Geology 228Applied Geophysics

Lecture 4

Seismic Refraction(Reynolds, Ch. 4-6)

Seismic Methods• Some fundamentals of seismic waves

– One dimensional wave equation– the solution of a plane wave in unbounded uniform

medium– amplitude, phase, frequency, wave number, wave

length, ... • Huygens’ Principle and some simulations• Wavefront and ray: from physical wave to geometric

wave• Seismic refraction• Field examples

Seismic Refraction• Snell’s law

– Incident angle, reflection angle, refraction angle– Reflection coefficient, transmission coefficient– Energy distribution

• Seismic refraction for a single horizontal layer• Seismic refraction for multiple horizontal layers• Seismic refraction for a single dipping layer• Seismic refraction tomography• Field examples

Wave energy is dissipating in the media. There are three major ways in energy dissipation, or attenuation. They are

• Geometry spreading (total energy conservation)

• Intrinsic absorption caused by material imperfection (conversion to other types)

• Diffraction caused by material heterogeneity (reflection, refraction, reverberation, etc.)

Seismic fundamentals• Some fundamentals of seismic waves

– One dimensional wave equation– Solution of a plane wave in unbounded

uniform medium– Amplitude, and phase – Frequency, and period– Wave number, and wave length

A simplified case for the wave equation is the plane wave propagating in 1 direction, say the x-direction. In thiscase, the wave equation can be written as

2

2

22

2 1tu

vxu

∂∂

=∂∂ (3.1)

One solution for a plane wave propagating in an unbounded, uniform medium can be expressed as

)cos(0 kxtuu += ω (3.2)

This plane wave can be viewed as the wave generated by a plane source occupying the entire yz-plane togenerate wave propagation in the x-direction. In this equation, u0 is the amplitude, ω is the angular frequency; kis called the wave number. We will show the relationship of k with respect to angular frequency ω bydemonstrating Eqn (3.2) does satisfy the 1-D wave equation (3.1). Taking the secondary derivative of u withrespect to space, here the x-coordinate, is

)cos(02

2

2

kxtukxu

+−=∂∂ ω

and putting the second derivative of u with respect to time on the right hand side of Eqn (3.1) gives

)cos(102

2

2

2

2 kxtuvt

uv

+−=∂∂ ωω

comparing the last 2 equation leads to

vk ω=

In time domain

Tf

fT

ff

112

2

==

==πωπω

In space domain

kk πλ

λπ 22

==

They are linked through the propagation velocity

vTv

k == λω

A seismic wave field movie• Uniform medium

– 2D– point source– impulse time function – a rich frequency content

Huygens’ Principle• In a wave field, all points with the same

phase construct a wave front;• Huygens’ Principle: Each point on a

particular wave front can be treated as a new source

• To illustrate this point let’s consider the following case.

More complicated case to show the relationship between wavefront and ray

• 3 layer model• v3>v2>v1• v3:v2:v1 = 9:4:3

Ray is the outward normal at each point of the wavefront

Rays in a two-layer model: the velocity in the upper layer increases linearly from 4.0 – 5.5 km/sec, over a thickness of 10 km (gradient 0.15/km/sec/km). The velocity in the lower layer increases linearly from 8.0-8.5 km/sec, over a thickness of 4 km (gradient 0.125 km/sec/km).

Snell’s law, Reflection coefficient, and

Transmission coefficient

Refraction from multiple horizontal layers

The refracted wave is the wave energy travels below the interface at the velocity on the second layer, but travel horizontally along the interface, then travel back to the geophone receiver planted on the surface. The refracted wave can only be received after the critical distance xcrit. To learn what is the critical distance, we have to know what is the critical angle icrit first. First we need introduce the Snell’s law.

2

2

1

1 sinsinv

iv

i=

Snell’s law

2

2

1

1 sinsinv

iv

i=

At the critical incident angle, there is no transmitted energy radiated in the second layer, so the refraction angle is 90 degrees, so we have

2

1

221

1

sin

1)2/sin(sin

vvithen

vvvi

crit =

==π

)arcsin(sin2

1

2

1

vvior

vvi critcrit ==

So that we can define the critical angle icrit as

and the critical distance xcrit is

2/12

1

2

1

2/12

2

1

2

1

1

11

]1)[(

2

])(1[(

)(2

cossin2tan2

−=

−=

==

vv

hvvvv

h

iihihx

crit

critcritcrit

When a plane wave impinging at a flat interface, both reflection and transmission occur. The quantitative description of the reflection and transmission relies on the reflection coefficient and transmission coefficient. The Reflection Coefficient R and the Transmission Coefficient T are, respectively, defined as

incident

transm

incident

reflect

AAT

AA

R

=

=

It can be demonstrated that the coefficients R and T are associated with the combination of the acoustic impedance. Acoustic impedance is defined as the product of the density and the velocity, I.e., Z = ρv. The reflection coefficient R in a general case is

i

i

nZZ

nZZR

α

α2

12

212

tan)1(1)/(

tan)1(1)/(

−−+

−−−=

where n = (v2 / v1)2 and αi is the angle of incidence of the wave ray.

For normal incidence, the reflection coefficient is

012

12 =+−

= iforZZZZR α

and the transmission coefficient is

1 2

2 1

20i

Z ZT for

Z Zα= =

+

0)()(

212

212 =

+−

= iR forZZZZE α

and the transmission energy coefficient is

Recall that the energy of wave motion is proportional to the amplitude, so for the reflection energy coefficient we have

0)(

42

12

21 =+

= iT forZZZZE α

These relations are applicable for the case of the incidence angel less than 15 degrees. Apparently, we have

1=+ TR EE

Seismic refraction• Seismic refraction only consider the first arrivals

- so simple and easy to use • The detection depth is about 1/4 to 1/10 of your

geophone spread

Seismic Refraction

Vertical GeophonesSource(Plate)

Rock: Vp2

oscilloscope

ASTM D 5777

Soil: Vp1

x1x2x3x4

t1t2t3t4

Note: Vp1 < Vp2

zR

Determine depthto rock layer, zR

Xcross

t

x

A piece of real data

The travel time to each geophone for the direct wave in the first layer is simply

1vxtdirect =

The travel time for the reflected wave for a 2-layer model can be derived as follows. Start with the ray path and the knowledge of Snell’s law we have

1

21

2

21

221

21

221

4

4)(

)2

()2

(

vhx

t

hxtv

hxtv

reflect

reflect

reflect

+=

+=

+=

The second equation in the last slide can be re-formatted as

144

4)(

21

2

21

221

21

221

=−

+=

hx

htv

hxtv

reflect

reflect

The second equation is in a standard format for expressing a hyperbola curve.

t

x

Another important concept is the cross-over distance Xcross. At Xcross, the refracted wave starts to take over to be the first arrive at a point.

The ray path and the travel time for the refracted wave for a 2-layer model can be derived as

121 vCR

vBC

vSBtrefract ++=

At the crossover distance xcross the travel times to the point are the same for the direct wave and the refracted wave, so we have

2/1

12

12

2/11212

21

2/121

22

2121

12

2

21

22

121

2

21

22

121

)(2

)])([(2)(2)(

2

2

vvvvxh

vvvvvvhvv

vvh

vvvvx

vvv

vh

vx

vx

vvv

vh

vx

vx

tt

cross

cross

crosscross

crosscross

refractdirect

+−

=

−+=−=−

−=−

−+=

=

Seismic Refraction

0.000

0.005

0.010

0.015

0.020 Tr

avel

Tim

e (s

econ

ds)

0 10 20 30 40 50 Distance From Source (meters)

Horizontal Soil Layer over Rock

Vp1 = 1350 m/s

1

Vp2 = 4880 m/s

1z

x2

V VV Vc

c p2 p1

p2 p1=

−+

Depth to Rock:zc = 5.65 m

xc = 15.0 m

x values

t va

lues

Travel times for the direct, reflected, and refracted waves for a 2-layer model

2

21

22

12

1

22

1

2

4

vvv

vh

vxt

vhxt

vxt

refract

reflect

direct

−+=

+=

=

nn

n

nn

nnk

n

k k

knn

nk

n

k k

kn

Tvxt

vvhTh

vhT

+=

−=

=

++

−

=

+=

∑

∑

)1()1(

1

1

)1(1

cos]cos

2[

cos2

θθ

θ

For multiple layers, the thickness of each layer for n>1 can be calculated from

For example, for the first 3 layers we have

34

324

2

214

1

13334

3

324

2

214

1

13

23

213

1

12223

2

213

1

12

2/121

22

211112

1

11

cos)]coscos(

2[]coscoscos[2

cos]cos

2[]coscos[2

)(cos2

θθθθθθ

θθθθ

θ

vvh

vhTh

vh

vh

vhT

vvhTh

vh

vhT

vvvvTh

vhT

+−=++=

−=+=

−==

Procedure to get the stratigraphic structure from the refraction x-t plot :

1, from the slope get the velocity in each layer;2, from the velocity to get the critical angle of the k-th interface

1)1(sin

=+ =

n

knk v

vθ

3, get the interception time Tn;4, get the thickness hn.

)11(211

2 du vvv+=

True velocity in the second layer:

Homework• Exercise: refraction over a 2 layer model

Xs (m) Ys (m) Xr (m) Yr (m) T (sec)36 8 36 8 0.00000036 8 38 8 0.00658736 8 41 8 0.01349436 8 44 8 0.01429836 8 47 8 0.01510136 8 50 8 0.01590436 8 53 8 0.01686836 8 56 8 0.01767136 8 59 8 0.01831436 8 62 8 0.01992036 8 65 8 0.02056336 8 68 8 0.02104536 8 71 8 0.022009

Seismic Travel Time (sec)

0.000000

0.005000

0.010000

0.015000

0.020000

0.025000

Tim

e (s

ec)

Seismic Refraction Surveys

• Measure: propagation of elastic wave through layers of the Earth

• Results: depth structure and velocities of elastic waves • Equipment: 48-channel portable StrataView

Geophones

Shots

Distance (m)

Intercept – Time MethodProcedure:

- First arrivals pick up.- Plot travel time versus distance - Calculate depths to layer interfaces:

21

12

1211 2 ⎟⎟

⎠

⎞⎜⎜⎝

⎛+−

=VVVVTZ

22

23

23

13

21

231

222

2

VV

VVVV

VVZTZ

−⎥⎥⎦

⎤

⎢⎢⎣

⎡ −−=

DISTANCE (m)

TRA

VEL

TIM

E (m

s)

Xc1 Xc2

1/V1

1/V21/V3

T1

T2

Delay Time Method – SIPwin Program

First arrivals pick up

Layer assignment Create depth-velocity model

Intercept Time Method

X-T plot of EE'

0102030405060

0 100 200 300 400 500

Offset (m)

Trav

el T

ime

(ms)

Travel time- Distance plot of line EE’ (top) and DD’ (bottom)

X-T plot of DD'

0

10

20

30

40

50

60

70

80

0 100 200 300 400 500 600 700 800 900 1000

Offset (m)

Trav

el time (m

s)

05

1015202530

0 100 200 300 400 500

Distance (m)

Depth (m

) 1963 m/s

4951 m/s

433 m/s

0

10

20

3040

50

60

70

0 100 200 300 400 500 600 700 800 900 1000

DISTANCE (m)

5800 m/s

1600 m/s

400 m/s

DE

PTH

(m)

Seismic Refraction ResultsProfile Parallel to the Tennis Courts

The Project

In central Japan, a 2300m long mountain tunnel was planned to be built in a Tertiary mudstone area. The rock condition was found to be quite different from the result of prior investigation, and the cutting face collapsed after construction was started. A High Resolution Seismic Refraction analysis is applied to evaluate the rock condition of a non-excavated section in detail. The principal objective of the HRSRanalysis was to detect the distribution and extent of weathered rock at the non-excavated section in order to modify the tunnel design and ensure construction safety.

Result and Interpretation

The existing data used in the HRSR analysis were collected at receiver intervals of 10m and maximum exploration depth of 150m. The figure shows the final velocity model and construction record. It can be seen that the collapsed zones correspond to the lower velocity areas (the green zones in the figure) relative to the surrounding area. This result of the HRSR analysis allowed precise prediction of the weathered rock zone and was very useful for modifying the tunnel design and ensuring construction safety at the non-excavated section.

Example of Geotechnical Applications

Seismic refraction tomography example: Locating DNAPL Traps in a Complex Shallow

Aquifer, Hill Air force Base, Utah

• Project of Dept of Geophysics, Rice University

• Reference: Zelt, C. A., "Lateral velocity resolution from 3-D seismic refraction data," Geophys. J. Int., 135 (1998): 1101-1112.

• http://terra.rice.edu/department/staff/morozov/emsp2000/60115.html

Seismic refraction tomography example: Locating DNAPL Traps in a Complex Shallow

Aquifer, Hill Air force Base, Utah

• Project of Dept of Geophysics, Rice University

• Reference: Zelt, C. A., "Lateral velocity resolution from 3-D seismic refraction data," Geophys. J. Int., 135 (1998): 1101-1112.

• http://terra.rice.edu/department/staff/morozov/emsp2000/60115.html

Map of the depth to the confining clay layer that underlies the area is based on geologic information collected at the site from monitoring wells and soil tests. Each data point is indicated with a blue dot. The map was constructed by interpolation between the data points. DNAPLscollect in the deep depression running through the center of the map. This area is the focus of the seismic imaging effort. The location of the three profiles from the 2-D survey are shown. The pipe is part of the remediation facilities.

First-arrival picks from the Line 2 slide hammer data plotted as a function of receiver position. There are 4173 picks from 62 shots. The picks from 3 shots have been colored to highlight the time-offset characteristics of the data; the position of the corresponding shots is indicated by the large colored circles on the distance axis.

Top: Final 2-D velocity model for the Line 2 sledge hammer data.

Bottom: Raypathsthrough the final model for the Line 2 sledge hammer data.

Final 2-D velocity model for the Line 2 combined rifle and shotgun data. Contour interval is 100 m/s; the 500, 1000 and 1500 m/s black contours are labeled. The known water table depth in the channel is indicated by the brown arrows. The pink dot indicates the depth (13.2 m) to the clay aquiclude from a well at 25 m distance.

Relative difference between the final model for the slide hammer data and the 1-D starting model. The approximate position of the channel inferred from the reflection images is indicated above, and the depth to the water table is also indicated. There is a good correlation between the low-velocity region in the tomographic model and the extent of the channel and its depth. Note the large velocity perturbations from the starting model, up to 65% in magnitude.