48 



RAY ACOUSTICS 



the circular arc bends upward; but if a is negative, 

 the circular arc bends downward. 



We can determine the center of the circle defining 

 the ray by a simple geometrical construction. Figure 

 7 shows the path of a ray leaving the projector at the 



VELOCITY 



Figure 7. Geometrical construction of ray path. 



angle Ot into a medium of constant negative velocity 

 gradient. The center of the circle is obtained by fol- 

 lowing the perpendicular to PQ down through the 

 medium a distance Co/a cos do. It is a simple conse- 

 quence of the geometry of the situation that this 

 center will lie on the horizontal Une a distance co/a 

 below the projector. For, from the illustration, 

 RO = (co/a)(cos Si/cos 6o), and di clearly equals do- 

 Similarly, if the constant velocity gradient is positive 

 so that the rays are bent upward, the centers of the 

 defining circles he on a horizontal Une a distance Co/a 

 above the projector. It is easily shown that the dashed 

 horizontal "line of centers" in Figure 7 is at the depth 

 where the velocity c would equal zero if the assumed 

 . linear gradient extended indefinitely. 



An approximate solution in the general case where 

 c is an arbitrary function of y can be obtained by re- 

 peated use of the solution for constant gradient. Even 

 a compUcated velocity-depth curve can be closely 

 approximated, as in Figure 8, by dividing the depth 

 interval into a relatively small number of segments 

 in each of which the velocity is assumed to change 

 linearly with depth. Within each layer the ray path 

 is an arc of a circle; and the total ray path is a con- 

 secutive series of such arcs. 



TRUE CURVE 



APPROXIMATING CURVE 



FiGUBE 8. Approximating velocity-depth cm-ve by a 

 succession of linear gradients. 



In practice, the path of the ray cannot be conven- 

 iently plotted as a sum of circular arcs because the 

 horizontal ranges are much greater than the depths of 

 interest, and therefore their scale must be contracted. 

 Instead, the ray is usually traced by calculating the 

 angles 6i and $2 at which it enters and leaves a given 

 layer, and the horizontal distance it travels in the 

 layer. This calculation is illustrated in Figure 9, 



Figure 9. Ray path in layer of linear gradient. 



where the top of the layer is at depth yi, the bottom 



is at depth 2/2, and the thickness of the layer is h. 



The ray leaves the projector at an angle do, enters 



the layer at the angle di, and leaves the layer at the 



angle 52- Then, by equation (26), 



[ciyi) cos dol 

 01 = arc cos 



(34) 



r cjyi) cos dp i 



02 = arc cos 



where c{yi) and 0(2/2) are calculated from equation 

 (31). 



