# Well-velocity survey

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 5.14a

Figure 5.14a shows data from a well-velocity survey tabulated on a standard calculation form. Calculate the average velocity and interval velocity, and plot graphs of time, average velocity, and interval velocity versus depth using a sea-level datum.

### Solution

Figure 5.14a shows only the measured data on a standard form whereas Figure 5.14b shows also calculated values. The successive columns in this form list

1 - Record number

2 - Source (shothole) location

3 - ${\displaystyle D_{gm}}$, geophone depth with respect to well datum

4 - ${\displaystyle D_{s}}$, depth of source

5 - ${\displaystyle t_{uh}}$, uphole time

6 -${\displaystyle t_{c}}$, arrival time at reference geophone

7 - ${\displaystyle {T}}$, arrival time at well geophone plus polarity and quality grades

8 - ${\displaystyle D_{gs}}$, geophone depth with respect to source elevation; ${\displaystyle D_{gs}=D_{gm}-D_{s}-\Delta e=D_{gm}-D_{s}-29}$

9 - ${\displaystyle {H}}$, horizontal distance of source from wellhead

10,11 - tangent, cosine of angle between straight raypath and vertical; ${\displaystyle \tan i={\rm {H}}/{\rm {D}}_{gs}}$

12 - ${\displaystyle T_{g}}$, vertical traveltime from source to geophone ${\displaystyle =T\cos i}$

13 - ${\displaystyle \Delta sd}$, source to datum elevation difference: ${\displaystyle \Delta {\rm {sd}}=D_{s}-D_{e}=D_{s}-78}$; a minus sign means that the shot was above datum

14 - time correction for ${\displaystyle \Delta {\rm {sd}}}$

15 - ${\displaystyle T_{gd}}$, vertical traveltime from datum to geophone

17 - ${\displaystyle D_{gd}}$, depth of geophone below datum

18 - ${\displaystyle \Delta D_{gd}}$, depth difference between successive geophone depths

19 - ${\displaystyle \Delta T_{gd}}$, time difference between successive geophone arrivals

20 - ${\displaystyle V_{i}}$ , interval velocity ${\displaystyle \Delta D_{gd}/\Delta T_{gd}}$

21 - ${\displaystyle T_{gs}}$, vertical traveltime from source to geophone ${\displaystyle =T\cos i}$

Figure 5.14a.  Data from a well-velocity survey.
Figure 5.14b.  Velocity calculations for data in Figure 5.14a.
Figure 5.14c.  Results of well velocity survey.

Depths below the datum are positive. The velocity used to correct ${\displaystyle \Delta sd}$ is ${\displaystyle V=D_{gs}/T_{gs}\approx 0.163/0.102\approx 1.60\ {\rm {km/s}}}$ (also obtainable from ${\displaystyle D_{gd}/T_{gd}}$). Note that the column headed ${\displaystyle \Delta {\rm {sd/V}}}$ is in milliseconds whereas all other times are in seconds. Column #16 headed ${\displaystyle T_{gs}/}$average is not used.

Figure 5.14c shows average velocity, interval velocity, and time plotted against depth.

## Problem 5.14b

How much error in average velocity and interval velocity values would result from (i) time-measurement errors of 1 ms, and (ii) depth-measurement errors of 1 m?

### Solution

1. A 1-ms time error produces an error in the average velocity of 0.1% to 1.5% and an error in the interval velocity of 1.5% to 8.3%.
2. A depth error of 1 m produces an error in the average velocity of 0.03% to 0.9% and an error in the interval velocity of 0.5% to 1.5%.
Table 5.14a. Data for least-squares calculation of ${\displaystyle V_{o}}$ and ${\displaystyle a}$.
${\displaystyle z\ ({\rm {m)}}}$ ${\displaystyle V_{i}\ ({\rm {{m/s})}}}$ ${\displaystyle \Delta z\ ({\rm {m)}}}$ ${\displaystyle z\ ({\rm {m)}}}$ ${\displaystyle V_{i}\ ({\rm {m/s)}}}$ ${\displaystyle \Delta \ z({\rm {m)}}}$
236 2200 255 1970 5400 65
441 2400 155 2038 3700 70
551 2600 65 2118 3900 90
661 2500 155 2228 6500 130
820 3100 165 2360 4000 135
973 2700 140 2506 4400 155
1101 3600 115 2643 4300 120
1203 2400 90 2783 4100 160
1350 4800 205 2928 4500 130
1508 4200 110 3043 4800 100
1603 2200 80 3145 5200 105
1730 7600 175 3243 6000 90
1878 5000 120

## Problem 5.14c

Determine ${\displaystyle V_{o}}$ and ${\displaystyle a}$ for a velocity-function fit to the data in (a) assuming the functional form ${\displaystyle V=V_{o}+az}$, where ${\displaystyle V}$ is the interval velocity and ${\displaystyle z}$ the depth.

### Solution

We can find ${\displaystyle V_{o}}$ and ${\displaystyle a}$ by (i) plotting the data and measuring the slope ${\displaystyle a}$ and intercept ${\displaystyle V_{o}}$ of the best-fit straight line, or (ii) using the least-squares method (see problem 9.33). The former method is difficult because of the large, irregularly spaced jumps in the curve, and therefore we shall use the latter method. We take ${\displaystyle z}$ as the depth in meters below datum to the center of each interval ${\displaystyle \Delta z}$ and give each data pair ${\displaystyle (V_{i},\;z)}$ the weight (see problem 9.33b) ${\displaystyle \Delta z}$. Using the data in Table 5.14a, we get ${\displaystyle V=2400+0.98z}$, as shown in Figure 5.14c.

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