1 / 49

Environmental and Exploration Geophysics I

Environmental and Exploration Geophysics I. Magnetic Methods ( III). tom.h.wilson tom.wilson@mail.wvu.edu. Department of Geology and Geography West Virginia University Morgantown, WV. Thus . H - monopole = . H - dipole.

ahanu
Download Presentation

Environmental and Exploration Geophysics I

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Environmental and Exploration Geophysics I Magnetic Methods (III) tom.h.wilson tom.wilson@mail.wvu.edu Department of Geology and Geography West Virginia University Morgantown, WV Tom Wilson, Department of Geology and Geography

  2. Thus .. H - monopole = H - dipole This yields the field intensity in the radial direction - i.e. in the direction toward the center of the dipole (along r). However, we can also evaluate the horizontal and vertical components of the total field directly from the potential. Tom Wilson, Department of Geology and Geography

  3. H Toward dipole center (i.e. center of Earth’s dipole field Vd represents the potential of the dipole. Tom Wilson, Department of Geology and Geography

  4. arc - length relationship HE is represented by the negative derivative of the potential along the earth’s surface or in the S direction. Tom Wilson, Department of Geology and Geography

  5. Evaluating the derivative along the surface Tom Wilson, Department of Geology and Geography

  6. -dV/dS Tom Wilson, Department of Geology and Geography

  7. Where M = pl and Let’s tie these results back into some observations made earlier in the semester with regard to terrain conductivity data. 32 Tom Wilson, Department of Geology and Geography

  8. The field along the dipole equator Given What is HE at the equator? … first what’s ?  is the angle formed by the line connecting the observation point with the dipole axis. So , in this case, is a colatitude or 90o minus the latitude. Latitude at the equator is 0 so  is 90o and sin (90) is 1. Tom Wilson, Department of Geology and Geography

  9.  is 90 The field at the pole At the poles,  is 0, so that What is ZE at the equator? Tom Wilson, Department of Geology and Geography

  10. Field at pole is twice that at the equator ZE at the poles …. The variation of the field intensity at the poles and along the equator of the dipole may remind you of the different penetration depths obtained by the terrain conductivity meters when operated in the vertical and horizontal dipole modes. Tom Wilson, Department of Geology and Geography

  11. Consider one of the Lab Questions …. compare the field of the magnetic dipole field to that of the gravitational monopole field Gravity:500, 1000, 2000m A more rapid decay Increase r by a factor of 4 reduces g by a factor of 16 Tom Wilson, Department of Geology and Geography

  12. A 4 fold increase in distance For the dipole field, an increase in depth (r) from 4 meters to 16 meters produces a 64 fold decrease in anomaly magnitude Thus the 7.2 nT anomaly (below left) produced by an object at 4 meter depths disappears into the background noise at 16 meters. 0.113 nT 7.2 nT Tom Wilson, Department of Geology and Geography

  13. Problem 7.1 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. On Tuesday during the last week of class, we’ll work through some problems that will help you review materials we’ve covered on magnetic fields. Some of the problems are not too much different from those we worked for gravitational fields and so will help initiate some review of gravity methods. Between now and next Tuesday complete the questions below. We will get you started today. Tom Wilson, Department of Geology and Geography

  14. Problem 7.1 The relationship between the potential and field intensity requires use of the minus sign. However, no minus sign is required when computing the derivative or gradient Tom Wilson, Department of Geology and Geography

  15. Again – just take the simple derivative to get the gradient (no negative sign) To answer this problem we must evaluate the horizontal gradient of the vertical component - or Take a minute and give it a try. Tom Wilson, Department of Geology and Geography

  16. A simple question (reference formula 7.2) Problem 7.2. The magnetic field intensity of a dipole is given as Offer a mathematical argument to show that the field intensity near one end of a long magnetic dipole (for example that produced by well casing) is nearly equal to that of an isolated magnetic pole. r- and r+refer to the distances from the point of observation to the negative and positive poles of the dipole, respectively. p denotes the pole strength. Hint: one of the poles will be at a much greater distance from the surface than the other. Tom Wilson, Department of Geology and Geography

  17. A simple geometrical approximation for “back of the envelope” computations … Can you detect the wall? 7.3 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. Tom Wilson, Department of Geology and Geography

  18. Magnetic effects of simple geometric shapes In-depth discussions of simple geometrical objects are presented on pages 456 to 482. For our purposes, however, we will avoid the heavy math and consider the general usage of half-max relationships in a manner similar to that developed for the analysis of gravitational fields produced by simple geometrical objects using diagnostic positions and depth index multipliers. The following problems illustrate some uses of these ideas. Tom Wilson, Department of Geology and Geography

  19. Is the wall detectable? Vertically Polarized Horizontal Cylinder Problem 7.3 Maximum field strength Remember this kind of formulization used in gravity General form Normalized shape term Tom Wilson, Department of Geology and Geography

  20. Problem 7.3 Vertically Polarized Horizontal Cylinder Maximum field strength recall 1.5m k=0 k = 0.001, FE= 55,000 nT k=0.001 0.5m What is R? 1.0m Approximate cross sectional area of rectangular wall in circular form and solve for R: i.e. Cross sectional area of wall = 1m x 0.5m Lastly – what is z? Tom Wilson, Department of Geology and Geography

  21. Another problem to consider: abandoned wells or something else Non text question: 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 equidimensionally shaped object (a sphere); or 2) to a long vertically oriented cylinder. Tom Wilson, Department of Geology and Geography

  22. Choose your SGO The well: Vertically Polarized Vertical Cylinder Look familiar? Tom Wilson, Department of Geology and Geography

  23. Equidimensional object? Vertical Magnetic Anomaly Vertically Polarized Sphere (see section 7.5.4) Simple geometrical objects As a function of x/z … The notation can be confusing at times. In the above, consider H = FE= intensity of earth’s magnetic field at the survey location. Tom Wilson, Department of Geology and Geography

  24. Non-text question Sphere or vertical cylinder Vertical cylinder sphere ZA is magnetic field intensity at some particular point of interest, for example at a point where Z is ½ the maximum value Zmax or ¼ or ¾. Tom Wilson, Department of Geology and Geography

  25. Diagnostic positions – vertical cylinder At what point (x/z) does the anomaly fall to ½ its maximum value? Let so Remember doing this with the gravity anomaly associated with a spherically distributed density distribution? Tom Wilson, Department of Geology and Geography

  26. The anomaly over a magnetized vertical cylinder has the same shape as gravity anomaly over a sphere and we could solve for z as x1/2 is referred to as a diagnostic position. It’s the distance (from the anomaly peak) to the point on the surface where the anomaly drops to ½ of its maximum value. 1.305 is referred to as the depth index multiplier. Tom Wilson, Department of Geology and Geography

  27. We could solve this equation for any diagnostic position For example – the point where the anomaly falls to 1/4th of it’s maximum value: then The diagnostic position is x¼ (the distance from the peak to the point where the anomaly drops to 1/4th its maximum value); the depth index multiplier is 0.81. Tom Wilson, Department of Geology and Geography

  28. Each diagnostic position gives you an estimate of z (depth to object). 1.2m 1.9m 2.8m Tom Wilson, Department of Geology and Geography

  29. This also gives you a way to test whether the anomaly is produced by a sphere (isolated dipole) or vertical cylinder (isolated pole). 1.2m 1.9m 2.8m Tom Wilson, Department of Geology and Geography

  30. The answer to the question “what kind of object is producing the anomaly” … is determined by finding which of the two possible objects under consideration give the most consistent estimates of z. Take a moment and answer that question in today’s handout. Tom Wilson, Department of Geology and Geography

  31. Another twist to the problem 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. Tom Wilson, Department of Geology and Geography

  32. Another non-text questionjust some equation manipulation … You have k = 0.1 and HE=55000nT, Zmax = 62.2nT. what else do you need and how do you get it? Tom Wilson, Department of Geology and Geography

  33. Back to the magnetics lab Recall, that in this lab we need to locate buried metallic drums containing radioactive waste. Simple excavation (back-hoeing) of the area is ruled out since accidental rupture of the drums would release radioactive materials into the local groundwater system and atmosphere. Clean up would be extremely hazardous, difficult, and expensive and might leave the local area uninhabitable for many years. Tom Wilson, Department of Geology and Geography

  34. The magnetic survey, by itself, would be unable to clearly delineate drum locations. Since the bedrock is magnetic, we have no way of differentiating between anomalies produced by bedrock and those produced by buried storage drums. ? Tom Wilson, Department of Geology and Geography

  35. Why gravity? Acquisition of gravity data allows us to estimate variations in bedrock depth across the profile. With this knowledge, we can directly calculate the contribution of bedrock to the magnetic field observed across the profile. Tom Wilson, Department of Geology and Geography

  36. Anomaly associated with buried metallic materials Computed magnetic field produced by bedrock Results obtained from inverse modeling Bedrock configuration determined from gravity survey The other day, we completed analysis of the gravity data. Bedrock depth has been determined. Tom Wilson, Department of Geology and Geography

  37. The drums are located in the areas where significant magnetic anomalies, unassociated with the bedrock, remain. Your attempts to model the remaining (residual) anomaly yield an excellent match to the data, but you wonder if the results are reliable – they don’t make sense. Beware of the flattened drum solution Tom Wilson, Department of Geology and Geography

  38. Working with the triangular shaped object estimate the location, number and depth of the buried drums. Where are the drums and how many are there? Tom Wilson, Department of Geology and Geography

  39. The triangular region must have a length and width large enough to contain whole drums. Questions?? Review from lab manual Where are the drums? Tom Wilson, Department of Geology and Geography

  40. Don’t fall victim to non-unique solutions From the bedrock from lab manual Tom Wilson, Department of Geology and Geography

  41. anomaly A collection of flattened drums easily explains the observed anomaly – but … Tom Wilson, Department of Geology and Geography

  42. Common sense tells you that the solution (right below) is the most likely one Tom Wilson, Department of Geology and Geography

  43. Use the area to estimate the number of drums buried in the cluster 4 square feet Area of one drum ~ What’s wrong with the format of this plot? The x and y scaling has to be 1:1! Tom Wilson, Department of Geology and Geography

  44. Given that magnetic field of the drum varies as the inverse cube of distance, you know you still have problems. Why? Gravity:500, 1000, 2000m A more rapid decay Increase r by a factor of 4 reduces g by a factor of 16 Tom Wilson, Department of Geology and Geography

  45. A 4 fold increase in distance For the dipole field, an increase in depth (r) from 4 meters to 16 meters produces a 64 fold decrease in anomaly magnitude Thus the 7.2 nT anomaly (below left) produced by an object at 4 meter depths disappears into the background noise at 16 meters. 0.113 nT 7.2 nT Tom Wilson, Department of Geology and Geography

  46. Magnetics lab Take a look at this and bring questions to class on Thursday. We’ll provide some additional review time on Tuesday, Dec. 2nd Tom Wilson, Department of Geology and Geography

  47. Enjoy the Thanksgiving break!See you on Dec. 3rd Items on the list …. • Gravity lab is due today • Magnetic papers are in the mail room • Work on problems 7-1 through 7-3. They will be due December 5th and returned for review on December 10th • Magnetic paper summaries are also due on December 5th • Magnetics lab summary and news article are due on the 10th • We will have two final exam review sessions: December 5th and December 10th. • Final is from 3-5pm on December 13th. Tom Wilson, Department of Geology and Geography

  48. Regular section submissionsMagnetics All those in the regular section submit paper copies of your paper summaries and lab reports. Tom Wilson, Department of Geology and Geography

  49. Writing Section reminders(electronic submissions only) • Self-reviewed magnetic methods paper summary (only one) is due on December 5th. • The self-reviewed magnetics article is due on December 10th. All those in the writing section submit their papers and lab electronically. Don’t forget to turn on track changes while doing your self-reviews. Only submit the self-reviewed file. Tom Wilson, Department of Geology and Geography

More Related