Gravity Methods (VII) wrap up

Transcription

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

2 Items on the list 6.3 due today Draft essay due Friday, 4 th by noon Due date for problems 6.5 and 6.9 moved to the 10 th Gravity lab due date delayed till Nov. 15 th due to election day recess. Begin reading Chapter 7 on Magnetics. We ll get into magnetic methods next week. Following Thanksgiving break we will wrap up magnetic methods on Nov. 8 th and Dec 1 st with some exam review on the 1 st and 6 th. Final, Wednesday, December 14 th from 11am-1pm in rm 35 Brooks.

3 For today Wrap up gravity lab demo Questions on handout problems 6.5 & 6.9 More on simple geometrical objects An in-class problem General applications

4 Gravity lab questions

5 Question from the lab Does the DC shift help? Why? What are Stewart s assumptions? + To use the plate approximation we have to honor the assumptions made by Stewart to develop the formula t=130g. Estimate the depth of this valley using the formula t=130g r. t=60ft??? 0-1 This representation follows Stewart s conceptualization of the problem Estimate the depth of this valley using the formula t=130g r. Shift on Shift off - -4 DC refers to a constant that is added to the calculations to shift them so they have approximately 0 average.

6 Present in an organized and sequential manner Disagreements we noted and worked with are discussed in the lab guide. Some obvious mismatches.

7 Valley width and valley length both must be incorporated to accurately model the field This part of the problem gets you to deal with the 3 dimensional aspects of buried valley geometry.

8 Undertake an additional inversion to produce a match between the observed and calculated gravity When you reduced the valley lengths, you eliminated the match between calculated and observed fields. To obtain a more accurate model you need to reinvert the transect.

9 After the inversion calculate drift thickness using the plate formula (t=130g) and compare valley! The valley depth increases, but the negative anomaly observed over this area does not. Discuss as requested above.

10 Present in an organized and sequential manner with labeled figures & captions. Organization is important!

11 Use questions to guide your discussion Respond to questions as indicated. Follow recommended organization of presentation. Use figures you've generated in GMSYS to illustrate your point. All figures should be numbered, labeled and captioned.

12 Diagnostic position X 1/ (see discussion of halfmaximum technique section 6.7.1) Last time we showed that when x 1/ /z = g g v max 1 1 x1/ 1 z 3/ x ½ is referred to as the diagnostic position, z/x 1/ is referred to as the depth index multiplier

13 You could solve for values of x r /z for other ratios of g v /g max. All solutions of z should be the same g g v max 3/4 1/ 1/4 Evaluation at multiple diagnostic locations does two things for you: allows you to obtain an average Z and helps test your assumption about anomaly origin. x x x z.. z... z If it s not a sphere, then the values of Z will differ significantly.

14 You get different depth index multipliers; BUT if the object is a sphere, you should get the same z! Diagnostic Position Depth Index Multiplier (g/gmax) 3/4 max 1/0.46 =.17 /3 max 1/0.56 = / max 1/0.77 = /3 max 1/1.04 = /4 max 1/1.4 = 0.81 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. z x 3 4 z x 1 z x 1 4 Z (depth index multiplier) times X at the diagnostic position

15 Two anomalies - one broader (longer wavelength) than the other. Which has deeper origins? Depth index multiplier for X 1/ is Depth index multiplier for X 3/4 is.17 What depth do you get? X 1/ X 3/4 Wilson, Department of Geology and Geography

16 For example, you ve got two anomalies. One is broader than the other. Depth index multiplier for X 1/ is *1.305= Depth index multiplier for X 3/4 is *.17 = What depth do you get? ~750 ~ *1.305=457 and.17*00=434 We could use the average as the depth

17 Due next Thursday: 6.5 and 6.9 as revised in class handout 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,, =.7gm/cm 3 ) and that the void centers are never closer to the surface than 100m. Given the 0.05mG measurement accuracy, the detection limit should be at least 0.1 mg. i.e. z 100m

18 Basic formula with some mixed units variations 3 G(4 / 3 R ) gmax Z 3 R for meters Z 3 R for feet Z 1/3 gmax Z R (4 / 3) G 1/3 gmax Z R (feet) gmaxz (feet) 3 (4 / 3 ) GR gmaxz (feet) R These constants (i.e or ) assume that depths and radii are in the specified units (feet or meters), and that density is always in gm/cm 3.

19 Revised 6.9 In a problem similar to the problem in the text, you re given three anomalies. These anomalies are assumed to be associated with three buried spheres. Determine their depths using the diagnostic position and depth index multiplier as discussed in class. Carefully consider where the anomaly drops to one-half of its maximum value. Assume a minimum value of Bouguer Anomaly (mgals) Distance from peak (m)

20 This technique can be applied to other geometrical objects: for example, the horizontal cylinder. What could the horizontal cylinder represent geologically? X z Cylinder with radius R and density R

21 The anomaly across the cylinder is also symmetrical X z r At surface distance x away from a point directly over the cylinder

22 The result shares similarity to that for the sphere (see equation 6.37 and excel table 6.7) g cyl x GR 1 Z See derivation in text z 1 gcyl g and 1 x z max g max 1 GR Z

23 For the diagnostic position of ½ g max Choose the position of interest and solve for the ratio x/z g cyl 1 x1/ gmax 1 z 1 1 x x 1/ 1 1/ 1 z x 1/ z 1 z x1 This tells us that the anomaly falls to ½ its maximum value at a distance from the anomaly peak equal to the depth to the center of the horizontal cylinder z

24 See class handout/worksheet Locate the points along the X/Z Axis where the normalized curve falls to diagnostic values - 1/4, 1/, etc. The depth index multiplier is just the reciprocal of the value at X/Z at the diagnostic position. X 1/4 X 1/3 Just as we did for X 1/ solve for X 3/4, etc X 3/4 X /3 X 1/ Z=X 1/ X times the depth index multiplier yields Z

25 Just as we did for the sphere, we ve derived depth index multipliers for several diagnostic positions For the cylinder we have Diagnostic Position Depth Index Multiplier 3/4 max 1/0.58 = 1.7 /3 max 1/0.71 = / max 1/1= 1 1/3 max 1/1.4 = 0.7 1/4 max 1/1.74 = 0.57 GR gmax Z R for meters Z R gmax Z R gmax Z R Z for feet 1/ (feet) (feet) With Z, you can then speculate on the density contrast or radius of the object in question. 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.

26 Horizontal Cylinder 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. This is the key idea associated with evaluation of diagnostic positions: if the estimated z s are similar for a certain assumed shape, that helps confirm your interpretation.

27 Can you tell which anomaly is produced by a horizontal cylinder and which, by the sphere? The depth to the center, Z, is the same for each Remember Z=1.305X 1/ for the sphere and Z=X 1/

28 Assume the anomaly below is produced by long horizontal tunnel What is the depth to the tunnel? What are the depth index multipliers? X 1/ ~100m DIM=1 Z= X 3/4 ~60m DIM=1.7 Z= X 1/ X 3/4

29 The example below illustrates the application in detail It s been worked up in the table below. What do you think? Diagnostic positions Multipliers Sphere Z Sphere Multipliers Cylinder X 3/4 = X /3 = X 1/ = X 1/3 = X 1/4 = Z Cylinder 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?

30 As we ve shown, we can estimate other properties of the buried object 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/cm 3. R gmaxz 8.5 g max Z 8.5R 3 1/ 3 (kilofeet) (kilofeet)

31 In-class activities precision matters use a ruler

32 Take a few minutes and determine what shaped object produces each anomaly

33 Just note that this approach has been developed for a number of simple geometrical shapes Diagnostic Position Depth Index Multiplier 3/4 max 1/0.86 = 1.16 /3 max 1/1.1 = / max 1/1.7= /3 max 1/.76 = /4 max 1/3.7 = 0.7 g G R A vertical cylinder or volcanic pipe 1 1 z x z L x 1/ 1/ R for meters Z 1 R for feet Z gmaxz1 R gmaxz R 1 1/ (feet) (feet)

34 For a given anomaly certain simple geometries can be assumed and tested Sphere or vertical cylinder Horizontal cylinder or vertical dyke A A A A A A

35 How about the anomaly below? Half plate or faulted plate 10 mg 0 mg

36 High angle fault: normal or reverse Fault is located at the anomaly inflection point Half plate or faulted plate 10 mg 0 mg Low High

37 Recall the use of this simple geometrical object used to obtain the topographic correction 1 sectors with R i =1100 and R o =00 Ring The butte fits into one sector Butte Consider how you would do this

38 Another application of simple geometrical objects and the residual Just for general discussion > (see 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/cm 3 A lot of these ideas carry over into the analysis of magnetic data Bouguer Anomaly (mgal) Problem Horizontal Position (km)

39 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

40 Just as with the graphical approach, the idea is to remove the regional so you can investigate the residual. With the residual anomaly you can answer the question: what is the depth?

41 Gravity model studies help us estimate the possible configuration of the continental crust across the region Derived from Gravity Model Studies

42 As always, consider the possibility of non-unique solutions Are alternative acceptable solutions possible?

43 Gravity applications span a variety of scales Shallow environmental applications Roberts, 1990

44 Topographic extremes Japan Archipelago North American Plate Pacific Plate Philippine Sea Plate Geological Survey of Japan

45 The Earth s gravitational field In the red areas you weigh more and in the blue areas you weigh less. g ~0.6 cm/sec North American Plate Pacific Plate Philippine Sea Plate Geological Survey of Japan

46 Gravity methods have applications over a wide range of scales

47 Items on the list 6.3 due today Draft essay due Friday, 4 th by noon Due date for problems 6.5 and 6.9 moved to the 10 th Gravity lab due date delayed till Nov. 15 th due to election day recess. Begin reading Chapter 7 on Magnetics. We ll get into magnetic methods next week. Following Thanksgiving break we will wrap up magnetic methods on Nov. 8 th and Dec 1 st with some exam review on the 1 st and 6 th. Final, Wednesday, December 14 th from 11am-1pm in rm 35 Brooks.