ON RIFLED GUNS. 17 



!/ = "■'■ x = !>'!'. 



whence * 



L—1/ — /1. p. '7=0 > , 4 >„ 



p = constant J 



in which a is the cotangent of the angle which an element of the helix makes with the axis of 



the piece. Making y=-l, 



a = ll < 30 > 



Equations (4)' and (-1)" treated after the manner of Eq. (4) and combined with Eqs. 

 (A), will give new values for T and L. 



(19) — But why are the indications of the plug, in excess of the pressure measures? Let 

 us see. The circumstances of the experiments are these, viz: A hole is made in the barrel of 

 the gun, in the direction of a line intersecting at right angles the axis. In this hole is 

 inserted a plug, tight enough to permit freedom of motion and yet prevent the sensible escape 

 of gas. The outer end of the plug terminates in an obtuse lanceolate cutter. In contact 

 with the point of this cutter is firmly secured a piece of thick copper plate, with its plane 

 face at right angles to the axis of the plug. When the gun is fired the expanding gas acts 

 upon the head of the plug and drives it into the plate, and the intensity of the action is 

 inferred to be measured by a dead weight which, by its simple pressure and without sensible 

 motion, will produce an equal cut. 



Denote by x the distance the cutter has penetrated the copper; by F^x) the corre- 

 sponding pressure of the expanding gas on the end. of the plug, in pounds; by F 2 (x) the corre- 

 sponding resistance of the copper; by W the weight of the plug and cutter, and by V their 

 common velocity; then will 



fl\ (x) dx - fF 2 (x) dx - —. V 2 = 0. 



The value of F,(x) begins with zero and when x is zero; it rapidly increases to a maximum, 

 and then rapidly diminishes as the missile progresses and x increases. The function denoted 

 by Fi must have a maximum. Such a function is 



Fi (.-) = A. sin Q >--^)i 



in which A denotes the maximum elastic force; a the value of x or copper penetration when 

 it occurs. And although this may not be the precise function which expresses the law in 

 question, it will be sufficient for the purposes of the illustration. 

 Let the law of copper resistance be 



F 2 (a;) = Ba; m , 

 and the work becomes 



f A. sin ( i n. X \ dx - f B a? dx - ^-. V 2 = ; 



A.^ri-cosfi,.nl-B^.-^--^.V 2 = 0. 

 - | V." «.'J m + 1 2g 



43 (329) 



