1888.] the Cross-sections of Pipes and Channels. 105 



A/to i i)\ l (1+ + a 2 )\ . 

 o>vQ + log I ^ J — al 



« = j — 5 • 



The shape of the curve depends on the relation between its para- 

 meter and its length, hence we must find the value of a which makes 

 u a maximum in the above expression. The problem seems therefore 

 to amount to this elementary statical question : — A flexible and 

 uniform chain is attached to two supports on the same horizontal 

 line ; required the distance between the supports so as to make the area 

 of the surface included between the chain and the horizontal line the 

 greatest possible ; or given the perimeter of a catenary to find the 

 chord, so that the area between itself and the curve shall be a maxi- 

 mum. The above expression gives — 



W( Z2 + a 3)_jkg( jJ-Z 



= VW+^) l0 ^\ a )~ 



d2u _ (SP + 2a 2 ) a? log ( Z + V ^ + a ^ -l(V + 2a?) </{P + a 2 ) 

 da 2 a (P + aJ)W ' 



If we write z — Ija, and make dujda = 0, we have 



log(* + + *-)) = 2 + g2 • 



By successive trials this equation may be satisfied by substituting 

 z = 2*4, whence I = 2 - 4a. This value substituted in the expression 

 for dPu/da 2 gives a negative result, and therefore the value of u is a 

 maximum when a = T \jl. When z = 2*4, -s/(l + z 2 ) = 2*6, and 

 log + ^(1 + z 2 )) = log 5 = 1-6094 nearly. 



*gf±3 - ±|# , 



With further approximation we should find therefore — 



