This completes the second Iteration. Since the new embedment 

 depth is beyond the increment, ML, centered about £ = 0, two 

 multiplication terms need to be included in the siommation portion of 

 the force equation. The calculation proceeds: 



F = W - D (v„. J^ - C (0) A (0.5) - c (0.5) A (0) 



b s zAt s e s e 



= 1000 - 2.5 (19.99995)^ - (20) (4.7124) - (26) (.5892) 



= -102.7 lb 



a = 'Yil'J = .01843 ft/sec^ 

 5576 



= 19.99995 - (.01843) (.01) 

 = 19.99977 ft/sec 



3At 



/n ^ .19.99977 + 19.99995 , , _, , 



= .60 ft 



g. This iterative procedure was followed through 125 increments 

 of At or 1.26 seconds of penetration time. At that point the velocity 

 became negative and the corresponding embedment depth was assumed to 

 be the point of maximum penetration. The penetration depth corresponding 

 to this point was 17.8 feet. A summary of some of the intermediate 

 results is given in the following table. The calculation procedure 

 was simplified through the use of a brief computer code. 



25 



