76 



PROCEEDINGS OP SECTION A. 



Taking the mean value of this product, we find that the mean driving 

 torque T is given by — 



T = - ^ I Ya,a. + Va.a, + Va,a3 + Ya.a^ + &c. | , 



which result, combined with the last obtained in section 11, gives the 

 power equation — 



wT = E^ + lit 

 as it ought. 



Fig. II. 



13. The solution obtained can be represented geometrically in an 

 interesting way as follows : — 



Take two lines OX, OY (Fig. II.) at right angles. Measure off 

 from OY in the positive direction the angles YOl = i,, 102 = )8„ 

 203 = ij, 304 = /?t, &c., where b„ ^,, b^, (3^, &c., are the angles 

 determined in section 8. 



In Oi' take Oa^ =: 2v;/p = 2 x excitiiig current. Produce 10 

 through to a, so that Oa, = — ^° . In 02 take Oa, = '. Produce 



30 through O to a;, so that Oa^ = — ', and so on where s„ cr,, 5„ o-^, &c., 

 are the quantities determined in section 8. 



