# Velocity versus depth from sonobuoy data

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Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 5 141 - 180 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 5.19

Determine velocity versus depth from Figure 5.19a assuming horizontal refractors. The direct wave that travels through the water (assume $V_{W}=1500\ {\rm {m/s}}$ ) can be used to give source-receiver distances.

### Background

A sonobuoy is a free-floating device that radios the outputs of hydrophones to a recording ship. The ship fires its sources as it sails away from the sonobuoy to achieve a refraction profile.

### Solution

Figure 5.19b shows the picked events $A$ , $B$ , $C$ , and the waterbreak. This is an old profile where navigation was not as accurate as today. The direct-arrival waterbreak forms a distinct first arrival out to about SP 120 and an alignment with about the same slope can be seen from about SP 140 to SP 190, but it does not quite align with the water break seen at shorter offsets. The disagreement may merely indicate that the recording ship speed varied (note slight slope changes in the waterbreak alignment) and/or the sonobuoy drifted during the recording. We determine 48.5 m/SP or 11.8 km for the maximum offset. The first trace is 400 m from the sonobuoy.

Three distinct refraction (headwave) events can be seen: event $A$ , which gives the first breaks beyond SP 150; $B$ , which gives the first breaks between SP 120 and 150, and $C$ , which has a projected arrival time at SP 240 of about 6.4 s. When $B$ is a first break, its velocity is about 2.5 km/s but when it is a second arrival (problem 6.12), its velocity is about 2.9 km/s (the difference may be due to change of dip); we take its velocity as 2.7 km/s. Thus apparent velocities and intercept times for these events are about 5300, 2700, and 2400 m/s and 2.8, 1.7, and 1.4 s, respectively. Bearing in mind the distance uncertainties and timing errors (since first cycles are not clear enough for timing), we get crude answers only.

We get the depth of water by estimating $t_{o}$ for the sea-floor reflection; since $t_{o}\approx 1.45\ {\rm {s}}$ , the water depth is about $1.5\times 1.5/2\approx 1.1$ km.

Next we calculate depths to the refracting horizons using equation (4.18a) for $C$ , equation (4.18d) for $B$ and $A$ . The shallowest refractor $C$ is probably the top of the first consolidated rock, the material above it being unconsolidated sediments. We assume that the velocity in the sediments is close to that of water, so we get the depth to $C$ as follows:

{\begin{aligned}\sin \theta _{c1}=(1.5/2.4)\;,\;\theta _{c1}=39^{\circ }\;,\ \cos \theta _{c1}=0.78,\\h_{C}=1.5\times 1.4/2\times 0.78=1.3\ {\rm {km}}.\end{aligned}} We get the distance between $B$ and $C$ using the intercept time difference:

{\begin{aligned}\sin \theta _{c2}=(2.4/2.7),\ \theta _{c2}=63^{0},\ \cos \theta _{c2}=0.45,\ t_{i}=1.7-1.4=0.3\ {\rm {s}},\\\Delta h=2.4\times 0.3/2\times 0.45=0.8\ {\rm {km}},\end{aligned}} so the depth to $B$ is $1.3+0.8=2.1\ {\rm {km}}$ .

For event $A$ , we get

{\begin{aligned}\sin \theta _{c3}=2.7/5.3\;,\;\theta _{c3}=31^{\circ }\;\cos \theta _{c3}=0.86.\end{aligned}} To get the distance from $B$ to $A$ ,

{\begin{aligned}t_{i}=2.8-1.7=1.1\ {\rm {s}},\;\Delta h=2.7\times 1.1/2\times 0.86=1.7\ {\rm {km}},\end{aligned}} depth to $A$ = depth to $B+1.7=3.8\ {\rm {km}}$ .

The first reflection (from the sea floor) occurs at 1.45 s and a multiple of the seafloor reflection arrives at about 2.9 s; events below this multiple are so confused that interpretation cannot be done.