Environmental and Exploration Geophysics II

tom.wilsontom.wilson@mail.wvu.edu

Department of Geology and GeographyWest Virginia University

Morgantown, WV

Gravity Methods (V)Gravity Methods (V)Simple geometrical objects and Simple geometrical objects and

problem discussionsproblem discussions

Problem 3: Prepare a drift curve for the following data and make drift corrections. Convert your corrected data to milligals. The data were collected by a gravimeter with dial constant equal to 0.0869mGals/scale division.

Station Time minutes Reading in dialDivisions

Base 11:30 0 762.71GN1 11:42 22 774.16GN2 12:14 54 759.72GN3 12:37 77 768.95GN4 12:59 99 771.02Base 13:10 110 761.18

GN4

In the 110 minute time elapsed between base station measurements gobs has decreased 1.53 scale divisions or 0.133 milliGals. Thus there is a -0.133mG drift over 110 minutes or a -0.0012 milliGal/minute drift.

Time (minutes)0 20 40 60 80 100 120

g (m

Gal

s)

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

Base Station Drift Line

Initial Base Station Reference

Initial Base StationMeasurement

Note that we can develop a simple equation to determine the value of g relative to the drift curve. The drift is just-

However, we know that this drop in g through time will increase differences that were initially positive relative to base and decrease values initially negative in relation to the base. Thus our equation should look like

Corrected final base station measurement works out as we suspect it should …

( ) 0.0012mG/min . (min)g t t

( ) Initial difference (mG)+ 0.0012mG/min . (min)g t t

( ) -0.133+ 0.0012mG/min .110(min) 0g t mG

1 2 3 4 5 6

time dial reading

Converted to milligals

relativedifference

Tide &Drift

Driftcorrected

Base Station 0 762.71 66.279499 0 0 0

1 22 774.16

2 54 759.72

3 77 768.95 66.821755 0.542256 -0.09307 0.6353259

4 99 771.01

Base Station 110 761.18 66.146542 -0.132957 -0.132957 0

….pbs2.xls

What’s the station elevation?

What’s the average elevation in Sector 1?

What’s the relative difference between the station elevation and the average elevation of sector 1?

200

520

520

280

2840

2002640 3 (0.03mG) 0.028mG

Determine the average elevation, relative elevation and T for all 8 sectors in the ring. Add these contributions to determine the total contribution of the F-ring to the terrain correction at this location.

We will also consider the F-ring contribution if the replacement density of 2.67 gm/cm3 is used instead of 2 gm/cm3 and the result obtained using equation 6-30

What did you get?

2222 2

1 21

02 zRzR oiiring RRGg

If you tried to use the preceding formula to calculate the sector effects explicitly - did you remember to convert from feet to meters?

Remember that these corrections were made assuming a replacement density of 2 gm/cm3. If you wish to estimate the effect of topography assuming another replacement density then you must adjust the total sum by a factor equal to the ratio of the desired density contrast to 2.0 gm/cm3. Hence, in the present example, use of a 2.67 gm/cm3 replacement density requires that the sum (___milligals) be factored by the ratio 2.67/2.00 = 1.33. Thus the total F-ring adjustment for 2.67gm/cm3 replacement density is ____ milligals.

Hand in on Nov. 10th

Examine the map at right. Note the regional and residual (or local) variations in the gravity field through the area.

The graphical separation method involves drawing lines through the data that follow the regional trend.

The green lines at right extend through the residual feature and reveal what would be the gradual drop in the anomaly across the area if the local feature were not present.

The residual anomaly is identified by marking the intersections of the extended regional field with the actual anomaly and labeling them with the value of the actual anomaly relative to the extended regional field.

-0.5 -0.5-1

After labeling all intersections with the relative (or residual ) values, you can contour these values to obtain a map of the residual feature.

0

Max =Min ~sign

Nettleton, 1971

Note that a particular anomaly, such as that shown below, could be attributed to a variety of different density distributions.

Note also, however, that there is a certain maximum depth beneath which this anomaly cannot have its origins.

gravity anomaly

anomaly

If there are no subsurface density contrasts - i.e. no geology, then the theoretical gravity equals the observed gravity and there is 0 anomaly.

Now let’s consider the significance of the corrected accelerations from a graphical point of view.

If there are density contrasts, i.e. if there are materials with densities different from the replacement density, then there will be an anomaly. That anomaly is the geology or site characteristics we are trying to detect.

What would be the maximum value of the residual anomaly over this model?Convert t=500 meters to feet.500 meters = 1640 feet.

g = -1640/130 milligals or-12.6 milligals

12 tan2

xg G tz

If we are only 700 meters from the edge - what would the computed depth be using the plate approximation?g = 9.5 milliGals.t = 130 g = 1245 feet

or 380 meters.

Again … remember - 1) the underlying assumption is that the drift valleys are much wider than they are deep, 2) that the expression t = 130g specifically uses the residual gravity to estimate drift thickness, 3) and in order to do that effectively, the reference value will have to be chosen carefully. If bedrock rises to the surface at some point, that might be a good point to assign a value of 0 to the residual (depends on how extensive the outcrop is).

We can often estimate the gravitational acceleration associated with complex objects such as dikes, sills, faulted layers, mine shafts, cavities, caves, culminations and anticline/syncline structures by approximating their shape using simple geometrical objects - such as horizontal and vertical cylinders, the infinite sheet, the sphere, etc.Estimates of maximum depth, density contrast, fault offset, etc. can often be made quickly and without the aid of a computer using simple relationships derived for simple geometrical forms.

Let’s start with one of the simplest of geometrical objects - the sphere

The “diagnostic position” is a reference location. It refers to the X location of points where the anomaly has fallen to a certain fraction of its maximum value, for example, 3/4 or 1/2.

In the above, the “diagnostic position” is X1/2, or the X location where the anomaly falls to 1/2 of its maximum value. The value 1.31 is referred to as the “depth index multiplier.” This is the value that you multiply the reference distance X1/2 by to obtain an estimate of the depth Z.

Diagnostic Position Depth Index Multiplier3/4 max 1/0.46 = 2.172/3 max 1/0.56 = 1.791/2 max 1/0.77 = 1.3051/3 max 1/1.04 = 0.961/4 max 1/1.24 = 0.81

3

max 2

3

2

3

2

1/32max

2max

3

(4 / 3 )

0.02793 for meters

0.00852 for feet

(feet)0.00852

(feet)0.00852

G RgZ

RZRZ

g ZR

g ZR

A table of diagnostic positions and depth index multipliers for the Sphere (see your

handout).

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.

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.

Diagnostic Position Depth Index Multiplier3/4 max 1/0.46 = 2.172/3 max 1/0.56 = 1.791/2 max 1/0.77 = 1.3051/3 max 1/1.04 = 0.961/4 max 1/1.24 = 0.81

What is Z if you are given

X1/3?

… Z = 0.96X1/3

In general you will get as many estimates of Z as you have diagnostic positions. This allows you to estimate Z as a statistical average of several values.We can make 5 separate estimates of Z given the diagnostic position in the above table.

You could measure of the values of the depth index multipliers yourself from this plot of the normalized curve that describes the shape of the gravity anomaly associated with a sphere.

Just as in the case of the anomaly associated with spherically distributed regions of subsurface density contrast, 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.

How would you determine the depth index multipliers from this graph?

X3/4 X2/3

X1/2X1/3X1/4

Z=X1/2

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. X times the depth index multiplier yields Z

Diagnostic Position Depth Index Multiplier3/4 max 1/0.58 = 1.722/3 max 1/0.71 = 1.411/2 max 1/1= 11/3 max 1/1.42 = 0.71/4 max 1/1.74 = 0.57

(feet) 01277.0

(feet) 01277.0

feetfor 01277.0

metersfor 0419.0

)2

2max

2/1max

2

2

2max

RZg

ZgR

ZR

ZR

ZRGg

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.

Nettleton, 1971

We should note again, that the depths we derive assuming these simple geometrical objects are maximum depths to the centers of these objects - cylinder or sphere. Other configurations of density could produce such anomalies.

This is the essence of the limitation we refer to as non-uniqueness. Our assumptions about the actual configuration of the object producing the anomaly are only as good as our geology.

That maximum depth is a depth beneath which a given anomaly cannot have its origins.

Problem-Determine which anomaly is produced by a sphere and which is produced by a cylinder.

Diagnosticpositions

MultipliersSphere

ZSphere MultipliersCylinder

ZCylinder

X3/4 = 0.95 2.17 2.06 1.72 1.63X2/3 = 1.15 1.79 2.06 1.41 1.62X1/2 = 1.6 1.305 2.09 1 1.6X1/3 = 2.1 0.96 2.02 0.7 1.47X1/4 = 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?

(kilofeet) 52.8

(kilofeet) 52.8

3

2max

3/12max

RZg

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.

Vertical Cylinder Diagnostic Position Depth Index Multiplier3/4 max 1/0.86 = 1.162/3 max 1/1.1 = 0.911/2 max 1/1.72= 0.581/3 max 1/2.76 = 0.361/4 max 1/3.72 = 0.27

2

1

2

1

1/ 2

max 1

max 12

0.01886 for meters

0.000575 for feet

(feet)0.000575

(feet)0.000575

RZ

RZ

g ZR

g ZR

We will spend more time on simple geometrical objects during the next lecture, but for now let’s spend a few moments and review the problems that were assigned last lecture. Pb. 3 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 void centers are never closer to the surface than 100m.

(feet) 00852.0

(feet) 00852.0

feetfor 00852.0

metersfor 02793.0

)3/4(

3

2max

3/12max

2

3

2

3

2

3max

RZg

ZgR

ZR

ZR

ZRGg

Begin by recalling the list of formula we developed for the sphere.

(feet) 02793.0

metersfor 02793.0

3/12max

2

3

ZgR

ZR

What are your givens?

& Z

Pb. 4: 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

Bou

guer

Ano

mal

y (m

Gal

)

-1.50

-1.25

-1.00

-0.75

-0.50

-0.25

0.00

Problem 5

residual

Regional

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.

In problem 5 your given three anomalies. These anomalies are assumed to be associated with three buried spheres. Determine their depths using the diagnostic positions and depth index multipliers we discussed in class today. 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)

Boug

uer A

nom

aly

(mG

als)

A.

C.B.

Nov. 10th Problems 1 & 2. Gravity lab

Nov. 15th Problems 4, 5, & 6.

Nov. 17thGravity paper summaries