106 Prof. H. Hennessy. Hydraulic Problems on [Apr. 19, 



But the depth In of the curve below its chord is y — a or In = § I. 

 In this inquiry I is the perimeter of the half curve, so that the total 

 perimeter, the chord, and the depth are respectively in the ratio of 

 the numbers 3, 2, and 1, or the chord of the catenary of greatest area 

 for a given perimeter is twice the depth, and the length of the curve 

 is three times the depth. The outline of this curve is readily shown 

 by attaching a fine chain of 3 units of length to supports at a distance 



Fig. 2. 



The chord AB = 2CD. 



The arc ADB of the catenary = 3CD. 



of 2 units. The catenary which would give a maximum hydraulic 

 mean depth for an open channel is therefore one whose depth is the 

 radius and chord the diameter of a semicircle. On substituting the 

 value of a found above in the equation for u, we shall find that the 

 hydraulic mean depth of the catenary under consideration is nearly 

 O'Sll or 0*155L, where L is the total perimeter of the curve. In a 

 semicircle the hydraulic mean depth is \r .== L/27T, or 0'159L 

 nearly, hence the hydraulic mean depth of the catenary of maximum 

 area is nearly equal to that of a semicircle of equal perimeter. But 

 a channel formed by the outline of such a catenary would when the 

 contained liquid falls, not be liable to so rapid a reduction of hydraulic 

 mean depth as in the semicircle. For small arcs of a circle it has 

 been shown that this is proportional to the square of the angle sub- 

 tended at the centre. In the catenary if is the angle made by the 

 tangent with the directrix, it is also the angle made by the radius 

 of curvature with the axis of y, which in this case coincides with the 

 axis of depth, and as 



y = a sec 0, x = a log (sec + tan 0), 



a 



tan 



[sec log (sec + tan 0) — tan 0] 



= r~~ r Hog tan i (tt + 0) - sin 0"1 



Bin L & - J ' J 



