Seismic interpretation

Quick refresher

Seismic data measure time. Reflections are tied (matched) to layers. For this we use a check-shot survey (a well survey that measures travel times to given depths) or a VSP (a denser version of it), and a synthetic seismogram (a trace calculated from the well logs). The sonic log (a well log of the speed of sound in the rock) is “faster” than seismic data by a few per cent. Without a correction for drift (a time difference that grows with depth), the tie goes off by a wavelet period or more.
A horizon is followed on one phase (a peak, a trough or a zero crossing) and checked by the mis-ties, which are the differences in time where two lines cross. Faults are found from breaks and offsets of reflections, diffractions, time slices and coherence cubes (cubes that show where neighbouring traces stop being similar).
Time is converted to depth with a velocity model. One average velocity is right only while the section above the target does not change sideways. A layer-by-layer model allows for changes in the thickness and velocity of the layers. In every case the map is tied to wells.
A high in time is not yet a high in depth. Thinning of a slow unit over a structure, salt, reefs and gas distort the relief. In the lesson's example, the average velocity of the well on the crest overstated the closure by a third.
Direct hydrocarbon indicators (signs of oil or gas that show up directly in the amplitudes) are the bright spot, the dim spot, the flat spot, polarity reversal and push-down. They are convincing when they conform to the structure contour in depth. Their false lookalikes are coals, tuning of thin beds, low-saturation gas and multiples.
A velocity error gives a depth error $\delta z \approx z\,\delta V/V$ . One per cent is tens of metres, which is comparable with the closure of a small trap. A flat spot fixes the contact by the time inside the reservoir, so the volume above it depends much less on velocities.

Key relationships

z = \dfrac{V_\text{avg}\,t}{2}

depth from the average velocity down to the horizon

z = \dfrac{V_1 t_1}{2} + \dfrac{V_2\,(t - t_1)}{2}

layer-by-layer conversion for two layers: $t_1$ is the time of the base of the first one

V_\text{int}^2 = \dfrac{V_n^2 t_n - V_{n-1}^2 t_{n-1}}{t_n - t_{n-1}}

the Dix formula: interval velocity from stacking velocities

z = \dfrac{V_0}{k}\left(e^{k t/2} - 1\right)

depth when the velocity grows linearly with depth: $V = V_0 + k z$

H = \dfrac{V_\text{g}\,\Delta t}{2}

height of a gas accumulation from the time between its top and the flat spot

\dfrac{\delta z}{z} \approx \dfrac{\delta V}{V}

depth error from velocity error

Common pitfalls

Fitting the synthetic to the section by shifting and stretching it, without first calibrating the sonic log with the check-shot survey.
Jumping to the neighbouring phase while following a horizon, across a fault or a zone of poor quality. The mis-tie where lines cross shows an error of a whole period.
Building a trap from a time structure map, or from one average velocity, without checking how the thickness and velocity of the layers above the target change.
Taking a bright spot as proof of commercial gas.
Forgetting that the depth error grows with distance from wells, and may be comparable with the closure.

After this lesson you can

tie reflections to layers using check-shot surveys, vertical seismic profiles (VSP) and synthetic seismograms, and allow for the drift of the sonic log
explain how reflection horizons and faults are followed on sections and in a cube, and check the correlation by the mis-ties where lines cross
convert time to depth with an average velocity, layer by layer and with a linear velocity law, find the interval velocity with the Dix formula, and explain why depths are tied to wells
recognise structural and stratigraphic traps and the main seismic facies on seismic data
check direct hydrocarbon indicators (a bright spot, a flat spot, and conformance to the structure contour) and name their false lookalikes
estimate how the error of velocities becomes an error of the closure and of the volume of a trap

What interpretation gives, and the order in which it is done

The interpreter receives a migrated section or cube (Seismic acquisition and processing) and well data: logs, formation picks (the depths of the tops of layers) and velocities. From them the interpreter builds a consistent picture of the subsurface. The main products of interpretation are: structure maps (maps of the shape of a layer surface) of the reflecting horizons, first in time and then in depth; a framework of faults; thickness maps, amplitude maps and maps of other attributes (numbers calculated from the seismic signal); maps of seismic facies (areas where the reflections look alike); the basis of the geological model; and a forecast of the section for planned wells, with the error of the depths.

The order of interpretation

1
Checking the data
Check the polarity and phase of the wavelet, the datum level (the reference level for times and depths), and the matching of surveys of different years: time shifts, and differences of phase and amplitude between them.
2
Well tie
Decide which reflection belongs to which layer, using the check-shot survey, the VSP and the synthetic seismogram.
3
Following horizons and faults
Follow them on sections and in the cube, and check them by the mis-ties where lines cross.
4
Time structure maps
Map the surface of each horizon in time.
5
Velocity model and structure maps
Convert time to depth and tie the maps to wells.
6
Traps
Find the closure, the spill point and the faults that act as seals (barriers that stop oil and gas from escaping).
7
Stratigraphy and amplitudes
Study the seismic facies and attributes, and look for direct hydrocarbon indicators.
8
Uncertainty
Make alternative versions of the maps and ranges of volumes, for estimating resources and for the well plan.

All the examples in this lesson use one assumed line, 12 km long, across an anticline (an arch-shaped fold of the layers). Beneath the surface lies a slow unit (2000 m/s) about 1150 m thick. Over the crest it thins to 1070 m. Below it, down to the top of the productive layer, lie shales with a velocity of 3200 m/s. The top of the layer is at a depth of 2440 m on the crest, and 60 m deeper 3 km from the crest. On the eastern limb a normal fault with a throw of 40 m displaces it. The sand layer is 45 m thick, and it is saturated with gas down to the gas–water contact (GWC) at 2475 m. Well W-1 was drilled on the crest, and well W-2 later on the western limb. This is the time section that was calculated for this model, in the interval of the layer:

CalculatedA calculated time section of the assumed line in the interval of the productive layer (vertical rays, Ricker wavelet with a peak frequency of 30 Hz; peaks are filled dark, troughs orange). Over the crest, the top of the gas sand gives a bright trough, and below it there is a flat reflection from the GWC. On the limbs, where the layer holds water, the reflections are weak. The fault at 8.2 km offsets the reflections by 25 ms.

This section shows almost everything that we shall talk about: the crest, the fault, the bright spot (a very strong reflection) and the flat spot (a level reflection from a fluid contact). But the section is in time. As the examples will show, neither the height of the crest on it nor the shape of the flat spot matches what lies in depth.

The well tie: which reflection is which layer

A well gives layers in depth, and seismic data give reflections in time. The bridge between them is the relation between time and depth at the well, together with a synthetic seismogram (a seismic trace that we calculate from the logs of the well; convolution model and synthetic). We get the relation in several ways.

Check-shot survey (velocity survey): A source stands at the wellhead. A receiver is lowered into the borehole and, at a series of depths, it records the time of the first arrival of the direct wave. This gives the time and the average velocity down to each depth. These are control points for the tie and for the velocity model.
Vertical seismic profile (VSP): This is the same, but the receivers stand close together, every 10–20 m, and record the whole wave. The downgoing and upgoing waves are separated. They give the wavelet at the well, tell primary reflections from multiples, give a trace for the tie right at the borehole, and forecast the section below the bottom of the well.
Sonic log: It gives the velocity every 15–20 cm, and the integrated interval transit time (the travel times of all the steps of the log added up from the top) is also a relation between time and depth (integrated time). But the wave of the log has a frequency hundreds of times higher, and it passes through only centimetres of rock at the borehole wall. So the log velocity is usually a few per cent higher than the seismic velocity, and the time from the log “drifts” away from the seismic time with depth. This is called drift.

So we first calibrate the sonic log. We stretch its time scale so that, at the depths of the check-shot survey, its time agrees with the seismic time. Between the control points we spread the correction smoothly. We calculate the synthetic from the calibrated curve and compare it with the traces near the well. After calibration we allow only a small overall shift, a few milliseconds, and very little stretch. A strong stretch lets us “tie” anything to anything. A match of the pattern of several reflections, not of just one, is a sign of a correct tie. We check the polarity on a reflection of known sign (wavelet and polarity).

Worked exampleSonic log drift in well W-1

Given (assumed data)

Check-shot survey (two-way time): 600 m — 0.600 s; 1070 m (H1) — 1.070 s; 2440 m (top of the layer) — 1.926 s
Sonic log recorded from 600 m: two-way time from the log: from 600 to 1070 m, 0.4563 s; from 1070 to 2440 m, 0.8395 s
Wavelet: Ricker wavelet with a peak frequency of 30 Hz; its apparent (dominant) period is about 26 ms

Solution

Time from the log, tied to the check-shot survey at 600 m: to H1, $0.600 + 0.4563 = 1.0563$ s. To the top of the layer, $1.0563 + 0.8395 = 1.8958$ s.
Drift: at H1, $1.070 - 1.0563 = 13.7$ ms. At the top of the layer, $1.926 - 1.8958 = 30.2$ ms.
By interval: in the slow unit, $13.7$ ms over 470 ms of seismic time is 2.9%. In the shales, $30.2 - 13.7 = 16.5$ ms over 856 ms is 1.9%. So the log velocities are higher than the seismic ones by about 3% and 2%.
The drift at the top of the layer, 30.2 ms, is larger than the apparent period of the wavelet.

Answer: Without calibration, the synthetic would put the top of the layer 30.2 ms higher than its reflection on the section. This is more than a period. The H1 reflection would be only 13.7 ms higher.

What it means. The shift is different at different depths, so one overall shift cannot match the synthetic to the section. If we fitted the top of the layer, we would move H1 by 16.5 ms the other way (figure below). In a section with several similar reflections, an error of one period means tying to the wrong layer. It leads to a wrong map and a wrong depth in the well plan. Calibration with the check-shot survey removes this ambiguity. The data in this example are assumed for illustration.

CalculatedThe tie at W-1: a trace of the section (left), the synthetic from the sonic log before calibration (centre) and after it (right). Dashed lines show the times of the reflections on the section: H1 and the top of the layer. Before calibration the synthetic arrives earlier than the section: by 14 ms at H1 and by 30 ms at the top of the layer. The shift grows with depth. After calibration both reflections match.

Following horizons and faults

A seismic horizon is a reflection that has been followed across an area and tied to a stratigraphic boundary. In Russian-language practice, horizons are given letter-number codes. For example, P1 is the top of the subsalt deposits of the Pre-Caspian Basin. We follow the same phase throughout: a peak, a trough or a zero crossing. On zero-phase data, a boundary where impedance (density times velocity) increases coincides with the centre of a peak (in SEG polarity). A boundary where impedance falls coincides with the centre of a trough. On 2D lines, a horizon is followed on each line and compared where lines cross. In a 3D cube we mark starting (seed) points, and automatic tracking (autotracking) carries the phase from trace to trace along the inlines and crosslines. The result is checked on arbitrary sections and on time slices.

The main check of the correlation is the mis-ties: the differences between the times of a horizon where lines cross. Lines of different years and different processing are shifted against each other in time and in phase. We find these systematic mis-ties and remove them before we build maps. If we go round a closed loop of several lines, the sum of the mis-ties around the loop must be close to zero, because the constant shifts of the lines cancel in such a sum. A sum equal to the wavelet period means that somewhere on the loop the correlation jumped to the neighbouring phase (a cycle skip).

Faults show up on seismic data by several signs (faults on maps and on seismic data):

a break and offset of reflections, seen for a package of reflections and not for one
bending of layers beside the fault plane (drag near the fault, and rollover at listric normal faults, which are faults whose plane flattens with depth)
diffractions from the cut edges of layers, when migration is incomplete
a zone of weak and chaotic reflections along the fault plane (the fault plane itself seldom gives a reflection)
on time slices of the cube, lines where the contours break off; on coherence maps (coherence, variance), lines where neighbouring traces stop being similar

We draw a fault plane through the break points on many sections. This gives the surface of the fault and its lines of intersection with the horizons on both sides of it. These lines are the fault polygons on the maps (fault polygon). Across a fault we correlate not the nearest similar reflection but a package of reflections, by its pattern, checking against wells and mis-ties. We estimate the throw of the fault from the offset of the horizon in time: $\Delta z \approx V\Delta t/2$ , where $V$ is the interval velocity of the unit above the horizon, if it is the same on both sides. Seismic data do not show faults with a throw smaller than the resolution.

Time structure maps and converting time to depth

A followed horizon is a set of times $t_0$ . When we draw contours through them (Structure maps and contouring), we get a time structure map, which is a structure map in time. For a well plan and for volumes we need depths, and for that we need velocities.

Velocities come from three sources. Wells (check-shot surveys, VSP and the calibrated sonic log) give accurate velocities, but only at separate points. Seismic velocities, which are stacking and migration velocities, exist everywhere. But they are the velocities that straighten traveltime curves (traveltime curve and velocities). They are affected by dips, by the anisotropy of shales (the velocity in them depends on direction, and the stacking velocity is usually higher than the vertical one) and by picking errors. Trends help between wells: velocity grows with depth because of compaction, and it depends on lithology (the type of rock). We get interval velocities from the seismic ones with the Dix formula:

The Dix formula

V_\text{int}^2 = \frac{V_n^2\,t_n - V_{n-1}^2\,t_{n-1}}{t_n - t_{n-1}}

where:

$V_\text{int}$ interval velocity between two reflections, m/s
$V_n,\ V_{n-1}$ root-mean-square velocities (in practice, stacking velocities) down to the lower and the upper reflection, m/s
$t_n,\ t_{n-1}$ two-way times of these reflections, s

Units: SI: m/s, s

Holds when

Horizontal uniform layers, and the stacking velocity equals the root-mean-square velocity.
Both reflections are on one vertical line, and the stacking velocities are picked without errors. In practice their errors grow many times over when they are subtracted.

The formula follows from the definition of the root-mean-square velocity: $V_n^2 t_n$ is the sum of $V_i^2\Delta\tau_i$ over all the layers down to the $n$ -th reflection, and the difference of two such sums leaves one layer. The relative error of the interval velocity is larger than the error of the stacking velocity by a factor of $V_n^2 t_n/(V_\text{int}^2\,\Delta t)$ (advanced section). The thinner the interval, the worse it is.

Worked exampleInterval velocity from stacking velocities and from a well

Given (assumed data)

Stacking velocities at W-1 (velocity analysis): 2030 m/s at H1 ( $t = 1.070$ s), 2650 m/s at the top of the layer ( $t = 1.926$ s)
From the check-shot survey in W-1: H1 at 1070 m, top of the layer at 2440 m; the interval velocity of the shales between them is 3201 m/s

Solution

The terms: $2650^2\cdot 1.926 = 13.525\cdot 10^6$ and $2030^2\cdot 1.070 = 4.409\cdot 10^6$ (m²/s² × s).
$V_\text{int}^2 = (13.525 - 4.409)\cdot 10^6/0.856 = 10.650\cdot 10^6$ , so $V_\text{int} = 3263$ m/s, which is 1.9% higher than the velocity in the well.
Depth of the top from the seismic velocities: $2030\cdot 1.070/2 + 3263\cdot 0.856/2 = 1086.1 + 1396.6 = 2482.7$ m, which is 42.7 m deeper than in the well.
Sensitivity: $V_n^2 t_n/(V_\text{int}^2\,\Delta t) = 13.525/(10.6472\cdot 0.856) = 1.48$ . An error of 1% in the stacking velocity at the top gives an error of 1.48% in the interval velocity.

Answer: The Dix formula gave 3263 m/s instead of 3201 m/s in the well, and the depth of the top from the seismic velocities came out 42.7 m greater than the true one.

What it means. The stacking velocities here are only 1.5–2% higher than the root-mean-square ones. This happens because of the anisotropy of shales and picking errors. Yet the depth is already almost 1.8% too great. An interval 856 ms thick is a thick one. For an interval of 0.2 s with the same velocities, the error multiplier would be about 6. So we use seismic velocities for the shape of the velocity field between wells, and we set its level from wells. The data in this example are assumed for illustration.

The ways of converting time to depth differ in how much detail they give of the section above the target.

Converting time to depth with an average velocity, and layer by layer

z = \frac{V_\text{avg}\,t}{2}, \qquad z = \frac{V_1 t_1}{2} + \frac{V_2\,(t - t_1)}{2}

where:

$z$ depth of the horizon below the datum, m
$t$ two-way time of the horizon, s
$V_\text{avg}$ average velocity down to the horizon: $2z/t$ at wells, and from a map between them, m/s
$t_1$ two-way time of the intermediate horizon, the base of the upper layer, s
$V_1,\ V_2$ interval velocities of the upper and the lower layer, m/s

Units: SI: m, s, m/s

Holds when

Rays are vertical: the time is reduced to the vertical, and dips are moderate.
The average velocity works while the thickness and velocity of the layers above the horizon change little sideways. The layer-by-layer method needs followed intermediate horizons and the interval velocity of each layer (from wells or by the Dix formula).
For more layers the layer-by-layer formula continues as a sum (depth from two-way time).

Converting time to depth with one average velocity carries all the distortions of the time map into depth if the section above the target changes. The layer-by-layer method takes them into account: thinning of a slow layer reduces the time but not the depth, and the layer-by-layer conversion “sees” this. After conversion we tie the map to wells. We carry the corrections at the wells over the area with a map of corrections (data of a structure map).

CalculatedAn assumed line: the depth of the base of the slow unit H1 (top), the depth of the top of the layer, true, and calculated with one average velocity of well W-1 (middle), and the time of the top of the layer and of the flat spot (bottom). The slow unit thins over the crest. So in time, and in conversion with one velocity, the crest looks higher above the limbs than it really is.

Worked exampleAverage velocity or layer-by-layer model: the closure of a trap

Given (assumed data)

W-1 (crest), times and picks: H1: $t = 1.070$ s, 1070 m; top of the layer: $t = 1.926$ s, 2440 m
Point A on the western limb, 3 km from the crest: H1: $t = 1.138$ s; top of the layer: $t = 1.989$ s
Depth of the spill point of the trap: known from the map: the closing contour passes through point A

Solution

Velocities at W-1: the average velocity to the top is $V_\text{avg} = 2440/(1.926/2) = 2533.7$ m/s. The interval velocities are $V_1 = 2\cdot 1070/1.070 = 2000$ m/s and $V_2 = (2440 - 1070)/(0.856/2) = 3201$ m/s.
Point A with the average velocity: $z = 2533.7\cdot 1.989/2 = 2519.8$ m. The closure is $2519.8 - 2440 = 79.8$ m.
Point A layer by layer: H1 is at $2000\cdot 1.138/2 = 1138.0$ m, and the top is at $1138.0 + 3201\cdot(1.989 - 1.138)/2 = 1138.0 + 1362.0 = 2500.0$ m. The closure is 60.0 m.
An even cruder way is to convert the time relief (the difference in time between point A and the crest), $1.989 - 1.926 = 63$ ms, with the interval velocity of the shales: $3201\cdot 0.063/2 = 100.8$ m.
The reason: at point A the slow unit is $1138.0 - 1070 = 68$ m thicker than on the crest. It adds $2\cdot 68\cdot(1/2000 - 1/3201) = 25.5$ ms to the time of the limb.

Answer: The layer-by-layer model gives a closure of 60.0 m. One average velocity gives 79.8 m (33% more), and converting the time relief with the interval velocity gives 100.8 m. Well W-2, which was drilled later at point A, met the top of the layer at 2500 m.

What it means. Of the 63 ms of time relief of the crest, 25.5 ms was made not by the structure but by the thinning of the slow unit over it. The average velocity of the well on the crest is “faster” than the limbs need, because on the crest there is more fast shale and less slow unit. So the closure and the volume of a trap that is filled right down to its spill point are overstated by about a third. If the thinning were greater, the structure in time might have no closure in depth at all. The rule: if a time high coincides with a thinning of the slow unit or a thickening of the fast unit above it, we check it with a layer-by-layer model. The data in this example are assumed for illustration.

Where velocity grows with depth because of compaction, the model $V = V_0 + kz$ is convenient. We fit the velocity $V_0$ at the top of the unit and its increase $k$ per metre to the wells, and then we find the depth from the time:

Depth when velocity grows linearly with depth

z = \frac{V_0}{k}\left(e^{k t/2} - 1\right)

where:

$z$ depth below the top of the unit (below the datum), m
$V_0$ velocity at the top of the unit, m/s
$k$ increase of velocity per metre of depth, s⁻¹
$t$ two-way time from the top of the unit, s

Units: SI: m, m/s, s⁻¹, s

Holds when

Velocity grows with depth linearly (a compacting unit of uniform composition), and the instantaneous velocity does not change sideways.
Vertical rays. For several units, each is described by its own $V_0$ and $k$ .

This is the integral of $dz/dt_1 = V_0 + kz$ over the one-way travel time $t_1 = t/2$ (advanced section). If we fit $V_0$ for each well and draw a map of it, changes from place to place are taken into account too, and $k$ sets the compaction with depth.

Traps on seismic data

A trap is first of all a closed shape. The reservoir layer beneath a seal must rise to the crest from all sides, or butt against a barrier such as a fault, a pinch-out (a layer thinning to nothing), an unconformity (an old erosion surface) or salt (Traps and timing). On seismic data we look for both parts: the shape of the reservoir and the signs of a barrier.

Traps on seismic data

Trap: Anticline, dome
What it looks like: a crest with closed contours on all sides
What to check: the closure and the spill point, in depth and not in time

Trap: Fault-bounded (tectonically sealed)
What it looks like: a limb that butts against a normal fault; the trap closes on three sides
What to check: whether the fault seals: juxtaposition of layers (which layers face each other across the fault) and the shale content of the fault zone (fault seal)

Trap: Salt-related
What it looks like: layers turned up at the wall of a dome, or under a salt overhang
What to check: the shape of the salt and the image beneath it: depth migration

Trap: Pinch-out, onlap
What it looks like: reflections of the reservoir converge and butt against a boundary (onlap)
What to check: whether there is a seal above and to the side; whether this is tuning (the reflections of a thin layer merging)

Trap: Beneath an unconformity
What it looks like: dipping layers cut off by an erosion surface
What to check: the seal above the unconformity

Trap: Reef, carbonate buildup
What it looks like: a mound with a structureless or chaotic core, draped by layers above
What to check: the porosity of the body, and the false uplift beneath it

Trap	What it looks like	What to check
Anticline, dome	a crest with closed contours on all sides	the closure and the spill point, in depth and not in time
Fault-bounded (tectonically sealed)	a limb that butts against a normal fault; the trap closes on three sides	whether the fault seals: juxtaposition of layers (which layers face each other across the fault) and the shale content of the fault zone (fault seal)
Salt-related	layers turned up at the wall of a dome, or under a salt overhang	the shape of the salt and the image beneath it: depth migration
Pinch-out, onlap	reflections of the reservoir converge and butt against a boundary (onlap)	whether there is a seal above and to the side; whether this is tuning (the reflections of a thin layer merging)
Beneath an unconformity	dipping layers cut off by an erosion surface	the seal above the unconformity
Reef, carbonate buildup	a mound with a structureless or chaotic core, draped by layers above	the porosity of the body, and the false uplift beneath it

Many traps are combined: a structure plus a pinch-out, a fault plus a change of facies. Stratigraphic traps are harder to see on seismic data than structural ones, and their risk is higher (Geological risk and chance of success).

We read the closure and the spill point, the deepest point down to which the trap holds hydrocarbons, on a structure map in depth (crest, closure, spill point). A closure smaller than the depth error is not a fact but a risk, and such a trap is described by a probability of existence. We compare the time when the trap formed with the time when oil and gas migrated (moved) into the area, using the thicknesses between horizons. Thinning over the structure, as in the lesson's example, shows that the structure grew while sediments were being deposited (Structural styles and tectonic history).

Seismic stratigraphy in brief

Reflections follow bedding surfaces. To a first approximation, each of those surfaces has the same age everywhere (it is isochronous), because it was once the sea floor. So the pattern of reflections records the history of deposition. The places where reflections end against a surface (onlap, downlap, toplap and erosional truncation) mark the boundaries of seismic units (sequences), as explained in the topic on stratigraphy (layer terminations and sequences). Inside a unit, the character of the reflections, the seismic facies, hints at the depositional environment.

Seismic facies and what they usually mean

Pattern of reflections: Parallel, continuous
What it usually means: steady deposition: shelf, deep basin

Pattern of reflections: Divergent, wedge-shaped
What it usually means: uneven subsidence, or growth of a structure or fault during deposition

Pattern of reflections: Clinoforms: inclined, sigmoid or oblique
What it usually means: a delta or shelf edge building out towards the basin

Pattern of reflections: Mounded
What it usually means: reefs and carbonate buildups, submarine fans, slump bodies

Pattern of reflections: Chaotic
What it usually means: slumps, strongly deformed units, salt

Pattern of reflections: Structureless (transparent)
What it usually means: uniform units: salt, massive shales, the core of a reef

Pattern of reflections	What it usually means
Parallel, continuous	steady deposition: shelf, deep basin
Divergent, wedge-shaped	uneven subsidence, or growth of a structure or fault during deposition
Clinoforms: inclined, sigmoid or oblique	a delta or shelf edge building out towards the basin
Mounded	reefs and carbonate buildups, submarine fans, slump bodies
Chaotic	slumps, strongly deformed units, salt
Structureless (transparent)	uniform units: salt, massive shales, the core of a reef

A seismic facies is a hint, not a diagnosis. Different environments give the same pattern, and we check the conclusion against core, logs and a regional model (Facies, facies successions and sedimentary logs).

In a 3D cube, the environment is often visible in plan view. On amplitude maps along a horizon, or in a narrow window below it, winding channels, lobes of submarine fans and the edges of reefs appear. These are direct data on the shape and size of bodies that wells see only at points (Reservoir architecture and performance).

Amplitudes and direct hydrocarbon indicators

The amplitude of a reflection depends on the contrast of impedances (impedance and reflection coefficient), and gas and light oil clearly lower the impedance of a porous rock. So an accumulation is sometimes visible in the amplitudes. These are direct hydrocarbon indicators (DHI). In Russian-language literature there is also a more general term, “anomaly of the deposit type” (the Russian abbreviation is ATZ). It is used for such anomalies not only in seismic data but also in other geophysical data.

Bright spot: A sharp strengthening of the negative reflection from the top of a reservoir whose impedance is lower than that of the seal, and falls even more because of gas.
Dim spot: A weakening of a positive reflection. The reservoir is denser than the seal, and gas reduces the contrast.
Polarity reversal: The reservoir is slightly denser than the seal, but with gas it is softer. The positive reflection above the accumulation is replaced by a negative one.
Flat spot: The reflection from the contact of gas (or oil) with water. It is positive, level in depth, and it cuts across the dipping reflections of the layers. It is the strongest indicator, because geology rarely makes flat boundaries that cut across bedding.
Push-down (sag): The reflections below an accumulation arrive later, because the wave spends longer in the slow gas-bearing layer.
Gas chimneys: Zones of weak, chaotic and delayed reflections above leaking traps, along which gas has gone upward.

The main check of an indicator is conformance to the structure contour. If the anomaly is made by an accumulation, its edge lies at one depth all round, on the contour of the contact, and the flat spot lies at the same depth. The check is made in depth, not in time: in time the flat spot is distorted just as everything else is.

Worked exampleThe flat spot, the edge of the bright spot and the depth of the GWC

Given (assumed data)

Above W-1 (crest): top of the layer at 2440 m; the flat spot is 26.4 ms below the reflection from the top
Velocity in the gas sand: 2650 m/s (from the calibrated log of W-1)
Western edge of the bright amplitude of the top: H1: $t = 1.123$ s; top: $t = 1.968$ s
Velocities from the previous example: $V_1 = 2000$ m/s, $V_2 = 3201$ m/s; the average velocity at W-1 is 2533.7 m/s

Solution

Height of the gas accumulation from the flat spot: $H = 2650\cdot 0.0264/2 = 35.0$ m, so the GWC is at $2440 + 35.0 = 2475.0$ m.
The edge of the bright spot, layer by layer: $2000\cdot 1.123/2 + 3201\cdot(1.968 - 1.123)/2 = 1123.0 + 1352.4 = 2475.4$ m. This is the same depth as the flat spot, and on the eastern limb too.
The same edge with one average velocity: $2533.7\cdot 1.968/2 = 2493.2$ m, which is 18.2 m deeper than the GWC.
In time, the flat spot under the crest is at $1.926 + 0.0264 = 1.9524$ s, and at the edge it is at 1.968 s. So it is arched upwards by 15.6 ms.

Answer: The flat spot and the edge of the bright spot coincide in depth, at 2475.0 and 2475.4 m. The indicator conforms to the structure contour, and the height of the gas accumulation is 35.0 m.

What it means. Both checks gave one contact, which is a strong argument for a gas accumulation. Two caveats. First, with one average velocity the edge would fall 18.2 m below the flat spot, and a correct indicator might be rejected as “non-conforming”. Second, in time the flat spot is not flat. From the edge to the crest the slow unit is thinner by 53 m. This lifts the contact under the crest by about 20 ms, while gas in the layer lowers it by 4.5 ms. The sum is about 15 ms, which is what the readings showed (advanced section). Only conversion to depth makes it level. The flat spot measures the height of the accumulation inside the layer, by the velocity of the gas sand, so the velocities above the target hardly affect it. The data in this example are assumed for illustration.

False direct indicators

What: Low-saturation gas (fizz gas: gas that fills only a small part of the pores)
How it is similar: a bright spot and a flat spot almost like those of an accumulation
How to tell it apart: hard to tell apart on seismic data; we try to estimate the density from AVO at large angles; a well decides

What: Coals, dense thin layers
How it is similar: strong reflections
How to tell it apart: they do not conform to the structure and are seen outside the trap; polarity

What: Tuning of a thin layer
How it is similar: stronger amplitude without a fluid
How to tell it apart: a wedge model, thicknesses from wells (tuning)

What: High-porosity water sand
How it is similar: a soft reservoir, a bright top
How to tell it apart: there is no flat spot and no conformance to the contour

What: Multiples, diagenetic boundaries
How it is similar: flat or cross-cutting reflections
How to tell it apart: stacking velocity, VSP, the pattern outside the structure

What	How it is similar	How to tell it apart
Low-saturation gas (fizz gas: gas that fills only a small part of the pores)	a bright spot and a flat spot almost like those of an accumulation	hard to tell apart on seismic data; we try to estimate the density from AVO at large angles; a well decides
Coals, dense thin layers	strong reflections	they do not conform to the structure and are seen outside the trap; polarity
Tuning of a thin layer	stronger amplitude without a fluid	a wedge model, thicknesses from wells (tuning)
High-porosity water sand	a soft reservoir, a bright top	there is no flat spot and no conformance to the contour
Multiples, diagenetic boundaries	flat or cross-cutting reflections	stacking velocity, VSP, the pattern outside the structure

A well-documented indicator is one that conforms to the contour, has a flat spot and has a suitable dependence of amplitude on angle (AVO). Such an indicator raises the geological chance of success of a prospect (Geological risk and chance of success). How much it raises it is decided by the statistics of the basin: how many wells on similar indicators were successful. Besides signs of an accumulation, amplitudes are used to forecast properties. Two tools do this. Seismic attributes are numbers calculated from the seismic signal, such as the average amplitude in a window, coherence and spectral components. Inversion is a calculation that turns the seismic data into the impedance of the rocks. We compare their results with porosity and thicknesses at wells (forecast of reservoir quality). With few wells, such links easily turn out to be chance, and we check them on wells that did not take part in the fitting. Repeat surveys over a producing accumulation (4D seismic) show, from changes in amplitudes and times, where water or gas has advanced.

Interpretation errors and the uncertainty of depths

Frequent sources of errors

Source: Velocity changes sideways: salt, reefs, gas, the thickness of slow units
What it looks like: false highs and lows in time
How to check: a layer-by-layer velocity model, depth migration, wells

Source: Multiples
What it looks like: “boundaries” that repeat the sea floor, or strong reflections that cut across bedding
How to check: stacking velocity, periodicity, VSP

Source: Side reflections on 2D lines
What it looks like: reflections that do not tie where lines cross
How to check: a grid of lines, a 3D survey

Source: Migration artefacts
What it looks like: “smile” arcs, crossing reflections
How to check: migration velocities, a comparison of variants

Source: Cycle skip
What it looks like: a mis-tie of a whole period, a step on the map
How to check: mis-ties around loops, synthetics, thicknesses

Source: Tuning of thin layers
What it looks like: the amplitude and the thickness from time are wrong
How to check: a wedge model

Source	What it looks like	How to check
Velocity changes sideways: salt, reefs, gas, the thickness of slow units	false highs and lows in time	a layer-by-layer velocity model, depth migration, wells
Multiples	“boundaries” that repeat the sea floor, or strong reflections that cut across bedding	stacking velocity, periodicity, VSP
Side reflections on 2D lines	reflections that do not tie where lines cross	a grid of lines, a 3D survey
Migration artefacts	“smile” arcs, crossing reflections	migration velocities, a comparison of variants
Cycle skip	a mis-tie of a whole period, a step on the map	mis-ties around loops, synthetics, thicknesses
Tuning of thin layers	the amplitude and the thickness from time are wrong	a wedge model

The depth error is made up of errors of following the horizon, of the tie, of the position after migration and, most of all, of the velocities. A velocity error of $\delta V/V$ shifts the depth by $z\,\delta V/V$ . It is small at a well, where the velocity is measured, and it grows away from it. We issue a map together with an estimate of its error. We calculate volumes for three cases (low, base and high), or by Monte Carlo simulation (many random trials). We must remember that the depths of the map and the contact, if both come from the same seismic data, depend on each other (dependent parameters, uncertainty of volume).

Worked exampleVelocity error and the volume of a trap

Given (assumed data)

Trap: a dome, nearly circular in plan, with a profile like the lesson's line: crest 2440 m, spill point 2500 m at 3 km from the crest
Velocity of shales far from W-1: known with an error of ±1.5% (assumption of the example); their thickness at the spill point is 1362 m (from the example with point A)
Flat spot: GWC 2475 m; the edge of the bright amplitude is 2.291 km from the crest. Errors: the edge ±0.1 km, the velocity in the gas sand ±3%

Solution

Depth error at the spill point: $1362\cdot 0.015 = 20.4$ m. The closure is $60 \pm 20.4$ m, that is from 39.6 to 80.4 m.
Rock volume of the trap filled to spill (the dome is a paraboloid): $\pi R^2 A/2 = \pi\cdot 3000^2\cdot 60/2 = 848.2$ million m³. If the velocity error stretches the relief evenly, the areas at corresponding contours do not change, and the volume is proportional to the closure: from 559.8 to 1136.6 million m³.
Volume above the GWC from the flat spot: the area inside the edge of the bright amplitude is $\pi\cdot 2.291^2 = 16.49$ km², the height is 35 m, and for a paraboloid dome the volume is half of their product (paraboloid cap): $16.49\cdot 35/2 = 288.6$ million m³.
Error of this estimate: an edge error of ±0.1 km gives an area from 15.08 to 17.96 km², and a velocity error of ±3% gives a height from 33.95 to 36.05 m. The volume is from 256.0 to 323.7 million m³.

Answer: Without a direct indicator, assuming filling to spill, the rock volume is from 559.8 to 1136.6 million m³. With a conforming flat spot it is from 256.0 to 323.7 million m³.

What it means. The flat spot did two things. It narrowed the range: instead of a twofold spread, about ±12%, because the contact and the area of the accumulation are taken directly from seismic data and not from velocities. And it showed that the trap is only partly filled: the volume above the GWC is almost three times smaller than for filling to spill. Without the indicator, the interpreter would most likely have set a range of contacts from the crest to the spill, and with it a range of volumes several times wide. The model is simplified: the dome is round, the velocity error stretches the relief evenly, and the volume is counted from the top only, as for a layer thicker than the accumulation. For a gas accumulation 35 m high this is so (the layer is 45 m thick). But when the trap is filled to spill, the base of the layer under the crest rises above the contact. Then the volume counted from the top is overstated: by almost 7% at a closure of 60 m, and by about a quarter at 80.4 m (thick and thin layer). Errors other than those of velocities and of the edge (following, tie, thickness of the layer) are not included here. The data in this example are assumed for illustration.

Terms and symbols

Seismic horizon: A reflection that has been followed across an area and tied to a stratigraphic boundary.
Check-shot survey, VSP: A measurement of the travel time from the surface to a receiver in a well. A VSP records the whole wave on a dense grid of depths.
Drift: The difference between the time from the sonic log and the time from the check-shot survey. It grows with depth.
Mis-tie: The difference between the times of a horizon where lines cross. The sum of mis-ties around a closed loop shows a cycle skip.
Time structure map: A structure map of a horizon in time $t_0$ .
Average and interval velocity: The average velocity is the ratio of the depth to half the time down to a horizon. The interval velocity is the velocity in a single layer.
Depth conversion: Making depth maps from time maps and a velocity model, and tying them to wells.
Seismic facies: The character of the reflections inside a seismic unit: shape, continuity, amplitude.
Direct hydrocarbon indicators: Bright, dim and flat spots, polarity reversal and push-down. They are convincing when they conform to the structure contour.
Conformance to the contour: The edge of an amplitude anomaly lies at one depth all round and coincides with the flat spot.

Common mistakes and misconceptions

Wrong

“A synthetic looks like the trace if we shift it by 30 ms and stretch it a little, so the tie is ready.”

In fact

First the sonic log is calibrated with the check-shot survey. After that only small shifts are allowed. A large shift and stretch will fit any synthetic to any reflection. In the lesson's example, the drift of 30 ms is more than the period of the wavelet.

Wrong

“A crest on the time structure map is the crest of the trap.”

In fact

Time depends on the velocities above the target. In the lesson's example, the thinning of the slow unit over the crest added 25.5 ms of the 63 ms of time relief, and with one average velocity the closure came out a third greater than the true one.

Wrong

“Interval velocities from the Dix formula are accurate: they are calculated on all the traces.”

In fact

The formula subtracts close numbers, and the errors of stacking velocities grow many times over in it. They grow more the thinner the interval. Besides, stacking velocities are affected by anisotropy and dips. In the lesson's example, the depth from them came out 43 m greater than in the well.

Wrong

“A flat spot must be flat on a time section.”

In fact

It is flat in depth. In time it is bent by the velocities above the layer and by the gas itself. In the lesson's example it is arched upwards by 15.6 ms.

Wrong

“A bright spot on a crest is a gas accumulation, so the volume can be counted to the spill point.”

In fact

A bright spot can come from low-saturation gas, coal or tuning. An accumulation is confirmed by the edge of the anomaly conforming to the contour and by a flat spot. The contact from it is often above the spill point. In the lesson's example, the volume is almost three times smaller than for filling to spill.

In depth: advanced detail

Derivation of the Dix formula, and why it amplifies errors

By the definition of the root-mean-square velocity, $V_n^2 t_n = \sum_{i \le n} V_i^2\,\Delta\tau_i$ . If we subtract the same for the $(n-1)$ -th reflection, we get $V_n^2 t_n - V_{n-1}^2 t_{n-1} = V_\text{int}^2 (t_n - t_{n-1})$ , which is the Dix formula. Both sums are large and close, and their difference is small. So the relative errors grow in it.

Differentiate $V_\text{int}^2\,\Delta t = V_n^2 t_n - V_{n-1}^2 t_{n-1}$ with respect to $V_n$ . This gives $2V_\text{int}\,\delta V_\text{int}\,\Delta t = 2V_n t_n\,\delta V_n$ , and so $\delta V_\text{int}/V_\text{int} = \left(V_n^2 t_n/(V_\text{int}^2\,\Delta t)\right)\cdot\delta V_n/V_n$ . When the interval velocity is close to the stacking velocity, the multiplier is about equal to the ratio of the total time to the time of the interval. For an interval of a tenth of the total time, a percentage error of the stacking velocity becomes almost ten per cent of error in the interval velocity. The error of the upper velocity $V_{n-1}$ enters with the opposite sign and the multiplier $V_{n-1}^2 t_{n-1}/(V_\text{int}^2\,\Delta t)$ . So errors of the two velocities with the same sign partly cancel each other, and errors with opposite signs add up.

Where the formula for velocity growing with depth comes from

Let the instantaneous velocity be $V = V_0 + kz$ . In the one-way travel time $t_1$ , the wave passes $dz = V\,dt_1 = (V_0 + kz)\,dt_1$ . If we separate the variables, $dz/(V_0 + kz) = dt_1$ , and integrating from the surface gives $\tfrac1k\ln\left((V_0 + kz)/V_0\right) = t_1$ . From this, $z = (V_0/k)\left(e^{k t_1} - 1\right)$ , and with the two-way time, $t_1 = t/2$ . For small $kt$ the exponential is close to $1 + kt/2$ , and the formula turns into $z \approx V_0 t/2$ , which is a constant velocity.

The average velocity down to depth $z$ , which is $2z/t$ , is less than the instantaneous velocity at that depth. For example, take $V_0 = 1800$ m/s, $k = 0.6$ s⁻¹ and $t = 2$ s. The depth is $3000\cdot(e^{0.6} - 1) \approx 2466$ m, the average velocity is about 2466 m/s, and the instantaneous one is $1800 + 0.6\cdot 2466 \approx 3280$ m/s. We fit $V_0$ and $k$ to the check-shot surveys of wells. If $V_0$ changes across the area, we draw a map of it.

Why the flat spot is not flat in time

The time of the GWC at a point of the line is made up of the time to the top of the layer and the time through the gas sand to the contact: $t_\text{GWC} = 2h_1/V_1 + 2(z_\text{top} - h_1)/V_2 + 2(z_\text{GWC} - z_\text{top})/V_\text{g}$ . Here $h_1$ is the thickness of the slow unit and $z_\text{top}$ is the depth of the top. Let us group the terms: $t_\text{GWC} = 2z_\text{GWC}/V_2 + 2h_1(1/V_1 - 1/V_2) + 2(z_\text{GWC} - z_\text{top})(1/V_\text{g} - 1/V_2)$ .

The first term is the same everywhere, because the contact is horizontal. The second term is smaller where the slow unit is thinner. At the edge of the lesson's accumulation (2.7 km) its thickness is about 1123 m, and above the crest it is 1070 m. So under the crest the contact is “lifted” by $2\cdot 53\cdot(1/2000 - 1/3200) \approx 19.9$ ms. The third term is the push-down under gas: the taller the gas accumulation, the later the contact arrives. At the crest, with a height of 35 m, it is $2\cdot 35\cdot(1/2650 - 1/3200) \approx 4.5$ ms, and at the edge it is zero. Together they give a “flat” spot that is arched upwards by $19.9 - 4.5 \approx 15$ ms. The same term also lifts the top of the layer in time. Between the crest and the far limbs, the difference in the thickness of the slow unit is 80 m, and this is 30 ms. Up to point A of the example on closure, it is 68 m and 25.5 ms. In other sections the sign can be the opposite: under a thick slow unit, or under a gas chimney, the contact sags in time.

Connections to other topics

Seismic acquisition and processing — how the section and the stacking velocities were obtained, migration and its limits.
Seismic waves, acoustic impedance and reflections — impedance, synthetics, tuning of thin layers and AVO.
Geophysical exploration methods — depth from two-way time, and false uplift beneath salt.
Faults — the geometry of faults, their signs and sealing.
Cross-sections from maps, map combination and faults on maps — fault polygons, thickness maps and the matching of maps.
Structure maps and contouring — drawing contours and tying seismic surfaces to wells.
Gross rock volume from maps — rock volume from a structure map, closure and spill point.
Traps and timing — types of traps and when they form.
Probabilistic volumetrics by Monte Carlo simulation — how the error of a map enters the distribution of resources.
Geological risk and chance of success — direct indicators and the quality of seismic data in the probability of success.

Check yourself

1
Three 2D lines form a triangle. The mis-ties of a horizon where the lines cross, going round the loop, are +6 ms, −3 ms and +30 ms. The dominant frequency of the reflection, measured from its apparent period, is 30 Hz. What does this mean?
Show the answer
The constant shifts of the lines cancel in the sum around a closed loop, so the sum must be close to zero. Here it is $6 - 3 + 30 = 33$ ms. This is exactly the apparent period of the reflection ( $1/30$ s; for a Ricker wavelet this is a peak frequency of about 23 Hz, not 30 as in the drift example). So on one of the legs of the loop the correlation jumped to the neighbouring phase. Most likely it is the third one, with a mis-tie of 30 ms. It is checked again, especially near faults and in zones of poor quality.
2
A horizon is offset by a fault by 32 ms. The interval velocity of the unit above it is 3400 m/s on both sides. What is the vertical throw of the fault?
1. A.About 27 m
2. B.About 109 m
3. C.About 54 m
4. D.It cannot be estimated from time
Show the answer
Correct answer: C. About 54 m
$\Delta z \approx V\Delta t/2 = 3400\cdot 0.032/2 \approx 54$ m. We would get 109 m if we forgot that the time is two-way. The estimate is right while the velocity above the horizon is the same on both sides. If one side has gas or a thicker slow unit, part of the offset in time is made by velocity, not by the fault.
3
The velocity in a unit grows with depth by the law $V = 1700 + 0.5\,z$ (m/s, with $z$ in metres). At what depth is a horizon with a two-way time of 2.0 s, and what is the average velocity down to it?
Show the answer
$z = (V_0/k)(e^{kt/2} - 1) = (1700/0.5)(e^{0.5} - 1) = 3400\cdot 0.6487 \approx 2206$ m. The average velocity is $2z/t = 2206$ m/s, and the instantaneous velocity at this depth is $1700 + 0.5\cdot 2206 \approx 2803$ m/s. If we used the instantaneous velocity at the horizon for the conversion, we would overstate the depth by more than a quarter.
4
There is a bright spot above the crest. On the western limb its edge coincides with the 2450 m contour, and on the eastern limb with the 2510 m contour. What follows from this?
1. A.It is a gas accumulation with a GWC at 2480 m, the average of the two depths
2. B.The indicator does not conform to the contour: we must check the depth conversion, a tilted contact, lithology and tuning
3. C.Nothing: conformance to the contour is checked in time
4. D.It proves that there is no accumulation
Show the answer
Correct answer: B. The indicator does not conform to the contour: we must check the depth conversion, a tilted contact, lithology and tuning
For a simple accumulation, the edge of the anomaly lies at one depth all round. A difference of 60 m says that something is wrong. It may be the velocity model (as in the lesson's example, where one average velocity moved the edge by 18 m), a tilted contact (tilted contacts), a change of lithology or the tuning of a thin layer. This is not proof that there is no accumulation, but it is not a confirmation either: the indicator needs to be checked.
5
Why is the interval velocity of a thin interval from the Dix formula unreliable? Estimate the error multiplier for an interval of 0.2 s below a reflection at 2.2 s with a stacking velocity of 2800 m/s, if the interval velocity is about 3600 m/s.
Show the answer
The multiplier is $V_n^2 t_n/(V_\text{int}^2\,\Delta t) = 2800^2\cdot 2.2/(3600^2\cdot 0.2) = 17.25/2.59 \approx 6.7$ . An error of 1% in the stacking velocity gives almost 7% error in the interval velocity. The formula subtracts two large close numbers, and a short interval divides their small difference by a small time.
6
A time structure map shows a closed high with a height of 40 ms. When can it disappear on a depth map?
1. A.If a slow unit thins over it, a fast one thickens, or a fast body (salt, a reef) lies above it, the high can be entirely a velocity effect
2. B.Only if the migration was wrong
3. C.Only with inclined rays
4. D.Never: a high in time is always a high in depth
Show the answer
Correct answer: A. If a slow unit thins over it, a fast one thickens, or a fast body (salt, a reef) lies above it, the high can be entirely a velocity effect
Time is the sum of the thicknesses of the layers divided by their velocities. Thinning of a slow unit, a fast body above the target, or thickening of a fast layer reduces the time without changing the depth of the target. In the lesson's example, such thinning gave 25.5 ms of the 63 ms of relief. If it had been greater, the high in time would have had no closure in depth.

Updated: 17 October 2026