276 Professor Bertram HopUnson [Jan. 2G, 



^l-Q of the time {\ second) which it takes the weight of the hammer- 

 head (a force of 1 lb.) to generate its motion. Thus the pressure 

 required for the rapid stoppage is as before, 400 lbs. 



We may take another instance essentially similar to the hammer 

 and nail, but differing greatly as regards scale. A 14-inch armour- 

 piercing shell weighs al;)Out 1400 lbs., and when moving at 1800 feet 

 per second possesses about 81,000 foot-tons of energy, or about 

 15,000,000 times as much as our hammer-head. Such a shell would 

 just pierce a plate of wrought iron 2^ feet thick, and the average 

 force which must be exerted to pull it up in that distance, which is 

 of course the pressure which it exerts on the plate, is 30,000 divided 

 by 2|, or about 12,000 tons. This is equivalent to some 80 tons on 

 the square inch. 



When a hammer strikes a nail, the force acting during the blow 

 is practically constant, and the average value obtained as above by 

 dividing the energy by the distance moved, or the momentum by the 

 time taken, is equal to the actual force exerted throughout the 

 impact. In many cases, however, this force is not constant, and it 

 is then necessary to divide the course of the impact into short 

 intervals either of space or of time, calculate the change of energy or 

 momentum in each, and add the result. A familiar instance is that 

 of two billiard balls. We may suppose one ball to strike the other 

 full with a velocity of 16 feet per second, which corresponds to a 

 fairly hard stroke. It simplifies the consideration of the problem, 

 if instead of one ball moving and the other at rest we suppose them 

 to be travelling in opposite directions with equal velocities of 8 feet 

 per second. At the instant when the lialls first touch tliere is no 

 pressure l^etween them, but as they continue to approach, each 

 flattens the other at the point of contact. The balls no longer touch 

 at a point, but over a circular area which rapidly increases in diameter. 

 Corresponding to any given amount of flattening or distance of 

 approach, there is of course a definite pressure which might be 

 measured by actually squeezing the balls together under known 

 forces and measuring the corresponding amount of approach. Or 

 the relation between pressure and distance could be calculated as was 

 done by Hertz. The results are shown in the curve (Fig. 1), and 

 the area of the curve connecting pressure and distance up to any 

 point gives the number of foot-pounds of energy destroyed. When 

 this is just equal to the original energy of the balls they will have 

 been reduced to rest, and in the case supposed, the distance of ap- 

 proach is then j^^-^ of an inch, and the total pressure between them 

 1300 lbs. This pressure is distributed over the circle of contact which 

 is ^ of an inch in diameter, and the average intensity of the pressure 

 is 27 tons per square inch. The distribution, however, is not uniform, 

 the pressure at the centre being \\ times the average. [The shape 

 of the balls at the moment of closest ai)proach was shown on a 

 drawing with a magnification of 100.] The balls arc then like 



