12 



ON RIFLED GUNS. 



V 2 =(1250, 2 

 - 4 =(3.1416) 4 

 h 2 



W = 100 

 2 §- = 64 

 » 2 = (:{,5) 2 

 P= (10,833 ) 2 



3.0969100x2 



I •(Z'/'Y 

 2\dt) 



f)S52 f i',5 



The work of the torsion strain will be that required to raise 5852,5 pounds through a 

 vertical distance of one foot in the time required for the projectile to pass from rest to the 



mouth of the gun. 



(14) — To compute the work in the missile, due to its motion of translation, we have 



W = 100 lbs 



# = 32 



2 = 2 

 V 2 = (1250) 2 



a c 

 a c 



MV 2 _ W 

 2 ~2g- 



V 2 



2441406 f P 



(15) — The torsion strain at mouth, orT- 11110 



Circumferential or C = 8422050 



log 

 a c 



T 

 C 



0,001319 



2.0000000 

 S.494S500 

 9.6989700 

 6.1938200 



6.3S76400 



4.0456870 

 3.0745822 



7.1202692 



So that the strain which twists the gun is but a little more than one one-thousandth part of 

 that which acts to split it. 



And again, 



Longitudinal strain, or L = 320S.9 

 Circumferential or C — S422050 



Ion 



L 



c : 



0,000381 



3.5063534 

 3 0745822 



0.5809356 



and the longitudinal strain is not quite four ten-thousandths that which acts to split. 



(16) — To integrate Eqs. (18) and (19) it will be necessary to have the law of continuity 

 that connects the different values of p v which is, as we have seen from Major Rodman's 

 experiments, variable. The integral function would enable us to find the actual work per- 

 formed by the torsion and longitudinal strains during the motion of the missile within the 

 gun. The work performed by the expansive action of the gas against the base of the pro- 

 jectile would also result from the integration of the expression 



jr. f> 2 . Pl . di/=-. /> 2 - F (>/) dy. 

 (324) 



