Mass carried forward by a Vortex. 609* 



From V, the corresponding value of b/a is found, and 



= tan- -j X x — ^aVtan 3 - I, where k = sin 0. 



These give % l5 V, b/a for a configuration in which the 

 boundary of the translated fluid cuts the equatorial plane at 



a distance x = a(l — k')/(l + k') = a tan 2 » from the centre. 



The boundary is drawn as before by taking a series of 

 values of Jc, greater than that chosen for the node, solving for 

 x from the equation ^ = ^ 1? and taking the corresponding- 

 point on the k circle. The equation however now is the 

 quartic 



say -; 4 — })z — q = 



•where z 2 = x/a and q is essentially positive since ^ is negative- 

 The left-hand member may be written 



here 



(* 



J +« 



z + b)Q 



: 2 -« 



z + b') = 



= 0. 







b + b'- 



-a 2 = 



--o, 









aV- 



-ab = 



bb' = 



-- -P, 

 : -q. 









2b' = 



-a" — 



p 



} 



a 









2b = 



:a 2 4 



P 





Hence 



Since a 2 — 4:b is negative, the first factor gives imaginary 

 roots and the real roots are 





in which the + sign has to be taken since the smaller root 

 will be found to give y imaginary. 

 The equation to find a is 



a* + 4tqa*=p*. 



