102  '      Dr.  W.  E.  Suinpner  on  the 
From  these  equations  it  can  be  shown,  with  the  help  of  the 
last  relations  o£  (25)  and  equivalent  relations  such  as 
2  e2e3 = eA  —  e2  —  e3  , 
that 
el2  +  012  =  e22  +  e22  =  e32  +  6s2  =  2e2,]        _     _     _     (27) 
where  3  e2  =  e2-re2-]-€2 ;  J 
or  e  is  the  square  root  o£  mean  square  o£  the  quantities 
€j,  e2,  e3,  and  may  be  said  to  measure  the  extent  to  which  the 
load-currents  are  "  out  of  balance."  For  small  values,  e  is 
essentially  the  same  as  the  arithmetic  mean  of  el5  e2,  and  e3. 
Since  algebraically  the  sum  of  these  quantities  is  zero,  it 
follows  that  the  greatest  is  equal  numerically  to  the  sum  of 
the  other  two,  and  thus  e  is  two  thirds  of  the  greatest  of  the 
quantities  els  e2,  e3. 
Now,  if  we  equate  to  A0  each  of  the  three  ratios  of  (22) 
and  multiply  up  and  reduce,  we  obtain  the  three  equations 
e0  cos  (j)  —  00  sin  </)  =  ex  cos  0  —  6X  sin  <f>, 
e0  cos((/)-60)-6>0sin(^-60)  =  62cos((/)-60)-6'2sin(^)-60),  [  (28) 
e0  cos(<£  +  60)  -  0O  sin(<£  +  60)  =  e3  cos(c£  +  60)  -  0Z  sin (0  +  60) . 
If  we  subtract  the  third  of  these  equations  from  the  second 
and  simplify  with  the  aid  of  (25)  and  (26),  we  get 
e0  sin  cj)  +  60  cos  <£  =  —  [ej  sin  <£  +  6X  cos  <f\ .     .     .     (29) 
By  squaring  this  equation  and  adding  the  square  of  the 
first  of  (28)  we  get 
€O84-0o2  =  €i2  +  012=2e2  by  (27);     .     .     .      (30) 
or  #0  is  necessarily  less  than  e2i 
If  we  choose  two  angles  j30  and  j3Y  such  that 
e0  =  00  tan  /30,       e:  =  0,  tan  /31? 
we  can  use  (30),  the  first  equation  of  (28),  and  (29)  to  prove 
sin  (£o-0)  =sin  (/3L -(/>), 
cos  (/30-<M  =  -  cos  (/3L-cf>)  ; 
these  simultaneous  equations  involve  the  relation 
or  /3o  =  7r  +  2  0—  ySj. 
