Computation of takeoff angles for diffracted multiple

From Figure  we can immediately compute the takeoff angle of the diffracted receiver ray as  

$$\alpha_r=\tan^{-1}\left[\frac{h_D+m_D-X_{diff}}{Z_{diff}}\right].$$ (78)

 

mul_sktch13 Figure 26 Sketch showing the geometry of the zero surface half-offset diffracted multiple from a dipping water-bottom.

In this equation the depth of the diffractor is not known, but it can be calculated from the geometry of Figure :  

$$Z_{diff}=\frac{\tilde{Z}_D}{\cos\varphi}+(X_{diff}-m_D)\tan\varphi$$ (79)

As we did for the diffracted multiple from the flat water-bottom, we can use the traveltime of the multiple at the zero surface-offset trace to compute $\tilde{Z}_D$, except that this time the computation is much more involved. Figure  shows the basic geometry. From triangle ABC we have  

$$[V_1(t_m(0)-t_{r_1}(0))] ^2=(2\tilde{Z}_D\cos\varphi+Z_{diff})^2+(2\tilde{Z}_D\sin\varphi+(X_{diff}-m_D))^2,$$ (80)

were t_r1 is the traveltime of the diffracted segment that, according to triangle DEF in Figure  is given by

 

[V₁t_r1(0)]²=Z_diff²+(X_diff-m_D)². (81)

 

mul_sktch14 Figure 27 Sketch to compute $\tilde{Z}_D$ in equations 51,  80 and 81.

Replacing equations 79 and 81 into equation 80 gives a quartic equation for $\tilde{Z}_D$ which can be solved numerically. Once $\tilde{Z}_D$ is known, we can easily compute Z_diff with equation 79 and therefore $\alpha_r$ with equation 78 in terms of the known quantities h_D, m_D, X_diff and t_m(0).

 

mul_sktch15 Figure 28 Sketch to compute the takeoff angle of the source ray from a diffracted multiple.

In order to compute $\alpha_s$, we apply the law of sines to triangle ABC in Figure  to get  

$$\sin(\alpha_s+2\varphi)=\frac{2\tilde{Z}_s\sin\varphi+(X_{diff}-m_D+h_D)}{V_1(t_m-t_{r_1})},$$ (82)

where $\tilde{Z}_s=\tilde{Z}_D-h_D\sin\varphi$ and V₁t_r1 is the length of the diffracted receiver ray and is given by  

$$V_1t_{r_1}=\sqrt{(h_D+m_D-X_{diff})^2+Z_{diff}^2}$$ (83)

Therefore, plugging equation 83 into equation 82 we get equation 50:

$$\alpha_s=\sin^{-1}\left[\frac{2\tilde{Z}_D\sin\varphi+(h_D+X... ...V_1t_m-\sqrt{(h_D+m_D-X_{diff})^2+Z_{diff}^2}}\right]-2\varphi.$$ (84)