344 Mr. J. Prescott on the 



when 7 is expressed in minutes. But since the term — X; is 

 involved on the right-hand side of (22), it follows that 



2 



= -13'"92 + 10'-76, 

 = -3'-16. 



That is, the rifle jumps downwards through an angle of 

 about 3''2. 



26. Considering the degree o£ accuracy of our data our 

 constants can be regarded as only approximately correct. 

 We shall take, m future, 



1 = 2100, u= 2890 feet per sec. . . . (30) 



The muzzle velocity given in the ' Musketry Regulations ' 

 is 2440 ; but this is obviously wrong, for a very brief con- 

 sideration of the range table will show that the downward 

 jump, which is the elevation of the rifle for zero range, is 

 something in the neighbourhood of 3'. This can be dis- 

 covered quickly by plotting y against the range and producing 

 the curve backwards to the point where the range is zero. 

 The value of the jump would make the angle of departure 5' 

 for the 200 yards range. But a projectile fired at this 

 angle with a velocity of 2440 feet per second would only 

 have a range of about 180 yards if there were no air- 

 resistance at all. The estimate of the jump given in the 

 'Musketry Regulations/ namely, between 4' and 5', makes 

 the case a great deal worse. It is very probable then that 

 our value of u is much better than the value in the 

 ' Musketry Regulations.' 



27. With our values of u Q and I the horizontal distance 

 traversed by the bullet before the law of resistance changes is 



X 1 = 21001og c ?g|g =2106 feet = 702 yards. . (31) 



Since the value of the velocity (u{) at which the change 

 takes place is not very definite, it will be more convenient 

 to take X x to be exactly 700 yds., and then by equation (5), 



Wl = 2890 6" 1 = 1063 (32) 



Now the equation for 7, when the range X is not greater 

 than 2100 feet, becomes 



2X 



X 7 = 14620 (*5Iob_l)_ 10-76 X. ... (33) 



