Motion of Viscous Liquids in Channels. 



33 



or, since 



^O + i) 6 15' 



iow 2 = ir~-27rS , — — rrytanh (n + 4)7r. 

 12b7r 2 5 n=o(™ + i) 



If now we take the latus-rectum to be 4a instead of 47r 2 , 

 F will be multiplied by (a/7T 2 ) 4 , so that 



F _ 128 _ 256 | 



-5tanh(n + i)7r. . . (7) 



Particular Triangular Section, 



3. The section is that in which one side of the channel is 

 vertical and the other inclined at an angle \tt to it (fig. 3). 



Fig. 3. 



The solution for this case can be derived from an expression 

 previously given *, but it is just as easy to verify directly 

 that all the conditions of the problem are satisfied by 



9 00. 



X=0+*/) ('*-*)-- 2 



7T tt=0 + ir sinh (2m + 1)tt 

 X {sinh (n + ^)(27T— <r+y)sin (n + £)(# + #) 

 — sinh(>i + i)(>-f y) sin (n + £)(#—#)}, . 



(8) 



the axes being as shown in fig. 3. The boundary conditions 

 are that % = on x = it and on x=—y, and that d^/dy = 

 on y = ; but instead of the latter we take ^ = on x=y> 

 again appealing to symmetry. We have taken 0A = 7r for 

 convenience. 



* Lond. Math. Soc, Records for March 13th, 1913. 

 Phil. Mag. S. 6. Vol. 28. No. 163. July 1914. D 



