10 



h x and h 2 being the heads i\ and r 2 respectively. The same 

 arguments apply as before for the cases of the axis of the screw 

 being vertical, horizontal, or inclined, and for particles at opposite 

 ends of the same diameter, &c, Subtracting one of these 

 equations from the other, and supposing r x and r 2 to be two radii 

 nearly equal, containing between them a ring on the screw disc 

 of infinitesimal breadth dr, we get 

 dh d 



By geometry, in Fig. 2, it appears that u 2 is equal to — gl + v ^ 



the angle A CD in the triangle A CD being a right angle, so 



r 2 w 2 

 that u 2 -K 2 = — ^,+v 2 -K 2 which is equal to 2r 2 wft by the 

 cos 2 ^ 



fundamental equation previously found. We have therefore, 



equating these and dividing across by 2 



dh d 

 — p-—=-—-r^o)il 

 6 dr dr 



But again, since the water is moving in a curved path (both 



longitudinally and circumferentially) we have also 



dh - d 2 r 



*dr= r(a —d7 



d 2 r 

 where —=» is the radial acceleration of a particle in its longi- 

 tudinal path, and can therefore be written otherwise 

 d dr d , . 



Tfdt =v di {vtme) > 



where 6 is the angle made by the boundary of the tube of flow 

 ASA', or any similar inner tube of flow at radius r, with the 

 axis of the screw. We can, on comparing the last two equations, 



get rid of — -, since we are not interested in the value of the 

 dr 



head, or pressure of the water hydrostatically, and, substituting 



d 2 r 

 for -— its value just mentioned in terms of v and tan# we get 



— r (r 2 o)tt) = rw 2 — v— (z>tan 6) 

 dr dx 



