1848.] Quantity of Iron necessary in a Tension Chain Bridge. 



BJ: hence the triangles are equal, and .\ Ek' = Bf. 



... Ek 1 + Ejl = Bf H- Ef— BE. 

 Hence, then, from the above 



EH%Eh + EJ%Eg = EB% Eb.—Q. E. D. 



We have thus proved the Proposition, which we began by enunciating, 

 in the case represented in fig. 2. But the same is true in any other 

 case. For (see fig. 4.) we may suppose the rods KG, GC taken away, 

 and others KM, MC, KL, LC put in their place, and the reasoning will 

 be precisely the same, and the result the same, however many subdivi- 

 sions be made. And therefore the property is universally true. 



The above demonstration is geometrical only ; but by the help 

 of analysis we may give the following proof which at once applies to 

 every case which can occur. 



2. Analytical Demonstration. 



Suppose EB, BC removed, and replaced by any number of rods 

 EF, FC; EG, GC; EH, HC ; EJ, JC ; EK, KC, &c. 



Let Q X Q. 2 9 3 be the angles which EF, EG, EH make with AB. 



e^e 2 e 3 * e j, ek, 



Let £ be the transverse section of iron in AB : 



S l S 2 S 3 ditto. . in EF, EG, EH, 



8^8 JSf ditto. . EJ,EK, 



Then, by hypothesis, £ S 1 S 9 8 3 . . . . S^S^S^, are proportional to 

 the tensions of those rods. 



Draw Cf, Cy, Ch, Cb, Cj, Ck, . . perpendicular to EF, EG, EH, 

 EB, EJ, EK, .... Join EC. Let EC = a : and BEC = d. 



Now because the tensions at E are in equilibrium ; 



,K8 =$ x Cos Q x -J- S 2 Cos 9 2 +. . + 8 X ' Cos 9/ +S 2 ' Cos 9. 2 ' -f ... 



o = S x Sin 0, + S 2 Sin 9 2 +. . + SJ Sin 0, — S 2 ' Sin Q 2 < — ... 



Then multiplying the first of these by a Cos d and the second by a 

 Sin d and subtracting. 



8. a Cos d = S, a Cos (9, + d) + S 2 a Cos (9, -f d) + ... 



+ V a Cos (B.'—d) + S 2 ' a Cos (Q 2 '—d) + . . . 



or S xM=z a S 1 *Ef+ S 2 x Eg +.,-,. + S,' x Ej + 8J X 

 Ek+... 



