# Reflections/diffractions from refractor terminations

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 6 181 - 220 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 6.15a

A horizontal refractor is located under a north-south seismic line at a depth of 1200 m. The overburden velocity is 2500 m/s and the refractor velocity is 4000 m/s. The refractor is terminated by a linear vertical fault (HF in Figure 6.15a) 3500 m from the source point. Determine the traveltime curves when the fault strikes: (i) east-west, (ii) north-south, (iii) N30${\displaystyle ^{\circ }}$W.

### Background

The traveltime curve for a horizontal refractor is given by equation (4.18a). The critical angle ${\displaystyle \theta _{c}={\rm {sin}}^{-1}(V_{1}/V_{2})}$. In Figure 6.15a the refraction does not exist in the interval SQ; the distance SQ is the critical distance ${\displaystyle x^{\prime }}$ where

 {\displaystyle {\begin{aligned}x^{\prime }=2h{\rm {\;tan\;}}\theta _{c}.\end{aligned}}} (6.15a)
Figure 6.15a.  Events involving refractor termination.

### Solution

We require the values of ${\displaystyle \theta _{c}}$ and ${\displaystyle x^{\prime }/2}$:

{\displaystyle {\begin{aligned}\theta _{c}={\rm {sin}}^{-1}(2.50/4.00)=38.7^{0};x^{\prime }/2=1200{\rm {\;tan\;}}38.7^{0}=0.96\ {\rm {km}}.\end{aligned}}}

The fault is located at ${\displaystyle H}$ which is more than 960 m from the source in all three cases, so refracted waves are involved.

Case (i). Fault perpendicular to the north-south line.

Events are the following (refer to Figure 6.15a):

1. the direct wave; a straight line through the source with slope (1/2.50) s/km;
2. an inline refraction, a straight line beginning at ${\displaystyle Q}$ with slope (1/4.00) s/km; it is tangent at ${\displaystyle Q}$ to the reflection hyperbola (curve 5);
3. a reflected head wave between ${\displaystyle S}$ and ${\displaystyle G}$ from reflection point ${\displaystyle H}$ with paths such as ${\displaystyle STHMP}$, a straight line extending from ${\displaystyle S}$ to ${\displaystyle G}$ with slope opposite to (2) and a larger intercept time;
4. a diffraction generated at ${\displaystyle H}$, a curve through ${\displaystyle F}$ symmetrical about the vertical and tangent to the head-wave curve (2) at ${\displaystyle U}$ (if prolonged beyond ${\displaystyle F}$) and to curve (3) at ${\displaystyle G}$;
5. a reflection, a hyperbola symmetrical about the vertical through ${\displaystyle S}$ and tangent to curve (2) at ${\displaystyle Q}$;
6. a reflected refraction such as ${\displaystyle STJKL}$ (if the impedance contrast at the fault is large enough to produce a recognizable reflection), a straight line extension of curve (3) to the right of ${\displaystyle G}$.

Case (ii). Fault parallel to the seismic line.

A plan view of the fault is shown to the left of the “line” in Figure 6.15b. The observed events are

1. the direct wave; a straight line passing through ${\displaystyle S}$ with slope (1/2.50) s/km;
2. an inline refraction;
3. a reflected refraction along paths such as STFMP shown in plan view in Figure 6.15b where the energy goes down from ${\displaystyle S}$ at the critical angle until the refractor is reached at ${\displaystyle T}$, then along ${\displaystyle TF}$ to the fault at ${\displaystyle F}$ where reflection occurs, after which the energy travels along ${\displaystyle FM}$ until a ray peels off at ${\displaystyle M}$ and travels up to the recorder at ${\displaystyle P}$, its traveltime curve is shown in Figure 6.15c; the curve is given by equation (4.18a) where we replace ${\displaystyle x}$ with the distance

{\displaystyle {\begin{aligned}SF+FP=2SF=2[y^{2}+(x/2)^{2}]^{1/2}=2[3.50^{2}+x^{2}/4]^{1/2},\end{aligned}}}

${\displaystyle x}$ here being the distance ${\displaystyle SP}$; the curve is a hyperbola (see Figure 6.15c);

4. no diffractions will be observed because there is no point source;
5. inline reflections; the reflection curves are the usual hyperbolas;
6. there is no reflected refraction such as STJKL in Figure 6.15a in case (i) because it could only exist within a distance of 0.96 km from the fault.

Figure 6.15b.  Travelpaths for faults parallel and at angle to seismic line.

Figure 6.15c.  Arrival time curve for the reflected refractions.

Case (iii). Fault at an angle to the seismic line.

This case is similar to case (ii). The fault in the plan view shown to the right of the “line” in Figure 6.15b strikes at the angle N30${\displaystyle ^{\circ }}$W. The observed events are

1. the direct wave;
2. an inline refraction;
3. a reflected refraction, a typical path being ${\displaystyle {ST}^{'}{F}^{'}{M}^{'}{P}}$,${\displaystyle T^{'}}$ and ${\displaystyle M^{'}}$ being equivalents of ${\displaystyle T}$ and ${\displaystyle M}$ in Figure 6.15b; its traveltime curve is shown in Figure 6.15c; to derive the traveltime curve for this event, we use image point ${\displaystyle I}$ for reflection in the fault, so ${\displaystyle IP}$ replaces ${\displaystyle x}$ in equation (4.18a). We get

{\displaystyle {\begin{aligned}t=IP/V_{2}+(2h{\rm {\;cos\;}}\theta _{c})/V_{1}\\=x^{2}+SI^{2}-2xSI{\rm {\;cos\;}}60^{\circ })^{1/2}/4.00+2\times 1.20{\rm {\;cos\;}}38.7^{\circ }/2.50\\=(1/4.00)(x^{2}-7.00x+49.00)^{1/2}+0.749.\end{aligned}}}

Thus, ${\displaystyle (4-2.96)^{2}=(x^{2}-7.00x+49.00)}$. The curve is a hyperbola (see Figure 6.15c);

4. no diffraction event;
5. a normal reflection;
6. same as in case (ii).

All traveltime curves are normal except (3).

## Problem 6.15b

Repeat for the east-west fault for a refractor that dips ${\displaystyle 10^{\circ }}$ to the north with the source to the south.

### Solution

In Figure 6.15d, ${\displaystyle \theta _{c}=38.7^{\circ }}$, ${\displaystyle SB=1.20}$ km, ${\displaystyle BH=3.50}$ km. We make frequent use of the law of sines to calculate distances:

{\displaystyle {\begin{aligned}BT=(1.20/{\rm {\;sin\;}}51.3^{\circ }){\rm {\;sin\;}}38.7^{\circ }=0.74\ {\rm {km}},\\ST=1.2{\rm {\;sin\;}}100^{\circ }/{\rm {\;sin\;}}51.3^{\circ }=1.51\ {\rm {km}},\\TH=QC=3.50-BT=2.76\ {\rm {km}},\\SQ=ST{\rm {\;sin\;}}77.4^{\circ }/{\rm {\;sin\;}}41.3^{\circ }=2.23\ {\rm {km}},\\TQ=ST{\rm {\;sin\;}}61.3^{\circ }/{\rm {\;sin\;}}41.3^{\circ }=2.01\ {\rm {km}},\\QU=QC{\rm {\;sin\;}}128.7^{\circ }/{\rm {\;sin\;}}41.3^{\circ }=3.26\ {\rm {km}},\\SU=SQ+QU=5.49\ {\rm {km}},\\CU=QC{\rm {\;sin\;}}10^{\circ }/{\rm {\;sin\;}}41.3^{\circ }=0.73\ {\rm {km}},\\HU=HC+CU=TQ+CU=2.74\ {\rm {km}},\\HG=HU{\rm {\;sin\;}}41.3^{\circ }/{\rm {\;sin\;}}61.3^{\circ }=2.06\ {\rm {km}},\\HD=1.20+3.50{\rm {\;sin\;}}10^{\circ }=1.81\ {\rm {km}},\\SG=SU-GU=SU-(HU{\rm {\;sin\;}}77.4^{\circ }/{\rm {\;sin\;}}61.3^{\circ })\\=5.49-3.05=2.44\ {\rm {km}},\\SD=3.50{\rm {\;cos\;}}10^{\circ }=3.45\ {\rm {km}}.\end{aligned}}}

Figure 6.15d.  Events involving termination of a dipping refractor (${\displaystyle \theta =}$ critical angle).

For the refraction,

{\displaystyle {\begin{aligned}t_{Q}=(ST-TQ)/2.50=1.41\ {\rm {s}},\\t_{U}=(ST+HU)/2.50+TH/4.00=2.39\ {\rm {s}}.\end{aligned}}}

For the reflected refraction,

{\displaystyle {\begin{aligned}t_{G}=(ST+HG)/2.50+TH/4.00=2.12\ {\rm {s}},\\t_{S}=2(ST/2.50+TH/4.00)=2.59\ {\rm {s}}.\end{aligned}}}

The diffraction from ${\displaystyle H}$ has its minimum traveltime at ${\displaystyle D}$:

{\displaystyle {\begin{aligned}t_{D}=(ST+HD)/2.50+TH/4.00=2.02\ {\rm {s}}.\end{aligned}}}

For the reflection,

{\displaystyle {\begin{aligned}t_{0}=2(1.20{\rm {\;cos\;}}10^{\circ })/2.50=0.95\ {\rm {s}},\\t_{Q}={\rm {same\ as}}\ t_{Q}\ {\rm {for\ the\ refraction}}\ =1.41\ {\rm {s}}.\end{aligned}}}

## Problem 6.15c

What effect will the manner of terminating the refractor have, that is, how will the amplitude of the reflected refraction depend on the dip of the terminating fault?

### Solution

Provided the impedance contrast across the fault is large enough, any abrupt termination of the refractor will generate a reflected refraction. The attitude of the terminating fault will have a relatively small effect on the amplitude provided that the dip of the fault is such that the angle of incidence is close to ${\displaystyle 90^{\circ }}$.

## Problem 6.15d

Most commonly a faulted refractor terminates against rock of lower acoustic impedance, but the opposite situation can also happen. What differences will this make?

### Solution

The nature of refractors is that they have high velocity, hence usually terminate against rocks of lower acoustic impedance and a reflected refraction has opposite phase to the incident refraction. However, if a refractor terminates against a higher impedance, they will have the same phase.

## Problem 6.15e

Extend the profile for part (a), case (i), an appreciable distance beyond the fault so as to plot the diffraction from the refractor termination. Assume uniform 2.50-km/s material beyond the refractor termination.

### Solution

The extensions of curves (1), (2), (4), and (5) are shown in Figure 6.15a. Events (3) and (6) do not exist to the right of the fault.

Figure 6.16a.  Refraction multiple.