Faults · MunaiHub

Quick refresher

The block above an inclined fault plane is the hanging wall, and the block below it is the footwall. In a normal fault the hanging wall has moved down (extension, the rock is pulled apart). In a reverse fault it has moved up (compression, the rock is squeezed). A thrust is a low-angle reverse fault. A strike-slip fault moves sideways along the strike (the compass direction of the fault plane).
In Anderson's model (which links the type of fault to the directions of the stresses), new normal faults dip at about 60°, thrusts at about 30°, and strike-slip faults are vertical. Real faults often follow old weak zones and depart from these angles.
The slip along the dip splits into the throw (the vertical offset) $t = D_d\sin\delta$ and the heave (the horizontal offset) $w = D_d\cos\delta$ . The slip is greatest in the middle of the fault and falls to zero at its tips.
In a vertical well, a normal fault “cuts out” part of the section, and a reverse fault repeats it. The thickness of the missing interval is the vertical separation of the bed, $s = t \pm w\,\mathrm{tan}\,\delta_b$ . With dipping layers it is not equal to the throw. And if the layers dip the same way as the fault but more steeply, the rule is reversed.
A fault can seal in two ways. Impermeable rock may stand against the reservoir across the fault (juxtaposition). Or the fault zone itself may pass fluid badly (clay smear, which is clay dragged along the fault plane; cataclasis, the crushing of grains; or cement). The measure of the second is the shale gouge ratio, SGR. Below an SGR of 15–20%, faults usually let hydrocarbons through.
A fault that holds a pressure difference for millions of years is a barrier for development too. But a fault with no pressure difference can still slow the flow between blocks during production. For development the transmissibility of a fault matters, not only its sealing.

Key relationships

t = D\sin r\,\sin\delta,\qquad w = D\sin r\,\cos\delta

throw and heave from the slip $D$ that points at angle $r$ to the strike of a fault plane with dip $\delta$

s = t \pm w\,\mathrm{tan}\,\delta_b,\qquad s_n = s\cos\delta_b

vertical separation of the bed (the missing interval in a vertical well) and stratigraphic separation; “+” if the layers dip towards the fault plane

SGR = \dfrac{\sum_i V_{sh,i}\,\Delta z_i}{t}\cdot 100\,\%

share of clay material that has slid past a point of the fault plane

\Delta p_f = 10^{\,SGR/27 - C}\ \text{bar},\qquad H = \dfrac{\Delta p_f}{(\rho_w - \rho_h)\,g}

empirical seal-failure envelope, and the height of column that the fault will hold above the point

Common pitfalls

Calling every missing interval a fault: an unconformity and the pinching out of layers give one too.
Taking the thickness of the missing interval for the throw when the layers are inclined.
Believing that a large throw is always a reliable seal and a small one is always a window for flow.
Using SGR thresholds from one area in another with no calibration, and without allowing for the way shale volume was calculated.
Labelling a fault with a single throw: at the tips it goes to zero, and there the layers stand against themselves.

After this lesson you can

tell normal faults, reverse faults and thrusts, strike-slip faults and oblique-slip faults apart, and link the type of fault to the stress regime (the directions in which the rock is squeezed or pulled)
split the slip on a fault into components along the dip and along the strike, and into a vertical and a horizontal offset (throw and heave), and find the vertical and the stratigraphic separation of a dipping bed
recognise a fault in a well from a missing or repeated part of the section, tell it from an unconformity (a gap in the rock record), and estimate the throw from the missing interval
assess the sealing of a fault from the juxtaposition of layers (which layers face each other across the fault) and the shale gouge ratio (SGR), and turn it into the height of accumulation that the fault will hold
estimate the number of faults below the resolution of seismic data, and explain the role of faults in traps, in migration and in dividing an accumulation into separate blocks

A fault and its parts

We recognise a fault by the offset: a layer boundary that we follow on one side turns up higher or lower on the other side, or is shifted sideways. In core (rock samples cut from the well) we see faults with an offset of a millimetre. On seismic sections we usually see faults with a throw of 10–20 m or more. In wells they show up as missing or repeated intervals of the section. Which faults get onto the map and into the model depends on the resolution of the data (the smallest feature that the data can show). Small faults are nearly always there, but we cannot see them (see the section on faults on maps).

Fault plane (fault surface): The surface along which the movement took place. Its position is set by the attitude, as for a layer. The attitude is the strike (the compass direction of a horizontal line on the plane) and the dip (the angle of its slope) (attitude). Large fault planes are rarely flat. They bend, branch, and consist of overlapping segments that step sideways one after another (en echelon).
Hanging wall and footwall: The block above an inclined fault plane “hangs” over it (the hanging wall). The block below it “lies” on it (the footwall). These names do not depend on the direction of movement. For a vertical fault they cannot be told apart, and we speak of the west and east wall, and so on. The wall that has moved up relative to the other is called upthrown, and the one that has moved down is called downthrown.
Fault zone: In natural faults the slip is spread over a zone of finite width. In its core are tectonites, rocks that have been crushed and ground during the movement: fault breccia (angular broken pieces), cataclasite (finely crushed rock) and fault gouge (a soft, ground-up, clay-like paste). The core is surrounded by a damage zone with fractures and small faults. The core can be from millimetres to metres wide, and the damage zone from tens of centimetres to a hundred metres or more.
Slickensides and slickenlines: Polished surfaces with grooves. In core and outcrops (places where rock shows at the surface) we use them to work out the direction of the last movement.

Types of faults by the direction of movement

The name of a fault comes from the direction of movement relative to the fault plane. If the walls have moved along the dip of the fault plane, it is a dip-slip fault. If they have moved along the strike, it is a strike-slip fault. If both, it is an oblique-slip fault: a normal–strike-slip or a reverse–strike-slip fault.

Main types of faults

Type: Normal fault
Movement: the hanging wall has moved down along the dip
What happens to the sequence: it becomes longer horizontally; in a vertical well part of the section is missing
Setting: extension: rifts (zones where the crust is stretched and breaks), passive margins (continental edges that are not plate boundaries), crests over salt domes

Type: Reverse fault
Movement: the hanging wall has moved up along the dip
What happens to the sequence: it becomes shorter; part of the section is repeated
Setting: compression: fold belts (mountain belts of folded rock), basin inversion (a basin that was stretched and is squeezed later)

Type: Thrust
Movement: a reverse fault with a gentle fault plane: shallower than 45°, often about 30°
What happens to the sequence: slices of rock are pushed over one another, sometimes for kilometres
Setting: fold-and-thrust belts and foreland basins (basins in front of them)

Type: Strike-slip fault
Movement: along the strike: right-lateral or left-lateral
What happens to the sequence: the vertical position of the layers hardly changes
Setting: plate boundaries, strike-slip zones inside plates

Type	Movement	What happens to the sequence	Setting
Normal fault	the hanging wall has moved down along the dip	it becomes longer horizontally; in a vertical well part of the section is missing	extension: rifts (zones where the crust is stretched and breaks), passive margins (continental edges that are not plate boundaries), crests over salt domes
Reverse fault	the hanging wall has moved up along the dip	it becomes shorter; part of the section is repeated	compression: fold belts (mountain belts of folded rock), basin inversion (a basin that was stretched and is squeezed later)
Thrust	a reverse fault with a gentle fault plane: shallower than 45°, often about 30°	slices of rock are pushed over one another, sometimes for kilometres	fold-and-thrust belts and foreland basins (basins in front of them)
Strike-slip fault	along the strike: right-lateral or left-lateral	the vertical position of the layers hardly changes	plate boundaries, strike-slip zones inside plates

Right-lateral (dextral): standing on one wall and facing the fault, we see the other wall moved to the right. Left-lateral (sinistral): it has moved to the left. Large low-angle thrusts that move rock by kilometres are called overthrusts, and the sheets of rock moved along them are called nappes (thrust sheets).

SchematicA normal fault, a reverse fault and a right-lateral strike-slip fault in block diagrams (schematic). Half-arrows show the movement of the walls relative to each other. Layers of the same colour on the two sides of the fault were once one layer.

The type of fault is linked to the stress regime. A new fault forms along the plane that is closest to slip. This plane contains the direction of the intermediate principal stress $\sigma_2$ (the principal stresses $\sigma_1$ , $\sigma_2$ and $\sigma_3$ are the greatest, middle and least pushes on the rock, at right angles to each other). The normal to the plane (the line at right angles to it) makes an angle $\theta = 45° + \varphi/2$ with $\sigma_1$ , where $\mathrm{tan}\,\varphi = \mu$ is the coefficient of friction (stress on a plane and friction). For rock that has not yet broken, this is the coefficient of internal friction, usually of the same order. So the plane itself runs at an angle $90° - \theta$ to $\sigma_1$ . Which of the principal stresses is vertical decides the type of fault. This is Anderson's theory of faulting.

Ideal fault angles in Anderson's model

Regime: extension
Vertical: $\sigma_1$
Fault and its angle: normal fault, dip angle
$\mu = 0.6$: 60°
$\mu = 0.85$: 65°

Regime: compression
Vertical: $\sigma_3$
Fault and its angle: thrust, dip angle
$\mu = 0.6$: 30°
$\mu = 0.85$: 25°

Regime: strike-slip
Vertical: $\sigma_2$
Fault and its angle: vertical strike-slip fault, angle with $\sigma_1$
$\mu = 0.6$: 30°
$\mu = 0.85$: 25°

Regime	Vertical	Fault and its angle	$\mu = 0.6$	$\mu = 0.85$
extension	$\sigma_1$	normal fault, dip angle	60°	65°
compression	$\sigma_3$	thrust, dip angle	30°	25°
strike-slip	$\sigma_2$	vertical strike-slip fault, angle with $\sigma_1$	30°	25°

The angles are calculated from $\theta = 45° + \tfrac12\,\mathrm{arctan}\,\mu$ for friction coefficients of 0.6 and 0.85, the limits of the usual range for rocks. For stress regimes in the layers, see the lesson on the fracture gradient.

This is a guide, not a law. Faults often form along weak zones that already exist, and they inherit their orientation. Listric normal faults flatten with depth. When the stress regime changes, old faults move again, now in a different direction: a normal fault, squeezed later, becomes a steep reverse fault. This is inversion. So we find the type of a fault from the displacement of layers, and we use Anderson's angles to check whether an interpretation is plausible.

Faults occur in groups. In extension, normal faults that dip towards each other bound downthrown blocks, called grabens, with uplifted blocks, called horsts, left between them. Normal faults of one direction tilt the blocks and make half-grabens. Small faults that dip the same way as the main fault are called synthetic, and those that dip the opposite way are called antithetic. A listric fault flattens with depth, and its hanging wall, as it slides down, bends into a rollover anticline (an arch next to the fault) (Folds, salt diapirs and their relationship to faults). In compression, thrusts run now along weak layers (flats), now cut across a package of rocks (ramps). They form fans of slices (imbricate fans) and duplexes (stacks of thrust slices) above a common detachment surface (a surface along which the upper rocks slide over the lower ones). Strike-slip faults compress or stretch the rocks at their bends. Next to them appear pull-apart basins (basins that open in a stretched bend) and “flower” structures (fans of faults that spread upwards from a strike-slip fault) (basins on strike-slip faults). Which combinations of faults are typical for each setting is covered in the lesson Structural styles and tectonic history.

Slip, throw and heave

The displacement on a fault is a vector (it has a size and a direction). It shows how far, and in which direction, a point on one wall has moved relative to the point on the other wall with which it used to coincide. Its length $D$ is the net slip. Its direction is set by the angle $r$ between the strike line of the fault plane and the slickenlines, measured in the fault plane (the rake). At $r = 90°$ the slip is purely along the dip, and at $r = 0$ it is pure strike-slip.

We split the slip along the dip into two more parts. The vertical part is the throw. It is what we usually mean when we say “the throw of the fault is 40 m”. The horizontal part, at right angles to the strike, is the heave. On a structure map of a normal fault, the heave equals the width of the strip where the layer is missing. For a reverse fault, it is the width of the strip where the layer is covered by itself.

Components of the slip on a fault

D_d = D\sin r,\quad D_s = D\cos r,\quad t = D_d\sin\delta,\quad w = D_d\cos\delta

where:

$D$ slip on the fault plane (net slip), m
$r$ angle in the fault plane between the strike line and the direction of slip (rake), °
$D_d,\ D_s$ components of the slip along the dip and along the strike of the fault plane, m
$\delta$ dip of the fault plane, °
$t$ throw (vertical offset), m
$w$ heave (horizontal offset), at right angles to the strike, m

Units: any equal units of length; angles in degrees

Holds when

The fault plane is flat within the area. The values refer to the given point of the fault, since the slip changes along it.
$t$ and $w$ are parts of the slip along the dip. The component $D_s$ is horizontal and runs along the fault.

The slip along the dip is the hypotenuse of a right triangle with legs $t$ and $w$ : $D_d^2 = t^2 + w^2$ . In a steep normal fault almost all the slip goes into the throw. In a gentle thrust it goes into the heave: at a dip of 30° the heave is 1.7 times the throw.

Worked exampleComponents of the slip from slickenlines

Given (assumed data)

Fault plane (from a borehole image log, a picture of the borehole wall, and seismic data): dip azimuth 120°, dip 58°, strike 30° and 210°
Slickenlines on core: angle with the strike $r = 65°$ ; the hanging wall has moved down
Slip on the fault plane $D$: 150 m, from the offset of the axis of a sand channel followed on both sides of the fault

Solution

Components in the fault plane: along the dip $D_d = 150\sin 65° = 135.95$ m, along the strike $D_s = 150\cos 65° = 63.4$ m.
The throw is $t = 135.95\cdot\sin 58° = 115.3$ m, and the heave is $w = 135.95\cdot\cos 58° = 72.0$ m.

Answer: The fault is a normal fault with a strike-slip part (the slip along the dip is larger than along the strike). The throw is about 115 m and the heave about 72 m, and there is also about 63 m of slip along the strike.

What it means. On a structure map, the top of the layer is broken by this fault in a strip about 72 m wide. The points that used to coincide are separated along the fault (along the 30–210° line) by about 63 m. So when we match layers across the fault, we look for the “pair” not directly opposite but shifted along the strike. Suppose we had not measured the direction of slip and had taken the fault for a pure normal fault. Then the slip for the same throw would come out as 136 m, which is 9% less than the true value. The data in this example are assumed for illustration.

The slip is not constant along a fault. It is greatest near the middle and falls to zero at the tips of the fault. The throw on maps changes in the same way. The greatest slip grows with the length of the fault. According to summaries of measurements, it is usually from 0.1% to 10% of the length. So a fault 5 km long moves layers by anything from a few to hundreds of metres. Large faults grow by joining segments. Between overlapping segments, relay ramps are left: tilted “bridges” of layer that link the walls. Fluid can cross through them even if both segments are impermeable.

If the fault moved while sediment was being deposited, the downthrown wall received more sediment, and the same layers are thicker in the hanging wall than in the footwall. Such growth faults are typical of deltas. The difference in thickness between the walls tells us when the fault was active. So it also tells us whether the fault could have sealed the accumulation, or served as a path for migration, while the trap was being filled (timing of trap formation).

A fault in a well: missing and repeated section

A vertical well that crosses a normal fault passes from the hanging wall into the footwall. At the crossing point, the hanging wall holds younger layers than the footwall does at the same depth. So the well never meets part of the section. This is a missing section. A reverse fault, on the contrary, puts the older layers of the hanging wall on top of the younger layers of the footwall, and the well passes through part of the section twice. This is a repeated section. This is how we find faults in correlation: we compare the section of the well with a reference section and look for missing or repeated markers (layers that are easy to recognise).

SchematicA normal fault in a section across the strike, drawn to the scale of the example below. The fault plane dips at 55°, and the layers dip at 8° towards it. The figure shows the slip on the fault plane

D

, the throw

t

, the heave

w

, and the vertical separation of the bed

s

, which is the interval that is missing in a vertical well.

Vertical and stratigraphic separation of an inclined bed

s = t \pm w\,\mathrm{tan}\,\delta_b,\qquad s_n = s\cos\delta_b = D_d\sin(\delta \pm \delta_b)

where:

$s$ vertical separation of the bed: the vertical distance between the parts of one bed in the two walls. In a vertical well it is the thickness of the missing interval (for a normal fault) or the repeated interval (for a reverse fault), m
$s_n$ stratigraphic separation: the same thickness, measured at right angles to the layers, m
$t,\ w$ throw and heave, m
$D_d$ slip along the dip of the fault plane, m
$\delta$ dip of the fault plane, °
$\delta_b$ dip of the beds in a section across the strike of the fault, °

Units: equal units of length; angles in degrees

Holds when

The sign “+” is for beds that dip towards the fault plane, and “−” for beds that dip the same way.
The slip is along the dip. The beds are planar (not curved) and run parallel to the strike of the fault. Otherwise we take the apparent dip of the beds (the dip seen in a section that is not at right angles to the strike) in the section across the fault. The strike-slip part of the slip also changes $s$ .
The well is vertical. In an inclined well, we convert the length of the interval along the hole to thickness at right angles to the beds, using the angle between the hole and the normal to the beds (inclined wells). The difference of vertical depths equals $s$ only for horizontal beds or for a hole along the strike. The stratigraphic separation at the crossing point does not depend on the well path.

The throw $t$ equals $s$ only for horizontal beds. If the beds dip the same way as the fault plane, less is missing than $t$ . If they dip towards it, more is missing. And if the beds dip the same way but more steeply than the fault plane ( $\delta_b > \delta$ ), then $s$ changes sign. A normal fault in a vertical well then repeats the section, and a reverse fault cuts it out. The rule “normal fault: missing section; reverse fault: repeated section” is true as long as the beds are gentler than the fault or dip towards it. The derivation is in the advanced section.

Worked exampleThrow of a normal fault from the missing interval

Given (assumed data)

Correlation with a reference vertical well: an interval is missing at a depth of 2030 m; its thickness in the reference well is 41 m (the dip of the beds there is the same)
Dip of the fault plane (from seismic data): 55°
Beds (from the image log): dip at 8°, strike parallel to the fault
Two cases: 1: the beds dip towards the fault plane; 2: they dip the same way

Solution

From $s = t \pm w\,\mathrm{tan}\,\delta_b$ and $w = t/\mathrm{tan}\,\delta$ we get $t = s/(1 \pm \mathrm{tan}\,\delta_b/\mathrm{tan}\,\delta)$ . Here $\mathrm{tan}\,8°/\mathrm{tan}\,55° = 0.09841$ .
Case 1: $t = 41/1.09841 = 37.33$ m, $w = 37.33/\mathrm{tan}\,55° = 26.14$ m, and the slip on the fault plane is $D_d = 37.33/\sin 55° = 45.57$ m.
Case 2: $t = 41/0.90159 = 45.48$ m, $w = 31.84$ m, $D_d = 55.51$ m.
The stratigraphic separation is the same in both cases: $s_n = 41\cos 8° = 40.60$ m.

Answer: The 41 m that are missing correspond to a throw of about 37 m if the beds dip towards the fault plane, and about 45 m if they dip the same way.

What it means. The difference of 22% comes from the gentle dip of the beds, only 8°, and it matters. We use the throw to estimate what will stand against the reservoir on the other side of the fault, and what the SGR will be (see the section on sealing). On a structure map this fault will break the top of the layer in a strip 26 or 32 m wide. If the beds dipped the same way more steeply than the fault plane, the same normal fault would repeat part of the section. The data in this example are assumed for illustration.

Besides correlation, a fault in a well shows itself by other signs:

Dipmeter and borehole images (a dipmeter measures the dip of layers in the well, and borehole images are pictures of the wall of the well). Near a fault the dips of layers change, because the beds bend along the fault plane (fault drag). We also see crushed zones, fractures and the fault plane itself (image logs).
Core: fault breccia, fault gouge, slickensides and bands of crushed grains.
Drilling: a change in rate of penetration and in torque in the crushed zone, mud losses in a conducting fault, and a sudden change in pore pressure if the fault separates blocks with different pressures.

What else gives a missing or repeated interval

Cause: Unconformity: a break in deposition with erosion (unconformities)
How it differs from a fault: the missing interval changes from well to well and grows towards the crest of an uplift, where the erosion is deeper; for a fault it is nearly constant along the stretch of the fault

Cause: Thinning and pinching out of layers
How it differs from a fault: the thickness falls gradually, the markers are kept, and the rock type changes

Cause: Overturned limb of a fold
How it differs from a fault: the interval repeats in reverse order: younger layers come below

Cause: Well path
How it differs from a fault: an inclined or horizontal hole can cross a layer twice with no fault at all; we check this against the directional survey and the dip of the layers

Cause	How it differs from a fault
Unconformity: a break in deposition with erosion (unconformities)	the missing interval changes from well to well and grows towards the crest of an uplift, where the erosion is deeper; for a fault it is nearly constant along the stretch of the fault
Thinning and pinching out of layers	the thickness falls gradually, the markers are kept, and the rock type changes
Overturned limb of a fold	the interval repeats in reverse order: younger layers come below
Well path	an inclined or horizontal hole can cross a layer twice with no fault at all; we check this against the directional survey and the dip of the layers

The signs support each other. A reliable conclusion about a fault rests on several wells and on seismic data.

In an inclined or horizontal well, what is missing or repeated also depends on the side from which the hole approaches the fault. A horizontal hole that crosses a normal fault from the footwall into the hanging wall jumps into younger layers. Crossing from the hanging wall into the footwall, it jumps into older ones. This matters when steering a hole inside a layer (Horizontal wells). The stratigraphic separation at the crossing point is the same as for a vertical well. We convert the length of the interval along the hole to thickness at right angles to the beds. For this we use the angle between the hole and the normal, taken from the directional survey (Directional survey calculations) and from the attitude of the beds (inclined wells).

Faults on maps, on seismic data and below its resolution

On a seismic section a fault shows itself as an offset and a break of reflections, a bend of the layers at the fault plane, and diffractions (waves scattered at the ends of broken layers). The fault plane itself usually gives no reflections, and we draw it along the ends of the reflections (Seismic interpretation). On a structure map, a normal fault is shown as a strip. Its edges are the lines where the horizon (the mapped surface of a layer) meets the fault plane in the footwall and in the hanging wall (the fault cutoffs). Inside the strip there is no horizon, and its width is the heave. For a reverse fault these lines swap places, the horizon in the strip is covered by itself, and there are two sets of structure contours (lines of equal depth). The contours break off at the fault and are offset. The throw changes smoothly along the fault, with no change of sign, and fades to nothing at the tips. How such maps and sections are built is covered in the lesson Cross-sections from maps, map combination and faults on maps.

Seismic data show only faults with a throw greater than their resolution. There are many more small faults. In many basins the distribution of faults by throw is close to a power law: the number of faults with a throw not less than $t$ falls as $t^{-c}$ . We fit such a law to the mapped faults and estimate how many are hidden below the resolution. This is an extrapolation, and it is risky. It assumes that the same law continues downwards. But near the limit of resolution seismic data miss some faults, and the data bend downwards.

Power-law distribution of faults by throw

N(t) = N_1\left(\frac{t_1}{t}\right)^{c},\qquad c = \frac{\ln(N_1/N_2)}{\ln(t_2/t_1)}

where:

$N(t)$ number of faults whose greatest throw is not less than $t$ in the area, —
$t$ greatest throw of a fault, m
$N_1,\ N_2$ numbers of faults with throw not less than $t_1$ and $t_2$ ; both throws are at or above the resolution limit, —
$c$ exponent, —

Units: throws in metres; the number of faults per area or per unit area; natural logarithms (the ratio of logarithms does not depend on the base)

Holds when

One population of faults: one setting and one stage of deformation.
The power law holds over the whole range of extrapolation.
Points below the resolution limit are not used for the fit, because the number of faults there is too low.

On a double logarithmic scale (both axes logarithmic) the law is a straight line with slope $-c$ . The exponent depends on the way of counting too: by line (profile, well), by area or by volume it is different. So we compare only exponents found in the same way.

CalculatedCumulative number of faults against greatest throw, for the example below (double logarithmic scale). The points are faults mapped from seismic data. The solid line is a power law through 18 faults with throw not less than 20 m and 7 faults with throw not less than 40 m. The dashed line is its continuation below the resolution limit. The two points to the left of the resolution limit lie below the line: not all small faults have been mapped.

Worked exampleHow many faults cannot be seen on seismic data

Given (assumed data)

Area: 60 km², a 3D seismic cube; resolution limit about 20 m
Mapped faults with greatest throw not less than 20 m: 18
…not less than 40 m: 7
Reservoir thickness: 25 m

Solution

The exponent is $c = \ln(18/7)/\ln 2 = 1.363$ .
Faults with throw not less than 5 m: $N(5) = 18\cdot(20/5)^{1.363}$ . Since $2^{c} = 18/7$ , we get $(20/5)^{c} = 4^{c} = (18/7)^2 = 6.61$ and $N(5) = 119$ , that is 2.0 per km².
The faults with a throw from 5 to 20 m, which cannot be seen on seismic data, number $119 - 18 = 101$ .
The fitted law gives $N(25) = 18\cdot 0.8^{1.363} = 13.3$ faults with a throw greater than the reservoir thickness, 25 m. In the list of mapped faults there are 14 of them.

Answer: There are about 101 faults with a throw of 5–20 m hidden below the resolution, which is 5.6 times more than the number mapped.

What it means. Only about 13–14 of the mapped faults offset the reservoir by more than its thickness, and they can divide the accumulation by juxtaposing layers. The invisible faults with a throw of 5–20 m do not cut the reservoir right through. But their zones, with crushed grains and clay material, lower the permeability across them. In the model we do not put them in one by one. We set them on average, by lowering permeability or with transmissibility multipliers (transmissibility multipliers). The estimate is uncertain: the exponent was found from two points, and a change of $c$ by 0.2 changes $N(5)$ by about 1.3 times. We check it against faults in core and on image logs. The data in this example are assumed for illustration.

The fault zone: a barrier or a channel

What matters for flow is what the rock has become in the fault zone. This depends first on the rocks through which the fault passed, and on the depth at which it moved.

What happens to rock in a fault zone

Host rock: Clean porous sandstones
Fault-zone rock: cataclastic bands: crushed and ground grains, millimetre-wide bands and networks of them
Permeability across the fault: orders of magnitude lower than the sandstone; the bands are often discontinuous

Host rock: Shaly sandstones, alternating sandstones and shales
Fault-zone rock: a mixture of grains and clay material; with a lot of clay, a continuous clay smear
Permeability across the fault: the lower the more clay material; a smear can be almost impermeable

Host rock: Shales
Fault-zone rock: fault gouge smeared along the fault plane
Permeability across the fault: very low

Host rock: Tight brittle rocks: carbonates, cemented sandstones
Fault-zone rock: breccias and fractures of the damage zone
Permeability across the fault: along the fault often higher than the host rock: the fault is a channel

Host rock	Fault-zone rock	Permeability across the fault
Clean porous sandstones	cataclastic bands: crushed and ground grains, millimetre-wide bands and networks of them	orders of magnitude lower than the sandstone; the bands are often discontinuous
Shaly sandstones, alternating sandstones and shales	a mixture of grains and clay material; with a lot of clay, a continuous clay smear	the lower the more clay material; a smear can be almost impermeable
Shales	fault gouge smeared along the fault plane	very low
Tight brittle rocks: carbonates, cemented sandstones	breccias and fractures of the damage zone	along the fault often higher than the host rock: the fault is a channel

These are qualitative tendencies. With burial (especially deeper than about 3 km), fault zones in sandstones become compacted and cemented, and their sealing grows. Fractures stay open better in brittle rocks, and where the fault is close to slip in the present stresses.

In sandstones the permeability of fault-zone rock can be orders of magnitude lower than that of the host rock, down to $10^{-4}$ mD. But neither a wide crushed zone nor a thin film of clay matters for flow by itself. What counts is the ratio of the permeability of the zone to its width (dimensionless fault transmissibility). A fault may conduct differently across and along. The core blocks flow across the fault, while the fractures of the damage zone pass fluid along it. So faults play a double role in a petroleum system: the same fault can be a seal for a layer and a path for vertical migration.

Active faults, and faults that become active again when pore pressure rises, are the main paths of vertical migration. Oil and gas rise along them from deep layers to shallow traps, and sometimes right to the surface. If a fault moved after a trap was filled, the accumulation can lose some of its hydrocarbons, and traps higher in the section can receive them (timing of trap formation). Clean sandstones broken by a network of cataclastic bands stay good reservoirs, but they become uneven. The bands slow flow across themselves, and a dense network of them can clearly worsen the communication between parts of the layer.

Does a fault seal: juxtaposition and the shale gouge ratio

A fault holds hydrocarbons in two ways. One is juxtaposition (the juxtaposition seal): across the fault from the reservoir there is rock that does not let fluid through, such as shale, salt or tight limestone. The other is the fault rock (the fault-rock seal): even where reservoir stands against reservoir, the fault zone with its small pores makes a capillary barrier, like a seal (capillary barrier). So the assessment of a fault answers two questions: what stands against the reservoir, and what fills the fault zone.

The first question is answered by the Allan diagram. We project onto the plane of the fault the lines where the layers meet the fault plane in both walls (the fault cutoffs). Where the permeable layers of the two walls overlap, there are juxtaposition windows, through which flow is possible. If the throw is greater than the thickness of the reservoir, the reservoir can stand entirely against shales. If it is smaller, the reservoir is partly juxtaposed with itself. The throw falls towards the tips of the fault, so there are nearly always windows there.

CalculatedAn Allan diagram for the section of the example below: the throw grows from zero at the tips of a fault 4 km long to 40 m in the middle. The sandstones of the footwall are light horizontal bands. The sandstones of the hanging wall are outlined bands, lowered by the amount of the throw. The windows, where sandstone stands against sandstone, are dark. At the tips each sandstone is juxtaposed with itself. Nearer the middle, the lower part of sandstone B stands against sandstone A of the hanging wall, and sandstone C stands against sandstones A and B.

The second question is answered in numbers by the share of clay material that has slid past a given point of the fault plane. During slip, material from all the layers that slid past the point is drawn into the fault zone. The more shale there is among them, the more clayey and tighter the rock of the zone. This share is the shale gouge ratio (SGR).

Shale gouge ratio: the share of clay material in the fault zone

SGR = \frac{\sum_i V_{sh,i}\,\Delta z_i}{t}\cdot 100\,\%

where:

$SGR$ share of clay material in the fault zone at a given point of the fault plane, %
$V_{sh,i}$ shale volume of layer $i$ from well logs (Determining shale volume), fraction
$\Delta z_i$ thickness of layer $i$ within the interval of height $t$ that has slid past the point: in the section of the footwall, from the point upwards by $t$ , m
$t$ throw of the fault at this point, m

Units: thicknesses and throw in the same units; $V_{sh}$ as a fraction

Holds when

The material of the interval that has slid past is fully mixed in the fault zone.
$V_{sh}$ is calculated in the same way as in the calibration data. SGR thresholds found with a different way of calculating shale volume must not be used.
We use the throw at the given point. Along the fault, SGR changes together with it.

SGR is decided not by the size of the throw itself but by the make-up of the interval that has slid past the point. From summaries of fields, noticeable sealing starts at an SGR of about 15–20%. Below this, faults usually let hydrocarbons through. Above it, they can hold a column that is taller the larger the SGR. The threshold is not universal, and we calibrate it on the fields of the area.

To turn SGR into the height of an accumulation, we use an empirical envelope (the seal-failure envelope, a limiting curve drawn from field data). From fields with known pressure differences across faults, we find the greatest pressure difference that a fault with a given SGR holds at a given burial depth. One widely used calibration is written like this:

Pressure difference that a fault holds, and the height of column above the point

\Delta p_f = 10^{\,SGR/27 - C}\ \text{bar},\qquad H = \frac{\Delta p_f}{(\rho_w - \rho_h)\,g}

where:

$\Delta p_f$ greatest pressure difference across the fault that it holds (1 bar = 0.1 MPa), bar
$SGR$ SGR at this point, %
$C$ constant that depends on the greatest burial depth: 0.5 for less than 3 km; 0.25 for 3 to 3.5 km; 0 for more than 3.5 km, —
$H$ how far below the point the free water level of the accumulation can fall before the fault at this point starts to let hydrocarbons through, m
$\rho_w,\ \rho_h$ densities of formation water and of the hydrocarbon in the reservoir, kg/m³
$g$ acceleration of gravity, 9.81, m/s²

Units: $\Delta p_f$ in bar (to calculate $H$ in SI, multiply by $10^5$ Pa/bar); $H$ in m; densities in kg/m³

Holds when

The envelope is an empirical upper bound from fields of one region. A real fault may hold less.
The growth of the pressure difference held with SGR is confirmed by data up to an SGR of about 40%. Above this value, buoyancy measurements show no further growth, and we do not extrapolate the envelope there.
Across the fault there is a water-bearing layer of the same hydrodynamic system, and the pressure difference equals the buoyancy of the column of hydrocarbons (buoyancy).
The rocks are water-wet. Calibrations for oil and for gas differ (because of the larger interfacial tension, gas needs a larger pressure difference to break through the same fault rock). Outside the region of the calibration, we check the envelope against local data.

The envelope grows exponentially: a rise of SGR by 27 percentage points raises the pressure difference held 10 times. With burial, fault-zone rock becomes compacted and cemented, and the same SGR holds more. As under a seal, the height of the column is inversely proportional to the density difference: the same fault holds a lower gas accumulation. Every point where the reservoir of the trap stands against permeable rock must hold the accumulation. The point that allows the shallowest free water level decides. This is usually the top of the “weakest” window, where buoyancy is greatest and SGR is lowest.

CalculatedA section at the fault of the example below with a throw of 30 m: columns of the footwall and the hanging wall (the hanging wall is lowered by 30 m), and the SGR along the fault plane. Sandstone B of the footwall stands against shales down to 2210 m. Deeper it stands against sandstone A (a window at 2210–2220 m, SGR about 30%). Below is a second window: sandstone C against sandstone B (SGR 20–23%). The dashed line is the 20% threshold.

Worked exampleWill the fault hold the accumulation: two throws

Given (assumed data)

Section from top to bottom (thickness, shale volume $V_{sh}$ ): seal: shales (0.80); sandstone A 14 m (0.12); shale 8 m (0.75); sandstone B 18 m (0.15); shale 5 m (0.70); sandstone C 25 m (0.10)
Trap: sandstone B in the footwall, with its crest at the fault: top at 2202 m; the structural spill point is 2290 m
Throw of the fault at the crest: case 1: 30 m; case 2: 10 m
Densities at reservoir conditions: water 1080, oil 780 kg/m³; the sandstones of the hanging wall contain water
Greatest burial depth: 2.2 km, so $C = 0.5$

Solution

Case 1, $t = 30$ m. Across from sandstone B (2202–2220 m), the hanging wall has the section that lies 30 m higher: first the seal, then sandstone A. The window where sandstone stands against sandstone runs from 2210 to 2220 m.
SGR at the top of the window: the interval 0–30 m of the section has slid past the point, so $(14\cdot 0.12 + 8\cdot 0.75 + 8\cdot 0.15)/30 = 29.6$ %.
Envelope: $\Delta p_f = 10^{29.6/27 - 0.5} = 3.95$ bar. The buoyancy of the oil grows by $300\cdot 9.81 = 2943$ Pa per metre of height, so $H = 394.7\cdot 10^3/2943 = 134$ m. The water level could drop to $2210 + 134 = 2344$ m, which is deeper than the spill point. So the spill limits the accumulation: the column is 88 m.
Case 2, $t = 10$ m. The top 2 m of sandstone B stand against the bottom of sandstone A, but shale has slid past the point there: SGR $= (2\cdot 0.12 + 8\cdot 0.75)/10 = 62.4$ %. From 2212 m, sandstone B stands against itself, and only sandstone has slid past these points: SGR $= 15$ %.
For the window “B against B”: $\Delta p_f = 10^{15/27 - 0.5} = 1.14$ bar and $H = 39$ m. The water level is no deeper than 2251 m, and the column is no more than 49 m. But suppose the fault at an SGR of 15% does not seal at all, as is usual below 15–20%. Then oil leaves through the top of this window, and the column is only 10 m.

Answer: With a throw of 30 m, the fault holds the accumulation up to the spill point (a column of 88 m). With a throw of 10 m, the column is from 10 to 49 m.

What it means. The smaller throw turned out worse. At 10 m, sandstone B is juxtaposed with itself, and only clean sandstone has slid past the window. At 30 m, a shaly interval and sandstone A stand against B, and the window is sealed by rock with an SGR of about 30%. Suppose the trap held gas (220 kg/m³). Then in case 1 the same pressure difference would hold only 47 m above the window, and the fault, not the spill, would limit the gas accumulation (for gas, though, the calibration is its own). So we assess sealing along the whole closing stretch of the fault, especially where the throw is small. We check the envelope, as an upper bound, against the contacts and pressures in the neighbouring blocks and in the fields of the area. The data in this example are assumed for illustration.

Faults in traps and in development

A trap at a fault (fault-bounded) is closed by the dip of the layers on three sides and by the fault on the fourth. The fault must hold the accumulation along the whole length of the closing stretch. The most vulnerable places are the windows near the tips and where the throw is small, the crossings of faults, and relay ramps. So for such prospects we assess the probability of fault sealing separately (Geological risk and chance of success): what share of the closed volume depends on the fault, and how reliable the SGR calibration is in the area.

In development the same fault decides whether the blocks of an accumulation communicate. Signs of separation show up even before production: different contacts and different pressure lines in neighbouring blocks (contacts from pressure measurements). A fault that has held a pressure difference for millions of years is a barrier in development too. The reverse is not true. If the pressures on the two sides are equal, the fault can still slow the flow strongly during production. Over geological time the pressure equalises even through a poorly conducting zone, but over the years of development it does not. How much a fault slows the flow is assessed from well tests and the production history (Partially communicating faults).

In the reservoir simulation model, faults are given as transmissibility multipliers between cells. We calculate them from the permeability and the width of the fault zone, which empirical relations link to SGR and to the slip. We refine them when the model is matched to the production history (History matching). Small faults and cataclastic bands, which cannot be drawn one by one, are included on average, by lowering the permeability of cells.

Order of assessing a fault

1
Geometry
The fault plane and the throw along the fault, from seismic data and wells. Where the throw is smaller than the reservoir thickness.
2
Juxtaposition
The Allan diagram: windows where reservoir stands against reservoir, and their position relative to the closure of the trap (the closed area where oil can collect).
3
Fault-zone rock
SGR from well logs and throw, the burial depth, and signs of cataclasis and cementation in core.
4
Calibration
Contacts and pressures in neighbouring blocks and in fields of the area: what pressure difference faults with a given SGR hold.
5
Uncertainty
A range of accumulation height for exploration, and a range of fault transmissibility for development. Both go into the risk assessment and into the model.

Terms and symbols

Fault: A surface or zone along which blocks of rock have moved relative to each other.
Fault plane: The surface of movement of a fault.
Hanging wall and footwall: The blocks above an inclined fault plane and below it.
Normal fault, reverse fault, thrust, strike-slip fault: A fault with a downthrown hanging wall, a fault with an upthrown hanging wall, a low-angle reverse fault, and a fault with movement along the strike.
Slip on the fault plane $D$: The full displacement of the walls relative to each other (net slip). It splits into components along the dip $D_d$ and along the strike $D_s$ .
Throw and heave $t$ , $w$: The vertical and the horizontal (at right angles to the strike) parts of the slip along the dip.
Vertical separation of a bed $s$: The vertical distance between the parts of a bed in the two walls. In a vertical well it is the thickness of the missing or repeated interval.
Stratigraphic separation $s_n$: The thickness of the missing interval, measured at right angles to the layers.
Damage zone and fault core: The zone of fractures and small faults around a fault, and its central part with tectonites.
Allan diagram: A projection of the layers of both walls onto the fault plane. It shows the juxtaposition windows.
SGR: The share of clay material that has slid past a point of the fault plane (shale gouge ratio).
Seal-failure envelope: An empirical relation that gives the greatest pressure difference a fault can hold from its SGR and burial depth.
Relay ramp: A tilted part of a layer between overlapping segments of a fault, which links its walls.

Common mistakes and misconceptions

Wrong

“The interval missing in a well is the throw of the normal fault.”

In fact

It is the vertical separation of the bed $s$ . With inclined beds it differs from the throw by $w\,\mathrm{tan}\,\delta_b$ , and with beds that dip the same way more steeply than the fault, a normal fault even repeats the section.

Wrong

“The larger the throw, the more reliably the fault seals.”

In fact

What matters is what stands against the reservoir and which layers have slid past the point of the fault plane. A small throw in clean sandstone gives a window with a low SGR. With a throw larger than the reservoir thickness, the reservoir may stand entirely against shales.

Wrong

“The pressures in the blocks are equal, so the fault does not affect development.”

In fact

Over millions of years pressure equalises even across a poorly conducting fault. During production, such a fault can slow the flow between blocks noticeably. We check this with well tests and the production history.

Wrong

“An SGR threshold of 20% (or 30%) is universal.”

In fact

Thresholds are empirical. They depend on the area, the burial depth, the fluid and the way shale volume is calculated, and we calibrate them on local data.

Wrong

“Part of the section is missing, so the well has crossed a normal fault.”

In fact

A missing section can also come from an unconformity or the pinching out of layers, and a repeated one from overturned layers or the path of the hole. We confirm a fault from several wells, image logs and seismic data.

In depth: advanced detail

Where the formulas for vertical and stratigraphic separation come from

Take a vertical section across the strike of the fault. The $x$ axis points in the direction of dip of the fault plane, and the depth $z$ points down. Let the top of the layer in the footwall be the straight line $z = z_0 \mp x\,\mathrm{tan}\,\delta_b$ (the sign “−” if the layer dips towards the fault plane, that is, sinks towards smaller $x$ ; “+” if it dips the same way). The hanging wall has moved along the dip of the fault plane by $D_d$ : by $w = D_d\cos\delta$ horizontally and by $t = D_d\sin\delta$ downwards. Its top is the same line shifted by this vector: $z = z_0 \mp (x - w)\,\mathrm{tan}\,\delta_b + t$ .

On one vertical line, the difference in depth of the two tops is $s = t \pm w\,\mathrm{tan}\,\delta_b$ . It does not depend on $x$ , so any vertical well sees the same missing interval. The thickness at right angles to the layers is $1/\cos\delta_b$ times smaller: $s_n = s\cos\delta_b = D_d\,(\sin\delta\cos\delta_b \pm \cos\delta\sin\delta_b) = D_d\sin(\delta \pm \delta_b)$ . The last form is easy to picture: the stratigraphic separation is the projection of the slip onto the normal to the layers, and the angle between the fault plane and the layers is $\delta + \delta_b$ or $\delta - \delta_b$ . If the layer is parallel to the fault plane ( $\delta_b = \delta$ , in the same direction), slip along it does not break the layer, and $s = s_n = 0$ .

In oblique slip, the strike-slip component $D_s$ changes the section only if the layer is inclined along the strike of the fault. Then $D_s\,\mathrm{tan}\,\delta_{b\parallel}$ is added to $s$ or subtracted from it, where $\delta_{b\parallel}$ is the apparent dip of the layer in the direction of the strike of the fault. The general case is easy to solve with vectors, as in the lesson on structural analysis. The difference in depth of two tops on a vertical line equals the vertical component of the slip (downwards) plus the scalar product (dot product) of the horizontal component of the slip with the vector of rise of the top. That vector points to where the top rises, and its length is $\mathrm{tan}\,\delta_b$ .

Limits of the SGR method, and other measures

Mixing. SGR assumes that the material of the interval that has slid past is completely mixed. In fact clay is smeared unevenly, and the continuity of the smear depends on the thickness of the source shale layer and on the slip. Another measure is the ratio of the throw to the thickness of the shale layer (the shale smear factor, SSF). From observations on faults of seismic scale, a continuous smear survives as long as this ratio is below about 4.
The way shale volume is calculated. The threshold at which sealing begins, found with one way of calculating $V_{sh}$ from well logs, shifts with another way. Calibration and prediction must use the same way.
Clean sandstones. Cataclasis (crushing of grains) and cementation can make fault-zone rock with a high entry pressure in clean sandstone. But there are few reliable examples of tall columns held only in this way. At an SGR below 15–20%, a fault is usually taken to be permeable.
Pressure difference and buoyancy. The pressure difference across a fault is made not only by the buoyancy of the column but also by the difference of water pressures in the blocks, if they are separated hydrodynamically. The common envelopes are built from the full pressure differences across the fault. It is more exact to calibrate on buoyancy, which is the pressure of the column above its contact. If there are hydrocarbons on both sides of the fault, the buoyancy is taken above the deeper of the two contacts.

Connections to other topics

Stress, strain and structural analysis — stresses, friction and the attitudes of planes.
Traps and timing — fault-bounded traps, closure and the timing of their formation.
Reservoir and seal — the capillary barrier and the height of accumulation that a seal holds.
Partially communicating faults — how the transmissibility of a fault shows in well tests, and how it is put into the model.
A sealing fault near the well: the method of images — an impermeable fault as a boundary of the layer in well testing.
Folds, salt diapirs and their relationship to faults — folds above faults, rollover anticlines and faults above salt domes.
Fractures and naturally fractured reservoirs — fractures with no displacement and fractured reservoirs.
Cross-sections from maps, map combination and faults on maps — how faults are shown on structure maps and sections.
Structural styles and tectonic history — combinations of faults in extension, compression and strike-slip settings.
Geological risk and chance of success — the probability of fault sealing in the assessment of a prospect.

Check yourself

1
In a vertical well, below 2650 m, a package of markers that has already been passed higher up repeats. The beds dip at 5° and the fault, from seismic data, at 50°. What kind of fault is it?
1. A.Reverse fault or thrust
2. B.Any: at such angles any fault gives a repetition
3. C.Normal fault
4. D.Strike-slip fault
Show the answer
Correct answer: A. Reverse fault or thrust
The beds are gentler than the fault, so the rule works: in a vertical well a repeated section means a reverse fault (a thrust). A normal fault would repeat the section only if the beds dipped the same way more steeply than the fault plane. A strike-slip fault hardly changes the vertical position of the beds.
2
A normal fault dipping at 50° has moved a bed by 80 m along the dip of the fault plane. Find the throw and the heave, and the thickness of the interval missing in a vertical well with horizontal beds.
Show the answer
$t = 80\sin 50° \approx 61.3$ m, $w = 80\cos 50° \approx 51.4$ m. With horizontal beds, the vertical separation of the bed equals the throw, so about 61 m of section is missing. On a map the fault will break the bed in a strip about 51 m wide.
3
Sandstone stands against sandstone at a fault. At point A the SGR is 12%, and at point B it is 35%. Where will the fault more likely hold oil?
1. A.At point A: there is less clay there, and the pores of the fault zone are not blocked
2. B.At point B
3. C.Equally: where sandstone meets sandstone the fault is always permeable
4. D.It depends only on the throw, not on the SGR
Show the answer
Correct answer: B. At point B
The more clay material is drawn into the fault zone, the smaller its pores and the higher the capillary barrier. At an SGR below 15–20% faults usually let hydrocarbons through. At 35% and burial of less than 3 km, the envelope allows a pressure difference of about $10^{35/27 - 0.5} \approx 6$ bar, which is 0.6 MPa.
4
In two blocks separated by a fault, the oil–water contacts are at 2350 and 2390 m, and the pressure in the water-bearing part of the first block is 0.6 MPa higher than in the second block at the same depth. What does this say about the fault and about development?
Show the answer
The fault separates the blocks. Over geological time neither the pressures nor the contacts have equalised, so the transmissibility of the fault zone is very low. The blocks were filled, and will be developed, separately. Each needs its own wells and its own estimate of hydrocarbons in place, and injection in one block will not support the pressure in the other.
5
Why do windows where sandstone stands against sandstone most often appear at the tips of a fault?
1. A.The throw goes to zero there and becomes smaller than the reservoir thickness
2. B.There is more clay material there
3. C.The fault plane is steeper there
4. D.The fault is younger there
Show the answer
Correct answer: A. The throw goes to zero there and becomes smaller than the reservoir thickness
The slip falls to zero at the tips of a fault. When the throw is smaller than the reservoir thickness, the reservoir partly stands against itself, and little shale slides past such points, so the SGR is low. So we check sealing along the whole closing stretch of the fault, not only in the middle.
6
Seismic data show 12 faults with a throw not less than 30 m. The exponent of the power law is $c = 1.2$ . How many faults with a throw not less than 10 m should we expect, and how many of them cannot be seen, if the resolution limit is 30 m?
Show the answer
$N(10) = 12\cdot 3^{1.2} \approx 12\cdot 3.74 \approx 45$ . About 33 faults with a throw of 10 to 30 m cannot be seen. The estimate is correct only if the power law continues below the resolution limit, and the exponent was found from faults above that limit.

Updated: 19 October 2026