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# DETECTION OF A PLANE LAYER VOLUME OF A GIVEN SLOPE

Sedimentary structures in seismic sections are sometimes compared to a cake whose every layer lies neatly upon another layer. Most Austrian pastry chiefs, however, would be horrified by the jumble of folded and broken layers that we find in most seismic sections. Nevertheless, at almost every location a small local image volume does resemble a cake (see Figure 1). The effect of large scale folding is negligible within such detailed image volumes and, despite the dip of the layers and their variable thickness, every culinary connoisseur will quickly recognize the fundamentally parallel-plane nature of such a subsurface piece.

 sigmoid Figure 1 The seismic section is folded and broken. But a small enough subvolume will approximate a plane layer volume almost everywhere.

The plane layer volume (our cake model) can be expressed mathematically as a 1-D function f(x,y,z) = g(nx x + ny y + nz z) where is the unit normal vector to all planes. Such a one-dimensional representation is much simpler than the general volume representation .

The two-dimensional example in Figure 2 and 3 illustrates the relationship between the normal vector of a plane layer volume and its gradient function. The image f(x,y) in Figure 2 consists of parallel lines of constant amplitude. Consequently, the image can be expressed as where is orthogonal to all lines of constant amplitude. Figure 3 displays the gradient vector by small arrows. The chain rule applied to proves what Figure 3 illustrates: the constant normal vector is everywhere parallel to the gradient . In general, we can formulate a simple (and maybe obvious) detection criterion: an image volume is a plane layer volume with normal , if (and only if) at all
 (1)

 puckin Figure 2 Input synthetic image: parallel lines of constant amplitudes.

 puckout Figure 3 Gradient vectors at various locations of the input image. (All gradient vectors should be parallel. I believe it is an artifact of my gradient plotting program that they are not.)

Equation (1) might remind the reader of the vector analysis identity .This identity states that the curl of any conservative vector field (a field that can be represented as the gradient of a scalar function f (potential function)) is zero everywhere. All scalar fields f, including plane layer volumes, satisfy this vector analysis identity. However, not all scalar fields f satisfy the cross product equation (1). In other words, the replacement of the left differential operator by a constant vector limits the set of scalar functions f to the functions that are plane layer volumes (with normal ).

For example, the image volume shown in Figure 4 satisfies the conservative field identity (since it is a scalar field f), but not the cross product expression (1) (as I will demonstrate in a later section). Similarly the radially-symmetric gravitational field of a point mass is a conservative field and satisfies the general curl expression, but defies the plane layer criterion (1).

However, the conservative field and the plane layer criterion are related if the scalar field is a plane layer volume. In the case of a plane layer volume, the left differential of the conservative field criterion yields the i-th component of the plane layer gradient .

Next: ESTIMATION OF THE SLOPE Up: Schwab: Cross Product Operator Previous: Introduction
Stanford Exploration Project
11/11/1997