DR WALLACE ON A FUNCTIONAL EQUATION. ^61 



Construction of Curves of Equilibeation by the Tables. 



59. Various problems may be proposed respecting the construction of a 

 catenary and equilibrated arches ; but of these, I believe the two which follow 

 are the most useful. 



Problem I. — A chain of a given length hangs freely between two points, which 

 are at a given distance in a horizontal line ; to find the position of its lowest 

 point, and the parameter of the catenary. 



Suppose the chain to be 100 feet in length, and the distance between the 

 points of suspension to be 60 feet. 



Appljdng our notation : in a catenary of which a is the parameter, x the 

 amplitude oi f{x) an ordinate, and ¥ {x) the corresponding arc of the curve, 

 there are given x = '^() feet, and F {x) = 50 feet; to find a and/(,2?)-<7. 



Assuming o^ to be = 1, the problem requires that a tabular value of x 



F (-x) .50 

 be found, which shall satisfy the condition — 5^=^=1.66667. Now, the quan- 

 tity — —, at first =1, increases continually: and it appears from our second 



= 1.63454, and to .^,= 1.85, — ^ = 1.67637, 

 = .03213; and i^ - IM = .04183 . 



X. x^ 



LILV -^ ? "'*' i-iAOU X, 



xij-^^j. ^:^ci 



table, that to a;o = 1.8, 



Xo 





FW 



X 



^0 



o 



Now, as an approximation, the first of these differences will be to the second 

 nearly as «— aro to x,—x^'. 



Therefore, 4183 .^21^ = x-x^-. x-x, = m : x-x^; 



3213 X .05 

 and «-«o= — 45^3 — = .03841, and a;=1.8 + 03841 =1.83841; 



and F(;.)=-^-=3.06402. 



This is the tabular value of F {x) when the parameter = 1, but to the parameter 

 a, we have <r= 1.83841 a, and F (^) =3.06402 «. In the catenary formed by the 



50 

 chain, F {x) = 50 feet; therefore, a= ^naArxo ~ 16.318 feet. Now, when a=l, 



f{x) is the secant of an arc 0, of which F (x) is the tangent : therefore, 

 tan <^= 3.06402, and = 71° 55' 30"; and /(^) = sec 0=3.22308, and/(a?)-l = 

 2.22308. Hence the distance between the lowest point of the chain and the line 



2.22308 X 50 

 joining the points of suspension, is o obaqo — ~ 36.277 feet. Now, the para- 



VOL. XIV. PART II. 6 I 



