# Interpreting uphole surveys

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 8 253 - 294 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem

Uphole surveys in five different (unrelated) areas give the uphole-time versus depth information in Table 8.17a. Explain the possible velocity layering for each case. How reliably are velocities and depths of weathering defined?

### Background

An uphole geophone is a geophone placed at or near a borehole for the purpose of recording the uphole time, that is, the time for a wave generated by a subsurface source to reach the surface.

### Solution

Because the uphole geophone is very close to the borehole, the raypath is essentially vertical.

Figures 8.17a, b, c, d, e are plots of the uphole times in Table 8.17a. The plotted data were approximated by series of straight lines and the velocities determined from the slopes of these lines. The thickness of the LVL is determined by the abrupt change in velocity at the base of the layer.

Table 8.17a. Uphole times versus depth in five areas.
Depth (m) Area $A\left(s\right)$ Area $B\left(s\right)$ Area $C\left(s\right)$ Area $D\left(s\right)$ Area $E\left(s\right)$ 5 0.012 0.011 0.012 0.008
8 0.020 0.010
10 0.025 0.023 0.024 0.018
12 0.024 0.027 0.020
15 0.030 0.031 0.022
18 0.028 0.034 0.030 0.030
21 0.034 0.036 0.033 0.031
25 0.036 0.032 0.035 0.032
30 0.039 0.035 0.039
35 0.037 0.039 0.036
40 0.046 0.044 0.044 0.042
50 0.051 0.044 0.048 0.047 0.043

The results for each area are given below.

Area A (Figure 8.17a):

$V_{1}=400\ {\rm {m/s}}$ , $V_{2}=1690\ {\rm {m/s}}$ , $D_{W}=11\ {\rm {m}}$ . All three values are accurately determined.

Area B (Figure 8.17b):

This is a three-layer situation; $V_{1}=440{\rm {m/s}}$ , $V_{2}=1660{\rm {m/s}}$ , $V_{3}=2200{\rm {m/s}}$ , $D_{W}=10{\rm {m}}$ . To get the thickness of the second layer, we note that the bases of the LVL and second layer correspond to uphole times of 23 and 35 ms. Since the path is vertical, we can determine the depth to the base of the second layer, $D_{2}$ , to be $D_{2}=10+\left(0.035-0.024\right)\times 1660=30\ {\hbox{m}}$ . Values are reasonably well determined.

Using the simpler two-layer solution, the depth of the 440 m/s layer would be picked as 12 m rather than 10 m, a 20% error, but the statics correction would have an error of only 2 ms.

Area C (Figure 8.17c)

$V_{1}=410\ \mathrm {m/s} ,V_{2}=750\ \mathrm {m/s} ,V_{3}=2500\ {\hbox{m/s}},D_{1}=10\ {\hbox{m}},D_{2}=10+(0.035-0.024)\times 750=18\ {\hbox{m}}$ . Values are moderately well determined.

As in the case of Area B, this area could be interpreted as a two-layer situation but with larger errors.

Area D (Figure 8.17d):

The time-depth curve can be interpreted either in terms of three or four layers. The three-layer solution assumes that the measurement at 15 m is in error and the four-layer solution honors the data more closely.

The three-layer solution is given by the dashed lines. Measured values are: $V_{1}=420\ {\hbox{m/s}},V_{2}=740\ {\hbox{m/s}},V_{3}=2060\ {\hbox{m/s}},D_{1}=6\ {\hbox{m}},D_{2}=6+\left(0.033-0.015\right)\times 740=19\ {\hbox{m}}$ .

The four-layer case is shown by the solid lines. Measured values are: $V_{1}=420\ {\hbox{m/s}}$ , $V_{2}=1070\ {\hbox{m/s}},V_{3}=390\ {\hbox{m/s}},V_{4}=1890\ {\hbox{m/s}},D_{1}=7\ {\hbox{m}},D_{2}=7+\left(0.022-0.015\right)\times 1070=14\ {\hbox{m}}$ , $D_{3}=14+\left(0.030-0.022\right)\times 390=17\,{\rm {m}}$ . This interpretation postulates a high-velocity layer $\left(V_{2}\right)$ within the LVL.

In both of these interpretations velocities and depths are questionable except for the velocity of the deepest layer.

Area E (Figure 8.17e):

Two-layer and three-layer solutions are possible. Assuming two layers (dashed-line), measured values are: $V_{1}=610\,{\rm {m/s}}$ , $V_{2}=2360\,{\rm {m/s}}$ , $D_{W}=17\,{\rm {m}}$ .

The three-layer solution (solid line) gives: $V_{1}=610\ {\hbox{m/s}}$ , $V_{2}=2910\ {\hbox{ms}}$ , $V_{3}=2140\ {\hbox{m/s}},D_{W}=18m,D_{2}=18+\left(0.035-0.031\right)\times 2910=30\ {\hbox{m}}$ .