Motion of a Spinning Projectile. 357 



50. Putting 



<f> = & mT (52) 



in equation (51), and dropping the terms on the right, the 

 equation for the complementary function becomes 



Tg + (/+l) g - |(/_l)^-^T+ f } r=o. 



Next, putting 



X=& * > ( 53 > 



our equation becomes 



Writing this, for the sake of shortness, 



^+-X=0, • (55) 



and then making the substitution 



h=pe ¥ ; ( 56 ) 



then equation (55) becomes 



d?p /dfiV , -,-dp dfi , . d?p „ 



(57) 



Before proceeding further with the solution of this equation, 

 it will be useful to get an idea of the magnitudes of the 

 quantities involved in equation (54). 



51. For the present we shall deal with the Mark VII. 

 bullet, for which 



d (diameter) = 0-303 inch=^ inch, 

 / = 2100 feet, 

 h = 4840 feet; 

 h (length) = 1-28 inches, 



(58) 



m =289C. 



the value of co being obtained from the knowledge that the 

 bullet makes one turn on its axis as it leaves the muzzle for 

 33 calibres of forward motion, that is, for 10 inches of 

 forward motion. We should need very accurate details of 



