﻿58 
  Prof. 
  T. 
  R. 
  Lyle 
  on 
  the 
  Theory 
  of 
  

  

  gives 
  us 
  as. 
  Proceeding 
  in 
  this 
  way, 
  we 
  obtain 
  in 
  succession 
  

   a4, 
  a3, 
  as, 
  ai, 
  a^, 
  which 
  represent 
  the 
  harmonics 
  of 
  x 
  and 
  f 
  

   correctly 
  as 
  regards 
  relative 
  phase 
  and 
  relative 
  amplitude. 
  

   But 
  as 
  «() 
  must 
  be 
  equal 
  to 
  twice 
  the 
  exciting 
  current 
  we 
  

   have 
  a 
  scale 
  for 
  our 
  diagram, 
  and 
  hence 
  obtain 
  a 
  complete 
  

   solution. 
  

  

  The 
  fact 
  that 
  for 
  a 
  practical 
  alternator 
  the 
  r 
  operators 
  may 
  

   be 
  taken 
  as 
  pure 
  numbers 
  (see 
  §§ 
  8, 
  22), 
  renders 
  this 
  method 
  

   of 
  solution 
  both 
  easy 
  and 
  expeditious. 
  

  

  15. 
  If 
  a 
  source 
  of 
  constant 
  e.m.f. 
  be 
  included 
  in 
  the 
  

   armature 
  circuit 
  as 
  well 
  as 
  in 
  the 
  field 
  circuit 
  of 
  the 
  simple 
  

   alternator 
  indicated 
  in 
  fig. 
  1, 
  equations 
  I. 
  § 
  1 
  become 
  

  

  7\v 
  + 
  "77 
  (Ix 
  + 
  m^ 
  COS 
  (ot) 
  = 
  e 
  

   p^ 
  + 
  -J 
  (X^ 
  -\- 
  mx 
  cos 
  (Dt) 
  = 
  7), 
  

  

  and 
  both 
  the 
  armature 
  and 
  tield 
  currents 
  will 
  now 
  contain 
  

   harmonics 
  of 
  all 
  orders, 
  odd 
  and 
  even. 
  In 
  this 
  case 
  we 
  

   assume 
  that 
  

  

  ^ 
  = 
  -^ 
  + 
  ai 
  -f 
  ag 
  + 
  aa 
  -)- 
  &c. 
  

  

  where 
  ao 
  is 
  the 
  vector 
  to 
  the 
  point 
  whose 
  polar 
  coordinates 
  

   are 
  2e/r, 
  7r/2, 
  and 
  a^ 
  is, 
  as 
  before, 
  the 
  vector 
  to 
  the 
  point 
  

   27j/p, 
  7r/2. 
  The 
  other 
  vectors 
  ai, 
  a2, 
  ag, 
  &c., 
  «!, 
  ag? 
  <^35 
  &c., 
  

   have 
  to 
  be 
  determined. 
  

  

  On 
  substituting 
  for 
  x 
  and 
  f 
  in 
  the 
  above 
  equations 
  it 
  will 
  

   be 
  found 
  that 
  the 
  odd 
  order 
  vectors 
  in 
  x 
  and 
  the 
  even 
  order 
  

   ones 
  in 
  f 
  are 
  determined 
  by 
  the 
  same 
  equations 
  (IV. 
  § 
  6) 
  as 
  

   when 
  e 
  = 
  0, 
  and 
  are 
  completely 
  independent 
  of 
  the 
  even 
  

   order 
  vectors 
  in 
  x 
  and 
  the 
  odd 
  order 
  ones 
  in 
  f 
  , 
  these 
  latter 
  

   depending 
  only 
  on 
  ^ 
  and 
  vanishing 
  with 
  e. 
  

  

  This 
  being 
  so, 
  ai, 
  0C2, 
  ag, 
  a^, 
  &c., 
  are 
  given 
  by 
  the 
  solution 
  

   already 
  obtained, 
  and 
  aj, 
  as, 
  "3, 
  a45 
  &c., 
  will 
  be 
  given 
  by 
  a 
  

   similar 
  solution 
  the 
  equations 
  expressing 
  which 
  may 
  be 
  

   written 
  down 
  from 
  synmjetry. 
  

  

  Thus 
  the 
  complete 
  solution 
  is 
  given 
  by 
  

  

  uq 
  = 
  — 
  Siaj 
  = 
  Si22<^2 
  = 
  —Si^sSsas 
  = 
  ^c, 
  

   ao 
  = 
  — 
  Xi^i 
  = 
  SjSsas 
  = 
  — 
  XiSs^gag 
  = 
  (fee, 
  

  

  where 
  ao 
  is 
  the 
  vector 
  to 
  2r)/p, 
  7r/2, 
  as 
  before, 
  

  

  and 
  Rq 
  is 
  the 
  vector 
  to 
  2elr, 
  7r/2. 
  

  

  