54 



EARTHQUAICES. 



Assuming that the displacement of an earth particle 

 at C equals C B, and at Cj equals c^ 6j, and at C^ equals 

 Cg 62, &c., and let these displacements cb, c^ h^, c^b^, &c., 

 for the sake of argument, vary inversely as r, r^ r^, &c. 



1^'. y> 



■A\X\ 



Fig. 9. 



The question is to determine where the horizontal 

 component c A of these normal motions is a maximum. 

 First observe that the triangle c c is similar to a, b, c. 



Also r = -T -^, and therefore the normal component 

 sm 6 



c, 61 at Cj is equal to C — ^— , 



Also Cj a, = Cj 6,, cos 0. 



. c, a = c 



sin 6 cos 6 



h 



c sin 20 

 h 



2 



and sin 2^ is greatest when 2(9 = 90° or ^ = 45°. 



That is to say, the horizontal component reaches a 

 maximum where the angle of emergence equals 45°. 



This question has been discussed on the assumption 

 that the amplitude of an earth particle varies inversely as 

 its distance from the origin of the shock. Should we, 

 however, assume that this amplitude varies inversely as 

 the square of the distance from the origin, we are led to 

 the result that the area of greatest disturbance is nearer 

 to the point where the angle of emergence is 56° 44' 9". 



