 
 changes, the points c' and C' must move on the two 
 
THE INDUCTION MOTOR. 9 
 
 half circles represented in Fig. 5. The angles A c' C 
 and C C' D are always right angles, and the ratio 
 
 C'D . 
 
 - - is constant. 
 
 We see from the figure that the current cannot in- 
 crease indefinitely, even if we assume the resistance to 
 be zero. It reaches a maximum when C' coincides 
 
 with D, or 
 
 I, = AD. 
 
 The leakage field, which is then represented by the 
 line A D, equals in intensity the main field, and the 
 remaining part, the secondary field, becomes zero. 
 
 In this case 
 
 /! AD 
 
 (h = (h OC - OC 
 
 PS Ps 
 
 Let us study now the condition for minimum current. 
 The minimum current flows when the motor is run- 
 ning without load. In this case 7 2 = 0, and therefore 
 /j = the magnetizing current, or 
 
 7 t = / = A C, and 
 
 I I, AC 
 
 d> oc oc oc 
 
 P P P 
 
 or, since < remains constant, we have the relation: 
 
 Maximum current A D p s 
 Minimum current AC p 
 
 reluctance of leakage path 
 
 reluctance of path of secondary field 
 
 The above considerations may be briefly stated as 
 follows: In induction motors the relation between the 
 
JO THE INDUCTION MOTOR. 
 
 electromotive force and the current can be represented 
 by a vector diagram. In this diagram, the vector 
 representing the current, which, as we know, varies 
 with the load, is determined by the fact that its end 
 point moves in a circle; the position of which is given 
 by the ratio 
 
 A D jOs _ reluctance of leakage path 
 
 AC p reluctance of path of secondary field 
 
 DETERMINATION OF INPUT, OUTPUT, TORQUE AND SLIP. 
 
 So far, we have investigated current magnitudes and 
 phase differences only. To utilize the diagram, prac- 
 tically, and to determine from it the operation and 
 commercial excellence of a motor, we must introduce 
 the friction, iron losses, and especially the electrical 
 losses. 
 
 The friction and iron losses can be considered con- 
 stant, since the speed of the motor and the field be- 
 tween no load and full load are practically constant. 
 The iron losses of the short-circuited secondary can be 
 neglected on account of the low frequency of its field. 
 We shall see later that the exciting field does not 
 remain quite constant, but is somewhat reduced by 
 the ohmic drop in the primary circuit. On the other 
 hand, the iron losses in the secondary increase slightly 
 with the load, due to increasing slip. These variations 
 will almost neutralize each other, so that the total 
 :'ron losses are approximately constant. 
 
 It is quite different with the copper losses. As is 
 well known, these are proportional to the square of 
 the current. For this reason, it is very difficult to 
 represent them directly in the diagram. We shall see 
 however, that their influence appears in another form, 
 
THE INDUCTION MOTOR. 11 
 
 viz., by weakening the field, whereby they may be 
 very readily considered. The ohmic drop of the pri- 
 mary winding I^xR^ (7 t = current, and R^ = ohmic 
 resistance of primary circuit) opposes the primary im- 
 pressed electromotive force, and consequently the field 
 does not need to induce a counter electromotive force 
 equal to the impressed, but one equal to the difference 
 between the impressed electromotive force and the 
 drop. It follows that the main field, inducing the 
 counter-electromotive force, and, therefore, also the 
 secondary field, will be decreased by an amount corre- 
 sponding to the ohmic drop. The diagram was, how- 
 ever, constructed on the assumption of a constant 
 main field. Yet the omission of the loss of potential, 
 due to resistance, does not change the correctness of 
 the diagram, since, instead of the drop, we can intro- 
 duce an equivalent field, which may finally be sub- 
 tracted from the resulting seconda^ field. 
 
 Since an electromotive force always corresponds to 
 a field at right angles to itself, and is proportional to 
 it, we may take the ohmic drop into consideration by 
 a reduction of the field proportional to and at right 
 angles to the drop, that is at right angles to the direc- 
 tion of the current. Thus, we arrive at correct results 
 if we consider the main field constant in our diagram, 
 and finally decrease the secondary field by an amount 
 corresponding to the drop in potential. 
 
 In the diagram, the primary current A C' may be 
 resolved into two components, A C and C C' t of which 
 the no load component A C causes a constant drop. 
 The other component C C' is variable, and causes a 
 drop in potential which appears in the form of a dimi- 
 nution of the field <p a proportional to and in quadrature 
 with this component. In other words, since part of 
 
12 
 
 THE INDUCTION MOTOR. 
 
 tha impressed electromotive force is used in overcoming 
 the ohmic resistance, the secondary field needs only 
 to be assumed smaller by an amount corresponding to 
 the ohmic drop. Let us call this part to be subtracted 
 
 FIG. 6. 
 
 C' E'\ then C f E' is proportional to CC'\ that is, the 
 point E r must move on the circumference of the second 
 circle shown in Fig. 6. This circle, therefore, indicates 
 that the secondary field is not represented as before 
 
THE INDUCTION MOTOR. 13 
 
 by DC', but by the line D E f The main field de- 
 creases, of course, by the same amount. However, 
 since we do not use the main field in further investiga- 
 tion, it is unnecessary to introduce the change in the 
 diagram. 
 
 Similarly, the loss in the short-circuited secondary 
 appears in the diagram as the line E' F'. Introducing 
 in this way the ohmic drop of both windings by a corre- 
 sponding reduction of the secondary field, we are in a 
 position to derive the various quantities, such as 
 torque, output, etc. 
 
 (a) The electrical input = 3 e i cos 6 = v 7 3 E i cos 0, 
 can be found, since the impressed electromotive force 
 is constant, by scaling the line c' C', that is, the energy 
 component of the primary current A C', 
 
 . ' . Input oc c' C' 
 
 (b) The torque is determined by the product of the 
 secondary field and the secondary current, and is pro- 
 portional to E' D x C' C, i.e., to the area of the triangle 
 C E' D (since E' D is the base, and C' C the altitude 
 of this triangle). One side, CD, remaining constant, 
 we can represent the area of the triangle by the altitude 
 E' e f , or the torque is proportional to the abcissa of 
 the point E f . 
 
 Now, there are a number of constant losses, such as 
 those due to bearing and air friction, iron losses and 
 copper losses at no load. These losses being constant, 
 manifest themselves in the diagram by a constant 
 diminution of the torque. In other words, the line rep- 
 resenting the torque, E' e' , is for example, diminished 
 by a length e 1 e^, proportional to the constant losses. 
 Therefore, drawing a line parallel to the axis of F, and 
 
14 THE INDUCTION MOTOR. 
 
 at a distance from it equal to e' e/, we find the actual 
 torque represented by the line e^' E', or 
 
 Torque oc^/ E f 
 
 (c) The output is given by considering the loss in 
 the short-circuited secondary E' F' as the reduced 
 altitude of the triangle C F f D, or 
 
 Output oc // F' 
 
 (d) The slip results from the fact that the currents 
 produced in the secondary are proportional to the 
 product of secondary field and the slip, or 
 
 secondary current C' C 
 Slip oc - -= - cc .. 
 
 secondary field E' D 
 
 E' C 
 But C' C oc E' C, and therefore, also, slip oc -rz 
 
 Draw in the diagram any straight line F/ C/, so 
 that the angle DE l f C l f = D E' C. D E' C is con- 
 stant for all loads, since all angles inscribed in the 
 same segment are equal, and we have 
 
 E'C = F/CV 
 E' D ~ F/ D 
 
 In the above proportion F/ D remains constant; then 
 E C _.,., 
 
 E'D 
 
 Therefore, the slip is proportional to / C\'. Thus 
 we have in E l f C\' a measure of the slip. 
 
 In order to find the value of the slip, we may deter- 
 mine the point where the slip is 100%, that is when 
 
THE INDUCTION MOTOR, 15 
 
 the rotor is stationary. In this case the output must 
 be zero, and the point F must coincide with D. This 
 happens when C' coincides with C k and the line C k D 
 is tangent to the circle O F . Draw the line 5 C k parallel 
 to E/C/; then the intercept s s' is proportional to 
 E/C/, and is in direct ratio to s C k (100% slip) or 
 slip = 5 S'. 
 
 Finally, the electrical efficiency may be determined. 
 
 The electrical loss in per cent, of the input is pro- 
 portional to 
 
 secondary current 2 C'C 2 C' C r . . , ~ /rYI 
 
 GO- ^-.oo [since c'C oo C'C x CD] 
 
 Input c'C' C' D L 
 
 Drawing a line C k f from the point C k at right angles 
 to A D, we obtain directly the above ratio (as in the 
 case of the slip) in the line 7-7-', and the electrical effi- 
 ciency in the remainder of the line 7- C k , or 
 
 Electrical Efficiency = C k f. 
 
 The diagram now gives us a complete determination 
 of all the essential characteristics of an induction motor. 
 
 At no load (Fig. 7), the point C is fully determined 
 by the no load losses, viz., the iron losses, friction and 
 windage, and the copper loss of the magnetizing cur- 
 rent. These losses are represented by an energy cur- 
 rent C C, which, combined with the magnetizing 
 current A C gives the no load current A C. 
 
 The torque and output are zero, and the slip very 
 small. With increasing load, the point C' moves along 
 the circumference of the circle O c . The phase difference 
 6 between the electromotive force and the current de- 
 creases, that is, the power factor increases. At the 
 point C', where A C' is tangent to the circle O r , the 
 
16 
 
 THE INDUCTION MOTOR. 
 
 phase difference is a minimum. The torque is e' E', 
 the output /' F', and the slip 5 S'. 
 
 The output increases with increasing slip, and reaches 
 
 1) 
 
 FIG. 7. 
 
 a maximum value at the point F" . If the slip still 
 further increases, the output diminishes; the input 
 still increases, but later also decreases, while the phase 
 difference rapidly increases. The output becomes 
 
THE INDUCTION MOTOR. 17 
 
 equal to zero when the slip equals 100%, and the point 
 5' coincides with the point C k . The total input is 
 then consumed in the motor. 
 
 PRACTICAL APPLICATION OF THE DIAGRAM. 
 
 In the preceding pages, we have derived fully and 
 definitely all characteristics chiefly from theoretical 
 considerations. It remains to show the agreement 
 between the theory and practical tests. If .we have 
 the completed motor before us, we need the following 
 data in order to determine its properties. 
 
 1. Ratio of the 
 
 Reluctance of leakage path 
 
 Reluctance of path of main field. 
 
 2. Resistance of the circuits. 
 
 3. Iron and friction losses. 
 
 4. Current, motor running light. 
 
 5. Current, motor blocked. 
 
 Some of these qualities, as, for instance, the reluct- 
 ances, cannot be directly measured. As we have deter- 
 mined in the preceding work, it would be possible to 
 measure these reluctances if the losses were negligible, 
 that is, if the electrical resistances were equal to zero. 
 Then, if we measure the primary current, with the sec- 
 ondary circuit open, and also the current with the 
 secondary circuit closed and stationary, we would have, 
 in the first case, the value A C, and in the second case 
 the value A D and 
 
 A D _ reluctance of leakage path 
 A C reluctance of path of main field 
 
 As shown, in consequence of the losses, the actual, 
 limiting points are C and C k , instead of C and D. 
 
18 THE INDUCTION MOTOR. 
 
 These points are easily obtained, and from them the 
 centre of the circle O c . We then find the points C 
 and D, and all other characteristic points in a similar 
 manner. It is only necessary to perform two tests 
 with ammeter and wattmeter. 
 
 (1) Motor running without load. The current A C 
 and the angle of phase difference 6 = C A B. 
 
 (2) With stationary and short-circuited rotor. The 
 current A C k and the angle of phase difference = C k A B. 
 
 If it is impossible to allow full short-circuit current 
 to flow, it is sufficient to make the test at a lower im- 
 pressed electromotive force. The real short-circuit 
 current is then proportional to the electromotive force 
 applied. If we represent the electromotive force in 
 direction by the line A B; the direction and magnitude 
 of the no load current by A C, and the direction and 
 magnitude of the short-circuit current by A C k , then 
 the center O c of the circle is determined, since it must 
 lie on a line at right angles to A B. If, by a third 
 test, the resistance R^ of the primary circuit is deter- 
 mined, the point E k can be found by making C k E k 
 proportional to 7\ R^ Thence we find the centers 
 OB and OF and the slip line 5 C k . 
 
 These two simple tests, one at no load, and one 
 with short-circuited and stationary rotor, give sufficient 
 data to indicate clearly all characteristic points, and 
 to construct the load curves. Since the only tests 
 necessary are on the motor running light, and with 
 the rotor blocked, the method renders any exact load 
 tests unnecessary. Moreover, the diagram gives at a 
 glance the reasons for the various characteristics of 
 the motor, and shows how the proportions need to be 
 changed to obtain any desired result. 
 
 For instance, in a motor with low electrical losses, 
 
THE INDUCTION MOTOR. 19 
 
 the line E k D which represents them will be very 
 small. Therefore, the point E k will be very high, and 
 the starting torque E k e k will be small. As is well 
 known, this may be avoided by inserting, by means 
 of slip rings, resistances in the secondary circuits. 
 
 If it is desired to avoid slip rings, the resistance of 
 the short-circuited secondary must not be made too 
 small. Then the points E k and C k are removed farther 
 from D. Moreover, the variation of the slip line 5 C k 
 and the slip is greater. 
 
 The motor will have, under load, greater speed 
 variations and low efficiency. 
 
 It is possible, however, to obtain good torque without 
 excessive losses, if the motor is so designed that the 
 normal load is reached at a point far beyond C f , that 
 is, if the field is made strong. But in this case the 
 magnetizing current is very large in proportion to the 
 energy current and cos 6 will be unfavorably small. 
 Such a motor will, moreover, be comparatively large. 
 The size of the motor is, in general, determined by the 
 maximum abscissa O c C" ' , The abscissae for normal 
 load c' C' is then chosen so that the desired relation 
 between normal load and maximum load is obtained. 
 The leakage constant of Rothert, which fixes for each 
 overload capacity a certain ratio between current and 
 field, corresponds with this point of view. 
 
 In the diagram this relation is expressed by the ratio 
 
 , , that is by the tangent of the angle C' D C. If 
 
 this angle becomes 45, the motor has its maximum energy 
 input. The smaller this angle, the larger the maximum 
 capacity of the motor. Rothert does not consider in 
 his constant the magnetizing current. The overload 
 capacity of a certain motor type is limited by the 
 
20 THE INDUCTION MOTOR. 
 
 magnetizing current, which increases so that the phase 
 difference becomes large and the power factor small. 
 
 The magnetizing current is determined by the air- 
 gap. If the air-gap is changed, the magnetizing current, 
 A C, changes; while the diagram above the point C 
 regains unchanged. 
 
 Therefore, we can say with Rothert that, disregarding 
 the magnetizing current, the diagrams for all multi- 
 phase motors are identical. 
 
 The maximum input C'" is proportional to the energy 
 current, that is, to the radius of the circle. In case of 
 double input capacity, for instance, the normal input 
 corresponds to one half of the radius. 
 
 It is possible to judge of the output of a motor ap- 
 proximately by the short circuit current. Since the 
 latter is nearly 'equal to the diameter of the circle, we 
 may employ the following rough approximation: 
 
 The normal capacity of a motor corresponds to about 
 one-quarter of its short circuit current, that is, if a 
 motor of 100 volts has a short-circuit current of 100 
 amperes its normal output corresponds to 
 
 iXlOOXlOOX\/3 = 4300 watts, or about 5 h.p., 
 
 assuming 75% efficiency, that is, the normal output 
 corresponds to one quarter of the current, and is pro- 
 portional to one quarter of the current necessary to 
 force the total field through the air-gap. 
 
 Obviously, with constant impressed electromotive 
 force, this current varies inversely as the square of the 
 turns, directly with the length of the leakage path, 
 and therefore with the diameter d, and inversely with 
 the width of the effective iron; hence 
 
 ~j X ~n 
 
 b m 2 
 
THE INDUCTION MOTOR. 21 
 
 For large output, therefore, d must be large and b 
 small.* 
 
 These interesting, and extremely simple, considera- 
 tions become, after a little practice, very serviceable. 
 
 Considering that w can be replaced by 
 
 electromotive force 
 
 and L by electromotive force X current i, we obtain 
 all values of the Rothert leakage constant, or 
 
 mi b 
 
 Good design in standard motors requires good effi- 
 ciency and low heating, and should further fulfil the 
 following conditions: maximum output of one and 
 one-half to two times the normal output ; starting 
 torque about equal to normal full load running torque 
 and moderate starting current; small "no load" cur- 
 rent, varying from one-third to one-fifth of the normal 
 full load current, depending on the size of the motor; 
 the power factor should be high and the slip small. 
 
 The above conditions being more or less conflicting, 
 
 * This statement does not agree with the results of modern re- 
 search made by Behn-Eschenburg and others. It has been 
 shown and generally recognized that the end connection leakage 
 is a considerable part of the total leakage, and that this kind of 
 leakage is large in motors which have a comparatively large di- 
 ameter and a small iron width ; therefore the output of a motor 
 is not always improved by simply making d larger and b smaller. 
 In very many cases, the opposite effect is obtained. (See Jour- 
 nal of the Institute of Electrical Engineers, Vol. 33, page 239, 
 April, 1904.) Paper by Behn-Eschenburg on Magnetic Disper- 
 sion in Induction Motors. ROWE-HELLMUND. 
 
22 
 
 THE INDUCTION MOTOR. 
 
 it is not always possible to satisfy them all at the 
 same time, especially in small motors. 
 
 The derivation of the circle diagram above given, contains a 
 simplification which leads to a slight inaccuracy which did not 
 appear in the original derivation, published in the Electrotech- 
 msche Zeitschrift, October 11, 1894, to which reference has al- 
 ready been made. 
 
 In consequence of the resistance of the primary winding, the 
 
 FIG. 8. 
 
 theoretical magnetizing current is not entirely wattless, and its 
 phase difference, with respect to the primary impressed electro- 
 motive force, is somewhat less than 90. Thus, the magnet- 
 izing current A C is not exactly at right angles to A B, and the 
 center of the circle O c is not exactly in the perpendicular erected 
 at A on A B. Practically, the inaccuracy introduced is so 
 small that in most cases it does not introduce a measurable 
 error. In very small motors, only, is the result of the in- 
 
THE INDUCTION MOTOR. 23 
 
 accuracy noticeable, and in these cases, the result is always 
 somewhat better than that shown by the diagrams. 
 
 In the first rough laying out of a motor, this refinement will 
 be found unnecessary, inasmuch as the use of the uncorrected 
 diagram introduces only a factor of safety, while in cases of 
 relatively high stator resistance the correction is easily made. 
 
 For the sake of completeness, the method of applying the 
 correction will be given: The original circle diagram on page 
 563, E.T.Z., 1894, had the form shown in Fig. 8. The influ- 
 ence of the primary resistance, in causing a shifting of the line 
 A D', representing the magnetizing current, has been much 
 exaggerated in the figure. 
 
 The theoretically exact position for the center of the circle 
 O c is determined as follows: The point C for the theoretical no 
 load condition (assuming no friction and no core losses), as 
 well as the point C Iv for the theoretical short circuit condition 
 (the resistance of secondary being equal to zero), must be 
 situated on a half circle, the diameter of which is the impressed 
 electromotive force A B. The center of the circle O c is on the 
 intersection of the tangent to the semi-circle A B at C and C Iv . 
 
 The center of the circle O c is not, therefore, exactly on the 
 perpendicular to A B, but on a line inclined to it by an angle e. 
 This angle is 
 
 < s = 2 C B A, and sin C B A = dj 
 
 In motors, as usually constructed, this ratio representing the 
 relation between the drop of potential caused by the magnet- 
 izing current and the impressed electromotive force, is so small 
 that the result is not influenced by a measurable amount. 
 Therefore, we shall use the approximate diagram in the fol- 
 lowing examples illustrating the application of the diagram. 
 
 APPLICATION OF THE DIAGRAM. 
 
 It may be well to exemplify the preceding considera- 
 tions by applying them to several practical cases. The 
 motors considered in the following were built by the 
 " Societe Electricite et Hydraulique " and were in- 
 tended to give a starting torque equal to twice the 
 full load running torque. 
 
24 THE INDUCTION MOTOR. 
 
 Two Horse Power Motor'. 
 
 1. At no load, 120 volts applied. 
 
 4.1 amperes, 
 210 watts, 
 
 2. Short-circuited secondary and stationary rotor, 120 
 v. applied. 
 
 55 amperes, 
 6940 watts, 
 
 3. Resistance of primary member per phase. 
 
 .344 ohm. 
 
 From the foregoing data, the diagram (Fig. 9) was 
 constructed. 
 
 From the watts at no load, we obtain the energy 
 current : 
 
 210 
 
 7 " = TW3 == ] ampere ' 
 
 First construct the no load current triangle A C C, 
 in which 
 
 A C = no load current =4.1 amperes 
 
 C C = energy current = 1 
 
 A C = magnetizing current = 4 
 
 The energy current C C represents the no load losses 
 by friction, hysteresis and eddy currents. These losses 
 are assumed to be constant for all loads. Therefore, 
 they must be allowed for by subtracting a constant 
 amount from the mechanical output and torque. This 
 can be accomplished in the diagram by drawing a line 
 x y through C parallel to A C and measuring the output 
 and torque from the line x y. At the point C', for 
 instance, the input is C f c f , the torque E f /, and the 
 output //F'. 
 
THE INDUCTION MOTOR. 
 
 25 
 
 120 Volts 
 
26 THE INDUCTION MOTOR. 
 
 From the stationary test we obtain an energy current 
 6940 
 
 120V3 
 
 33.5 amperes, 01 
 
 a 33.5 _ 
 
 cos 6 = = .61 
 oo 
 
 This value enables us to determine the point C k , 
 since A C k is proportional to the current with rotor 
 blocked, or 55 amperes, and cos C k A R equals the power 
 factor, or .61. 
 
 From these points C and C k , it is easy to find the 
 center O c , and the corresponding circle C C C k D. 
 
 If now the line C k D is drawn, then the circle giving 
 the mechanical output is determined, since we know 
 that the output of the short circuited and stationary 
 rotor corresponding to the point C k is zero. Therefore, 
 the intersection of the line C k D with the circle must 
 fall at the point D. The line C k D must therefore be 
 tangent to the output circle at D. Hence, the center 
 E -and the circle E' itself is determined. 
 
 As we have seen, the mechanical output derived 
 from this circle must be. decreased by an amount equal 
 to the no load losses which are represented by the 
 parallel line xy, or, in other words, the output is E' e v 
 etc. 
 
 Finally, in order to find the third circle for the torque, 
 we must remember that the line C k D represents the 
 electrical losses of locked rotor. These losses divide 
 themselves into two parts, primary losses and secondary 
 losses, which are proportional, respectively, to the 
 reduced resistance in each member. 
 
 Since the total field A D corresponds to an electro- 
 
THE INDUCTION MOTOR. 
 
 27 
 
 motive force of 120 volts, we find in the same scale the 
 loss vector at short circuit. 
 
 C k D X 
 
 = 73 volts. 
 
 In the primary member, of which the resistance per 
 phase is .344 ohm, and in which the short circuit current 
 is 55 amperes, the drop of potential is 
 
 V3X55X.344 = 32.7 volts. 
 
 4 H.P. 
 
 If we therefore make C k E k equal to C k D multiplied 
 
 oo 1-7 
 
 by '=^-, we know that the torque circle must pass 
 7o 
 
 through the point E k , and therefore this 'circle, and its 
 center O E are determined. 
 
 Finally, it remains to draw the line C k s from C k at 
 right angles to E D, from which may be found the slip 
 for any load. At no load the slip is zero, and at full 
 load 5%. 
 
 The diagram now shows all the characteristic proper- 
 ties of the motor. It is possible to transfer, by means 
 
28 THE INDUCTION MOTOR. 
 
 of a compass, all of the various values directly to a 
 rectangular system, as shown in Fig. 10. 
 
 The normal output of the motor is two horse power 
 at 78% efficiency, 86% power factor and 5% slip. 
 The maximum output is more than double the normal 
 full load value. The maximum power factor is .90. 
 Finally, the distance between C k and the line X Y gives 
 
 FIG. 11. 
 
 a starting torque of four synchronous horse power, the 
 starting current being 55 amperes. The motor thus 
 meets the requirements usually assumed. 
 
 The above motor is a small one ; in the diagram for which 
 the center Oc should be shifted from the line at right angles to 
 A B, as was shown in the note on page 16, it may be here shown 
 that this correction is very small and that it may be practically 
 neglected. 
 
THE INDUCTION MOTOR. 29 
 
 We have: 
 
 Impressed electromotive force .............. 120 volts 
 
 Magnetizing current ...................... . 4 amperes 
 
 Resistance of stator winding per phase ....... 344 ohm 
 
 and therefore 
 
 The angle e is shown in Fig. 11, and the dotted circle shows 
 the corrected position of the circle. 
 
 If one calculates results from this corrected circle, it will be 
 seen that the maximum power factor has increased from . 9 
 to .91, and that other values, such as efficiency and maximum 
 output, are unchanged. The difference in power factor is 
 smaller than can be observed with ordinary measuring instru- 
 ments. 
 
 As a matter of fact, only in very small low speed motors is a 
 difference noticeable, and in these cases the correction may 
 be made. Exact calculations are anyhow difficult in the case 
 of small motors, and since we know that the actual values are 
 somewhat better than those given by the diagram, it is hardly 
 necessary to make a correction. 
 
 SEVEN HORSE POWER MOTOR. 
 
 Tests on a seven horse power motor gave the fol- 
 lowing data: 
 
 1. At no load, 110 volts per phase. 
 / = 6.3 amperes per phase, 
 
 A = 100 watts per phase. 
 
 2. Locked and short-circuited rotor, 110 volts per 
 phase. 
 
 / = 118 amperes per phase, 
 A = 6670 watts per phase. 
 
 3. Resistance per phase = .142 ohm. 
 
 From these results, the diagram shown in Fig. 12 
 was constructed. 
 
30 
 
 THE INDUCTION MOTOR. 
 
 The no load energy current is 
 
 Iw = r = -91 ampere. 
 
 110 Volts 
 
 Output 
 
 Torque 
 
 Electrical Input c 
 C 
 
 A 
 
 I'O 20 30 40 ' 50 60 
 <-0.9 Amp. 
 
 FIG. 12. 
 
 Draw the current triangle A C C, in which 
 A C no load current = 6.3 amperes 
 . C C = energy current = .91 " 
 A C = magnetizing current = 6.2 amperes. 
 
THE INDUCTION MOTOR. 31 
 
 For the no load losses, draw the line H Y, parallel 
 to A C, passing through the point C. The mechanical 
 output, as we know, is measured from the line h y, 
 since the constant no load losses have always to be 
 subtracted. The test of the stationary and short- 
 circuited rotor, gives us the energy current, 
 
 -rr- = 60.7 amperes = i cos 6 or 
 
 60.7 
 cos 6 = = .515 power factor. 
 
 llo 
 
 Thus we find the point C k of the diagram. The center 
 OE of the input circle can now be easily found by con- 
 struction. The point OP is obtained by drawing through 
 D a perpendicular to C k D until it intersects the line 
 OE OF in the desired point OF. In other words, the line 
 D C k is a tangent to the circle of mechanical output. 
 It now remains to find the third, or torque circle. 
 This is done by dividing the line C k D which represents 
 the electrical losses at short circuit in the ratio of the 
 resistances of the primary and secondary circuits. 
 The third circle then passes through this point. 
 
 In the diagram, the field per phase corresponds to a 
 potential of 110 volts, and the loss vector at short 
 circuit, measured in the same scale, is 
 
 C*DX110 
 
 If the primary winding, which has a resistance of .142 
 ohms per phase, we have 118 amperes at short circuit, 
 therefore the drop of potential is 118 X .142 = 16.8 volts. 
 
 To find the intersection of the third circle with C k D, 
 
32 
 
 THE INDUCTION MOTOR. 
 
 10 H.P. 
 
 D no Volts 
 
 Torque \\^-^/-- 
 Electrical Input c^-^^^ 
 
 A 
 
 10 20 30 40 50 60 70 80 90 100 
 
 FIG. 14. 
 
THE INDUCTION MOTOR. 
 
 33 
 
 we must lay off this value = C k E k on C k D in the ratio 
 of 16.8 to 57. 
 
 Finally, erect a perpendicular from C k on the line 
 OE D and obtain the slip for any load. At no load the 
 slip is very small; at full load only 6.6%. 
 . Fig. 13 shows the characteristic curves of the motor 
 as functions of the output. The full load output is 
 seven horse power. At this load, the efficiency is 
 
 8 9 10 11 12 13 14 15 H.P 
 
 85.5%, and the power factor 90%. The power factor 
 reaches a maximum of about 91% at a load of about 
 nine horse power. The motor starts with a torque 
 of 16.7 synchronous horse-power, and has a maximum 
 output of 15.4 horse-power. 
 
 TEN HORSE POWER MOTOR. 
 
 Figs. 14 and 15 give the phase diagram and the 
 characteristic curves of a 10 horse power motor. 
 
34 THE INDUCTION MOTOR. 
 
 The tests gave the following results: 
 
 1. At no load, 110 volts applied per phase. 
 1 = 8 amperes, 
 
 A = 197 watts. 
 
 2. Rotor locked and short-circuited. 
 110 volts applied. 
 
 / = 162.5 amperes, 
 A = 9 kw. 
 
 3. Resistance per phase = .123 ohm. 
 
 Fig. 14 is the diagram constructed from these data. 
 Fig. 15 shows the characteristic curves. The following 
 results are obtained at full load of 10 horse power: 
 
 Efficiency = 83% 
 
 Current = 29.5 amperes 
 
 Electrical Input = 2.95 kw. 
 
 Power Factor = .915 
 
 Slip = 6% 
 
 Maximum output = 21 horse power. 
 
 THE INDUCTION MOTOR AS GENERATOR. 
 
 It is a well known fact that an induction motor, if 
 driven above synchronism, will act as a generator 
 analagously to the direct current shunt machine. It 
 is interesting that the circle diagram gives also light 
 on this use of the motor, if the circles are completed 
 on the left hand side of the line A D as in Fig. 16. The 
 slip is now negative, and what was before mechanical 
 output, is now mechanical input. Further, the elec- 
 trical input now becomes the electrical output. 
 
 We see that the generator gives now exactly the same 
 current for the same phase difference, as when employed 
 as a motor. In other words, the electrical energy is 
 the same in both cases. The mechanical input is, of 
 course, considerably larger in the case of the gen- 
 
THE INDUCTION MOTOR. 
 
 35 
 
 FIG. 16. 
 
 3 Input 4 H.P. 
 
 FlG. 17. 
 
36 THE INDUCTION MOTOR. 
 
 erator than the mechanical output of the motor. It 
 is easily seen that the difference amounts to twico 
 the losses, since now all losses add themselves to the. 
 electrical output, while in the motor they were sub- 
 tracted from the electrical input. 
 
 From a theoretical standpoint, the curves become 
 exceedingly interesting if drawn, not as functions to 
 the mechanical output, but as functions to the elec- 
 trical input; that is, as functions of the values which 
 correspond to the energy currents which are the ab- 
 scissae of the diagram (Fig. 18). 
 
 The curves thus obtained are of regular mathematical 
 forms, and are simple closed curves. The curves of 
 torque and mechanical output are ellipses. The power 
 factor curve is a function of an angle, etc. 
 
 There are three essentially different parts of the 
 curves to be considered: 
 
 1. As a motor, (slip to 1) C >C k . 
 
 2. As a generator j slip < ) _^_^k 
 
 3. As a generator ( slip > 1 I 
 
 With these general diagrams and curves, all essential 
 points of the use of the present method are exhausted. 
 All complicated details of the motor are based on two 
 phenomena, magnetic leakage and electric resistance. 
 These are essentially the only values which determine 
 the operation of the motor, and it remains for the 
 designer to choose the proper proportions. 
 
 SINGLE PHASE MOTORS. 
 
 The fundamental theory of the polyphase motor is 
 much simplified by the introduction of the principle 
 of the rotating field, but the theory of the single-phase 
 motor is somewhat more difficult. 
 
THE INDUCTION MOTOR. 
 
 37 
 
38 
 
 THE INDUCTION MOTOR. 
 
 The single-phase motor possesses a single phase 
 winding for a simple alternating current (Fig. 19), and 
 can therefore produce a simple alternating field in the 
 direction of the coil axis only. Such a field, of course, 
 will not cause rotation of a short-circuited armature, 
 without commutator or similar devices. Therefore it 
 seems, at first thought, somewhat astonishing, that 
 such a motor can operate at all, and indeed, in its 
 
 FIG. 19. 
 
 simplest form such a motor will not start; however, 
 after the motor has been started by any means what- 
 soever, it then operates exactly like a polyphase motor. 
 This phenomenon is explained by the fact that in the 
 single phase motor, rotating at a certain speed, there 
 is a field caused in the secondary which rotates with it. 
 Between this field and the ampere turns of the single 
 phase winding, a torque is established in the same 
 manner as between the field of a short-circuited second- 
 
THE INDUCTION MOTOR. 39 
 
 ary and one of the two or three phase windings of 
 the polyphase motor. The torque, however, is pulsat- 
 ing, since there is no second phase, and it becomes a 
 maximum, falls to zero and again reaches a maximum 
 varying with the alternations of the field. 
 
 In the case of induction motors, single as well as poly- 
 phase, we may consider the short-circuited secondary 
 as a field magnet of a synchronous motor. The only 
 difference is that the field does not rotate in the in- 
 duction motor in synchronism with the armature, but 
 has a small relative rotation with respect to the arma- 
 ture, corresponding to the slip. 
 
 In the short-circuited secondary of a single-phase 
 motor there exists, just as in the polyphase motor, a 
 constant rotating field notwithstanding the fact that 
 the current exciting winding produces only a simple 
 alternating field. This is explained by the character- 
 istic property of the short-circuited winding. We 
 know that any short-circuited winding tends strongly to 
 keep up the field interlinked with it, since any change 
 in this field causes current in the short-circuited wind- 
 ing which opposes such change. This counteracting 
 effect is the larger the lower the resistance of the wind- 
 ing and the greater the frequency. 
 
 If we consider the secondary of a single phase motor 
 revolving synchronously, the primary alternating field 
 will not appear as an alternating field in the secondary, 
 since in the same time, during which the field intensity 
 starting from zero reaches a maximum and decreases 
 again to zero, the secondary has turned through 180, 
 and the primary alternating field will therefore appear 
 as a pulsating field, and will have always the same 
 direction in regard to the secondary member. 
 
 Each pulsation of the secondary field causes, how- 
 
40 
 
 THE INDUCTION MOTOR. 
 
 ever, in the secondary winding wattless currents which 
 diminish the pulsation to a negligible minimum, i.e,, 
 they keep the pulsating field of the short-circuited 
 secondary practically constant. In the positions where 
 the field corresponding to the exciting current tends 
 to exceed this constant value, the secondary ampere 
 turns demagnetize. In other positions they furnish 
 the magnetizing current. 
 
 FIG. 20. 
 
 FIG. 22. 
 
 FIG. 23. 
 
 \ 
 
 FIG. 24. 
 
 In order to obtain a still clearer understanding of 
 this phenomena, let us consider the four characteristic 
 positions: 
 
 In Figs. 20-23, let a b be the primary winding, and 
 c the secondary, which rotates synchronously. 
 
 The field caused by the current 7, will then be as 
 shown in Fig 24. 
 
THE INDUCTION MOTOR. 
 
 41 
 
 In the positions 1 and 3, the maximum value of the 
 current acts once in one direction and once in the 
 other. Since, however, the rotor has meanwhile turned 
 through 180, its field will have in the positions 1 and 3 
 the same direction, and will have a characteristic 
 shown by the dotted line. The above described action 
 of the rotor now appears, and prevents the field from 
 
 FIG. 25. 
 
 being strongly pulsating, the pulsations being smoothed 
 out. Therefore, we obtain a constant field rotating 
 with the rotor. 
 
 It is now possible to picture the action of the rotor 
 as follows: Let 7 t be the magnetizing ampere turns of 
 the exciting member, / 2 those of the rotor; then these 
 two obviously must combine to form the ampere turns 
 
42 THE INDUCTION MOTOR. . 
 
 1 1 which produce the constant field. This is true, if, 
 for the moment, we neglect the leakage. 
 
 If we assume again that the alternating current is a 
 sine function, and that 7, is the maximum value of this 
 current, the motor being unloaded, then the instan- 
 taneous value of the ampere turns of the exciting 
 member at no load is proportional to 
 
 /! oc / t sin a 
 
 If we consider the short-circuited secondary to be 
 stationary, and the exciting winding to rotate, then 
 the constant field must also be stationary, This field 
 is produced by the ampere turns of the rotating exciting 
 member, and these ampere turns vary according to a 
 sine law. Therefore, the field of the short-circuited 
 secondary must have a component proportional to 
 
 T . 7-2 7 1 ~~ 2 cos a 
 
 /jSinaoc T^sm 2 a: oc 7 t - - - 
 
 The field of the short-circuited secondary must, 
 however, be constant, and therefore the resulting 
 ampere turns must be proportional to /, where / is the 
 magnetizing current. The difference between the mag- 
 netizing ampere turns and those active in the exciting 
 member must, therefore, be produced in the short-cir- 
 cuited secondary. 
 
 Thus, if / 2 is the value of the current in the secondary 
 we have 
 
 / 2 = 7- (1 -cos 2 a) 
 
 - *-k 4 ~ cos 2 a 
 
THE INDUCTION MOTOR. 3 
 
 The currents in the secondary can only be periodic 
 functions of the time, that is, there can be no constant 
 terms. Thus, we have, 
 
 7 2 ~ cos 2 a 
 
 Therefore, the ampere turns of the primary member 
 are 
 
 A T l oc - i cos 2 a 
 and those of the secondary are 
 
 A T 2 oc ~ cos 2 a, 
 and the magnetizing ampere turns 
 
 These equations indicate that in the single-phase 
 motor the magnetizing current / is half of the no load 
 current. The four characteristic positions are now 
 represented in Fig. 26. 
 
 The sum of the two fields gives a straight line corre- 
 sponding to the amplitude ~. 
 
44 THE INDUCTION MOTOR. 
 
 The single fields are represented by the full and 
 dotted waves. The rotating field is not quite con- 
 stant, but is slightly pulsating. It is obvious, how- 
 ever, that this may be neglected. 
 
 Another approximation previously made will now be 
 corrected. Owing to the presence of leakage, the ratio 
 
 No load current 
 
 is not exactlv two. 
 
 Magnetizing current 
 
 In the position of Figs. 27 and 28 the secondary 
 fields are the same. In Fig. 27 it is induced by the 
 
 > sin a oc 3i sin 2 ar I (Z-cos 5 
 
 primary winding by the magnetizing current I = I ^ 7 2 , 
 or, introducing the reluctance of the secondary 
 
 In the other positions (Fig. 28), the secondary field 
 is produced by the ampere turns of the short-circuited 
 secondary. The reluctance will be smaller, since the 
 field now has two paths, the main path and the leakage 
 path. If we call the reluctance in this case p f then 
 
THE INDUCTION MOTOR. 
 
 45 
 
 FIG. 28. 
 
46 THE INDUCTION MOTOR. 
 
 We have seen that the secondary field remains almost 
 constant, and therefore we have 
 
 = , or, since 
 P P 
 
 -j,y 
 
 No load current 
 Magnetizing current 
 
 By substitution of the simple reluctances p^ p s and p 2 
 the ratio can appear in a somewhat different form, in 
 which only known values appear. It will be seen then 
 that p' is little different from p and that the ratio is 
 almost two. 
 
 The phenomena in single-phase motors are thus as 
 follows: 
 
 In the unloaded single-phase motor, there is formed 
 in the short-circuited secondary, as in the polyphase 
 motor, a field which rotates with the short-circuited 
 secondary. This field when it coincides with the axis 
 of the primary winding is produced by the latter. 
 At right angles to this position, it is produced by the 
 no load currents of the short-circuited secondary. The 
 latter are wattless and are produced by the primary 
 member. Therefore, the no load current equals twice 
 the magnetizing current, or, more accurately, the no 
 
 load current is - times the magnetizing current. 
 
THE INDUCTION MOTOR. 47 
 
 Obviously, it is the same as far as the primary winding 
 is concerned, whether the field in the position at right 
 angles to the primary winding is produced by current 
 in the short-circuited secondary, or by current in the 
 winding of a second-phase, as in a two-phase motor. 
 
 Therefore, the single phase primary acts with the 
 same torque on a short-circuited secondary as that of 
 one-phase of a two-phase motor. From these reflec- 
 tions, we are able to obtain the diagram for the single- 
 phase motor from that of the polyphase. 
 
 In the polyphase motor we have 
 
 A C oc / = magnetizing current. 
 In the single phase motor we have 
 A C = / 
 
 or since p is little different from p' 
 AC = 21 
 
 The ratio- ^ is therefore no longer -r p = , as 
 /L u A \s p 
 
 in the polyphase motor, but 
 
 A_D = ^ 
 
 A C ~ 2/> 
 
 This ratio is equal to that of a polyphase motor 
 having double the reluctance. 
 
 One must, however, not carry the analogy too far. 
 For instance, we must not expect to obtain a good 
 single-phase motor by running a polyphase motor on a 
 single-phase circuit by simply not using the second-phase. 
 Such a motor would give bad results, and, on account 
 
48 THE INDUCTION MOTOR. 
 
 of the omission of the second phase, the output would 
 not much exceed one-half the output of the polyphase 
 motor of the same dimensions. In a properly designed 
 single-phase motor, however, the relatively poor per- 
 formance is not so marked. There are certain consider- 
 ations in the design of the single-phase motor which 
 materially reduce various dimensions; consequently a 
 good single-phase motor is little larger than a polyphase 
 motor of equal output, and its efficiency is not much 
 lower. These matters of design, however, are beyond 
 the scope of this publication, and I reserve the discussion 
 of them for some future time. 
 
 The main object in the preceding pages has been to 
 reduce to practice my diagram, and to present it in a 
 clear and easily understood form. If I have succeeded 
 in making clearer a complicated part of our alternating 
 current theory, the purpose of this work is fulfilled. 
 
UNIVERSITY OF CALIFORNIA LIBRARY 
 BERKELEY 
 
 Return to desk from which borrowed. 
 This book is DUE on the last date stamped below. 
 
 ENGINEERING LIBRARY 
 
 MAY 2 2 1950 
 
 MAR 2 9 1951 
 
 LD 21-100m-9/48(B399sl6)476 
 
YC 33497 
 
 838991 
 
 H4 
 
 THE UNIVERSITY OF CALIFORNIA LIBRARY