128 



Prohlems. 



Fig. 3. 



Put the area ABCD=E. 



ahcd=Y. 



efgh=G. 



length =L. 



solidity=S. 



Then S=iL . (E4-F+4G).* Or, 



To the area of the two ends, add four times the area of the mid- 

 dle cross section : multiply by one sixth of the length. 



It is evident that the above solution holds true, whether the incli- 

 nation of the surface at each end be in the same ratio or not. 



The solutions of this problem given in Major Long's Rail Road 

 Manual, and also by Mr. Charles Potts, in a pamphlet on canal cut- 

 ting, are correct only when the surface inclines in the same ratio at 

 each end of the given distance. 



In the construction of wing walls for culverts, 

 bridges, Sec, the form represented in Fig. 3, 

 has been adopted by some engineers, as com- 

 bining the greatest advantages. The large por- 

 tion of the curve presents its convex surface 

 to the pressure of the embankment, and the 

 smaller curve allows the water and floating 

 bodies to pass more readily without injuring 

 the walls. ; 



The distance AB, BC and the smaller radius E- 

 AD are usually given. The problem is to find 

 the corresponding radius EF. 



Find the angle BCA. The angle DAF=BCA, (Euc. Book I, 

 Prop. 29 ;) and the angle DFA=DAF, (Euc. Book I, Prop. 5.)t 

 Therefore, (by Euc. B. I, Prop. 32,) 180-DFA+DAF=ADF; 

 and (Euc. Book I, Prop. 29) DEG=ADF. Then, as log. sin. of 

 GED : DG : : radius : ED. ED - FD=EF. Or, algebraically, 

 Put FD=AD=a. 

 DG=AB=6. 

 BC-AD=GC=c. 

 EC=EF=a:. 



• Day's Mensuration, Part III, p. 37. 



■f It may be well to add, for the sake of practical men, who have no time for mathe- 

 matical investigations, and as a truth which is essential to their clear understanding 

 of the problem, that the point F, which is the intersection of the lines AC, DE, is 

 also, in fact, the point oi contact of the two circles. — Co7n. 



