Gravity Methods (IV)

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

2 Possible employment opportunities Robert McCue, President of Raptor Consulting, Inc., will be visiting the Department and presenting on career opportunities at noon Thursday (11/3). The presentation will be in Brooks 325. Information on Raptor Consulting is online at: Their focus is on well site geology and geosteering There will be pizza.

3 Also Intern 2017 (Geology) Department: EQT Prod - Geology See - Description - Intern Geologist The Geology Intern works with EQT's Geology Department in Pittsburgh for the summer of The intern will work on one of our Geology Teams with a Geologist as a mentor. Responsibilities include the following: Logs evaluation and correlation of formation tops Generates geologic prognoses for drilling and well planning Updates geologic maps used in the well planning process Completes a mapping project that coincides with the team s objectives Required Experience Currently enrolled, full-time student in a bachelor's degree program in Geology Completion of at least junior year and returning to school at the end of the internship preferred Experience with wireline logs and Geographix digital mapping system preferred EQT Corporation and its subsidiaries is an EEO/Affirmative Action employer -- M/F/Disability/Veteran

4 Today More details on corrections a) Free air b) Plate c) Topography d) Tide and drift Some in-class problems! Hand in any redo work before leaving Hand out lab guide for gravity lab and review as time permits

5 What did you get? Brief comment on Tuesday s exercise

6 Predicting g anywhere on earth What physical characteristics of the observation point and it s surroundings influence the observed acceleration of gravity?

7 The gravity anomaly Remember all those terms in the theoretical or predicted gravity equation? Subtract them from the observed g to get the anomaly. The gravity anomaly gobs g t The geologist/geophysicist is then left with the task of interpreting/modeling the anomaly in terms of geologically reasonable configurations of subsurface intervals.

8 What s the geologist interested in? What is this gravity anomaly associated with? air =0 gm/cm 3 wb =2.55gm/cm 3 ls =2.71gm/cm 3 bu =2.75gm/cm 3 cave =0 gm/cm 3

9 What is this anomaly produced by? wb =0 gm/cm 3 air =0 gm/cm 3 ls =2.71gm/cm 3 bu =2.75gm/cm 3 cave =0 gm/cm 3

10 No anomaly implies homogeneous subsurface geology. The anomaly below(from Stewart).. If the observed values of g behave according to our ideal model then there are no lateral density contrasts of unknown origin in the subsurface so there is no interesting geology only a series of homogeneous layers But in this case -> In the till covered areas of Wisconsin differences in g are in part related to variations on depth to bedrock.

11 We can estimate the geological properties of the subsurface from the anomaly Figure below is based on gravity anomalies presented by Stewart Because we know something about the local geology Good location for a water well

12 Let s discuss the background on Stewart s simple formula for making a rough estimate of depth to bedrock He starts by assuming that glacial valleys are much wider than they are deep and he uses and infinite plate to approximate the anomaly associated with them. Assuming this G = x 10 cm gm s 2 gvalley 2Gt G 4.192x10 cm ( gm s ) Thus g plate = x 10-7 t Gals, where t is valley depth in centimeters and, the density contrast between drift and bedrock in gm/cm 3. This is also x 10-4 mgals since there are 10 3 milligals per Gal. 4

13 g plate 2Gt Also if we want to allow the user to input thickness (t) in meters, we have to introduce a factor of 100 into the above relationship. So that when we enter a value of 1meter for t we want the computer or calculator to realize that and convert it to 100cm. That way all input values of t in meters are converted to the required units of centimeters. This would change x 10-4 t mgals (in the previous slide) to x 10-2 t or t mgals. Pretty simple!

14 Bottom line revised equation mixed units gplate t Where is in gm/cm 3 and t is in meters These equations can be as confusing as they are helpful. If you don t mix units, you ll get the wrong answer!

15 Stewart does something similar g t 130 or t 130g where t is in feet This expression comes directly from g plate 2Gt Stewart has also solved it using a constant density =0.6 gm/cm 3. (He s actually assuming a negative density contrast of 0.6. So keep in mind that positive t represents depth beneath the surface.) He has also included the factor which transforms centimeters to feet so that the user can input t in units of feet. g is in units of milligals. He s made a couple assumptions. What are they? Valley s have infinite extent and density contrast is constant.

16 Don t forget what this formula represents it provides the acceleration of a flat plate, infinite in extent, with constant thickness of t g plate 2Gt t 130g Consider this with much less vertical exaggeration of the section 557 ft 10,000 ft Vertical exaggeration is still about 2:1

17 So are valleys infinitely wide? g plate 2Gt t 130g Consider this with much less vertical exaggeration of the section 557 ft 10,000 ft If valleys are 10 times wider than they are thick not too bad of an approximation, but shape is lenticular not rectangular.

18 While not completely accurate, it be used to make a rough approximation? gvalley 2Gt t 130g 557 ft 10,000 ft Simple geometrical objects are often used in gravity interpretation to represent complex objects for the purpose of getting a rough approximation. There s more on this topic later in the chapter.

19 T=130g continued As noted earlier -8 3 G = x 10 cm gm s 2 and G 4.192x10 cm ( gm s ) Thus g plate = 2Gt = x 10-7 t in units of cm/s 2 or Gals. This is also x 10-4 t mgals as mentioned earlier Add in the constant density of 0.6gm/cm 3 and Conversion of 1 foot into 30.48cm, and you get g plate = 2Gt = t or t=130.4g 3

20 With some expanded discussion for reference But, if we want to allow the user to input thickness in feet, we have to introduce a factor of (i.e. our input of 1 foot has to be multiplied by 30.48cm/foot) to convert the result to centimeters. 4 Where 2 Gt 4.192x10 t milligals of gm/cm 3 and cm respectively, and and t are in units This would change the above to t milligals, where and t have units of gm/cm 3 and ft, respectively, giving another form of the equation with mixed units and variable density. Note that if we then fix the density = 0.6 gm/cm 3 then we have g = t in mgals and thus t =(1/ )g

21 Again for reference Stewart s formula is developed as follows - g 2G t plate t 4 (4.192x10 (0.6)(30.48) In this case we're interested to solve for t given a value for g thus g t= 130g Just remember that g is the residual gravity anomaly. We ll talk more about residual gravity later. The residual gravity is basically another form of the anomaly for which the contributions associated with the regional geology have been removed. t

22 Problems 6.1 and 6.2 5

23 Problems 6.1 and 6.2 Corrections vs. effects g freeair R cos z The free air effect is often simplified by ignoring the influence of latitude and z g R or t freeair & the Bouguer plate effect of g plate 2Gt When we calculate the theoretical gravity, the free air term is subtracted and the plate term is added. When we are correcting the observed gravity to obtain the anomaly, the free air is added and the Bouguer plate term is subtracted.

24 g g g anom obs t The elevation correction is: g ( ) h in milligals E In this case we also have mixed units where is in units of gm/cm 3 and h is in meters Correction refers to the changes we introduce into the observed gravity to get the anomaly. Note the minus sign for g t changes the sign of the free air and Bouguer plate terms in the elevation correction.

25 Another reminder g g g anom obs t g g g ( ) g g g g anom obs n FA B t TideDrift The observation is corrected by subtracting the terms in the theoretical gravity. Carry the minus sign through g g g ( ) g g g g anom obs n FA B t TideDrift g g g ( ) ( g g ) g g anom obs n FA B t TideDrift The elevation correction g e

26 Problems 6.1 and 6.2

Topographic correction correction g ( ) g g g g n FA B t TideDrift g B may seem like a pretty unrealistic approximation of the topographic surface. It is.

27 Topographic correction correction g ( ) g g g g n FA B t TideDrift g B may seem like a pretty unrealistic approximation of the topographic surface. It is. You had to scrape off all mountain tops above the observation elevation and fill in all the valleys when you made the plate correction. See figure 6.3 5

28 Recall the effect of reintroducing topography onto our flat plate

29 We estimate the effect of topography by approximating topographic features as ring-sectors whose thickness (z) equals the average elevation of topographic features within them.

30 1/ 2 1/ gring 2G Ro Ri Ri z Ro z R i = inner radius of the ring R o = outer radius of the ring z = thickness of the ring (average elevation of the topographic features inside the sector of interest) g g ring sector number of sectors (n) For a derivation, see Burger et al. (2006) Using the units simplification discussed earlier gsector r2 r1 r1 z r2 z n / 2

31 The topographic effect g t is always negative, the correction positive. Again, this may seem like a crude approximation of actual topography. But topographic compensation is a laborious process and if done in detail the estimate is fairly accurate. We can increase the detail of our computation depending on the accuracy needed in a given application. Now we use digital elevation data and let the computer do a very detailed computation. But the principle is the same We ll discuss methods used to compute the topographic effect more in the next lecture. The last term we will look at incorporates the effects of tide and instrument drift. g g ( ) g g g g t n FA B t TideDrift

32 The topography surrounding the station in problem 6.2 is very simple Topography surrounding you consists of a broad plain extending to great distances in all directions You are here

33 The formula for the disk This is the formula for that disk 1/ 2 1/ gring 2G Ro Ri Ri z Ro z Plug in for the thickness z, inner and outer radii, R i and R o, respectively, and density

34 Problem 6.3, will not be due till later, but a good idea to cover these concepts in the text To make it simple, we ll assume that the base station value is 0 mgals for reference.

35 Tide and instrument drift We are used to thinking in terms of ocean tides. The ocean surface rises and falls under the influence of the combined gravitational attraction of the sun and moon. The solid earth also deforms in response to the differential pull of the sun and the moon. The change in surface elevation in addition to their gravitational pull on the gravimeter spring can be significant and these tidal effects must be incorporated into our estimate of theoretical gravity. Berger (1992)

36 The gravimeter is just a mechanical system. Its parts - while simple - change over time. The spring for example, subjected to the constant tug of gravity will experience permanent changes in length over time. These changes fall under the heading of instrument drift. Berger (1992) 2

In general the influence of tide and drift on the theoretical gravity is estimated by direct and repeated measurement of gravitational acceleration at the same place (a base station) over and over

37 In general the influence of tide and drift on the theoretical gravity is estimated by direct and repeated measurement of gravitational acceleration at the same place (a base station) over and over again, throughout the duration of your survey. Usually during a survey a base station is reoccupied every couple hours or so. The drift curve is constructed from these measurements and measurements of acceleration made at other stations are corrected relative to the drift curve.

38 Gravity observations made during survey 4 Station 2 Reoccupied base station milligals Base station Station Base S1 S2 Base TIME (am) Is the acceleration of gravity measured at 9am the same as that measured at the base station an hour earlier?

39 The influence of drift M i l l i g a l s Base station +1 mg Station 1-1 mg Station 2 Reoccupied base station Tide & Drift Curve Base S1 S2 Base TIME (am)

40 Questions? In this example, the acceleration at station 1 (S1) is 1 milligal less than that at the base station - not the same. At station 2, the acceleration is only 1 milligal greater, not 3 milligals greater. M i l l i g a l s mg -1 mg Base S1 S2 Base TIME Tide and Drift Curve 4

41 Theoretical or predicted gravity g g ( ) g g g g t n FA B t TideDrift Any questions about the model we ve proposed to explain the gravitational acceleration at an arbitrary point on the surface of our theoretical (but geologically unrealistic) earth? g g g anom obs t

As geologists we expect there will be considerable subsurface density contrast associated, for example, with structure - or stratigraphy, drift thickness, caves, trenchs In preparing our gravity

42 As geologists we expect there will be considerable subsurface density contrast associated, for example, with structure - or stratigraphy, drift thickness, caves, trenchs In preparing our gravity data, we start by computing the theoretical gravity but usually find that the theoretical gravity we compute at a given latitude and elevation does not equal the observed gravity at that location.

43 The gravity anomaly gobs gt Geology inferred from modeling the anomaly An anomaly exists - and therein lies the geology.

44 Complete and turn in problems 6.1 and 6.2 next class. 6.1 If gravity determination is made at an elevation of 400 m, what is the value of the free-air correction (assuming sea level as the datum)? What is the Bouguer correction (assuming a 2.5 gm/cm3 reduction density)? 6.2 A gravity station at an elevation of 0 m is located in the center of an erosional basin. The floor of the basin has virtually no relief. The plateau escarpment is located at a distance of 450 m from the gravity station. The surface of the plateau has a relatively constant elevation of 400 m. Will the terrain correction be necessary? Assume a density of 2.5 gm/cm3. Hint: 2 1 1/ 2 ring o i i o g G R R R z R z where Ri is the inner radius of the ring R0 is the outer radius of the ring z is the thickness of the ring

45 We will discuss the in-class drift problem and problem 6.3 more next class so look over and bring your questions time dial reading Converted to milligals relative difference Tide & Drift Drift corrected Base Station Base Station See handout and bring questions to class on the 3 rd

46 In-class problems: free air, plate and terrain

47 Time permitting - bring up GM-SYS and have a look at the data from Stewart s paper Hand out Copy the folder Stewart to your network drive, bring up GM-SYS, and open the file Stewart.sur Follow along in the lab guide 4

48 t 130g g t 130 Edge effects Valley Predicted anomaly = 600/130 or 4.62 mg g=-3.12mg g=-4.25mg 1000 wide Valley 5000 Depth= The Glacial Valley The 3.12 milligal anomaly implies a valley depth of only 406 feet. The 4.25 milligal anomaly implies 550foot bedrock depth. We have errors of 32% and 8.3% in these two cases.

49 The Valley and the Plate g drift 2Gt density contrast t drift thickness If the valley width is much greater than its thickness, then the gravitational acceleration due to the drift is approximated by the infinite plate

50 If the valleys do not extend to infinity how will this change the observed gravity?

51 Due dates In the writing section, 1 st draft essay 2, also due on the 3 rd. Problems due next class Problem 6.3 due Nov. 3 rd. The gravity lab will be due on Thursday, November 10 th. We will clear up any questions you have about the gravity lab on the 3 rd and 8 th. Review the remainder of the chapter 6 (past page 378).

Intro to magnetic methods

Environmental and Exploration Geophysics I Intro to magnetic methods tom.h.wilson tom.wilson@mail.wvu.edu Department of Geology and Geography West Virginia University Morgantown, WV Items on the list Gravity