﻿Problem of the Weir. 779 



Substituting in (6) we find 



7s = { 1 + k(8 — -cr) } "% 



which after rationalizing- gives 



*«r 3 -(H-«5)«r 2 -hl = 0, .... (12) 



as a cubic to determine -sr in terms of k and 5. 



It may be shown that the roots of (12) are given by 



tsr=-( v /3/2).(l + «8)-*sec^, . . (13) 



where 6 is any value satisfying 



cos 0= (3^/3/2) .Xl-b*S)- 3/2 . • • • (14) 



An inspection of (12) will show that the roots are distributed 

 as follows : — 



— X> <^'Sr 1 <0 <:: OT2< 1<£<!'373<8 + l//c. 



There are thus two positive roots and one negative root. 

 The last does not appear to have a physical interpretation, 

 but the question arises — which of the two positive roots is to 

 be chosen? If we put 8 = 1 in (12) there results 



(fir-lX^-isr— 1)=0, .... (15) 



of which the only physical solution is -sr= 1 *. In accordance 

 with the work of §3 we write 8 = 1 + e and ot=1 + 97, where 

 e and 7j are small and vanish together. Substituting in (12) 

 and neglecting squares, &c, of e and rj, we have 



so that 7] is positive or negative according as k ^ 2. 



[The expansion of 77 in terms of e proceeds as follows : — 



K 3fC 2 



, k 3 ( 4«: + 10) 5k 4 (k2 + 8k+7) 



' a: — 2 (a:— 2) d («— 2) & (^-2) 



Ck 5 (k 3 -3k 2 + 126/c-59) - 



This becomes divergent when k = 2, i. e., when U 2 = <//», in 

 accordance with (7) § 3, because when /c = 2, (15) becomes 

 (ot — 1) 2 (2'bt + 1) = 0, and there is a double root -57 = 1. The 

 series is useful for calculation when e is small or when ic is 

 great or small.] 



* Appendix A. 



