332 



Mr. A. Mallock. 



Putting v = 2000 f.s., and I = 2*3 ft., 

 we have p = 860,000 f.s.s., % = 0-0023 sees. 



An acceleration of 860,000 is about 27,000^, so that a uniform force 

 of 27,000 times its own weight, or 835 lbs., would give the 215-grain 

 shot its observed velocity in the actual length of the barrel. 



With a uniform force, the pressure curve in terms of space is the 

 same, of course, as if expressed in terms of time ; but for any other 

 case we must, for the purpose of this paper, express the pressure curve 

 (which experiment would give in terms of the distance travelled by 

 the shot in the barrel) in terms of time. 



The pressure at time t being p, we have 



dv , dv dv els civ . 7 7 



p = — ; also — = — — = v -t , and pels = v dv 

 r at at ds at ds 



v = \J(2)pds) (6); 



and t = 1 J S ( • (7). 



If we take the case of the pressure decreasing uniformly with the 

 travel of the shot, it is easy to show by (5) and (6) (although the 

 analogy with the force acting on a pendulum or spring at once suggests 

 it), that the velocity and position of the shot are : — 



I I - cost 



\/ '/•-•• ^ 



v VpMnt^y7« (9), 



Pc = V -f (10), 



(n) - 



With the before-mentioned values for Z, v, and w, p = 1*174 x 

 10 6 f.s.s. and % = 0-00171 second. 



One more case by way of example will suffice. Let the pressure 

 decrease uniformly with the time so that 



p - & { l -i) (12) - 



From this we get 



t' 2 



v =pot-p — (13), 



