Exercise_2_30092010_answers.256

8/17/2019 Exercise_2_30092010_answers.256

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The fastest reflection from the interface between the two layers arrives after 0.4

seconds.

= 2

∙ ℎ� 12

(22 12)= 358

The headwave (critical refracted) from the interface between the two layers does not

reach the surface before a distance of 358 m away from the shot (offset = 358 m).

= 2 ∙ ℎ� (2 + 1)(2 1) = 894 The headwave takes over the direct wave at a distance of 894 m form the shot.

c.) Taking the traveltime formulas it is possible to plot t vs x, since all other quantities are

given.

direct wave: = reflected wave: = √ +4 headwave: = + 2

Note the following characteristics of the diagram:


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-

The headwave does not appear on the diagram before the critical distance

-

The reflected approximates the direct wave with increasing offset

-

The headwave arrivals, when they appear at all, always arrive earlier than the

reflected wave arrivals.

-

The headwave takes over the direct wave at the crossover distance.

2. To find the depth to bedrock in a dam-site survey, traveltimes were measured from the

shot-point to 12 geophones laid out on a straight line through the shotpoint. The offsets x

range from 15 to 180 m. Determine the depth of overburden from the data in the Table.

Hints: Determine velocities, cross-over point or intercept time, and use them in formulas

ANSWER:

First a traveltime diagram was created using the given X-T values.

In the diagram you can discriminate between two zones of different slopes, i.e. different

velocities, here a direct wave (red) and a headwave (green). By regression, formulas for the two

straight lines could be found and the slope could be determined. The inverse of the slopes are

the velocity of the upper and lower layer respectively. The intersection time of the two lines is

the crossover point:

X(m) 15 30 45 60 75 90 105 120 135 150 165 180T(ms) 19 29 39 50 59 62 65 68 72 76 78 83


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= 41.54−8.50.687−0.226 = 71.74 The general formula for the crossover distance is

= 2 ℎ � 2+

12 1 Transposed to h:ℎ =

2 � 2 11 + 2 = 25.47

3. Given is the following Layered earth:

a.) What is the root-mean-square (RMS) velocity in reflection surveying, and how is it

related to the interval velocity and to stacking velocity?

b.) Determine the RMS-velocity for each layer.

c.) Determine from the RMS-velocities the interval velocities for each separate layer.

ANSWERS:

a.) For small offsets the stacking velocity is equal to the RMS velocity. The stacking

velocity can be obtained from the normal moveout (NMO) of the reflection hyperbola

associated to the particular reflector.

Δ = 0 ≈ 2 ≈ 2 if ≪ with z the depth of reflectorThe true

is determined by the following formula:

, = � ∑ ,2 =1∑ =1


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with , the root-mean square velocity down to the n’th interface, , the intervalvelocity of the i’th layer and the one way travel time through the i’th layer.Using Dix’s formula one can obtain from the RMS velocities the interval velocities.

= � ,2 ,−12 −1 −1 With the interval velocity of the n’th interval and ,2 , and ,−12 , −1 ,respectively, the root-mean square velocity and reflected ray travel times to the (n-

1)’th and n’th reflector.

b.) The one way travel times through the layers are:

1 = ℎ11 = 0.025 s2 = = 0.026 s 3 = = 0.043 s 4 = = 0.050 sUsing the RMS formula the root-mean square velocities for the four layers are:

,1 = � 1211 = 600 / ,2 = � 121 + 2221 + 2 = 1156/

,3 = � 121 + 222 + 3231 + 2 + 3 = 1773/

,4 = � 121 + 222 + 323 + 4241 + 2 + 3 + 4 = 2272 / c.) From the RMS velocities and with the Dix formula it is possible to derive the interval

velocities. The results can be proved directly since the interval velocities are the

velocities given in the exercise. The travel times down to the i’th reflector are:

1 = 1 = 0.025

2=

1+

2= 0.051

3 = 1 + 2 + 3 = 0.094 4 = 1 + 2 + 3 + 4 = 0.144


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and the interval velocities:

1 = � 6002 ∙ 0.0250.025

= 600

2 = � 11562 ∙ 0.051 6002 ∙ 0.0250.051 0.025 = 1508 ≈ 1500 3 = � 17732 ∙ 0.094 11562 ∙ 0.051

0.094 0.051 = 2299 ≈ 2300 = � 22722 ∙ 0.144 17732 ∙ 0.094

0.144 0.094 = 2993 ≈ 3000

4. Try to distinguish the following seismic events in the picture shown below:

• direct wave,

• two refracted waves,

• two reflected waves that come from the same interface as the refracted waves.

Try to obtain the velocity of the different layers using the direct wave and the refracted

waves.

Hints: Look for cross-over and critical points/ distances and use logic


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ANSWER:

: Direct Wave

: Reflected at 1 / 2, asymptotic to direct wave

: Reflected at 2 / 3, asymptotic to hadwave 1 / 2



: Headwave from 1 / 2, overtaken by headwave

2 / 3 before its cross over distance

: Headwave from 2 / 3

: Headwave from 3 / 4

1.

2.

3.

4.

5.

0.12/1, ≈crit X

5.13/2, ≈crit X

0.34/3, ≈

crit X


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From the slopes of the direct wave and the refracted waves the velocities of layer 1 to 4

can be calculated:

1

1=0 4.80

4.8

≈0 2.70

4.8

= 0.56

⇒

1 ≈1.78

12 = 1.2 4.81.2 6.0 ≈ 1.1 2.91.2 6.0 = 0.375 ⇒ 2 ≈ 2.66 13 = 2.4 6.02.4 6.0 ≈ 1.6 2.52.4 6.0 = 0.25 ⇒ 3 ≈ 4.00

1

4 = 3.

6 6.

03.6 6.0 ≈2.3

2.6

3.6 6.0 = 0.125 ⇒ 4 ≈ 8.00

Exercise_2_30092010_answers.256

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