A rock has bulk modulus K = 20 GPa, shear modulus μ = 9 GPa and density 2400 kg/m³. What is its P-wave speed, to the nearest 10 m/s?

About 3650 m/s, from the square root of (K + 4μ/3) divided by the density

About 2890 m/s, from the square root of K divided by the density

About 3480 m/s, from the square root of (K + μ) divided by the density, adding the full shear modulus

About 3650 m/s, from the square root of (K + 4μ/3) divided by the density

About 1940 m/s, from the square root of μ divided by the density

A Vibroseis record is cross-correlated with the transmitted sweep. What does this do to the data?

It compresses each long sweep response into a short zero-phase wavelet with the sweep bandwidth

It extends the usable bandwidth beyond the sweep band by restoring the frequencies that the earth has absorbed

It converts the source signature into a minimum-phase wavelet so that predictive deconvolution can follow

It removes multiples, because the sweep carries a built-in repetition period

Which of the following statements about seismic noise are correct?

Stacking N traces that carry the same signal and independent random noise improves the signal-to-noise ratio in amplitude by √N

A noise train is sampled without spatial aliasing only when the group interval is at most half its apparent wavelength

Stacking N traces that carry the same signal and independent random noise improves the signal-to-noise ratio in amplitude by √N

A multiple has a higher stacking velocity than a primary that arrives at the same zero-offset time

A noise train is sampled without spatial aliasing only when the group interval is at most half its apparent wavelength

A geophone array cannot reduce ground roll, because ground roll travels vertically and adds in step across the array

A noise test shows ground roll with an apparent velocity of 400 m/s and frequencies up to 40 Hz. Using the rule that a train is free of spatial aliasing when the sampling interval does not exceed half its apparent wavelength, what is the largest group interval that samples it unaliased?

5 m, because the apparent wavelength is 10 m and half of it is allowed

10 m, because the interval must be one apparent wavelength

5 m, because the apparent wavelength is 10 m and half of it is allowed

20 m, because only the lowest frequency of the train has to be sampled

2.5 m, because the interval must be a quarter of the apparent wavelength

A 2D line is recorded with 120 channels at a group interval of 20 m, shooting every 40 m. The shot interval is then changed to 80 m with everything else unchanged. What happens?

The fold halves from 30 to 15, and the CMP spacing stays at 10 m

Both the fold and the CMP spacing are halved, because the survey becomes coarser

The fold doubles to 60, because fewer shots mean less overlap of the spreads and cleaner gathers

The fold is unchanged at 30, because it depends only on the number of channels

The fold halves from 30 to 15, and the CMP spacing stays at 10 m

For a flat two-layer earth with V1 = 1500 m/s, V2 = 2500 m/s and h = 60 m, which statement about the head wave is correct?

It first exists at 90 m but becomes the first arrival only beyond 240 m

It first exists at 240 m, where it also becomes the first arrival

It first exists at 90 m but becomes the first arrival only beyond 240 m

It first exists at 240 m but becomes the first arrival from 90 m

It exists at all offsets, since the interface is flat, and arrives first beyond 90 m

Forward and reverse refraction shooting over a gently dipping interface give apparent head-wave speeds of 2400 m/s down-dip and 3600 m/s up-dip. Taking cosδ ≈ 1, what is the true speed of the refractor?

2880 m/s, the harmonic mean of the two apparent speeds

3000 m/s, the arithmetic mean of the two apparent speeds

2940 m/s, the geometric mean rounded to the nearest ten

2880 m/s, the harmonic mean of the two apparent speeds

3600 m/s, because the up-dip value is the true speed

A planar reflector dips at 30° in a medium of true speed 3000 m/s. For a CMP gather at small offsets the NMO (stacking) velocity of this reflector is closest to

3460 m/s, because the stacking velocity is V/cosδ

2600 m/s, because dip lowers the apparent speed by cosδ

3000 m/s, because the stacking velocity is independent of dip

6000 m/s, because the reflection path doubles on a dipping interface

3460 m/s, because the stacking velocity is V/cosδ

Which of the following statements about reflection travel-times and velocities are correct?

In a layered earth the RMS velocity to a given time is never less than the average velocity to that time

For the same velocity and offset, the normal moveout is larger for a shallower reflector

In a layered earth the RMS velocity to a given time is never less than the average velocity to that time

For the same velocity and offset, the normal moveout is larger for a shallower reflector

The interval velocity of a layer is the difference of two RMS velocities divided by the difference of their times

On a dipping reflector the stacking velocity is smaller than the true medium speed

A weathered layer thickens smoothly along a 6 km stretch of a line, and the statics are left uncorrected. The geophone spread is 2.4 km long. What is the most likely effect on the stacked section?

A false structural high or low, because each gather shifts almost uniformly and still stacks coherently

Loss of stack coherence and broken events, because every trace of a single gather is shifted by a different amount

An error in the stacking velocity only, with the reflector times unchanged

No visible effect, since NMO correction removes surface-related time shifts

A shale (ρ = 2.2 g/cm³, Vp = 2500 m/s) overlies a sandstone (ρ = 2.5 g/cm³, Vp = 3500 m/s). The normal-incidence P-wave reflection coefficient of the interface is closest to

0.23, from the acoustic impedances 5500 and 8750

0.17, from the velocity contrast alone, ignoring density

0.06, from the density contrast alone

0.23, from the acoustic impedances 5500 and 8750

0.59, from the ratio of impedances minus one

Which of the following statements about static corrections and CMP geometry are correct?

A static correction is a single time shift for the whole trace that depends on the surface position of the source or receiver

Residual statics are estimated from the data themselves in a surface-consistent way, for example by correlating traces with a pilot trace

A static correction is a single time shift for the whole trace that depends on the surface position of the source or receiver

Residual statics are estimated from the data themselves in a surface-consistent way, for example by correlating traces with a pilot trace

In a CMP gather the NMO correction is the same for all traces, so that they can be summed

The CMP spacing of a 2D line equals the shot interval

Seismic Methods I: Elastic Waves, Noise, Acquisition, Travel-Times, CMP Stacking and Statics

This is the first of two chapters for the Geology and Geophysics (GG) paper's Part B2 heading “Seismic Methods”. It answers the first half of that heading: “Elastic properties of earth materials; Reflection, refraction, generation, and propagation of elastic waves, velocity – depth models, seismic noise and noise profile analysis, … CDP stacking charts, binning, … static and dynamic corrections”. It teaches wave speeds, sources, noise and the noise test, fold and binning, refraction and reflection travel-times, velocity-depth models and statics. The numericals are wave speed, fold, refraction depth, crossover distance, NMO, Dix velocity, a static shift and a reflection coefficient. Chapter 2 takes up processing and imaging.

1. Elastic properties, wave speeds and seismic sources

A seismic wave is a small, reversible strain that travels through rock, so its speed is set by the elastic moduli and the density. With bulk modulus K, shear modulus μ (the Lamé constant of rigidity) and density ρ: Vp = √[(K + 4μ/3)/ρ] for the compressional (P) wave, whose particle motion is along the ray, and Vs = √(μ/ρ) for the shear (S) wave, whose particle motion is across it. Since K is positive, Vp is always larger than Vs, and Vp/Vs = √[2(1 − ν)/(1 − 2ν)] in terms of Poisson's ratio ν; for ν = 0.25 this is √3 ≈ 1.73. A fluid has μ = 0, so Vs = 0: shear waves do not cross a liquid or gas-filled void, and gas reduces Vp by cutting K.

What raises velocity: compaction and effective pressure (so depth and age), cementation and lower porosity. For a fluid-filled rock of porosity φ the time-average (Wyllie) rule is 1/V = φ/Vfluid + (1 − φ)/Vmatrix. What lowers it: porosity, fractures, gas, and unconsolidated weathered material (a few hundred m/s).
Propagation and reflection: at an interface sinθ/V is conserved (Snell's law, the ray parameter p), amplitudes fall by geometrical spreading and absorption, and at normal incidence the reflection coefficient is R = (Z2 − Z1)/(Z2 + Z1), where the acoustic impedance is Z = ρ·V.
Land sources: dynamite in shot holes drilled below the weathered layer, weight drop or hammer for shallow work, and the Vibroseis truck, which sends a long frequency-swept signal (a sweep of several seconds, for example 8 to 80 Hz). The recorded trace is cross-correlated with the sweep, which compresses it to a short, zero-phase (Klauder) wavelet whose bandwidth is the sweep bandwidth.
Marine source and receivers: the air gun releases compressed air whose bubble then oscillates (the bubble pulse), so guns are fired in tuned arrays. Land receivers are geophones (particle velocity); marine receivers are hydrophones (pressure) in a towed streamer.

⚠️ Vp and Vs both rise with rigidity, but only Vp feels the fluid

Replacing brine by gas lowers K and the density. Vp falls sharply, while Vs, which depends on μ and ρ only, stays almost unchanged (it rises slightly because ρ falls). A fall in Vp with no change in Vs is therefore a fluid signature, not a lithology change.

2. Seismic noise and noise-profile analysis

Signal is the primary reflection energy that images the subsurface; everything else is noise. Noise is coherent (repeatable from trace to trace, so it forms recognisable events) or random (incoherent, from wind, traffic, instruments and power lines). The two are fought differently: coherent noise by arrays, filters and moveout discrimination, random noise by stacking.

Common noise types on a land or marine shot record
Noise	Character	Usual remedy
Ground roll (Rayleigh waves)	Slow, low-frequency, strong, dispersive; a fan of steep linear events	Geophone arrays, low-cut filter, f-k filter
Direct wave, refractions, guided waves	Linear events with a constant apparent velocity	Mute, f-k filter
Air wave	Sound in air at about 330 m/s; linear and narrow-band	Mute, deeper charges
Multiples (water-bottom reverberation, peg-legs)	Periodic repeats of a primary; velocity lower than the primary at the same arrival time	Predictive deconvolution, velocity filtering, Radon transform
Power-line, wind, traffic, instrument noise	Random, or a single 50 Hz line in India	Notch filter, stacking, quiet recording windows

Noise-profile analysis is done before production. A noise test (walkaway or noise spread) records a few shots into a long line of closely spaced single geophones, with different charge sizes, hole depths and array types. The velocity and frequency band of each noise train are read from the record or its f-k spectrum, and group interval, array length, source type and filter bands are chosen to suppress it. The central aliasing rule is that a train of apparent velocity V and frequency f has apparent wavelength λa = V/f, and it is sampled without spatial aliasing only if the group interval is at most λa/2. Ground roll of 300 m/s at 20 Hz has λa = 15 m, so it is aliased at a normal 25 m interval, and the array must remove it before sampling.

Arrays sum N geophones at spacing d into one trace. Reflections arrive almost vertically, with a very long apparent wavelength, and add in step; a noise train of short apparent wavelength is cancelled, with the first null at λa = N·d. Stacking does the same on the CMP gather: N traces with the same signal and independent noise improve the signal-to-noise ratio by √N, so a 36-fold stack gains a factor 6.

3. Acquisition geometry, CMP fold, stacking charts and binning

In 2D land work a line of receiver groups at group interval Δg records every shot, and the shot point moves along the line by the shot interval Δs. The spread is split (shot in the middle) or end-on (shot at one end). A common midpoint (CMP) gather is the set of traces whose source and receiver share the same midpoint; after NMO correction they are summed into one stacked trace. Midpoints fall at half the group interval, so the CMP spacing is Δg/2. For a flat reflector the common midpoint is also a common reflection point (the CDP of older usage); for a dipping reflector the reflection points of a gather are spread up-dip (dip smear), so “CMP” is the correct name.

Fold is the number of traces in a CMP gather. With N channels, group interval Δg and shot interval Δs, fold = N·Δg/(2·Δs). Moving the shot up by one group interval each time gives N/2; every second station gives N/4. Worked example: 96 channels, Δg = 25 m, Δs = 50 m give fold = 96 × 25/(2 × 50) = 24, with CMP spacing 12.5 m. Fold ramps up over about half a spread length at the ends of a line, so lines are extended beyond the target.

A stacking chart (CDP chart) plots offset against midpoint. Each shot is a diagonal line of points, each CMP gather a vertical column at one midpoint, and the fold is the number of points in the column. In 3D surveys the midpoints of all shot-receiver pairs are sorted into rectangular bins, and the traces in a bin form the gather. For orthogonal geometry (receiver lines across source lines) the bin is half the receiver interval by half the source interval, and the fold is the number of traces in it. The bin must also be small enough for dipping events not to alias on the stacked section: Δx ≤ V/(4·fmax·sinθ), because a dip of angle θ has a time dip 2·sinθ/V.

⚠️ Fold is not half the channel count in general

N/2 holds only when the shot moves by exactly one group interval. The general rule divides by the ratio of shot interval to group interval: doubling Δs halves the fold, doubling Δg doubles it.

4. Refraction travel-times: head waves, intercept time and critical distance

Take a flat interface at depth h between a slow layer V1 above and a faster layer V2 below. At the critical angle θc = sin⁻¹(V1/V2) the transmitted ray runs along the interface at V2, and it keeps shedding energy upward at θc as a head wave. The arrival at offset x is: direct wave T = x/V1; head wave T = x/V2 + ti, where the intercept time ti = 2h·cosθc/V1 is the value of the head-wave line extended back to x = 0. The head-wave line has slope 1/V2 on a time-distance plot, so V2 is read from its inverse slope and h from ti: h = ti·V1/(2·cosθc) = ti·V1·V2/(2·√(V2² − V1²)).

Two distances matter. The critical distance xc = 2h·tanθc is the nearest offset at which the head wave exists at all, because below it the ray hits the interface at less than θc. The crossover distance xco = 2h·√[(V2 + V1)/(V2 − V1)] is where the head wave first overtakes the direct wave in time, so only beyond xco does it appear as a first break. The crossover is always farther than the critical distance. Worked example: V1 = 1500 m/s, V2 = 2500 m/s, h = 60 m. Then sinθc = 0.6, cosθc = 0.8, tanθc = 0.75; ti = 2 × 60 × 0.8/1500 = 64 ms; xc = 2 × 60 × 0.75 = 90 m; xco = 2 × 60 × √(4000/1000) = 240 m. Check at 240 m: the direct wave takes 240/1500 = 0.160 s, and the head wave 240/2500 + 0.064 = 0.160 s.

Dipping interface. Shooting down-dip gives a smaller apparent speed Vd = V1/sin(θc + δ), up-dip a larger one Vu = V1/sin(θc − δ). The true speed is V2 = 2·cosδ/(1/Vd + 1/Vu), the harmonic mean for small dip, and reciprocal times (A to B and B to A) must be equal, which checks the picks. Irregular refractors need the delay-time, plus-minus or generalised reciprocal methods, or refraction tomography.

⚠️ Refraction needs velocity to increase with depth

A layer slower than the one above it sends no head wave, so it is invisible (the blind zone), and a thin layer whose head wave never becomes a first arrival is a hidden layer. Both make depths computed from the picked layers wrong without any sign in the data. Reflection seismics has no such restriction, which is why it is used for deep structure.

5. Reflection travel-times, NMO, Dix equation and velocity-depth models

For a flat reflector at depth h in a medium of speed V, the reflected path from a source to a receiver at offset x has length √(x² + 4h²), so T² = T0² + x²/V², where T0 = 2h/V is the two-way time at zero offset. On a record this is a hyperbola. The delay of the far trace relative to the zero-offset time is the normal moveout, ΔT = T − T0 ≈ x²/(2·V²·T0) for x much smaller than the depth. It grows with the square of the offset and is larger for shallow (small T0) and slow reflectors. The NMO correction subtracts ΔT trace by trace using the right velocity, flattening the hyperbola so that the traces can be summed. Worked example: T0 = 0.8 s, x = 1200 m, V = 2000 m/s gives T = √(0.64 + 0.36) = 1.0 s, so ΔT = 0.2 s.

For a dipping reflector (dip δ) the moveout is T² = T0² + x²·cos²δ/V², with T0 the zero-offset time at the midpoint, so the stacking velocity V/cosδ exceeds the true speed. For a stack of horizontal layers the same form holds at small offset with the RMS velocity, V²rms = Σ Vi²·ti / Σ ti, where ti are the interval times. Given the RMS velocities at two zero-offset times, the Dix equation recovers the interval velocity of the layer between them: V²int = (V²n·Tn − V²n−1·Tn−1)/(Tn − Tn−1). Worked example: Vrms = 2000 m/s at T = 0.5 s and 2400 m/s at T = 1.0 s give V²int = (2400² × 1.0 − 2000² × 0.5)/0.5 = 7.52 × 10⁶, so Vint ≈ 2742 m/s, higher than either RMS value, as it must be for a layer that raises the RMS velocity.

The velocities of a layered earth
Velocity	Definition	Where it comes from
Interval	Thickness of a layer ÷ one-way time in it	Sonic log, VSP, Dix conversion
Average	Total depth ÷ total one-way time	Check-shot or VSP; time-to-depth conversion
RMS	Time-weighted root mean square of interval velocities; at least the average velocity	Stacking velocity at small offset

A velocity-depth model turns times into depths. The simplest is layer-cake (constant interval velocities), improved by a linear gradient V = V0 + k·z, in which rays are circular arcs and the one-way time to depth z is t = (1/k)·ln(1 + k·z/V0). Models come from refraction picks, Dix conversion of stacking velocities, well logs, check-shots and tomography, and are tied to the well time-depth curve.

⚠️ RMS is not the average velocity, and Dix is not a plain average

RMS velocity weights the fast layers more, so it exceeds the average velocity unless the earth is uniform. The Dix equation inverts the weighting: never take the simple difference of RMS values divided by the time difference. Small picking errors also blow up in thin layers, so interval velocities from Dix are noisy.

6. Static and dynamic corrections

Dynamic corrections (NMO) change with offset and with arrival time. Static corrections are a single time shift applied to the whole trace, and depend only on the surface position of the source or receiver. They remove the effect of surface elevation and of the slow, variable weathered layer, so that every trace looks as if shot and recorded on one flat datum. Left uncorrected, statics make reflections wobble from trace to trace, spoil the stack and create false structure.

For a station at elevation E over weathering of thickness h and speed Vw, with the sub-weathering speed Vb used down to a datum at elevation Ed below the base of the weathering, the one-way vertical time to the datum is t = h/Vw + (E − h − Ed)/Vb, and the static shift applied is minus this time (the data move earlier). The shot and receiver terms are added. Worked example: h = 6 m, Vw = 500 m/s, E = 116 m, Ed = 100 m, Vb = 2000 m/s give t = 12 ms + 10/2000 s = 12 + 5 = 17 ms, so the shift is −17 ms. The weathering thickness and speed come from uphole surveys (a buried source timed at the surface) and from refraction analysis of the first breaks.

Elevation and refraction statics: elevation statics use a known weathering model; refraction statics (delay-time or generalised linear inversion) estimate the weathering from first breaks along the line.
Residual statics: what is left is estimated from the data themselves, surface-consistently (one value per source and one per receiver station), by cross-correlating each trace with a pilot trace, or by maximising stack power. Velocities are re-picked afterwards.
Long and short wavelength: statics that vary slowly along the line (longer than the spread) produce false structure; rapid ones (shorter than the spread) destroy stack coherence. A shift larger than half the dominant period causes cycle skipping, and the correlation then locks on the wrong cycle.

Key takeaways

Vp = √[(K + 4μ/3)/ρ] and Vs = √(μ/ρ), so Vp > Vs and Vs = 0 in fluids; Vp/Vs = √[2(1 − ν)/(1 − 2ν)]. Gas lowers Vp strongly and leaves Vs almost unchanged. Normal-incidence R = (Z2 − Z1)/(Z2 + Z1).
Coherent noise is fought by arrays, filters and moveout, random noise by stacking (gain √N). No spatial aliasing needs Δx ≤ λa/2, and a noise test fixes group interval and array length.
Fold = N·Δg/(2·Δs) and CMP spacing = Δg/2. A stacking chart plots offset against midpoint; in 3D the midpoints are binned, half the receiver interval by half the source interval for orthogonal geometry.
Refraction: h = ti·V1/(2·cosθc), xc = 2h·tanθc, xco = 2h·√[(V2 + V1)/(V2 − V1)]. Velocity must increase with depth; blind and hidden layers are its blind spots.
Reflection: T² = T0² + x²/V², ΔT ≈ x²/(2·V²·T0), V_NMO = V/cosδ for dip δ, and Dix V²int = (V²n·Tn − V²n−1·Tn−1)/(Tn − Tn−1). RMS velocity is never below average velocity.
A static is one time shift per trace, t = h/Vw + (E − h − Ed)/Vb to the datum, while NMO depends on offset and time. Long-wavelength statics fake structure.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A rock has bulk modulus K = 20 GPa, shear modulus μ = 9 GPa and density 2400 kg/m³. What is its P-wave speed, to the nearest 10 m/s?
1. About 2890 m/s, from the square root of K divided by the density
2. About 3480 m/s, from the square root of (K + μ) divided by the density, adding the full shear modulus
3. About 3650 m/s, from the square root of (K + 4μ/3) divided by the density
4. About 1940 m/s, from the square root of μ divided by the density
Show answer
Answer: C — About 3650 m/s, from the square root of (K + 4μ/3) divided by the density
The P-wave modulus is K + 4μ/3 = 20 + 12 = 32 GPa, so Vp = √(32 × 10⁹/2400) = √(1.333 × 10⁷) ≈ 3651 m/s. The tempting 2890 m/s is the speed in a fluid of the same K (no rigidity term), and 1940 m/s is the S-wave speed √(μ/ρ). Adding μ instead of 4μ/3 is also wrong.
A Vibroseis record is cross-correlated with the transmitted sweep. What does this do to the data?
1. It compresses each long sweep response into a short zero-phase wavelet with the sweep bandwidth
2. It extends the usable bandwidth beyond the sweep band by restoring the frequencies that the earth has absorbed
3. It converts the source signature into a minimum-phase wavelet so that predictive deconvolution can follow
4. It removes multiples, because the sweep carries a built-in repetition period
Show answer
Answer: A — It compresses each long sweep response into a short zero-phase wavelet with the sweep bandwidth
Correlating the record with the sweep compresses the long, spread-out response into a short, symmetric, zero-phase (Klauder) wavelet whose spectrum is that of the sweep. It cannot create frequencies the sweep did not contain, and it does not touch multiples. The tempting claim of a minimum-phase output is wrong, since the correlation wavelet is symmetric about zero time.
Which of the following statements about seismic noise are correct?
1. Stacking N traces that carry the same signal and independent random noise improves the signal-to-noise ratio in amplitude by √N
2. A multiple has a higher stacking velocity than a primary that arrives at the same zero-offset time
3. A noise train is sampled without spatial aliasing only when the group interval is at most half its apparent wavelength
4. A geophone array cannot reduce ground roll, because ground roll travels vertically and adds in step across the array
Show answer
Answer: A — Stacking N traces that carry the same signal and independent random noise improves the signal-to-noise ratio in amplitude by √N; C — A noise train is sampled without spatial aliasing only when the group interval is at most half its apparent wavelength
Independent noise adds in power while the signal adds in amplitude, which gives the √N gain, and the Nyquist-type rule Δx ≤ λa/2 is the spatial aliasing limit. A multiple has travelled a longer, slower path through the shallow layers, so its stacking velocity is lower than that of a primary at the same time, which is how stacking and Radon filters discriminate it. Ground roll has a short apparent wavelength along the surface, so an array of suitable length cancels it while reflections, with a very long apparent wavelength, add up.
A noise test shows ground roll with an apparent velocity of 400 m/s and frequencies up to 40 Hz. Using the rule that a train is free of spatial aliasing when the sampling interval does not exceed half its apparent wavelength, what is the largest group interval that samples it unaliased?
1. 10 m, because the interval must be one apparent wavelength
2. 5 m, because the apparent wavelength is 10 m and half of it is allowed
3. 20 m, because only the lowest frequency of the train has to be sampled
4. 2.5 m, because the interval must be a quarter of the apparent wavelength
Show answer
Answer: B — 5 m, because the apparent wavelength is 10 m and half of it is allowed
The shortest apparent wavelength is λa = V/f = 400/40 = 10 m, and the stated rule allows an interval of at most λa/2 = 5 m. The 10 m choice is the tempting one-wavelength error, and the highest frequency, not the lowest, sets the limit because it has the shortest wavelength. The 2.5 m choice is stricter than the rule requires.
A 2D line is recorded with 120 channels at a group interval of 20 m, shooting every 40 m. The shot interval is then changed to 80 m with everything else unchanged. What happens?
1. Both the fold and the CMP spacing are halved, because the survey becomes coarser
2. The fold doubles to 60, because fewer shots mean less overlap of the spreads and cleaner gathers
3. The fold is unchanged at 30, because it depends only on the number of channels
4. The fold halves from 30 to 15, and the CMP spacing stays at 10 m
Show answer
Answer: D — The fold halves from 30 to 15, and the CMP spacing stays at 10 m
Fold = N·Δg/(2·Δs) = 120 × 20/(2 × 40) = 30 before, and 120 × 20/(2 × 80) = 15 after. The CMP spacing is Δg/2 = 10 m and does not involve the shot interval. Fold depends on the shot interval as well as on the channel count, so the claim that only the channels matter fails, and fewer shots mean fewer traces per midpoint, not more.
A 2D land survey records 240 channels at a group interval of 12.5 m with a shot every 50 m. Using fold = (number of channels × group interval)/(2 × shot interval), the nominal CMP fold is ____.
Numerical answer — type the value.
Show answer
Answer: 30
Fold = 240 × 12.5/(2 × 50) = 3000/100 = 30. The common slip is to quote N/2 = 120, which holds only when the shot moves by one group interval each time.
For a flat two-layer earth with V1 = 1500 m/s, V2 = 2500 m/s and h = 60 m, which statement about the head wave is correct?
1. It first exists at 240 m, where it also becomes the first arrival
2. It first exists at 90 m but becomes the first arrival only beyond 240 m
3. It first exists at 240 m but becomes the first arrival from 90 m
4. It exists at all offsets, since the interface is flat, and arrives first beyond 90 m
Show answer
Answer: B — It first exists at 90 m but becomes the first arrival only beyond 240 m
sinθc = 1500/2500 = 0.6, so tanθc = 0.75 and the critical distance is 2 × 60 × 0.75 = 90 m, the nearest offset at which a head wave exists. The crossover distance is 2h√[(V2 + V1)/(V2 − V1)] = 120 × √4 = 240 m; only beyond it does the head wave beat the direct wave and show as a first break. Treating the crossover as the start of the head wave, or the critical distance as the first-break point, swaps the two distances.
In a flat two-layer refraction survey the head-wave line has an intercept time of 80 ms. The upper layer has V1 = 1800 m/s and the lower V2 = 3000 m/s. The depth of the interface is ____ m.
Numerical answer — type the value.
Show answer
Answer: 90
cosθc = √(1 − (1800/3000)²) = √(1 − 0.36) = 0.8, so h = ti·V1/(2·cosθc) = 0.080 × 1800/(2 × 0.8) = 144/1.6 = 90 m. Using ti·V1/2 = 72 m (forgetting cosθc) is the usual error.
Forward and reverse refraction shooting over a gently dipping interface give apparent head-wave speeds of 2400 m/s down-dip and 3600 m/s up-dip. Taking cosδ ≈ 1, what is the true speed of the refractor?
1. 3000 m/s, the arithmetic mean of the two apparent speeds
2. 2940 m/s, the geometric mean rounded to the nearest ten
3. 2880 m/s, the harmonic mean of the two apparent speeds
4. 3600 m/s, because the up-dip value is the true speed
Show answer
Answer: C — 2880 m/s, the harmonic mean of the two apparent speeds
V2 = 2·cosδ/(1/Vd + 1/Vu) with cosδ ≈ 1 is the harmonic mean: 2 × 2400 × 3600/(2400 + 3600) = 2880 m/s. The arithmetic mean of 3000 m/s is the tempting error, since it is the slownesses (1/V) that average, not the speeds. Neither apparent speed is the true speed except for a horizontal interface.
A single flat reflector gives T0 = 0.8 s at zero offset in a medium of constant speed 2000 m/s. Using the exact relation T² = T0² + x²/V², what is the normal moveout at offset 1200 m?
1. 100 ms
2. 200 ms
3. 225 ms
4. 400 ms
Show answer
Answer: B — 200 ms
T = √(0.8² + (1200/2000)²) = √(0.64 + 0.36) = 1.0 s, so ΔT = 1.0 − 0.8 = 0.2 s = 200 ms. The small-offset formula x²/(2V²T0) gives 225 ms, which is the tempting figure, but the stem asks for the exact relation, and the offset here is not small compared with the depth of 800 m.
Stacking analysis gives RMS velocities of 2100 m/s at a two-way time of 0.6 s and 2500 m/s at 1.4 s. Using the Dix equation, the interval velocity of the layer between the two times, to the nearest m/s, is ____.
Numerical answer — type the value.
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Answer: 2762
V²int = (2500² × 1.4 − 2100² × 0.6)/(1.4 − 0.6) = (8.75 × 10⁶ − 2.646 × 10⁶)/0.8 = 7.63 × 10⁶, so Vint = 2762 m/s. Dividing the RMS difference by the time difference (500 m/s) or averaging the two RMS values is not the Dix equation.
A planar reflector dips at 30° in a medium of true speed 3000 m/s. For a CMP gather at small offsets the NMO (stacking) velocity of this reflector is closest to
1. 2600 m/s, because dip lowers the apparent speed by cosδ
2. 3000 m/s, because the stacking velocity is independent of dip
3. 6000 m/s, because the reflection path doubles on a dipping interface
4. 3460 m/s, because the stacking velocity is V/cosδ
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Answer: D — 3460 m/s, because the stacking velocity is V/cosδ
For a dipping reflector T² = T0² + x²cos²δ/V², so the NMO velocity is V/cosδ = 3000/0.866 ≈ 3464 m/s, which is higher than the true speed. Multiplying by cosδ (2600 m/s) inverts the relation, and the stacking velocity is not dip-free.
Which of the following statements about reflection travel-times and velocities are correct?
1. In a layered earth the RMS velocity to a given time is never less than the average velocity to that time
2. For the same velocity and offset, the normal moveout is larger for a shallower reflector
3. The interval velocity of a layer is the difference of two RMS velocities divided by the difference of their times
4. On a dipping reflector the stacking velocity is smaller than the true medium speed
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Answer: A — In a layered earth the RMS velocity to a given time is never less than the average velocity to that time; B — For the same velocity and offset, the normal moveout is larger for a shallower reflector
The time-weighted RMS of interval velocities cannot fall below their time-weighted mean, which is the average velocity, and ΔT ≈ x²/(2V²T0) grows as T0 falls, so shallow events bend more. The Dix equation works on V²·T, not on V, so a simple difference quotient is wrong, and a dip raises the stacking velocity to V/cosδ rather than lowering it.
A weathered layer thickens smoothly along a 6 km stretch of a line, and the statics are left uncorrected. The geophone spread is 2.4 km long. What is the most likely effect on the stacked section?
1. A false structural high or low, because each gather shifts almost uniformly and still stacks coherently
2. Loss of stack coherence and broken events, because every trace of a single gather is shifted by a different amount
3. An error in the stacking velocity only, with the reflector times unchanged
4. No visible effect, since NMO correction removes surface-related time shifts
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Answer: A — A false structural high or low, because each gather shifts almost uniformly and still stacks coherently
A variation much longer than the spread shifts all traces of one CMP gather by nearly the same amount, so the stack stays coherent but the whole event moves, mimicking structure. Short-wavelength statics (shorter than the spread) are the ones that destroy coherence. NMO depends on offset and time, not on surface position, and cannot absorb a static; it also changes the apparent velocity only slightly here.
A receiver station at surface elevation 125 m sits on a weathered layer 9 m thick of speed 600 m/s. The sub-weathering speed is 2000 m/s and the datum is at 100 m elevation, below the base of the weathering. Using t = h/Vw + (E − h − Ed)/Vb, the magnitude of this station's static correction is ____ ms.
Numerical answer — type the value.
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Answer: 23
Weathering: 9/600 = 0.015 s = 15 ms. Below it: E − h − Ed = 125 − 9 − 100 = 16 m, and 16/2000 = 0.008 s = 8 ms. Total = 15 + 8 = 23 ms, applied as a shift of −23 ms. Using Vb for the whole 25 m (12.5 ms) ignores the slow layer.
A shale (ρ = 2.2 g/cm³, Vp = 2500 m/s) overlies a sandstone (ρ = 2.5 g/cm³, Vp = 3500 m/s). The normal-incidence P-wave reflection coefficient of the interface is closest to
1. 0.17, from the velocity contrast alone, ignoring density
2. 0.06, from the density contrast alone
3. 0.23, from the acoustic impedances 5500 and 8750
4. 0.59, from the ratio of impedances minus one
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Answer: C — 0.23, from the acoustic impedances 5500 and 8750
R = (Z2 − Z1)/(Z2 + Z1) with Z = ρV: Z1 = 2.2 × 2500 = 5500 and Z2 = 2.5 × 3500 = 8750, so R = 3250/14250 = 0.228. Using velocities alone (0.17) or densities alone (0.06) leaves out one factor of the impedance, and the ratio minus one is not a reflection coefficient. R is positive because impedance increases downward.
Which of the following statements about static corrections and CMP geometry are correct?
1. A static correction is a single time shift for the whole trace that depends on the surface position of the source or receiver
2. Residual statics are estimated from the data themselves in a surface-consistent way, for example by correlating traces with a pilot trace
3. In a CMP gather the NMO correction is the same for all traces, so that they can be summed
4. The CMP spacing of a 2D line equals the shot interval
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Answer: A — A static correction is a single time shift for the whole trace that depends on the surface position of the source or receiver; B — Residual statics are estimated from the data themselves in a surface-consistent way, for example by correlating traces with a pilot trace
A static depends only on where the source or receiver sits, and residual statics are decomposed into source and receiver terms from the data. NMO depends on offset and time, which is why it flattens a hyperbola trace by trace; it is different for every trace of the gather. The CMP spacing is half the group interval and is independent of the shot interval.