346 THE INDUCTION MOTOR. 



By addition we get : Z, + P + P = Z sin - \ sin 



(6 - 120) - 4 sin (B + 120). 

 = Z {sin - i [sin (B + 120) + sin (6 - 120)]} 

 = Z {sin (9 - 2 sin cos 120} 

 = Z {sin + | sin 0} v cos 120 = - i. 



= -4- Z sin 0. 



On evaluating the vertical components we get : 

 for 2,, zero i.e., 0. 



z z ,PZ, = OZ 2 (- sin 60) = - Z sin (B - 120) ^ 



A 



z s , PZ 3 = OZ 3 sin 60 = + Z sin (B + 120) 



Adding, 

 0+PZ,+.PZ S =Z sin (0+ 120) -sin (0-120) 



= Z - 2 cos ^ sin 120 



= Z V^cos 



= -o- Z cos 0. 



The three fields are thus equivalent to two fields at 

 right angles to each other, whose values are : 



3 3 



-a- Z sin and 4- Z cos 0. 



The resultant field formed of these two will be : 

 Z r = ~ Z Vsin 2 + cos 2 = -|z. 



Hence the resultant field is constant in value and 



3 

 equal to -jy- the maximum value of each field taken singly. 



Further, if a is the angle which this resultant field 

 makes with the vertical 



sum of horizontal components 2 



tan a = 5 r- i = ~o ~ = tan v. 



sum oi vertical components 6_ ~ 



A 



hence a = 6, and the angle of the resultant field will vary 

 at the same rate as the phase of the separate fields, and 

 the resultant field will complete one revolution in the 



