C. 15-degree algorithm in one-layer model

This example shows distortions that are connected with application of a wave-field continuation operator which is not kinematically equivalent to the classical Eiconal equation (see the next chapter).

First step: The time t₀ in a true model is the same as in the example:

$$\tau_{0} = {2 \sin \phi \over v} x + {2 h_{0} \cos \phi \over v}.$$

Second step: If we apply 15-degree algorithm with the continuation velocity $v_{c}={v^{\prime} \over 2}$, then discontinuities propagate in accordance to the Eiconal equation

$$\tau _{x}^{2} - {4 \over v^{\prime}} \tau _{z} = {4 \over {v^{\prime}}^{2}}$$

(see equation (98) from Chapter 8 at $v=v_{c}={v^{\prime} \over 2}$).

The solution of this equation which satisfies boundary condition $\tau \vert _{z=0}=t_{0}$ is:

$$\tau^{(-)}(x,z)={2h_{0}\cos \phi \over v} + {\sin \phi \over... ...over v}\right) }^{2}\sin^{2}\phi \right] {2z \over v^{\prime}}.$$

Third step: From the condition $\tau^{(-)}=0$ we find location of the reflector image

$$z= {({v^{\prime} \over v}) \cos \phi \over 1 - {1\over 2} {({v^{\prime} \over v})}^{2}\sin^{2} \phi}(h_{0}+x\tan \phi )$$

when $v^{\prime}=v$

$$z={\cos \phi \over 1 - {1\over2 } \sin^{2}\phi } (h_{0}+x\tan \phi ).$$

The factor

$${\cos \phi \over 1 - {1 \over 2}\sin^{2}\phi } \simeq 1-{1\over 8}\sin^{4} \phi$$

so the depth distortion is very small for subhorizontal reflectors.