# User:Ageary/Chapter 5

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

## 5.1 Maximum porosity versus depth

What physical fact determines the “limit-of-porosity” line in Figure 5.1a? What is implied for measurements to the right of this line?

Background

The porosity ${\displaystyle \phi }$ of a rock is the pore volume per unit volume. The pore space is generally filled with salt water except near the surface, where air may be present, and in petroleum deposits, where the pore spaces contain oil and/or gas.

Pore spaces are usually sufficiently interconnected so that the fluid pressure approximates that of a fluid column extending to the surface; this is called normal pressure. The weight of the rock column exerts an overburden pressure. The differential, effective, or net pressure on the rock matrix is the overburden pressure less the interstitial fluid pressure. However, if there is no communication between the pore spaces and the surface, the interstitial fluid pressure may be greater than the normal pressure causing the differential pressure (and velocity) to be lower than usual for a given depth—a situation known as overpressure.

Figure 5.1a.  Factors affecting velocity (after Ziegler and Spotts, 1978).

Solution

The line marking the limit of porosity is fixed by the strength of the rock matrix that is presumed to be carrying the major part of the weight of the overburden. The overburden weight is partially supported by the interstitial fluid; in overpressured formations the effective stress on the rock’s matrix structure will be smaller than normal, equal to the stress that would be experienced at a shallower depth. The porosity and velocity will thus be that at the shallower depth and such points may fall to the right of the line.

Figure 5.2a.  S- and P-wave velocities (after Pickett, 1963).

## 5.2 Relation between lithology and seismic velocities

Figures 5.2a and 5.2b are based on different experimental data and plotted differently. Show the compatibility of these figures.

Solution

Dashed lines have been drawn on Figure 5.2a to approximate the data for the three lithologies and these have been transferred to Figure 5.2b. The limestone line lies within its domain, while the dolomite and sand lines lie at the edges of their domains. The sandstone data in Figure 5.2a show the greatest scatter. The concept that ${\displaystyle Vs/Vp}$ provides a basis for distinguishing these three lithologies is generally supported.

Figure 5.2b.  S- and P-wave velocities.
Figure 5.3a.  Histogram of rock densities (from Grant and West, 1965).

## 5.3 Porosities, velocities, and densities of rocks

5.3a Assume that sandstone is composed only of grains of quartz, limestone only of grains of calcite, and shale of equal quantities of kaolinite and muscovite. For sandstone, limestone, and shale saturated with salt water ${\displaystyle (\rho =1.03\ {\rm {g/cm}}^{3})}$, what porosities are implied by the densities shown in Figure 5.3a? (Mineral densities are: ${\displaystyle \rho _{\rm {quartz}}=2.68}$; ${\displaystyle \rho _{\rm {calcite}}=2.71}$; ${\displaystyle \rho _{\rm {kaolinite}}=2.60}$; ${\displaystyle \rho _{\rm {muscovite}}=2.83}$, all in g/cm${\displaystyle ^{3}}$.)

Background

Gardner et al. (1974) plotted the log of velocity against the log of density for sedimentary rocks and obtained the empirical relation known as Gardner’s rule:

 {\displaystyle {\begin{aligned}\rho =aV^{1/4}\end{aligned}}} (5.3a)
Table 5.3a. Rock densities and porosities.
Porosity
Rock Density range Density av. Mineral density Max Av. Min
Ss 2.00–2.60 g/cm${\displaystyle ^{3}}$ 2.35 g/cm${\displaystyle ^{3}}$ 2.68 g/cm${\displaystyle ^{3}}$ 41% 20% 5%
Ls 2.20–2.75 2.55 2.71 30 10 0
Sh 1.90–2.70 2.40 2.72 48 19 0

${\displaystyle \rho }$ being in g/cm${\displaystyle ^{3}}$ and ${\displaystyle a=0.31}$ or 0.23 when ${\displaystyle V}$ is in m/s or ft/s, respectively. The rule is valid for the major sedimentary rock types, but not for evaporites or carbonaceous rocks (coal, lignite).

When a porous rock is saturated with a fluid, its density ${\displaystyle \rho _{s}}$ is given by the equation

 {\displaystyle {\begin{aligned}\rho _{s}=\phi \rho _{f}+\left(1-\phi \right)\rho _{m},\end{aligned}}} (5.3b)

${\displaystyle \phi }$ being the porosity, ${\displaystyle \rho _{f}}$ and ${\displaystyle \rho _{m}}$ the densities of the fluid and rock matrix, respectively.

Solution

The density ranges in Table 5.3a were obtained from Figure 5.3a. The mineral densities are for ${\displaystyle \phi =0}$, the values for shale being averages for kaolinite and muscovite.

We solve equation (5.3b) for ${\displaystyle \phi }$, obtaining

 {\displaystyle {\begin{aligned}\phi ={\frac {\rho _{m}-\rho _{s}}{\rho _{m}-\rho _{f}}}.\end{aligned}}} (5.3c)
Figure 5.3b.  P-wave velocities from different sources. (1) Grant and West, 1965; (2) Kearey and Brooks, 1984; (3) Lindseth, 1979; (4) Mares, 1984; (5) Sharma, 1997; (6) Sheriff and Geldart, 1995; (7) Waters, 1987.

The histogram in Figure 5.3a does not encompass the complete range of samples and the range limits have been picked somewhat arbitrarily. Porosity in rocks ranges from about 50% to 0%. The upper limits of the density range sometimes exceed the mineral densities, hence heavier minerals must be present in the rocks; in these cases we assume that ${\displaystyle \phi =0}$. We take ${\displaystyle \rho _{f}=1.03\ {\rm {g/cm}}^{3}}$ as the fluid density.

5.3b What velocities would be expected for the density values in Table 5.3a according to Gardner’s rule? Plot these on Figure 5.3b.

Solution

We solve equation (5.3a) for the velocity ${\displaystyle V}$, obtaining

{\displaystyle {\begin{aligned}V=(\rho /a)^{4}=(\rho /0.31)^{4}\times 10^{-3}=0.11\rho ^{4}\ {\rm {km/s}}.\end{aligned}}}

The velocities in Table 5.3b are plotted as triangles on Figure 5.3b.

Table 5.3b. Velocities obtained from densities in Table 5.3a.
Rock ${\displaystyle V_{\hbox{min}}\left({\hbox{km/s}}\right)}$ ${\displaystyle V_{\hbox{av}}\left({\hbox{km/s}}\right)}$ ${\displaystyle V_{\hbox{max}}\left({\hbox{km/s}}\right)}$
Ss 1.8 (41%) 3.4 (20%) 6.8 (0%)
Ls 2.6 (30%) 4.6 (10%) 6.8 (0%)
Sh 1.4 (48%) 3.4 (19%) 6.8 (0%)
 Note: The values in parentheses are the porosities.

5.3c From Figure 5.3c, what densities would you expect at 7500 ft and how do these compare with Figures 5.3d and 5.3e from offshore Louisiana?

Figure 5.3c.  Porosity versus depth (from Atkins and McBride, 1992).
Figure 5.3d.  Average sand density between 7000 and 8000 ft (courtesy Geophy. Develop. Corp.).
Figure 5.3e.  Average shale density between 7000 and 8000 ft (courtesy Geophy. Develop. Corp.).

Solution

Using ${\displaystyle \rho _{\rm {quartz}}=2.68\ {\rm {g/cm}}^{3}}$ and ${\displaystyle \phi =31\%}$ from Figure 5.3c, equation (5.3c) gives ${\displaystyle \rho =2.19\ {\rm {g/cm}}^{3}}$, which is in accord with Figure 5.3d. Using ${\displaystyle \rho _{\rm {shale\ minerals}}=2.71}$, equation (5.3c) gives ${\displaystyle \rho =2.22\ {\rm {g/cm}}^{3}}$, which is slightly lower than most values in Figure 5.3e.

## 5.4 Velocities in limestone and sandstone

Assume that the velocity in calcite is 6.86 km/s and in quartz 5.85 km/s. What velocities should be expected for 10, 20, and 30% porosities in (a) limestone composed only of calcite; (b) sandstone composed only of quartz? Where do these values plot on lithology versus velocity curves (Figure 5.4a) and porosity versus velocity curves (Figure 5.4b)? Assume that the pore fluid is water with velocity 1.55 km/s.

Background

The velocity in a porous rock depends primarily on the matrix velocity, the porosity, and the nature of and velocity in the pore fluid. The empirical time-average equation [of the same form as equation (5.3b)] relates the specific transit time (reciprocal of the velocity) or slowness to the volume fraction of pore space ${\displaystyle \phi }$ and the remaining volume ${\displaystyle \left(1-\phi \right)}$:

 {\displaystyle {\begin{aligned}1/V=\phi /V_{f}+\left(1-\phi \right)/V_{m},\end{aligned}}} (5.4a)

where ${\displaystyle V}$, ${\displaystyle V_{f}}$, and ${\displaystyle V_{m}}$ are the velocities in the saturated rock, the fluid, and the rock matrix, respectively.

Solution

Substituting ${\displaystyle V_{f}=1.55\ {\rm {km/s}}}$ in equation (5.4a) yields ${\displaystyle V={\frac {1.55}{\phi +(1-\phi )(1.55/V_{m})}}}$. For ${\displaystyle \phi }$ values of 10, 20, and 30%, we obtain the velocities in Table 5.4a.

Figure 5.4a.  P-wave velocities (various sources of data).
Figure 5.4b.  Velocity versus porosity (after Wyllie et al., 1958). The limestone values (plotted as crosses on both Figures 5.4a and 5.4b) are slightly high and the sandstone values (plotted as circles) are reasonable. The line in Figure 5.4b is the time-average equation.

Table 5.4a. Velocities calculated from porosities.
Rock ${\displaystyle V_{m}}$ ${\displaystyle \phi =10\%}$ ${\displaystyle \phi =20\%}$ ${\displaystyle \phi =30\%}$
Limestone 6.86 km/s 5.11 km/s 4.07 km/s 3.38 km/s
Sandstone 5.85 4.58 3.76 3.19

## 5.5 Dependence of velocity-depth curves on geology

5.5a Why do the velocity-depth curves for various areas shown in Figures 5.5a and 5.5b depart from each other? Incorporate your knowledge of the geology of the various areas in your answer.

Solution

Velocities are increased not only by compaction with depth of burial but also by cementation and other factors attendant upon age. They are also affected by lithology and other factors. The U.S. Gulf Coast and offshore Venezuela sediments are predominantly young siliciclastics that have never been buried deeper than they are now. Hence their velocities generally relate to the maximum porosities such as shown in Figure 5.1a. Similar reasoning applies to the shallower portions of the offshore U.S. East Coast and the Gulf of Alaska curves of Figure 5.5b. The high-velocity values in the upper part of the Gulf of Alaska curve probably indicate limestone or volcanics. The Texas Gulf Coast-1 curve penetrated Cretaceous rocks containing limestone at fairly shallow depths whereas the Gulf Coast-2 well did not encounter this section until a depth of 4.25 km. The Illinois Basin and Permian Basin wells contain much older, higher-velocity rocks including limestone.

Figure 5.5a.  Velocity-depth, U.S. Gulf Coast.

5.5b Plot the shale and limestone values from Figure 5.5c for depths of 1000 and 2000 m on the velocity-depth Figure 5.5a. How do they compare?

Figure 5.5b.  Velocity-depth, selected wells.

Solution

The values read from Figure 5.5c are:

 shale at 1000 m 2.2 km/s shale at 2000 m 2.8 limestone at 1000 4.2 limestone at 2000 5.4
Figure 5.5c.  Velocity/depth relations (from Jankosky, 1970).

These points are plotted as triangles on Figure 5.5a for regions where the sections are sand-shale. The shale values fit nicely, but the limestone values are much too large, as one would expect. The limestone values are also plotted on Figure 5.5b where they fit in nicely with data for regions that are mostly carbonate.

## 5.6 Effect of burial history on velocity

5.6a Assume a subsiding area where there has been no uplift and a shale that is normally pressured until it reaches a depth of burial of 1.40 km, at which point fluid communication is cut off, that is, interstitial fluid can no longer escape. If it is at a depth of 2.00 km, what velocity and fluid pressure would you expect? If at a depth of 3.00 km?

Background

The normal fluid pressure at depth ${\displaystyle z}$ is ${\displaystyle z\rho _{f}}$, where ${\displaystyle \rho _{f}}$ is the fluid density. The fluid pressure gradient is ${\displaystyle \partial {\mathcal {P}}_{f}/\partial z=\rho _{f}\approx 10}$ MPa/km, whereas the pressure gradient due to the rock overburden is about 22.5 MPa/km. The differential pressure gradient is therefore approximately 12 MPa/km.

Solution

At a depth of 1.40 km, the velocity is about 2.5 km/s (see Figure 5.6a) and the fluid pressure ${\displaystyle {\mathcal {P}}_{f}}$ is about 10 MPa/km ${\displaystyle \times }$ 1.4 km = 14 MPa, while the pressure on the matrix ${\displaystyle {\mathcal {P}}_{m}\approx 22.5\times 1.4=32\ {\rm {MPa}}}$, the differential pressure being 18 MPa. The porosity will be about 30% (see Figure 5.1a). If the shale is cut off at a depth of 1.4 km and the depth is then increased to 2.0 km, the differential pressure and velocity will not change greatly.

At 2.0 km, the overburden pressure ${\displaystyle {\mathcal {P}}_{m}=22.5\times 2.0=45\ {\rm {MPa}}}$, the fluid pressure will be about ${\displaystyle {\mathcal {P}}_{f}\approx 45-18=27\ {\rm {MPa}}}$. At 3.0 km, ${\displaystyle {\mathcal {P}}_{m}\approx 22.5\times 3.0=68}$ MPa and ${\displaystyle {\mathcal {P}}_{f}\approx 68-18=50\ {\rm {Mpa}}}$.

Figure 5.6a.  Velocity/depth relations (Venezuela data from Gregory, 1977). Triangles show data from problem 5.8, squares are from problem 5.9b.

5.6b Assume a shale buried to 3.00 km and then uplifted to 2.00 km, being normally pressured all the time. What velocity and fluid pressure would you expect?

Solution

The velocity at 3.00 km is about 3.2 km/s from Figure 5.6a and the fluid pressure about 30 MPa. Burial to this depth will have reduced the porosity to about 25% (see Figure 5.1a) and very little of the porosity will be recovered upon uplift. Hence the velocity will be larger than for a situation where the rock had not been buried deeper.

5.6c Assume the shale in part (a) is buried to 3.00 km and then uplifted to 2.00 km without fluid communication being established. What velocity and fluid pressure would you expect? What if the shale is uplifted to 1.00 km?

Solution

Since the differential pressure was not changed, we expect about the same result at 2.00 km as in part (a). At 2.00 km, ${\displaystyle {\mathcal {P}}_{f}\approx 20\ {\rm {MPa}}}$. When the shale is then uplifted to 1.40 km, its pressure will again be normal. If further uplifted to 1.00 km without fluid communication being established, the porosity will remain about 32%. The velocity and differential pressure will be about the same as they were at 1.40 km, but ${\displaystyle {\mathcal {P}}_{f}}$ will now be ${\displaystyle 22.5\times 1.00-18=4.5\ {\rm {MPa}}}$ (instead of the usual value of 10 Mpa); the shale will be under-pressured and will have approximately the same velocity as at 1.40 km.

## 5.7 Determining lithology from well-velocity surveys

By comparing the velocity-depth relations in Figures 5.6a and 5.7a, what can you deduce about the nature of the rocks in the well for Figure 5.7a?

Figure 5.7a.  Plot of well-velocity survey.

Solution

We superimpose the sand-shale curve from Figure 5.6a on Figure 5.7a (the dashed line) and note that the data are nearly parallel. It appears that the geologic section represented by Figure 5.7a is normally pressured, mainly clastics, and has not been uplifted significantly (i.e., the velocity is about what is expected for normal compaction).

## 5.8 Reflectivity versus water saturation

What shale velocities are consistent with the Figure 5.8a oil-sand data?

Solution

We determine velocities and reflectivities for two values of water saturation for each of the three depths, neglecting differences between sand and shale densities. Setting ${\displaystyle \rho _{ss}=\rho _{sh}}$, equation (3.6a) becomes {\displaystyle {\begin{aligned}R=\left(V_{ss}-V_{sh}\right)/\left(V_{ss}+V_{sh}\right),\\V_{sh}=V_{ss}\left(1-R\right)/\left(1+R\right).\end{aligned}}}

Figure 5.8a.  Effect of water saturation on velocity (i) and reflectivity (ii) (after Domenico, 1974). Solid curves are for gaseous, dashed curves for liquid hydrocarbons.
Table 5.8a. Oil-sand data from Figure 5.8a.
Depth Water saturation ${\displaystyle R}$ ${\displaystyle V_{ss}}$ ${\displaystyle V_{sh}}$
610 m 30% –0.08 1.7 km/s 2.0 km/s
70 –0.03 1.8 1.9
1830 30 –0.10 2.3 2.8
70 –0.07 2.4 2.8
3050 30 –0.10 2.7 3.3
70 –0.07 2.8 3.2

From Figures 5.8a(i,ii) we obtain the data in Table 5.8a for oil sands. These are plotted as triangles on Figure 5.6a.

## 5.9 Effect of overpressure

5.9a Figure 5.6a shows velocity versus depth for normally pressured shales. How do the Figure 5.9a velocities above and below the top of the abnormal pressure zone compare with the sand and shale velocity-depth curves of Figure 5.6a? What depth corresponds to normal pressure for the top overpressure? What porosity would you expect for the overpressured shale?

Solution

The transit times above and below the “top of the overpressure zone” are roughly ${\displaystyle (125\ \mu s/{\hbox{ft}})}$ and ${\displaystyle (175\ \mu s/{\hbox{ft}})}$, respectively, that is, velocities of 8.0 and 5.7 kft/s (2.4 and 1.7 km/s). According to Figure 5.6a, the shale velocity at 6000 ft (2.1 km) should be higher (about 8.8 kft/s); the shale is probably somewhat undercompacted. The 5.7 kft/s velocity of the overpressured shale corresponds to a depth of about 1000 ft (300 m) and the porosity of the overpressured shale is probably close to 50%.

5.9b Plot the velocities for 100% water saturated sands from Figure 5.8a(i) on Figure 5.6a. How do they compare?

Solution

The values for 100% water saturated sands from Figure 5.8a(i) are listed below and plotted as triangles on Figure 5.6a. They lie slightly below the Gulf Coast sand curve.

 Depth 0.61 km 1.83 3.05 Velocity 1.9 km/s 2.5 3

The data from Figure 5.9a are plotted as squares on Figure 5.6a.

## 5.10 Effects of weathered layer (LVL) and permafrost

5.10a Assume that raypaths have angles of approach of ${\displaystyle 10^{\circ }}$, ${\displaystyle 20^{\circ }}$, ${\displaystyle 30^{\circ }}$, ${\displaystyle 40^{\circ }}$, and ${\displaystyle 60^{\circ }}$ in the subweathering where the velocity is 2400 m/s. For a weathered layer 10 m thick with velocity 500 m/s, how do travel-times through the weathered layer compare with that for a vertically traveling ray? What are the horizontal components of the raypaths in the LVL?

Solution

Referring to Figure 5.10a,

{\displaystyle {\begin{aligned}\sin \theta _{1}=\left(V_{1}/V_{2}\right)\sin \theta _{2}=0.208\sin \theta _{2},\\\Delta t=0.010/\left(0.500\cos \theta _{1}\right){\rm {s,}}\\\Delta x=\left(10\tan \theta _{1}\right){\rm {m.}}\end{aligned}}}

Figure 5.9a.  Effect of overpressure on sonic and resistivity logs for an offshore Gulf of Mexico well (from MacGregor, 1965).
Table 5.10a. Traveltimes and ${\displaystyle \Delta x}$ for raypaths in the LVL and permafrost.
Incident In low-velocity layer In permafrost
${\displaystyle \theta _{2}}$ ${\displaystyle \theta _{1}}$ ${\displaystyle \Delta x}$ (m) ${\displaystyle \Delta t}$ (ms) ${\displaystyle \theta _{1}}$ ${\displaystyle \Delta x}$ (m) ${\displaystyle \Delta t}$ (ms)
${\displaystyle 0^{\circ }}$ ${\displaystyle 0.0^{\circ }}$ 0.0 20.0 ${\displaystyle 0.0^{\circ }}$ 0 27.8
${\displaystyle 10^{\circ }}$ ${\displaystyle 2.1^{\circ }}$ 0.4 20.0 ${\displaystyle 15.1^{\circ }}$ 27 28.8
${\displaystyle 20^{\circ }}$ ${\displaystyle 4.1^{\circ }}$ 0.7 20.1 ${\displaystyle 30.9^{\circ }}$ 60 32.4
${\displaystyle 30^{\circ }}$ ${\displaystyle 6.0^{\circ }}$ 1.1 20.1 ${\displaystyle 48.6^{\circ }}$ 113 42.0
${\displaystyle 40^{\circ }}$ ${\displaystyle 7.7^{\circ }}$ 1.4 20.1 ${\displaystyle 74.6^{\circ }}$ 363 104.6
${\displaystyle 42^{\circ }}$ ${\displaystyle 8.0^{\circ }}$ 1.4 20.1 ${\displaystyle 90.0^{\circ }}$
${\displaystyle 60^{\circ }}$ ${\displaystyle 10.4^{\circ }}$ 1.8 20.3

Substituting the values of ${\displaystyle \theta _{2}}$, we get the results in Table 5.10a. The traveltimes in the LVL vary by only 0.5% over most of the range of ${\displaystyle \theta _{2}}$, and, even for ${\displaystyle \theta _{2}=60^{\circ }}$, change by only 1.5%.

5.10b For permafrost 100 m thick with a velocity of 3600 m/s, answer the questions in part (a).

Solution

We repeat the calculations of part (a) changing ${\displaystyle V_{1}}$ to 3.60 km/s and layer thickness to 100 m. The results are also shown in Table 5.10a. Because rays now have large horizontal components, the changes in ${\displaystyle \Delta x}$ and ${\displaystyle \Delta t}$ are considerable. This large ray bending makes corrections for permafrost very difficult. If ${\displaystyle \theta _{2}>42^{\circ }}$, upcoming waves are totally reflected.

Figure 5.10a.  Raypath bending at LV base.

## 5.11 Horizontal component of head waves

In the early days of refraction exploration for salt domes, sketches of expected raypaths indicated that the angle of approach to the surface should have a large horizontal component, but measurements showed very little horizontal component. Controversy arose over whether the travelpaths could be as drawn. Explain the apparent discrepancy.

Solution

The raypath bending at the base of the low-velocity layer made travel through the LVL nearly vertical, as illustrated in problem 5.10a so the horizontal component at the surface was very small.

## 5.12 Stacking velocity versus rms and average velocities

5.12a Assume six horizontal layers, each 300 m thick and with constant velocity (Figure 5.12a). The successive layers have velocities of 1.5, 1.8, 2.1, 2.4, 2.7, and 3.0 km/s. Ray-trace through the model to determine offset distances and arrival times for rays that make angles of incidence at the base of the 3.0 km/s layer of ${\displaystyle 0^{\circ }}$, ${\displaystyle 10^{\circ }}$, ${\displaystyle 20^{\circ }}$, and ${\displaystyle 30^{\circ }}$. Calculate stacking velocity for each angle and compare with the average velocity ${\displaystyle {\bar {V}}}$ and the rms velocity ${\displaystyle V_{\rm {rms}}}$.

Background

Average velocity ${\displaystyle ({\bar {V}})}$ and rms velocity ${\displaystyle (V_{\rm {rms}})}$ were discussed in problem 4.13 [see equations (4.13a,b)].

In the common-midpoint (CMP) technique, a number of traces are obtained with different source-geophone distances (offsets, see problem 4.1) but the same midpoint. After correcting for NMO (and for dip if necessary), they are added together (stacked), the number of traces added together being the multiplicity. The velocity used to remove the NMO is the stacking velocity ${\displaystyle V_{s}}$. If we use equation (4.1c) to remove the NMO, that is, if we assume a single horizontal constant-velocity layer, the velocity ${\displaystyle V}$ in equations (4.1a,c) becomes ${\displaystyle V_{s}}$. The ${\displaystyle x^{2}-t^{2}}$ plot of equation (4.1a) is a straight line with slope ${\displaystyle 1/V_{s}^{2}}$; thus,

 {\displaystyle {\begin{aligned}V_{s}^{2}=x^{2}/\left(t^{2}-t_{o}^{2}\right)\approx x^{2}/\left(2t\Delta t\right)\;,\;V_{s}\approx x/(2t\Delta t)^{1/2},\end{aligned}}} (5.12a)

where ${\displaystyle t_{o}}$ and ${\displaystyle t}$ are the two-way traveltimes at the origin and at offset ${\displaystyle x}$ while ${\displaystyle \Delta t=t-t_{o}}$. When the velocity changes with depth, the ${\displaystyle x^{2}-t^{2}}$ plot is curved but the curvature is generally small enough that the best-fit straight line gives reasonably accurate results. For horizontal velocity layering and small offsets, ${\displaystyle V_{s}\approx V_{\rm {rms}}}$.

Figure 5.12a.  Model of 300-m-thick layers.

Solution

We use Snell’s law to calculate the raypath angles ${\displaystyle \theta _{i}}$ in each layer. The two-way time in a layer is ${\displaystyle t_{i}=600/V_{i}\cos \theta _{i}}$ and the offset in a layer is ${\displaystyle x_{i}=600\tan \theta _{i}}$. The values in Table 5.12a have been calculated without regard to the number of significant figures to illustrate the sensitivity of the calculations. The average velocities ${\displaystyle {\bar {V}}}$ along the respective raypaths have also been calculated for comparisons.

The calculations for the intermediate layer boundaries assume that reflections are generated at each boundary. Traveltime differences, shown in parentheses in Table 5.12b, are very small for most of the situations, and, especially where the differences are less than 20 ms, are not very reliable for calculating ${\displaystyle V_{s}}$. A general rule for ${\displaystyle V_{s}}$ calculations, that the offset should be comparable to the depth, is not reached for any of these situations.

Table 5.12a. Calculation of ${\displaystyle V_{s}}$, ${\displaystyle {\bar {V}}}$, and ${\displaystyle V_{\rm {rms}}.}$
layer 1 layer 2 layer 3 layer 4 layer 5 layer 6
${\displaystyle \theta _{i}}$ ${\displaystyle 0^{\circ }}$ ${\displaystyle 0^{\circ }}$ ${\displaystyle 0^{\circ }}$ ${\displaystyle 0^{\circ }}$ ${\displaystyle 0^{\circ }}$ ${\displaystyle 0^{\circ }}$
${\displaystyle t_{i}\ (s)}$ 0.400 0.333 0.286 0.250 0.222 0.200
${\displaystyle \Sigma t_{i}\ (s)}$ 0.400 0.733 1.019 1.269 1.491 1.691
${\displaystyle x_{i}\ (m)}$ 0 0 0 0 0 0
${\displaystyle {\bar {V}}\ (m/s)}$ 1500 1640 1770 1890 2010 2130
${\displaystyle V_{\rm {rms}}\ (m/s)}$ 1500 1640 1780 1920 2060 2190
${\displaystyle t_{o}^{2}({\rm {s}}^{2})}$ 0.1600 0.5373 1.0384 1.6104 2.2231 2.8595
${\displaystyle \theta _{i}}$ ${\displaystyle 5.0^{\circ }}$ ${\displaystyle 6.0^{\circ }}$ ${\displaystyle 7.0^{\circ }}$ ${\displaystyle 8.0^{\circ }}$ ${\displaystyle 9.0^{\circ }}$ ${\displaystyle 10.0^{\circ }}$
${\displaystyle t_{i}\ (s)}$ 0.402 0.325 0.288 0.252 0.225 0.203
${\displaystyle \Sigma t_{i}\ (s)}$ 0.402 0.737 1.025 1.277 1.502 1.705
${\displaystyle x_{i}\ (m)}$ 52 63 73 84 95 106
${\displaystyle \Sigma x\ (m)}$ 52 105 189 273 368 473
${\displaystyle \Delta t\ (s)}$ 0 0.077 0.111 0.143 0.181 0.218
${\displaystyle V_{s-10}\ (m/s)}$ * 1400* 1700* 1900* 2029 2170
${\displaystyle {\bar {V}}_{10}\ (m/s)}$ 1500 1636 1767 1892 2013 2131
${\displaystyle \theta _{i}}$ ${\displaystyle 9.8^{\circ }}$ ${\displaystyle 11.8^{\circ }}$ ${\displaystyle 13.9^{\circ }}$ ${\displaystyle 15.9^{\circ }}$ ${\displaystyle 17.9^{\circ }}$ ${\displaystyle 20.0^{\circ }}$
${\displaystyle t_{i}\ (s)}$ 0.406 0.340 0.294 0.260 0.224 0.213
${\displaystyle \Sigma t_{i}\ (s)}$ 0.406 0.746 1.041 1.301 1.534 1.747
${\displaystyle x_{i}\ (m)}$ 104 126 146 171 194 218
${\displaystyle \Sigma x\ (m)}$ 104 230 378 549 743 961
${\displaystyle \Delta t\ (s)}$ 0.0695 0.1386 0.2128 0.2867 0.3606 0.4388
${\displaystyle V_{s-20}\ (m/s)}$ 1496 1658 1775 1914 2060 2190
${\displaystyle {\bar {V}}_{20}\ (m/s)}$ 1500 1637 1768 1894 2017 2137
${\displaystyle \theta _{i}}$ ${\displaystyle 14.5^{\circ }}$ ${\displaystyle 17.5^{\circ }}$ ${\displaystyle 20.5^{\circ }}$ ${\displaystyle 23.6^{\circ }}$ ${\displaystyle 26.7^{\circ }}$ ${\displaystyle 30.0^{\circ }}$
${\displaystyle t_{i}\ (s)}$ 0.413 0.349 0.305 0.273 0.248 0.231
${\displaystyle \Sigma t_{i}\ (s)}$ 0.413 0.762 1.067 1.340 1.589 1.820
${\displaystyle x_{i}\ (m)}$ 155 189 224 262 302 346
${\displaystyle \Sigma x\ (m)}$ 155 344 568 830 1132 1478
${\displaystyle \Delta t\ (s)}$ 0.103 0.208 0.316 0.430 0.549 0.673
${\displaystyle V_{s-30}\ (m/s)}$ 1508 1652 1795 1928 2060 2196
${\displaystyle {\bar {V}}_{30}\ (m/s)}$ 1500 1637 1770 1898 2024 2147
 *Not enough significant figures to calculate with sufficient accuracy.
Table 5.12b. Calculated values of ${\displaystyle V_{\rm {rms}},}$ ${\displaystyle V_{s}}$, and ${\displaystyle {\bar {V}}.}$
layer 1 layer 2 layer 3 layer 4 layer 5 layer 6
${\displaystyle {\bar {V}}\ (m/s)}$ 1500 1635 1770 1890 2010 2130
${\displaystyle V_{\rm {rms}}\ (m/s)}$ 1500 1643 1780 1920 2060 2190
Stacking velocity calculations:
${\displaystyle V_{s-10}\ (m/s)}$ * (0) 1370 (4) 1707 (6) 1924 (8) 2029 (9) 2170 (14)
${\displaystyle V_{s-20}\ (m/s)}$ 1496 (6) 1658 (13) 1775 (22) 1914 (32) 2060 (43) 2190 (56)
${\displaystyle V_{s-30}\ (m/s)}$ 1508 (13) 1652 (29) 1795 (48) 1928 (71) 2060 (98) 2196 (129)
Average velocity along raypaths:
${\displaystyle {\bar {V}}_{10}\ (m/s)}$ 1500 1636 1767 1892 2013 2131
${\displaystyle {\bar {V}}_{20}\ (m/s)}$ 1500 1637 1768 1894 2017 2137
${\displaystyle {\bar {V}}_{30}\ (m/s)}$ 1500 1637 1770 1898 2024 2147
 *Not enough significant figures to calculate with sufficient accuracy. Values in parentheses are traveltime differences.

Table 5.12c. Calculation of raypaths for dipping layers.
layer 1 layer 2 layer 3 layer 4 layer 5 layer 6
${\displaystyle \theta _{i}}$ ${\displaystyle 10.0^{\circ }}$ ${\displaystyle 12.0^{\circ }}$ ${\displaystyle 14.1^{\circ }}$ ${\displaystyle 16.1^{\circ }}$ ${\displaystyle 18.2^{\circ }}$ ${\displaystyle 20.0^{\circ }}$
${\displaystyle t_{i}\ (s)}$ 0.400 0.333 0.286 0.250 0.222 0.200
${\displaystyle x_{i}\ (m)}$ 106 128 151 173 197 221
${\displaystyle \theta _{i}}$ ${\displaystyle 20.0^{\circ }}$ ${\displaystyle 24.2^{\circ }}$ ${\displaystyle 28.6^{\circ }}$ ${\displaystyle 33.2^{\circ }}$ ${\displaystyle 38.0^{\circ }}$ ${\displaystyle 43.1^{\circ }}$
${\displaystyle t_{i}\ (s)}$ 0.426 0.366 0.326 0.303 0.288 0.274
${\displaystyle x_{i}\ (m)}$ 219 270 328 393 469 561
${\displaystyle \theta _{i}}$ ${\displaystyle 30.0^{\circ }}$ ${\displaystyle 37.0^{\circ }}$ ${\displaystyle 44.5^{\circ }}$ ${\displaystyle 54.0^{\circ }}$ ${\displaystyle 64.1^{\circ }}$ ${\displaystyle 90.0^{\circ }}$
${\displaystyle t_{i}\ (s)}$ 0.462 0.417 0.401 0.424 0.511 *
${\displaystyle x_{i}\ (m)}$ 346 452 591 620 1245 *
 *A head wave is generated at the base of layer 5.

Table 5.12d. Calculation of stacking velocity ${\displaystyle V_{s}}$ (m/s).
${\displaystyle \theta _{i}}$ layer 1 layer 2 layer 3 layer 4 layer 5 layer 6
${\displaystyle 10^{\circ }}$ 1500 1630 1750 1860 1960 2050
${\displaystyle 20^{\circ }}$ 1500 1640 1750 1880 1990 2110
${\displaystyle 30^{\circ }}$ 1500 1650 1790 1920 2100 *

We note that the stacking velocity ${\displaystyle V_{s}}$ increases with the offset ${\displaystyle x}$. The calculated velocities are summarized in Table 5.12b.

5.12b Repeat part (a) for the case where rays make angles of incidence at the free surface of ${\displaystyle 0^{\circ }}$, ${\displaystyle 10^{\circ }}$, ${\displaystyle 20^{\circ }}$ and ${\displaystyle 30^{\circ }}$.

Solution

Figure 5.12b.  Dipping model.

The case where ${\displaystyle \theta _{1}=0^{\circ }}$ is the same as that for ${\displaystyle \theta _{6}=0^{\circ }}$ so that we need to calculate only for ${\displaystyle \theta _{1}=10^{\circ },20^{\circ },30^{\circ };}$ the results are given in Table 5.12c.

We now calculate a stacking velocity for reflections for each layer for each of the angles (Table 5.12d).

As before, we note that the stacking velocity ${\displaystyle V_{s}}$ increases with the offset ${\displaystyle x}$.

5.12c Assume the 300-m-thick layers dip ${\displaystyle 20^{\circ }}$ as shown in Figure 5.12b and determine arrival times for a zero-offset ray and one that leaves the free surface at an angle of ${\displaystyle 30^{\circ }}$ and is reflected at ${\displaystyle B}$.

Solution

The raypath for a zero-offset trace makes a ${\displaystyle 20^{\circ }}$ angle in the updip direction at the surface and ${\displaystyle 0^{\circ }}$ angles at all of the interfaces so that after reflection the raypath will return to the sourcepoint. The traveltimes are the same as calculated in part (a).

A ray that leaves the free surface at ${\displaystyle 30^{\circ }}$ in the updip direction is incident on the ${\displaystyle V_{1}/V_{2}}$ interface at ${\displaystyle 10^{\circ }}$ and thus makes the same angles with other interfaces as calculated for the ${\displaystyle 10^{\circ }}$ case in part (b). The time spent in each of the layers will also be the same as in part (b) but the ${\displaystyle x_{i}}$ distances are now measured along the bedding planes. Thus, to determine the locations of the source and the emergent location, these have to be corrected by ${\displaystyle \cos 20^{\circ }}$ The geometry is shown in Figure 5.12c. We have from part (b), e = 435 m, g = 488 m, one-way time from top of ${\displaystyle V_{1}}$ layer to ${\displaystyle B=0.875\ {\rm {s}}}$, time from ${\displaystyle B}$ to the base of ${\displaystyle V_{1}}$ layer 0.672 s.

Figure 5.12c.  Geometry of problem 5.12c.

{\displaystyle {\begin{aligned}{\rm {a}}+\ {\rm {b}}=300/\sin 20^{\circ }=877\ {\rm {m}},\\{\rm {c}}=\ {\rm {g}}\left(\sin 110^{\circ }/\sin 60^{\circ }\right)=488\left(0.934/0.866\right)=526\ {\rm {m}},\\{\rm {k}}=\ {\rm {g}}\left(\sin 20^{\circ }/\sin 60^{\circ }\right)=488\left(0.342/0.866\right)=193\ {\rm {m}},\\\Delta t_{1}={\rm {k}}/1500=0.128\ {\rm {s}},\\{\rm {e}}+\ {\rm {f}}=300/\tan 20^{\circ }=300/0.364=824\ {\rm {m}},\\{\rm {f}}=824-435=389\ {\rm {m}},\\{\rm {j}}={\rm {f}}\left(\sin 20^{\circ }/\sin 80^{\circ }\right)=389\left(0.342/0.985\right)=135\ {\rm {m}},\\\Delta t_{2}={\rm {j}}/1500=0.090\ {\rm {s}},\\{\rm {b}}=877-389=488\ {\rm {m}},\\{\rm {traveltime}}=t=0.875+\Delta t_{1}+672+\Delta t_{2}=1.765\ {\rm {s}},\\{\rm {source{\text{-}}receiver\ distance}}={\rm {b}}+{\rm {c}}=488+526=1014\ {\rm {m}}.\end{aligned}}}

The source is farther from the zero-offset location than the emergent point, so that the data are not suitable for stacking velocity calculations unless a DMO correction (Sheriff and Geldart, 1995, section 9.10.2) has been applied. Calculating arrival times for dipping reflections for split-dip situations is often done by trial and error.

## 5.13 “Quick-look” velocity analysis; Effects of errors

5.13a Velocity analysis usually results in a plot of stacking velocity against traveltime. Bauer (private communication) devised a “quick look” method of determining the interval velocity, assuming horizontal layering and that the stacking velocity equals the average velocity. The method is shown in Figure 5.13a. A box is formed by the two picks between which the interval velocity is to be picked; the diagonal that does not contain the two picks when extended to the velocity axis gives the interval velocity. Prove that the method is valid and discuss its limitations.

Figure 5.13a.  Interval velocity determination.

Solution

We extend the diagonal of the box as shown in Figure 5.13a, thus giving ${\displaystyle V_{m}}$. The interval velocity is given by

 {\displaystyle {\begin{aligned}V_{i}=\Delta z/\Delta t=\left(V_{2}t_{2}-V_{1}t_{1}\right)/\left(t_{2}-t_{1}\right)\\=\left[\left(V_{1}+\Delta V\right)t_{2}-V_{1}t_{1}\left]/\Delta t=\right[V_{1}\left(t_{2}-t_{1}\right)+t_{2}\Delta V\right]/\Delta t\\=\left(V_{1}\Delta t+t_{2}\Delta V\right)/\Delta t=V_{1}+t_{2}\left(\Delta V/\Delta t\right).\end{aligned}}} (5.13a)

In Figure 5.13a the triangle with apices at the points ${\displaystyle V_{m}}$, ${\displaystyle V_{1}}$, and ${\displaystyle A}$ is similar to the triangle with sides ${\displaystyle \Delta V}$ and ${\displaystyle \Delta t}$, so we have

{\displaystyle {\begin{aligned}\Delta V/\Delta t=\left(V_{m}-V_{1}\right)/t_{2}.\end{aligned}}}

Substituting in equation (5.13a), we get

{\displaystyle {\begin{aligned}V_{i}=V_{1}+\left(V_{m}-V_{1}\right)=V_{m}.\end{aligned}}}

Thus the method gives the interval velocity provided the stacking velocity equals the average velocity. For horizontal velocity layering the stacking velocity is often about 2% higher than the average velocity, but the two may differ considerably if the reflectors are dipping.

5.13b This method can be used to see the influence of measurement error. Discuss the sensitivity of interval-velocity calculations to

1. errors in picking velocity values from this graph,
2. errors in picking times,
3. picking events very close together, and
4. picking events late.

Solution

1. Errors in ${\displaystyle V_{1}}$ or ${\displaystyle V_{2}}$ change the slope of the diagonal and hence change ${\displaystyle V_{m}}$ and ${\displaystyle V_{i}}$; the error is proportional to ${\displaystyle t/\left(t_{2}-t_{1}\right)}$ where ${\displaystyle t}$ is either ${\displaystyle t_{1}}$ or ${\displaystyle t_{2}}$.
2. Changes in ${\displaystyle t_{1}}$ or ${\displaystyle t_{2}}$ has an effect similar to that in (i).
3. When both ${\displaystyle \Delta V}$ and ${\displaystyle \Delta t}$ are small, the slope of the diagonal and ${\displaystyle V_{i}}$ are very sensitive to errors.
4. Picking each event late by the same amount ${\displaystyle \Delta \tau }$ will increase ${\displaystyle V_{i}}$ by an amount proportional to ${\displaystyle \Delta \tau .}$

We now derive mathematical expressions for the changes in (i) to (iv).

i) Equation (5.13a) is

{\displaystyle {\begin{aligned}V_{1}=(V_{2}t_{2}-V_{1}t_{1})/(t_{2}-t_{1}).\\{\rm {Thus}},\qquad \qquad \qquad \qquad {\rm {d}}V_{i}/{\rm {d}}V_{1}=-t_{1}/(t_{2}-t_{1})=-t_{1}/\Delta t,\\{\rm {and}}\qquad \qquad \qquad \qquad {\rm {d}}V_{i}/{\rm {d}}V_{2}=t_{2}/(t_{2}-t_{1})=t_{2}/\Delta t.\end{aligned}}}

Hence the error is directly proportional to either ${\displaystyle t_{1}}$ or ${\displaystyle t_{2}}$ and inversely proportional to ${\displaystyle (t_{2}-t_{1})}$; it increases rapidly as ${\displaystyle t_{1}}$ approaches ${\displaystyle t_{2}}$.

ii) {\displaystyle {\begin{aligned}{\frac {{\rm {d}}V_{i}}{{\rm {d}}t_{1}}}=-{\frac {V_{1}}{\left(t_{2}-t_{1}\right)}}+{\frac {\left(V_{2}t_{2}-V_{1}t_{1}\right)}{(t_{2}-t_{1})^{2}}}={\frac {-V_{1}\left(t_{2}-t_{1}\right)+\left(V_{2}t_{2}-V_{1}t_{1}\right)}{(t_{2}-t_{1})^{2}}}\\={\frac {(V_{2}-V_{1})t_{2}}{(t_{2}-t_{1})^{2}}}.\end{aligned}}}

Likewise,

{\displaystyle {\begin{aligned}{\frac {{\rm {d}}V_{i}}{{\rm {d}}t_{2}}}={\frac {-\left(V_{2}-V_{1}\right)t_{1}}{(t_{2}-t_{1})^{2}}}.\end{aligned}}}

Since ${\displaystyle t_{2}>t_{1}}$, errors in ${\displaystyle t_{1}}$ are more serious than errors in ${\displaystyle t_{2}}$; also, the errors increase rapidly as ${\displaystyle t_{1}}$ approaches ${\displaystyle t_{2}}$.

iii) Let ${\displaystyle t_{2}=t_{1}+\Delta t}$, ${\displaystyle \Delta t\approx 0}$. Then using equation (5.13a),

{\displaystyle {\begin{aligned}V_{t}=\left[V_{2}\left(t_{1}+\Delta t\right)-V_{1}t_{1}\right]/\Delta t=V_{2}+\left(V_{2}-V_{1}\right)\left(t_{1}/\Delta t\right).\end{aligned}}}

In general ${\displaystyle t_{1}\gg \Delta t}$, so the error in ${\displaystyle V_{i}}$ depends mainly on the factor ${\displaystyle t_{1}/\Delta t}$.

iv) We assume that both events are late by the same amount ${\displaystyle \Delta \tau }$. Then equation (5.13a) becomes

{\displaystyle {\begin{aligned}V_{i}+\Delta V_{i}=\left[V_{2}\left(t_{2}+\Delta \tau \right)-V_{1}\left(t_{1}+\Delta \tau \right)\right]/\left(t_{2}-t_{1}\right)\\=V_{i}+\left(V_{2}-V_{1}\right)\Delta \tau /\left(t_{2}-t_{1}\right),\\{\rm {so}}\qquad \qquad \qquad \qquad \Delta V_{i}=\Delta \tau \left[\left(V_{2}-V_{1}\right)/\left(t_{2}-t_{1}\right)\right]=\Delta \tau \tan \zeta ,\end{aligned}}}

where ${\displaystyle \tan \zeta }$ is usually fairly constant over a moderate range of depths; in this case the error in ${\displaystyle V_{i}}$ is proportional to the error ${\displaystyle \Delta \tau .}$

## 5.14 Well-velocity survey

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.

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

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.

## 5.15 Interval velocities from ${\displaystyle X^{2}-T^{2}}$ analyses

An ${\displaystyle X^{2}-T^{2}}$ survey gives the stacking velocity results in Table 5.15a. Calculate the interval velocities.

Table 5.15a. ${\displaystyle X^{2}-T^{2}}$ survey results.
${\displaystyle i}$ ${\displaystyle z\ ({\hbox{km}})}$ ${\displaystyle t_{i}\ ({\hbox{s}})}$ ${\displaystyle V_{s}\ ({\hbox{km/s}})}$
1 1.20 1.100 2.18
2 2.50 1.786 2.80
3 3.10 1.935 3.20
4 4.10 2.250 3.64

Background

In equation (4.13a), the sum ${\displaystyle \mathop {\sum } \nolimits _{i=1}^{n}\Delta t_{i}}$ is the traveltime down to the base of the ${\displaystyle n^{th}}$ layer, that is, ${\displaystyle t_{n}}$. Therefore equation (4.13a) can be written

 {\displaystyle {\begin{aligned}V_{\rm {rms}}^{2}t_{n}=\mathop {\sum } \limits _{i=1}^{n}V_{i}^{2}\Delta t_{i}.\end{aligned}}} (5.15a)

If we write ${\displaystyle V_{L}}$ and ${\displaystyle t_{L}}$ for the rms velocity and traveltime down to the base of the ${\displaystyle n^{th}}$ layer, ${\displaystyle V_{U}}$ and ${\displaystyle t_{U}}$ for the same quantities down to the top of the layer, we can get an equation for the interval velocity ${\displaystyle V_{n}}$ in the ${\displaystyle n^{th}}$ bed by subtracting expressions for ${\displaystyle V_{L}}$ and ${\displaystyle V_{U}}$ obtained from equation (5.15a). The result is

 {\displaystyle {\begin{aligned}V_{n}^{2}=\left(V_{L}^{2}t_{L}-V_{U}^{2}t_{U}\right)/\Delta t_{n}.\end{aligned}}} (5.15b)

Solution

We assume horizontal velocity layering so that the values of ${\displaystyle V_{s}}$ in Table 5.15a are approximate values of ${\displaystyle V_{\rm {rms}}}$ (see Table 5.12b) and therefore we can use equation (5.15b) to find interval velocities. We take 2.18 km/s as the interval velocity from the surface to the first reflector. We now calculate the interval velocities ${\displaystyle V_{2}}$ to ${\displaystyle V_{4}}$:

{\displaystyle {\begin{aligned}V_{2}^{2}=\left(V_{L}^{2}t_{L}-V_{U}^{2}t_{U}\right)/\Delta t_{2}=\left(2.80^{2}\times 1.786-2.18^{2}\times 1.100\right)/0.686,\\V_{2}^{2}=3.57\ {\rm {km/s}},\\V_{3}^{2}=\left(3.20^{2}\times 1.935-2.80^{2}\times 1.786\right)/0.149,\;V_{3}=6.25\ {\rm {km/s}},\\V_{4}^{2}=\left(3.64^{2}\times 2.250-3.20^{2}\times 1.935\right)/0.315\;,\;V_{4}=5.63\ {\rm {km/s}}.\end{aligned}}}

## 5.16 Finding velocity by the ${\displaystyle X^{2}-T^{2}}$ method

Determine the velocity by the ${\displaystyle X^{2}-T^{2}}$ method using the data given in Table 5.16a, ${\displaystyle t_{A}}$ being for a horizontal reflector and ${\displaystyle t_{B}}$ for a reflector dipping ${\displaystyle 10^{\circ }}$ toward the source. (The two reflections are observed on different records.)

Background

Using equation (4.3a), we have for a dipping horizon,

{\displaystyle {\begin{aligned}(Vt)^{2}=(x\cos \xi )^{2}+(2h_{c})^{2},\end{aligned}}}

where ${\displaystyle x}$ is the offset, ${\displaystyle \xi }$ is the dip, and ${\displaystyle h_{c}}$ the slant depth at the midpoint between source and receiver.

Table 5.16a. ${\displaystyle X^{2}-T^{2}}$ data.
${\displaystyle x}$ (km) ${\displaystyle t_{A}}$ (s) ${\displaystyle t_{B}}$ (s) ${\displaystyle x}$ (km) ${\displaystyle t_{A}}$ (s) ${\displaystyle t_{B}}$ (s) ${\displaystyle x}$ (km) ${\displaystyle t_{A}}$ (s) ${\displaystyle t_{B}}$ (s)
0.0 0.855 0.906 1.4 1.005 0.977 2.8 1.330 1.202
0.1 0.856 0.902 1.5 1.017 0.991 2.9 1.360 1.234
0.2 0.858 0.898 1.6 1.037 1.004 3.0 1.404 1.253
0.3 0.864 0.898 1.7 1.068 1.019 3.1 1.432 1.272
0.4 0.868 0.899 1.8 1.081 1.037 3.2 1.457 1.296
0.5 0.874 0.902 1.9 1.105 1.058 3.3 1.487 1.304
0.6 0.882 0.903 2.0 1.118 1.066 3.4 1.513 1.334
0.7 0.892 0.909 2.1 1.151 1.083 3.5 1.548 1.356
0.8 0.900 0.916 2.2 1.166 1.102 3.6 1.580 1.377
0.9 0.906 0.922 2.3 1.203 1.121 3.7 1.610 1.407
1.0 0.930 0.932 2.4 1.237 1.127 3.8 1.649 1.415
1.1 0.945 0.943 2.5 1.255 1.158 3.9 1.674 1.438
1.2 0.950 0.950 2.6 1.283 1.177 4.0 1.708 1.459
1.3 0.979 0.965 2.7 1.304 1.195
Table 5.16b. Data used in ${\displaystyle X^{2}-T^{2}}$ calculations.
${\displaystyle x^{2}}$ ${\displaystyle t_{A}^{2}}$ (${\displaystyle x\cos \xi )^{2}}$ ${\displaystyle t_{B}^{2}}$ ${\displaystyle x^{2}}$ ${\displaystyle t_{A}^{2}}$ (${\displaystyle x\cos \xi )^{2}}$ ${\displaystyle t_{B}^{2}}$
0.00 0.731 0.00 0.821 4.41 1.325 4.28 1.173
0.01 0.733 0.01 0.814 4.84 1.360 4.69 1.214
0.04 0.736 0.04 0.806 5.29 1.447 5.13 1.257
0.09 0.746 0.09 0.806 5.76 1.530 5.59 1.270
0.16 0.753 0.16 0.808 6.25 1.575 6.06 1.341
0.25 0.764 0.24 0.814 6.76 1.646 6.56 1.385
0.36 0.778 0.35 0.815 7.29 1.700 7.07 1.426
0.49 0.796 0.48 0.826 7.84 1.769 7.60 1.445
0.64 0.817 0.62 0.839 8.41 1.850 8.16 1.523
0.81 0.821 0.79 0.850 9.00 1.971 8.73 1.570
1.00 0.865 0.97 0.869 9.61 2.051 9.32 1.618
1.21 0.893 1.17 0.889 10.24 2.123 9.93 1.680
1.44 0.902 1.40 0.902 10.89 2.211 10.56 1.700
1.69 0.968 1.64 0.931 11.56 2.289 11.21 1.780
1.96 1.010 1.90 0.965 12.25 2.396 11.88 1.839
2.25 1.034 2.18 0.962 12.96 2.496 12.57 1.896
2.56 1.075 2.48 1.008 13.69 2.592 13.28 1.980
2.89 1.141 2.80 1.030 14.43 2.719 14.00 2.002
3.24 1.169 3.14 1.075 15.21 2.802 14.75 2.068
3.61 1.221 3.50 1.119 16.00 2.917 15.52 2.129
3.61 1.250 3.88 1.136
Figure 5.16a.  ${\displaystyle X^{2}-T^{2}}$ plot.

Solution

The values of ${\displaystyle (x^{2},\;t_{A}^{2})}$ and ${\displaystyle [(x\cos \xi )^{2},\;t_{B}^{2}]}$ are tabulated in Table 5.16b and plotted in Figure 5.16a. The best-fit lines in Figure 5.16a determined by eye give

{\displaystyle {\begin{aligned}V_{A}=2.71\ {\rm {km/s}},h_{A}=1.16\ {\rm {km}};V_{B}=3.35\ {\rm {km/s}},h_{B}=1.49\ {\rm {km}}.\end{aligned}}}

Using the least-squares method (see problem 9.33), we get

{\displaystyle {\begin{aligned}V_{A}=2.71\ {\rm {km/s}},h_{A}=1.15\ {\rm {km}};V_{B}=3.38\ {\rm {km/s}},h_{B}=1.51\ {\rm {km}}.\end{aligned}}}

## 5.17 Effect of timing errors on stacking velocity, depth, and dip

5.17a Given that the trace spacing in Figure 5.17a is 50 m, determine the stacking velocity, dip, and depth at approximately 0.5, 1.0, 1.5, 2.0, and 2.4 s.

Solution

Using an enlarged version of Figure 5.17a, we measured arrival times on the center traces and on traces symmetrically located left and right of the center, limiting the offsets to where we felt we could pick the events with confidence. The measurements give us ${\displaystyle t_{o}}$, ${\displaystyle \Delta t_{\rm {NMO}}}$, and ${\displaystyle \Delta t_{d}/\Delta x}$ and we must find ${\displaystyle z}$ and ${\displaystyle \xi }$; for this we need the velocity ${\displaystyle {\bar {V}}}$. We can find ${\displaystyle V_{s}}$ from equation (5.12a) and, since the dip is small, ${\displaystyle V_{s}\approx {\bar {V}}}$ and we can therefore use equation (4.2b) to get an approximate value of ${\displaystyle \xi }$. The equations are

Figure 5.17a.  A split-dip record (courtesy Grant Geophysical).
 ${\displaystyle t_{o}}$ (s) 0.533 0.975 1.575 2.008 2.417 offset (m) 575 m 1125 1425 1625 2375 ${\displaystyle t_{\rm {left}}}$ (s) 0.633 1.150 1.658 2.058 2.533 ${\displaystyle t_{\rm {right}}}$ (s) 0.608 1.142 1.758 2.150 2.575 ${\displaystyle t_{\rm {NMO}}}$ (s) 0.0875 0.171 0.133 0.096 0.137 ${\displaystyle V_{s}}$ (m/s) 1885 1950 2200 2615 2920 ${\displaystyle z}$ (m) 502 950 1730 2625 3530 ${\displaystyle \Delta t_{d}/\Delta x}$(ms/m) 0.0435 0.0071 0.0702 0.0566 0.0177 ${\displaystyle \sin \xi }$ 0.0410 0.0070 0.0772 0.0740 0.0258 ${\displaystyle \xi }$ ${\displaystyle 2.3^{\circ }}$ ${\displaystyle 0.4^{\circ }}$ ${\displaystyle 4.4^{\circ }}$ ${\displaystyle 4.2^{\circ }}$ ${\displaystyle 1.5^{\circ }}$

{\displaystyle {\begin{aligned}\Delta t_{\rm {NMO}}=(t_{\rm {left}}+t_{\rm {right}})/2-t_{o}\\V_{s}=x/(2t_{o}\Delta t_{\rm {NMO}})^{1/2},\\\sin \xi =(V_{s}/2)(\Delta t_{d}/\Delta x)\\z=V_{s}t_{o}/2.\end{aligned}}}

The calculated results for five reflections are shown in Table 5.17a.

5.17b What problems or ambiguities do you have in picking these events? How much uncertainty is there in your ability to pick times and how much uncertainty does this introduce into the velocity, depth, and dip calculations?

Solution

There are clearly different families of events interfering with each other on this record, which we have not attempted to sort out. The axes of symmetry of some of the data shift to the left with depth, indicating dip to the right. The event at 2.417 s may be a multiple. Clearly many more events could be picked.

We timed the centers of the black peaks, and this involves 5-10 ms ${\displaystyle (\approx 0.5-2\%)}$ uncertainty in this case. At a work station where a best-fit curve can be used to smooth-out noise, uncertainty can be reduced appreciably, and measurements can be accurate to 1 ms. Measurements of ${\displaystyle V_{s}}$ and ${\displaystyle \xi }$ are based on time differences and their errors are probably about 5%. If the offsets had been longer, measured differences would have been larger, giving better accuracy, but then uncertainties in event continuity and interference with other events might have increased the errors. In calculating depths, the onset of reflections should be measured so time measurements are probably 20 ms (1 to 4%) late. This may introduce 1% error in ${\displaystyle V_{s}}$, but other errors involved in ${\displaystyle V_{s}}$ are probably more important, including the assumption that it is the correct velocity to use. Dips are almost certainly underestimated by the use of ${\displaystyle V_{s}}$, which does not allow for the fact that the velocity at the reflector is usually larger than ${\displaystyle V_{s}}$ because of the usual increase of velocity with depth.

## 5.18 Estimating lithology from stacking velocity

5.18a A velocity analysis at SP 100 of Figure 5.18a yields the ${\displaystyle V_{s}-t}$ plot shown in Figure 5.18b. Pick stacking velocity versus time pairs and calculate interval velocities.

Figure 5.18a.  Unmigrated seismic section (courtesy Grant Geophysical).

Solution

Velocity-time pairs are listed in Table 5.18a. Depth ${\displaystyle z=V_{s}t/2}$ and interval velocities ${\displaystyle V_{i}=2\Delta z/\Delta t}$. In Table 5.18a, ${\displaystyle V_{i}}$ values are for the intervals above the reflection picks.

Table 5.17a. Time-velocity pairs.
${\displaystyle t}$ (s) ${\displaystyle V_{s}}$ (m/s) ${\displaystyle z}$ (m) ${\displaystyle V_{i}}$ (m/s)
0.22 2450 270 2450
0.38 2650 495 2800
0.58 2800 800 3050
0.72 2750 990 2715
1.12 3150 1765 3875
1.35 3600 2450 4890
1.58 4100 3280 7215
2.22 4550 5050 5530
Figure 5.18b.  A velocity analysis.

The quality of Figure 5.18b does not permit accurate picking, but the NMO correction is fairly tolerant of errors. We did not tabulate time-velocity pairs for the events at (0.97, 2600), (1.48, 3000), (1.79, 3050), and (1.90, 3000), because we thought these events were multiples. Interpreters usually ignore stacking velocity values lower than those at shallower depths. Structure, faulting, and other features can distort velocity analyses.

5.18b What can you tell about the lithology from this?

Solution

The interval velocity values are plotted on Figure 5.18c with a smooth curve for Tertiary clastics plotted as a reference. The interval-velocity curve is everywhere above the reference, and above 1 km it is roughly parallel to but higher than the reference. The section above 1.8 km is probably mainly clastics whose velocity is increased, perhaps by age, cementation, uplift, or the presence of carbonates. The higher velocities below 1.8 km suggest carbonates. The velocity from 2.5 to 3.3 km is unreasonably high. While overlying reflections are mostly strong and continuous, data quality deteriorates here.

5.18c If the section that is present in the syncline but is missing over the anticline consists of poorly consolidated rocks, what values would you expect for a velocity analysis at SP 45?

Figure 5.18c.  Interval velocity versus depth.

Solution

Poorly consolidated rocks will have lower velocities than at SP 100 and this will lower the measured stacking velocities. Interval velocities between the same reflections, which will be deeper at SP 45, will probably be slightly higher than for the same intervals at SP 100.

5.18d Note the downdip thinning of the section to the left of SP 100 between 0.75 to 1.25 s. Suggest an explanation.

Solution

If the interval thickness remains constant, traveltime through it will decrease with increasing depth because the rocks have higher velocities as they are buried deeper and thus are under greater pressure.

## 5.19 Velocity versus depth from sonobuoy data

Determine velocity versus depth from Figure 5.19a assuming horizontal refractors. The direct wave that travels through the water (assume ${\displaystyle V_{W}=1500\ {\rm {m/s}}}$) 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.

Figure 5.19a.  Sonobuoy refraction profile.

Solution

Figure 5.19b shows the picked events ${\displaystyle A}$ , ${\displaystyle B}$, ${\displaystyle C}$, 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.

Figure 5.19b.  Identification of events on Figure 5.19a.

Three distinct refraction (headwave) events can be seen: event ${\displaystyle A}$, which gives the first breaks beyond SP 150; ${\displaystyle B}$, which gives the first breaks between SP 120 and 150, and ${\displaystyle C}$, which has a projected arrival time at SP 240 of about 6.4 s. When ${\displaystyle B}$ 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 ${\displaystyle t_{o}}$ for the sea-floor reflection; since ${\displaystyle t_{o}\approx 1.45\ {\rm {s}}}$, the water depth is about ${\displaystyle 1.5\times 1.5/2\approx 1.1}$ km.

Next we calculate depths to the refracting horizons using equation (4.18a) for ${\displaystyle C}$, equation (4.18d) for ${\displaystyle B}$ and ${\displaystyle A}$. The shallowest refractor ${\displaystyle C}$ 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 ${\displaystyle C}$ as follows:

{\displaystyle {\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 ${\displaystyle B}$ and ${\displaystyle C}$ using the intercept time difference:

{\displaystyle {\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 ${\displaystyle B}$ is ${\displaystyle 1.3+0.8=2.1\ {\rm {km}}}$.

Figure 5.19c.  Profile interpretation.

For event ${\displaystyle A}$, we get

{\displaystyle {\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 ${\displaystyle B}$ to ${\displaystyle A}$,

{\displaystyle {\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 ${\displaystyle A}$ = depth to ${\displaystyle B+1.7=3.8\ {\rm {km}}}$.

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.

## 5.20 Influence of direction on velocity analyses

Given orthogonal dip and strike seismic lines, will velocity analyses at line intersections yield the same values?

Background

Reflecting points on a CMP gather in the dip direction are located updip by distances which increase with offset [see equation (4.11e)] so that a CMP stack smears the data. A DMO correction (Sheriff and Geldart, 1995, section 9.10.2) remedies this.

Solution

A line in the strike direction should yield reliable velocity values, but depths will be slant distances perpendicular to the bed, and reflecting points will be located updip. A commonmidpoint gather in the dip direction will not involve common reflecting points, and measurements will yield erroneous values unless DMO corrections have been made. The velocity analyses will probably yield somewhat different values.

## 5.21 Effect of time picks, NMO stretch, and datum choice on stacking velocity

5.21a Because velocity analysis is not made on the wavelet onset, how will this effect stacking-velocity values?

Background

A causal wavelet has zero amplitude for negative time values, that is, when ${\displaystyle t<0}$.

A normal-moveout correction is subtracted from the arrival times of a reflection to compensate for the increase of raypath distance with offset. The normal-moveout equation (4.1c) has the factor ${\displaystyle V^{2}t_{o}}$ in the denominator. While the value of ${\displaystyle t_{o}}$ is the same for all traces, it generally cannot be measured and the traveltime ${\displaystyle t}$ is used instead of ${\displaystyle t_{o}}$. Also, the velocity usually increases with traveltime, and hence the correction is smaller than it should be. This effectively lowers the frequency, an effect called normal-moveout stretch.

Because of variable weathering (LVL) and the fact that the source and geophones are at various elevations, seismic traces are usually time shifted to effectively locate them on a horizontal datum surface below which conditions are assumed to be constant; this is called applying static corrections (see problem 8.18).

Solution

Equation (4.1a) gives for the velocity

{\displaystyle {\begin{aligned}V=(1/t)(x^{2}+4h^{2})^{1/2}.\end{aligned}}}

Because the onset of a wave is not obvious, measured times are slightly larger than those associated with the reflector depths ${\displaystyle h}$, and the calculated velocities will be slightly smaller.

5.21b What will be the effect of NMO stretch?

Solution

All parts of a wavelet should be corrected for the time and velocity that is appropriate for the wavelet onset, but instead they are corrected for a delayed time and the velocity appropriate to it. The undercorrection will increase with offset, making the measured velocity too low.

5.21c What will be the effect if the datum is appreciably removed from the surface?

Solution

The objective of corrections to a datum is to be able to treat the data as if the sources and geophones were all located at the same elevation and there are no horizontal velocity changes below the datum. It effectively corrects arrival times, but it does not change horizontal locations to account for the changes in datum depth. Because migration is often done assuming the same vertical velocity at all locations, errors in horizontal location will create errors in the migrated location of events. The datum should be sufficiently deep in the section so that horizontal velocity changes below the datum are so small that they do not affect the results. Note that corrections can be made for depths below the datum level so that deeper corrections do not imply a deep datum. In fact, the datum sometimes is above the ground level.

A datum that is appreciably deeper than the surface makes time measurements too small and calculated velocities too large. The horizontal component of raypaths is usually small near the surface but becomes larger as depths increase. Thus geophones actually located on the datum would involve smaller horizontal components of the distances to features that are not vertically below and thus create errors when the data are migrated.