Determining static corrections from first breaks

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 8.19a

Figure 8.19a shows the first arrivals (first breaks) at geophone stations 100 m apart from sources 25 m deep at each end of the spread. The geophone group at each end is not recorded because of hole noise. The uphole time is on the third trace from the right. Elevations for each group are given at the top. Weathering velocity is 500 m/s. The valley midway between the sources produces a change in the firstbreak slope, as if two refractors were involved, which is not the case. How can we be sure?

Figure 8.19a.  First breaks on reflection record.
Table 8.19a. Values observed from Figure 8.19a.
${\displaystyle x}$ ${\displaystyle E}$ ${\displaystyle \Delta _{t}}$ ${\displaystyle t_{A}}$ ${\displaystyle t_{B}}$ ${\displaystyle t_{A}^{*}}$ ${\displaystyle t_{B}^{*}}$
0 1125 433 433
100 1130 —10 66 398 56 388
200 1127 —4 106 357 102 353
300 1128 —6 151 317 145 311
400 1125 0 184 270 184 270
500 1120 +10 218 220 228 230
600 1120 +10 257 172 267 182
700 1125 0 313 143 313 143
800 1133 —16 367 118 351 102
900 1138 —26 418 83 392 57
1000 1140 —30 466 436

Background

The source instant ${\displaystyle \left(t=0\right)}$ or time break is the sharp deflection on the third trace from the right. Each time division is 10 ms.

Hole noise is caused by reverberations within the shothole and by material ejected from the shothole and falling back to the earth when an explosive charge is detonated.

Solution

We correct first-break readings for elevation by taking as the reference datum the elevation of the left-hand source point ${\displaystyle A}$ and adding or subtracting 2 ms (=1 m/500 m/s) for each meter below or above the datum. The corrected times ${\displaystyle t_{A}^{*}}$ and ${\displaystyle t_{B}^{*}}$ are given in Table 8.19a where ${\displaystyle t_{A}}$ and ${\displaystyle t_{B}}$ are first-break times for sources at ${\displaystyle A}$ and ${\displaystyle B}$ at offset ${\displaystyle x}$ from source ${\displaystyle A}$, All times are in milliseconds, distances in meters.

The plots of the corrected times in Figure 8.19b give straight lines whose slopes have an average value of 2390 m/s for ${\displaystyle V_{H}}$ whereas the plots of the uncorrected times suggest a 3-layer situation with a low-velocity layer in between two higher-velocity layers. The corrected times fit a straight line in each case which is strong evidence that there is only one high-velocity layer.

Problem 8.19b

Determine the weathering thickness at the two sourcepoints from the uphole times.

Figure 8.19b.  Plots of first breaks in Figure 8.19a.

Solution

To find ${\displaystyle D_{W}}$ at ${\displaystyle A}$ and ${\displaystyle B}$ from ${\displaystyle t_{uh}}$, we have

{\displaystyle {\begin{aligned}t_{uh}=D_{W}/V_{W}+\left(D_{S}-D_{W}\right)/V_{H}\\{\mbox{or}}\qquad \qquad D_{W}=\left(t_{uh}-D_{S}/V_{H}\right)/\left(1/V_{W}-1/V_{H}\right).\end{aligned}}}

The uphole times are 0.025 s and 0.049 s at ${\displaystyle A}$ and ${\displaystyle B}$, ${\displaystyle D_{S}=25}$ m at both ${\displaystyle A}$ and ${\displaystyle B}$, ${\displaystyle V_{W}=500}$ m/s ${\displaystyle V_{H}=2390}$ m/s; thus ${\displaystyle D_{W}=9.2}$ m at ${\displaystyle A}$ and 24 m at ${\displaystyle B}$.

Problem 8.19c

What correction ${\displaystyle \Delta t_{0}}$ should be applied to reflection times at the two source-points for a datum of 1125 m?

Solution

Applying equation (8.18c), we have

{\displaystyle {\begin{aligned}{\mbox{at A,}}\qquad \Delta t_{0}=2(1125-25-1125/2390+0,025=0.004\quad {\rm {s}},\\{\mbox{and at B,}}\qquad \Delta t_{0}=2\left(1140-25-1125\right)/2390+0.049=0.041\quad {\rm {s}}.\end{aligned}}}

Table 8.19b. Calculating ${\displaystyle D_{W}}$ and ${\displaystyle \Delta t_{g}}$ for each geophone.
${\displaystyle x_{A}}$ ${\displaystyle t_{AC}+tBC}$ ${\displaystyle t_{w}}$ ${\displaystyle D_{W}}$ ${\displaystyle E_{g}}$ ${\displaystyle (\Delta t_{g})_{A}}$ ${\displaystyle (\Delta t_{g})_{B}}$
0 10 1125 32
100 464 23 12 1130 31 37
200 463 22 11 1127 29 35
300 468 25 12 1128 32 38
400 454 18 9 1125 25 31
500 438 10 5 1120 16 22
600 429 6 3 1120 13 19
700 456 19 10 1125 25 31
800 485 34 17 1133 41 47
900 501 42 21 1138 49 55
1000 24 1140 55
 Times are in milliseconds and distances in meters.

Problem 8.19d

Calculate the weathering thickness and the time correction for each geophone station.

Solution

In Table 8.19b the second column is the sum of the uncorrected times at geophone ${\displaystyle C}$ from Table 8.19a. To get ${\displaystyle t_{w}}$ we require the quantity ${\displaystyle \left(AB/V_{H}\right)=\left(1000/2390\right)=0.418\ {\hbox{s}}}$; we now get ${\displaystyle t_{w}=\left[\left(t_{AC}+t_{BC}\right)-0.418\right]/2}$. Next, ${\displaystyle D_{W}=500\ t_{w}}$ (except for the source points where ${\displaystyle D_{W}}$ is obtained from the uphole times [part (b)]. The weathering correction for a geophone group is given by equation (8.18g), that is,

{\displaystyle {\begin{aligned}\Delta t_{g}=t_{w}+\left(E_{C}-D_{W}-E_{D}\right)/V_{H},\end{aligned}}}

where ${\displaystyle D_{W}=V_{W}t_{w}}$, ${\displaystyle t_{w}}$ being given by equation (8.18e), namely,

{\displaystyle {\begin{aligned}t_{w}={\frac {1}{2}}[t_{AC}+t_{BC}-(AB/V_{H}].\end{aligned}}}

This correction is equivalent to placing the geophone group at C on the datum. To locate the sources on the datum also, we must add to ${\displaystyle \Delta t_{g}}$ the time from the source down to the datum, that is, ${\displaystyle \left(E_{S}-D_{S}-E_{D}\right)/V_{H}}$; this amounts to 0 and 4 ms at sources ${\displaystyle A}$ and${\displaystyle B}$, respectively. The column headed ${\displaystyle (\Delta t_{g})_{A}}$ gives the corrections for the arrival times at the given offset for source ${\displaystyle A}$ while that headed ${\displaystyle (\Delta t_{g})_{B}}$ gives the corrections for source ${\displaystyle B}$.

Problem 8.19e

Plot corrected reflection arrival times in an ${\displaystyle X^{2}}$-${\displaystyle T^{2}}$ plot and determine the depth, dip, and average velocity to the reflector giving the reflection at 0.30 and 0.21 s in Figure 8.19a.

Table 8.19c. ${\displaystyle X^{2}}$-${\displaystyle T^{2}}$ data for profile from A.
${\displaystyle x_{A}}$ ${\displaystyle x_{A}^{2}}$ ${\displaystyle t_{A}}$ ${\displaystyle (\Delta t_{g})_{A}}$ ${\displaystyle t_{AC}}$ ${\displaystyle t_{AC}^{2}}$
100 ${\displaystyle 1\times 10^{4}}$ 0.303 31 0.272 0.0740
200 4 0.305 29 0.276 0.0762
300 9 0.317 32 0.285 0.0812
400 16 0.325 25 0.300 0.0900
500 25 0.334 16 0.318 0.1011
600 36 0.341 13 0.328 0.1076
700 49 0.355 25 0.330 0.1089
 Distances are in meters, traveltimes in seconds, corrections in milliseconds. Underlined values are doubtful.
Table 8.19d. ${\displaystyle X^{2}}$-${\displaystyle T^{2}}$ data for profile from B.
${\displaystyle x_{B}}$ ${\displaystyle x_{B}^{2}}$ ${\displaystyle t_{B}}$ ${\displaystyle (\Delta t_{g})_{B}}$ ${\displaystyle t_{BC}}$ ${\displaystyle t_{BC}^{2}}$
100 ${\displaystyle 11\times 10^{4}}$ 0.211 55 0.156 0.0243
200 4 0.224 47 0.177 0.0313
300 9 0.237 31 0.206 0.0424
400 10 0.251 19 0.232 0.0538
500 25 0.272 22 0.258 0.0666

Solution

Tables 8.19c and 8.19d list the offsets and their squares, the uncorrected times and the correction for each (from Table 8.18a), the corrected times ${\displaystyle t_{AC}}$ and their squares. Table 8.19c is for source ${\displaystyle A}$ , Table 8.19d is for source ${\displaystyle B}$. The ${\displaystyle X^{2}-T^{2}}$ data are plotted in Figure 8.19c. Drawing the best-fit straight lines, we measure the intercepts on the ${\displaystyle t^{2}}$-axis and the slopes. Assuming that the dip is small so that the factor ${\displaystyle {\rm {\;cos\;}}\xi }$ in equation (4.3a) can be neglected, the reciprocals of the slopes give the velocities squared. The measured results are:

{\displaystyle {\begin{aligned}V_{A}=2.36\,\mathrm {km/s} ,\;t_{i}=0.15\,\mathrm {s} ,\;h_{A}=180\,\mathrm {m} ;\\V_{B}=2.98\,\mathrm {km/s} ,\;t_{i}=0.27\,\mathrm {s} ,\;h_{B}=400\,\mathrm {m} ;\\\mathrm {dip} \ \xi ={\tan }^{-1}\left[\left(400-180\right)/1000\right]=12^{\circ }.\end{aligned}}}

Figure 8.19c.  ${\displaystyle X^{2}}$-${\displaystyle T^{2}}$ graph.