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

Problems we’ve been working on … Tom Wilson, Department of Geology and Geography

Questions? Tom Wilson, Department of Geology and Geography

The first problem relates to our discussions of the dipole field and their derivatives What is the horizontal gradient in nT/m of the Earth’s vertical field (Z E ) in an area where the horizontal field (H E ) equals 20,000 nT and the Earth’s radius is 6.3 x 10 8 cm.

Tom Wilson, Department of Geology and Geography Recall that horizontal gradients refer to the derivative evaluated along the surface or horizontal direction and we use the form of the derivative discussed earlier for the potential. The negative sign is NOT needed when computing the gradient.

Tom Wilson, Department of Geology and Geography To answer this problem we must evaluate the horizontal gradient of the vertical component - or See Equation 7.20

Evaluate the horizontal gradient Tom Wilson, Department of Geology and Geography Since  is co-latitude, the direction of increasing  is southward (in the northern hemisphere). As we travel from pole to equator Z E decreases, thus the gradient is negative.

Tom Wilson, Department of Geology and Geography 4. 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. What is z? What is I?

Tom Wilson, Department of Geology and Geography is a function of the unit-less variable x/z The vertical field is often used to make a quick estimate of the magnitude of an object. This is fairly accurate as long as i is 60 or greater Dipole/sphere Horizontal cylinder Vertical cylinder

Tom Wilson, Department of Geology and Geography Vertically polarized sphere or dipole Vertically polarized horizontal cylinder

Considerable difference in magnitude of Tom Wilson, Department of Geology and Geography For the dipole For the horizontal cylinder

Tom Wilson, Department of Geology and Geography 4. 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 equidemensional shaped object (a sphere); or 2) to a long vertically oriented cylinder.

Tom Wilson, Department of Geology and Geography The map view clearly indicates that consideration of two possible origins may be appropriate - sphere or vertical cylinder.

Tom Wilson, Department of Geology and Geography In general one will not make such extensive comparisons. You may use only one of the diagnostic positions, for example, the half-max (X 1/2 ) distance for an anomaly to quickly estimate depth if the object were a sphere or buried vertical cylinder…. Burger limits his discussion to half-maximum relationships. Breiner, 1973 X 1/2 = Z/2 X 1/2 = 0.77Z X 1/2 = Z X 1/2 = Z/2

Tom Wilson, Department of Geology and Geography Just as an aside: The sample rate you use will depend on the minimum depth of the objects you wish to find. Your sample interval should probably be no greater than X 1/2. But don’t forget that equivalent solutions with shallower origins do exist!

Tom Wilson, Department of Geology and Geography We’ll make quick work of it an use only three diagnostic positions (red above)

Tom Wilson, Department of Geology and Geography Again, we can get by with only three diagnostic positions (red above)

Tom Wilson, Department of Geology and Geography Determine depths (z) assuming a sphere or a cylinder and see which assumption yields consistent estimates. It’s all about using diagnostic positions and the depth index multipliers for each geometry.

Tom Wilson, Department of Geology and Geography Sphere vs. Vertical Cylinder; z = __________ Diagnostic positions Multipliers Sphere Z Sphere Multipliers Cylinder Z Cylinder X 3/4 = X 1/2 = X 1/4 = The depth diagnostic distance 0.9 X 3/ X 1/ X 1/

Tom Wilson, Department of Geology and Geography Diagnostic positionsMultipliers Sphere Z Sphere Multipliers Cylinder Z Cylinder X 3/4 = 1.6 meters X 1/2 = 2.5 meters21.31 X 1/4 = 3.7 meters Sphere or cylinder? g max g 3/4 g 1/2 g 1/

Tom Wilson, Department of Geology and Geography 5. Given that derive an expression for the radius, where I = kH E. 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 H E =55000nT. Z max (62.2nT from graph below) is the maximum vertical component of the anomalous field produced by the vertical casing. Algebraic manipulation

Tom Wilson, Department of Geology and Geography Follow the recommended reporting format. Specifically address points mentioned in the results section, above.

Tom Wilson, Department of Geology and Geography

Where are the drums?

Tom Wilson, Department of Geology and Geography

anomaly

Tom Wilson, Department of Geology and Geography

4 square feet Area of one drum ~ Make sure the scale of your graph is 1:1

Tom Wilson, Department of Geology and Geography …. compare the field of the magnetic dipole field to that of the gravitational monopole field Gravity:500, 1000, 2000m Increase r by a factor of 4 reduces g by a factor of 16

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

Tom Wilson, Department of Geology and Geography Again - follow the recommended reporting format. Specifically address listed points (1-5).

Tom Wilson, Department of Geology and Geography You are asked to run a magnetic survey to detect a buried drum. What spacing do you use between observation points? Sampling issues – for leisure consideration … Jump to last slide for reminders

Tom Wilson, Department of Geology and Geography How often would you have to sample to detect this drum? X 1/2 =Z/2

Tom Wilson, Department of Geology and Geography …. how about this one? The anomaly of the drum drops to ½ at a distance = ½ the depth.

Tom Wilson, Department of Geology and Geography Sampling does depend on available equipment! As with the GEM2, newer generation magnetometers can sample at a walking pace.

Tom Wilson, Department of Geology and Geography Remember, the field of a buried drum can be approximated by the field of a dipole or buried sphere. X 1/2 for the sphere (the dipole) equals one-half the depth z to the center of the dipole. The half-width of the anomaly over any given drum will be approximately equal to its depth Or X 1/2 =Z/2

Tom Wilson, Department of Geology and Geography

Feel free to discuss these problems in groups, but realize that you will have to work through problems independently on the final.

Tom Wilson, Department of Geology and Geography General Review this coming Thursday Turn in your magnetics lab report Thursday, December 10 th. Exam, Friday December 17 th ; 3-5pm