Environmental and Exploration Geophysics II

Gravity Methods (VII) wrap up tom.h.wilson [email protected] Department of Geology and Geography West Virginia University Morgantown, WV

Tom Wilson, Department of Geology and Geography

Items on the list

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Problems 6.1-6.3 are due today

• • • •

Also hand in the terrain calculation for one of the f-ring sectors Gravity lab will be due on Tuesday November 18th Problems 6.5 and 6.9 will be due on Thursday, Nov. 20 th Magnetic papers will be in the mail room this Thursday. Begin reading Chapter 7 on Magnetics.

•

We’ll get into magnetics next Tuesday and Thursday. Following Thanksgiving break we will continue with magnetic methods on December 2nd and 4th and begin some final review on the 4th with a wrap-up review session on the 9th. Tom Wilson, Department of Geology and Geography

One last look at the gravity lab Abstract: a brief description of what you did and the main result(s) you obtained (~200 words). Background: Provide some background on the purpose of the survey, the data we’re analyzing... All of this would come from Stewart’s paper. Explain the approximation he uses and answer question 1 below in this section to illustrate his approach. Results: Describe how you tested the model proposed by Stewart (the one we are reworking in the lab). Include answers to questions 2 through 4 below in this results discussion. Conclusions: Summarize the highlights of results obtained in the forgoing modeling process.

Tom Wilson, Department of Geology and Geography

Questions on the questions? 1. The residual gravity plotted in Figure 5 of Stewart's paper (also see illustrations in this lab exercise) has both positive and negative values. Assume that an anomaly extends from +2milligals to -2 milligals. Use the plate approximation (i.e. Stewart’s plate formula) and estimate the depth to bedrock? What do you need to do to get a useful result? Residuals of any kind usually fluctuate about zero mean value. What would you guess Stewart must have done to the residual values before he computed bedrock depth?

The model is simplified to consist of two layers: one with a -0.6 gm/cm3 density (the till) relative to the second (bedrock) with a 0 gm/cm3 density. Actual densities are 2 and 2.6 gm/cm3, but for modeling purposes the results are equivalent in a relative sense.

Also note that Stewart’s equation, t=130gr requires densities of -0.6 and 0 gm/cm3. The formula is derived assuming a density contrast of 0.6 gm/cm3, where g is expressed in milliGals and t is obtained in units of ft. Use of the formula actually assumes |gr|. Tom Wilson, Department of Geology and Geography

Relative anomalies don’t vary This representation follows Stewart’s conceptualization of the problem

2

Estimate the depth of this valley using the formula t=130gr. 260ft??

-2

Tom Wilson, Department of Geology and Geography

0 -1

-4

Estimate the depth of this valley using the formula t=130gr. > 546’

The only possible anomalies are negative and valley depth is approximated using the plate formula gp=2Gt or t=130gr

0 -1

-4

Tom Wilson, Department of Geology and Geography

Estimate the depth of this valley using the formula t=130gr. > 546’

We’ve emphasized that the formula provides only a rough approximation since the valleys are not infinitely wide.

In your write-up answer the following questions and refer to them by number for identification. 2. At the beginning of the lab you made a copy of GMSYS window showing some disagreement between the observations (dots) and calculations (solid line) across Stewart's model (section XX' Figure 7). As we did in class and in the lab manual, note a couple areas along the profile where this disagreement is most pronounced, label these areas in your figure for reference. In your lab report discussion offer an explanation for the cause(s) of these differences? Assume that the differences are of geological origin and not related to errors in the data.

Disagreements we noted are discussed in the lab guide.

Tom Wilson, Department of Geology and Geography

In your write-up answer the following questions and refer to them by number for identification. 3. With a combination of inversion and manual adjustments of points defining the till/bedrock interface, you were able to eliminate the significant differences between observed and calculated gravity. Your model is incorrect though since the valleys do not extend to infinity in and out of the cross section. Use the 2 ¾ modeling option to reduce the extents of the valleys in and out of the section to 1000 feet. Make the changes to the Y+ and Y- blocks and then apply. Take a screen capture to illustrate the reduction in g associated with the glacial valleys. Make a screen capture of this display showing the new calculation line and the dashed gray values associated with the infinite valleys. Include this figure in your report and discuss your results.

This part of the problem gets you to deal with the 3 dimensional aspects of buried valley geometry.

Tom Wilson, Department of Geology and Geography

Note differences between infinitely long valleys and those extending in and out of the section only 0.1 km.

Tom Wilson, Department of Geology and Geography

Refer to answers by question number in your lab report 4. Use Stewart's formula t = 130g and estimate the depth to bedrock at the x location of 7920 feet along the profile. Does it provide a reliable estimate of bedrock depth in this area? Explain in your discussion. Your answer here summarizes issues addressed in question 1 and 3. You can do it on the infinitely long valleys or the ones you limit to 1000’ (just indicate which). But do it with and without shifting and then compare! Remember, to shift, RC on the data display pane and select DC Shift – Absolute. 5. Lastly, describe the model you obtained and comment on how it varies from the starting model taken from Stewart. Answers to question 5 reasonably go in the conclusions. Tom Wilson, Department of Geology and Geography

Use questions to guide your discussion

• The questions in the lab guide provide discussion points for your lab report.

• Use figures you've generated in GMSYS to illustrate your point.

• All figures should be numbered, labeled and captioned.

Tom Wilson, Department of Geology and Geography

Back to simple geometrical shapes with some review .. We’ve developed the shape of the anomaly produced by a g vert sphere.

    3 G (4 / 3 R )  1   3   2 Z2  2  x   z 2  1    

Also considered the idea of diagnostic positions in addition to the half-max gvert relation discussed in z the text.

X

r Sphere with radius R and density 

Tom Wilson, Department of Geology and Geography

Diagnostic position X1/2 (see discussion of halfmaximum technique section 6.7.1)

gv 1 1   3/ 2 2 g max 2  x  1/ 2  2  1  z  x½ is referred to as the diagnostic position, 1/x1/2 is referred to as the depth index multiplier

Tom Wilson, Department of Geology and Geography

We solve for x1/2/z and find that x1/2/z = 0.766.

Also noted that we could have several diagnostic positions (see handout

gv g max 3/4 1/2

1/4

Evaluation at multiple diagnostic locations does two things for you: allows you to obtain an average Z and helps test your assumption about anomaly origin. If it’s not a sphere, then the values of Z will differ significantly.

Tom Wilson, Department of Geology and Geography

The other day we developed these tables of diagnostic positions and depth index multipliers Diagnostic Position (g/gmax) 3/4 max 2/3 max 1/2 max 1/3 max 1/4 max

Depth Index Multiplier 1/0.46 = 2.17 1/0.56 = 1.79 1/0.77 = 1.305 1/1.04 = 0.96 1/1.24 = 0.81

Note that regardless of which diagnostic position you use, you should get the same value of Z. Each depth index multiplier converts a specific reference X location distance to depth.

Z  (depth index multiplier) times X at the diagnostic position Tom Wilson, Department of Geology and Geography

For example, you’ve got two anomalies. One is broader than the other. Which has deeper origins? Depth index multiplier for X1/2 is 1.305

Depth index multiplier for X3/4 is 2.17

What depth do you get? X1/2 X3/4

m Wilson, Department of Geology and Geography

For example, you’ve got two anomalies. One is broader than the other. Depth index multiplier for X1/2 is 1.305 750*1.305=

Depth index multiplier for X3/4 is 2.17 450*2.17 =

What depth do you get? ~750 ~450

Tom Wilson, Department of Geology and Geography

To be handed in next Thursday: Problems 6.5 and 6.9 Problem 6.5 What is the radius of the smallest equidimensional void (such as a chamber in a cave & think of it more simply as an isolated spherical void) that can be detected by a gravity survey for which the Bouguer gravity values have an accuracy of 0.05 mG? Assume the voids are in limestone and are air-filled (i.e. density contrast, , = 2.7gm/cm3) and that the void centers are never closer to the surface than 100m.

i.e. z ≥ 100m

Tom Wilson, Department of Geology and Geography

For problem 6.5: You are given Z and , solve for R G (4 / 3 R 3 ) g max  Z2 R3  0.02793 2  for meters Z R3  0.00852 2  for feet Z 1/3  g max Z 2  R  (4 / 3)  G     1/3

 g max Z  R  0.00852     2

Tom Wilson, Department of Geology and Geography

(feet)

g max Z 2   (4 / 3 )GR3 g max Z 2   0.00852 R 3

(feet) (feet)

These constants (i.e. 0.02793 or 0.00852) assume that depths and radii are in the specified units (feet or meters), and that density is always in gm/cm3.

Problem 6.9 In a problem similar to problem 6.9 (Burger et al.) you’re given three anomalies. These anomalies are assumed to be associated with three buried spheres. Determine their depths using the diagnostic position and depth index multiplier as discussed in class. Carefully consider where the anomaly drops to one-half of its maximum value. Assume a minimum value of 0. 0.5

Bouguer Anomaly (mGals)

0.45 0.4 0.35

A.

0.3

C.

B.

0.25 0.2 0.15 0.1 0.05 0 -1500

-1000

-500

0

500

Distance from peak (m)

Tom Wilson, Department of Geology and Geography

1000

1500

We’ll discuss one more simple geometrical shape: the horizontal cylinder What could the horizontal cylinder represent geologically? X

z

Cylinder with radius R and density 

R

Tom Wilson, Department of Geology and Geography

X

z

r At surface distance x away from a point directly over the cylinder

Tom Wilson, Department of Geology and Geography

The result shares similarity to that for the sphere (see equation 6.37 and excel table 6.7)

g cyl

  2 G  R 2  1    x 2  1 Z  z 2  g cyl

Tom Wilson, Department of Geology and Geography

and

g max

  1   g max  2 x   1 2  z 

2 G  R 2  Z

We can ask the same kinds of questions we asked regarding the sphere. For example, g cyl

Where does

g max



1 2

1 1  2 2 x1/2  1 z2 2 x1/2 2 1  2 z

2 x1/2

z

2

1

2 x1/2 1 z

x1  z 2

This tells us that the anomaly falls to ½ its maximum value at a distance from the anomaly peak equal to the depth to the center of the horizontal cylinder Tom Wilson, Department of Geology and Geography

See class handout/worksheet Locate the points along the X/Z Axis where the normalized curve falls to diagnostic values - 1/4, 1/2, etc. The depth index multiplier is just the reciprocal of the value at X/Z at the diagnostic position. X times the depth index multiplier yields Z

X3/4X2/3 X1/4 X1/3 0.58

Just as we did for X1/2 solve for X3/4, …etc. 0.58

0.71 1.0

1.42 Tom Wilson, Department of Geology and Geography

X1/2

1.74

0.71

Z=X1/2

Just as we did for the sphere, we’ve derived depth index multipliers for several diagnostic positions Diagnostic Position 3/4 max 2/3 max 1/2 max 1/3 max 1/4 max

G 2R 2  g max  Z R2  0.0419  for meters Z R2  0.01277  for feet Z 1/ 2

 g max Z   (feet) R   0 . 01277     g max Z   (feet) 0.01277 R 2 Tom Wilson, Department of Geology and Geography

For the cylinder we have Depth Index Multiplier 1/0.58 = 1.72 1/0.71 = 1.41 1/1= 1 1/1.42 = 0.7 1/1.74 = 0.57

With Z, you can then speculate on the density contrast or radius of the object in question. Again, note that these constants (i.e. 0.02793) assume that depths and radii are in the specified units (feet or meters), and that density is always in gm/cm3.

Horizontal Cylinder Just as was the case for the sphere, objects which have a cylindrical distribution of density contrast all produce variations in gravitational acceleration that are identical in shape and differ only in magnitude and spatial extent.

When these curves are normalized and plotted as a function of X/Z they all have the same shape. It is that attribute of the cylinder and the sphere which allows us to determine their depth and speculate about the other parameters such as their density contrast and radius.

Tom Wilson, Department of Geology and Geography

Can you tell which anomaly is produced by a horizontal cylinder and which, by the sphere? The depth to the center, Z, is the same for each

Remember Z=1.305X1/2 for the sphere and Z=X1/2

Tom Wilson, Department of Geology and Geography

Assume the anomaly below is produced by long horizontal tunnel – What is the depth to the tunnel?

What are the depth index multipliers?

X1/2~100m DIM=1 Z= X3/4~60m DIM=1.72 Z=

X1/2

Tom Wilson, Department of Geology and Geography

X3/4

Use diagnostic positions to differentiate between source geometries It’s been worked up in the table below. What do you think? Diagnostic positions X3/4 = 0.95 X2/3 = 1.15 X1/2 = 1.6 X1/3 = 2.1 X1/4 = 2.5

Multipliers Sphere 2.17 1.79 1.305 0.96 0.81

ZSphere 2.06 2.06 2.09 2.02 2.03

Multipliers Cylinder 1.72 1.41 1 0.7 0.57

ZCylinder 1.63 1.62 1.6 1.47 1.43

Which estimate of Z seems to be more reliable? Compute the range. You could also compare standard deviations. Which model - sphere or cylinder - yields the smaller range or standard deviation?

Tom Wilson, Department of Geology and Geography

As we’ve shown, we can estimate other properties of the buried object To determine the radius of this object, we can use the formulas we developed earlier. For example, if we found that the anomaly was best explained by a spherical distribution of density contrast, then we could use the following formulas which have been modified to yield answer’s in kilofeet, where Z is in kilofeet, and  is in gm/cm3.

1/ 3

2 g Z R   max   8.52   

 

g max Z 2 8.52 R

Tom Wilson, Department of Geology and Geography

3

(kilofeet)

(kilofeet)

In-class activity 1

Tom Wilson, Department of Geology and Geography

Just note that this approach has been developed for a number of simple geometrical shapes Diagnostic Position 3/4 max 2/3 max 1/2 max 1/3 max 1/4 max

Depth Index Multiplier 1/0.86 = 1.16 1/1.1 = 0.91 1/1.72= 0.58 1/2.76 = 0.36 1/3.72 = 0.27

R2  0.01886  for meters Z1 R2  0.000575  for feet Z1 1/ 2

  1 1   g  G  R 2    1/ 2 1/ 2 2 2 2 2   z  x   z  L  x     Tom Wilson, Department of Geology and Geography

 g max Z1  R   0.000575  g max Z1   0.000575 R 2

(feet) (feet)

For a given anomaly certain simple geometries can be assumed and tested Horizontal cylinder or vertical dyke

Sphere or vertical cylinder

A

A’

A

A’

Tom Wilson, Department of Geology and Geography

A

A’

How about the anomaly below?

Half plate or faulted plate 10 mG

0 mG Tom Wilson, Department of Geology and Geography

Fault is located at the anomaly inflection point Half plate or faulted plate 10 mG

Low 

Tom Wilson, Department of Geology and Geography

High angle fault: normal or reverse

0 mG

High 

Consider another in-class problem

12 sectors with Ri=1100 and Ro=2200

Ring

The butte fits into one sector

Tom Wilson, Department of Geology and Geography

Butte

We’ll return to some examples as time permits, but work through the two in-class problems before leaving Just for general discussion > (see 6.8, Burger et al.): The curve in the following diagram represents a traverse across the center of a roughly equidimensional ore body. The anomaly due to the ore body is obscured by a strong regional anomaly. Remove the regional anomaly and then evaluate the anomaly due to the ore body (i.e. estimate it’s deptj and approximate radius) given that the object has a relative density contrast of 0.75g/cm 3 Bouguer Anomaly (mGal)

Problem 5 0.00 -0.25 -0.50 -0.75 -1.00 -1.25 -1.50 0.0

0.5

1.0

1.5

Horizontal Position (km) Tom Wilson, Department of Geology and Geography

2.0

You could plot the data on a sheet of graph paper. Draw a line through the end points (regional trend) and measure the difference between the actual observation and the regional (the residual). You could use EXCEL or PSIPlot to fit a line to the two end points and compute the difference between the fitted line (regional) and the observations.

residual

Tom Wilson, Department of Geology and Geography

Just as with the graphical approach, the idea is to remove the regional so you can investigate the residual.

Tom Wilson, Department of Geology and Geography

Gravity model studies help us estimate the possible configuration of the continental crust across the region

Derived from Gravity Model Studies Tom Wilson, Department of Geology and Geography

Are alternative acceptable solutions possible?

Tom Wilson, Department of Geology and Geography

Gravity applications span a variety of scales

Shallow environmental applications

Roberts, 1990 Tom Wilson, Department of Geology and Geography

Topographic extremes Japan Archipelago

North American Plate

Pacific Plate

Philippine Sea Plate

Tom Wilson, Department of Geology and Geography

Geological Survey of Japan

The Earth’s gravitational field In the red areas you weigh more and in the blue areas you weigh less.

North American Plate

g ~0.6 cm/sec2

Pacific Plate

Philippine Sea Plate

Tom Wilson, Department of Geology and Geography

Geological Survey of Japan

Gravity methods have applications over a wide range of scales

Tom Wilson, Department of Geology and Geography

Items on the list

•

Problems 6.1-6.3 are due today

• • • • •

Also hand in the terrain calculation for one of the f-ring sectors Turn in the two in-class problems before leaving Gravity lab will be due on Tuesday, November 18th Problems 6.5 and 6.9 will be due on Thursday, Nov. 20 th Magnetic papers will be in the mail room this Thursday. Begin reading Chapter 7 on Magnetics.

•

We’ll get into magnetics next Tuesday and Thursday. Following Thanksgiving break we will continue with magnetic methods on December 2nd and 4th and begin some final review on the 4th with a wrap-up review session on the 9th. Tom Wilson, Department of Geology and Geography