# Using a velocity function linear with depth

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 4 79 - 140 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 4.17a

When the velocity increases linearly with depth according to the relation

 {\begin{aligned}V=V_{0}+az,\end{aligned}} (4.17a)

$a$ being constant, show that

 {\begin{aligned}x=(1/pa)(\cos i_{0}-\cos i),\end{aligned}} (4.17b)

 {\begin{aligned}t=(1/a)\ln \left({\frac {\tan i/2}{\tan i_{0}/2}}\right),\end{aligned}} (4.17c)

where $V=$ velocity, $V_{0}=$ velocity at depth $z=0$ , $x=$ source-geophone distance, $p=(\sin i)/V=$ raypath parameter, $t=$ arrival time, and $i=$ angle of incidence.

### Background

When the velocity is a function of depth only, as in equation (4.17a), expressions for the offset $x$ (see problem 4.1) and traveltime $t$ can be found by dividing the medium into horizontal layers, each of infinitesimal thickness (Figure 4.17a) and then integrating. We have

{\begin{aligned}\Delta x_{n}=\Delta z_{n}\tan i_{n},\end{aligned}} {\begin{aligned}\Delta t_{n}=\Delta z_{n}/(V_{n}\cos i_{n}).\end{aligned}} Also, using equation (3.1a), we have

{\begin{aligned}\sin i_{n}/V_{n}=\sin i_{0}/V_{0}=p.\end{aligned}} In the limit we get the following integrals for $x$ and $t$ :

 {\begin{aligned}x=\int _{0}^{z}\tan i\ \mathrm {d} z,\quad t=\int _{0}^{z}\mathrm {d} z/(V\cos i).\end{aligned}} (4.17d)

When the velocity increases monotonically with depth, a ray must eventually return to the surface (see Figure 4.20a). For horizontal velocity layering the raypaths are symmetrical about the deepest point. Figure 4.17a.  Raypaths where $V=V(z)$ .

### Solution

In equation (4.17d) we substitute $u=pV=\sin i$ , ${\rm {d}}z={\rm {d}}V/a={\rm {d}}u/pa$ from equation (4.17a), also $\tan {i}=u/(1-u^{2})^{1/2}$ . Thus, we get

{\begin{aligned}x&={\frac {1}{pa}}\int _{u_{0}}^{u}{\frac {d{\rm {d}}u}{[1-u^{2}]^{1/2}}}={\frac {1}{2pa}}\int _{u_{0}}^{u}{\dfrac {{\rm {d}}(u^{2})}{(1-u^{2})^{1/2}}}={\frac {1}{pa}}(1-u^{2})^{1/2}\left|\right._{u}^{u_{0}}\\&=(1/pa)(\cos i_{0}-\cos i);\\t&={\frac {1}{a}}\int _{u_{0}}^{u}{\frac {{\rm {d}}u}{u(1-u^{2})^{1/2}}}={\frac {1}{a}}\ln \left[{\frac {u}{1+(1-u^{2})^{1/2}}}\right]\left|\right._{u_{0}}^{u}\\&={\frac {1}{a}}\ln \left[\left({\frac {\sin i}{\sin i_{0}}}\right)\left({\frac {1+\cos i_{0}}{1+\cos i}}\right)\right]={\frac {1}{a}}\ln \left({\frac {\tan i/2}{\tan i_{0}/2}}\right).\end{aligned}} ## Problem 4.17b

Show that the angle of incidence $i$ and the depth $z$ can be written

 {\begin{aligned}i&=2\tan ^{-1}(e^{at}\tan i_{0}/2),\end{aligned}} (4.17e)

 {\begin{aligned}z&=(\sin i-\sin i_{0})/pa.\end{aligned}} (4.17f)

### Solution

From equation (4.17c) we get

{\begin{aligned}e^{at}=\left({\frac {\tan i/2}{\tan i_{0}/2}}\right),\end{aligned}} {\begin{aligned}{\hbox{hence}}\qquad \qquad i=2\tan ^{-1}(e^{at}\tan i_{0}/2).\end{aligned}} Solving equation (4.17a) for $z$ gives

{\begin{aligned}z=(V-V_{0})/a=(\sin i-\sin i_{0})/pa.\end{aligned}} ## Problem 4.17c

Given the velocity function $V=1.60+0.600\ z$ km/s, find the depth and offset of the point of reflection and the reflector dip when $t_{0}=4.420$ s and $\Delta t/\Delta x=0.155$ s/km. What interpretation would you give the result?

### Solution

First, we note that $t$ in the preceding equations is one-way time, so we take $t_{0}=2.210$ s, $V_{0}=1.60$ km/s, $a=0.600$ . To get $p$ , we find the angle of approach $i_{0}$ [see equation (4.2d)] for $\Delta t/\Delta x=0.155$ s/km:

{\begin{aligned}i_{0}&=\sin ^{-1}(1.60\times 0.155)=14.4^{\circ },\\i&=2\tan ^{-1}[e^{at}\tan(i_{o}/2)]=2\tan ^{-1}(e^{0.600\times 2.21}\tan 7.2^{\circ })=50.9^{\circ }.\end{aligned}} The dip of the reflector $\xi$ equals $i$ , 50.9$^{\circ }$ in this case, because, in order for the ray to return to the source, it must have been incident on the reflector at right angles.

Because of the unusually steep dip, this event might be from a fault plane or it might represent a steeply dipping bed in an area of highly folded sediments.

## Problem 4.17d

If the ray continued without reflection, when and where would it emerge? What moveout would be observed at the recording point? Calculate the maximum depth of penetration.

### Solution

[Equations (4.20a,b,c) could be used here instead of the following.]

Because the path is symmetrical about the midpoint, the angle of approach at the receiver equals $(\pi -i_{0})=(180^{\circ }-14.4^{\circ })=165.6^{\circ }$ .

From equations (4.17b,c) we have

{\begin{aligned}x&=(1/pa)[\cos i_{0}-\cos \ (\pi -i_{0})]=2(V_{0}/a)\cot i_{0}\\&=(2\times 1.60/0.600)\cot 14.4^{\circ }=20.8\ \mathrm {km} ,\\t&=2(1/0.600)\ln \left({\frac {\tan(\pi /2-i_{0}/2)}{\tan i_{0}/2}}\right)=3.33\ \ln \ (\cot ^{2}(i_{o}/2))]\\&=6.66\ \ln \ (\cot 7.2^{\circ })=6.66\ \mathrm {\ } {In}\ (\tan 82.8^{\circ })=13.8\ \mathrm {s} .\end{aligned}} Because of symmetry, the moveout on emergence would be $-0.155$ s/km. To find the maximum depth of penetration $h_{m}$ , we use the fact that the ray is traveling horizontally at the depth $h_{m}$ , that is, $i=\pi /2$ . From part (c) we have $i_{0}=14.4^{\circ }$ , so $p=\sin i_{0}/V_{0}=\sin 14.4^{\circ }/1.60=0.155$ . Equation (4.17f) now gives

{\begin{aligned}z=h_{m}=(\sin 90^{\circ }-\sin 14.4^{\circ })/(0.155\times 0.600)=8.08\ \mathrm {km} .\end{aligned}} 