212 PROCEEDINGS OF THE AMERICAN ACADEMY., 



sure on a ring of breadth AB (Figure 4). Similarly the piston exerts 

 a pressure P 2 equivalent to that on a ring CD. The free liquid at the 

 inner end exerts P 3 on the ring BE. Since the liquid escapes steadily 

 without acceleration, we have 



2F+ P 2 = P 1 + P S . 



The effective force on the piston is F + P 



a 



F + P 2 



P l + P 3 + P 2 



We now can calculate P 1 and P 2 without any assumption as to the 

 distribution of pressure in the crack if we assume only that at every 

 point the radial displacement is proportional to the pressure at that 

 point. This gives 



P 1 = 2ttR jpdr, 



where r x is the value of r at the end ABE of the cylinder, and r 2 at the 

 end CD. R is the average of r x and r 2 . But 



r 2 — r = Cp, 

 dr = - Cdp, 



JrPc 

 pdp 

 Pa 



= 2.0^ = 2.^10* 



That is, P x is equal to th,e pressure exerted by the total internal pres- 

 sure P on a ring of half the breadth of AB. Similarly, P 2 is the pressure 

 on a ring of half the breadth of CD. If now we put R equal original 

 radius of piston, and R + A# equal original radius of cylinder, 



AB = 5.3 x 10- 7 x (R + AP) x P, 



CD = 3.5 X 10- 7 X RxP, 



BE = AR+ (2.1 x 10- 7 - 5.3 x 10- 7 ) x R x P, 



= Atf - 3.2 X 10- 7 x RxP. 



IT J? J P OR (2 - G + L8 ~ 3 - 2 ) 10 ~ 7 X R + AR n 



Hence, F+P 2 = 2ttR— '- X P 



ttR (~ + 1.2 X 10- 7 X R] X P. 



