Gravity Methods (V) Environmental and Exploration Geophysics II

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

Tom Wilson, Department of Geology and Geography
Tuesday survey Tom Wilson, Department of Geology and Geography

Orhtophoto of the site before it was covered Tom Wilson, Department of Geology and Geography

Questions about the Homework?
6.1 a) Free air correction b) Bouguer plate correction 6.2 Terrain correction 6.3 due today Tom Wilson, Department of Geology and Geography

Questions? 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 Tom Wilson, Department of Geology and Geography

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. Also determine the F-ring contribution if the replacement density of 2.67 gm/cm3 is used instead of 2 gm/cm3 and compare to the result obtained using the ring equation. Tom Wilson, Department of Geology and Geography

In your submission, show complete computations for Sector 5.
Equation 6-30 In your submission, show complete computations for Sector 5. Due on the 11th Tom Wilson, Department of Geology and Geography

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. -0.5 -1 ? After labeling all intersections with the relative (or residual ) values, you can contour these values to obtain a map of the residual feature. Questions? Due on the 11th Tom Wilson, Department of Geology and Geography

Bouguer anomaly Regional anomaly - Residual anomaly = Tom Wilson, Department of Geology and Geography

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  Tom Wilson, Department of Geology and Geography

Edge Effects Gravity Lab Due Thursday - Nov. 13th ...
We’ll spend time reviewing the calculations, edge effects and questions asked as part of the lab exercise next Tuesday (11th). Tom Wilson, Department of Geology and Geography

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? Tom Wilson, Department of Geology and Geography

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 Tom Wilson, Department of Geology and Geography

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. Tom Wilson, Department of Geology and Geography

What is the only kind of anomaly you can have when the density contrasts in your model are either 0 or negative? -0.6gm/cm3 0 gm/cm3 Tom Wilson, Department of Geology and Geography

The anomalies have to be negative. The next question is how far do you have to shift the anomalies to make t =130 g work? -0.6gm/cm3 0 gm/cm3 ? The anomaly here is but to get 130g = 520 feet what would it have to be? Tom Wilson, Department of Geology and Geography

We obtained a good match between the computations and observations. Now shift the calculations down and apply Stewart’s formula. -4.3mGals Estimated depth is 130 x (-4.3) = 559’ -0.6gm/cm3 0 gm/cm3 Depth = 630feet Tom Wilson, Department of Geology and Geography

Read through the Gravity lab exercise and spend some more time working with the ideas in the lab and in Stewart’s paper. We’ll come back in on Tuesday and go over the lab requirements and format for the write-up. The lab will be due on Thursday. Tom Wilson, Department of Geology and Geography

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. Tom Wilson, Department of Geology and Geography

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. Tom Wilson, Department of Geology and Geography

Let’s start with one of the simplest of geometrical objects - the sphere - Go to 31 Tom Wilson, Department of Geology and Geography

gmax Tom Wilson, Department of Geology and Geography

Divide through by gmax Tom Wilson, Department of Geology and Geography

Shape of the anomaly is independent of the size of the sphere that produced it. Tom Wilson, Department of Geology and Geography

At what point does the anomaly fall off to one-half of its maximum value? Tom Wilson, Department of Geology and Geography

Let the ratio g/gmax = ½ and solve for X/Z Tom Wilson, Department of Geology and Geography

X1/2 /Z = 0.766 implies that Z can be expressed in terms of X1/2
Tom Wilson, Department of Geology and Geography

X measures distance from the anomaly peak, and is NOT an absolute reference along the profile line. The location of the peak corresponds to X=0 Tom Wilson, Department of Geology and Geography

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. Tom Wilson, Department of Geology and Geography

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. Tom Wilson, Department of Geology and Geography

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. Tom Wilson, Department of Geology and Geography

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. Tom Wilson, Department of Geology and Geography

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 Tom Wilson, Department of Geology and Geography

Given X1/2 what is Z? Tom Wilson, Department of Geology and Geography

Go to 42 Tom Wilson, Department of Geology and Geography

Now compute variation of g across the cylinder
and consider only the vertical component. take vertical component Tom Wilson, Department of Geology and Geography

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. Tom Wilson, Department of Geology and Geography

How would you determine the depth index multipliers from this graph? 1.72 1.41 1 0.7 0.57 Tom Wilson, Department of Geology and Geography

X3/4 X2/3 X1/2 X1/3 X1/4 Z=X1/2 0.7 0.57 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 Tom Wilson, Department of Geology and Geography

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. Tom Wilson, Department of Geology and Geography

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 Tom Wilson, Department of Geology and Geography

In-class problem- Determine which anomaly is produced by a sphere and which is produced by a cylinder. Tom Wilson, Department of Geology and Geography

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? Tom Wilson, Department of Geology and Geography

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. Tom Wilson, Department of Geology and Geography

Vertical Cylinder ? A A' Tom Wilson, Department of Geology and Geography

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. Tom Wilson, Department of Geology and Geography

Begin by recalling the list of formula we developed for the sphere.
Tom Wilson, Department of Geology and Geography

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 Tom Wilson, Department of Geology and Geography

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 Tom Wilson, Department of Geology and Geography

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. C. B. Tom Wilson, Department of Geology and Geography

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. Tom Wilson, Department of Geology and Geography

Gravity Methods (V) Environmental and Exploration Geophysics II

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