Primary sedimentary structures

Quick refresher

A bed is bounded by bedding surfaces — pauses in deposition or changes of conditions. Laminae a few millimetres thick make up its internal structure. The line between a bed and a lamina is a convention: about 1 cm.
Grains move when the Shields number $\theta = \tau/((\rho_s - \rho)\,g\,d)$ (the pull of the flow on the bed compared with the weight of the grains) exceeds the critical value — about 0.03–0.1 for sand and 0.04–0.055 for gravel and pebbles. The Froude number $\text{Fr} = U/\sqrt{g h}$ separates subcritical (slow, tranquil) flow (below 1) from supercritical (fast, rapid) flow (above 1).
As the flow gets faster, sand finer than about 0.6–0.7 mm goes through this sequence of bedforms (shapes built on the sand bed): ripples, then dunes, then upper-stage plane bed, then antidunes. In very fine sand, finer than about 0.1 mm, there are no dunes. In coarser sand, lower-stage plane bed stays at first instead of ripples.
Ripples leave small-scale cross-lamination (sets up to 4–5 cm thick). Dunes leave large-scale cross-bedding (sets from decimetres to metres thick). A set is one body of parallel cross-laminae. Straight crests give planar cross-bedding, and curved crests give trough cross-bedding. Cross-laminae dip downcurrent, in the direction the current flowed.
A graded bed (coarse at the bottom, fine at the top) is deposited from a flow that is dying away. Casts of scours and grooves are preserved on the bases of sand beds. Deformation structures and the burrows of organisms form in sediment that is still soft. Many structures show which side of a bed is its top.
Palaeocurrent azimuths (the compass directions of ancient currents) are averaged as vectors, not as numbers. On an image of the borehole wall, an inclined plane appears as a sinusoid (a wave-shaped curve): $\tan\delta = h/d$ ( $h$ is its peak-to-trough height, $d$ the borehole diameter), and the dip azimuth is given by the lowest point of the sinusoid.

Key relationships

\theta = \dfrac{\tau}{(\rho_s - \rho)\,g\,d},\quad \tau = \rho g h S

the Shields number and the bed shear stress of a uniform flow

\text{Fr} = \dfrac{U}{\sqrt{g h}}

the Froude number: below 1 the flow is subcritical, above 1 supercritical

\tan\zeta = \dfrac{V_{agg}}{V_{mig}}

the angle of climb of ripples: the ratio of the rates of deposition and of bedform migration

\tan\delta = \dfrac{h}{d}

dip of a plane from the sinusoid on the image of a vertical borehole: $h$ is the peak-to-trough height of the sinusoid, $d$ the borehole diameter (not the flow depth and not the grain size, as in the Shields number)

Common pitfalls

Confusing ripples (a height of centimetres) with dunes (decimetres and more), and small-scale cross-bedding with large-scale.
Averaging azimuths arithmetically: the mean of 350° and 10° is 0°, not 180°.
Reading the palaeocurrent in a tilted formation without removing the structural tilt: a tilt across the palaeocurrent of even 10° can turn the azimuth of the cross-laminae by about 20°.
Treating a massive sandstone as one deposited “without a history”: the bedding may have been erased by organisms or by liquefaction (sediment turning into a fluid slurry), and sometimes it is visible only on a tomogram (a CT scan).
Taking a line for a direction: grooves and wave ripples do not say in which of the two directions the current went.

After this lesson you can

distinguish a bed from a lamina, describe the boundaries and thickness of beds on a standard scale, and explain why bedding makes permeability anisotropic
estimate from the Shields and Froude numbers whether grains move and in what regime the flow runs, and read the bedform phase diagram: ripples, dunes, plane bed, antidunes
recognise the structures made by currents, waves and wind, the graded beds of gravity flows, sole marks, soft-sediment deformation and bioturbation, and explain what each says about the environment and about where the top and base of a bed are
find the palaeocurrent direction from cross-bedding and sole marks, average azimuths as vectors and remove the structural tilt
recognise the structures in core and on borehole images and find the dip of laminae from the sinusoid on the unrolled image

Beds, laminae and their boundaries

We divide a sedimentary sequence into beds. A bed is a body of rock that differs from its neighbours in composition, grain size, colour or structure, and is bounded by bedding surfaces. A bedding surface marks a pause in sedimentation or a change of conditions. The pause may last minutes (between two gusts of the current) or thousands of years. Inside a bed we often see laminae: layers about a millimetre thick that differ in grain size or in the content of mica, clay or organic matter.

Thickness of beds and laminae

Name: Very thick beds
Thickness: more than 1 m

Name: Thick beds
Thickness: 30–100 cm

Name: Medium beds
Thickness: 10–30 cm

Name: Thin beds
Thickness: 3–10 cm

Name: Very thin beds
Thickness: 1–3 cm

Name: Thick laminae
Thickness: 3–10 mm

Name: Thin laminae
Thickness: less than 3 mm

Name	Thickness
Very thick beds	more than 1 m
Thick beds	30–100 cm
Medium beds	10–30 cm
Thin beds	3–10 cm
Very thin beds	1–3 cm
Thick laminae	3–10 mm
Thin laminae	less than 3 mm

One of the commonly used scales. Russian-language core descriptions speak of a massive, thick-, medium- and thin-bedded or micro-bedded structure; the limits of these words are defined in the key to the description.

When we describe a bed, we note the character of its base and top. A boundary may be sharp and flat; erosional, cutting into the underlying laminae, often with pebbles of eroded rock at the base; or gradational, when one sediment changes into another without a clear line. An erosional base shows a strong current. A gradational transition shows a smooth change of conditions.

Laminae within a bed may be horizontal (parallel to the base), inclined (at an angle to it — see the section on cross-bedding) or wavy. In interbedded sand and clay (layers of the two alternating) we distinguish three types: flaser bedding (lenses of clay in sand), wavy bedding (continuous wavy layers of both) and lenticular bedding (lenses of sand in clay). Such interbedding arises where the current alternately strengthens and carries sand, then weakens and lets mud settle — for example, on tidal flats.

How a current transports and deposits grains

A current of water or air carries grains in two ways. Coarse grains roll and bounce along the bottom. This is the bedload. Fine grains are picked up by the eddies of the turbulent (swirling) flow and stay in suspension for a long time. This is the suspended load. The boundary between them is not fixed: in a flood, a river lifts sand into suspension too. Grains are deposited when the flow weakens and can no longer hold them. Fine mud settles only in almost still water (settling velocity by Stokes' law).

A grain starts to move when the flow pulls it harder than its weight and its support on the neighbouring grains can hold it. We express the force of the flow at the bed as the bed shear stress $\tau$ (the drag of the flow along the bed, per unit area). We compare it with the weight of a grain in water through the dimensionless Shields number.

The Shields number and the bed shear stress

\theta = \frac{\tau}{(\rho_s - \rho)\,g\,d},\qquad \tau = \rho\,g\,h\,S

where:

$\theta$ Shields number, —
$\tau$ bed shear stress of the flow, Pa
$\rho_s,\ \rho$ density of the grains (2650 for quartz) and of water (about 1000), kg/m³
$g$ gravitational acceleration, 9.81, m/s²
$d$ grain diameter, m
$h$ flow depth, m
$S$ slope of the water surface, m/m

Units: SI

Holds when

$\tau = \rho g h S$ holds for steady uniform flow in a wide channel.
$\tau$ includes the resistance of the bedforms themselves, so only a part of the stress acts on the grains. The Shields number for the grains therefore comes out too high (an upper estimate).

The numerator is the force with which the flow pulls on a unit area of the bed. The denominator is the weight, in water, of a layer of grains one grain thick. A grain moves when $\theta$ exceeds the critical value $\theta_{cr}$ .

The critical Shields number (Soulsby–Whitehouse approximation)

\theta_{cr} = \frac{0.30}{1 + 1.2\,D_*} + 0.055\left(1 - e^{-0.020\,D_*}\right),\qquad D_* = d\left[\frac{g\,(\rho_s/\rho - 1)}{\nu^2}\right]^{1/3}

where:

$\theta_{cr}$ critical Shields number — the threshold of motion, —
$D_*$ dimensionless grain size, —
$\nu$ kinematic viscosity of water; at 20 °C about $1.0\cdot10^{-6}$ , m²/s

Units: SI

Holds when

Non-cohesive grains (grains that do not stick together) of one size on a flat bed.
The experimental data are widely scattered — at the same $D_*$ the threshold varies severalfold, so the result is only a guide.

For sand $\theta_{cr}$ is about 0.03–0.1 (the upper values for very fine sand), for gravel and pebbles 0.04–0.055 (about 0.055 for fragments coarser than about 8 mm). Fine particles are harder to move than their weight suggests. They hide in the viscous sublayer at the bed (a thin layer of slow, sticky water), and clay particles also stick together. So compacted mud can resist a current that rolls gravel along, and mud can be deposited only in very quiet water.

The Froude and Reynolds numbers of the flow

\text{Fr} = \frac{U}{\sqrt{g\,h}},\qquad \text{Re} = \frac{U\,h}{\nu}

where:

$\text{Fr}$ Froude number, —
$\text{Re}$ Reynolds number of the flow, —
$U$ mean flow velocity, m/s
$h$ flow depth, m

Units: SI

Holds when

$U$ and $h$ are the averages over the cross-section of a wide flow.
Flow in an open channel is laminar (smooth) at Re below about 500 and turbulent at Re above about 2000 (based on the depth or the hydraulic radius).

Fr compares the flow velocity with the speed of waves on the water surface. At Fr below 1, the flow is subcritical (tranquil): disturbances on the surface can travel upstream. At Fr above 1, it is supercritical (rapid). Re compares the push of the flow's inertia with the drag of its viscosity: low values mean smooth flow, high values mean turbulent flow. Rivers are almost always turbulent, with Re of hundreds of thousands and more, and Fr is usually 0.1–0.5. Mountain streams and floods on alluvial fans (fan-shaped heaps of sediment at the foot of mountains) may be supercritical.

Worked exampleWhat a river carries, and in what regime

Given (assumed data)

Depth $h$: 1.5 m
Slope of the water surface $S$: 0.0003 (0.3 m per kilometre)
Mean velocity $U$: 0.9 m/s
Grains: quartz 2650 kg/m³: sand 0.25 mm and pebbles 16 mm
Water: 1000 kg/m³, $\nu = 1.0\cdot10^{-6}$ m²/s

Solution

Shear stress at the bed: $\tau = 1000\cdot 9.81\cdot 1.5\cdot 0.0003 = 4.41$ Pa.
Sand: $\theta = 4.41/(1650\cdot 9.81\cdot 0.25\cdot 10^{-3}) = 1.09$ . The dimensionless size is $D_* = 6.3$ , the threshold $\theta_{cr} = 0.0415$ : the flow exceeds it by a factor of $1.09/0.0415 \approx 26$ .
Pebbles: $D_* = 405$ , $\theta_{cr} = 0.0556$ , that is, $\tau_{cr} = 0.0556\cdot 1650\cdot 9.81\cdot 0.016 = 14.4$ Pa — more than 4.41 Pa: the pebbles lie still ( $\theta = 0.017$ ).
Regime: $\text{Fr} = 0.9/\sqrt{9.81\cdot 1.5} = 0.23$ — the flow is subcritical; $\text{Re} = 0.9\cdot 1.5/10^{-6} = 1{,}350{,}000$ — turbulent.

Answer: The sand moves with a large margin: its Shields number is 26 times the threshold. The pebbles are stationary. The flow is subcritical (Fr ≈ 0.23) and turbulent.

What it means. Such a river drives sand along the bed, most likely as dunes, and lifts some of it into suspension. The pebbles stay on the channel floor as a lag (a layer of coarse leftovers) until a flood increases the depth and the slope. At Fr of about 0.2 there are no antidunes or upper-stage plane bed, so the deposit will show the cross-bedding of dunes and ripples (next section). We calculated the shear stress from the total resistance of the channel, and part of that resistance comes from the bedforms themselves. So the Shields numbers for the grains are an upper estimate. This makes the conclusion about the pebbles only more reliable. The data in this example are assumed for illustration.

On slopes, sediment can also move by itself, under gravity, as sediment gravity flows. We distinguish several kinds as the share of water grows. In falls, slides and slumps (collapses and slips of the slope), the mass moves as a whole. In debris flows, the fragments are held by a viscous clay-rich mass. In grain flows on steep slopes, the grains are held up by collisions. In liquefied flows, they are supported by pore water escaping upwards. Turbidity currents are clouds of suspension that turbulence carries down the slope for tens and hundreds of kilometres. Their deposits are described below.

Bedforms and cross-bedding

A sandy bed under a flow rarely stays flat. The grains that the current rolls along the bottom gather into repeating ridges, called bedforms. A bedform lives on the surface of the bed and changes with the current. In the sediment it leaves a structure: cross-bedding, parallel laminae and so on. That is why the structures can tell us what the current was like.

Ripples: ridges up to 3–4 cm high and up to 60 cm long. They appear in sand finer than about 0.6–0.7 mm in a weak current; their size depends on the grain size and hardly at all on the depth of the flow.
Dunes (called megaripples in older literature): ridges from 0.6 m to hundreds of metres long and from a few centimetres to a few metres high, in large rivers up to ten metres. Their size grows with the depth of the flow: the height of dunes is usually from a tenth to a sixth of the depth.
Plane bed: lower-stage — in coarse sand in a weak current; upper-stage — in a strong current, when the ridges are washed out and a thin layer of grains moves over the whole bed.
Antidunes: gentle waves of the bed, linked with standing waves on the water surface at a Froude number of about 1. They may move upstream as well as downstream, and are rarely preserved in deposits.

Ripples and dunes are asymmetric. They have a gentle stoss side, up which the current drives the grains, and a steep lee side. In ripples and many dunes, the slope of the lee side is close to the angle of repose of sand (the steepest slope at which loose sand stays in place), about 30–35°. In large river dunes it is much gentler. Behind the crest, the flow separates from the bed. Grains roll over the crest and avalanche down the lee side, and the bedform advances. Each former position of the lee side is preserved as an inclined lamina. This is how cross-bedding (cross-stratification) arises.

Which bedform appears depends first of all on the flow velocity and the grain size. A bedform phase diagram (also called a bedform stability diagram) summarises flume experiments (experiments in laboratory channels) and observations in rivers. At a given depth, as the velocity grows, fine and medium sand develops ripples, then dunes, then upper-stage plane bed and antidunes. In sand coarser than about 0.6–0.7 mm, there are no ripples: after motion begins, the bed first stays flat and is then covered with dunes. In very fine sand, on the contrary, the dune field narrows, and ripples are replaced by plane bed directly.

SchematicBedforms as a function of grain size and flow velocity at a constant depth. Schematic. The position of the boundaries depends on the depth and temperature of the water. As the depth increases, they shift to higher velocities.

Ripples and dunes belong to the lower flow regime. Upper-stage plane bed and antidunes belong to the upper flow regime. Antidunes need a Froude number of about 1. But in deep rivers, the transition from dunes to plane bed may occur at Fr noticeably below 1. So the flow regimes and the Froude numbers coincide only roughly.

Bedforms and the structures they leave

Bedform: Ripples
Structure in the deposit: small-scale cross-lamination
Scale: sets up to 4–5 cm thick

Bedform: Dunes
Structure in the deposit: large-scale cross-bedding
Scale: sets from decimetres to metres

Bedform: Lower-stage plane bed
Structure in the deposit: horizontal lamination in coarse sand
Scale: laminae a few grains thick

Bedform: Upper-stage plane bed
Structure in the deposit: horizontal lamination; on bedding planes, a fine primary current lineation
Scale: thin laminae; the rock splits along them

Bedform: Antidunes
Structure in the deposit: gentle cross-bedding, often inclined upcurrent
Scale: decimetres; rarely preserved

Bedform	Structure in the deposit	Scale
Ripples	small-scale cross-lamination	sets up to 4–5 cm thick
Dunes	large-scale cross-bedding	sets from decimetres to metres
Lower-stage plane bed	horizontal lamination in coarse sand	laminae a few grains thick
Upper-stage plane bed	horizontal lamination; on bedding planes, a fine primary current lineation	thin laminae; the rock splits along them
Antidunes	gentle cross-bedding, often inclined upcurrent	decimetres; rarely preserved

The shape of the crests decides the kind of cross-bedding. Straight, long crests are two-dimensional (2D) bedforms. They give planar (tabular) cross-bedding: the sets are bounded by almost parallel planes, and the laminae inside are planar. Curved, tongue-shaped crests are three-dimensional (3D) bedforms, formed in a stronger current. They scour hollows in front of them, and the sets become trough cross-bedding. In a section across the current, these are concave lenses. Along the current, the laminae are curved and flatten smoothly towards the base. In both cases the laminae dip downcurrent, and this is the basis for finding palaeocurrents.

A package of inclined laminae between two bounding surfaces is called a set. Several sets one above another make a coset. Reactivation surfaces cut off (truncate) laminae inside a set. They mark pauses and changes of the current, for example a fall and rise of the water level. Sets with opposite dips of laminae, one above another, make herringbone cross-bedding. It is a sign of a current that changes direction, such as a tidal one.

If grains settle quickly, each new bedform rises over the previous one and has no time to erode it. This makes climbing ripples and the climbing lamination they leave. The angle at which the crests climb is the angle of climb $\zeta$ . It shows how many times more slowly the bed was building up than the bedforms were moving.

The angle of climb of ripples

\tan\zeta = \frac{V_{agg}}{V_{mig}}

where:

$\zeta$ angle of climb — the inclination to the bedding of the line joining successive positions of the crests, °
$V_{agg}$ aggradation rate of the bed (how fast it builds up) — the thickness of sediment deposited per unit time, cm/min
$V_{mig}$ migration rate of the bedform, cm/min

Units: any identical velocity units

Holds when

The bedforms move at a constant speed and keep their shape.
The angle is measured in a section parallel to the current: in an oblique section it appears smaller.

If $\zeta$ is less than the slope of the stoss side, that side is eroded and only the lee-side laminae are preserved (subcritical climb); if it is greater, the stoss-side laminae are preserved as well and pass continuously over the crests (supercritical climb).

Worked exampleRate of deposition from climbing ripples

Given (assumed data)

Angle of climb measured in core along the current: 12°
Ripples: height 1.5 cm, length of the stoss side 12 cm
Ripple migration rate: from 0.5 to 2 cm/min (an assumption — an order of magnitude)

Solution

Slope of the stoss side: $\arctan(1.5/12) = 7.1°$ . The angle of climb of 12° is greater, so the stoss-side laminae are preserved too.
Ratio of the rates: $V_{agg}/V_{mig} = \tan 12° = 0.213$ — for every centimetre the bedform advances, the bed rises by 2.1 mm.
At $V_{mig}$ = 0.5 cm/min: $V_{agg} = 0.5\cdot 0.213 = 0.106$ cm/min, that is, $0.106\cdot 60 \approx 6.4$ cm/h; at 2 cm/min, 0.425 cm/min, about 25.5 cm/h.

Answer: The bed rose by 6.4–25.5 cm per hour: a bed a few tens of centimetres thick could have been deposited in a few hours.

What it means. Such rates occur when a lot of sediment is dropped quickly: on the falling stage of a flood, in breaches of levees (the raised natural banks beside a river), and in turbidity currents. We do not know the migration rate of ripples in an ancient rock, so the result is only an order of magnitude. What we can find reliably is the ratio of the rates and the very fact of rapid deposition. The data in this example are assumed for illustration.

Structures made by waves and wind

A wave on the water surface hardly carries water forward. The water particles move in closed orbits, and the size of the orbits decreases rapidly with depth. The bed “feels” the wave as long as noticeable motion reaches it, down to about half the wavelength (the distance from one crest to the next). This depth is called the wave base.

Decay of wave motion with depth

\frac{d_z}{d_0} = e^{-2\pi z/L},\qquad L = \frac{g\,T^2}{2\pi}

where:

$d_z/d_0$ the ratio of the diameter of the orbit of water particles at depth $z$ to the diameter at the surface, —
$z$ depth below the still-water level, m
$L$ wavelength in deep water, m
$T$ wave period, s

Units: SI

Holds when

Waves of small height in deep water (depth greater than about $L/2$ ).
In shallow water the orbits flatten into ellipses, and at the bed itself the water sways back and forth.

At a depth of $L/2$ , the amplitude of the motion is $e^{-\pi} \approx 4$ % of that at the surface. This is the conventional wave base. Whether grains move there also depends on the wave height and the grain size. So the actual base of ordinary waves is often shallower, while long-period storm waves reach much deeper.

Worked exampleThe wave base of ordinary and storm waves

Given (assumed data)

Ordinary waves: period 6 s
Storm waves: period 10 s

Solution

Length of ordinary waves: $L = 9.81\cdot 6^2/(2\pi) = 56.2$ m; base $L/2 \approx 28$ m.
Length of storm waves: $L = 9.81\cdot 10^2/(2\pi) = 156$ m; base $\approx 78$ m.
Amplitude of the orbits at a depth of $L/2$ : $e^{-\pi} = 0.043$ of that at the surface; at a depth of $L/4$ : $e^{-\pi/2} = 0.21$ .

Answer: Ordinary waves affect the bed down to about 28 m, storm waves down to 78 m.

What it means. Between the two bases, the waves leave the bed undisturbed in fair weather. Mud settles there, and burrowing organisms live there. In a storm the waves rework the bed and deposit sand with hummocky cross-stratification. This is how interbedding of bioturbated clays (clays stirred up by burrowing animals) and storm sands arises. Below the storm wave base there are no wave structures. The actual depths also depend on the wave height and the grain size, so the numbers are only a guide. The data in this example are assumed for illustration.

Wave ripples are formed by the back-and-forth swaying of water at the bed. They are almost symmetrical, with sharp crests that often branch. Their ratio of length to height is usually smaller than in current ripples, which means wave ripples are relatively steeper. The laminae inside them dip in both directions, converge in a herringbone pattern or drape the ripple from above. The crests of wave ripples run perpendicular to the direction in which the waves travelled — usually parallel to the shore. So they give a line, but not a direction.

In severe storms, waves and currents act together. On the bed between fair-weather wave base and storm wave base they form low mounds and hollows a few metres across. In section, these leave hummocky cross-stratification (HCS): gentle laminae, now convex, now concave, converging and diverging at low angles above an erosional base. If mainly the hollows are preserved, the stratification is called swaley (SCS). This is one of the main signs of storm deposits on a shelf (the shallow sea floor near a coast).

Wind carries sand mainly in leaps (saltation). Aeolian ripples (ripples made by wind) are very low and elongated, and their length is set by the length of the grain leap. Desert dunes — barchans, transverse, linear and star dunes — are from metres to hundreds of metres high. They leave the largest cross-bedded sets, from metres to tens of metres thick, with laminae of grain fall and grain flow and with erosion surfaces between the sets. Aeolian (wind-blown) and fluvial (river) deposits are described in the topic on continental environments, coastal and shelf deposits in the topic on marine environments.

Graded beds, gravity flows and sole marks

When a current carrying grains of different sizes weakens, the coarsest grains settle first, followed by ever finer ones. The result is a graded bed: coarse at the bottom, fine at the top — normal grading. The base of such a bed is usually erosional: at the start of the event the flow was strong and cut into the bed. Inverse grading — fine at the bottom, coarse at the top — is less common, for example in grain flows, where the large grains are pushed upwards by collisions.

Deposits of gravity flows

Flow: Debris flow
What supports the grains: the strength of the clay-rich mass
Signs of the deposit: poorly sorted, fragments “float” in clay, no grading or large fragments at the top

Flow: Grain flow
What supports the grains: collisions between grains
Signs of the deposit: thin beds a few centimetres thick with sharp boundaries, inverse grading

Flow: Liquefied flow
What supports the grains: pore water escaping upwards
Signs of the deposit: dish structures and water-escape pillars, almost no grading

Flow: Turbidity current
What supports the grains: turbulence
Signs of the deposit: graded beds with an erosional base and sole marks on it; the Bouma sequence

Flow	What supports the grains	Signs of the deposit
Debris flow	the strength of the clay-rich mass	poorly sorted, fragments “float” in clay, no grading or large fragments at the top
Grain flow	collisions between grains	thin beds a few centimetres thick with sharp boundaries, inverse grading
Liquefied flow	pore water escaping upwards	dish structures and water-escape pillars, almost no grading
Turbidity current	turbulence	graded beds with an erosional base and sole marks on it; the Bouma sequence

Schematic. In nature, the flows change from one kind into another on their way down the slope.

The deposit of a turbidity current is called a turbidite. In the ideal case it forms a Bouma sequence, which reflects how the flow dies away:

The Bouma sequence (from the bottom up)

Division: Ta
Structure: massive or graded sand
How it forms: rapid fallout from suspension without rolling along the bed

Division: Tb
Structure: horizontal lamination
How it forms: upper-stage plane bed

Division: Tc
Structure: small-scale cross-lamination, climbing ripples, convolute lamination
How it forms: ripples in the waning current

Division: Td
Structure: fine horizontal lamination of silt and clay
How it forms: settling from the tail of the flow

Division: Te
Structure: clay
How it forms: settling of suspension and ordinary accumulation of mud between events

Division	Structure	How it forms
Ta	massive or graded sand	rapid fallout from suspension without rolling along the bed
Tb	horizontal lamination	upper-stage plane bed
Tc	small-scale cross-lamination, climbing ripples, convolute lamination	ripples in the waning current
Td	fine horizontal lamination of silt and clay	settling from the tail of the flow
Te	clay	settling of suspension and ordinary accumulation of mud between events

Complete sequences are rare. Near the source, the lower divisions are thicker and the upper ones are eroded by the next flow. Far away, only the upper ones remain. Where turbidites accumulate and how submarine fans (fan-shaped bodies of sediment on the deep sea floor) are built is described in the topic on marine environments.

A strong current cuts into the bed at the start of the event. The large cuts are channels and scours. The small ones are the marks of eddies and of objects dragged by the current along the bed. They are cut into the top of the underlying clay and filled with the sand of the new bed. So they are preserved as convex casts on the base (the sole) of the sand bed. In Russian-language literature, such sole marks are called “hieroglyphs”.

Flute casts: tongue-shaped bulges with a blunt, deep end facing upcurrent and a gently widening tail pointing downcurrent; they give the direction.
Groove casts: straight long ridges made by objects dragged by the current along the bed; they show the line of the current but not which way it flowed.
Prod and bounce marks: short casts of objects that struck the bed: the asymmetric prod marks give the direction, the symmetrical bounce marks only the line.

Soft-sediment deformation and traces of life

Fresh sediment is loose and full of water, so it deforms easily: under its own weight, when it is shaken, when more sediment is deposited quickly on top of it, or on a slope. Soft-sediment deformation arises before the sediment becomes rock. It is usually confined to one bed or package of beds, and this is how it differs from tectonic deformation.

Load casts and flame structures: heavy sand is pressed into soft clay in rounded bags or “pillows”, and the clay is squeezed upwards between them in sharp tongues. The tongues are often bent downcurrent.
Convolute lamination: laminae crumpled into small folds within an undistorted bed. It is a trace of liquefaction (sediment turning into a fluid slurry) during or immediately after deposition, and is common in turbidites.
Dish and pillar structures and sand volcanoes: traces of water squeezed upwards out of compacting sand.
Slump folds (slumps): packages of beds crumpled by sliding down a slope. They are overturned downslope and show its direction.
Sand dykes and sills (injectites): sand forced under excess pressure into fractures in the overlying clays, as vertical sheets (dykes) or horizontal ones (sills). In some basins such bodies form reservoirs in their own right and paths for crossflow (flow from one layer to another) through the seal.
Desiccation cracks: a polygonal network of cracks in clay dried in air, filled with sand from above. They are a sign of drying out and a reliable way to tell which side is the top.

Many bottom-dwelling animals burrow, crawl over the bed and feed on the sediment. Their traces are called trace fossils (ichnofossils): burrows, tracks, trails and imprints. The mixing of sediment by organisms is called bioturbation. We classify traces first of all by the behaviour that left them: dwelling burrows, feeding burrows, crawling and grazing traces, resting traces, and escape traces (made by animals digging out from under rapidly deposited sediment).

The set of traces depends on water energy, sedimentation rate, firmness of the bed, oxygen and food. So sets of traces, called ichnofacies, are a clue to the environment. Vertical dwelling burrows in clean sand mean mobile shallow water (the Skolithos ichnofacies). Horizontal feeding burrows in silts and clays mean a quiet shelf below fair-weather wave base (Cruziana). Complex burrows in the muds of quiet, poorly oxygenated water belong to Zoophycos. Patterned grazing traces mean the deep-sea floor (Nereites). Ichnofacies tell us about conditions, not directly about depth: “deep-water” conditions occur in shallow stagnant basins as well.

The bioturbation index

Index: 0
Reworked by organisms: nothing
What is seen: bedding distinct, no burrows

Index: 1
Reworked by organisms: up to 5%
What is seen: isolated burrows

Index: 2
Reworked by organisms: 5–30%
What is seen: burrows noticeable, bedding distinct

Index: 3
Reworked by organisms: 30–60%
What is seen: many burrows, bedding disturbed in places

Index: 4
Reworked by organisms: 60–90%
What is seen: bedding visible only in fragments

Index: 5
Reworked by organisms: 90–99%
What is seen: bedding almost obliterated

Index: 6
Reworked by organisms: 100%
What is seen: sediment completely mixed

Index	Reworked by organisms	What is seen
0	nothing	bedding distinct, no burrows
1	up to 5%	isolated burrows
2	5–30%	burrows noticeable, bedding distinct
3	30–60%	many burrows, bedding disturbed in places
4	60–90%	bedding visible only in fragments
5	90–99%	bedding almost obliterated
6	100%	sediment completely mixed

A scale commonly used in core description; in different versions the limits of the lower classes differ slightly.

For a reservoir, bioturbation works both ways. When animals mix sand with clay, they spoil the sorting and plug the pores, so permeability falls. When they destroy thin clay laminae and make vertical burrows filled with sand, they do the opposite: they connect beds and raise the vertical permeability. In carbonates, networks of burrows sometimes become the main paths of flow. So we enter the bioturbation index and the type of burrows in the core description, and take them into account when we identify rock types for a reservoir model.

Palaeocurrents and way-up indicators

The direction of ancient currents, or palaeocurrents, shows where the material came from and how the sand bodies are elongated. Channel sands are elongated along the palaeocurrent, and coastal bars along the ancient shoreline. We use this to choose where to site exploration and production wells and how to orient horizontal wellbores.

Palaeocurrent indicators

Structure: Cross-bedding
What is measured: dip azimuth of the cross-laminae, axes of troughs
Gives: direction

Structure: Current ripples
What is measured: azimuth of the steep side
Gives: direction

Structure: Flute casts
What is measured: axis; the blunt end points upcurrent
Gives: direction

Structure: Pebble imbrication (flat pebbles overlapping like roof tiles)
What is measured: dip of flat pebbles — upcurrent
Gives: direction

Structure: Groove casts, primary current lineation
What is measured: strike
Gives: line only

Structure: Wave ripples
What is measured: strike of the crests
Gives: line only (the waves travelled perpendicular to it)

Structure	What is measured	Gives
Cross-bedding	dip azimuth of the cross-laminae, axes of troughs	direction
Current ripples	azimuth of the steep side	direction
Flute casts	axis; the blunt end points upcurrent	direction
Pebble imbrication (flat pebbles overlapping like roof tiles)	dip of flat pebbles — upcurrent	direction
Groove casts, primary current lineation	strike	line only
Wave ripples	strike of the crests	line only (the waves travelled perpendicular to it)

The azimuths of individual sets are scattered. Within a channel, the laminae dip in different directions around the main one, and curved bedforms give a whole fan of azimuths. So we measure tens of sets and average them. The arithmetic mean of the azimuth numbers will not do for this. The true mean of 350° and 10° is 0°, but plain arithmetic gives 180°. Instead we add the azimuths as unit vectors (arrows of length 1).

Mean azimuth and mean resultant length

\bar\theta = \operatorname{atan2}\Big(\sum_i \sin\theta_i,\ \sum_i \cos\theta_i\Big),\qquad \bar R = \frac{1}{n}\sqrt{\Big(\sum_i \sin\theta_i\Big)^2 + \Big(\sum_i \cos\theta_i\Big)^2}

where:

$\theta_i$ azimuths of individual measurements, from north clockwise, °
$\bar\theta$ mean azimuth — the direction of the sum of the unit vectors, °
$\bar R$ mean resultant length: from 0 (scatter in all directions) to 1 (all measurements coincide), —
$n$ number of measurements, —

Units: degrees; $\bar R$ is dimensionless

Holds when

The measurements are of one kind of structure in one formation; indicators of different kinds are averaged separately.
The distribution is unimodal (it has one peak). If there are two directions, as with ebb and flood tides, the overall mean is meaningless.
The azimuths are taken after the structural tilt of the beds has been removed (advanced section).

The sum of the sines is the east component, and the sum of the cosines is the north component. The function atan2 returns the angle with the signs of both sums taken into account, so the angle falls in the correct quadrant (quarter) of the circle.

Worked exampleThe mean palaeocurrent direction from ten sets

Given (assumed data)

Dip azimuths of cross-laminae, °: 340, 355, 5, 12, 20, 28, 35, 48, 60, 15

Solution

Arithmetic mean of the numbers: $918/10 = 91.8°$ — pointing east, although almost all the measurements are directed north.
Sum of the sines (east component): 3.12; sum of the cosines (north component): 8.69.
Mean azimuth: $\bar\theta = \operatorname{atan2}(3.12;\ 8.69) = 20°$ .
Length of the sum: $R = \sqrt{3.12^2 + 8.69^2} = 9.23$ ; $\bar R = 9.23/10 = 0.92$ .

Answer: The palaeocurrent was directed towards 20° — north-north-east; the scatter is small:

\bar R

= 0.92.

What it means. A channel body with such a palaeocurrent is most likely elongated from south-south-west to north-north-east. It is logical to site delineation wells (wells that mark out the edges of a discovery) along this line and on both sides of it. A high $\bar R$ (the measurements are tightly grouped) shows a well-defined direction. A simple test helps to check that it is not accidental (advanced section). The data in this example are assumed for illustration.

CalculatedA rose diagram (a circular histogram) of the azimuths of the example. The petals show the number of measurements in 30° sectors, the dots on the circle are individual measurements, and the arrow is the mean azimuth.

Many structures show which side of a bed was its top. This matters when the sequence is steeply tilted or overturned in a fold, and in core when we do not know how it lay:

in a graded bed with normal grading, the coarse part is at the bottom
cross-laminae are truncated at the top and flatten smoothly downwards, towards the base of the set
flute and groove casts and load casts are at the base of a sand bed and are convex downwards
desiccation cracks taper downwards and are filled with material from the overlying bed
escape traces of organisms run from bottom to top through a bed deposited all at once

The topic Stratigraphy and geologic time covers the principles for working out the order of beds and their age.

Structures in core and on borehole images

Core is the only direct sample of the structures of a reservoir. After the core is slabbed (sawn lengthwise), the flat surface is photographed and described: beds and their boundaries, structures, the bioturbation index, and grain size by interval, usually at a scale of 1:20–1:50 (core scanning and description). X-ray tomography (CT scanning) shows bedding even where the rock looks uniform on the surface. Core has one limitation: its small diameter. A large cross-bedded set metres wide is seen in it only as a fragment. The azimuth of the laminae is unknown unless the core is oriented. We orient core with a special tool while it is being cut, or by matching it with an image of the same interval.

Borehole imagers build a picture of the borehole wall. Electrical micro-imagers press pads with dozens of small electrodes against the wall. They record variations of resistivity (how strongly the rock resists electric current) in millimetre detail. In a borehole about 200 mm in diameter, they cover most of the circumference (logging tools). Classical electrical imagers need a conductive water-based mud. Acoustic scanners use a rotating transducer. They record the amplitude and travel time of the reflected signal around the whole circumference. They work in any mud, but see details more coarsely. Ordinary logs, with a resolution of tens of centimetres, do not distinguish individual laminae.

We unroll the image of the wall onto a flat plane. Azimuth runs along the horizontal axis, from north through east, south and west back to north. Depth runs along the vertical axis. An inclined plane, such as a bed boundary or a cross-lamina, cuts the cylindrical borehole along an ellipse. On the unrolled image it becomes a sinusoid (a smooth wave). Its lowest point lies on the side towards which the plane dips. The steeper the dip, the greater its peak-to-trough height.

Dip of a plane from the sinusoid on the unrolled borehole image

\tan\delta = \frac{h}{d}

where:

$\delta$ dip angle of the plane, °
$h$ peak-to-trough height of the sinusoid — the difference between the depths of its lowest and highest points, m
$d$ borehole diameter from the caliper log, m

Units: any identical length units

Holds when

The borehole is vertical and round. In an inclined borehole this gives the dip relative to the borehole axis, and the true dip is found by recalculating with the inclination and azimuth of the borehole (attitude (strike and dip)).
The plane cuts the whole borehole; its dip azimuth is the azimuth of the lowest point of the sinusoid.

A horizontal bed gives a straight line. At a dip of 45°, the peak-to-trough height of the sinusoid equals the borehole diameter. From hundreds of sinusoids we build a plot of dips against depth, called a “tadpole plot”. Each circle is plotted at the dip angle, and its tail points to the dip azimuth.

Worked exampleDip of bedding and cross-laminae from an image

Given (assumed data)

Borehole diameter from the caliper log: 216 mm
Sinusoid 1 (bed boundary): peak-to-trough height 19 mm, lowest point at azimuth 200°
Sinusoid 2 (a cross-lamina in the set above): peak-to-trough height 105 mm, lowest point at azimuth 40°

Solution

Bed boundary: $\delta = \arctan(19/216) = 5.0°$ towards azimuth 200° — this is the structural dip of the formation.
Cross-lamina: $\delta = \arctan(105/216) = 25.9°$ towards azimuth 40°.
Rotating the formation so that the bed boundary becomes horizontal (advanced section), we get a dip of the lamina of 30.7° towards azimuth 37°.

Answer: The formation is tilted by 5.0° to the south-south-west, and the flow that deposited this set went to the north-east — towards an azimuth of about 37°.

What it means. Gentle sinusoids repeat the structural dip. Steep sinusoids of a different azimuth inside a set are cross-laminae. Here the structural tilt is small. The correction turned the azimuth of the lamina by only 3°, but it increased the dip by almost 5°, to an angle close to the angle of repose of sand. From such measurements in many sets we build a rose diagram of palaeocurrents, as in the previous section. The data in this example are assumed for illustration.

CalculatedThe unrolled wall of a 216 mm borehole for the data of the example. Gentle sinusoids are the boundaries of beds and of the set (5° towards azimuth 200°). Steep ones are cross-laminae inside the set (26° towards azimuth 40°). The peak-to-trough height

h

and the lowest point are shown on one lamina.

On a plot of dips, the structures give patterns that you can learn to recognise. Beds of constant dip repeat the structural tilt. Inside a cross-bedded set, the dip of the laminae decreases downwards, because the laminae flatten towards the base of the set. The azimuth stays roughly constant: this is the palaeocurrent azimuth for that set. Bioturbated intervals give a mottled image without continuous sinusoids. Massive sandstones give a uniform image. Erosion surfaces appear as sinusoids that cut off the underlying laminae.

The topic on facies and sedimentary logs describes how structures combine through a section into facies and their successions. The topic on well correlation describes how the sections of wells are correlated.

Terms and symbols

Bed and lamina: A body bounded by bedding surfaces, and a millimetre-thick layer within a bed; the dividing line is about 1 cm.
Bedforms: Ripples, dunes, antidunes and plane bed, which a current creates on a sandy bed.
Cross-bedding: Laminae inclined to the bed boundaries; the trace of the lee sides of migrating bedforms; small-scale from ripples, large-scale from dunes.
Set and coset: Cross-laminae between two bounding surfaces; several sets one above another.
Shields number $\theta$: The ratio of the bed shear stress of the flow to the weight of a submerged layer of grains; grains move at $\theta > \theta_{cr}$ .
Froude number Fr: $U/\sqrt{gh}$ ; below 1 the flow is subcritical, above 1 supercritical.
Graded bedding: A decrease (normal) or increase (inverse) of grain size upwards through a bed.
Bouma sequence: The ideal section of a turbidite: Ta — massive sand, Tb — horizontal lamination, Tc — small-scale cross-lamination, Td — fine laminae, Te — clay.
Sole marks (“hieroglyphs”): Marks on the bases of sand beds: flute casts, groove casts, prod marks, load casts.
Bioturbation: Mixing of sediment by burrowing organisms; the index runs from 0 to 6.
Palaeocurrent: The direction of an ancient current, found from cross-bedding, sole marks and pebble imbrication.
Borehole image log: An image of the borehole wall from resistivity or sound reflection; inclined planes on it are sinusoids.

Common mistakes and misconceptions

Wrong

“Cross-laminae dip upcurrent.”

In fact

Cross-laminae are former lee sides of bedforms, so they dip downcurrent. Flat pebbles of imbrication and, often, the laminae of antidunes are inclined upcurrent.

Wrong

“Azimuths can be averaged like ordinary numbers.”

In fact

For 350° and 10° the arithmetic mean gives 180° — the opposite direction. Azimuths are added as unit vectors (the sum of the sines and the sum of the cosines).

Wrong

“A massive sandstone was deposited without any bedding.”

In fact

A lack of structure may be primary (rapid fallout from suspension, as in division Ta of a turbidite) or secondary (the bedding has been destroyed by organisms or liquefaction). Sometimes the bedding is visible only on a tomogram or an image.

Wrong

“The peak-to-trough height of a sinusoid on the image of an inclined borehole immediately gives the true dip.”

In fact

In an inclined borehole the sinusoid gives the dip relative to the borehole axis. We find the true dip and azimuth by recalculating with the inclination and azimuth of the well.

Wrong

“Bioturbation always makes a reservoir worse.”

In fact

It can mix in clay and spoil the sorting, but it can also destroy clay laminae and create vertical flow paths — then the vertical permeability grows.

In depth: advanced detail

How the structural tilt is removed

Beds that built up almost horizontally were later tilted by tectonics. To find the original direction of the cross-laminae, we rotate the formation back in our minds: about the strike line of the beds (a horizontal line along the bed), through the angle of their dip, until the beds are horizontal. The most convenient way is to rotate the normal vector of the plane (the arrow standing at right angles to it).

The normal to a plane and its rotation

\mathbf n = (\sin\alpha\sin\delta,\ \cos\alpha\sin\delta,\ \cos\delta),\qquad \mathbf n' = \mathbf n\cos\beta + (\mathbf k\times\mathbf n)\sin\beta + \mathbf k\,(\mathbf k\cdot\mathbf n)(1 - \cos\beta)

where:

$\mathbf n$ the unit upward normal to the plane: components to the east, to the north and up, —
$\delta,\ \alpha$ dip angle and dip azimuth of the plane, °
$\mathbf k$ the unit vector along the strike of the beds, chosen so that the rotation returns the beds to the horizontal, —
$\beta$ dip angle of the beds — the angle of rotation, °

Units: dimensionless vectors, angles in degrees

Holds when

The formation was tilted by rotation about a horizontal axis, without folds with an inclined hinge.
The beds were deposited horizontally. On the slopes of deltas and in dunes, an original dip of a few degrees remains as an error of the method.

After the rotation the dip of the lamina is $\delta' = \arccos n'_z$ and the azimuth is $\alpha' = \operatorname{atan2}(n'_x,\ n'_y)$ . Image-processing software and stereographic nets (a graphical method on a circular chart) do the same calculation.

Worked exampleHow much the tilt of the formation distorts the azimuth of cross-laminae

Given (assumed data)

Cross-lamina measured today: dip 25° towards azimuth 40°
Tilt of the formation: a) 20° towards azimuth 40° — along the palaeocurrent; b) 10° towards azimuth 130°; c) 20° towards azimuth 130° — across it

Solution

Case a: after the rotation the lamina dips at 5° towards azimuth 40°: the azimuth has not changed, the angle has decreased by 20°.
Case b: 26.8° towards azimuth 20° — the azimuth has shifted by 20°.
Case c: 31.6° towards azimuth 4° — a shift of 36°.

Answer: A tilt along the palaeocurrent changes only the dip of the laminae. A tilt across it changes above all their azimuth: by 20° at 10° and by 36° at 20°.

What it means. That is why we always remove the structural tilt, even if it seems small. We cannot guess its effect in advance: what matters is not only the angle but also the direction of the tilt relative to the palaeocurrent. The data in this example are assumed for illustration.

River depth from the thickness of cross-bedded sets

A dune that migrates while the bed builds up only slowly (weak aggradation) leaves only the lower part of its height, because the next dune cuts off the top. Flume experiments and observations in rivers have shown two things. The mean height of dunes is about 2.9 times the mean thickness of the sets they leave. And the height of dunes is usually from a sixth to a tenth of the flow depth. So we can estimate the depth of the palaeochannel (the ancient river channel) from core.

Dune height and flow depth from the thickness of sets

h_d \approx 2.9\,\bar s\ \ (2.2\ldots 3.6),\qquad H \approx (6\ldots 10)\,h_d

where:

$\bar s$ mean thickness of the dune cross-bedded sets in the interval, m
$h_d$ mean height of the dunes, m
$H$ flow depth, m

Units: any identical length units

Holds when

The sets were left by dunes under weak aggradation, and their thicknesses are markedly scattered (the coefficient of variation is about 0.9 ± 0.3).
The ratio of depth to dune height differs between rivers. In large rivers dunes may be lower than a tenth of the depth. So the estimate is rough, within about a factor of two.

The depth of the channel is the first clue to the thickness of a single channel sand body. We estimate the width of channels from other relationships (fluvial deposits).

Worked exampleRiver depth from the sets in core

Given (assumed data)

Mean thickness of the trough cross-bedded sets in a channel sandstone: 0.16 m (from 14 sets)

Solution

Dune height: $h_d \approx 2.9\cdot 0.16 = 0.46$ m, within the range from 0.35 to 0.58 m.
Flow depth: from $6\cdot 0.46 = 2.8$ to $10\cdot 0.46 = 4.6$ m.
Allowing for the scatter of both coefficients, from 2.1 to 5.8 m.

Answer: The river was about 2.8–4.6 m deep (the extreme estimates are 2.1–5.8 m).

What it means. A single channel body of such a river probably has a thickness of about the channel depth: 3–5 m. Thicker sandstones in a section are made up of several stacked channels. The estimate is rough, but it distinguishes a shallow river from a large one, where the sands of one channel would be tens of metres thick. The data in this example are assumed for illustration.

Could the mean direction be accidental?

Even randomly scattered azimuths give some mean vector. The Rayleigh test checks that the direction is real. For $n$ independent measurements, we compare the quantity $z = n\bar R^2$ with what a uniform scatter would give. For $n$ of ten and more, the probability of getting such a $z$ or a greater one by chance is about $e^{-z}$ . The direction is considered significant at the 5% level (less than a 5% chance that it is accidental) if $z$ is greater than about 3. For large $z$ , this estimate is far too high (for ten measurements, by tens of times). That is, it is cautious: the true probability is even lower.

Worked exampleTesting the palaeocurrent of the example

Given (assumed data)

Measurements: $n$ = 10, $\bar R$ = 0.92

Solution

$z = 10\cdot 0.92^2 = 8.5$ — much greater than 3.
The probability of a chance coincidence is no more than $e^{-8.5} \approx 0.0002$ .
For comparison: with the same 10 measurements and $\bar R$ = 0.5 we would get $z = 2.5$ and $e^{-2.5} \approx 0.08$ — the direction could not be considered proven.

Answer: The palaeocurrent direction is significant: a random scatter would give such a concentration with a probability of no more than 0.0002.

What it means. The measurements must be independent. Several laminae of one set are in effect one measurement, so we average them by set first. The data in this example are assumed for illustration.

Connections to other topics

Sediment texture: grain size, sorting and shape — size, sorting and shape of grains in laminae; the settling velocity of particles.
Sedimentary rocks and their classification — names of the rocks that make up the beds.
Facies, facies successions and sedimentary logs — how structures combine into facies and successions through a section.
Continental environments: aeolian and fluvial — structures and bodies of fluvial and aeolian sands.
Coastal, shallow-marine and deep-marine environments — storm beds, tidal flats and turbidites.
Reservoir heterogeneity and permeability averaging — how bedding and interbeds create heterogeneity and anisotropy of permeability.
Core analysis — description, scanning and depth-matching of core.
Stress, strain and structural analysis — attitude (strike and dip), true and apparent dip.

Check yourself

1
Inside a sand bed the laminae are inclined at 25° towards the east, and the sets are about 3 cm thick. What is this structure, and where did the current flow?
1. A.Small-scale ripple cross-lamination; current towards the west
2. B.Wave ripples; the direction cannot be found
3. C.Large-scale dune cross-bedding; current towards the west
4. D.Small-scale ripple cross-lamination; current towards the east
Show the answer
Correct answer: D. Small-scale ripple cross-lamination; current towards the east
Sets thinner than 4–5 cm are left by ripples, and cross-laminae — former lee sides of bedforms — dip downcurrent, so the current flowed towards the east. Wave ripples would give laminae inclined in both directions.
2
Sand of mean size 0.4 mm, and a current that gradually strengthens from the threshold of motion at constant depth. In what order will the bedforms change?
1. A.Dunes → ripples → plane bed
2. B.Ripples → dunes → upper-stage plane bed → antidunes
3. C.Plane bed → dunes → ripples → antidunes
4. D.Antidunes → dunes → ripples
Show the answer
Correct answer: B. Ripples → dunes → upper-stage plane bed → antidunes
In sand finer than about 0.6–0.7 mm ripples appear first, then dunes, then the bed flattens (upper-stage plane bed), and at a Froude number of about 1 antidunes form. Lower-stage plane bed instead of ripples occurs only in coarser sand.
3
Which of the indicators gives only the line of the palaeocurrent but not its direction?
1. A.Flute casts
2. B.Groove casts
3. C.Pebble imbrication
4. D.Cross-bedding
Show the answer
Correct answer: B. Groove casts
A groove is left by an object dragged along the bed, and it looks the same for a current in either direction along it. In flute casts the blunt end faces upcurrent, cross-laminae dip downcurrent, and flat pebbles are inclined upcurrent.
4
A sand bed is crossed by continuous clay films 1 mm thick after every 10 mm of sand. The permeability of the sand is 500 mD, of the films 0.01 mD. Estimate the horizontal and vertical permeability.
Show the answer
Along the layers we take the average weighted by thickness: $(10\cdot 500 + 1\cdot 0.01)/11 \approx 455$ mD. Across them we take the harmonic average: $11/(10/500 + 1/0.01) = 11/100.02 \approx 0.11$ mD, that is, about four thousand times less. If the films are discontinuous, as in flaser bedding, the flow bypasses them through the gaps, and the vertical permeability is orders of magnitude higher than this estimate. What decides is not only the share of clay but also how continuous the films are.
5
On an image of a vertical borehole 0.22 m in diameter, the sinusoid of a cross-lamina has a peak-to-trough height of 0.13 m and its lowest point is at azimuth 250°. What are the dip of the lamina and the direction of the current, if the structural tilt is negligible?
Show the answer
$\tan\delta = 0.13/0.22 \approx 0.59$ , $\delta \approx 31°$ . The dip azimuth is given by the lowest point of the sinusoid, 250°. The laminae dip downcurrent, so the flow went to the west-south-west (about 250°). If there is a noticeable structural tilt, we must remove it first.
6
Why do beds of sand with hummocky cross-stratification alternate with bioturbated clays in shelf sections?
Show the answer
Such interbedding accumulates between fair-weather wave base and storm wave base. In calm weather no wave motion reaches that depth. Mud settles, and burrowing organisms rework it. In a storm, long-period waves reach the bed, erode it and deposit sand with hummocky cross-stratification above an erosional base. Each “sand–clay” pair is one storm and the calm that followed.

Updated: 19 October 2026