150 CONTINUOUS GIEDER. [CHAP 3T. 



and a parabolic segment C E B. [PI. 16, Fig. 64.] If in the 

 first force polygon B'D' is the total load upon the span, the 

 line O A' parallel to the end tangent A G divides B' D' in the 

 same ratio as the end of the load divides the span, or denoting 

 the length B,, C by I. 



B'A' :B'D'::/8Z: 



The intersection G of the end tangents lies in the vertical 

 G H, which halves B F. Since the triangle B G T is similar to 

 O A' B', we have 



BT:B'A' :::; 

 or since B' D' =p I, B' A' =p = p I j. 



It is therefore easy to construct B T as in Fig. 64, where 



a 



B! D x p I, A! B! = B' A' = p I j, and pole distance = /3 ; 



then B.J D 2 = B T. 



If B T is thus found, we can easily, when A and B are given, 

 construct the end tangents, and then construct the first equili- 

 brium curve itself. 



2. Make G K = | GI, then the triangle C X B is equal to 

 the parabolic area CEB. If, then, through K we draw a 

 parallel to C B, intersecting C G in L, and through L the ver- 

 tical L M, the triangle A L B is equal to the entire simple 

 moment area. This last is therefore proportional to L M, or 

 '$11 = I x LM; hence 2ft = L M. It is easily proved 

 also that F M = F B. Thus, as G K is of G I, G L is 

 of GO; hence M H is of F H, and therefore F M is f of 

 F H, or f of F B = | F B. L M can therefore be easily 

 drawn. 



3. Let N be the middle of A B. Make N O equal to F N. 

 Then the centre of gravity of the triangle A C B is in the ver- 

 tical through O, while the centre of gravity of the parabola is 

 in H G. If through L we draw a parallel to A B, intersecting 

 the vertical through C in P, the two areas are to each other 

 as F C to C P. If in the vertical through O we make O Q = 

 i C P, then, since I H = -J C F, we have 



IH:OQ::FC:CP. 



