# User:Ageary/Chapter 4

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

## 4.1 Accuracy of normal-moveout calculations

4.1a Show that, for constant velocity ${\displaystyle V}$, the traveltime ${\displaystyle t}$ for the reflection path SCR in Figure 4.1a is

 {\displaystyle {\begin{aligned}t=\left(1/V\right)(x^{2}+4h^{2})^{1/2}.\end{aligned}}} (4.1a)

Background

Where the velocity is constant, raypaths are straight lines. In Figure 4.1a, a wave travels from the source ${\displaystyle S}$ to the receiver ${\displaystyle R}$ after being reflected at ${\displaystyle C}$, the angle of incidence ${\displaystyle \alpha }$ being equal to the angle of reflection. The image point I (or virtual source) is the point on the perpendicular from ${\displaystyle S}$ to the reflector as far below the reflector as ${\displaystyle S}$ is above. The line IR is equivalent to the actual path SCR.

The difference between the traveltime ${\displaystyle t_{0}}$ for a receiver at the source ${\displaystyle S}$ and the traveltime ${\displaystyle t}$ for a receiver at an offset ${\displaystyle x}$ (${\displaystyle x}$ being the source-to-geophone separation) is called the normal moveout, ${\displaystyle \Delta t_{\rm {NMO}}}$. We can get an approximate value of ${\displaystyle \Delta t_{\rm {NMO}}}$ by expanding equation (4.1a) in an infinite series to get ${\displaystyle t}$, then subtracting ${\displaystyle t_{0}}$.

An expression ${\displaystyle (1+\phi )^{n}}$, ${\displaystyle \left|\phi \right|<1}$, can be expanded as a binomial series [see, e.g., Sheriff and Geldart, 1995, equation (15.40)]:

 {\displaystyle {\begin{aligned}(1+\phi )^{n}&=1+n\phi +{\frac {n\left(n-1\right)}{2!}}\phi ^{2}+{\frac {n\left(n-1\right)\left(n-2\right)}{3!}}\phi ^{3}+\cdots \\&\quad +{\frac {n\left(n-1\right)\ldots \left(n-r+1\right)}{r!}}\phi ^{r}+\cdots .\end{aligned}}} (4.1b)

This series converges for ${\displaystyle \left|\phi \right|<1}$ and is infinite except when ${\displaystyle n}$ is a positive integer.

While the velocity is assumed to be constant in this problem, equations such as (4.1 a,c) are also used when the velocity varies (usually increasing with depth), ${\displaystyle V}$ being replaced by a suitable velocity such as the average velocity ${\displaystyle {\bar {V}}}$ (see problem 4.13), the root-mean-square velocity ${\displaystyle V_{\rm {rms}}}$ (problem 4.13), or the stacking velocity ${\displaystyle V_{s}}$ (problem 5.12).

Figure 4.1a.  Geometry of NMO.

Solution

The virtual path ${\displaystyle IR}$ equals ${\displaystyle SCR}$ and thus equals ${\displaystyle Vt}$. Since ${\displaystyle SI}$ is normal to the ${\displaystyle x}$-axis, we have

{\displaystyle {\begin{aligned}(Vt)^{2}=x^{2}+(2h)^{2},\end{aligned}}}

i.e., the ${\displaystyle (t,x)}$ curve is an hyperbola. Taking the square root, we get equation (4.1a).

4.1b Show that when ${\displaystyle 2h>x}$ the normal moveout is approximately

 {\displaystyle {\begin{aligned}\Delta t_{\rm {NMO}}\approx x^{2}/2V^{2}t_{0}\approx x^{2}/4Vh.\end{aligned}}} (4.1c)

Solution

We write equation (4.1a) as

{\displaystyle {\begin{aligned}t=\left(2h/V\right)[1+(x/2h)^{2}]^{\frac {1}{2}}.\end{aligned}}}

If ${\displaystyle 2h>x}$, we use equation (4.1b) to expand this expression to get the series

 {\displaystyle {\begin{aligned}t={\frac {2h}{V}}\left\{1+{\frac {1}{2}}\left({\frac {x}{2h}}\right)^{2}+\left[{\frac {1}{2}}\left(-{\frac {1}{2}}\right){\frac {1}{2!}}\left({\frac {x}{2h}}\right)^{4}\right]+\cdots \right\}.\end{aligned}}} (4.1d)

Neglecting terms higher than ${\displaystyle (x/2h)^{2}}$ (i.e., taking the first approximation) and noting that ${\displaystyle t_{0}=2h/V}$, we get

 {\displaystyle {\begin{aligned}t\approx t_{0}\left[1+{\frac {1}{2}}(x/2h)^{2}\right].\end{aligned}}} (4.1e)

Then, ${\displaystyle \Delta t_{\rm {NMO}}=t-t_{0}\approx t_{0}x^{2}/8h^{2}=x^{2}/2V^{2}t_{0}=x^{2}/4Vh}$.

4.1c Calculate the normal moveout ${\displaystyle \Delta t_{\rm {NMO}}}$ for geophones 600, 1200, and 3600 m from the source for a reflection at ${\displaystyle t_{0}=2.358}$ s, given that the velocity ${\displaystyle {}=2.90}$ km/s. What is the depth ${\displaystyle h}$?

Solution

From equation (4.1c) we have for ${\displaystyle x=600}$ m:

{\displaystyle {\begin{aligned}\Delta t_{\rm {NMO}}=x^{2}/2V^{2}t_{0}=0.600^{2}/\left(2\times 2.90^{2}\times 2.358\right)=9.10\ {\rm {ms}}=9\ {\rm {ms}}.\end{aligned}}}

Because the NMO varies as ${\displaystyle x^{2}}$, the value for ${\displaystyle x=1200}$ m will be 4 times that for ${\displaystyle x=600}$ m, that is, 36 ms, and for ${\displaystyle x=3600}$ m ${\displaystyle (x\approx h)}$, ${\displaystyle \Delta t_{\rm {NMO}}=328}$ ms. We have ${\displaystyle h=(2.358/2)2.90=3420}$ m.

4.1d Typical uncertainties in measurements of ${\displaystyle x}$, ${\displaystyle V}$, and ${\displaystyle t_{0}}$ be 0.6 m, 0.2 km/s, and 5 ms. Calculate the corresponding uncertainty in ${\displaystyle \Delta t_{\rm {NMO}}}$. What do you conclude about the accuracy of ${\displaystyle \Delta t_{\rm {NMO}}}$ calculations?

Solution

The uncertainty in ${\displaystyle x}$ is about 0.1%, that in ${\displaystyle V}$ is about 7%, and that in ${\displaystyle t_{0}}$ is about 0.2%. The uncertainties in ${\displaystyle x^{2}}$ and ${\displaystyle V^{2}}$ are 0.2% and 14%. Since the three factors are multiplied or divided to get ${\displaystyle \Delta t_{\rm {NMO}}}$, the uncertainties add so that the uncertainty in ${\displaystyle \Delta t_{\rm {NMO}}}$ is about ${\displaystyle 14.4\%\approx 14\%}$. This uncertainty is due mainly to the error in ${\displaystyle V}$.

4.1e Show that a more accurate normal moveout can be written

 {\displaystyle {\begin{aligned}\Delta t_{\rm {NMO}}^{*}\approx \Delta t_{\rm {NMO}}\left(1-\Delta t_{\rm {NMO}}/2t_{0}\right).\end{aligned}}} (4.1f)

How much difference is there between ${\displaystyle \Delta t_{\rm {NMO}}^{*}}$ and ${\displaystyle \Delta t_{\rm {NMO}}}$ for ${\displaystyle x=1200}$ and 3600 m? Taking into account the uncertainties in ${\displaystyle x}$, ${\displaystyle V}$, ${\displaystyle t_{0}}$, when is this equation useful?

Solution

If we go to the second approximation in the expansion of equation (4.1d), equation (4.1e) becomes

{\displaystyle {\begin{aligned}t=t_{0}\left[1+{\frac {1}{2}}\left({\frac {x}{2h}}\right)^{2}-{\frac {1}{8}}\left({\frac {x}{2h}}\right)^{4}\right].\end{aligned}}}

{\displaystyle {\begin{aligned}{\hbox{Then,}}\qquad \Delta t_{\rm {NMO}}^{*}&=t-t_{0}={\frac {t_{0}}{2}}\left({\frac {x}{2h}}\right)^{2}-{\frac {t_{0}}{8}}\left({\frac {x}{2h}}\right)^{4}\\&={\frac {t_{0}}{2}}\left({\frac {x}{2h}}\right)^{2}\left[1-{\frac {1}{4}}\left({\frac {x}{2h}}\right)^{2}\right]=\Delta t_{\rm {NMO}}\left[1-{\frac {1}{4}}\left({\frac {x}{2h}}\right)^{2}\right]\\&=\Delta t_{\rm {NMO}}\left[1-\left({\frac {x^{2}}{4Vh}}\right)\left({\frac {V}{4h}}\right)\right]=\Delta t_{\rm {NMO}}\left(1-\Delta t_{\rm {NMO}}/2t_{0}\right).\end{aligned}}}

Equation (4.1f) reduces the uncertainty by ${\displaystyle \Delta t_{\rm {NMO}}/2t_{0}=(x/4h)^{2}}$. To affect the result significantly, this increase in accuracy should be at least 1%, i.e., ${\displaystyle (x/4h)^{2}\geq 0.01}$ or ${\displaystyle x\geq 0.4}$ h. Accordingly, equation (4.1f) is useful when the offset ${\displaystyle x}$ is greater than about one-half the reflector depth. When ${\displaystyle x=1200}$ m, the difference in ${\displaystyle \Delta t_{\rm {NMO}}^{*}}$ is less than 1 ms but for ${\displaystyle x=3600}$ m, the difference is 22 ms. However, the effect of velocity errors in real situations is usually more important than the approximation error.

## 4.2 Dip, cross-dip, and angle of approach

4.2a Show that for a bed dipping in the direction of the ${\displaystyle x}$-axis (see Figure 4.2a), equation (4.1a) becomes

 {\displaystyle {\begin{aligned}(Vt)^{2}=x^{2}+4h^{2}+4hx\sin \xi .\end{aligned}}} (4.2a)

Background

If the normal from the origin to a plane has direction cosines ${\displaystyle (\ell ,m,n)}$ and length ${\displaystyle h}$, the equation of the plane is ${\displaystyle \ell x+my+nz=h}$ (see Sheriff and Geldart, 1995, problem 15.9b).

Solution

In Figure 4.2a, the traveltime ${\displaystyle t}$ at ${\displaystyle R}$ equals ${\displaystyle IR/V}$ (assuming constant velocity). We use the cosine law to express IR in terms of IS and SR. This gives

{\displaystyle {\begin{aligned}(Vt)^{2}&=x^{2}+(2h)^{2}-4hx\cos \left({\frac {\pi }{2}}+\xi \right)\\&=x^{2}+4h^{2}+4hx\sin \xi ,\end{aligned}}}

Figure 4.2a.  Raypaths for dipping reflector.

where ${\displaystyle \xi }$ is the dip. This equation is also used when ${\displaystyle V}$ is not constant by replacing ${\displaystyle V}$ by the average velocity ${\displaystyle {\bar {V}}}$, but because ${\displaystyle V}$ usually increases with depth, this underestimates ${\displaystyle \xi }$.

4.2b For two receivers spaced a distance ${\displaystyle \Delta x}$ away from the source in opposite directions, show that to the first approximation the dip is given by

 {\displaystyle {\begin{aligned}\sin \xi =\left({\frac {V}{2}}\right)\left({\frac {\Delta t_{d}}{\Delta x}}\right),\end{aligned}}} (4.2b)

where ${\displaystyle \Delta t_{d}}$ is the difference in traveltimes at the two receivers.

Solution

We rewrite equation (4.2a) replacing ${\displaystyle x}$ with ${\displaystyle \Delta x}$ and get for the traveltime at ${\displaystyle R}$

 {\displaystyle {\begin{aligned}t_{1}={\frac {2h}{V}}\left(1+{\frac {(\Delta x)^{2}+4h\Delta x\sin \xi }{4h^{2}}}\right)^{1/2}.\end{aligned}}} (4.2c)

Expanding and taking the first approximation gives

{\displaystyle {\begin{aligned}t_{1}\approx t_{0}\left(1+{\frac {(\Delta x)^{2}+4h\Delta x\sin \xi }{8h^{2}}}\right).\end{aligned}}}

Figure 4.2b.  Dip moveout on zero-offset sections.

Next we take a receiver at a distance ${\displaystyle -\Delta x}$ to the left of the source to get the traveltime ${\displaystyle t_{2}}$. Subtracting the expressions for ${\displaystyle -\Delta x}$ from that for ${\displaystyle +\Delta x}$ gives the result

{\displaystyle {\begin{aligned}\Delta t_{d}=t_{1}-t_{2}=t_{0}\left(\Delta x\sin \xi \right)/h=\left(2\Delta x/V\right)\sin \xi ,\end{aligned}}}

{\displaystyle {\begin{aligned}{\hbox{so}}\qquad \qquad \qquad \sin \xi ={\frac {V}{2}}\left({\frac {\Delta t_{d}}{\Delta x}}\right),\end{aligned}}}

where ${\displaystyle (\Delta t_{d}/\Delta x)=dip}$ moveout.

4.2c Figure 4.2b represents a vertical section in the direction of dip of a bed CD, ${\displaystyle S_{1}}$ and ${\displaystyle S_{2}}$ being sources on the surface. Show that for receivers located at the sources, equation (4.2b) can be used provided ${\displaystyle \Delta t_{d}}$ is the difference between the two-way traveltimes at the two sources.

Solution

In Figure 4.2b the 2-way traveltime at ${\displaystyle S_{2}}$ is ${\displaystyle t_{02}}$. The wavefront arrives at ${\displaystyle S_{1}}$ after a further time interval of ${\displaystyle \left(t_{01}-t_{02}\right)=\Delta t_{d}}$. Because the traveltimes are 2-way, the distance ${\displaystyle AS_{1}={\frac {1}{2}}V\Delta t_{d}}$, so

{\displaystyle {\begin{aligned}\sin \xi =\left({\frac {1}{2}}V\Delta t_{d}\right)\left/\right.\Delta x={\frac {V}{2}}\left({\frac {\Delta t_{d}}{\Delta x}}\right).\end{aligned}}}

4.2d In Figure 4.2c, CA is a wavefront approaching two receivers ${\displaystyle A}$ and ${\displaystyle B}$ on the surface. Derive the following expression for the angle of approach ${\displaystyle \alpha }$:

 {\displaystyle {\begin{aligned}\sin \alpha =V\left({\frac {\Delta t}{\Delta x}}\right),\end{aligned}}} (4.2d)

where ${\displaystyle \Delta t}$ is the difference in traveltimes at ${\displaystyle A}$ and ${\displaystyle B}$, and ${\displaystyle V}$ is the near-surface velocity.

Figure 4.2c.  Angle of approach of a wave.

Solution

If the wavefront ${\displaystyle CA}$ arrives at ${\displaystyle A}$ at time ${\displaystyle t}$, it will arrive at ${\displaystyle B}$ at time ${\displaystyle t+\Delta t}$ where ${\displaystyle CB=V\Delta t.}$ Therefore the angle of approach ${\displaystyle \alpha }$ is given by

 {\displaystyle {\begin{aligned}\sin \alpha =V\left(\Delta t/\Delta x\right).\end{aligned}}} (4.2e)

A variation of this equation is

 {\displaystyle {\begin{aligned}V_{\alpha }=\Delta x/\Delta t=V/\sin \alpha ,\end{aligned}}} (4.2f)

that is, the distance between geophones on the surface divided by the difference in their arrival times is the apparent velocity ${\displaystyle V_{\alpha }}$.

4.2e Discuss the relationship between equation (4.2b) as applied in parts (b) and (c) and equation (4.2e).

Solution

In equation (4.2b) in part (b) ${\displaystyle \Delta t_{d}}$ is a time difference between two rays generated by the same source and is therefore a time interval due to the dip of the bed and the fact that the two receivers are a distance ${\displaystyle 2\Delta x}$ apart.

When equation (4.2b) is used in part (c), ${\displaystyle \Delta t_{d}}$ is a two-way time difference between waves generated by two different sources a distance ${\displaystyle \Delta x}$ apart, each wave returning directly to the source where it was generated. In Figure 4.2c, ${\displaystyle CA}$ is a plane wave that travels from ${\displaystyle C}$ to ${\displaystyle B}$ in time ${\displaystyle \Delta t}$. Equation (4.2d) would be identical with equation (4.2b) if the source were at the midpoint of AB, ${\displaystyle \Delta x}$ in equation (4.2d) taking the place of ${\displaystyle 2\Delta x}$ in equation (4.2b).

A further difference between the equations is related to the velocity ${\displaystyle V}$. In equation (4.2b) ${\displaystyle V}$ is assumed to be constant; if it is not, an average velocity for the section between the surface and the reflector is used. In equation (4.2e), ${\displaystyle V}$ is the average velocity over the short interval ${\displaystyle CB}$; this distance is usually small enough that velocity variations can be neglected.

4.2f Show that the quantity ${\displaystyle \left(\Delta t_{d}/\Delta x\right)}$ can be considered as a vector or component of a vector, according as ${\displaystyle \Delta t_{d}}$ corresponds to the total dip or a component of dip.

Figure 4.2d.  3D view of reflection path.

Solution

Referring to Figure 4.2d, we take the ${\displaystyle z}$-axis as positive vertically downward and consider a plane reflector which intersects the three axes at ${\displaystyle M}$, ${\displaystyle N}$, and ${\displaystyle P}$, the strike being at the angle ${\displaystyle \Xi }$ to the ${\displaystyle x}$-axis. The vertical depth to the plane is ${\displaystyle z}$, which is a function of ${\displaystyle x}$ and ${\displaystyle y}$. We write (see Sheriff and Geldart, 1995, Section 15.1.2c)

 {\displaystyle {\begin{aligned}\nabla z={\frac {\partial z}{\partial x}}\mathbf {i} +{\frac {\partial z}{\partial y}}\mathbf {j} .\end{aligned}}} (4.2g)

It is shown in Sheriff and Geldart, 1995, problem 15.6a that ${\displaystyle \nabla z}$ is perpendicular to the strike (because contours on the reflector are parallel to the strike). It is therefore in the direction of dip, that is, OA in Figure 4.2e is the horizontal projection of the total dip moveout. If we take the ${\displaystyle x'}$-axis in this direction, then

{\displaystyle {\begin{aligned}\nabla z=\left({\frac {\partial z}{\partial x{'}}}\right)\mathbf {i} '=\left(\tan \xi \right)\mathbf {i} '\approx \left(\sin \xi \right)\mathbf {i} ',\end{aligned}}}

where ${\displaystyle \mathbf {i} '}$ is a unit vector along the ${\displaystyle x'}$-axis and ${\displaystyle \sin \xi }$ is given by equation (4.2b) with ${\displaystyle x'}$ replacing ${\displaystyle x}$. Thus we can regard the dip as a vector whose magnitude ${\displaystyle \sin \xi }$ is given by equation (4.2b), ${\displaystyle \Delta t_{d}}$ being measured in the direction of dip.

We now take the ${\displaystyle x}$-axis and a unit vector ${\displaystyle \mathbf {i} }$ along ${\displaystyle OB}$ at an angle ${\displaystyle \beta }$ to ${\displaystyle OA}$ in Figure 4.2e. The component of dip [apparent dip—see part (g)] in this direction is the projection

Figure 4.2e.  Dip and strike from nonperpendicular measurements.

(or component) of ${\displaystyle \nabla z}$ in this direction (see Sheriff and Geldart, 1995, problem 15.6c), so the quantity given by equation (4.2b) when applied to measurements along this axis is the component of the total dip ${\displaystyle \nabla z}$ in this direction. Since ${\displaystyle \Delta t_{d}/\Delta x}$ is multiplied by the scalar quantity ${\displaystyle \left(V/2\right)}$ to get dip, we can consider ${\displaystyle \left(\Delta t_{d}/\Delta x\right)}$ as a vector or component of a vector depending on whether ${\displaystyle \Delta t_{d}}$ is measured using a spread in the direction of dip or at an angle to this.

4.2g When a profile is not in the direction of dip, the value given by equation (4.2b) is less than the true dip and is called the apparent dip. Show how to calculate the true dip and the direction of strike from values of the apparent dip measured along the ${\displaystyle x}$- and ${\displaystyle y}$-axes.

Solution

The direction cosines ${\displaystyle (\ell ,m,n)}$ of the normal to the reflector (see Figure 4.2d) are the cosines of the angles ${\displaystyle \theta _{1}}$, ${\displaystyle \theta _{2}}$, and ${\displaystyle \theta _{3}}$, ${\displaystyle \theta _{3}}$ being the angle of dip ${\displaystyle \xi }$ and ${\displaystyle \Xi }$ the angle of strike relative to the ${\displaystyle x}$-axis. The coordinates of the image point ${\displaystyle I}$ are ${\displaystyle (2h\ell ,2hm,2hn)}$. For a point ${\displaystyle R}$ on the ${\displaystyle x}$-axis with coordinates ${\displaystyle \left(x,0,0\right)}$, we have

{\displaystyle {\begin{aligned}(IR)^{2}&=(Vt_{R})^{2}=(x-2h\ell )^{2}+(2hm)^{2}+(2hn)^{2}\\&=x^{2}+4h^{2}-4h\ell x\end{aligned}}}

(note that ${\displaystyle \ell ^{2}+m^{2}+n^{2}=1}$). Expanding this result as in part (b) and subtracting the values for two geophones located a distance ${\displaystyle \Delta x}$ on either side of the source, we arrive at the result

{\displaystyle {\begin{aligned}\ell =V\left({\frac {\Delta t_{x}}{2\Delta x}}\right).\end{aligned}}}

Repeating the above procedure for a spread along the ${\displaystyle y}$-axis we get

{\displaystyle {\begin{aligned}m=V\left({\frac {\Delta t_{y}}{2\Delta y}}\right).\end{aligned}}}

 {\displaystyle {\begin{aligned}{\hbox{But}}\qquad \qquad \qquad \sin \xi &=(1-n^{2})^{1/2}=(\ell ^{2}+m^{2})^{1/2}\\&={\frac {V}{2}}\left[\left({\frac {\Delta t_{x}}{\Delta x}}\right)^{2}+\left({\frac {\Delta t_{y}}{\Delta y}}\right)^{2}\right]^{1/2}.\end{aligned}}} (4.2h)

Thus, ${\displaystyle \sin \xi }$ is ${\displaystyle \left(V/2\right)}$ times the hypoteneuse of a right-angled triangle whose sides are the two apparent dips.

To get the angle of strike ${\displaystyle \Xi }$, we use the equation of the plane ${\displaystyle NOM}$ in Figure 4.2d, that is, ${\displaystyle \ell x+my+nz=h}$ (see Sheriff and Geldart, 1995, problem 15.9b), and set ${\displaystyle z=0}$. This gives the equation of the line NM, namely ${\displaystyle \left(\ell x+my\right)=h}$, from which we get ${\displaystyle OM=h/\ell }$, ${\displaystyle ON=h/m}$, so ${\displaystyle \tan \Xi =\left(h/m\right)/\left(h/\ell \right)}$ or

 {\displaystyle {\begin{aligned}\tan \Xi =\left({\frac {\Delta t_{x}}{\Delta x}}\right)/\left({\frac {\Delta t_{y}}{\Delta y}}\right).\end{aligned}}} (4.2i)

4.2h If the two spreads in part (g) are at an angle ${\displaystyle \alpha \neq \pi /2}$ to each other, show how to calculate the total dip and the strike direction.

Solution

We take ${\displaystyle OB}$ and ${\displaystyle OC}$ in Figure 4.2e equal to the components of dip moveout along the ${\displaystyle x}$- and ${\displaystyle y'}$-axes. Then,

{\displaystyle {\begin{aligned}\ell &=\left(V/2\right)\left(\Delta t_{x}/\Delta x\right)=OB=OA\cos \beta ,\\m&=\left(V/2\right)\left(\Delta t_{y}/\Delta y\right)=AB=OA\sin \beta ,\\{\frac {V}{2}}\left({\frac {\Delta t_{y}^{'}}{\Delta y{'}}}\right)&=OC=OA\cos \left(\alpha -\beta \right)\\&=\left(OA\cos \beta \right)\cos \alpha +\left(OA\sin \beta \right)\sin \alpha ,\end{aligned}}}

 {\displaystyle {\begin{aligned}{\hbox{or}}\qquad \qquad \left(\Delta t'_{y}/\Delta y'\right)=\left(2/V\right)\left(\ell \cos \alpha +m\sin \alpha \right).\end{aligned}}} (4.2j)

We measure the value of the left-hand side of equation (4.2j) and the apparent dip moveout along the ${\displaystyle x}$-axis, which gives us ${\displaystyle \ell }$, so we can calculate ${\displaystyle m}$, given both ${\displaystyle \ell }$ and ${\displaystyle m}$. We can now find the total dip and the angle of strike as in part (${\displaystyle g}$).

In Figure 4.2e, ${\displaystyle OB}$ and ${\displaystyle OC}$ are the projections of the total dip ${\displaystyle OA}$ onto the ${\displaystyle x}$- and ${\displaystyle y'}$-axes. The perpendiculars to ${\displaystyle OB}$ at ${\displaystyle B}$ and ${\displaystyle OC}$ at ${\displaystyle C}$ must thus pass through ${\displaystyle A}$, the terminus of the total-dip vector. Hence we can solve the problem graphically as shown in Figure 4.2e to give the magnitude and direction of the true dip moveout.

4.2i Verify that the result in part (h) can be obtained graphically by combining the apparent dips as in Figure 4.2e.

Solution

From part (f) we see that ${\displaystyle OB}$ in Figure 4.2e is the projection of the total dip moveout onto the ${\displaystyle x}$-axis and that ${\displaystyle OC}$ is the projection onto the ${\displaystyle y'}$-axis. Therefore the lines ${\displaystyle AB}$ and ${\displaystyle AC}$ must be perpendiculars to ${\displaystyle OB}$ and ${\displaystyle OC}$. Thus, if we are given ${\displaystyle OB}$ and ${\displaystyle OC}$, drawing perpendiculars at ${\displaystyle B}$ and ${\displaystyle C}$ locates point ${\displaystyle A}$, so ${\displaystyle OA}$ is the total dip moveout and the strike is normal to ${\displaystyle OA}$.

## 4.3 ${\displaystyle X^{2}-T^{2}}$ relationship for a dipping bed

Show that, for a dipping reflector and constant velocity, equation (4.2a) becomes (see Gardner, 1947)

 {\displaystyle {\begin{aligned}(Vt)^{2}=(2x\cos \xi )^{2}+4h_{c}^{2},\end{aligned}}} (4.3a)

where ${\displaystyle h}$ in equation (4.2a) is replaced by ${\displaystyle h_{c}}$, the slant depth at the midpoint ${\displaystyle M}$ between source ${\displaystyle S}$ and receiver ${\displaystyle R}$, and ${\displaystyle t=t_{SR}}$ in Figure 4.3a.

Solution

Equation (4.2a) is based on Figure 4.2a where the receiver is down dip from the source, the offset being ${\displaystyle x}$; in Figure 4.3a the up-dip receiver ${\displaystyle R}$ is offset ${\displaystyle 2x}$ from source ${\displaystyle S}$, so that the dip ${\displaystyle \xi }$ is negative; thus equation (4.2a) becomes

{\displaystyle {\begin{aligned}(Vt)^{2}=(2x)^{2}+(2h)^{2}-4h\left(2x\right)\sin \xi .\end{aligned}}}

Replacing ${\displaystyle h}$ with ${\displaystyle h_{c}}$ where ${\displaystyle h=h_{c}+x\sin \xi }$, we obtain

Figure 4.3a.  Geometry for dipping bed.

{\displaystyle {\begin{aligned}(Vt)^{2}&=4x^{2}+4(h_{c}+x\sin \xi )^{2}-8x\left(h_{c}+x\sin \xi \right)\sin \xi \\&=4x^{2}+4\left(h_{c}^{2}+2xh_{c}\sin \xi +x^{2}\sin ^{2}\xi \right)-8xh_{c}\sin \xi -8x^{2}\sin ^{2}\xi \\&=4x^{2}\left(1+\sin ^{2}\xi -2\sin ^{2}\xi \right)+4h_{c}^{2}\\&=(2x\cos \xi )^{2}+4h_{c}^{2}.\end{aligned}}}

## 4.4 Reflector dip in terms of traveltimes squared

4.4a Using the dip-moveout equation (4.2b) and the result of problem 4.3, verify that

{\displaystyle {\begin{aligned}\tan \xi =t/(t_{SR}^{2}-t_{0}^{2})^{1/2},\end{aligned}}}

where ${\displaystyle \xi ={\rm {dip}}}$, ${\displaystyle t=t_{SM}-t_{MR}}$, ${\displaystyle t_{SR}=}$ traveltime for path ${\displaystyle {\rm {SR}}'{\rm {R}}={\rm {S}}'{\rm {R}}'{\rm {R}}}$, ${\displaystyle t_{0}=2h_{c}/V=}$ traveltime to receiver at ${\displaystyle M}$ (Figure 4.4a).

Solution

Since ${\displaystyle t}$ in problem 4.3 equals ${\displaystyle t_{SR}}$ here, we have

{\displaystyle {\begin{aligned}(Vt_{SR})^{2}=(2x\cos \xi )^{2}+4h_{c}^{2}=(2x\cos \xi )^{2}+(Vt_{0})^{2},\end{aligned}}}

{\displaystyle {\begin{aligned}{\hbox{or}}\qquad \qquad \cos \xi =\left(V/2x\right)(t_{SR}^{2}-t_{0}^{2})^{1/2}.\end{aligned}}}

From equation (4.2b) we get

{\displaystyle {\begin{aligned}\sin \xi =\left(V/2\right)\left(\Delta t_{d}/\Delta x\right)=\left(V/2x\right)\left(t_{SM}-t_{MR}\right),\end{aligned}}}

{\displaystyle {\begin{aligned}{\hbox{or}}\qquad \qquad \tan \xi =\left(t_{SM}-t_{MR}\right)/(t_{SR}^{2}-t_{0}^{2})^{1/2}=t/(t_{SR}^{2}-t_{0}^{2})^{1/2}.\end{aligned}}}

Figure 4.4a.  Geometry for dipping bed.

4.4b Using equation (4.2a), show that

{\displaystyle {\begin{aligned}\sin \xi =V^{2}\left(t_{SM}^{2}-t_{MR}^{2}\right)/8h_{c}x.\end{aligned}}}

Solution

Equation (4.2a) gives

{\displaystyle {\begin{aligned}(Vt_{SM})^{2}=x^{2}+4h_{c}^{2}+4h_{c}x\sin \xi ,\\(Vt_{MR})^{2}=x^{2}+4h_{c}^{2}+4h_{c}x\sin \xi ,\\{\hbox{thus}}\qquad \qquad \sin \xi =V^{2}\left(t_{SM}^{2}-t_{MR}^{2}\right)/8h_{c}x.\end{aligned}}}

4.4c Under what circumstances is the result for part (b) the same as equation (4.2b) and also consistent with part (a)?

Solution

The result in part (b) can be written

{\displaystyle {\begin{aligned}\sin \xi =\{[V(t_{MS}-t_{MR})]/2x\}\{[V(t_{MS}+t_{MR})]/4h_{c}\}.\end{aligned}}}

The expression in the first curly brackets is the same as the right-hand side of equation (4.2b). Hence, for the above to be the same as equation (4.2b), we must have ${\displaystyle V(t_{MS}+t_{MR})/4h_{c}=1}$ that is,

{\displaystyle {\begin{aligned}V(t_{MS}+t_{MR})=4h_{c}=2Vt_{0},\end{aligned}}}

{\displaystyle {\begin{aligned}{\hbox{so}}\qquad \qquad \qquad \qquad t_{0}={\frac {1}{2}}(t_{MS}+t_{MR}).\end{aligned}}}

In part (a) we got ${\displaystyle \tan \xi }$ by finding ${\displaystyle \sin \xi }$ and ${\displaystyle \cos \xi }$. The expression for ${\displaystyle \cos \xi }$ involves no approximation, so the only approximation is that used in the derivation of equation (4.2b) to get ${\displaystyle \sin \xi }$. Thus, results in (a) and (c) are consistent provided we take ${\displaystyle t_{0}={\frac {1}{2}}(t_{MS}+t_{MR})}$.

## 4.5 Second approximation for dip moveout

The expressions for dip in terms of dip moveout, equation (4.2b), involves the approximation of dropping higher-order terms in the expansion of equation (4.2a). What is the effect on equation (4.2b) if an additional term is carried in this expansion? What is the percentage change in dip?

Solution

In problem 4.2b we obtained the dip equation (4.2b) by taking the first approximation of equation (4.2a), that is, using

{\displaystyle {\begin{aligned}t&=(1/V)(x^{2}+4h^{2}+4hx\sin \xi )^{1/2}\\&=t_{0}[1+r(r+2\sin \xi )]^{1/2},\end{aligned}}}

where ${\displaystyle r=x/2h}$. Expanding and taking the second approximation gives

 {\displaystyle {\begin{aligned}t=t_{0}[1+(r/2)(r+2\sin \xi )-(r^{2}/8)(r+2\sin \xi )^{2}].\end{aligned}}} (4.5a)

If we take two offsets, ${\displaystyle \Delta x}$ and ${\displaystyle -\Delta x}$, and let ${\displaystyle r=|\Delta x/2h|}$, then

{\displaystyle {\begin{aligned}\Delta t_{d}&=t_{1}-t_{2}=t_{0}(2r\sin \xi -r^{3}\sin \xi )\\&=t_{0}[\Delta x/h-(\Delta x/h)^{3}/8]\sin \xi \\&=(t_{0}\Delta x/h)\sin \xi [1-(\Delta x/h)^{2}/8]\\&=(2/V)(\Delta x\sin \xi )[1-(\Delta x/h)^{2}/8],\end{aligned}}}

{\displaystyle {\begin{aligned}{\hbox{so}}\qquad \qquad \sin \xi =(V/2)\left({\frac {\Delta t_{d}}{\Delta x}}\right)[1+(2\Delta x/h)^{2}/32].\end{aligned}}}

Comparing this result with equation (4.2b), we see that the second approximation increases the calculated dip by the fraction ${\displaystyle (2\Delta x/h)^{2}/32}$, that is, by the approximate percentage

{\displaystyle {\begin{aligned}3(2\Delta x/h)^{2}=3({\hbox{spread length/depth}})^{2}.\end{aligned}}}

## 4.6 Calculation of reflector depths and dips

In Figure 4.6a assume that the depth to 1.0 s is 1500 m and that the interval velocity between 1.0 and 1.4 s is 3300 m/s, and the trace spacing is 100 m. Calculate the depths and dips of the four picked reflectors.

Figure 4.6a.  Portion of a reflection section (from Fitch, 1976).

Solution

There are 126 trace intervals corresponding to 12.6 km between the end traces. We have timed four reflections (on an enlargement) at the end traces to the nearest millisecond, giving the depths:

{\displaystyle {\begin{aligned}A&=1.075\ \mathrm {s} \to 1500+(3300/2)\times 0.075=1624\ \mathrm {m} ;\quad A'=1.045\ \mathrm {s} \to 1574\ \mathrm {m} ;\\B&=1.200\ \mathrm {s} \to 1830\ \mathrm {m} ;\quad B'=1.184\ \mathrm {s} \to 1804\ \mathrm {m} ;\\C&=1.308\ \mathrm {s} \to 2008\ \mathrm {m} ;\quad C'=1.302\ \mathrm {s} \to 1998\ \mathrm {m} ;\\D&=1.347\ \mathrm {s} \to 2073\ \mathrm {m} ;\quad D'=1.347\ \mathrm {s} \to 2074\ \mathrm {m} .\end{aligned}}}

The dips of the four reflections are given by tan ${\displaystyle \xi =\Delta z/\Delta x}$:

{\displaystyle {\begin{aligned}\tan \xi _{\rm {A}}&=(1624-1574)/12.60=3.97\ \mathrm {m/km} ,\qquad \xi _{\rm {A}}=0^{\circ }14';\\\tan \xi _{\rm {B}}&=2.06\ \mathrm {m/km} ,\qquad \xi _{\rm {B}}=0^{\circ }7';\\\tan \xi _{\rm {C}}&=0.79\ \mathrm {m/km} ,\qquad \xi _{\rm {C}}=0^{\circ }3';\\\tan \xi _{\rm {D}}&=-0.08\ \mathrm {m/km} ,\qquad \xi _{\rm {D}}=0^{\circ }0'.\end{aligned}}}

The section is thickening to the right. Although the seismic reflections are nearly flat, the small dip can be measured.

Note that all four reflections have slight bending and changes in character about one-third of the way from the left end, suggesting that something unresolvable is happening here, perhaps a very small fault.

## 4.7 Plotting raypaths for primary and multiple reflections

A well encounters a horizon at a depth of 2.7 km with a dip of ${\displaystyle 7^{\circ }}$. Sources are located 2 km downdip from the well and a geophone is placed at depths of 1000 and 2600 m. Plot the raypaths and calculate the traveltimes for the primary reflection from the 3-km horizon and its first multiple. Assume ${\displaystyle V=3.0}$ km/s.

Figure 4.7a.  Constructing raypath for a multiple from a dipping bed.

Background

Image points were used in Figures 4.1a and 4.2a to find travel paths in constant velocity situations. Figure 4.7a illustrates their use for the more complicated situation of multiples. The image point ${\displaystyle I_{1}}$ gives the raypaths for the wave after reflection at the dipping horizon. When the wave is then reflected at the surface ${\displaystyle I_{1}}$ is equivalent to a new source, so the image point ${\displaystyle I_{2}}$ is located on the perpendicular from ${\displaystyle I_{1}}$ to the surface as far above the surface as ${\displaystyle I_{1}}$ is below. If this multiple is reflected a second time at the horizon, the image point ${\displaystyle I_{3}}$ is located on the perpendicular to the horizon as far below the horizon as ${\displaystyle I_{2}}$ is above it.

Solution

With the origin at the source, the vertical depth of the reflector beneath the source is ${\displaystyle z=2700+200\tan 7^{\circ }=2725}$ m. The perpendicular from the source to the reflector meets the reflector at ${\displaystyle x=2725\sin 7^{\circ }}$, ${\displaystyle z=2725\cos 7^{\circ }}$, i.e., at ${\displaystyle (-330,2700)}$, and the source image for the primary ${\displaystyle I_{1}}$ is at ${\displaystyle x}$, ${\displaystyle z=(-660,5400)}$. The image for the surface multiple ${\displaystyle I_{2}}$ (reflected at the reflector and then at the surface) is at ${\displaystyle -660,-5400}$.

The arrival time of the primary reflection at the phone 1000 m deep is

{\displaystyle {\begin{aligned}(460^{2}+4400^{2})^{1/2}/3000=1.475\ \mathrm {s} ,\end{aligned}}}

and the surface multiple arrives at

{\displaystyle {\begin{aligned}(460^{2}+6400^{2})^{1/2}/3000=2.139\ \mathrm {s} .\end{aligned}}}

The arrival time of the primary at the phone 2600 m deep is

{\displaystyle {\begin{aligned}(460^{2}+2800^{2})^{1/2}/3000=0.948\ \mathrm {s} ,\end{aligned}}}

and the multiple arrives at

{\displaystyle {\begin{aligned}(460^{2}+8000^{2})^{1/2}/3000=2.671\ \mathrm {s} .\end{aligned}}}

The source and geophone stations in the well are drawn to scale but displaced to the right in Figure 4.7b. The raypath to the 2600 m phone for the primary (the solid line) and for the surface multiple (the dashed line) are shown. The reflection point moves updip as the distance between the phone and the reflector increases, i.e., as the phone becomes shallower. The reflection points for the multiples lie still further updip.

Figure 4.7b.  Raypaths for well survey.
Table 4.7a. Reflection traveltimes.
Traveltimes
Depth Primary Multiple
1000 m 1.475 s 2.139 s
266 0.946 2.671

The traveltimes for intermediate depths are calculated in the same way. The results are tabulated in Table 4.7a.

## 4.8 Effect of migration on plotted reflector locations

4.8a Figure 4.8a shows a hand-migrated section having the same horizontal and vertical scales. The steepest dips at a depth of about 1500 m below datum are around ${\displaystyle 45^{\circ }}$ to the left of the central uplift and ${\displaystyle 55^{\circ }}$ to the right. If the velocity is 2500 m/s, what are the dip moveouts and the horizontal distances between the migrated reflection points and the points of observation?

At a depth around 2500 m, the steepest dips are about ${\displaystyle 40^{\circ }}$ to the left and about ${\displaystyle 55^{\circ }}$ to the right, the latter extending to a depth of about 4000 m. If the velocity is 3500 m/s, what are the dip moveouts and horizontal displacements of these reflections?

Background

After reflection events have been identified on an unmigrated seismic record section and their arrival times and dip moveouts measured, they can be migrated to place them at the reflector locations. When the dip moveout is zero, reflection points are located directly below the source, but otherwise they are located updip. Numerous methods are available for migrating events. The simplest hand-migration method is to assume constant velocity (usually the average velocity${\displaystyle {\overline {V}}}$ calculate source-reflector distances, and then swing arcs centered at the sources with radii equal to these distances.

Figure 4.8a.  Hand-migrated seismic section.
Figure 4.8b.  Migrated section in Wyoming thrust belt (from Harding et al., 1983).

Solution

Equation (4.2b) states that ${\displaystyle \sin \xi =(V/2)\Delta t_{d}/\Delta x}$, where ${\displaystyle \xi }$ is the dip and ${\displaystyle \Delta t_{d}/\Delta x}$ is the dip moveout. For ${\displaystyle V=2500}$ m/s,

{\displaystyle {\begin{aligned}\Delta t_{d}/\Delta x=2\sin \xi /2500=800\sin \xi \ \mathrm {ms/km} .\end{aligned}}}

For ${\displaystyle \xi =45^{\circ }}$, the dip moveout is 566 ms/km, and for ${\displaystyle \xi =55^{\circ }}$, it is 655 ms/km. For constant velocity, horizontal displacement is ${\displaystyle \Delta x=z\tan \xi }$. For the depth 1500 m, ${\displaystyle x=1500}$ m for ${\displaystyle \xi =45^{\circ }}$ and 2140 m for ${\displaystyle \xi =55^{\circ }}$.

For the depth 2500 m and ${\displaystyle V=3500}$ m/s, the dip moveouts are

{\displaystyle {\begin{aligned}\Delta t_{d}/\Delta x&=571\sin \xi =367\ \mathrm {ms/km} \quad \mathrm {for} \quad \xi =40^{\circ },\\&=468\ \mathrm {ms/km} \quad \mathrm {for} \quad \xi =55^{\circ }.\end{aligned}}}

The horizontal displacements are, respectively, 2100 and 3570 m.

4.8b How far horizontally did selected reflections migrate in Figure 4.8b? This section has been plotted so that the scale is approximately 1:1 over the depth of principal interest, 10 to 20 kft.

Solution

Although at first glance the band of energy in the thrust sheet in the central third of the section looks like parallel events, careful examination shows apparent downdip thinning. An increase of velocity with depth can produce this effect.

Figure 4.8c.  Portion of Figure 4.8b.

From the depth scale in Figure 4.8b, we estimate the average velocity to 1 s is 10 kft/s and from 1 s to 1.8 s is 12.5 kft/s. Event ${\displaystyle A}$ has a dip moveout of 45 ms/kft, dips about ${\displaystyle 15^{\circ }}$, and extends downward from about 7 to 10 kft. Assuming straight rays, this gives a horizontal displacement of ${\displaystyle z\tan \xi =1900}$ to 2700 ft. The shallow continuation of ${\displaystyle A}$, event ${\displaystyle B}$, has dip moveout of 80 ms/kft and dips about ${\displaystyle 25^{\circ }}$; it extends from about 2 to 7 kft and has horizontal displacements of 930 to 3250 ft. If the dip change from ${\displaystyle A}$ to ${\displaystyle B}$ is abrupt, the events may overlap before migration.

Event ${\displaystyle C}$ has about the same dip moveout as event ${\displaystyle A}$ and the dip extends from about 17 to 20 kft with horizontal displacements of 4600 to 5400 ft. Event ${\displaystyle D}$ with about the same dip moveout as event ${\displaystyle B}$ extends from about 9 to 17 kft with horizontal displacements of 4200 to 7900 ft.

Allowing for raypath curvature would decrease these displacements.

## 4.9 Resolution of cross-dip

4.9a Sources ${\displaystyle B}$ and ${\displaystyle C}$ are, respectively, 600 m north and 500 m east of source ${\displaystyle A}$. Traveltimes to zero-offset geophones at ${\displaystyle A}$, ${\displaystyle B}$, and ${\displaystyle C}$ for a certain reflection are 1.750, 1.825, and 1.796 s. What are the dip and strike of the horizon, the average velocity being 3.25 km/s?

Solution

The dip moveouts are ${\displaystyle 0.075/0.600=0.125}$ s/km to the north and ${\displaystyle 0.046/0.500=0.092}$ s/km to the east. We take the ${\displaystyle x}$-axis east-west (see Figure 4.9a). If we apply equation (4.2b) as in problem 4.2c, we get

{\displaystyle {\begin{aligned}\sin \xi _{y}&=(3.25/2)(0.075/0.60)=0.203\\&=m;\qquad \xi _{y}=11.7^{\circ };\\\sin \xi _{x}&=(3.25/2)(0.046/0.50)=0.150\\&=\ell ;\qquad \xi =8.6^{\circ }.\end{aligned}}}

Figure 4.9a.  Geometry of dip measures.

Using equations (4.2g,h), we find that

{\displaystyle {\begin{aligned}\sin \xi &=(\ell ^{2}+m^{2})^{1/2}=(0.150^{2}+0.203^{3})^{1/2}=0.252,\quad \xi =14.6^{\circ };\\\mathrm {strike} \ \Xi &=\tan ^{-1}(\ell /m)=\tan ^{-1}(0.150/0.203)\\&=36.4^{\circ }\ {\hbox{with respect to the}}\ x{\hbox{-axis}}{}={\hbox{N}}53.6^{\circ }{\hbox{W}}.\end{aligned}}}

4.9b What are the changes in dip and strike if line AC has the bearing N80${\displaystyle ^{\circ }}$E?

Solution

We use equation (4.2j) and Figure 4.2e to obtain the dip moveout along the east-west ${\displaystyle x}$-axis when the measured moveout along the ${\displaystyle x'}$-axis with bearing N80${\displaystyle ^{\circ }}$E is 0.092 s/km. Equation (4.2j) gives

{\displaystyle {\begin{aligned}{\frac {3.25}{2}}\left({\frac {0.046}{0.500}}\right)&=0.150\\&=\ell \cos 10^{\circ }+0.203\sin 10^{\circ };\end{aligned}}}

Figure 4.9b.  Geometry with change of direction.

thus,

{\displaystyle {\begin{aligned}\ell =0.150/\cos 10^{\circ }-0.203\tan 10^{\circ }=0.117.\end{aligned}}}

{\displaystyle {\begin{aligned}{\hbox{So}}\qquad \qquad \sin \xi =(0.117^{2}+0.203^{2})^{1/2}=0.234,\quad \xi =13.6^{\circ },\\\tan \Xi =0.117/0.203=0.576,\end{aligned}}}

and the strike angle ${\displaystyle \Xi =30.0^{\circ }}$, that is, N60.0${\displaystyle ^{\circ }}$W. Thus, the ${\displaystyle 10^{\circ }}$ change in line direction changes the dip about ${\displaystyle 1.0^{\circ }}$ and the strike about ${\displaystyle 6^{\circ }}$.

4.9c Solve part (b) graphically.

Solution

The graphical construction, shown reduced in Figure 4.9c, yields dip moveout of 0.145 s/km and strike of N61${\displaystyle ^{\circ }}$W.

The dip moveout 0.145 gives a dip 13.6${\displaystyle ^{\circ }}$. Thus the graphical solution gives almost exactly the same values as in part (b).

Figure 4.9c.  Graphical solution.

## 4.10 Cross-dip

4.10a Two intersecting spreads have bearings N10${\displaystyle ^{\circ }}$E and N140${\displaystyle ^{\circ }}$E. If the first spread shows an event at ${\displaystyle t_{0}=1.760}$ s with dip moveout of 56 ms/km down to the south and the same event on the second spread has a dip moveout of 32 ms/km down to the northwest. Assume average velocity of 3.00 km/s.

1. Find the true dip, depth, and strike.
2. What are the values if the dip on the second spread is southeast?

Solution

We first give numerical solutions, then graphical solutions.

i) We take the ${\displaystyle x}$-axis in the N40${\displaystyle ^{\circ }}$W direction, the ${\displaystyle y}$-axis in the N130${\displaystyle ^{\circ }}$W direction, and the ${\displaystyle y'}$-axis in the N170${\displaystyle ^{\circ }}$W direction (see Figure 4.10a). We now use equation (4.2j) to calculate ${\displaystyle (\Delta t/\Delta y)}$ for ${\displaystyle \alpha =130^{\circ }}$.

{\displaystyle {\begin{aligned}56=32\cos 130^{\circ }+(\Delta t_{y}/\Delta y)\cos 50^{\circ },\end{aligned}}}

Figure 4.10a.  Dips northwest and southwest.

hence, ${\displaystyle (\Delta t_{y}/\Delta _{y})=100}$ ms/km; total dip moveout${\displaystyle {}=(32^{2}+100^{2})^{1/2}=105}$ ms/km, dip ${\displaystyle \xi =\sin ^{-1}(3.00\times 0.105/2)=9.1^{\circ }}$,

{\displaystyle {\begin{aligned}\mathrm {strike} \ \Xi &=\tan ^{-1}(32/100)\\&=17.7^{\circ }\ \mathrm {relative\ to} \ x-\mathrm {axis} \\&=\mathrm {N} (40^{\circ }-17.7^{\circ })\mathrm {W} =\mathrm {N} 22.3^{\circ }\mathrm {W} ,\end{aligned}}}

depth ${\displaystyle h={\frac {1}{2}}\times 3.00\times 1.76=2.64}$ km (normal to bed).

ii) We take the ${\displaystyle x}$-axis in the S10${\displaystyle ^{\circ }}$W direction (see Figure 4.10b) where ${\displaystyle \alpha =50^{\circ }}$, so

{\displaystyle {\begin{aligned}32=56\cos 50^{\circ }+(\Delta t_{y}/\Delta y)\sin 40^{\circ },(\Delta t_{y}/\Delta y)=-5.2\ \mathrm {ms/km} ,\end{aligned}}}

The positive ${\displaystyle y}$-axis is toward S80${\displaystyle ^{\circ }}$E, so the minus sign means that the ${\displaystyle y}$-component of dip is in the N80${\displaystyle ^{\circ }}$W direction.

{\displaystyle {\begin{aligned}{\hbox{Total dip moveout}}&=(56^{2}+5.2^{2})^{1/2}\\&=56.2\ \mathrm {ms/km} \end{aligned}}}

{\displaystyle {\begin{aligned}{\hbox{so}}\qquad \qquad \mathrm {dip} \ \xi =\sin ^{-1}\left({\frac {1}{2}}\times 0.056\times 3.00\right)=4.8^{\circ },\end{aligned}}}

{\displaystyle {\begin{aligned}{\hbox{and}}\qquad \qquad \mathrm {strkie} \ \Xi =\tan ^{-1}[56/(-5.2)]=-84.7^{\circ }.\end{aligned}}}

Figure 4.10b.  Dips southeast and southwest.

The strike is measured relative to the negative direction of the ${\displaystyle x}$-axis as shown in Figure 4.2d where both dip components are positive; when the ${\displaystyle y}$-component of dip is negative, as it is here, the strike line goes from ${\displaystyle M}$ to a point on the negative ${\displaystyle y}$-axis. Referring to Figure 4.10b, we see that the strike line is rotated counter-clockwise ${\displaystyle 84.7^{\circ }}$ from the negative direction. Graphical solutions for (i) and (ii) are shown in Figure 4.10c.

Figure 4.10c.  Graphical solutions to cross-dip problem.

4.10b Calculate the position of the reflecting point (migrated position) for each spread in (i) as if the cross information had not been available and each had been assumed to indicate total moveout; compare with the result of part (a). Would the errors be more or less serious if the calculations were made for the usual situation where the velocity increases with depth?

Solution

We find the coordinates of the migrated reflecting points assuming the velocity is constant at the value of the average velocity. We take the ${\displaystyle x}$-, ${\displaystyle y}$-, and ${\displaystyle z}$-axes positive to the south, west, and downward, respectively, the source being at the origin.

In (i) the spread along the ${\displaystyle x}$-axis with bearing N10${\displaystyle ^{\circ }}$E has dip moveout down to the south and west (Figure 4.10b), hence

{\displaystyle {\begin{aligned}\xi =\mathrm {dip} =\sin ^{-1}(3.0\times 0.056/2)=4.8^{\circ }.\\h={\hbox{depth normal to bed}}\approx {\hbox{vertical depth}}={\frac {1}{2}}\times 3.00\times 1.760=2.64\ \mathrm {km} .\end{aligned}}}

Since the dip is mainly south and west, the reflection point is shifted north and east along the spread direction, the distance ${\displaystyle 2640\sin 4.8^{\circ }=221}$ m. Resolving this along the ${\displaystyle x}$- and ${\displaystyle y}$-axes, we get ${\displaystyle x=-221\cos 10^{\circ }=-218}$ m and ${\displaystyle y=-221\sin 10^{\circ }=-38}$ m. The vertical depth is ${\displaystyle 2640\times \cos 4.8^{\circ }=2630}$ m, giving coordinates (${\displaystyle -218,-38,2630}$).

For the spread bearing S140${\displaystyle ^{\circ }}$E (Figure 4.10b),

{\displaystyle {\begin{aligned}\xi =\sin ^{-1}(3.0\times 0.032/2)=2.8^{\circ }\ \mathrm {S} 40^{\circ }\mathrm {E} .\end{aligned}}}

The reflecting point is ${\displaystyle 2640\times \sin 2.8^{\circ }=129}$ m north and west of the source. Thus, ${\displaystyle x=-129\cos 40^{\circ }=-99}$ m, ${\displaystyle y=129\sin 40^{\circ }=83}$ m, ${\displaystyle z=2640\cos 2.8^{\circ }=2640}$ m, so the coordinates are ${\displaystyle (-99,83,2640)}$.

Taking into account cross-dip, the total dip is ${\displaystyle 9.1^{\circ }}$ down to the south and west. The horizontal displacement of the reflecting point is ${\displaystyle 2640\sin 9.1^{\circ }=418}$ m in the direction ${\displaystyle \mathrm {N} (22^{\circ }+90^{\circ })\mathrm {E} =\mathrm {N} 112^{\circ }\mathrm {E} (\mathrm {since} \Xi =\mathrm {N} 22^{\circ }\mathrm {W} )}$. Thus, ${\displaystyle x=-418\cos(180^{\circ }-112^{\circ })=-157}$ m, ${\displaystyle y=-418\sin(180^{\circ }-112^{\circ })=-388}$ m, ${\displaystyle z=2640\cos 9.1^{\circ }=2610}$ m. The coordinates are now ${\displaystyle (-157,-388,2610)}$. The change in ${\displaystyle z}$ is small ${\displaystyle (30\ \mathrm {m} \approx 1\%)}$ but the ${\displaystyle x}$- and ${\displaystyle y}$-coordinates vary considerably, both percentagewise and in absolute values.

The errors become more serious when the velocity increases with depth because these calculations are based on the average velocity ${\displaystyle {\bar {V}}}$ rather than the interval velocity ${\displaystyle V_{i}}$, which is usually greater than ${\displaystyle {\bar {V}}}$.

## 4.11 Variation of reflection point with offset

4.11a Equation (4.3a) for an offset geophone can be written

 {\displaystyle {\begin{aligned}(Vt)^{2}=(2h_{c})^{2}+(2s\cos \xi )^{2},\end{aligned}}} (4.11a)

where ${\displaystyle 2s}$ is the offset and ${\displaystyle h_{c}}$ is the slant depth at the midpoint between the source ${\displaystyle S}$ and receiver ${\displaystyle G}$ (see Figure 4.11a). The point of reflection ${\displaystyle R(x_{1},z_{1})}$ is displaced updip the distance ${\displaystyle \Delta L}$ from the zero-dip position ${\displaystyle P(x_{0},z_{0})}$. Show that the coordinates of a point ${\displaystyle x,z}$ on the line ${\displaystyle IG}$ must satisfy the relation

 {\displaystyle {\begin{aligned}(2s-x)/(s+h\ell )=z/hn=k,\end{aligned}}} (4.11b)

where ${\displaystyle I}$ is the image point, ${\displaystyle \ell ,n}$ are the direction cosines of ${\displaystyle SI}$, ${\displaystyle h}$ is the slant depth at the source, and ${\displaystyle k}$ is a parameter fixing the location of a point on ${\displaystyle IG}$.

Solution

Referring to Figure 4.11a, ${\displaystyle (\ell ,n)}$ are the direction cosines of ${\displaystyle SI}$ where

{\displaystyle {\begin{aligned}\ell =\sin \xi ,\quad n=\cos \xi ,\quad \xi =\tan ^{-1}(\ell /n).\end{aligned}}}

To get the coordinates of ${\displaystyle A(x,z)}$, a point on ${\displaystyle IG}$, we draw ${\displaystyle AB}$ and ${\displaystyle IC}$ perpendicular to ${\displaystyle TG}$. Then, using the similar triangles ${\displaystyle ABG}$ and ${\displaystyle ICG}$, we have ${\displaystyle AB/BG=IC/CG}$, that is,

 {\displaystyle {\begin{aligned}z/(2s-x)=2hn/2(s+h\ell ),\qquad \mathrm {so} \ z/hn=(2s-x)/(s+h\ell ),\end{aligned}}} (4.11c)

${\displaystyle x}$ being the horizontal distance from ${\displaystyle S}$. If we write

{\displaystyle {\begin{aligned}k=z/hn=(2s-x)/(s+h\ell ),\end{aligned}}}

Figure 4.11a.  Displacement of reflection point for offset geophone.

we can vary ${\displaystyle k}$ to get different points on ${\displaystyle IG}$.

4.11b Verify the following relations:

 {\displaystyle {\begin{aligned}x_{1}=x_{0}-s^{2}\ell n^{2}/h_{c},\quad z_{1}=z_{0}-s^{2}\ell ^{2}n/h_{c},\end{aligned}}} (4.11d)

 {\displaystyle {\begin{aligned}{\hbox{and}}\qquad \qquad \Delta L=RP=-(s^{2}/2h_{c})\sin 2\xi .\end{aligned}}} (4.11e)

Solution

To get ${\displaystyle R(x_{1},z_{1})}$, the point of intersection of ${\displaystyle IG}$ and ${\displaystyle PT}$, we first find the equation of ${\displaystyle PT}$; the line ${\displaystyle PT}$ has slope ${\displaystyle \tan \xi }$ and passes through ${\displaystyle T(-h/\ell ,0)}$, so the equation is

 {\displaystyle {\begin{aligned}z=x\tan \xi +h/n=(\ell /n)x+h/n=(\ell xh)/n.\end{aligned}}} (4.11f)

We now solve equations (4.11c) and (4.11f) as simultaneous equations. Eliminating ${\displaystyle z}$ gives

{\displaystyle {\begin{aligned}z=hn((2s-x_{1})/(s+h\ell ))=(\ell x_{1}+h)/n,\end{aligned}}}

{\displaystyle {\begin{aligned}{\hbox{so}}\qquad \qquad hn^{2}(2s-x_{1})=(s+h\ell )(\ell x_{1}+h).\end{aligned}}}

Using the equations ${\displaystyle (\ell ^{2}+n^{2})=1}$, ${\displaystyle h_{c}=(h+s\ell )}$ ; this reduces to

 {\displaystyle {\begin{aligned}x_{1}=h(s-2s\ell ^{2}-h\ell )/h_{c}.\end{aligned}}} (4.11g)

 {\displaystyle {\begin{aligned}{\hbox{Also}}\qquad \qquad x_{0}=s-h_{c}\ell =s-h-s\ell ^{2}=sn^{2}-h\ell ,\end{aligned}}} (4.11h)

{\displaystyle {\begin{aligned}{\hbox{so}}\qquad \qquad \Delta x=x_{1}-x_{0}&=[h(s-2s\ell ^{2}-h\ell )-(sn^{2}-h\ell )/h_{c}]/h_{c}\\&=[hs-h\ell (h+2s\ell )+(h+s\ell )(h\ell -sn^{2})]/h_{c}\\&=[hs-h\ell (h+2s\ell )+h^{2}\ell -hsn^{2}+hs\ell ^{2}-s^{2}\ell n^{2}]/h_{c}\\&=-s^{2}\ell n^{2}/h_{c}.\end{aligned}}}

From equation (4.11f) we get

{\displaystyle {\begin{aligned}z_{1}&=(x_{1}\ell +h)/n=h\ell [(s-2s\ell ^{2}-h\ell )+h_{c}h]/h_{c}n\\&=[h\ell (2s-2s\ell ^{2}-h\ell )+h^{2}]/h_{c}n.\end{aligned}}}

Since ${\displaystyle z_{0}=h_{c}n}$,

{\displaystyle {\begin{aligned}\Delta z&=z_{1}-z_{0}=\left[h\ell (2s-2s\ell ^{2}-h\ell )+h^{2}-h_{c}^{2}n^{2}\right]/h_{c}n\\&=[h\ell (2s-2s\ell -h\ell )+h^{2}-n^{2}(h^{2}+2hs\ell +s^{2}\ell ^{2})]/h_{c}n\\&=[2hs\ell -2hs\ell ^{3}+h^{2}(1-\ell ^{2}-n^{2})-2hs\ell n-s^{2}\ell ^{2}n^{2}]/h_{c}n\\&=[2hs\ell (1-\ell ^{2}-n)-s^{2}\ell ^{2}n^{2}]/h_{c}n=-S^{2}\ell ^{2}n/h_{c}.\end{aligned}}}

We now have

{\displaystyle {\begin{aligned}\Delta L=[(\Delta x)^{2}+(\Delta z)^{2}]^{1/2}=-s^{2}\ell n/h_{c}=-(s^{2}/2h_{c})\sin 2\xi .\end{aligned}}}

## 4.12 Functional fits for velocity-depth data

Given the velocity-depth data shown in Figure 4.12a, what problems would you expect when using simple functional fits in different areas?

Figure 4.12a.  Velocity-depth relations in various areas.

Solution

The problems are of several kinds. A velocity function involves averaging and thus involves problems wherever the data do not fit the form of the function or where a mixture of data families (lithologies) is involved.

1. A single velocity function cannot represent variations caused by major changes in depositional conditions such as changes from siliciclastic to carbonate deposition or vice versa. An example is the Texas Gulf Coast-2 well: the rocks below about 4.2 km are predominantly carbonate whereas above this they are mainly sand-shale siliclastics. To fit this situation with some sort of average function will give velocities that are too high in the siliciclastic portion and too low in the carbonate section. Problems will also be encountered because of the velocity inversion at about 2.9 km where a thick, low-velocity marine shale is encountered. A thick shale section may also restrict compaction and the release of interstitial water, causing overpressuring and consequent abnormally low velocities. Other areas encounter similar problems.
2. Similar situations are created by major tectonic changes, for example, stresses that produced folding and/or thrusting in a deeper section may have no longer been active during the deposition of a shallower section, resulting in major velocity changes. This is especially apt to be the case where rocks representing some geologic ages are missing because of nondeposition or an erosional unconformity.
3. Velocity functions will also not work well where conditions along the section vary considerably, for example, where an unconformity has removed different amounts of the section, so that the velocity-depth relation changes rapidly in the horizontal direction.
4. Extrapolation of a velocity function beyond the area where the velocity data were obtained will usually give poor results. Consider again the Texas Gulf Coast-2 data and imagine what velocity a function would yield if measurements below some depth (perhaps 2.0 km) had not been available? Functions for all of the areas shown will give unreasonably high velocities if extrapolated well below their data bases. Because the velocity of sedimentary rock is often influenced mainly by compaction and consequent loss of porosity, the rate of increase of velocity with depth should decrease with depth; most velocity functions do not allow for this.
5. Where the functional fit is determined by a least-squares or similar algorithm, the distribution of data will affect the fit.
Figure 4.12b.  Average velocity versus depth.

## 4.13 Relation between average and rms velocities

4.13a Both the average velocity ${\displaystyle {\bar {V}}}$ and the root-mean-square velocity ${\displaystyle V_{\rm {rms}}}$ can be regarded as weighted averages along a particular raypath. Discuss the difference between the two velocities from this point of view.

Background

For a bed of constant velocity ${\displaystyle V}$ extending from the surface down to a horizontal reflector at a depth ${\displaystyle h}$, the two-way traveltime is ${\displaystyle t=2h/V}$. If the velocity is not constant and we assume vertical raypaths, we obtain an average velocity ${\displaystyle {\bar {V}}}$ by dividing ${\displaystyle 2h}$ by the two-way traveltime, in effect replacing the actual section by a single constant-velocity layer. When the section consists of a series of horizontal constant-velocity layers, we find ${\displaystyle {\bar {V}}}$ by summing the traveltimes through each layer, and dividing the total thickness by this sum.

Dix (1955) showed that better results could be obtained by using the root-mean-square velocity ${\displaystyle V_{\rm {rms}}}$ calculated according to the formula,

 {\displaystyle {\begin{aligned}V_{\rm {rms}}^{2}=\sum _{i=1}^{n}V_{i}^{2}\Delta t_{i}/\sum _{i=1}^{n}\Delta t.\end{aligned}}} (4.13a)

This formula applies to horizontal velocity layering as the offset goes to zero.

Fermat’s principle of stationary time states that a wave travels from one point to another along a path for which the traveltime is stationary, usually a minimum, compared to that along adjacent paths. This principle can be used to derive laws involving the geometrical aspects of wave travel (see problem 6.3).

Solution

For horizontal layering, ${\displaystyle {\bar {V}}}$ for a given depth is found by dividing the depth by the total one-way traveltime for a ray traveling vertically down to the required depth. This can be expressed as

 {\displaystyle {\begin{aligned}{\bar {V}}=h/t=(\Sigma \Delta h_{i})/\Sigma (\Delta h_{i}/V_{i}),\quad \mathrm {or} \quad h/{\bar {V}}=\Sigma (\Delta h_{i}/V_{i}),\end{aligned}}} (4.13b)

where ${\displaystyle \Delta h_{i}}$ and ${\displaystyle V_{i}}$ are the thickness and velocity of the the ${\displaystyle i^{\rm {th}}}$ th bed and ${\displaystyle h}$ is the total thickness. Thus, ${\displaystyle 1/{\bar {V}}}$ is the weighted average of the reciprocal of velocity when the weights are the layer vertical thicknesses.

Since the denominator of equation (4.13a) is the total one-way traveltime, we can write the equation in the form

{\displaystyle {\begin{aligned}V_{\rm {rms}}^{2}t=\sum \limits _{i=1}^{n}V_{t}^{2}\Delta t_{i},\end{aligned}}}

from which we see that ${\displaystyle V_{\rm {rms}}^{2}}$ is the weighted average of ${\displaystyle V_{i}^{2}}$, the weights being the one-way vertical traveltimes through each bed.

4.13b Calculate ${\displaystyle {\bar {V}}}$ and ${\displaystyle V_{\rm {rms}}}$ down to each of the interfaces in Figure 4.13a. Why do they differ (give a geometrical explanation)?

Figure 4.13a.  Model for the calculation of ${\displaystyle {\bar {V}}}$ and ${\displaystyle V_{\rm {rms}}}$.

Solution

${\displaystyle {\bar {V}}}$ is found by calculating the one-way time to pass through each layer, summing the time increments to a given depth and dividing the sum into the depth; ${\displaystyle V_{\rm {rms}}}$ is found from equation (4.13a). The values are shown in Figure 4.13a.

The average velocity ${\displaystyle {\bar {V}}}$ is calculated for a straight-line vertical path and is obtained from the weighted average of ${\displaystyle 1/V_{i}}$, the weights being the layer thicknesses, whereas ${\displaystyle V_{\rm {rms}}}$ gives more weight to high-velocity beds. This is consistent with the fact that the least-time path between two points involves changes in path direction that increase travel in the high-velocity layers. Thus, using ${\displaystyle V_{\rm {rms}}}$ somewhat compensates for the failure of straight-line paths to account for ray bending at interfaces.

4.13c Plot ${\displaystyle {\bar {V}}}$ and ${\displaystyle V_{\rm {rms}}}$ versus depth and versus traveltime and determine the best-fit straight lines for the four cases. What are the main problems in approximating data with functional fits?

Figure 4.13b.  Linear fits of velocity-depth and velocity-time data.

Solution

Figure 4.13b is the plot of ${\displaystyle {\bar {V}}}$ and ${\displaystyle V_{\rm {rms}}}$ versus depth and two-way traveltime. Least-squares fits (see Sheriff and Geldart, 1995, problem 15.13) are also tabulated on Figure 4.13b.

Use of any functional form involves approximations. The main problems in approximating data with functional fits are the following:

1. We must select a functional form to fit the data and the form selected may be either too simple or too complex. A simple form is easy to determine and use but it may not represent the data with sufficient precision. Use of a more complex form increases the accuracy but at the expense of increased computation time.
2. Single points far from the line of the function have a disproportionate effect on the determination of the line.
3. When the data are not evenly spaced, data bunched together have less influence on the best-fit line than more widely spaced data.
4. The accuracy of functional fits changes with offset because the relative contributions of high-velocity members increases with offset.
Table 4.14a. Depth calculations using various velocity functions.
Function ${\displaystyle t_{0}\to }$ 1.00 s 2.00 s 2.10 s 3.10 s
i) ${\displaystyle {\bar {V}}}$ 2.00 km/s 2.50 km/s 2.67 km/s 3.10 km/s
${\displaystyle z}$ 1.00 km 2.50 km 2.80 km 4.80 km
ii) ${\displaystyle V_{\rm {rms}}}$ 2.00 km/s 2.55 km/s 2.81 km/s 3.24 km/s
${\displaystyle z}$ 1.00 km 2.55 km 2.95 km 5.02 km
iii) ${\displaystyle {\bar {V}}}$ 2.02 km/s 2.54 km/s 2.59 km/s 3.12 km/s
${\displaystyle z}$ 1.01 km 2.54 km 2.72 km 4.84 km
iv) ${\displaystyle V_{\rm {rms}}}$ 2.03 km/s 2.62 km/s 2.68 km/s 3.27 km/s
${\displaystyle z}$ 1.02 km 2.62 km 2.81 km 5.07 km

## 4.14 Vertical depth calculations using velocity functions

Assuming flat bedding, calculate depths corresponding to ${\displaystyle t_{0}=1.0}$, 2.0, 2.1, and 3.1 s using the velocity functions determined in problem 4.13b,c. What depth errors are created?

Solution

The velocity functions determined are

1. average velocity ${\displaystyle {\bar {V}}}$ versus depth (in problem 4.13b),
2. rms-velocity ${\displaystyle V_{\rm {rms}}}$ versus depth (in problem 4.13b),
3. the best-fit ${\displaystyle {\bar {V}}}$ versus depth function (in problem 4.13c),
4. the best-fit ${\displaystyle V_{\rm {rms}}}$ versus traveltime function (in problem 4.13c).

Using these, we obtain the depths in Table 4.14a.

No depth errors are present in (i) because ${\displaystyle t_{0}}$ and ${\displaystyle {\bar {V}}}$ were derived from the the given data. The errors in calculated depths in (ii), (iii), and (iv) are tabulated in Table 4.14b. Using ${\displaystyle V_{\rm {rms}}}$ gives ${\displaystyle z}$-values 2–5% too large. The best-fit depth function in (iii) gives the best results overall while the best-fit traveltime function in (iv) has errors of the same order of magnitude as those in (ii).

Table 4.14b. Errors in depth calculations.
1.00 km 2.50 km 2.80 km 4.80 km
ii) ${\displaystyle V_{\rm {rms}}}$ 0.0% 2.5% 5.4% 4.6%
iii) best-fit ${\displaystyle {\bar {V}}}$ 1.0% 1.6% –2.9% 0.8%
iv) best-fit ${\displaystyle V_{\rm {rms}}}$ 2.0% 4.8% 0.4% 5.6%

## 4.15 Depth and dip calculations using velocity functions

4.15a Repeat the calculations of problem 4.14 assuming horizontal velocity layering and dip moveout of 104 ms/km, and find the dips.

Background

While velocity generally follows the layering, especially in structurally deformed areas, isovelocity surfaces may not parallel interfaces. Where the section has not been uplifted significantly, isovelocity surfaces are apt to be nearly horizontal in spite of structural relief.

Solution

The depths will be the same as those calculated in problem 4.14 except that ${\displaystyle z}$ is now slant depth. Vertical depths are ${\displaystyle z_{v}=z\cos \xi }$, where the dip ${\displaystyle \xi }$ is given by ${\displaystyle \sin ^{-1}(V\Delta t_{d}/2\Delta x)=\sin ^{-1}(0.052\ V)}$, ${\displaystyle V}$ usually being either ${\displaystyle {\bar {V}}}$ or ${\displaystyle V_{\rm {rms}}}$. Using the values of ${\displaystyle t_{0}}$, ${\displaystyle {\bar {V}}}$ and ${\displaystyle V_{\rm {rms}}}$ from Table 4.14a, we obtain the results in Table 4.15a.

Table 4.15a. Calculated depths and dips.
Time Velocity Slant Dip Vert. depth
${\displaystyle t_{0}}$ ${\displaystyle V}$ depth ${\displaystyle \xi }$ ${\displaystyle z\cos \xi }$
i) Assuming average velocities:
1.00 s 2.00 km/s 1000 m 5.97${\displaystyle ^{\circ }}$ 990 m
2.00 2.50 2500 7.47 2480
2.10 2.67 2800 7.98 2770
3.10 3.10 4800 9.28 4740
ii) Assuming rms velocities:
1.00 s 2.99 km/s 1000 m 5.97${\displaystyle ^{\circ }}$ 990 m
2.00 2.55 2550 7.62 2530
2.10 2.81 2950 8.40 2920
3.10 3.24 5020 9.70 4950
iii) Assuming best-fit depth function:
1.00 s 2.02 km/s 1010 m 6.03${\displaystyle ^{\circ }}$ 1000 m
2.00 2.54 2540 7.59 2520
2.10 2.59 2720 7.74 2700
3.10 3.12 4840 9.34 4780
iv) Assuming best-fit traveltime function:
1.00 s 2.03 km/s 1020 m 6.00${\displaystyle ^{\circ }}$ 1010 m
2.00 2.62 2620 7.83 2800
2.10 2.68 2810 8.01 2780
3.10 3.27 5270 9.79 5000

4.15b Trace rays assuming (i) the velocity layering given in Figure 4.13a, and (ii) that the velocity is constant at the values of ${\displaystyle {\bar {V}}}$ and ${\displaystyle V_{\rm {rms}}}$ listed in Figure 4.13b. Find the arrival times and reflecting points of reflections at each of the interfaces.

Solution

i) We first calculate the angle of approach ${\displaystyle \alpha }$ [using equation (4.2d)] and then use Snell’s law to find the other angles:

{\displaystyle {\begin{aligned}\alpha =\theta _{1}&=\sin ^{-1}\left(2.00\times {\frac {0.104}{2}}\right)=6.0^{\circ },\\\sin \theta _{2}&=(V_{2}/V_{1})\sin \theta _{1},\\\theta _{2}&=\sin ^{-1}[(3.00/2.00)\sin 6.0^{\circ }]=9.0^{\circ },\\\theta _{3}&=\sin ^{-1}[(6.00/2.00)\sin 6.0^{\circ }]=18.3^{\circ },\\\theta _{4}&=\sin ^{-1}[(4.00/2.00)\sin 6.0^{\circ }]=12.1^{\circ },\\t_{A}&=2z/(V_{1}\cos \theta _{1})=2.000/(2.00\cos 6.0^{\circ })\\&=1.006\ \mathrm {s} ,\\t_{B}&=1.006+2\times 1.500/3.00\cos 9.0^{\circ }\\&=2.018\ \mathrm {s} ,\\t_{C}&=2.018+2\times 0.300/6.00\cos 18.3^{\circ }\\&=2.123\ \mathrm {s} ,\\t_{D}&=2.123+2\times 2.000/4.00\cos 12.1^{\circ }\\&=3.146\ \mathrm {s} .\end{aligned}}}

Next we find ${\displaystyle x}$-coordinates of intersections of rays and interfaces:

{\displaystyle {\begin{aligned}x_{A}&=z\sin 6.0^{\circ }=0.105\ \mathrm {km} ,\\x_{B}&=0.105+0.156=0.261\ \mathrm {km} ,\\x_{C}&=0.261+0.031=0.292\ \mathrm {km} ,\\x_{D}&=0.292+0.209=0.501\ \mathrm {km} .\end{aligned}}}

Figure 4.15a.  Raypath.

ii) Assuming ${\displaystyle {\bar {V}}=0.288z+1.77}$ (see Figure 4.13b) and the given depths,

{\displaystyle {\begin{aligned}V_{A}=0.288+1.77=2.058\ \mathrm {km/s} ,\ \alpha =6.1^{\circ },\\t_{A}=2z/(V_{1}\cos \theta _{1})=0.977\ \mathrm {s} ,\\x_{A}=2z\sin 6.1^{\circ }=0.107\ \mathrm {km} ;\\V_{B}=0.72+1.77=2.49\ \mathrm {km/s} ,\alpha =7.4^{\circ },\\t_{B}=5.00/2.562\cos 7.4^{\circ }=1.968\ \mathrm {s} ,x_{B}=0.324\ \mathrm {km} ;\\V_{C}=0.806+1.77=2.576\ \mathrm {km/s} ,\alpha =7.7^{\circ },\\t_{C}=5.600/2.660\cos 7.7^{\circ }=2.124\ \mathrm {s} ,x_{C}=0.365\ \mathrm {km} ;\\V_{D}=1.382+1.77=2.940\ \mathrm {km/s} ,\alpha =9.4^{\circ }\\t_{D}=9.600/3.310\cos 9.4^{\circ }=3.146\ \mathrm {s} ,x_{D}=0.787\ \mathrm {km} .\end{aligned}}}

Assuming ${\displaystyle V_{\rm {rms}}=0.325z+1.75}$ and the given depths,

{\displaystyle {\begin{aligned}V_{A}=0.325+1.75=2.075\ \mathrm {km/s} ,\alpha =6.2^{\circ }\\t_{A}=2z(V_{1}\cos \theta _{1})=2.00/(2.075\cos 6.2^{\circ })=0.970\ \mathrm {s} \\x_{A}=2z\sin 6.2^{\circ }=0.108\ \mathrm {km} ;\\V_{B}=0.812+1.75=2.562\ \mathrm {km/s} ,\alpha =7.7^{\circ }\\t_{B}=5.000/2.562\cos 7.7^{\circ }=2.018\ \mathrm {s} ,x_{B}=0.335\ \mathrm {km} ;\\V_{C}=0.910+1.75=2.660\ \mathrm {km/s} ,\alpha =8.0^{\circ },\\t_{C}=5.600/2.660\cos 8.0^{\circ }=2.123\ \mathrm {s} ,x_{C}=0.390\ \mathrm {km} ;\\V_{D}=1.560+1.75=3.310\ \mathrm {km/s} ,\alpha =9.9^{\circ },\\t_{D}=9.600/3.310\cos 9.9^{\circ }=3.146\ \mathrm {s} ,x_{D}=0.825\ \mathrm {km} .\end{aligned}}}

Assuming ${\displaystyle {\bar {V}}=0.526t+1.49}$ and times in part (i):

{\displaystyle {\begin{aligned}t_{A}=1.006\ \mathrm {s} ,{\bar {V}}=2.019\ \mathrm {km/s} ,\alpha =6.0^{\circ },z_{A}=2.042\ \mathrm {km} ,x_{A}=0.213\ \mathrm {km} ,\\t_{B}=2.018\ \mathrm {s} ,{\bar {V}}=2.551\ \mathrm {km/s} ,\alpha =7.6^{\circ },z_{B}=5.194\ \mathrm {km} ,x_{B}=0.683\ \mathrm {km} ,\\t_{C}=2.123\ \mathrm {s} ,{\bar {V}}=2.607\ \mathrm {km/s} ,\alpha =7.8^{\circ },z_{C}=5.586\ \mathrm {km} ,x_{C}=0.751\ \mathrm {km} ,\\t_{D}=3.146\ \mathrm {s} ,{\bar {V}}=3.145\ \mathrm {km/s} ,\alpha =9.6^{\circ },z_{D}=10.029\ \mathrm {km} ,x_{D}=1.618\ \mathrm {km} .\end{aligned}}}

Assuming ${\displaystyle V_{\rm {rms}}=0.595\mathrm {t} +1.43}$ and times in part (i):

{\displaystyle {\begin{aligned}t_{A}=1.006\ \mathrm {s} ,V_{\rm {rms}}=2.029\ \mathrm {km/s} ,\alpha =6.1^{\circ },z_{A}=2.053\ \mathrm {km} ,x_{A}=0.215\ \mathrm {km} ,\\t_{B}=2.018\ \mathrm {s} ,V_{\rm {rms}}=2.631\ \mathrm {km/s} ,\alpha =7.9^{\circ },z_{B}=5.360\ \mathrm {km} ,x_{B}=0.726\ \mathrm {km} ,\\t_{C}=2.123\ \mathrm {s} ,V_{\rm {rms}}=2.693\ \mathrm {km/s} ,\alpha =8.1^{\circ },z_{C}=5.774\ \mathrm {km} ,x_{C}=0.800\ \mathrm {km} ,\\t_{D}=3.146\ \mathrm {s} ,V_{\rm {rms}}=3.302\ \mathrm {km/s} ,\alpha =9.9^{\circ },z_{D}=10.454\ \mathrm {km} ,x_{D}=1.784\ \mathrm {km} .\end{aligned}}}

The use of any functional form involves approximation, so it is not surprising that values of the depths and horizontal displacements depend upon the way in which they are calculated.

## 4.16 Weathering corrections and dip/depth calculations

4.16a Figure 4.16a shows part of a seismic record where the geophone group spacing is 50 m and the offset to the nearest group is 50 m and that to the far group 600 m. What is the apparent velocity of the first breaks?

Background

The weathered layer or low-velocity layer (LVL) is a shallow layer that usually extends from the surface downward for 4 to 50 m and is characterized by low velocity of the order of 250 to 1000 m/s. Often the base of the LVL is near to or coincident with the water table, and many of the pore spaces in the LVL are filled with air. The LVL is important because of the high absorption in it and because its low (and frequently quite variable) velocity has considerable effect on the traveltime. The velocity change at the base of the LVL is usually large, making it a good reflector and therefore important in the generation of multiples (see problem 3.8) and in P- to S-wave conversion. The large velocity change at its base also bends raypaths from below the LVL so that they are nearly vertical within the LVL regardless of their direction below the LVL.

Figure 4.16a.  Seismic record (courtesy Chevron).

The first energy from a source to arrive at the geophone groups is called the first break. When the source is below the base of the LVL, as in Figure 4.16b, first-break raypaths are almost parallel to the base of the LVL until they are refracted upwards at approximately the critical angle ${\displaystyle \theta _{c}}$ (see problem 4.18).Writing ${\displaystyle V_{W}}$ and ${\displaystyle V_{H}}$ for the velocities in the LVL and just below it, ${\displaystyle \sin \theta _{c}=V_{W}/V_{H}}$.

Figure 4.16b.  Refraction weathering profile.

Solution

The slope of the first breaks is about 0.200 s/600 m, or a velocity of 3000 m/s.

4.16b Assuming that the source is just below the base of the LVL and that the LVL velocity is 500 m/s, how thick is the LVL?

Solution

The uphole time is about 31 ms, so the depth of the base of the LVL is ${\displaystyle D_{w}=500\times 0.031=16}$ m.

4.16c Arrival times at the sourcepoint for two reflections are given as 0.475 and 0.778 s; what is the average velocity to these reflectors?

Solution

The normal moveout equation [see equation (4.1c)] is ${\displaystyle \Delta t_{\rm {NMO}}=x^{2}/2V^{2}t_{0}}$, so ${\displaystyle V=(x^{2}/2t_{0}\Delta t_{\rm {NMO}})^{1/2}}$. The NMO measure gives about 0.040 ms/600m = 0.067 ms/km for the 0.475 s reflection and about 0.037 ms/600m = 0.062 ms/km for the 0.778 s reflection; this gives velocities of 2.4 km/s and 1.9 km/s, respectively.

4.16d For these reflections, the arrival-time differences between the far traces in opposite directions from the source point are given as ${\displaystyle +0.005}$ s for both reflections. What are the reflector dips?

Solution

For dip moveouts of 0.005 s/600m = 0.0083 s/km, equation (4.2b) gives for the 0.475 s reflection, ${\displaystyle \sin \xi =(2.4/2)\times 0.0083=0.010}$, ${\displaystyle \xi =0.6^{\circ }}$. For the deeper reflection we get ${\displaystyle \sin \xi =(1.9/2)\times 0.0083}$, ${\displaystyle \xi =0.5^{\circ }}$.

4.16e What is the dominant frequency of these reflections (approximately)?

Solution

Counting the number of cycles in 0.1 s at the arrival times of the two reflections, we get approximately 80 and 60 Hz.

## 4.17 Using a velocity function linear with depth

4.17a When the velocity increases linearly with depth according to the relation

 {\displaystyle {\begin{aligned}V=V_{0}+az,\end{aligned}}} (4.17a)

${\displaystyle a}$ being constant, show that

 {\displaystyle {\begin{aligned}x=(1/pa)(\cos i_{0}-\cos i),\end{aligned}}} (4.17b)

 {\displaystyle {\begin{aligned}t=(1/a)\ln \left({\frac {\tan i/2}{\tan i_{0}/2}}\right),\end{aligned}}} (4.17c)

where ${\displaystyle V=}$ velocity, ${\displaystyle V_{0}=}$ velocity at depth ${\displaystyle z=0}$, ${\displaystyle x=}$ source-geophone distance, ${\displaystyle p=(\sin i)/V=}$ raypath parameter, ${\displaystyle t=}$ arrival time, and ${\displaystyle i=}$ angle of incidence.

Background

When the velocity is a function of depth only, as in equation (4.17a), expressions for the offset ${\displaystyle x}$ (see problem 4.1) and traveltime ${\displaystyle t}$ can be found by dividing the medium into horizontal layers, each of infinitesimal thickness (Figure 4.17a) and then integrating. We have

{\displaystyle {\begin{aligned}\Delta x_{n}=\Delta z_{n}\tan i_{n},\end{aligned}}}

{\displaystyle {\begin{aligned}\Delta t_{n}=\Delta z_{n}/(V_{n}\cos i_{n}).\end{aligned}}}

Also, using equation (3.1a), we have

{\displaystyle {\begin{aligned}\sin i_{n}/V_{n}=\sin i_{0}/V_{0}=p.\end{aligned}}}

In the limit we get the following integrals for ${\displaystyle x}$ and ${\displaystyle t}$:

 {\displaystyle {\begin{aligned}x=\int _{0}^{z}\tan i\ \mathrm {d} z,\quad t=\int _{0}^{z}\mathrm {d} z/(V\cos i).\end{aligned}}} (4.17d)

When the velocity increases monotonically with depth, a ray must eventually return to the surface (see Figure 4.20a). For horizontal velocity layering the raypaths are symmetrical about the deepest point.

Figure 4.17a.  Raypaths where ${\displaystyle V=V(z)}$.

Solution

In equation (4.17d) we substitute ${\displaystyle u=pV=\sin i}$, ${\displaystyle {\rm {d}}z={\rm {d}}V/a={\rm {d}}u/pa}$ from equation (4.17a), also ${\displaystyle \tan {i}=u/(1-u^{2})^{1/2}}$. Thus, we get

{\displaystyle {\begin{aligned}x&={\frac {1}{pa}}\int _{u_{0}}^{u}{\frac {d{\rm {d}}u}{[1-u^{2}]^{1/2}}}={\frac {1}{2pa}}\int _{u_{0}}^{u}{\dfrac {{\rm {d}}(u^{2})}{(1-u^{2})^{1/2}}}={\frac {1}{pa}}(1-u^{2})^{1/2}\left|\right._{u}^{u_{0}}\\&=(1/pa)(\cos i_{0}-\cos i);\\t&={\frac {1}{a}}\int _{u_{0}}^{u}{\frac {{\rm {d}}u}{u(1-u^{2})^{1/2}}}={\frac {1}{a}}\ln \left[{\frac {u}{1+(1-u^{2})^{1/2}}}\right]\left|\right._{u_{0}}^{u}\\&={\frac {1}{a}}\ln \left[\left({\frac {\sin i}{\sin i_{0}}}\right)\left({\frac {1+\cos i_{0}}{1+\cos i}}\right)\right]={\frac {1}{a}}\ln \left({\frac {\tan i/2}{\tan i_{0}/2}}\right).\end{aligned}}}

4.17b Show that the angle of incidence ${\displaystyle i}$ and the depth ${\displaystyle z}$ can be written

 {\displaystyle {\begin{aligned}i&=2\tan ^{-1}(e^{at}\tan i_{0}/2),\end{aligned}}} (4.17e)

 {\displaystyle {\begin{aligned}z&=(\sin i-\sin i_{0})/pa.\end{aligned}}} (4.17f)

Solution

From equation (4.17c) we get

{\displaystyle {\begin{aligned}e^{at}=\left({\frac {\tan i/2}{\tan i_{0}/2}}\right),\end{aligned}}}

{\displaystyle {\begin{aligned}{\hbox{hence}}\qquad \qquad i=2\tan ^{-1}(e^{at}\tan i_{0}/2).\end{aligned}}}

Solving equation (4.17a) for ${\displaystyle z}$ gives

{\displaystyle {\begin{aligned}z=(V-V_{0})/a=(\sin i-\sin i_{0})/pa.\end{aligned}}}

4.17c Given the velocity function ${\displaystyle V=1.60+0.600\ z}$ km/s, find the depth and offset of the point of reflection and the reflector dip when ${\displaystyle t_{0}=4.420}$ s and ${\displaystyle \Delta t/\Delta x=0.155}$ s/km. What interpretation would you give the result?

Solution

First, we note that ${\displaystyle t}$ in the preceding equations is one-way time, so we take ${\displaystyle t_{0}=2.210}$ s, ${\displaystyle V_{0}=1.60}$ km/s, ${\displaystyle a=0.600}$. To get ${\displaystyle p}$, we find the angle of approach ${\displaystyle i_{0}}$ [see equation (4.2d)] for ${\displaystyle \Delta t/\Delta x=0.155}$ s/km: