THE SPHAEROBOLUS GUN 343 



1 77-5 30 



M = X X — = 0- 00008 units 



28,000,000 1 1 



i.e. a Sphaerobolus projectile shot to a maximum height of 14-5 



feet has, on leaving the gun, an initial momentum of 0- 00008 units. 



(6) Kinetic energy of the projectile when leaving the gun. The 



kinetic energy of the projectile can be found from the equation : 



K.E. = | mv 2 



where K.E. = the kinetic energy, m the mass, and v the velocity. 



Therefore for our projectile which was shot up 14-5 feet high and 



which therefore had an initial velocity of 30 feet per second 



1 78 30 x 30 



KK = 2 X 2^000^00 X 1 

 = 0-0012 foot poundals 

 00012 0-0012 



or 



Ys = A mv 2 



9 32 



= 0- 000037 foot pounds 



i.e. the kinetic energy of a projectile of Sphaerobolus when just 

 leaving a gun which can discharge it to a maximum height of 14-5 

 feet is 0- 0012 foot poundals or 0- 000037 foot pounds. 



(7) Average force of the propelling peridium when acting upon the 

 projectile. It is known that 



where F is the average force exerted on a projectile before it leaves 

 the gun, s is the distance that the projectile goes in the gun whilst 

 still subjected to its pressure, m is the mass of the projectile, and v 

 is the initial velocity of the projectile on leaving the gun. Applying 

 this equation to the Sphaerobolus gun in the case of the discharge 

 of a projectile to the maximum observed height of 14-5 feet, we 

 have : 



S " 3mm ' = 3048 f6et>m = 2M>0O^ro lbs " andB = 30feet P erSeCOnd ' 



1 78 30 x 30 3048 



Therefore P = - X ^^ X — p- X -^ 



= 0-13 poundals 



0-13 



or = = 0004 lbs. 



