[PDF] Geophysik Refracted waves correspond to energy





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Geophysik

Refraction seismics - the basic formulae

1. Two-layer case

We consider the case where a layer with thickness h and velocity v1 is situated over a halfspace with

velocity v2. A receiver is located at a distance D from the source, which itself is located at the surface.

What signals will we measure, if a seismic source is generating energy (e.g. an explosion)? Here we will

only consider the direct waves, reflections and refractions but no take into account multiple reverberations which would be recorded in nature (but often neglected in the processing).

The geometry of the problem looks like this:

Refraction profile:

i

Direct wave

ReflectionRefraction

Depth h

D Figure 1:Geometry of reflection/refraction experiment. There are three arrivals recorded at greater distances: the direct wave, the reflection from the discontinuity at depth h and the refracted wave. v 1 v2v1Before we try to determine the structure from observed travel times we have to understand the forward

problem: how can we determine the travel time of the three basic rays as a function of the velocity structure and the distance from the source. The most important ingredient we need is Snell"s law

Snell"s Law

22
11 sinsin vi vi= (1)

relating the incidence angle i in layer 1 with velocity v1 to the transmission angle i in layer 2 with velocity

v

2. Both angles are measured with respect to the vertical. Let us derive the arrival times for the three

types separately:

1.1 The direct wave

Geophysik Geophysik

http://www.geophysik.uni-muenchen.de

This is the easy one! In a layered medium the direct wave travels straight along the surface with velocity

v

1. At distance D clearly the travel time tdir will be:

travel time direct wave 1/vtdirD= (2)

1.2 The reflected wave

To calculate the reflected wave we need to do a little geometry. The length of the path the ray travels in

layer 1 is obviously related to the distance in a non-linear way. The travel time for the reflection is given

by travel time reflected wave 22

1)2/(2hvtrefl+D= (3)

In refraction seismology this arrival is often of minor interest, as the distances are so large that the

reflected wave has merged with the direct wave. Note that this has the form of a hyperbola.

1.3 The refracted wave

As we can easily see from the figure above the refracted wave needs a more involved treatment. Refracted waves correspond to energy which propagates horizontally in medium 2 with the velocity v2. This can only happen if the emergence angle i2 is 90°, i.e. critical angle 21

221sin190sinsin

vvivvvi cc==

°= (4)

where ic is the critical angle. So in order to calculate the travel time we need to consider rays which

impinge on the discontinuity with angle ic. From elementary geometry it follows that the arrival time trefr of

the refracted wave as a function of distance D is given by

Travel time refracted wave

221cos2

vtvvihti refrc refr

D+=D+= (5)

which is a straight line which crosses the time axis D=0 at the intercept time tirefr and has a slope 1/v2.

1.4 Travel time curves - the forward problem

Now we can put things together and calculate - for a given velocity model - the arrival times and plot

them in a travel-time diagram. Example:

The model parameters are:

h=30km v

1=5km/s

v

2=8km/s

This could correspond to a very

simple model of crust and upper mantle and the discontinuity would be the Moho. The distance at which the refracted arrival overtakes the direct arrival can be used to determine the layer depth.

According to ray theory there is a

minimal distance at which the refracted wave can be observed, this is called the critical distance (see below).

0 50 100 150 200 250 300 350 400 450 500 0

20 40
60
80
100
120
T I m e (s)

Distance (km)

Refracted wave

Direct wave

Reflected wave

Intercept time

Figure 2: Travel-time diagram for the two-layer case.

1.5 Critical distance and overtaking distance

Two concepts are useful when determining the depth of the top layer. The critical distance is the

distance at which the refracted wave is first observed according to ray theory (in real life it is observed

already at smaller distances, this is due to finite-frequency effects which are not taken into account by

standard ray theory). The critical distance Dc is from basic geometry critical distance ccihtan2=D (5)

where the critical angle ic is given by equation (4). If we equate the arrival time of the direct wave and

the refracted wave and solve for the distance we obtain the overtaking distance. It is given by overtaking distance

12122vvvvhu-+=D . (6)

1.6 Determining the structure from travel-time diagrams: the inverse problem

The problem: determine the velocity depth model from the observed travel times (Figure 2). We proceed

as follows: a. Determine v1 from the slope (1/ v1 ) of the direct wave. b. Determine v2 from the slope (1/ v2 ) of the refracted wave. c. Calculate the critical angle from v1 and v2. d. Read the intercept time ti from the travel-time diagram. e. Determine the depth h using equation (5), thus ci itvhcos21= . (7) or f. Read the overtaking distance from the travel-time diagram, and calculate h using equation (6).

2. Three-layer case

The three layer case is important for many realistic problems, particularly for near surface seismics,

where often a low velocity weathering layer is on top of the bedrock. In principle we follow the same

reasoning as before but through the additional layer the algebra is a little more involved. We have to

introduce a slightly different nomenclature to take into account the different layers. The incidence angles

will have two indices, the first index stands for the layer in which the angle is defined and the last index

corresponds to the layer in which the ray is refracted (see Figure 3). The equation for the direct waves is

of course the same as in the two-layer case. The same is true for the refraction from layer 2 but we show it to demonstrate the nomenclature.

2.1 The refraction from layer 2

The arrival time t2 of the refraction from layer 2 is given by 22
21121
2 cos2 vtvvihtiD+=D+= (8)

and - using the intercept time from the diagram - will allow us to determine the depth h1 of the topmost

layer.

Refraction profile 3-layer case

i 12 h 1 D Figure 3:Geometry of 3-layer refraction experiment. v 1 v 2 v12.2 The refraction from layer 3

Due to Snell©s law we have

3333
223
113

1sinsinsin

vvi vi vi=== (9)

we use this relation and basic trigonometry to derive the arrival time t3 of the refracted wave in layer 3

33
32232
1131
3

3cos2cos2

vtvvih vihti t iD +=D++= (10)

and again this is a straight line with the intercept time ti3 which can be read from the travel time diagram.

2.3 Determining the velocity depth model for the 3-layer case

As before our data is a diagram with the travel-times of the direct wave, the refraction from layer 2 and

the refraction from layer 3 (provided we were able to read the arrivals in the seismograms). To determine the velocities and the thicknesses of layers 1 and 2 we proceed as follows: a. Determine the velocities v1-3 from the slopes (1/v1-3) in the travel-time diagram. b. Read the intercept time ti2 for the refraction from layer 2. c. Determine thickness h1 - using equation (8) such that 122
1

1cos2itvh

i = , where 21
12 arcsinvvi= (11) d. Read the intercept time ti3 for the refraction from layer 3. e. Calculate with the already determined values h1 an intermediate intercept time t*

11313*

cos2 vihtti-= , where 31
13 arcsinvvi= (12) f. Using t* calculate the thickness h2 of layer 2 23
2

2cos2itvh= , where

32
23
arcsinvvi= (13) ti2ti31/v11/v2 1/v3 Figure 4: Travel-time diagram for the 3-layer case

In the model shown in Figure 4 the velocites are v1=3.5km/s, v2=5km/s, v3=8km/s. The layer thicknesses

are h1=10km and h2=25km.

3. Reduced time

In refraction seismology as well as in global seismology we often find travel-time diagrams where

reduced time is used. In principle this means that the refraction arrival of interest is approximately

horizontal in the travel-time diagram. This can be achieved by doing the following transformation redredvtt

D-= (14)

where vred is the reduction velocity. How can we determine the real velocity from the travel-time diagrams in reduced form? a. Choose a distance D0 and read the reduced travel time tr0 from the diagram for the desired arrival. b. Calculate the velocity using i redrtvtv -D+ D= 0 00 (15) where ti is the intercept time. Note that the intercept time does not change when using reduced time! Figure 4: Travel-time diagram for the 3-layer case in reduced form for the same model as before. vred=v3 tr0 D 0 To determine he velocity-depth structure from a travel-time diagram in reduced form you can - after having calculated the real velocities using equation (15) - follow the steps given in section 2.3.

4. Inclined 2-layer case

So far we have only considered plane layers with no structural variation along the profile. In this chapter

we consider the case where a high-velocity layer is inclined with an inclination angle a (see Figure 5).

The most important difference to thw previous examples (2-layer and 3-layer cases) is, that we now

perform two experiments, one shooting at the near end and one shooting at the far end of the region of

interest. Note that for the previous examples - due to symmetry - we would have observed the same travel time curves. For the case of an inclined layer this is no longer the case!

Let us develop the forward problem, i.e. calculating the travel times of the direct and refracted waves for

a given model. With seismic velocities v1 and v2 and inclination angle a the travel time of the refracted

waves are

D+=D-+=D+=D++=

2 112
11 1 )sin(cos21 )sin(cos2 v tvi vihtv tvi viht i cc refri cc refraa

where the (-) sign stands for the refracted arrival with smaller intercept time and the (+) sign for the

refraction with larger intercept time, ic is the critical angle at the interface and h+ and h- are defined

according to Figure 5. Note that - as in all previous cases - the arrival of the direct wave is at time

t dir=D/v1. An example for the travel time curves that will be observed for a model with a=8deg, v

1=1.2km/s and v2=4km/s is shown in Figure 6.

But how can we determine the model properties from the observed arrival times (the inverse problem)?

Here is how you should proceed:

a. Determine the velocities v1 and v2+/- from the slopes in the travel-time diagram. b. Use the following relations to determine a and v2: 21
2121
21
arcsin)sin(arcsin)sin( v vivvivvivviccccaaaa aaaaa=--+==-++2 )()(sin2)()( 1 2 iiivviii c. Read the intercept times ti+ and ti- from the travel time diagram. Determine the distances from the layer interface as ccitvhitvh ii cos2cos2 11 d. You can now graphically draw the layer interface by drawing circles around the profile ends with the corresponding heights h+/- and tangentially connecting the circles at depth.quotesdbs_dbs28.pdfusesText_34
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