1. Gravity Methods (IV) Environmental and Exploration Geophysics II
tom.h.wilson
Department of Geology and Geography
West Virginia University
Morgantown, WV

2. Questions? 200 520 2840 520 280 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

3. 2640
200
3 (0.03mG)
0.0279mG
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 the ring equation.

4. In your submission, show complete computations for Sector 5.
Equation 6-30
In your submission, show complete computations for Sector 5.

5. Graphical Separation of the Residual
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. -1 ?
-0.5 -1
-0.5
-0.5
After labeling all intersections with the relative (or residual ) values, you can contour these values to obtain a map of the residual feature.
Questions?

6. Bouguer anomaly
Regional anomaly -
Residual anomaly =

7. Stewart obtains the general relationships -or
Where Z, is the drift thickness at station , and
R is the value of the gravity residual at station 

8. Gravity Lab Due Thursday - Nov. 8th ...
Edge Effects
Gravity Lab Due Thursday - Nov. 8th ...

9. See equation 6.47
Think about how it might be used to estimate an “edge effect.”
The gravitational field associated with the half-plate drops from a maximum to minimum across the edge of the plate.
What would be the minimum value of the residual anomaly over this model?

10. To use t=130g, Convert t=500 meters to feet.
The gravitational field would drop from 0 to a minimum at an infinite distance from the plate edge. The minimum would just be -2Gt
Note that the density contrast in the model at right is identical to that assumed by Stewart.
To use t=130g, Convert t=500 meters to feet.
500 meters = 1640 feet, then
g = -1640/130 milligals or
-12.6 milligals

11. 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.
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.

12. Simple Geometrical Objects
The question of edge effects addressed above takes advantage of simple geometrical objects, the plate and half plate, to answer questions about possible anomaly magnitudes associated with the problem at hand.

13. Simple Geometrical Objects
We make simplifying assumptions about the geometry of 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. to estimate the scale of an anomaly we might be looking for or to estimate maximum depth, density contrast, fault offset, etc. without the aid of a computer.

14. Let’s start with one of the simplest of geometrical objects - the sphere -
Go to 31

21. gmax

22. Divide through by gmax

28. Highlight the fact that X measures distance from the anomaly peak, and is NOT an absolute reference along the profile line.

29. The “diagnostic position” is a reference location
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.

30. 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.

31. 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 – to Z.
Note that these constants (e.g ) assume that depths and radii are in the specified units (feet or meters), and that density is always in gm/cm3.

32. 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.

33. You could measure off 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.
2.17
1.79
1.305
0.96
0.81

36. Go to 42

40. Now compute variation of g across the cylinder
and consider only the vertical component.
take vertical component

45. Just as was the case for the sphere, 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.

46. How would you determine the depth index multipliers from this graph?
1.72
1.41 1
0.7
0.57

47. The depth index multiplier is just the reciprocal of the value at X/Z.
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
X3/4
X2/3
X1/2
X1/3
X1/4
0.57
0.7
Z=X1/2

48. Again, note that these constants (i. e
Again, note that these constants (i.e ) assume that depths and radii are in the specified units (feet or meters), and that density is always in gm/cm3.

49. 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 an anomaly of given wavelength cannot have a physical origin.
Maximum Depth
Nettleton, 1971

50. In-class problem-
Determine which anomaly is produced by a sphere and which is produced by a cylinder.

51. 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?

52. Z is in kilofeet, and  is in gm/cm3.
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.

53. Vertical Cylinder ? A A'

54. Part 2, Gravity Problem Set
Due Nov. 13th
Part 2, Gravity Problem Set
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 100.

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

56. Pb.4 (or 6.8, Burger et al.): 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

57. You could plot the data on a sheet of graph paper
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
Regional

58. In problem 5 (similar to problem 6. 9, Burger et al
In problem 5 (similar to problem 6.9, Burger et al.) you’re 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. A. B. C.

59. Due Dates Hand in gravity lab next Thursday, Nov. 8th .
Turn in Part 1 (problems 1 & 2) of gravity problem set 3, Thursday, November 8th. Remember to show detailed computations for Sector 5 in the F-Ring for Pb. 2.
Turn in Part 2 (problems 3-5) of gravity problem set 3, Tuesday, November 13th.
Gravity paper summaries, Thursday, November 15th.