•I Quantify of Tron necessary in a Tension Chain Bridge. [Jan. 



EH 



Hence tension of EF= tension of EA x 



1 1 ion of iron in EF = section of EA X 

 Also tension of EG = tension of EA x 

 .-. section of iron in EG = section of EA x 



EB 

 EH 

 ~EB 



EJ 

 ~EB 



EJ 



EB 



I [(Mice the quantity of iron in EF, FC, EG, G C == quantity in Eh 



and Eg 



= Eh% section of EF -f- Ey x section of EG 



, „j / Eh x EH -f Eg % EJ \ 



sss section of LA { ' J ^ \ 



I EB 1 



But by a property, (which we shall prove below, and which we 

 defer at present in order not to interrupt this demonstration) — 



If Ell, EJ represent the magnitudes and directions of two forces of 

 which the magnitude and direction of the resultant is EB, and from 

 any point C perpendiculars be drawn upon these three directions, (pro- 

 duced if necessary,) as Oh, Ccj, Cb : then EH % Eh -\- EJ x Eg = 

 EB x Eb. 



This being assumed the calculation above gives — 



Quantity of iron in EF, FC, EG, GC = Eb. X section of EB 



= quantity of iron in EB and BC. — Q. E. D. 



AVe shall now demonstrate the property we have just assumed. 



Tlu- lines in (fig. 3) are the same as in (fig. 2), except that in ad- 

 dition ///.-, Jj, are drawn at right angles to EC and meeting EB in k' 

 and/. Now the triangles EHk, ECh are similar. 



.'. EH: Ek : : EC : Eh 



.'. EH% Eh = EC*Ek. 

 So also from the similar triangles Ekk>, EbC we have 



Eh : Ek 1 : : Eb : EC. 



.*. EC^Ek = Eb%Ek'. 



Hence EH x Eh = Eb x Ek 1 . 

 In precisely the same manner 



EJ x Eg = Eb* Ej'. 

 No* in the triangles /•////<', BJf the angles are equal, and EH = 



