Transcript No Slide Title

Environmental and Exploration Geophysics II
Gravity Methods (III)
tom.h.wilson
[email protected]
Department of Geology and Geography
West Virginia University
Morgantown, WV
Model of the theoretical gravity
gt  gn ( )  gFA  gB  gt  gTide&Drift
•Normal gravity
•Elevation effect
•The effect of material beneath the
station - the plate effect
•Topographic or terrain effect
•Tide and Instrument drift effects
More on the plate and terrain
corrections In dealing with the derivation of the
Bouguer plate effect you may have
realized that the trick to integration
was in how one defined the
volume element. This remains true in computing the acceleration of
a ring. The approach to the terrain correction rests on the analytical
expression derived for the acceleration due to gravity of the ring - in
particular a given sector of the ring.
We start by deriving the acceleration over a disk with 0 inner radius
and outer radius R0. Our starting point should be familiar by now -
dm
g  G 2
r
and assuming that we have constant density throughout the disk
dm   dV
dV
g   G 2
s
g   G
dV
s2
dV  rd drdz
rd drdz
g   G
s2
Go to 15
s2  r 2  z 2
g   G
rd drdz
r2  z2


and again we take the vertical component
gV   G 
rd drdz
cos 
2
2
r z


gV   G 
rd drdz
cos 
2
2
r z

rd drdz
z
gV   G  2
r  z2 r2  z2

gV   G 

rd drzdz
r
2
z
2

3/ 2


1
2
Some missing steps in the results
of integration …….
Notice that we have three differential components, dr, dz,
and d, so three separate integrations are implied


rdr
2
h2  Ro
dz
g  G  d  z 
0
3 
h1  0
2 2 2

r z 


First consider
2
0 d
2
0
d   0  2  0  2
2
Which yields -


rdr
h2  Ro
dz
g  2 G  z 
3 
h1  0
2 2 2

r z 


Now take a few minutes and evaluate Ro

r 0
rdr
r 2  z 2 2
3
.
du
du
let u  r  z then
 2r and dr  .
dr
2r
2
2
Substitution yields
r du
y   3/ 2
u 2r
1 du
or 
2 u 3/ 2
Use the power rule to obtain -
1 du
1/ 2
 2 u 3/ 2  u
u

1/ 2
 r z
2
2

1/ 2
ro

r
1
2
z
2

1
1
 
z
2
0
r
1
2
o
z
2
1
2
Ro
Substitute
into
to obtain

r
1

1
2 2 2
z

1

z
0
1


1
2 2 2
Ro z


rdr
dz
g  2 G  h 2 z  R 0
3 
h1  0
2
2

r  z 2 


h2
h2
h1
h1 R 2  z 2
o
g  2 G  dz 2 G 

zdz
2
1
h2
2G  dz 2G z hh2  2G h2  h1 
1
h1
 2G
h2

zdz

h1 R 2  z 2
o

h2
1

 2 2 2
 2G  r o  z
  2G 

 h1



2
1
2
2

h
Ro 2
 
1
2
2
2

h
Ro 1

1 
2


gcylinder

 2 G   h2  h1 


 
2 1/ 2
2

Ro  h1

1/ 2 
2
2

Ro  h2

Consider the following•What happens when R0 goes to ?
g=2G (h) where h is just the thickness of the
plate and could be z or just z using the notation of
the text. This is just the Bouguer plate correction.



g drift  2Gt

 density contrast
t drift thickness
This is just the plate correction. At infinity
the effect of ring and plate are the same.

The foregoing approach represents another
way to derive the plate correction
g BouguerPla te  2Gt
- and also to determine the effect of a ring.


gcylinder  2G h2  h1 


2
2
Ro  h1
In the above equation where Ro is not , but the
outer radius of the cylinder. Rewrite the equation
substituting Ri for 0, where Ri is the inner radius.
Note also, that h1 = 0, and h2 = z, i.e., the point of
observation is right on the top of the cylinder.
1 
2 2
2
Ro  h2 
 
1
2

Ro

r  Ri
r
rdr
2
 z2


3
You could also show that subtracting the acceleration due to the inner
smaller cylinder (radius ri) from the outer cylinder (radius ro) yields the
acceleration of a ring with width r0-ri (or R0-Ri).
.
2
Ro

Hint -
Ri
rdr
r 2  z 2 2
.
1

3
Ro

r
1

1
2 2 2
and eventually -
z

Ri


1
2 2 2
Ri z

1

1
2 2 2
Ro z
gring

 2 G   R0  Ri 


2
2
Ri  z
 
1/ 2
2
2
Ro  z

1/ 2 


Remember we want
to approximate
topographic features
by ring sectors
because it’s easy to
compute the effect of
a ring sector on the
observed gravity.
In practice, the topography
surrounding a particular
observation point is divided
into several rings (usually A
through M) and each ring into
several sectors.
The F-ring extends from 1280
to 2936 feet and is divided
into 8 sectors. The average
elevation in each sector is
estimated and it’s contribution
to the acceleration at the
observation point is
computed.
Let’s spend a few moments
working through a simple
example to illustrate how the
terrain correction is applied.
In areas where the terrain is too complex to estimate the average
elevation visually, one can compile averages from the elevations
observed at several points within a sector.
As you might expect - this was a laborious process.
Hammer Table
What’s the station elevation?
2840 feet
What’s the average elevation in Sector 1? 2640 feet
What’s the relative difference between the
station elevation and the average elevation
of sector 1? 200 feet
2600
As the legend in the Hammer table notes, the
value for T is in hundredths of a milligal and
has been calculated assuming a replacement
density of 2 gm/cm3.
Thus the contribution to the topographic effect
from the elevation differences in this sector is
0.03 milligals.
Note that the elevation difference is reported in
a range and that the value is not exact for that
specific difference - in this case 200 feet.
The value could be computed more precisely
using the ring formula we developed earlier in
class.
Station elevation = 2840 feet
2640
200
3 (0.03mG) 0.0279 mG
For next class - determine the average elevation, relative elevation
differenceand 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.
Also determine the F-ring contribution if the replacement density of 2.67
gm/cm3 is used instead of 2 gm/cm3.
How?
This just requires multiplication of the
results obtained assuming 2 gm/cm3 by
the ratio 2.67/2 or 1.34.
Now that we’ve described all the corrections and gained some
experience and familiarity with their computation, let’s consider
the concept of the gravity anomaly. What is a gravity anomaly?
In general an anomaly is considered to be the difference between
what you actually have and what you thought you’d get. In
gravity applications you make an observation of the acceleration
due to gravity (gobs) at some point and you also calculate or make
a prediction about what the gravity should be at that point (gt).
The prediction assumes you have a homogeneous earth homogeneous in the sense that the earth can consist of concentric
shells of differing density, but that within each shell there are no
density contrasts. Similar assumptions are made in the
computation of the plate and topographic effects. gt then, in most
cases, is an imperfect estimate of acceleration. Some anomaly
exists.
ganom= gobs - gt
This is a simple definition, but there are several different types of
anomalies, which depend on the degree to which the theoretical
gravity has been estimated.
For example, in a relatively flat area close to sea-level we might
only include the elevation effect in the computation of gt. This
would also be standard practice in ocean surveys. In general tide
and drift effects are always included. In this case, the anomaly
(ganom) is referred to as the free-air anomaly (FAA).
gFAA  gobs  gt
g FAA  g obs   g n ( )  g FA  gTide& Drift 
gFAA  gobs  gn ( )  gFA  gTide&Drift
When only the elevation and plate effects are included in
the computation of theoretical gravity, the anomaly is
referred to as the simple Bouguer anomaly or just the
Bouguer anomaly. The combined corrections are often
referred to as the elevation correction.
gBA  gobs  gt
g BA  g obs   g n ( )  g FA  g B  gTide& Drift 
gBA  gobs  gn ( )  gFA  gB  gTide&Drift
When all the terms, including the terrain effect are included in the
computation of the gravity anomaly, the resultant anomaly is referred to
as the complete Bouguer anomaly or the terrain corrected Bouguer
anomaly (gTBA).
gt  gn ( )  gFA  gB  gT  gTide&Drift
gTBA  gobs  gt
gTBA  gobs  gn ( )  gFA  gB  gT  gTide&Drift
In this form -
gTBA  gobs  gn ( )  gFA  gB  gT  gTide&Drift
The different terms in the theoretical
gravity are referred to as corrections.
Thus -
gFA is referred to as the free-air correction
gB is referred to as the Bouguer plate correction
gT is referred to as the terrain correction, and
gTide and Drift is referred to as the tide and drift correction
There is one additional anomaly we need to discuss. This anomaly is
known as the residual anomaly. It could be the residual Bouguer
anomaly, or the residual terrain corrected Bouguer anomaly, etc. The
issue here is the concept of the residual. Recall from your reading of
Stewart’s paper, that he is dealing with the residual Bouguer anomaly.
What is it?
Most data contain long wavelength and short wavelength patterns or
features such as those shown in the idealized data set shown below.
milligals
20
15
10
5
0
0
20
40
60
Distance
80
100
Long wavelength features
are often referred to as the
regional field. The regional
variations are highlighted
here in green.
milligals
20
15
10
5
0
0
20
40
60
80
100
80
100
The residual is the
difference between the
anomaly (whichever it is)
and the regional field.
milligals
Distance
4
3
2
1
0
-1
-2
-3
-4
0
20
40
60
Distance
Just as a footnote, we shouldn’t loose sight of the fact that in all
types of data there is a certain amount of noise. That noise could
be in the form of measurement error and vary from meter to meter
or operator to operator. It could also result from errors in the
terrain corrections (operator variability) and the accuracy of the
tide and drift corrections.
5
5
3
3
residual
Noise
If we could separate out the noise (below), we might see the much
cleaner residual next to it.
1
-1
1
-1
-3
-3
-5
-5
0
20
40
60
Distance
Noisy Signal
80
100
0
20
40
60
80
100
Distance
Signal after noise
attenuation filtering
Stewart makes his estimates of valley depth from the residuals.
You shouldn’t be concerned too much if you don’t understand the
details of the method he used to separate out the residual, however,
you should appreciate in a general way, what has been achieved.
There are larger scale structural features that lie beneath the drift
valleys and variations of density within these deeper intervals
superimpose long wavelength trends on the gravity variations
across the area. These trends are not associated with the drift layer.
The potential influence of these deeper layers is hinted at in one of
Stewart’s cross sections.
Bouguer anomaly
Regional anomaly
Residual anomaly
=
If one were to attempt to model the Bouguer anomaly
without first separating out the residual, the interpreter
would obtain results suggesting the existence of an
extremely deep glacial valley that dropped off to great
depths to the west. However, this drop in the Bouguer
anomaly is associated with the deeper distribution of
density contrasts.
Let’s spend a few minutes and discuss one method for
determining the residual gravity anomaly.
The method we will discuss is referred to as a graphical
separation method.
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.
-1
-0.5
-0.5
Geology 252
Environmental and Exploration Geophysics I
In-Class Exercise - Determining residual acceleration
Use the graphical construction approach and
estimate the residual anomaly superimposed on
the regional gravity gradient.
What is the maximum value of the residual
anomaly?
What is the minimum value of the residual
anomaly?
Is the anomaly positive or negative?
Bring your results to lab
Your result will reveal the gravitational effects of an isolated
shallow body. It will be easier for you to evaluate the significance of
this feature when it is isolated from the regional variations on which
it is superimposed.
One thing that you should realize and that we will comment on more
fully in lab and in later lecture discussions, is the concept of the
non-uniqueness of potential field (gravity and magnetic) solutions.
The idea is expressed nicely in the following figure taken from
Nettleton, 1971
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 t convert
relative difference to absolute differences.
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.
Next time Thursday (Nov. 11th) - Hand in problems 1-3.
In the next lecture we will cover basic ideas related to the use of
simple geometrical objects in gravity interpretation.
The extra credit gravity exercise due on Thursday (Nov. 11th)
Gravity lab is due on Nov. 16th
Gravity paper summaries are due on Nov. 18th