Environmental and Exploration Geophysics II
Extracting Physical Properties from the
Shot Record and
The Two Layer Refraction Problem
tom.h.wilson
tom.wilson@mail.wvu.edu
Department of Geology and Geography
West Virginia University
Morgantown, WV
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
1) label all plotted curves,
2) label all relevant points and
3) Note (comment on) basic relationships
between events observed in the time-distance
plots (see following)
Tom Wilson, Department of Geology and Geography
•Do Exercise II and III using Excel.
•In exercise II comment on the origins of the
differences in the two reflection hyperbola? What is
their relationship to corresponding direct arrivals.
Tom Wilson, Department of Geology and Geography
For Exercise III, explain the differences observed in the arrival
times of the reflection and diffraction observed in the shot record.
Why does the diffraction event drop below the reflection?
Tom Wilson, Department of Geology and Geography
For the Barrel Interpreter
The following references will be provided on a sign out basis
Brown AAPG Memoir 42, 1986 3D Interp
Brown, 2004, more recent volume on 3D interp (6th ed)
Neidel, 1979, strat modeling
Yilmaz, 2001, seismic data analysis, volumes 1 and 2
Lines and Newrick, 2004, fundamentals of seismic interp.
Biondi, 2007, Concepts and apps in 3D imaging
…
We can be more focused once we know where the project
You will first want to research the AAPG
area is …
bulletin and Journal Geophysics for
For now the basics …
publications related to the project area
Tom Wilson, Department of Geology and Geography
SOFAR Channel – sound fixing and ranging channel
Tom Wilson, Department of Geology and Geography
Absorption
When we set a spring in motion, the spring oscillations
gradually diminish over time. In the same manner, we expect
that as a seismic wave propagates through the subsurface,
energy will be consumed through the process of friction and
there will be conversion of mechanical energy to heat energy.
We guess the following - there will be a certain loss of amplitude
dA as the wave travels a distance dr and that loss will be
proportional to the initial amplitude A.
i.e.
dA  AS dr
Tom Wilson, Department of Geology and Geography
dA(r )   AS dr
dA(r )
 dr
AS
 is a constant referred to
as the attenuation factor
In order to solve for A as a function of distance traveled (r)
we will have to integrate this expression In the following
discussion,let
A0  AS
A dA
r
A0 A  0dr
ln A  ln A0  r
Tom Wilson, Department of Geology and Geography
ln A  ln A0  r
A
ln
 r
A0
ln
e
A
A0
 er
A
 er
A0
A  A0er
Tom Wilson, Department of Geology and Geography
Mathematical Relationship
A(r )  A0er
Graphical Representation
Tom Wilson, Department of Geology and Geography
The physical significance of 
A(r )  A0er
 - the attenuation factor is also a function of additional terms  is wavelength, and Q is the absorption constant


Q
1/Q is the amount of energy dissipated in one
wavelength () - that is the amount of mechanical
energy lost to friction or heat.
Tom Wilson, Department of Geology and Geography


Q
 is also a function of interval velocity, period and frequency
Tom Wilson, Department of Geology and Geography

r
A(r )  A0e Q



A(r )  A0e QV
r
 is just the reciprocal of the frequency so we
can also write this relationship as

f
A(r )  A0e QV
Tom Wilson, Department of Geology and Geography
r
Smaller Q translates into higher energy loss or amplitude decay.
Tom Wilson, Department of Geology and Geography
A(r )  A0e

f
QV
r
increase f and decrease A
  Qf x 
 A
dB  20 log    20 log  e


 A0 


Higher frequencies are attenuated to a much
greater degree than are lower frequencies.
Tom Wilson, Department of Geology and Geography
When we combine divergence and absorption we get the
following amplitude decay relationship
A0 r
A(r ) 
e
r
The combined effect is rapid amplitude decay as the
seismic wavefront propagates into the surrounding
medium.
We begin to appreciate the requirement for high
source amplitude and good source-ground coupling to
successfully image distant reflective intervals.
Tom Wilson, Department of Geology and Geography
But we are not through - energy continues to be
dissipated through partitioning - i.e. only some of the
energy (or amplitude) incident on a reflecting surface
will be reflected back to the surface, the rest of it
continues downward is search of other reflectors.
The fraction of the incident amplitude of the
seismic waves that is reflected back to the surface
from any given interface is defined by the reflection
coefficient (R) across the boundary between layers
of differing velocity and density.
Z 2  Z1
R

Ainc Z1  Z 2
Arefl
Tom Wilson, Department of Geology and Geography
Z 2  Z1
R

Ainc Z1  Z 2
Arefl
Z1 and Z2 are the impedances of the bounding layers.
 2V2  1V1

1V1   2V2
Tom Wilson, Department of Geology and Geography
Z 2  Z1
R
Z1  Z 2
The transmitted wave amplitude T is
T 1  R
Z1  Z 2  Z 2  Z1 

T
 
Z1  Z 2  Z1  Z 2 
2 Z1
T
Z1  Z 2
Tom Wilson, Department of Geology and Geography
At a distance of 100 m from a source, the
amplitude of a P-wave is 0.1000 mm, and at a
distance of 150 m the amplitude diminishes to
0.0665 mm. What is the absorption coefficient
of the rock through which the wave is
traveling?
(From Robinson and Coruh, 1988)
Tom Wilson, Department of Geology and Geography
Set up the equation that needs to be solved in the
foregoing problem. Write it down on a piece of paper
and hand in.
Write down the Excel equation you would use to
solve for the diffraction time-distance relationship
as portrayed in ray trace exercise II
Tom Wilson, Department of Geology and Geography
The critical distance and the crossover distance.
xcrit  2h1 tan c
To determine the crossover
distance set the direct
arrival time equal to the
critical refraction arrival
time and solve for Xcross
Tom Wilson, Department of Geology and Geography
This is a new one, but pretty simple – see 3.2.4
xcross
Tom Wilson, Department of Geology and Geography
V2  V1
 2h1
V2  V1
We now have several equations which contain
quantities that we can measure directly from the shot
record and use to determine layer thickness - h1.
h1
?
Tom Wilson, Department of Geology and Geography
• Reflection time intercept
• Refraction time-intercept
• Crossover distance
• Critical Distance
Tom Wilson, Department of Geology and Geography
The two-layer refraction problem (see 3.3.1)
Tom Wilson, Department of Geology and Geography
Time =distance traveled/velocity
Tom Wilson, Department of Geology and Geography
For the details see 3.3.1
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
Snell’s law for multiple layers
sin 1 sin  2 1


V1
V2
V3
C
Tom Wilson, Department of Geology and Geography
The velocity triangle
Tom Wilson, Department of Geology and Geography
Expressing trig functions in terms of velocities
Tom Wilson, Department of Geology and Geography
V32  V12
The end result, where cos 1 
V3
and …
x 2h1 cos 1 2h2 cos  c
time= 

V3
V1
V2
2h2
x 2h1
2
2
time= 
V3  V1 
V32  V22
V3 V3V1
V3V2
dx 1

dt V3
Tom Wilson, Department of Geology and Geography
• What is the critical distance
• What is the relationship of the
reflection from the base of layer 2 to the
critical refraction from the top of layer 2
Tom Wilson, Department of Geology and Geography
As x gets larger and larger the reflection from the base of
layer 2 and the refraction across the top converge
Tom Wilson, Department of Geology and Geography
Tom Wilson, Department of Geology and Geography
In the three layer problem the number of possible terms
that could potentially be measured directly from the shot
record includes -
•V1, V2 and V3
•two reflection time intercepts
•two refraction time intercepts
•one crossover distance, and
•two critical distances
Tom Wilson, Department of Geology and Geography
How would you determine the thickness of layer 2 (h2)?
• From reflection arrivals
• From refraction events?
Tom Wilson, Department of Geology and Geography
What variables can be determined from an
analysis of the shot record
V1, V2, V3, ti1, & ti2, where the ti’s
refer to the reflection time intercepts
V2  ti 2  ti1 
V1ti1
h1 
& h2 
2
2
Tom Wilson, Department of Geology and Geography
What variables can be determined from an
analysis of the shot record
V1, V2, V3, ti1, & ti2, where the ti’s
refer to the refraction time intercepts
h1 
V1V2
2 V22  V12
ti1 &

2h1
2
2
h2 
t 
V3  V1 
2
2  i2
V3V1
2 V3  V2 

V3V2
Tom Wilson, Department of Geology and Geography
1) We have assumed that our layers have
successively higher and higher velocity.
What happens if we have a velocity inversion let’s say V2 is less than V1 and V3?
2) Another assumption we have made here is
that the refraction from the top of the third
layer, for example, will actually show itself,
and not get buried somewhere beneath the
earlier refraction and reflections.
This can happen if the 2nd layer is too thin.
Tom Wilson, Department of Geology and Geography
 Today turn in problem 2.6, 2.7 and 2.12
 Please read through Chapter 3, pages 95 to top
of 114.
 Chapter 4, pages 149 to 164 (as assigned
previously.
• Continue working Exercises I-III and bring
questions to class next Monday (due next
Wednesday)
• Also consider the attenuation problem and be
prepared to discuss further on Monday (due
next Wednesday).
Tom Wilson, Department of Geology and Geography