The median grain size of a sand is φ50 = 2.5. What is its median diameter and its Udden–Wentworth class?

0.18 mm, fine sand, since d = 2^(−2.5) mm lies between 0.125 and 0.25 mm

0.35 mm, medium sand, since d = 2^(−1.5) mm lies between 0.25 and 0.5 mm

0.18 mm, fine sand, since d = 2^(−2.5) mm lies between 0.125 and 0.25 mm

0.18 mm, very fine sand, since sands finer than 0.25 mm are called very fine

5.7 mm, pebble, since d = 2^(2.5) mm lies between 4 and 64 mm

Which of the following statements are correct?

Larger φ values correspond to smaller grains

A positive Folk and Ward skewness indicates a tail of fine grains

Larger φ values correspond to smaller grains

A positive Folk and Ward skewness indicates a tail of fine grains

A graphic standard deviation σI of 0.3 indicates poor sorting

The arithmetic mean of palaeocurrent azimuths 350° and 10° correctly gives the mean direction, 180°

A river 2.0 m deep flows at a mean speed of 4.0 m/s. Take g = 9.8 m/s² and ν = 1.0 × 10⁻⁶ m²/s, and use Re = Uh/ν and Fr = U/√(gh). Which description is correct?

Subcritical (Fr ≈ 0.90) and turbulent (Re ≈ 8 × 10⁶)

Supercritical (Fr ≈ 1.8) and turbulent (Re ≈ 8 × 10⁶)

Subcritical (Fr ≈ 0.90) and laminar (Re ≈ 8 × 10²)

Supercritical (Fr ≈ 1.1) and laminar (Re ≈ 8 × 10³)

In the Stokes regime the settling velocity of a sphere, with density and viscosity fixed, is proportional to

the square of the diameter, d²

the square root of the diameter, √d

the square of the diameter, d²

In a flume with sand of about 0.3 mm at constant depth, the flow velocity is raised steadily from below the threshold of motion until the Froude number exceeds 1. In which order do the bedforms appear?

ripples, dunes, upper plane bed, antidunes

dunes, ripples, antidunes, upper plane bed

ripples, antidunes, dunes, upper plane bed

upper plane bed, dunes, ripples, antidunes

ripples, dunes, upper plane bed, antidunes

A 25 cm thick sandstone bed has an erosional base with flute casts. From the base upward it shows parallel lamination, ripple cross-lamination, thin parallel-laminated silt and then mud, with no massive graded division at the base. Which interpretation is best?

A turbidity-current deposit with Ta missing, from a decelerating and more distal flow

A debris flow, because the bed is thick, has an erosional base and a sharp contact with the mud above

A turbidity-current deposit with Ta missing, from a decelerating and more distal flow

A fair-weather beach deposit, because parallel lamination and ripples are both made by waves

A grain flow on a dune slip face, because the bed has a sharp base and shows internal parallel lamination

A sandstone has framework grains of quartz 88%, feldspar 7% and lithic fragments 5%, and 4% of the rock is clay matrix. In the Pettijohn–Dott scheme it is a

subarkose, because it is an arenite with 75–95% quartz and more feldspar than lithics

quartz arenite, because it is very rich in quartz and has little matrix between the grains

arkose, because it contains feldspar, the diagnostic mineral of arkosic sandstones

subarkose, because it is an arenite with 75–95% quartz and more feldspar than lithics

wacke, because it contains some clay matrix between the framework grains

A limestone consists of ooids that touch one another, with lime mud filling part of the space between them. In Dunham's classification it is a

packstone, because the grains support each other and lime mud is present between them

grainstone, because the rock is made of ooids and shows high-energy deposition with winnowed sediment

wackestone, because lime mud is present among more than 10% grains

boundstone, because the ooids are cemented together by early carbonate cement

Which of the following statements about facies and cycles are correct?

Walther's law links lateral facies to vertical order only where no break, such as an unconformity, intervenes

A fining-upward sequence of channel lag, cross-bedded sand and mud is typical of a meandering-river point bar

Walther's law links lateral facies to vertical order only where no break, such as an unconformity, intervenes

A fining-upward sequence of channel lag, cross-bedded sand and mud is typical of a meandering-river point bar

A coarsening-upward profile, with sand at the top and mud at the base, is the typical signature of a meandering-river point bar

Autocyclic cycles are driven by Milankovitch changes in insolation and for that reason correlate between basins

In a carbonate succession, thin cycles that each represent about 20 kyr of precession occur in bundles of five. A bundle most likely records

about 100 kyr of short eccentricity, giving the 5:1 bundling ratio

about 41 kyr of obliquity, which is the next orbital period after precession

about 400 kyr of long eccentricity, because bundles of five always mark that period

about 20 kyr of autocyclic channel switching, repeated five times by chance

about 100 kyr of short eccentricity, giving the 5:1 bundling ratio

The youngest concordant detrital zircon U–Pb ages in a sandstone cluster at 520 Ma, with no younger grains. What does this tell us?

The sandstone cannot be older than about 520 Ma, so 520 Ma is a maximum depositional age

The sandstone was deposited at 520 Ma exactly, because zircon crystallises at the time of deposition

The sandstone cannot be older than about 520 Ma, so 520 Ma is a maximum depositional age

The sandstone cannot be younger than about 520 Ma, so 520 Ma is a minimum depositional age

The source terrane was metamorphosed at 520 Ma, because detrital zircon records only metamorphism

Which of the following statements about sedimentary basins are correct?

A foreland basin is an asymmetric wedge, deepest next to the thrust belt, where the lithosphere flexes under the thrust load

A passive-margin basin subsides mainly by thermal contraction after an earlier stage of fault-controlled stretching

A foreland basin is an asymmetric wedge, deepest next to the thrust belt, where the lithosphere flexes under the thrust load

A passive-margin basin subsides mainly by thermal contraction after an earlier stage of fault-controlled stretching

A foreland basin is deepest at its outer margin, far from the thrust load, and shallows towards the orogen

Pull-apart basins form at restraining bends of strike-slip faults, where compression thickens the crust and builds uplift

A river 1.5 m deep flows over a bed of 2 mm quartz gravel with a water-surface slope of 0.0002. Take ρ = 1000 kg/m³, ρs = 2650 kg/m³, g = 9.8 m/s², τ₀ = ρgh·S, a Shields parameter θ = τ₀/[(ρs − ρ)gD] and a critical value θc = 0.045. Will the gravel move?

Yes, because τ₀ ≈ 2.9 Pa gives θ ≈ 0.09, above the critical value

No, because τ₀ ≈ 2.9 Pa gives θ ≈ 0.03, below the critical value

No, because the flow is subcritical, so no grain can move whatever the stress

Yes, because τ₀ ≈ 2.9 Pa gives θ ≈ 0.09, above the critical value

Yes, because any turbulent flow moves 2 mm gravel whatever the Shields parameter

Sedimentology: Texture and Statistics, Fluid Flow and Structures, Sedimentary Rocks, Facies and Cycles, Provenance and Basins

This chapter answers Part B1, “Sedimentology”, of the Geology and Geophysics (GG) paper: “Texture, structure, sedimentary processes, physical principles related to fluid flow; Statistical analysis in sedimentology; Petrology of common sedimentary rocks; Sedimentary facies and environments, cyclicities in sedimentary succession; Provenance and basin analysis.” It teaches grain size, the phi scale and graphic statistics, circular statistics for palaeocurrents, fluid flow (Reynolds and Froude numbers, the Shields criterion, Stokes' law), bedforms, sedimentary structures and gravity flows, the classification of sandstones, mudrocks and carbonates, facies models and cycles, provenance tools and basin analysis. The numericals are phi conversion, Folk and Ward sorting, the vector mean of palaeocurrents, Stokes settling velocity, Bouma-bed arithmetic and basin subsidence.

1. Texture, grain size and statistical analysis

Texture comprises grain size, sorting, shape (sphericity and roundness) and fabric. Grain size follows the Udden–Wentworth scale: gravel above 2 mm, sand from 2 mm to 1/16 mm (0.0625 mm), silt from 1/16 to 1/256 mm and clay below. The phi scale is logarithmic: φ = −log₂ d (d in mm), so d = 2^(−φ), and larger φ means finer grains. 2 mm is −1φ, 1 mm is 0φ, 0.5 mm is 1φ, 0.25 mm is 2φ, 0.125 mm is 3φ and 0.0625 mm is 4φ, so sand spans −1 to 4φ. Sand is sieved at half-phi intervals; silt and clay are sized by settling (Stokes) or laser diffraction; grains measured in thin section give apparent diameters that underestimate the true size.

From the cumulative curve read the percentiles φ5, φ16, φ25, φ50, φ75, φ84 and φ95. The Folk and Ward graphic measures are: mean Mz = (φ16 + φ50 + φ84)/3; sorting (inclusive graphic standard deviation) σI = (φ84 − φ16)/4 + (φ95 − φ5)/6.6; skewness SkI = (φ16 + φ84 − 2φ50)/[2(φ84 − φ16)] + (φ5 + φ95 − 2φ50)/[2(φ95 − φ5)]; kurtosis KG = (φ95 − φ5)/[2.44(φ75 − φ25)]. Sorting classes in φ: below 0.35 very well sorted, 0.35–0.50 well, 0.50–0.71 moderately well, 0.71–1.00 moderately, 1.00–2.00 poorly, 2.00–4.00 very poorly, above 4.00 extremely poorly sorted. KG near 1 is mesokurtic, well above 1 leptokurtic (peaked) and well below 1 platykurtic.

⚠️ Phi runs backwards

A larger φ means a smaller grain, so a positive skewness in φ units means a tail of fines (excess fine material), not of coarse grains. A sorting value σI of 0.3 is very well sorted: a smaller σI means better sorting.

Maturity measures how far a sand has been worked. Textural maturity (Folk) runs from immature (clay matrix above about 5%) through submature (clay-free but poorly sorted) and mature (well sorted, not well rounded) to supermature (well sorted and well rounded). Compositional maturity is high in quartz-rich sands from which feldspar and rock fragments have gone, and the ZTR index (zircon, tourmaline, rutile) rises with recycling.

Palaeocurrent data are circular and cannot be averaged arithmetically: 350° and 10° average to 0°, not 180°. First restore tilted beds to horizontal, then use the vector mean θ̄ = atan2(Σ sin θᵢ, Σ cos θᵢ) and the mean resultant length R̄ = √[(Σ sin θᵢ)² + (Σ cos θᵢ)²]/n, which runs from 0 (no preferred direction) to 1 (all readings identical). Rose diagrams show the distribution: unimodal for most rivers, bimodal and opposed (herringbone) for tides. Rayleigh's test asks whether R̄ exceeds what random scatter would give.

2. Fluid flow, entrainment, settling and bedforms

A fluid has density ρ and dynamic viscosity μ (Pa·s); kinematic viscosity is ν = μ/ρ. Water at about 20 °C has ρ ≈ 1000 kg/m³ and μ ≈ 1.0 × 10⁻³ Pa·s. Water and air are Newtonian; debris flows and muds are non-Newtonian (Bingham: they have a yield strength). The Reynolds number Re = ρUh/μ = Uh/ν compares inertial and viscous forces; in open channels (h = flow depth) flow is laminar below about 500 and turbulent above about 2000, and natural flows are almost always turbulent (U = 1 m/s, h = 1 m gives Re ≈ 10⁶). The Froude number Fr = U/√(gh) compares flow speed with the speed √(gh) of a shallow-water wave: Fr < 1 is subcritical (waves can travel upstream), Fr > 1 supercritical, and the change from supercritical to subcritical is a hydraulic jump.

For steady uniform flow the bed shear stress is τ₀ = ρgh·S (S = slope) and the shear velocity is u* = √(τ₀/ρ). Grains move when the Shields parameter θ = τ₀/[(ρs − ρ)gD] exceeds a critical value θc, about 0.03–0.06 for sand and gravel in turbulent flow. The Hjulström diagram shows the entrainment velocity lowest for sand of roughly 0.1–1 mm; finer silt and clay need higher velocities because cohesion binds them. Grains travel by traction, saltation or suspension, and stay suspended when turbulent eddy velocities, of order u*, exceed the settling velocity.

Stokes' law gives the settling velocity of a small smooth sphere in laminar flow: vs = (ρs − ρf) g d² / (18 μ). It holds for particle Reynolds number Re_p = ρf vs d/μ below about 1 (quartz in water finer than about 0.1 mm). Because vs varies as d², doubling the diameter quadruples the velocity. A 2 µm clay particle needs about three days to fall 1 m, so clay deposits only in still water or after flocculation (clumping in saline water). Coarser grains follow a drag law in which vs rises only as √d, and angular or platy grains settle slower than spheres.

Bedforms in sand and the structures they leave
Bedform	Flow regime	Structure preserved
Current ripples	Lower	Small cross-lamination, in sand finer than about 0.6 mm
Dunes	Lower	Planar or trough cross-bedding; foresets dip downcurrent
Upper plane bed	Upper	Parallel lamination with parting lineation
Antidunes	Upper, Fr near or above 1	Low-angle cross-stratification, rarely preserved
Wave ripples, hummocky bedforms	Oscillatory or combined flow	Symmetrical ripples; hummocky cross-stratification in storm beds

At a fixed depth, with sand of about 0.3 mm, raising the velocity gives no motion, then ripples, dunes, upper plane bed and finally antidunes.

3. Sedimentary structures and sediment gravity flows

Common sedimentary structures and their meaning
Structure	Process	What it tells
Graded bedding	Normal grading from a waning flow; inverse grading in grain flows	Way-up and flow type
Cross-bedding	Migration of ripples and dunes	Palaeocurrent; bipolar (herringbone) means tides
Flaser, wavy, lenticular bedding	Alternating sand ripples and mud settling	Tidal flats; flaser is sandy, lenticular is muddy
Flute casts, groove casts	Scours in mud filled by sand; flute steep end faces upcurrent	Flow direction (grooves give axis only)
Mud cracks	Desiccation	Subaerial exposure
Load casts, convolute lamination	Unequal loading, dewatering of wet sediment	Rapid deposition, soft-sediment deformation

Sediment gravity flows are driven by their own weight. Debris flows are cohesive, matrix-supported and poorly sorted, with floating clasts, ungraded or inversely graded; grain flows are supported by grain collisions on steep slopes; turbidity currents are turbulence-supported suspensions that deposit graded turbidites. Debris flows are Bingham fluids, turbidity currents behave as Newtonian density currents.

The Bouma sequence has five divisions. Ta is massive or graded sand; Tb parallel lamination (upper-regime plane bed); Tc ripple and convolute lamination (lower regime); Td upper parallel lamination of fine sand and silt; Te mud (pelite and hemipelagite). It records a decelerating flow, so proximal beds are thick, coarse and Ta-rich, while distal beds are thin, fine and begin at Tc or Td, with the lower divisions missing. Flute and groove marks usually occur on the base. Bed thickness is the sum of the divisions.

🧠 Turbidite or debris flow?

Grading with lamination and ripples in a fining-upward bed points to a turbidity current. A poorly sorted bed with clasts floating in a muddy matrix, no grading and no Bouma divisions is a debris flow.

4. Petrology of common sedimentary rocks

Gravel rocks are conglomerate (rounded clasts) and breccia (angular clasts); a clast-supported rock is an orthoconglomerate and a matrix-supported one a paraconglomerate, and diamictite is a poorly sorted mixture, glacial or from mass flow. Sandstones are classified from the framework grains quartz (Q), feldspar (F) and lithic fragments (L), recalculated to 100, and the matrix (grains finer than about 0.03 mm): in Dott's scheme arenites have under 15% matrix, wackes 15–75%, and more than 75% is mudrock.

Main sandstone types
Rock	Framework	Meaning
Quartz arenite	Quartz above 95%	Very mature, recycled or long worked, as on stable cratons
Subarkose, sublitharenite	Quartz 75–95%; feldspar over lithics, or lithics over feldspar	Moderately mature
Arkose	Quartz below 75%, feldspar over lithics	First-cycle granite or gneiss source, rapid erosion and burial
Litharenite	Quartz below 75%, lithics over feldspar	Orogenic source of volcanic, sedimentary or metamorphic fragments
Greywacke (wacke)	Matrix over 15%, poorly sorted	Immature; commonly turbidity-current deposits

Mudrocks (silt and clay) are shale if fissile and mudstone if massive; clay minerals (kaolinite, illite, smectite, chlorite) record weathering and diagenesis, and black shales hold organic matter preserved in oxygen-poor bottom water. Carbonates are named by Dunham's textures: mudstone (under 10% grains, mud-supported), wackestone (over 10% grains, mud-supported), packstone (grain-supported with mud), grainstone (grain-supported, no mud, so high energy) and boundstone (organically bound, as in reefs). Folk names the allochems (ooids, peloids, bioclasts, intraclasts) and whether they sit in micrite (lime mud) or sparite (cement). Aragonite and high-Mg calcite are metastable and invert to low-Mg calcite, and dolomitisation replaces Ca by Mg.

Chemical and biochemical rocks include evaporites (gypsum, anhydrite, halite) from evaporated brine in arid basins; chert (microcrystalline silica, often from radiolarian or diatom ooze); banded iron formation (alternating iron-oxide and silica bands, mainly Precambrian); phosphorite; and coal, from buried plant matter (peat, lignite, bituminous coal, anthracite).

5. Facies, environments and cyclicity

A facies is a rock body distinguished by lithology, structures and fossils, and a facies association is the set of facies that occur together in one environment. Walther's law: facies that lie side by side in a continuous environment also follow each other vertically, provided there is no break such as an unconformity. A vertical profile is therefore read as a lateral model.

Environments and their vertical profiles
Environment	Typical vertical profile	Diagnostic features
Meandering river	Fining upward: channel lag, cross-bedded point-bar sand, floodplain mud	Erosive base, lateral accretion, palaeosols
Braided river	Coarse sand and gravel, little mud, stacked channel fills	Weak fining, sheet-like bodies
Delta	Coarsening upward: prodelta mud, delta-front sand, mouth bar, delta plain	Progradation, Gilbert-type foresets
Tidal flat	Fining upward from sand flat to mud flat	Flaser, wavy, lenticular bedding; herringbone cross-bedding
Turbidite fan	Thickening-upward lobes; thinning-upward channel fills	Bouma beds, sole marks
Aeolian dune	Large-scale cross-bedding in thick sets	Well-sorted, rounded, frosted sand
Glacial	Diamictite, laminated mud with dropstones	Striated clasts, varves

Cyclicity. Autocyclic cycles come from processes inside the system (channel avulsion, delta-lobe switching, tidal-flat progradation) and are local and irregular. Allocyclic cycles are driven from outside (tectonic pulses, climate, eustatic sea level) and correlate across basins; cyclothems (Pennsylvanian coal measures) alternate marine and non-marine strata with sea level. Milankovitch orbital cycles pace allocyclic cycles: precession about 19–23 kyr, obliquity about 41 kyr, eccentricity about 100 kyr (and 400 kyr). The test is bundling: about five precession cycles per eccentricity cycle (5:1). The time per cycle is the interval divided by the number of cycles.

6. Provenance and basin analysis

Provenance asks what and where the source was. The tools are framework petrography on Dickinson's QFL triangles (continental block: quartz-rich; magmatic arc: feldspar and volcanic lithics; recycled orogen: quartz with sedimentary and metamorphic lithics and chert), heavy minerals (zircon, tourmaline and rutile are stable; garnet, staurolite and kyanite come from metamorphic sources; chromite from ultramafic rocks), element ratios such as Th/Sc and Zr/Sc, and palaeocurrents. The chemical index of alteration, CIA = 100 × Al₂O₃/(Al₂O₃ + CaO* + Na₂O + K₂O) in moles (CaO* is silicate-bound CaO), is about 50 for fresh upper crust and rises towards 100 as feldspar alters to kaolinite. Detrital zircon U–Pb ages give a spectrum of source ages, and the youngest cluster sets a maximum depositional age.

Basin types
Basin	Mechanism	Fill and Indian example
Rift	Fault-bounded extension, half-grabens	Alluvial, lacustrine; Gondwana coal basins of India
Passive margin	Thermal subsidence after rifting	Wedge thickening seawards
Foreland	Flexure under thrust-sheet load	Asymmetric, deepest at the orogen; Siwalik molasse
Pull-apart	Releasing bend on a strike-slip fault	Narrow, deep, rapidly filled

Subsidence arithmetic. Sedimentation rate is thickness divided by time (1 m/Myr = 1 mm/kyr), and rates measured over shorter intervals look higher because gaps count (the Sadler effect). Sediment shrinks on burial as porosity falls, so thickness must be decompacted. By Airy isostasy, for a basin filled to sea level with sediment of mean density ρs, the tectonic subsidence is Y = S(ρm − ρs)/(ρm − ρw), where S is the sediment thickness and ρm and ρw the mantle and water densities. Backstripping removes the sediment load to recover the tectonic part, and after rifting the thermal subsidence decays roughly exponentially with time (McKenzie model).

Key takeaways

φ = −log₂ d (mm), larger φ is finer. Folk and Ward: Mz = (φ16 + φ50 + φ84)/3 and σI = (φ84 − φ16)/4 + (φ95 − φ5)/6.6, with σI below 0.35 very well sorted. Positive skewness in φ means excess fines. Palaeocurrents need the vector mean, never the arithmetic mean.
Re = ρUh/μ separates laminar from turbulent flow; Fr = U/√(gh) separates subcritical from supercritical. Grains move when the Shields parameter exceeds about 0.03–0.06. Stokes: vs = (ρs − ρf)gd²/(18μ), valid for Re_p below about 1, and vs ∝ d².
Raising velocity gives ripples, dunes, upper plane bed, antidunes. Bouma Ta to Te records a waning turbidity current; a matrix-supported, ungraded bed is a debris flow. Flutes show direction, grooves only the axis, herringbone cross-bedding means tides.
Sandstones: quartz arenite above 95% quartz, arkose feldspar-rich, litharenite lithic-rich, wacke over 15% matrix. Dunham: mudstone, wackestone, packstone, grainstone, boundstone. Evaporites, chert, banded iron formation and coal are chemical or biochemical rocks.
Walther's law links lateral and vertical facies if there is no break. Rivers fine upward, deltas and shorelines coarsen upward. Autocyclic cycles are local, allocyclic ones regional; Milankovitch periods are about 19–23, 41 and 100 kyr, with five precession cycles per eccentricity cycle.
Provenance uses QFL, heavy minerals, geochemistry, CIA and detrital zircon ages, whose youngest cluster gives a maximum depositional age. Rift, passive-margin, foreland and pull-apart basins differ in mechanism. Airy subsidence for a basin filled to sea level is Y = S(ρm − ρs)/(ρm − ρw).

Practice questions (18)

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

The median grain size of a sand is φ50 = 2.5. What is its median diameter and its Udden–Wentworth class?
1. 0.35 mm, medium sand, since d = 2^(−1.5) mm lies between 0.25 and 0.5 mm
2. 0.18 mm, fine sand, since d = 2^(−2.5) mm lies between 0.125 and 0.25 mm
3. 0.18 mm, very fine sand, since sands finer than 0.25 mm are called very fine
4. 5.7 mm, pebble, since d = 2^(2.5) mm lies between 4 and 64 mm
Show answer
Answer: B — 0.18 mm, fine sand, since d = 2^(−2.5) mm lies between 0.125 and 0.25 mm
φ = −log₂ d, so d = 2^(−φ) = 2^(−2.5) = 0.177 mm, which lies between 0.125 and 0.25 mm, the fine-sand class. Very fine sand is 0.0625–0.125 mm, so the option that gives the right diameter with the finer class is the tempting slip. Using d = 2^φ inverts the scale and gives a pebble, and 2^(−1.5) would be φ = 1.5.
Which of the following statements are correct?
1. Larger φ values correspond to smaller grains
2. A positive Folk and Ward skewness indicates a tail of fine grains
3. A graphic standard deviation σI of 0.3 indicates poor sorting
4. The arithmetic mean of palaeocurrent azimuths 350° and 10° correctly gives the mean direction, 180°
Show answer
Answer: A — Larger φ values correspond to smaller grains; B — A positive Folk and Ward skewness indicates a tail of fine grains
φ increases as grain size decreases, so positive skewness in φ units means excess fines. σI below 0.35 is very well sorted, so 0.3 is the opposite of poor sorting. Azimuths are circular: 350° and 10° have a vector mean of 0° (north), and the arithmetic value of 180° points the opposite way.
From the cumulative curve of a sand, φ5 = 0.5, φ16 = 1.2, φ50 = 2.1, φ84 = 3.3 and φ95 = 4.1. Use the Folk and Ward inclusive graphic standard deviation σI = (φ84 − φ16)/4 + (φ95 − φ5)/6.6 to find the sorting value. Give the answer to two decimal places.
Numerical answer — type the value.
Show answer
Answer: 1.07
(φ84 − φ16)/4 = (3.3 − 1.2)/4 = 2.1/4 = 0.525 and (φ95 − φ5)/6.6 = (4.1 − 0.5)/6.6 = 3.6/6.6 = 0.5455. Sum σI = 1.0705, which rounds to 1.07, so the sand is poorly sorted (1.00–2.00). The median φ50 is not used in σI.
Cross-bed azimuths, already corrected for tilt, measure 350°, 10°, 30° and 50°. Use the vector mean θ̄ = atan2(Σ sin θ, Σ cos θ) to find the mean palaeocurrent direction, in degrees to the nearest whole degree, within the range 0 to 360.
Numerical answer — type the value.
Show answer
Answer: 20
Σ sin θ = −0.1736 + 0.1736 + 0.5000 + 0.7660 = 1.2660 and Σ cos θ = 0.9848 + 0.9848 + 0.8660 + 0.6428 = 3.4784. θ̄ = atan(1.2660/3.4784) = 20.0°. The readings are symmetric about 20° (−30°, −10°, +10°, +30°). The arithmetic mean, (350 + 10 + 30 + 50)/4 = 110°, is wrong because azimuths are circular.
A river 2.0 m deep flows at a mean speed of 4.0 m/s. Take g = 9.8 m/s² and ν = 1.0 × 10⁻⁶ m²/s, and use Re = Uh/ν and Fr = U/√(gh). Which description is correct?
1. Subcritical (Fr ≈ 0.90) and turbulent (Re ≈ 8 × 10⁶)
2. Supercritical (Fr ≈ 1.8) and turbulent (Re ≈ 8 × 10⁶)
3. Subcritical (Fr ≈ 0.90) and laminar (Re ≈ 8 × 10²)
4. Supercritical (Fr ≈ 1.1) and laminar (Re ≈ 8 × 10³)
Show answer
Answer: A — Subcritical (Fr ≈ 0.90) and turbulent (Re ≈ 8 × 10⁶)
Fr = 4.0/√(9.8 × 2.0) = 4.0/4.43 = 0.90, below 1, so the flow is subcritical. Re = 4.0 × 2.0/10⁻⁶ = 8 × 10⁶, far above 2000, so it is turbulent. A fast river is not necessarily supercritical, which is the tempting error: the criterion compares speed with the wave speed √(gh) = 4.43 m/s. Fr ≈ 1.8 would need a depth of only 0.5 m.
Use Stokes' law, vs = (ρs − ρf) g d²/(18 μ). A quartz grain with ρs = 2650 kg/m³ and diameter 0.05 mm settles in water with ρf = 1000 kg/m³ and μ = 1.0 × 10⁻³ Pa·s. Take g = 9.8 m/s². What is its settling velocity in mm/s? Give the answer to two decimal places.
Numerical answer — type the value.
Show answer
Answer: 2.25
d = 0.05 mm = 5 × 10⁻⁵ m, so d² = 2.5 × 10⁻⁹ m². vs = (1650)(9.8)(2.5 × 10⁻⁹)/(18 × 10⁻³) = 4.0425 × 10⁻⁵/0.018 = 2.246 × 10⁻³ m/s = 2.25 mm/s. The particle Reynolds number is about 0.11, so Stokes' law is valid.
In the Stokes regime the settling velocity of a sphere, with density and viscosity fixed, is proportional to
1. the diameter d
2. the square root of the diameter, √d
3. the square of the diameter, d²
4. the cube of the diameter, d³
Show answer
Answer: C — the square of the diameter, d²
Stokes' law is vs = (ρs − ρf) g d²/(18 μ), so vs ∝ d²: doubling the diameter quadruples the velocity. The √d dependence belongs to the turbulent drag regime of coarser grains, and d³ is the dependence of the grain's weight, not of its velocity.
In a flume with sand of about 0.3 mm at constant depth, the flow velocity is raised steadily from below the threshold of motion until the Froude number exceeds 1. In which order do the bedforms appear?
1. dunes, ripples, antidunes, upper plane bed
2. ripples, antidunes, dunes, upper plane bed
3. upper plane bed, dunes, ripples, antidunes
4. ripples, dunes, upper plane bed, antidunes
Show answer
Answer: D — ripples, dunes, upper plane bed, antidunes
With increasing velocity and stress, fine to medium sand passes through the lower flow regime (ripples, then dunes) to the upper regime (plane bed, then antidunes at Fr near 1). The other orders put upper-regime forms before lower-regime ones, or reverse ripples and dunes.
A 25 cm thick sandstone bed has an erosional base with flute casts. From the base upward it shows parallel lamination, ripple cross-lamination, thin parallel-laminated silt and then mud, with no massive graded division at the base. Which interpretation is best?
1. A debris flow, because the bed is thick, has an erosional base and a sharp contact with the mud above
2. A turbidity-current deposit with Ta missing, from a decelerating and more distal flow
3. A fair-weather beach deposit, because parallel lamination and ripples are both made by waves
4. A grain flow on a dune slip face, because the bed has a sharp base and shows internal parallel lamination
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Answer: B — A turbidity-current deposit with Ta missing, from a decelerating and more distal flow
Flute casts with a sequence of Tb (parallel lamination), Tc (ripples), Td (laminated silt) and Te (mud) is a Bouma sequence lacking Ta, as expected for a waning, more distal turbidity current. A debris flow is matrix-supported and ungraded without internal lamination, which is the tempting alternative because of the thickness and erosional base. Wave deposits do not carry flute casts or a full upward-fining turbidite succession, and a grain flow shows no such lamination series.
A complete turbidite bed is 120 cm thick. Its Ta division is 45 cm thick, Tb is 25 cm, Tc is 20 cm and Td is 10 cm, and the rest of the bed is Te mud. A section contains 18 such beds. What is the total thickness of Te mud in the 18 beds, in metres? Give the answer to one decimal place.
Numerical answer — type the value.
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Answer: 3.6
Te = 120 − (45 + 25 + 20 + 10) = 120 − 100 = 20 cm = 0.20 m per bed. For 18 beds the total is 18 × 0.20 = 3.6 m.
A sandstone has framework grains of quartz 88%, feldspar 7% and lithic fragments 5%, and 4% of the rock is clay matrix. In the Pettijohn–Dott scheme it is a
1. quartz arenite, because it is very rich in quartz and has little matrix between the grains
2. arkose, because it contains feldspar, the diagnostic mineral of arkosic sandstones
3. subarkose, because it is an arenite with 75–95% quartz and more feldspar than lithics
4. wacke, because it contains some clay matrix between the framework grains
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Answer: C — subarkose, because it is an arenite with 75–95% quartz and more feldspar than lithics
Matrix of 4% is below 15%, so the rock is an arenite. Quartz of 88% lies between 75 and 95%, and feldspar (7%) exceeds lithics (5%), giving a subarkose. Quartz arenite needs more than 95% quartz and is the tempting answer for a quartz-rich rock. Arkose needs quartz below 75%, and a wacke needs more than 15% matrix.
A limestone consists of ooids that touch one another, with lime mud filling part of the space between them. In Dunham's classification it is a
1. packstone, because the grains support each other and lime mud is present between them
2. grainstone, because the rock is made of ooids and shows high-energy deposition with winnowed sediment
3. wackestone, because lime mud is present among more than 10% grains
4. boundstone, because the ooids are cemented together by early carbonate cement
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Answer: A — packstone, because the grains support each other and lime mud is present between them
Grains in mutual contact make the rock grain-supported, and the lime mud between them separates a packstone from a grainstone, which has no mud. A wackestone is mud-supported, with grains floating in the mud, which does not fit touching ooids. Boundstone requires organic binding, as in reefs, not simply cement.
Which of the following statements about facies and cycles are correct?
1. Walther's law links lateral facies to vertical order only where no break, such as an unconformity, intervenes
2. A fining-upward sequence of channel lag, cross-bedded sand and mud is typical of a meandering-river point bar
3. A coarsening-upward profile, with sand at the top and mud at the base, is the typical signature of a meandering-river point bar
4. Autocyclic cycles are driven by Milankovitch changes in insolation and for that reason correlate between basins
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Answer: A — Walther's law links lateral facies to vertical order only where no break, such as an unconformity, intervenes; B — A fining-upward sequence of channel lag, cross-bedded sand and mud is typical of a meandering-river point bar
Walther's law holds only for conformable successions, and a point bar fines upward from lag to floodplain mud. A coarsening-upward profile is the signature of progradation, as in deltas and shorelines. Cycles driven by orbital forcing are allocyclic, the external kind, and are the ones that correlate between basins; autocyclic cycles are local.
In a carbonate succession, thin cycles that each represent about 20 kyr of precession occur in bundles of five. A bundle most likely records
1. about 41 kyr of obliquity, which is the next orbital period after precession
2. about 400 kyr of long eccentricity, because bundles of five always mark that period
3. about 20 kyr of autocyclic channel switching, repeated five times by chance
4. about 100 kyr of short eccentricity, giving the 5:1 bundling ratio
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Answer: D — about 100 kyr of short eccentricity, giving the 5:1 bundling ratio
Five cycles of 20 kyr make 100 kyr, the period of short eccentricity, and the 5:1 bundling of precession cycles within eccentricity cycles is the diagnostic Milankovitch signature. A 400 kyr cycle would need about twenty precession cycles per bundle, and obliquity has a period of about 41 kyr, not 100.
The youngest concordant detrital zircon U–Pb ages in a sandstone cluster at 520 Ma, with no younger grains. What does this tell us?
1. The sandstone was deposited at 520 Ma exactly, because zircon crystallises at the time of deposition
2. The sandstone cannot be older than about 520 Ma, so 520 Ma is a maximum depositional age
3. The sandstone cannot be younger than about 520 Ma, so 520 Ma is a minimum depositional age
4. The source terrane was metamorphosed at 520 Ma, because detrital zircon records only metamorphism
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Answer: B — The sandstone cannot be older than about 520 Ma, so 520 Ma is a maximum depositional age
Detrital zircons crystallised in the source rocks and were eroded and deposited later, so the rock cannot be older than its youngest grain: 520 Ma is a maximum age, and deposition may be much younger. Treating the age as the time of deposition is the tempting error. It is not a minimum age, and zircon records igneous as well as metamorphic crystallisation.
A basin is filled to sea level with 3000 m of sediment of mean density 2400 kg/m³. Use Airy isostasy, Y = S(ρm − ρs)/(ρm − ρw), where S is the sediment thickness, ρm = 3300 kg/m³ is the mantle density and ρw = 1000 kg/m³ is the water density. What is the tectonic subsidence Y in metres, to the nearest whole metre?
Numerical answer — type the value.
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Answer: 1174
Y = 3000 × (3300 − 2400)/(3300 − 1000) = 3000 × 900/2300 = 1173.9 m, which rounds to 1174 m. Only about 39% of the sediment thickness is tectonic subsidence; the rest is accommodated by the isostatic response to the sediment load.
Which of the following statements about sedimentary basins are correct?
1. A foreland basin is an asymmetric wedge, deepest next to the thrust belt, where the lithosphere flexes under the thrust load
2. A passive-margin basin subsides mainly by thermal contraction after an earlier stage of fault-controlled stretching
3. A foreland basin is deepest at its outer margin, far from the thrust load, and shallows towards the orogen
4. Pull-apart basins form at restraining bends of strike-slip faults, where compression thickens the crust and builds uplift
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Answer: A — A foreland basin is an asymmetric wedge, deepest next to the thrust belt, where the lithosphere flexes under the thrust load; B — A passive-margin basin subsides mainly by thermal contraction after an earlier stage of fault-controlled stretching
Flexure under the thrust load makes the foreland basin deepest next to the orogen, and a passive margin subsides thermally after rifting. Pull-apart basins open at releasing bends, where extension thins the crust, not at restraining bends, which build uplift.
A river 1.5 m deep flows over a bed of 2 mm quartz gravel with a water-surface slope of 0.0002. Take ρ = 1000 kg/m³, ρs = 2650 kg/m³, g = 9.8 m/s², τ₀ = ρgh·S, a Shields parameter θ = τ₀/[(ρs − ρ)gD] and a critical value θc = 0.045. Will the gravel move?
1. No, because τ₀ ≈ 2.9 Pa gives θ ≈ 0.03, below the critical value
2. No, because the flow is subcritical, so no grain can move whatever the stress
3. Yes, because τ₀ ≈ 2.9 Pa gives θ ≈ 0.09, above the critical value
4. Yes, because any turbulent flow moves 2 mm gravel whatever the Shields parameter
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Answer: C — Yes, because τ₀ ≈ 2.9 Pa gives θ ≈ 0.09, above the critical value
τ₀ = 1000 × 9.8 × 1.5 × 0.0002 = 2.94 Pa. θ = 2.94/(1650 × 9.8 × 0.002) = 2.94/32.34 = 0.091, which is greater than 0.045, so the gravel moves. A subcritical Froude number does not prevent entrainment, and the entrainment criterion is the Shields threshold, not simply turbulence. The option that gives θ ≈ 0.03 uses the same inputs but misplaces the arithmetic.