Form of the Trailing Aerial. 735 



the normal pressure. Hence the tension cannot become 

 infinite without the weight, and therefore the length, o£ the 

 wire being infinite too. This is a physical proof that the 

 integral in (2) tends to infinity as sin tends to cot a. 

 Hence it is clear that there is no point of inflexion at a finite 

 distance along the wire, but the manner of approach to 

 infinity, whether by a rectilinear asymptote or by a curvi- 

 linear branch, does not seem obvious*. Bat this scarcely 

 affects the practical problem. When the wire is unweighted, 

 the constant C = 0, the wire hangs straight at the angle 

 given by sin<9 = cota, the above equations become illusory, 

 ••and it is necessary to return to the stage of the problem 

 anterior to the formation of equation IV. 



7. When, as in § 3 above, the first power of p only is 

 retained, it is possible to express the rectangular coordinates 

 of a point on the curve in terms of the single variable 6. 

 Xiet the axis of x be taken horizontal in the direction of 

 flight, and the axis of y vertical. Then 



x _ f\ 



'C = v 



-■fl 



ds . a 

 p-sin0 



[1+jp sec 2 sin 0+p log tan (i?r + 40)] sec 6 ton 6 dQ 



= sec#+£>[Jtan 3 0— tan 0+ sec log tan (J7r + £0)1 

 and 



y _ C ds 



cos 



_Cds 



o-Jc 



= i [l+psec 2 0sin0-|-jt>logtan(£7r + i0)] sec 6 dO 



= logtan(l7T + ^)+ip[tan 2 + {logtan(l7r + ^)} 2 ]. 



When p is altogether neglected, these expressions give, as 

 they should, the common catenary 



x , v 



7^ = cosh — . 



They are not impossibly complicated. By the tabulation 

 of three functions of besides sec — log is of course 



* Examination shows that the perpendicular from the point 0=0 on 

 the tangent to the curve at infinity (sin # = cot «), which can be expressed 

 by a definite integral, is finite; though it becomes infinite in the particular 

 case of the catenary (*=45°). Hence in general the curve does approach 

 a rectilinear asymptote. 



3E 2 



