3<50 ON SUSPENSION BRIDGES. 



a dz 



From equation (3) dyzz JTaTr^* Integrating 

 (y = 0, when z = o) 



y=a.hyp.log.l±J^±!!)--C7) 



Substituting for z in equation (7) its value 

 derived from (6) gives 



y = a. hyp. log. ^ ^ - - (8) 



From equation (4), tangent of inclination at 

 C to horizon = — (9) 



From equation (5) tension at C=n/ («"+=*) - - (10) 



Cor. 2nd. The formulae of the uniform 

 catenary may be rendered rather more simple, 

 by putting some multiple of a for the variable 

 quantity. Thus suppose ma = z, the cases of 

 the last corollary give 



x=ia(»y{\ +m2)--l\ y-a. hyp. I. Tm -f V (1 + m^)\ 



X v/(l -f m^)-! ^ ^ r 1 



.*. — = ^^ — > — :-> tanejent of ele- 



y hyp. log. Qm+ V(l + m2)^ ^ 



vation at C = »w, tension at C = a n/(1 + w^). 



If m = 1, X =z. 4142a, y = .8813a, 

 '~ ~ "8813"' •*• y^ 2.127 X. Tangent of ele- 



if 



