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Problem 1 The corrections can be larger than the anomaly Stat.Time T Dist. (m) Elev. (m) Reading (dial units) Base reading at time T Drift corr’d anom.

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Presentation on theme: "Problem 1 The corrections can be larger than the anomaly Stat.Time T Dist. (m) Elev. (m) Reading (dial units) Base reading at time T Drift corr’d anom."— Presentation transcript:

1 Problem 1 The corrections can be larger than the anomaly Stat.Time T Dist. (m) Elev. (m) Reading (dial units) Base reading at time T Drift corr’d anom. (gu) LC (gu) FAC (gu) BC (gu) Free air anom (gu) Boug. anom. (gu) BS0805002934.2000 108352010.372931.32934.49-12.10-0.1632.00-11.7319.748.01 208444012.622930.62934.57-15.05-0.3238.95-14.2823.589.3 308556015.322930.42934.68-16.23-0.4847.28-17.3430.5713.23 409038019.402927.22934.76-28.67-0.6359.87-21.9530.578.62 BS0918002934.90 Note that the most significant part of the free-air anomaly appear to due to the attraction of the extra material beneath the survey stations, and that when the Bouguer correction is applied the remaining anomaly is quite small and positive – i.e. the rocks below these stations are slightly denser than those below the base station.

2 Problem 2 The width of the anomaly is related to the depth of burial of the target body Set K = 2GΔρ2πb 2 and determine Δg at x = 0 and x = x 1/2 at x = 0 Δg max = Kz/(0 + z 2 ) and at x = x 1/2 Δg max /2 = Kz/(x 1/2 2 + z 2 ) equating these two we get: 1/2 x Kz/(0 + z 2 ) = Kz/(x 1/2 2 + z 2 ) Hence 2z 2 = x 1/2 2 + z 2 and z max = x 1/2

3 Problem 3 x (m) )g (gu)x (m))g (gu) -4-0.008-4-0.015 -3-0.014-3-0.025 -2-0.03-2-0.051 -0.102-0.1275 0-0.510-0.255 1-0.1021-0.1275 2-0.032-0.051 3-0.0143-0.025 4-0.0084-0.015

4 Why is it Z max ? If the body were deeper, the width of the anomaly would increase. But you could obtain the same gravity anomaly with other bodies placed at shallower levels in the section

5 Lecture 2Gravity 2 Formulae for a horizontal slab Burial depth on width of anomaly Modelling gravity data. Non-uniqueness Calculation of the total mass excess/deficit Filtering/processing of gravity data Problems 4-5

6 Gravity anomaly referred to as: Delta g or Δg or δg all interchangeable

7 Infinite horizontal slab Example, Δρ = 400 kgm -3 t = 50m δg = 2 x 3.14 x 400 x 6.67 x 10 -11 x 50 = 8.38 x 10 -6 ms -2 = 8.38 gu Independent of depth to slab

8 Thin semi-infinite slab

9 δg = 0 Change occurs more rapidly when slab is closer to the surface Sets of theoretical curves are very useful = 8.38 gu

10 Semi-infinite thick horizontal slab Start with a simple model – gradually make it more sophisticated

11 Modelling and non-uniqueness An infinite number of subsurface density models will fit the gravity data equally as well Can we constrain Δρ? Can drill holes constrain parts of the model? Do we have geological knowledge that limits the range of possible density models? Do we have other geophysical data that can distinguish between density models?

12 Forward modelling (by nature is often subjective) Guess model, see if it fits the observed data, change model until it does fit the data

13 Forward modelling example matics of crater formation Younger, thinner Proterozoic crust is adjacent to with the thickened Archean craton, NE India

14 Modelling and non-uniqueness Four models of Chicxulub (a,b,c from gravity, magnetic and borehole data)

15 Numerical modelling crater formation Morgan et al., 2000 Collins et al., 2002

16 Forward modelling cont. Debate on shape of stratigraphic uplift Shape tells us about the kinematics of crater formation

17 Inverse modelling – strive to make objective Set up an automated procedure to test a wide range of models and chose a range that fit the data reasonably well. e.g. allow density to vary between 0.05 and 0.2 in steps of 0.01 g/cm 3 and allow z (burial depth) to vary between 2 and 5 km in steps of 0.1 km

18 Mass excess/deficit Gauss’s theorum: The integral of the divergence of a vector field over a region of space is equal to the outward normal component of the field over the surface enclosing the region. The mass excess calculation is unique Meaning: For any particular body, the total gravity anomaly is the same i.e the area (or volume) under the curve is the same In 2D the mass excess per unit length is:

19 In 3D this would be: Me = 1/2πG x ∑ Δg x Δa where Δa is the area ∑ Δg = Δg 1 + Δg 2 + Δg 3... etc

20 Regional versus residual gravity anomalies Regional – long wavelength features due to deep seated structures Residual – short wavelength features due to shallow structures

21 Filtering of gravity data to enhance signal of interest

22 z = 0 z = -s z = s z = 2s We measure g on surface z = 0 at stations that are a distance x apart We can calculate the value of g on any surface we like z = s, -s etc g m = k 1 g 0 + K 2 (g 1 +g -1 ) + K 3 (g 2 + g -2 )… etc k i depends on s and x g gmgm g0g0 g1g1 g2g2 g3g3 g -1 g -2 x

23

24 The gradient of gravity data enhances the residual anomalies.

25 Bouguer gravityHorizontal gradient of Bouguer gravity


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