Motion of a Spinning Projectile, 343 



Since the data from which I has been calculated are not 

 very accurate, only three figures being available in the 

 angles 7, it will be good enough if we take the round 

 number 2100 for I. Then, by equation (26), 



1200 600 



2820 = re 2100 (e 2100 __ 1) 2^ 

 =«*(«»- l) f , 

 from which r= 14562. 



Next, by the first of equations (25), 



10170 ~ 4800 = 300? + reHe* - 1), 



that is, 



300^ = 5370-8527= -3157, 



q = -10-52. 

 Lastly, by the first of equations (24) , 



4800 =p-h 600? +re* r , 



=^-6314-25786, 

 p= -14672. 



This should be equal to — r, and it is near enough to give 

 confidence in our results and theory. We shall take the 

 value of r to be the mean of the results for r and — p, 

 namely 14620 approximately. Now, recalculating g with 

 these values of r and p, the first of equations (24) gives 



4800 = 600? + 14620 (*♦— 1), 

 = 600? + 11255, 

 ?=-.- 10-76. 



25. If the angle 7 were expressed in radians instead of 

 minutes, then our constants p, q, r would each have to be 



rrr 



multiplied by in o n /y Consequently, from equation (14), 



in which 7 is expressed in radians, 



gl 2 vr 14620 it 



W ^ 10800 ~~ 10800 ' 

 Therefore. 



u Q := 2888. 



Again, if there were no jump, it is clear from (14) that 

 2 

 £ would be equal to — jr, and this relation remains true 



