Transcript problem 7-3 - West Virginia University

Environmental and Exploration Geophysics I
Magnetic Methods (II)
tom.h.wilson
[email protected]
Department of Geology and Geography
West Virginia University
Morgantown, WV
Preparations for today - cover background
reading pages 389-420 of Chapter 7 and prepare
questions for problems 1 through 3 on page 449.
In today’s class presentation we continue to
develop key quantitative relationships in the use of
magnetic methods, and as we do, we will tie them
into homework problems 1 through 3.
Problems 7-1 through 7-3 will be Tuesday after
Thanksgiving break (Dec. 3rd).
As noted the other day, the potential is the
integral of the force over a displacement path.
m1m2
V    Fdr    2 dr
r
From above, we obtain a
basic definition of the
potential (at right) for a unit
positive test pole (mt).
V
m
r
Thus - F (sometimes also referred
to as H) can be easily derived from
the potential simply by taking the
derivative of the potential
dV m
F (or H )  
 2
dr r
re: problem 7-2
Consider the field at a point along the axis of a dipole. The
dipole in this case could be a buried well casing.
The field has vector properties however, in this case
vectors are collinear and its easy to determine the net effect.
In terms of the
potential we can write
V
m m

r
r
In the case at right, r+ is much
greater than r- , thus in
V
m m

r
r
V
m
r
Thus the potential near either end of a long dipole behaves like
the potential of an isolated monopole. If we are looking for
abandoned wells, we expect to find anomalies similar to the
gravity anomalies encountered over buried spherical objects.
21
Considering the general case of the dipole field we have to
take direction into account. However, in the case where
the distance to the center of the dipole is much greater
than the length of the dipole we can still treat the problem
of computing the potential as one of scalar summation
since the directions to each pole fall nearly along parallel
lines.
If r is much much greater than l then the angle 
between r+ and r- approaches 0 and r, r+ and rcan be considered parallel and the differences in
lengths r+ and r- from r equal to plus or minus the
projections of l/2 into r.
r+
r
r-
m m
Vdipole 

r
r
Recognizing that pole strength of the
negative pole is the negative of the positive
pole and that both have the same absolute
value, we rewrite the above as
Vdipole  
m m

r r
re: problem 7.1 -
Vdipole  
m
r  l cos
2

m
r  l cos
2
Converting to common denominator yields
ml cos
Vdipole 
r2
See Eqs. 7-12
where ml = M
From the previous discussion , the field intensity H (or F
as in Berger) is just
dV

dr
since V    Fdr, F  
dV
dr
H - monopole =
dVm
d m m

   2
dr
dr  r  r
H - dipole
dVd
d  ml cos 

 

2
dr
dr  r

2ml cos

r3
This yields the field intensity in the radial direction i.e. in the direction toward the center of the dipole.
However, we can also evaluate the horizontal and
vertical components of the total field directly from
the potential.
27
ZE  
dVd
d  ml cos 
 

dr
dr  r 2 
Vd represents the
potential of the dipole.
HE is represented by the negative derivative of the
potential along the earth’s surface or in the S direction.
d  ml cos 



2
rd  r

dV
M sin 

 HE 
ds
r3
Where M = ml
and
dV 2M cos
ZE  

dr
r3
Let’s tie these results back into some
observations made earlier in the semester
with regard to terrain conductivity data.
32
Given
M sin 
HE 
r3
What is HE at the equator? … first what’s ?
 is the angle formed by the line connecting the
observation point with the dipole axis. So , in
this case, is a colatitude or 90o minus the
latitude. Latitude at the equator is 0 so  is 90o
and sin (90) is 1.
M
HE  3
r
At the poles,  is 0, so that
HE  0
What is ZE at the equator?
2M cos
ZE 
r3
 is 90
ZE  0
2M cos
ZE 
r3
ZE at the poles ….
2M
ZE  3
r
The variation of the field intensity at the pole and
along the equator of the dipole may remind you of
the different penetration depths obtained by the
terrain conductivity meters when operated in the
vertical and horizontal dipole modes.
Z E max
2M
 3  2 H E max
r
Recall rule-of-thumb values for penetration
depths have the vertical dipole exploration
depth twice that of the horizontal dipole.
& Problem 7-1
Berger discusses computation of vertical and
horizontal gradients of magnetic field intensity.
The gradient is just the rate of change in some
direction - i.e. it’s just a derivative.
How would you evaluate the vertical gradient of the
horizontal component of the earth’s magnetic field?
Representing the earth’s
horizontal field in dipole form as
M sin 
HE 
r3
The vertical gradient is just the variation
with change of radius or
dH E
dr
dH E d M sin 

?
3
dr
dr r
dH E
M sin 
3
 3
  HE
4
dr
r
r
What is the horizontal gradient of the vertical
component of the earth’s magnetic field (ZE) ?
[see problem 7.1]
Recall
2M cos
ZE 
r3
Recall the horizontal gradient operator?
Evaluate
d
d

dS rd
dZ E dZ E

dS
rd
d 2M cos
2 M sin 

3
rd
r r3
r
2 M sin 
2H E


3
r r
r
(i.e. Eq. 7 - 20)
An important point to carry away from the preceding
discussion is the distinction between the horizontal and
vertical gradient operators.
Horizontal gradient operator 
d
d

dS rd
Vertical (or radial) gradient operator 
d
dr
Thus if you were asked “what is the horizontal gradient of
the horizontal component?” you are required to evaluate
dH E dH E

dS
rd
46
Problem 7.3
For problem 7-3, the main field has inclination i of 50o. The
following equations are needed for Problem 3. ZA is the
vertical component of the anomalous field, and HA, the
horizontal.
4 3


R
kF

2
E  sin i 

 
3
z
3xz
3


ZA 

cosi   1 (7 - 36)
 2
2
2 3/ 2
2 1/ 2  2
2 1/ 2
(x  z )
 ( x  z )
 (x  z )
 
4 3

 R kFE  cosi 
2
 

3
z
3
xz
3


HA 
 1   2
tan i  (7 - 37)
 2
2
2 3/ 2
2 1/ 2
2 1/ 2
(x  z )
 ( x  z )

  (x  z )
Eqns 7-36 and 7-37 are in general form and can be
used for arbitrary inclination i.
Before we address problem 7-3 in more detail, let’s
go back and look at the ideas discussed by Berger
pertaining to equations 7-15 through 7-19. In
problem 7-3 you will want to make use of Eq. (7-18)
FAT  Z A sin i  H A cos i
(7 - 18)
Consider the type of measurement provided
by the proton precession magnetometer
M
GF
f 
F
2L
2
FE  23.4874 f
The proton
precession
magnetometer
measures the scalar
magnitude of the
earth’s main field.
In this diagram FET is is the vector sum of the earth’s main field
and the anomalous field associated with a buried dipole field. The
proton precession magnetometer measures the magnitude of FET.
Magnetic Elements for your location
F is known
In most applications the anomalous field FA
is much smaller than the main field FE.
In this case, the magnetic
anomaly is just the difference
between the measured field (FET)
at some point and the predicted
value of the earth’s main field
(FE) at that point.
The magnetic anomaly is
T  FET  FE .
T is a scalar quantity
not a vector quantity.
53
FA
When FA (the anomalous field) is small, we consider the
difference T = FET - FE to be equivalent to the projection
of vector FA onto the direction of the main field.
In the case where FA is large the projection
FAT is significantly different from T.
Let’s zoom in for a closer look at the tip of FE. The magnetic anomaly is
T  FET  FE .
T is a scalar quantity
not a vector quantity.
Horizontal line
parallel to
earth’s surface
FET
T
i is the inclination of the earth’s main magnetic field.  is
the angle of FA relative to the earth’s main field FE.
T  FET  FE  FAT
FAT is the projection of FA onto the direction of the main field FE, and
is considered equal to T, the scalar difference between FE and FET.
Consider the significance of the terms
FA cos  and FA sin  in the previous expression.
The horizontal and vertical projections of FA
The horizontal and vertical projections of FA
appear in the expansion of FAT = FAcos(-i).
In summary - FAT is an approximation of T,
the scalar difference obtained from
measurements of the total field (FET) made by
the proton precession magnetometer.
FAT  Z A sin i  H A cos i
(7 - 18)
For the purposes of modeling we work backwards.
Given a certain object, we compute the
horizontal (HA) and vertical (ZA) components of
the anomaly (see equations 7-36 and 7-37, for
example) and combine them to obtain FAT - the
anomaly we obtain from the proton precession
magnetometer measurements.
Back to problem 7-3. In problem 7-3 you are asked to
estimate the detectability of a buried magnetized stone
wall. You are also instructed to approximate the effect of the
wall using the equations for the sphere. The inclination in
the area is given as 50o, but the total field is not given. As
you can see in equations 7-36 and 7-37 we need the total
field FE in order to solve for ZA and HA. How might you
determine what FE is given the inclination?
inclination
Main field intensity
Guess FE ~ 43000 nT
Do we have everything we need? If we are going to
approximate the effect of the wall using a sphere we
need to decide what the radius of the sphere will be.
The cross sectional area of the wall is 1/2 m2 (0.5m x
1m). A cross section through the center of the sphere
will have area R2. Let R2 = 0.5m2 and solve for the
equivalent radius R. This yields an R of 0.4m. From the
diagram below, we can see that z (depth to center of
wall or sphere) should be 1.75m.
You now have everything you need to calculate ZA and HA.
4 3


R
kF

2
E  sin i 

 
3
z
3xz
3


ZA 

cosi   1 (7 - 36)
 2
2
2 3/ 2
2 1/ 2  2
2 1/ 2
(x  z )
 ( x  z )
 (x  z )
 
Eqn 7-36 (R=0.4, z=1.75, F=43000,i=50)
8
7
6
Note that some texts
 .... 
 ( x2  z 2 ) 


In the denominator
instead of 

....
2
2 1/ 2 
(
x

z
) 

ZA (nT)
have
5
4
3
2
1
0
-1
-6
-4
-2
0
x (meters)
2
4
6
4 3

  R kFE  cos i 
 

3x 2
3xz
3


HA 

1

tan
i
 2
  2
  (7-37)
2
2 3/ 2
2 1/ 2
2 1/ 2
(x  z )

  (x  z )
 ( x  z )
Eqn 7-37 (R=0.4, z=1.75, F=43000,i=50)
4
Note that some texts
 .... 
 ( x2  z 2 ) 


In the denominator
instead of 

....
2
2 1/ 2 
 (x  z ) 
2
HA (nT)
have
3
1
0
-1
-2
-3
-6
-4
-2
0
x (meters)
2
4
6
FAT  H A cos i  Z A sin i
(see Eq. 7-18)
Total from Berger's Eqns (R=0.4, z=1.75, F=43000,i=50)
7
6
FA (nT)
5
4
3
2
1
0
-1
-6
-4
-2
0
2
4
6
x (meters)
The maximum anomaly associated with the sphere is
over 6nT and easily detected by a proton precession
magnetometer which can be read with an accuracy of
1nT. (Notice that at 0 meters FAT is 4.5 nT.)
Some comments on Simple Geometrical Objects - Magnetics Style
surface
The orientation of H and Hr has changed. We now have to think in
terms of their relation to a buried dipole rather than to the earth’s main
dipole field. ZA and HA refer to components of the buried dipole field
measured relative to the earth’s surface.
surface
ZA is the vertical component of the anomalous dipole
field - vertical with respect to the earth’s surface.
What are cos and sin?
Capitol versus lower case Z
HA is the horizontal component - component
oriented along the earth’s surface associated
with the dipole field of a buried object.
Recall from class last Thursday (see also
Equation 7-5 (Burger) that the intensity of
magnetization equals the magnetic dipole
moment per unit volume or
M
I
V
and also,
I  kFE .
M  IV  kFEV
Thus
M (2 z 2  x 2 )
ZA  2
( x  z 2 )5 / 2
Is equivalent to equation 7-35 in the text.
Substitute M=IV into
the above to obtain
IV (2 z 2  x 2 )
ZA  2
( x  z 2 )5 / 2
Recall that the volume of the sphere is
4 3
R I (2 z 2  x 2 )
ZA  3 2
( x  z 2 )5 / 2
7 - 35
4 3
R
3
hence,
where I = kFE
An easy approximation
Equations 7-36 and 7-37 are fairly cumbersome
equations to solve. The foregoing approximations
made assuming a vertically polarized sphere allow
one to make a quicker estimate of the field. A
comparison of results would be informative.
M (2 z 2  x 2 )
ZA  2
( x  z 2 )5 / 2
3Mxz
HA  2
 0 at x  0
2 5/ 2
(x  z )
Solve these two equations at x=0 and
compare ZA and HA to the values obtained
in the preceding detailed analysis.
Due Thursday, Dec. 5th
Since the bedrock
is magnetic, we
have no way of
differentiating
between anomalies
produced by
bedrock and those produced by buried storage drums.
Acquisition of gravity data allows us to estimate variations
in bedrock depth across the profile. With this knowledge,
we can directly calculate the contribution of bedrock to the
magnetic field observed across the profile.
With the information on bedrock configuration we can clearly
distinguish between the magnetic anomaly associated with
bedrock and that associated with buried drums at the site.
You may find that the autocalculations become a little
erratic. If that happens, just run forward again.
The spurious peaks in the calculation disappear.
How many drums are represented by the triangularshaped object you entered into your model?
Plot the corner coordinates for the
triangular shaped object you derived at
1:1 scale and compute the area.
How many drums?
• For Tuesday, December 3rd be prepared to
discuss problems 7-6, and 7-8.
• Finish reading Chapter 7; we’ll talk about simple
geometrical representations in magnetic problem
solving.
• Also be prepared to ask questions about the final
exam. (Reminder - the final exam will be from 35pm on Friday afternoon, Dec. 13th)