634     Messrs.  Hawthorne  and  Morton  on  Deflexions  caused 
If  w  is  the  weight  of  unit  length  of  wire  and  T  the  horizontal 
tension, 
T  =  wc; 
3T3  3T3 
dT= — 5-0.  2dx  =  x   0  „  x  (increase  of  horizontal  span). 
2w2x°  zwz  >'* 
We  shall  write  /"for 
x 
3T3 
2lD2#S" 
Let  the  flexible  poles  he  numbered  from  an  anchor-pole. 
Let  xr  be  the  deflexion  of  the  top  of  the  rth  pole,  and  let 
the  break  occur  beyond  the  nth  pole.  We  shall  take  first 
the  simpler  case  where  there  is  only  one  wire  suspended,  or 
where,  if  there  are  several  wires,  these  all  break  together. 
Denote  the  horizontal  tensions  in  successive  intervals  by 
Toi?  T12, . .  .  T(n_i),i,  and  the  common  value  before  the  break 
occurs  by  T.     Then  we  have 
Toi         =T->„ 
T12         =T—f(x2—x1), 
T(n  _ l)n  =  T  —  f(xn  -  Xn  - 1) . 
Toi— Ti2=#a?j, 
Tl2  —  T23  =  $T2, 
±n—2,  »■— 1       *-n  —  1,  n==<Pxn~U 
On  eliminating  the  tensions  T01,  Tn,  &c,  we  get  the  following 
equations  connecting  consecutive  deflexions,  <in  which  for 
shortness  we  write 
2+^  =  «, 
T      ; 
r~h> 
X2  —  €LX\ 
=  0, 
xz—ax2-\-x1 
=o, 
oc n  —axn_1  +  xn_ 
•  2  =  0, 
{a~-l)xn-xn_x 
=b. 
