gravity methods (vii) wrap up - west virginia...

Tom Wilson, Department of Geology and Geography

Environmental and Exploration Geophysics II

tom.h.wilson

[email protected]

Department of Geology and Geography

West Virginia University

Morgantown, WV

Gravity Methods (VII) wrap up

Items on the list


• 6.3 due today

• Draft essay 2 due Friday, 4th by noon

• Due date for problems 6.5 and 6.9 moved to the 10th

• Gravity lab due date delayed till Nov. 15th due to election

day recess.

• Begin reading Chapter 7 on Magnetics.

We’ll get into magnetic methods next week. Following Thanksgiving break we will wrap up magnetic methods on Nov. 28th and Dec 1st

with some exam review on the 1st and 6th. Final, Wednesday, December 14th from 11am-1pm in rm 325 Brooks.

For today


• Wrap up gravity lab demo

• Questions on handout problems 6.5 & 6.9

• More on simple geometrical objects

• An in-class problem

• General applications

Gravity lab questions


Question from the lab – Does the DC shift help? Why?

What are Stewart’s assumptions?


+2

-2

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

t=260ft???

0

-4

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

-1

This representation follows Stewart’s

conceptualization of the problem

DC refers to a constant that is added to the calculations

to shift them so they have approximately 0 average.

Shift on Shift off

To use the plate approximation we have

to honor the assumptions made by

Stewart to develop the formula t=130g.

Present in an organized and sequential manner


Disagreements we

noted and worked

with are discussed

in the lab guide.

Some obvious

mismatches.

Valley width and valley length – both must be

incorporated to accurately model the field


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

Undertake an additional inversion to produce a match

between the observed and calculated gravity


When you reduced the valley lengths, you eliminated the

match between calculated and observed fields. To obtain

a more accurate model you need to reinvert the transect.

After the inversion calculate drift thickness

using the plate formula (t=130g) and compare


The valley depth increases, but the negative anomaly observed

over this area does not. Discuss as requested above.

valley!

Present in an organized and sequential manner

with labeled figures & captions. Organization is important!


Use questions to guide your discussion


• Respond to questions as indicated.

• Follow recommended organization of presentation.

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

• All figures should be numbered, labeled and captioned.

Diagnostic position X1/2 (see discussion of half-

maximum technique section 6.7.1)


3/ 22

max1/ 2

2

1 1

21

vg

g x

z

x1/2/z = 0.766.

x½ is referred to as the

diagnostic position, z/x1/2 is referred to as the depth index multiplier

Last time we showed that when

You could solve for values of xr/z for other ratios of

gv/gmax. All solutions of z should be the same


max

vg

g

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.

3 1 14 2 4.. ......

x x x

z z z

You get different depth index multipliers; BUT if the

object is a sphere, you should get the same z!


Diagnostic Position

(g/gmax)

Depth Index Multiplier

3/4 max 1/0.46 = 2.17

2/3 max 1/0.56 = 1.79

1/2 max 1/0.77 = 1.305

1/3 max 1/1.04 = 0.96

1/4 max 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.

(depth index multiplier) times at the diagnostic positionZ X

34

12

14

z

x

z

x

z

x

Two anomalies - one broader (longer wavelength) than

the other. Which has deeper origins?


X1/2X3/4

Depth index multiplier

for X1/2 is 1.305


for X3/4 is 2.17

What depth do you get?

For example, you’ve got two anomalies. One is

broader than the other.


~750 ~450


for X1/2 is 1.305


for X3/4 is 2.17

What depth do you get?

750*1.305=

450*2.17 =

350*1.305=457 and 2.17*200=434

We could use the average as the depth

Due next Thursday: 6.5 and 6.9 as revised in

class handout


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. Given the 0.05mG measurement accuracy, the detection limit should be at least 0.1 mG.

i.e. z ≥ 100m

Basic formula with some mixed units

variations


3

2

3

2

0.02793 for meters

0.00852 for feet

R

Z

R

Z

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.

2

max

3 (feet)

(4 / 3 )

g Z

GR

1/32

max (4 / 3)

g ZR

G

3

max 2

(4 / 3 )G Rg

Z

1/32

max (feet)0.00852

g ZR

2

max

3 (feet)

0.00852

g Z

R

Revised 6.9


In a problem similar to the problem in the text, 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

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

-1500 -1000 -500 0 500 1000 1500

Distance from peak (m)

Bo

ug

uer

An

om

aly

(m

Gals

)

This technique can be applied to other geometrical

objects: for example, the horizontal cylinder.


Cylinder with radius R and density

X

z

R

What could the horizontal cylinder

represent geologically?

The anomaly across the cylinder is also symmetrical


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

X

z r

The result shares similarity to that for the sphere

(see equation 6.37 and excel table 6.7)


2

2

2

2 1

1cyl

G Rg

xZz

2

max

2 G Rg

Z

max 2

2

1

1cylg g

xz

and

See derivation in text

For the diagnostic position of ½ gmax …


max

1

2

cylg

g

Choose the position of interest and solve for the ratio x/z

2

1/22

1 1

21

x

z

2

1/22 1 2

x

z

2

1/22 1

x

z

2

1/2 1x

z

12

x z

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

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

X1/2X1/3X1/4

Z=X1/20.71

0.58

0.71

1.0

1.42

1.74

0.58

Just as we did

for X1/2 solve for

X3/4, …etc.

Just as we did for the sphere, we’ve derived depth

index multipliers for several diagnostic positions


Diagnostic Position Depth Index Multiplier

3/4 max 1/0.58 = 1.72

2/3 max 1/0.71 = 1.41

1/2 max 1/1= 1

1/3 max 1/1.42 = 0.7

1/4 max 1/1.74 = 0.57

(feet) 01277.0

(feet) 01277.0

feetfor 01277.0

metersfor 0419.0

2

2

max

2/1

max

2

2

2

max

R

Zg

ZgR

Z

R

Z

R

Z

RGg

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.

With Z, you can then speculate

on the density contrast or radius

of the object in question.

For the cylinder we have

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.

This is the key idea associated with evaluation of diagnostic

positions: if the estimated z’s are similar for a certain

assumed shape, that helps confirm your interpretation.

Can you tell which anomaly is produced by a

horizontal cylinder and which, by the sphere?


Remember Z=1.305X1/2 for the sphere

and Z=X1/2

The depth to the center, Z, is the same for each

Assume the anomaly below is produced by long

horizontal tunnel – What is the depth to the tunnel?


X3/4X1/2

What are the depth

index multipliers?

X1/2~100m

X3/4~60m

Z=

Z=

DIM=1

DIM=1.72

The example below illustrates the application

in detail


Diagnostic

positions

Multipliers

Sphere

ZSphere Multipliers

Cylinder

ZCylinder

X3/4 = 0.95 2.17 2.06 1.72 1.63

X2/3 = 1.15 1.79 2.06 1.41 1.62

X1/2 = 1.6 1.305 2.09 1 1.6

X1/3 = 2.1 0.96 2.02 0.7 1.47

X1/4 = 2.5 0.81 2.03 0.57 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?

It’s been worked up in the table

below. What do you think?

As we’ve shown, we can estimate other properties of

the buried object


(kilofeet) 52.8

(kilofeet) 52.8

3

2max

3/12

max

R

Zg

ZgR

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.

In-class activities – precision matters – use a ruler


Take a few minutes and determine what

shaped object produces each anomaly


Just note that this approach has been developed for a

number of simple geometrical shapes


Diagnostic Position Depth Index Multiplier

3/4 max 1/0.86 = 1.16

2/3 max 1/1.1 = 0.91

1/2 max 1/1.72= 0.58

1/3 max 1/2.76 = 0.36

1/4 max 1/3.72 = 0.27

2

1

2

1

1/ 2

max 1

max 1

2

0.01886 for meters

0.000575 for feet

(feet)0.000575

(feet)0.000575

R

Z

R

Z

g ZR

g Z

R

2

1/ 2 1/ 222 2 2

1 1g G R

z x z L x

A vertical

cylinder or

volcanic pipe

For a given anomaly certain simple geometries

can be assumed and tested


Sphere or vertical cylinderHorizontal cylinder or

vertical dyke

A A’A A’

A A’


Half plate or faulted plate

How about the anomaly below?

10 mG

0 mG

Fault is located at the anomaly inflection point


Half plate or faulted plate

10 mG

0 mG

High Low

Hig

h a

ngle

fau

lt: n

orm

al

or

reve

rse

Recall the use of this simple geometrical object used to

obtain the topographic correction


Butte

12 sectors with Ri=1100 and Ro=2200

Ring

The butte fits

into one sector

Consider how you

would do this

Another application of simple geometrical objects and

the residual


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/cm3

Horizontal Position (km)

0.0 0.5 1.0 1.5 2.0

Bouguer

Anom

aly (

mG

al)

-1.50

-1.25

-1.00

-0.75

-0.50

-0.25

0.00

Problem 5

A lot of these ideas carry

over into the analysis of

magnetic data


residual

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.

Just as with the graphical approach, the idea is to

remove the regional so you can investigate the residual.


With the residual

anomaly you can

answer the

question: what is

the depth?


Derived from Gravity Model Studies

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

As always, consider the

possibility of non-unique

solutions


Are alternative acceptable solutions

possible?

Gravity applications span a variety of scales


Roberts, 1990

Shallow environmental applications

Topographic extremes


Japan Archipelago

Pacific Plate

North

American

Plate

Philippine

Sea Plate

Geological Survey of Japan


The Earth’s gravitational field

In the red areas you weigh more and

in the blue areas you weigh less.

Pacific Plate

North

American

Plate

Philippine

Sea Plate

Geological Survey of Japan

g ~0.6 cm/sec2

Gravity methods have applications over a wide

range of scales


Items on the list


We’ll get into magnetic methods next week. Following Thanksgiving break we will wrap up magnetic methods on Nov. 28th and Dec 1st

with some exam review on the 1st and 6th. Final, Wednesday, December 14th from 11am-1pm in rm 325 Brooks.

• 6.3 due today

• Draft essay 2 due Friday, 4th by noon

• Due date for problems 6.5 and 6.9 moved to the 10th

• Gravity lab due date delayed till Nov. 15th due to election

day recess.

• Begin reading Chapter 7 on Magnetics.

gravity methods (vii) wrap up - west virginia...

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