Determining static corrections from first breaks

**Problems in Exploration Seismology and their Solutions**
Series	Geophysical References Series
Title	Problems in Exploration Seismology and their Solutions
Author	Lloyd P. Geldart and Robert E. Sheriff
Chapter	8
Pages	253 - 294
DOI	http://dx.doi.org/10.1190/1.9781560801733
ISBN	ISBN 9781560801153
Store	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.**
$x$	$E$	$\Delta _{t}$	$t_{A}$	$t_{B}$	$t_{A}^{*}$	$t_{B}^{*}$
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 $\left(t=0\right)$ 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 $A$ and adding or subtracting 2 ms (=1 m/500 m/s) for each meter below or above the datum. The corrected times $t_{A}^{*}$ and $t_{B}^{*}$ are given in Table 8.19a where $t_{A}$ and $t_{B}$ are first-break times for sources at $A$ and $B$ at offset $x$ from source $A$ , 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 $V_{H}$ 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 $D_{W}$ at $A$ and $B$ from $t_{uh}$ , we have

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

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

Problem 8.19c

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

Solution

Applying equation (8.18c), we have

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

**Table 8.19b. Calculating $D_{W}$ and $\Delta t_{g}$ for each geophone.**
$x_{A}$	$t_{AC}+tBC$	$t_{w}$	$D_{W}$	$E_{g}$	$(\Delta t_{g})_{A}$	$(\Delta t_{g})_{B}$
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 $C$ from Table 8.19a. To get $t_{w}$ we require the quantity $\left(AB/V_{H}\right)=\left(1000/2390\right)=0.418\ {\hbox{s}}$ ; we now get $t_{w}=\left[\left(t_{AC}+t_{BC}\right)-0.418\right]/2$ . Next, $D_{W}=500\ t_{w}$ (except for the source points where $D_{W}$ is obtained from the uphole times [part (b)]. The weathering correction for a geophone group is given by equation (8.18g), that is,

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

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

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

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 $\Delta t_{g}$ the time from the source down to the datum, that is, $\left(E_{S}-D_{S}-E_{D}\right)/V_{H}$ ; this amounts to 0 and 4 ms at sources $A$ and $B$ , respectively. The column headed $(\Delta t_{g})_{A}$ gives the corrections for the arrival times at the given offset for source $A$ while that headed $(\Delta t_{g})_{B}$ gives the corrections for source $B$ .

Problem 8.19e

Plot corrected reflection arrival times in an $X^{2}$ - $T^{2}$ 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. $X^{2}$ - $T^{2}$ data for profile from A.**
$x_{A}$	$x_{A}^{2}$	$t_{A}$	$(\Delta t_{g})_{A}$	$t_{AC}$	$t_{AC}^{2}$
100	$1\times 10^{4}$	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. $X^{2}$ - $T^{2}$ data for profile from B.**
$x_{B}$	$x_{B}^{2}$	$t_{B}$	$(\Delta t_{g})_{B}$	$t_{BC}$	$t_{BC}^{2}$
100	$11\times 10^{4}$	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 $t_{AC}$ and their squares. Table 8.19c is for source $A$ , Table 8.19d is for source $B$ . The $X^{2}-T^{2}$ data are plotted in Figure 8.19c. Drawing the best-fit straight lines, we measure the intercepts on the $t^{2}$ -axis and the slopes. Assuming that the dip is small so that the factor ${\rm {\;cos\;}}\xi$ in equation (4.3a) can be neglected, the reciprocals of the slopes give the velocities squared. The measured results are:

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

Figure 8.19c.

X^{2}

-

T^{2}

graph.

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Determining static corrections from first breaks

Determining static corrections from first breaks

Problem 8.19a

Background

Solution

Problem 8.19b

Solution

Problem 8.19c

Solution

Problem 8.19d

Solution

Problem 8.19e

Solution

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