Geology 5660/6660Applied Geophysics

This Week:

• No new lab assignment…

• But we’ll go over the previous labs

06 Feb 2014

© A.R. Lowry 2014For Fri 07 Feb: Burger 167-199 (§4.2-4.3)

Your lab assignment (due at 7:30 on 30 Jan):

• Install the Burger et al software on a pc/laptop you use• Model using Refract the following, using 20 stations, 5m shot offset, 5m spacing:

a 9.4 m layer (V = 2 km/s) over a 5 km/s infinite space a 3 m layer (V = 2 km/s) over a 3 m layer (V = 1 km/s) over a 5 km/s infinite space.

• Compare and contrast the travel times derived from these models. How would you distinguish these two Earth structures based on the observed data (i.e., the circles)?• Model the same two models using Reflect (be sure to check “show reflections”)• What, if any, additional information does this give you?

Lab write-ups should include plots, numericaldata if relevant, and explanatory backgroundwhere appropriate!

Visually, the observations(i.e., the circlesin these plots!)appear verysimilar, insomuchas one can tell bylooking…

The lines are modelrepresentations (&the first “directarrival” lines are abug!)

In fact they’re almost identical!

And if you look at times, they are identical to “standard” sig-dig

The reflections onthe other hand areslightly different (asone would expect,because the reflectionseismic method doesnot have the samelimitation on sensinglow velocity layers).

But differences aresubtle and may not beclear how one woulduse, because layersare thin and reflectionwavelets superpose

€

Arfl

Ai

=ρ 2V2 − ρ1V1

ρ 2V2 + ρ1V1

≡Z2 − Z1

Z2 + Z1

≈A12 = −0.333Ai

A23 = 0.667Ai

€

Arfr

Ai

=2ρ1V1

ρ 2V2 + ρ1V1

≡2Z1

Z2 + Z1

≈A12 =1.333Ai

A23 = 0.4Ai

However it’s mucheasier to pick outthe differences if weuse a shorter waveletpulse! Because nowthe wavelets are notsuperposed.

Hence, resolution ofthe seismic reflectionmethod depends onfrequency content ofthe returns!

(1) Seismic data above were collected on a farm near Gosport, IN. The faint lines demark 0.01 s intervals; darker lines are at 0.05 s. (a) Pick (as best you can) the time of the first arrival on each trace. (Hint: The furthest first arrival is at < 0.1 s). (b) Geophone distances are (0,4,8, 12,16,20,26,32,38,44,50,56,62,68,74,80,86,92,96,100,104,108,112, 116) m from the source. Enter the times & distances into Refract and model the velocity structure assuming 0° dip on all layer boundaries. Do any features of the data suggest nonuniform layer thickness?

Lab 2: Part 1. Travel-Time

0.01 s

0.02 s

0.03 s

0.04 s

0.05 s

One subtlety: The positive deflection at 4 m would imply a(nonphysical) velocity of 200 m/s (< 330 m/s for sound in air!) The true first arrivals are negative (polarity problem) so hard to see. Picking the positive gives wrong V, so wrong thickness, in the first layer, and hence wrong thickness (but not wrong V)in deeper layers.

Hints: Helps to blow up to large scale; more accurate if youcreate and use a ruler…

Gosport Best Fit RMS = 1.08 ms

Lab 2: Part 2. Amplitudes in Reflect

(2) This plot shows output from the program Reflect. Assume the distance between traces corresponds to a displacement amplitude of 1. Do amplitudes reflect expected relations for geometrical spreading? For attenuation? Why or why not?

1/r

Direct

Spherical Spreading:

1/r

€

A =A0e−

πfr

QV

Attenuation:Cylindrical Spreading:

€

1 r

Refracted

AirReflected

Different arrivals have different amplitudes within eachtrace, but do not exhibit expected distance dependence…

(3) The plot @ left shows only the reflected wave for the same model. (a) Derive a relation for angle of incidence of the reflected P-wave as a function of distance. (b) The image at right (from a website application) shows Zoeppritz amplitudes for the reflected P. Are these amplitude relations exhibited in the model at left? Why or why not?

Lab 2: Part 3. Amplitudes in Reflect

x

h

€

tanθ =x

2h

€

=tan−1 x

2h

⎛

⎝ ⎜

⎞

⎠ ⎟

14°27°

36°

Trace at 9 m should be ~ 5 timesthe amplitude of trace at 6 m!

Must be doing something though!Probably uses reflection and transmission coefficients, but notspreading, attenuation or Zoeppritz.

(4) Amplitudes shown below were measured from first arrivals in the Gosport data (Part 1). (a) Are these consistent with what you might expect from your travel-time analysis? Why or why not? (Hint:

Think about thepropagation pathof each arrival!) (b)Estimate frequencyof the first-arrivingwave for each arrivalfrom the plot (Part 1).(c) Correct theamplitudes for the effects of geometricalspreading. Then, separately for eachlayer, estimate thequality factor Q ofanelastic attenuation.

Lab 2: Part 4. Amplitudes from observations

(Hint: Don’t correct for geometrical spreading if r = 0!)

(Hint: Don’t try to compare refracted arrival amplitudes to r = 0 amplitude!)

• Must first recognize that there are three different arrivals with three different values of Q; two types of spreading• Layer 1 spreading correction is • Layer n (= 2, 3) spreading correction is

• Invert the relation:

(Since amplitudemeasurements arenoisy, I used allarrivals from a layer & averaged)

1 2 3

€

ΔAi1 = A0

1 1−1

ri

⎛

⎝ ⎜

⎞

⎠ ⎟

€

ΔAin = A0

n 1

r0n

−1

rin

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

V = 1330 m/sf = 250 Hz

Q = 9.5

V = 3510 m/sf = 125 Hz

Q = 6.6

€

A =A0e−

πfr

QV

€

Q = −πfr

V ln AA0

⎛ ⎝ ⎜

⎞ ⎠ ⎟X

X

46510015.1

(d) Assuming that both your velocity and your attenuation structure estimates are correct, look through your text (andwhatever other literature you may be able to find, but be sureto include references!) for examples of soils/sediments/rocksthat match your estimates of VP and Q. Does the combinationof these two properties reduce the ambiguities in possible interpretations? What interpretation of the data would make the most geological sense?

Crustal Rocks

1 2 3

The velocities are consistent witha wide range of possibilities but Layer 1 is clearly dry soil, alluviumor dry sand; Layer 2 is either the same material as Layer 1 (but below the water table) or a changeto a different type of alluvium or shale; and Layer 3 is shale,sandstone, limestone or dolomite.ALL the quality factors are extremely low (soil Q = 2–20!), ruling out consolidated sed rx in Layer 3. Given the geologic environment, L1 = dry soil; L2 is most likely water table, & L3 is either shale or (more likely)glacial-derived clay.