92 PROCEEDINGS OP SECTIOX A. 



Eotating OE' forward through 90° gives us k of section 25, and 

 completing the parallelogram a^OK, its diagonal is a^-\- k in the line OY'. 

 Knowing the motor circuits we can determine s„ 5„ o-,, /8„ s^,b , &c., 

 and then construct for a„ a„ a,, a^, &c., exactly as in section 13, unless 

 that in this construction the vector a„ + k takes the place of a^, in 

 section 13. 



Now the mechanical torque developed hu the machine is (see 

 section 12) — 



\ Va^a^ -{- Yaitt, -\- Ya,a3 -{- Ya,a^ -\- &c. [ 



— — ' into the sum of the areas, attending to signs, of the triangles 



a^Oai, ajOa,, a,Oa„ &c. 



But the triangles a,Oa,, a^Oa, , &c., are all essentially negative 



[their sum is equal to — - B a^ + k", see section 25], so that if the 



machine is to develop mechanical power and run as a motor the phase 

 of OE' must be such that the area of the triangle a„Oa, is positive (as 

 it is in Fig. IV.), and numerically greater than the sum of ajOa,, 

 a.Oa,, etc. 



The power supplied by the source is — 



= 2-^. '^^"•OE'cosa.OE- 



(1)731 - - . 



=. —7- a, K sm a, Ok 



4 ' 



— — X area of triangle a,OK 



And the power developed by the motor — 



= ojT = ^-i_ X sum of triangles a^Oa^, a,Oa,, a,Oa„ &c., attending to 



signs. Hence the efficiency is equal to — 



a^Oa, + a,Oa, + a.Oa^ + &c. 



a.O/c 



By rotating the vector from O to the middle point of a„a, back- 

 wards through 90°, and doubling, we obtain OE,, which represents in 

 amplitude and phase 2/owi times the first harmonic of the total e.m.f. 

 of the motor. (Sea section 13.) This vector can now be compared 

 with OE' which represents 2/oj??2 times the applied e.m.f. 



Fig. lY. easily explains how by increasing the exciting current of 

 an A.C. motor the phase of the armature current is advanced relative 

 to that of the applied e.m.f. 



