SEISMIC METHODS 753 



or 



T = TF^+2y^^^^ (103) 



It is evident that the travel-time curve vi^ill consist of as many segments 

 as there are layers of different velocity, each segment corresponding to a 

 wave path along an interface. Each segment is a straight line, and the 

 reciprocal of the slope of each segment equals the velocity in the cor- 

 responding layer. Moreover, it is possible either from the time-intercepts 

 or the critical distances to compute the depths to the layers. 



The thicknesses of the layers may be determined successively by 

 solving for /j„ ; that is, since 



n — 1 



7 = ^ + 2^^^^ + 



Vn + X ^ y^k 



n — 1 



COS <Ln[ ^^ Vj, 



where 



r»-H-S^^^ (104) 



Tn+1 — 



2 V f^n + l/ 



(105) 



is one-half the intercept time of the (n-hl)th segment of the travel-time 

 curve. 



Numerical Illustration 



Suppose the velocities and intercept-times shown in the following table 

 have been determined from a travel-time curve.* 



Layer » F„ (ft./sec.) Ton (sec.)** t„ (sec.) 







.102 .051 



.252 .126 



.418 .209 



.576 .288 



Let ak,n be the inclination in the kth layer of a ray which meets the interface 

 between the (n-l)th and the «th layer at the critical angle so that 



* In layer 1 the refracted wave and the direct wave are identical. 

 ** Ton is the time at which the ray reaches the nth segment. 



