APPLICATIONS

An obvious application for the estimation of the normal vector is the automatic measurement of the local dip in a seismic image. Such dip information should be useful when optimizing for the correct migration velocity. For example, the dip information could be useful for tying seismic reflections to dip meter measurements in a well.

Additionally, the cross product expression can be used to remove a dominant plane layer contribution from an image volume. First, we estimate the normal vector ${\bf n}$ for the given cube using equation (2). Second, we remove the plane layer contribution by computing equation (1) for the estimated ${\bf n}$. As we will see, this removal does not, unfortunately, yield a simple scalar residual.

Let us assume the dominant component of parallel planes in an image volume $f({\bf x})$ is $h({\bf x})$ and its normal is ${\bf n}$. Consequently, f can be expressed as a sum of the parallel planes h and the remaining image r where $f({\bf x}) = h({\bf x}) + r({\bf x})$.The application of the cross product filter (1) to f yields

$$\begin{eqnarray} {\bf n} \times {\nabla f} &=& {\bf n} \times {\nabla (h + r)} \... ...f n} \times {\nabla r} \\ &=& {\bf n} \times {\nabla r} \nonumber\end{eqnarray}$$ (3)

since ${\bf n} \times {\nabla h} = 0$.

We would prefer to yield $r({\bf x})$, a scalar volume of the difference f and its dominant set of parallel planes. Instead, we computed the expression ${\bf n} \times {\nabla r}$, which consists of a three component vector at each output location $({\bf x})$. Not only is the output's size tripled, its meaning is rather obscure: the vector at each image point $({\bf x})$ is the cross product of the local gradient $\nabla f({\bf x})$ and the normalized, global gradient ${\bf n}$.

 

$${\bf n} \times \nabla f = \left[ \begin{array} {c} 0 \\ 0 \... ...} \partial_x \\ \partial_y \\ \partial_z \end{array}\right] f$$ (4)

Each row is a finite difference operator in a plane, e.g., the third row is $(n_y \partial_x - n_x \partial_y)$.

One possible approach (but not the only one) to transform ${\bf n} \times \nabla f({\bf x})$ back to the original space of $f({\bf x})$ is to use the adjoint operator of ${\bf n} \times {\nabla}$:

$$\begin{eqnarray} f_{fil}({\bf x}) & = & ({\bf n} \times {\nabla})' ({\bf n} ... ...{xz} + (n_x^2 + n_y^2) d_{zz} - 2 n_x n_y d_{xy} ] f \nonumber\end{eqnarray}$$

I do not know of an interpretation of this expression, such as a rotated and weighted Laplacian operator. A geometric interpretation of the equation does not offer any additional insights, either.