a. The distance from a shot point in which no refracted energy is recorded is termed the.

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Ursula Chandler
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1 GE 3040 Midterm #2 4/08/2009 Name 1. Fill in the blank: (20 pts) a. The distance from a shot point in which no refracted energy is recorded is termed the. b. In a two-layer case, the upward moving wave front generated when the angle of incident is the critical angle is known as a wave. c. Total reflection occurs when the angle of incident is (greater than, less than) the critical angle. d. The phenomena of ground state splitting of electron energy levels in the presence of a magnetic field is known as the. e. The proportionality constant for Hooke's law for normal stress is known as the. f. states that the first arrival at a geophone travels the least time path. g. or radial scattering of incident seismic energy occurs at abrupt discontinuities along interfaces, or structures whose radius of curvature is shorter than the wavelength of the incident wave. h. Particle motion in this surface wave is transverse to the direction of propagation of the wave:. i. The bulk modulus is the proportionality constant for Hooke's law for stress. j. In a 3-layer earth problem where each layer s velocity is greater than the overlying layer and the second layer is rather thin so refracted arrivals are not recorded, the depth to the V 3 layer is (over, under) estimated. k. High frequency body waves attenuate (faster, slower) than low frequency body waves as a function of distance? l. This seismic wave alternately compresses and dilates the rock it passes through:

2 m. This process provides a means of obtaining second layer refraction times between the end-on shot and the original cross-over distance: n. The velocity of a seismic wave is directly proportional to the of the rock and inversely proportional to the of the rock. o. Removal of the IGRF field from magnetometer data is referred to as the correction. p. A seismic survey wishes to record frequencies at 100 Hz and possibly higher, therefore the natural frequency of the geophone for this survey should be (greater than, less than) 100 Hz. q. Seismic waves undergo attenuation due to and. r. In the Northern hemisphere, the maximum positive value of the ΔT anomaly caused by dipole-like source magnetized by induction is located to the (north, south) of the source of the anomaly. 2. What component of Earth's magnetic field is most commonly measured by each during a ground magnetic survey and is the measurement a scalar or vector quantity? (10 pts) i. proton precession magnetometer ii. fluxgate magnetometer iii. alkali vapor magnetometers iv. gradiometers v. dip needle

3 3. What is a total field anomaly, i.e. what is(are) the assumptions made in modeling these anomalies? (5 pts) 4. Using the diagrams below showing 2D structures explain why large magnetic anomalies observed over sedimentary basins are generally thought to arise from lithology contrasts within the basement and not from local uplifts on the basement? What interpretation technique is commonly used to determine depth to the source? Draw a diagram showing the pertinent elements in the depth determinations explaining each. (10 pts)

$critically refracted wave on (a) the 4-5 interface, and (b) on the 2-3 interface. For case (a) sketch the proper ray path on the diagram.$ (10 pts) x shot V 1 = 500 m/sec V 2 = 2000 m/sec V 3 = 2000 m/sec V 4 = 1500 m/sec V 5 = 4000 m/sec 6.

(10 pts) x shot V 1 = 500 m/sec V 2 = 2000 m/sec V 3 = 2000 m/sec V 4 = 1500 m/sec V 5 = 4000 m/sec 6.

latitude (I = 60 o ), and at the North magnetic pole (I = 90 o ).

4 5. Using the information on the diagram below, what must the angle of incidence of a seismic ray be on the 1-2 interface to cause a critically refracted wave on (a) the 4-5 interface, and (b) on the 2-3 interface. For case (a) sketch the proper ray path on the diagram. (10 pts) x shot V 1 = 500 m/sec V 2 = 2000 m/sec V 3 = 2000 m/sec V 4 = 1500 m/sec V 5 = 4000 m/sec 6. The diagrams below are total field magnetic anomaly maps associated with a buried dipole at the equator (I = 0 o ), at a mid northern latitude (I = 60 o ), and at the North magnetic pole (I = 90 o ). In the space at the bottom of the diagrams indicate the latitude each is from. Also indicate which anomaly is also the horizontal field anomaly and which anomaly is also the vertical field anomaly. For the mid-latitude anomaly indicate the north direction. (10 pts)

5 7. Questions pertain to Table 1 below. (5 pts) a. What is the drift corrected reading for station 2+00 N with respect to the nine o'clock base reading? b. What is the drift corrected reading for station 4+00 S with respect to the nine o'clock base reading? c. Assuming that the value of the IGRF field in the survey area is nt, calculate the total field anomaly for station 2+00 N. Table 1 Total Field Magnetometer Survey Station Reading Time Base : N : N : N : N :55 Base : S : S : S : S :50 Base :00

6 8. From the time distance plot shown below, determine the first, second layer apparent velocities and the true velocity of the second layer, the dip and dip direction of the V 1 V 2 interface, and the vertical depth directly below the both shot points. If the dip you calculated was 10 o greater, what would the up and down dip apparent velocities be for this problem? (12 pts) Arrival time (ms) Distance (m)

7 9. Draw three different geologic scenarios (subsurface configurations) that would give rise to the time-distance graph shown below. Do not use hidden or low velocity layers as part of your scenario. How would an off-end shot or shots placed on the left hand side of the spread help resolve the dilemma? (8 pts)

8 10. The magnetic anomalies shown below result from a magnetometer surveys over an east/west lava tunnel on one of the islands of the Galapagos which is near the equator. Determine the depth to the center and radius of the tunnel. Earth s total field intensity is 36,000 gammas and the susceptibility of the basalt is cgs units. Is the ΔZ anomaly correct? Explain. North is to the right. (5 pts) anomaly strength (gammas) distance (m) dz dh dt

9 11. Interpretation of the seismic data shown below indicates a horizontal interface between the V 1 and V 2 layers but with different depths to the second layer below each shot point. Calculate the dip of the V 1 - V 2 interface connecting the horizontal portions of the interface at each end and the depth to the V 2 layer directly under the geophone located 240 m from shot point A. Assume a planar interface. (5 pts)

10 Equation Sheet B r = 2(μ o m/4πr 3 ) cos ρ = 2(μ o m/4πr 3 ) sin λ; B θ = (μ o m/4πr 3 ) sin ρ = (μ o m/4πr 3 ) cos λ B = B r 2 + B θ 2 = (μ o m/4πr 3 )(4 cos 2 ρ + sin 2 ρ) 1/2 tan I = 2 cotρ = 2 tan λ B = μ o H + μ o J I; B = μh μ = μ o μ r μ r = 1 + k μ o = 4π x 10-7 webers/am X 2 + Y 2 + Z 2 = T 2 X 2 + Y 2 = H 2 H= T cos I; Z = T sin I; Z = H tan I tan D = Y/X; X = H cos D; Y = H sin D 2D Dipping Sheet ΔH x = 2ksinδ [(Z o sinδ - H o sinαcosδ) ln(r 2 r 3 /r 1 r 4 ) - (Z o cosδ + H o sinαsinδ)(θ 1 - θ 2 - θ 3 + θ 4 )] ΔZ = 2ksinδ [(H o sinαsinδ + Z o cosδ) ln(r 2 r 3 /r 1 r 4 ) - (H o sinαcosδ - Z o sinδ)(θ 1 - θ 2 - θ 3 + θ 4 )] ΔT = ΔH x cosβcosi o + ΔZsinI o Isolated Pole ΔH = - C m m x / r 3 ΔZ = C m m z / r 3 ΔT = C m m [zsini o - xcosi o ] / r 3 Line of Dipoles ΔH = 2 C m m [(x 2 - z 2 )cosi o sinα - 2xzsinI o ] / r 4 ΔZ = 2 C m m [(z 2 - x 2 )sini o - 2xzcosI o sinα] / r 4 ΔT = 2 C m m [(x 2 - z 2 )(cos 2 I o sinαcosβ - sin 2 I o ) - 4xzsinI o cosi o sinα] / r 4 Isolated Dipole ΔH = C m m [(3x 2 - r 2 )cosi o - 3xzsinI o ] / r 5 ΔZ = C m m [(3z 2 - r 2 )sini o - 3xzcosI o ] / r 5

11 ΔT = C m m [3(xcosI o - zsini o ) 2 - r 2 ] / r 5 Half-width Rules - Vertical field Dipole z = 2x 1/2 Pole z = 1.3x 1/2 Line of poles z = x 1/2 Line of dipoles z = 2x 1/2 Half width rules - Horizontal Field Dipole z = 2.5x 1/2 Pole z = 1.3 1/2 Line of poles z = x 1/2 Line of dipoles z = 2x 1/2 i = R p sin R s = (V s1 /V p1 )sin i sin r s = (V s2 /V p1 )sin i sin r p = (V p2 /V p1 )sin i i = R s sin R p = (V p1 /V s1 )sin i sin r s = (V s2 /V s1 )sin i sin r p = (V p2 /V s1 )sin i z 1 = (x c1 /2) (V 2 V 1 )/(V 2 + V 1 ) z 2 = (x c2 /2) (V 3 V 2 )/(V 3 + V 2 ) z 1 = (t i /2)(V 1 V 2 )/ (V 2 2 V 1 2 ) z 2 = ½[t i3 2z 1 (V 2 3 V 2 1 ) 1/2 ] (V 2 V 3 ) V 1 V 3 (V 2 3 V 2 2 ) 1/2 t r2 = x/v 2 + 2zcos i c /V 1 t r3 = x/v 3 + 2z 1 cos i 1 /V 1 + 2z 2 cos i 2 /V 2 t rn = x/v N + j=1 N-1 (2z j cos i j )/V j where i j = sin -1 (V j /V N ) ΔT D = ½ (t D1 + t D2 t r ) z D = ΔT D V 1 /cos i c t r downdip = 2z a cos i c /V 1 + x sin(i c + γ)/v 1 t r updip = 2z b cos i c /V 1 + x sin(i c - γ)/v 1 i c = ½ [sin -1 (V 1 /V 2u ) + sin -1 (V 1 /V 2d )] γ = ½ [sin -1 (V 1 /V 2u ) - sin -1 (V 1 /V 2d )] z a = V 1 t ia /2cos i c h a = z a /cos γ z b = V 1 t ib /2cos i c h b = z b /cos γ

GE 2400 Test #2 3/26/03. Name

GE 2400 Test #2 3/26/03. Name GE 2400 Test #2 3/26/03 Name 9. Fill in the blank. a. minerals have a negative magnetic susceptibility. b. Ground surveys that use gradiometer magnetometers generally measure the. c. Horizontal derivatives