Dip and offset together

Restatement of ellipse equations

Recall equation (8.9) for an ellipse centered at the origin.

$$0 \eq {y^{2}\over{A^{2}}} + {z^{2}\over{B^{2}}} -1 .$$ (19)

where

$$A \eq v_{\rm half} t_h ,$$ (20)

$$B^2 \eq A^2 - h^2 .$$ (21)

The ray goes from the shot at one focus of the ellipse to anywhere on the ellipse, and then to the receiver in traveltime $t_h$. The equation for a circle of radius $R=t_0 v_{\rm half}$ with center on the surface at the source-receiver pair coordinate $x=b$ is

$$R^2 \eq (y - b)^2 + z^{2} ,$$ (22)

where

$$R \eq t_0 v_{\rm half}.$$ (23)

To get the circle and ellipse tangent to each other, their slopes must match. Implicit differentiation of equation (8.19) and (8.22) with respect to $y$ yields:

$$0 \eq {y \over{A^2}} + {z \over{B^2}} {dz \over dy}$$ (24)

$$0 \eq (y-b) + z {dz \over dy}$$ (25)

Eliminating $dz/dy$ from equations (8.24) and (8.25) yields:

$$y \eq {b\over 1 - {B^{2}\over{A^{2}}}} .$$ (26)

At the point of tangency the circle and the ellipse should coincide. Thus we need to combine equations to eliminate $x$ and $z$. We eliminate $z$ from equation (8.19) and (8.22) to get an equation only dependent on the $y$ variable. This $y$ variable can be eliminated by inserting equation (8.26).

$$R^2 \eq B^2 \left( {A^2 - B^2 - b^2 \over{A^2 - B^2}} \right).$$ (27)

Substituting the definitions (8.20), (8.21), (8.23) of various parameter gives the relation between zero-offset traveltime $t_0$ and nonzero traveltime $t_h$:

$$t_0^2\eq \left(t_h^2-{h^{2}\over v_{\rm half}^2}\right) \left(1-{b^2\over h^2}\right).$$ (28)

As with the Rocca operator, equation (8.28) includes both dip moveout DMO and NMO.

Dip and offset together