ON SUSPENSION BRIDGES. ' 359 



dx 



-J— = the tangent of the inclination of the curve 



at C to the horizon, 



.'. Tangent of inclination at C = — j (4). 



From equation (3) we have -j— = — - — ^\ 



dz g^ 



where -^ is the secant of the inclination at C 



to the horizon i and as this quantity is the ratio 

 of the force or tension at C to that at B, and 

 a the tension at B 



.-. tension at C z= v(d» + w' ) (5) 



The conclusions (4) and (5) are equivalent 

 to those in the catenarian polygon above, since 

 here the weight w is as the tangent, and tjie 

 tension at C as the secant of the inclination of 

 the curve at C to the horizon. 



Cor. 1. li w = z, or the catenary be the 

 ordinary one in which the curve is of uniform 

 weight, we have a = tension at B in lengths 

 of the curve. Substituting then z for w in the 

 equations above we have : 



% 28 dz 



From equation (2) dx-^ .^^^^ y Integrating 

 (« = 0, when zzzo) 



.«=:V(a*+i«)-a (6) 



