# Dip, cross-dip, and angle of approach

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 4 79 - 140 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 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 ${\displaystyle \xi }$.

## Problem 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.

## Problem 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}}}

## Problem 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 ${\displaystyle V_{\alpha }}$.

## Problem 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.

## Problem 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.

## Problem 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)

## Problem 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.

## Problem 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 ${\displaystyle OA}$.