Environmental and Exploration Geophysics I

Tom Wilson, Department of Geology and Geography

Environmental and Exploration Geophysics I

[email protected]

Department of Geology and GeographyWest Virginia University

Morgantown, WV

Magnetic Methods Magnetic Methods (V)(V)

Problems we’ve been working on …


Questions?



The first problem relates to our discussions of the dipole field and their derivatives.

7.1. What is the horizontal gradient in nT/m of the Earth’s vertical field (ZE) in an area where the horizontal field (HE) equals 20,000 nT and the Earth’s radius is 6.3 x 108 cm.


3

sin Thus

r

MH E

3

cos2

r

M

dr

dVZE

2

cosE

dV dV d plH

ds rd rd r

Recall that horizontal gradients refer to the derivative evaluated along the surface or horizontal direction and we use the form of the derivative discussed earlier for the potential.

1 d

r d

The negative sign is NOT needed

when computing the gradient.


To answer this problem we must evaluate the horizontal gradient of the vertical component -

1E

dZ

r dor

3

1 2 cosd M

r d r

See Equation 7.20

Evaluate the horizontal gradient


3

2 cosE EdZ dZ d M

dS rd rd r

3 8

2 sin 2 220,000

6.3 10E

MH nT

r r r x cm

Since is co-latitude, the direction of increasing is southward (in the northern hemisphere). As we travel from pole to equator ZE decreases, thus the gradient is negative.


4. A buried stone wall constructed from volcanic rocks has a susceptibility contrast of 0.001cgs emu with its enclosing sediments. The main field intensity at the site is 55,000nT. Determine the wall's detectability with a typical proton precession magnetometer. Assume the magnetic field produced by the wall can be approximated by a vertically polarized horizontal cylinder. Refer to figure below, and see following formula for Zmax.

Background noise at the site is roughly

5nT.

What is z?

What is I?


-0.2

0.0

0.2

0.4

0.6

0.8

1.0

Nor

mal

ized

Res

pons

es

Relative Response Functions

-3 -2 -1 0 1 2 3

X/Z (no units)

Vertical C

ylinderSphere

Horizontal Cylinder

maxZZ A is a function of the unit-less variable x/z

The vertical field is often used to make a quick estimate of the magnitude of an object. This is fairly accurate as long as i is 60 or

greater

2/5

2

2

2

2

1

2

2

1

zx

zx

2/32

2

)1(

1

zx

22

2

2

2

1

1

zx

zx

Dipole/sphere

Horizontal cylinder

Vertical cylinder


Vertically polarized sphere or dipole

Vertically polarized horizontal cylinder

3

3

max38

z

kFRZ

E

2

2

max2

z

IRZ

EkFI Remember

2Area R

AreaR

Considerable difference in magnitude of


2

0.001 55,000 55

0.50.4

1.5 0.4 1.9

EI kF x nT nT

Area mR m

z m

3

max 3 3

88.38 0.064 55 0.5363 0.078

1.9 6.86

ER kF x xZ nT

z

2

max 2

2 ER kFZ

z

For the dipole

For the horizontal cylinder


4. In your survey area you encounter two magnetic anomalies, both of which form nearly circular patterns in map view. These anomalies could be produced by a variety of objects, but you decide to test two extremes: the anomalies are due to 1) a concentrated, roughly equidemensional shaped object (a sphere); or 2) to a long vertically oriented cylinder.


The map view clearly indicates that consideration of two possible origins may be appropriate - sphere or vertical

cylinder.


In general one will not make such extensive comparisons. You may use only one of the diagnostic positions, for example, the half-max (X1/2) distance for an anomaly to quickly estimate depth if the object were a sphere or buried vertical cylinder….

Burger limits his discussion to half-maximum relationships.

Breiner, 1973

X1/2 = Z/2

X1/2 = 0.77Z

X1/2 = Z

X1/2 = Z/2


Just as an aside: The sample rate you use will depend on the minimum depth of the objects you wish to find.

Your sample interval should probably be no greater than X1/2.

But don’t forget that equivalent solutions with shallower origins do exist!


3

max 3

83

R kHZ

z

2/5

2

2

2

2

max1

2

2

1)(

z

x

z

x

Z

xZ A

We’ll make quick work of it an use only three diagnostic positions (red above)


2

2

maxz

IRZ

2/32

2max )1(

1

z

xZ

Z A

Again, we can get by with only three diagnostic positions (red above)


Unknown Anomaly

0

2

4

6

8

10

12

14

16

-4 -3 -2 -1 0 1 2 3 4

Distance in meters

Inte

nsi

ty (

nT

)

Determine depths (z) assuming a sphere or a cylinder and see which assumption yields consistent estimates.

It’s all about using diagnostic positions and the depth index multipliers for each geometry.


Sphere vs. Vertical Cylinder; z = __________

Diagnostic positions

MultipliersSphere

ZSphere Multipliers Cylinder

ZCylinder

X3/4 =

X1/2 =

X1/4 =

The depth

Unknown Anomaly

0

2

4

6

8

10

12

14

16

-4 -3 -2 -1 0 1 2 3 4

Distance in meters

Inte

nsi

ty (

nT

)

2.86 3.13.35

1.95 2.032.00

2.17 1.310.81

3.18 21.37

diagnostic distance

0.9

X3/4

1.55

X1/2

2.45

X1/4

2.863.13.35

1.952.032.00


Diagnostic positions MultipliersSphere

ZSphere Multipliers Cylinder

ZCylinder

X3/4 = 1.6 meters 3.18 2.17

X1/2 = 2.5 meters 2 1.31

X1/4 = 3.7 meters 1.37 0.81

Sphere or cylinder?

Another Unknown Anomaly

-1

0

1

2

3

4

5

-5 -4 -3 -2 -1 0 1 2 3 4 5Distance in meters

Inte

ns

ity

(n

T)

5.01

5.0

5.07

gmax

g3/4

g1/2

g1/4

5.0855.1

3.473.283.00


5. Given that derive an expression for the radius,

where I = kHE. Compute the depth to the top of the casing for the anomaly shown below, and then estimate the radius of the casing assuming k = 0.1 and HE =55000nT. Zmax (62.2nT from graph below) is the maximum vertical component of the anomalous field produced by the vertical casing.

Abandoned well

0

10

20

30

40

50

60

70

-15 -10 -5 0 5 10 15

Distance in meters

Inte

nsi

ty (

nT

)

2

max 2

R IZ

z

Algebraic manipulation


Follow the recommended reporting format.Specifically address points mentioned in the results section, above.



Where are

the drums?



anomaly



Distance along profile

180 190 200 210 220 230

Dep

th

10

15

20

25

30

35

Outline of Drum ClusterDerived from the magnetics model

1

2TotalArea Base x Height

Drums

Total AreaN

Area of one Drum

4 square feet

Area of one drum ~

Make sure the scale of your graph is 1:1


…. compare the field of the magnetic dipole field to that of the gravitational monopole field

0

0.02

0.04

0.06

0.08

0.1

0.12

-1500 -1000 -500 0 500 1000 1500

Gravity:500, 1000, 2000m

2

1 varies

rasieldMonopole f

Increase r by a factor of 4 reduces g by a factor of 16

3

cos2

r

MZE

3

2

r

MZE


3

1 varies

rasfieldDipole

-1

0

1

2

3

4

5

6

7

8

-5 -3 -1 1 3 5

Distance in meters

Inte

nsi

ty (

nT

)

0

0.05

0.1

0.15

-10 -5 0 5 10

Distance in meters

Inte

nsi

ty (

nT

)

For the dipole field, an increase in depth (r) from 4 meters to 16

meters produces a 64 fold decrease in anomaly magnitude

7.2 nT0.113 nT

Thus the 7.2 nT anomaly (below left) produced by an object at 4 meter depths disappears into the background noise at 16 meters.


Again - follow the recommended reporting format.Specifically address listed points (1-5).


You are asked to run a magnetic survey to detect a buried drum.

What spacing do you use between observation points?

Sampling issues – for leisure consideration …Jump to last slide for reminders


How often would you have to sample to detect this drum?

X 1/2=Z/2


…. how about this one?The anomaly of the drum drops to ½ at

a distance = ½ the depth.


Sampling does depend on available equipment!

As with the GEM2, newer generation magnetometers can

sample at a walking pace.


Remember, the field of a buried drum can be approximated by the field of a dipole or buried sphere. X1/2 for the sphere (the dipole) equals one-half the depth z to the center of the dipole. The half-width of the anomaly over any given drum will be approximately equal to its depth

Or X1/2 =Z/2



Feel free to discuss these problems in groups, but realize that you will have to work through problems independently on the final.


General Review this coming Thursday

Turn in your magnetics lab report Thursday, December 10th.

Exam, Friday December 17th; 3-5pm

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Environmental and Exploration Geophysics I

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