Transcript No Slide Title

Note that a particular anomaly, such as that shown below, could be
attributed to a variety of different density distributions.
gravity anomaly
Note also, however, that
there is a certain
maximum depth
beneath which this
anomaly cannot have its
origins.
Nettleton, 1971
Now let’s consider the significance of the corrected
accelerations from a graphical point of view.
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.
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.
Problems taken from Berger (1992)
Questions about problem 1?
1. If a gravity determination is made at an elevation of 152.7m, what is
the value of the free air correction (assuming that sea-level is the
datum)?
Also, what is the Bouguer correction assuming a 2.5 gm/cm3 reduction
density?
Note that the Bouguer plate and free-air corrections are often combined
into the elevation correction -
Elevation Correction  gFA  gB
Elevation Correction  0.3086 z  0.04193 z
In metric units (mks)
Problem 2: A gravity station at 0m is located in the center of an
erosional basin. The floor of the basin has virtually no relief. An
escarpment of a plateau is located at a distance of 450 m from the
gravity station. The surface of the plateau has a relatively
constant elevation of 400m. Will terrain corrections be
necessary?
Assume that the material above sea level that forms this vast
plateau has a density of 2.5gm/cm3.
Note that this problem is nicely set up to employ direct
computation of the terrain effect using a ring of inner radius
450m and outer radius of .
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 dial
Divisions
0
762.71
22
774.16
Base
GN1
11:30
11:42
GN2
12:14
54
759.72
GN3
12:37
77
768.95
GN4
12:59
99
771.02
Base
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.
Station
Time
minutes Reading in dial
Divisions
0
762.71
22
774.16
Base
GN1
11:30
11:42
GN2
12:14
54
759.72
GN3
12:37
77
768.95
GN4
12:59
99
771.02
Base
13:10
110
761.18
Difference
(mG)
0
+0.54mG
-0.133mG
GN4
Let’s take the value at station 3 (GN3). Its value in milliGals
relative to the base station is 0.54 milliGals. However, we know
that the gravity at the base and therefore throughout the area has
been falling at a rate of -0.0012mG/minute. Thus we suspect that
the actual difference between the base station and station 3 must
be larger. Calculating the drift over a 77 minute period, we find
that the gravity at station 3 must, in fact be 77min x
0.0012mG/min or 0.092mG greater than reported.
Actual difference relative
to the base station is
+0.63mG.
0.5
0.54mG
0
0.092mG
-0.133
Base
1
2
3
4
Base
Note that we can develop a simple equation to determine
the value of g relative to the drift curve. The drift is just-
g (t )    0.0012mG/min .t (min)
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
g (t )  Initial difference (mG)+  0.0012mG/min .t(min)
Of course it works - look at the final base station measurement
g(t )  -0.133+  0.0012mG/min .110(min)  0mG
Station
Time
minutes Reading in dial
Relative
Corrected
Divisions
difference (mg)
value
0
762.71
0
22
774.16
Base
GN1
11:30
11:42
GN2
12:14
54
759.72
GN3
12:37
77
768.95
GN4
12:59
99
771.02
Base
13:10
110
761.18
+0.54mG
+0.63mG
-0.133mG
GN4
I suggest that you let the first measurement at the base station - gbase = 0.
Also note that there is nothing absolute about 0 scale divisions or, for that
matter, the 0 milliGals reference point.
There’s no point working with the scale divisions when it is accelerations
we are after.
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
2840
520
280
2640
200
3 (0.03mG)
0.028mG
What did you get?
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
Sector
Average
Elevation
Relative
Difference
T
(1/100ths milligals)
1
2
3
4
5
6
7
8
2640
2320
2320
2560
2640
240
2760
2840
200
520
520
280
200
440
80
0
3
? (guess 17)
? (guess 17)
6

g ring  2G  R0  Ri 


Ri2  z 2
*Sector effect
computed from
Equation 6-30
0.028
0.18
0.18
0.055
1 
2 2
 R  z 
1
2
2
o

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
(0.6065 milligals) be factored by the ratio 2.67/2.00 = 1.33. Thus the total fring adjustment for 2.67gm/cm3 replacement density is 0.81 milligals.
Hand in on Nov. 16th
Sector
1
2
3
4
5
6
7
8
Average
Elevation
Relative
Difference
T
(1/100 milligals)
2640
2320
2320
2560
2640
240
2760
2840
200
520
520
280
200
440
80
0
3
? (guess 17)
? (guess 17)
6
ths
*Sector effect
computed from
Equation 6-30
0.028
0.18
0.18
0.055
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.
After labeling all
intersections with the
relative (or residual )
values, you can contour
these values to obtain a
map of the residual feature.
0
-1
-0.5
-0.5
Max = 0
Min ~-1.8
negative
Gravity Anomaly across t he Edge of a Burried Valley

1  x  
g  2G t   tan  
 z 
2
What would be the maximum
value of the residual anomaly
over this model?
g (milliGals)
0.0
-2.0
-4.0
g1/2 = 6.35mGal
-6.0
g3/4 = 9.5mGal
-8.0
-12.0
-3000
-2000
-1000
Convert t=500 meters to feet.
1000
2000
3000
Model
250
Depth (meters)
0
-12.6 milligals
0
x (meters)
500 meters = 1640 feet.
g = -1640/130 milligals or
ganom = -11.7mGal
X3/4 = 700m
-10.0
Surface
-250
z = 700m
-500
t = 500m
-750
-1000
-3000
3
 = -600 kg/m
3
 = 0 kg/m
-2000
-1000
0
x (meters)
1000
2000
3000
0.0
g (milliGals)
If we are only 700 meters
from the edge - what would
the computed depth be using
the plate approximation?
Gravity Anomaly across the Edge of a Burried Valley
g = 9.5 milliGals.
-2.0
-4.0
g1/2 = 6.35mGal
-6.0
g3/4 = 9.5mGal
-8.0
X3/4 = 700m
-10.0
t = 130 g = 1245 feet
-12.0
-3000
-2000
-1000
0
1000
2000
3000
x (meters)
or 380 meters.
Model
0
Depth (meters)
Note how the reference point
becomes super critical here.
In order to get total depth
rather than depth relative to
the top of the valley wall, all
these anomaly values need to
be shifted down 11.3
milligals
250
Surface
-250
z = 700m
-500
 = 0 kg/m
t = 500m
-750
-1000
-3000
3
 = -600 kg/m
3
-2000
-1000
0
x (meters)
1000
2000
3000
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.
A table of diagnostic positions and depth index
multipliers for the Sphere (see your handout).
Diagnostic Position
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.
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

 0.00852 
g max Z 2
 
0.00852 R 3
(feet)
(feet)
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
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
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?
Locate the points
along the X/Z Axis
where the
normalized curve
falls to diagnostic X1/4 X1/3
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
X3/4X2/3
X1/2
Z=X1/2
Diagnostic Position
3/4 max
2/3 max
1/2 max
1/3 max
1/4 max
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
G 2R 2 )
g max 
Z
 0.0419
2
R
 for meters
Z
R2
 0.01277
 for feet
Z
1/ 2
 g maxZ 

R  
0
.
01277




g maxZ
 
0.01277 R 2
(feet)
(feet)
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.
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.
Nettleton, 1971
Problem-
Determine which anomaly is produced by a sphere
and which is produced by a cylinder.
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?
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 maxZ 2
8.52 R
3
(kilofeet)
(kilofeet)
Vertical Cylinder
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
 g max Z1 
R

 0.000575 
g max Z1
 
0.000575R 2
(feet)
(feet)
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. 4 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 100.
Begin by recalling the list of formula we developed for the sphere.
g max 
G (4 / 3R3 )
 0.02793
Z2
R3
Z
 0.00852
2
R
3
Z2
 for meters
 0.02793
 for feet
1/ 3
 g
Z 2 
max

R
 0.00852  


 
g maxZ 2
0.00852 R
3
(feet)
(feet)
R3
Z
2
 for meters
1/ 3
 g
Z 2 
max

R
 0.02793  


(feet)
Pb. 5: 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
Bouguer Anomaly (mGal)
P roblem 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)
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
In problem 6 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.
0.5
Bouguer Anomaly (mGals)
0.45
0.4
0.35
0.3
A.
0.25
B.
0.2
C.
0.15
0.1
0.05
0
-1500
-1000
-500
0
500
Distance from peak (m)
1000
1500
Nov. 11th
Problems 1-3.
Interactive modeling exercises: up to 10
points extra credit
Residual separation exercise: up to 5 points
extra credit
Nov. 16th
Terrain correction
Gravity lab
Nov. 18th
Problems 4, 5, & 6.
Gravity paper summaries