# Depth structure maps

Series Investigations in Geophysics Öz Yilmaz http://dx.doi.org/10.1190/1.9781560801580 ISBN 978-1-56080-094-1 SEG Online Store

We now have a set of time horizon maps (Figure 9.4-4) and a corresponding set of interval velocity maps (Figure 9.4-7). Recall from introduction to earth imaging in depth that, in principle, a time-migrated image can be converted to a depth section by mapping the amplitudes along image rays. This notion also can be employed to convert time horizons into depth horizons. The process is done layer by layer starting with the shallowest horizon.

A comprehensive mathematical discussion on image-ray tracing is given by Hubral and Krey . Refer to Figure 9.4-8 for a brief description of image-ray tracing. Consider an image ray that departs the nth surface at point Sn with coordinates (xn, yn, zn) and emerges at the right angle at the earth’s surface at point S0 with coordinates (x0, y0, z = 0). Our goal is to determine the coordinates of the output point (xn, yn, zn) on the depth map from image-ray depth conversion of the input point (x0, y0, tn) on the time map. To achieve this goal, we want to trace the image ray from the point of emergence (x0, y0, 0) back to its point of departure (xn, yn, zn).

Suppose that the first n − 1 horizons have already been converted to depth, and that next we want to convert the nth horizon to depth. Since the earth model is known for the first n − 1 layers, then we know the coordinates of the intersection point S1 of the image ray with the first layer, (x1, y1, z1), where by definition of the image ray, x1 = x0 and y1 = y0. By using Snell’s law, we can determine the direction of the ray as it departs point S1 reaching point S2 on the next surface. As the image ray moves from one surface to the next, we add up the time it takes to travel. When it reaches the (n − 1)st layer, the elapsed two-way time tn−1 is

 $t_{n-1}=2\sum _{k}^{n-1}{\frac {\Delta s_{k}}{v_{k}}},$ (2)

where vk is the interval velocity of the kth layer, and Δsk is the distance between the intersection points of the image ray, Sk−1 and Sk, on the (k − 1)st and kth surfaces given by

 $\Delta s_{k}=\left[(x_{k}-x_{k-1})^{2}+(y_{k}-y_{k-1})^{2}+(z_{k}-z_{k-1})^{2}\right]^{1/2}.$ (3)

Now, we examine the situation when the image ray departs the (n − 1)st layer at point Sn−1 on the way to the nth surface. Again, by Snell’s law we know the direction of the ray. We also know the elapsed time tn − tn−1 from the (n − 1)st surface to the nth surface since we know the total elapsed time tn−1 from equation (2) and the total elapsed time tn from the input time horizon read at point (x0, y0, z = 0). Finally, we know the interval velocity vn of the nth layer from the interval velocity map. Therefore, we can calculate the elapsed distance Δsn along the raypath as it departs the point Sk−1 on the (k − 1)st surface in the direction dictated by Snell’s law. The quantity Δsn is given by

 $\Delta s_{n}={\frac {v_{n}}{2}}(t_{n}-t_{n-1}).$ (4)

Finally, the coordinates of the point Sn that we need to know to perform the time-to-depth conversion are given by

 $x_{n}=x_{n-1}+\Delta s_{n}\cos \ \alpha ,$ (5a)

 $y_{n}=y_{n-1}+\Delta s_{n}\cos \ \beta ,$ (5b)

 $z_{n}=z_{n-1}+\Delta s_{n}\cos \ \gamma ,$ (5c)

where α, β, and γ are the directional cosines of the ray at point Sn−1. The directional cosines are known by the application of Snell’s law at point Sn−1 with known coordinates (xn−1, yn−1, zn−1).

To summarize, given the depth and interval velocity maps for the first n − 1 horizons, and the time and interval velocity maps for the nth horizon, we can trace an image ray associated with the time tn(x0, y0) on the time map and derive the depth value zn(xn, yn) on the depth map.

Figure 9.4-9 shows the depth maps derived from image-ray depth conversion of the time maps shown in Figure 9.4-4 using the interval velocity maps shown in Figure 9.4-7. As for the time maps, the usual way to post the depth maps is to contour them (Figure 9.4-10).

The depth maps are compatible with the time maps; nevertheless, there can be subtle differences because of velocity variations that would give rise to the departure of image rays from the vertical. To quantify the differences between vertical-ray and image-ray trajectories, and thus to quantify the differences between the time maps and depth maps, we can calculate the modulus Δdn of the lateral displacement vector between the points S0 and Sn as

 $\Delta d_{n}={\sqrt {(x_{n}-x_{0})^{2}+(y_{n}-y_{0})^{2}}},$ (6a)

and create the displacement modulus maps shown in Figure 9.4-11. Note that the most significant displacement between the vertical rays and image rays is at the fault zones.

The displacement vector also has a directional azimuth ϕn which is given by

 $\phi _{n}=\tan ^{-1}{\frac {y_{n}-y_{0}}{x_{n}-x_{0}}}$ (6b)

as measured from the inline x direction. The displacement azimuth maps are shown in Figure 9.4-12. Again, note that the most significant azimuthal variations are along the fault zones.