38 Mr. T. H. Blakesley on Magnetic Lag. 



FB is one of the components of AB. 



The other component (i. e. that arising from the machine's 



-pi 



proper electromotive force) is AF. Hence AF=m— • 



It follows that the electromotive force E, which the machine 

 is exerting, may he thus determined by means of the dynamo- 

 meter observations. 



AF 2 =AB 2 + BF 2 + 2ABBFcosABC, 



=? = M{ + C4 a ) W + 8mli 4 ^ cos 0, 



r x 2 \n 2 r%/ n 2 r ± 



.-. E 2 = n 2 I x 2 + rf I 2 2 ~ + 2r ,r 2 ™ IA cos 0, 



= n 2 2A ai + ~ r 2 2 2B* 2 + 4^r 2 ^ C« 3 , 



= 2 { n 8 A«x + r£ ~ Ba 2 + 2^ 2 ™ C* 3 } ■ 



Another interesting magnitude is AC, or the total impressed 

 magnetic force. 



AC 2 = AB 2 + BC 2 -2ABBCcos0, 



= m 2 Ii + n 2 1 2 2 — 2mnI 1 I 2 cos 0, 



= 2m 2 Aa x + 2n 2 Bu 2 — 4m?iCa 3? 



= 2 { m 2 A«! + n 2 B* 2 - 2mnC« 3 } . 



By means of this we may calculate what current should be 

 passed through the primary circuit, the secondary being open, 

 to produce the same state in the core. 



But perhaps the most interesting point to men of science 

 and to civil engineers is the question of power. We may 

 approach it thus perhaps in the simplest way. 



By dropping a perpendicular from F upon AB produced 

 we easily see that 



AF cos BAF = BF cos ABC + AB. 



Multiplying through by AB we have 



AF AB cos BAF= AB BF cos ABC + AB 2 . 



