Sediment texture: grain size, sorting and shape

Quick refresher

In Russian-language petrography the size, sorting, shape and packing of grains are called the struktura of a rock (in English, texture). Bedding and other features of how the rock is put together are called the tekstura (in English, sedimentary structures).
The Udden–Wentworth scale doubles at each step. The class boundaries are …, 1/16, 1/8, 1/4, 1/2, 1, 2 mm, …, and sand runs from 1/16 to 2 mm. The units $\varphi = -\log_2 d$ ( $d$ in millimetres) make the scale uniform: every class is exactly one φ unit wide, and the larger the φ, the finer the grains.
Grain sizes follow roughly a log-normal distribution (a bell shape on a log scale). So we calculate the statistics in φ units from the cumulative curve: the mean size $M_z$ , the sorting $\sigma_I$ (below 0.5 is good, above 1 is poor) and the skewness $Sk_I$ (plus means an excess of fine material, minus an excess of coarse).
Sieve, laser, settling and thin section measure different “diameters” of the same grain, so compare only results from the same method.
We describe shape at three levels: axial ratios and sphericity, roundness of the corners, and the appearance of the surface. Sphericity and roundness are independent: a rounded disc does not become spherical.
The packing porosity of equal grains does not depend on their size, but it falls when fine grains fill the pores of the coarse ones. Permeability grows roughly as the square of the grain size, and it falls as sorting gets worse.

Key relationships

\varphi = -\log_2 d,\qquad d = 2^{-\varphi}

size in φ units; $d$ is in millimetres

M_z = \dfrac{\varphi_{16} + \varphi_{50} + \varphi_{84}}{3}

mean size by the Folk–Ward method

\sigma_I = \dfrac{\varphi_{84} - \varphi_{16}}{4} + \dfrac{\varphi_{95} - \varphi_{5}}{6.6}

sorting: the standard deviation of sizes in φ units

k \approx 760\,d_g^{\,2}\,e^{-1.31\,\sigma}

permeability of a loose sand in darcies; $d_g$ is the geometric mean diameter, mm

Common pitfalls

Confusing φ units with porosity. In this lesson, grain size in phi units is written $\varphi$ and porosity is written $\phi$ .
Forgetting that the φ scale is inverted: 3 φ is a fine sand (0.125 mm), and −1 φ is gravel (2 mm).
Comparing descriptive terms from different systems. A “well sorted” sand when screens are chosen ( $D_{40}/D_{90}$ up to 3) and a “well sorted” sand for a geologist ( $\sigma_I$ up to 0.5) are not the same thing.
Taking “rounded” as a synonym of “round”. Roundness describes how smooth the corners are, not how close the grain is to a sphere.
Naming a rock by grain size without checking the scale. A grain of 0.08 mm is a very fine sand on the Wentworth scale, but a silt on a scale with the sand boundary at 0.1 mm.

After this lesson you can

convert grain size from millimetres to φ units and back, name the class on the Udden–Wentworth scale, and remember that Russian-language decimal scales draw the boundaries differently
explain what sieve, laser and sedimentation analysis and the analysis of thin sections actually measure, and why their results differ
find the mean size, sorting, skewness and kurtosis by the Folk and Ward method from a cumulative curve, and give the descriptive term for each (such as “well sorted”)
describe grain shape (the Zingg class, sphericity and roundness) and explain why these are independent properties
explain how packing and the share of fine grains change porosity, and estimate the initial permeability of a loose sand from its mean size and sorting

What a sediment's texture includes

A clastic sediment is made of grains brought by water, wind or ice, and the space between them. Texture describes the grains themselves and how they are arranged. Here are its components, in the order in which they are usually described in core:

Grain size: The main feature: it decides whether a rock is called a conglomerate, sandstone, siltstone or clay (classes by size).
Sorting: How close the grain sizes are, from “all the grains are almost the same” to “a mixture from clay to pebbles”. In Russian-language descriptions a well-sorted rock is also called “uniform-grained” and a poorly sorted one “varied-grained”.
Shape and roundness: The ratio of the length, width and thickness of a grain; how smooth its corners and edges are; and the appearance of its surface (shiny, matt, or with crystals grown on it).
Packing and fabric: How the grains touch one another, whether they hold up the framework themselves or “float” in a fine filling, and whether they are lined up in one direction.
Matrix and cement: Fine clastic filling deposited together with the grains, and minerals that grew in the pores after burial (framework, matrix and cement).

Texture records the conditions of transport. The strength of the current decides which grains it carries away and which it leaves behind. Long transport sorts and rounds the grains. A flow that dumps its whole load at once, such as a mudflow or a submarine slide, leaves a mixture of all sizes. The sequence “clay washed out → grains sorted → grains rounded” is called the textural maturity of a sand. The further a sediment has moved along this sequence, the longer and harder currents, waves or wind reworked it (the two “maturities” of a sandstone).

Stages of textural maturity of a sand

Stage: Immature
Features: more than 5% clay matrix; the grains are poorly sorted and angular

Stage: Submature
Features: less than 5% clay, but sorting is moderate or poor ( $\sigma_I$ greater than 0.5)

Stage: Mature
Features: almost no clay; good sorting ( $\sigma_I$ less than 0.5); the grains are still angular

Stage: Supermature
Features: no clay; good sorting; the grains are rounded

Stage	Features
Immature	more than 5% clay matrix; the grains are poorly sorted and angular
Submature	less than 5% clay, but sorting is moderate or poor ( $\sigma_I$ greater than 0.5)
Mature	almost no clay; good sorting ( $\sigma_I$ less than 0.5); the grains are still angular
Supermature	no clay; good sorting; the grains are rounded

This is Folk's scheme. The sands of beaches and desert dunes are usually mature or supermature; the deposits of mudflows and glaciers are immature. If the features contradict one another (for example, rounded grains in a clay matrix), that is a clue: material of different origins has been mixed. An example is the rounded sand of ancient dunes that a mudflow redeposited.

A petroleum engineer needs texture for several reasons:

Texture sets the initial porosity and permeability of a sand. Compaction and cement change them only later (see the last section of the lesson).
The grain-size distribution of the formation tells you which gravel and screens to use against sand production (sieve analysis of formation sand).
Texture, together with sedimentary structures, helps to work out the depositional environment, and from it the shape and extent of sand bodies.
Fines migration and sand production while the well is producing depend on the share of fine grains and the strength of the framework (Sand production: types, prediction and cost of the problem).

Grain size: the Udden–Wentworth scale and φ units

The sizes of fragments in sediments differ by a factor of millions, from fractions of a micrometre for clay particles to metres for boulders. So size scales are built not with equal steps but by multiplication. On the Udden–Wentworth scale each boundary is twice the one before it. The main boundaries are 2 mm (gravel and sand), 1/16 mm (sand and silt) and 1/256 mm (silt and clay). Sand is divided further into five classes at 1, 1/2, 1/4 and 1/8 mm.

What counts as the size of an irregular grain depends on how you measure it. A grain passes through the square opening of a sieve with its two smaller axes facing the opening. So the sieve size is close to the intermediate axis of the grain (its middle dimension). For a pebble, you measure the same axis with callipers. Other methods give an equivalent diameter: the diameter of a sphere of the same volume, or of a sphere that settles at the same speed (next section).

Grain size in φ units

\varphi = -\log_2\frac{d}{d_0},\qquad d = d_0\cdot 2^{-\varphi}

where:

$\varphi$ grain size in phi units, —
$d$ grain diameter, mm
$d_0$ reference size, 1 mm; it makes the argument of the logarithm dimensionless (a plain number without units), mm

Units: $d$ and $d_0$ in millimetres; $\varphi$ is dimensionless

Holds when

The logarithm is to base 2: $\log_2 x = \ln x/\ln 2 \approx 3.322\,\log_{10} x$ .
The minus sign is chosen so that sand and finer particles have positive φ values. For gravel and coarser fragments, φ is negative.

One φ unit is a change of size by a factor of two, so all the Wentworth classes have the same width, 1 φ. In φ units the size distribution of a natural sediment is close to normal (bell-shaped), so you can use the usual statistics: the mean, standard deviation and skewness.

The Udden–Wentworth scale

Class: Boulders
Size, mm: more than 256
φ: less than −8

Class: Cobbles
Size, mm: 64–256
φ: −6 to −8

Class: Pebbles
Size, mm: 4–64
φ: −2 to −6

Class: Granules
Size, mm: 2–4
φ: −1 to −2

Class: Very coarse sand
Size, mm: 1–2
φ: 0 to −1

Class: Coarse sand
Size, mm: 0.5–1
φ: 1–0

Class: Medium sand
Size, mm: 0.25–0.5
φ: 2–1

Class: Fine sand
Size, mm: 0.125–0.25
φ: 3–2

Class: Very fine sand
Size, mm: 0.0625–0.125
φ: 4–3

Class: Silt
Size, mm: 0.0039–0.0625
φ: 8–4

Class: Clay
Size, mm: less than 0.0039
φ: more than 8

Class	Size, mm	φ
Boulders	more than 256	less than −8
Cobbles	64–256	−6 to −8
Pebbles	4–64	−2 to −6
Granules	2–4	−1 to −2
Very coarse sand	1–2	0 to −1
Coarse sand	0.5–1	1–0
Medium sand	0.25–0.5	2–1
Fine sand	0.125–0.25	3–2
Very fine sand	0.0625–0.125	4–3
Silt	0.0039–0.0625	8–4
Clay	less than 0.0039	more than 8

Silt is divided into four classes, each 1 φ wide: coarse (4–5 φ), medium, fine and very fine (7–8 φ). A rock is named after its most common class: a “fine-grained sandstone” is a sandstone in which fine sand grains are the most numerous.

Russian-language core descriptions often use decimal scales, and there is no single standard one. A typical version uses the boundaries 1, 0.5, 0.25 and 0.1 mm. They divide sand into very coarse-, coarse-, medium- and fine-grained. Some scales count particles of 0.05–0.1 mm as very fine sand, and others count them as coarse silt. The boundary between silt and clay is drawn at 0.01 or 0.005 mm. The classes coarser than 0.25 mm (very coarse-, coarse- and medium-grained) match the Wentworth ones (very coarse, coarse and medium sand). The finer ones are shifted: the “fine-grained” sand of the decimal scale is 0.1–0.25 mm, not 0.125–0.25 mm, and a grain of 0.08 mm is either a very fine sand or a coarse silt.

Worked exampleConverting sizes to φ units and class names

Given (assumed data)

A sand grain from drill cuttings (intermediate axis): 0.35 mm
Modal size in a grain-size report: 3.6 φ
A pebble from a conglomerate (intermediate axis): 24 mm

Solution

The sand grain: $\varphi = -\log_2 0.35 = -\ln 0.35/\ln 2 = 1.51$ , which is between 1 and 2 φ: a medium sand. On a decimal scale (0.25–0.5 mm) it is also medium-grained.
The size from the report: $d = 2^{-3.6} = 0.082$ mm, that is, about 82 µm, which is between 3 and 4 φ: a very fine sand on the Wentworth scale. On a decimal scale whose sand boundary is at 0.1 mm, this is already a coarse silt. If the boundary is at 0.05 mm, it is a very fine sand.
The pebble: $\varphi = -\log_2 24 = -4.58$ , which is in the pebble class (−2 to −6 φ).

Answer: 0.35 mm is 1.51 φ (medium sand); 3.6 φ is 0.082 mm (very fine sand, and on some Russian-language scales a coarse silt); 24 mm is −4.58 φ (a pebble).

What it means. The φ scale turns multiplication into addition: “twice as fine” means “one φ unit more”. So it is convenient to calculate means and spreads of sizes in φ. The second result shows why a core description states the scale: a grain of 0.08 mm falls into sand or silt depending on the scale, and the name of the rock changes with it, from sandstone to siltstone. The data in this example are assumed for illustration.

How grain size is measured

For the analysis, loose sand (or a rock broken down into separate grains) is sorted into size fractions, and the share of each is found. Do not crush the grains in the process: fragments of grains would shift the result towards small sizes. A cemented sandstone cannot be broken into grains without losses, so its grain sizes are measured in thin section.

Methods of grain-size analysis

Method: Sieving
Range: from about 0.04 mm to tens of millimetres
What it measures: the mass of grains between neighbouring sieves; the size is close to the intermediate axis
Limitations: the fines are washed out on the 0.063 mm sieve; flat grains can slip through an opening on edge

Method: Laser diffraction
Range: from fractions of a micrometre to 2–3 mm
What it measures: the volume share of “equivalent spheres” that scatter light as the grains do
Limitations: flat particles look larger; less clay is found than by settling

Method: Sedimentation (pipette, hydrometer)
Range: finer than 0.063 mm
What it measures: the diameter of a sphere that settles at the same speed
Limitations: slow; the particles are separated with a dispersant, otherwise they stick together

Method: Thin section, image analysis
Range: grains in a cemented rock
What it measures: visible sections of grains (the number of grains, not their mass)
Limitations: a random cut is smaller than the grain, so the sizes must be converted to sieve sizes

Method	Range	What it measures	Limitations
Sieving	from about 0.04 mm to tens of millimetres	the mass of grains between neighbouring sieves; the size is close to the intermediate axis	the fines are washed out on the 0.063 mm sieve; flat grains can slip through an opening on edge
Laser diffraction	from fractions of a micrometre to 2–3 mm	the volume share of “equivalent spheres” that scatter light as the grains do	flat particles look larger; less clay is found than by settling
Sedimentation (pipette, hydrometer)	finer than 0.063 mm	the diameter of a sphere that settles at the same speed	slow; the particles are separated with a dispersant, otherwise they stick together
Thin section, image analysis	grains in a cemented rock	visible sections of grains (the number of grains, not their mass)	a random cut is smaller than the grain, so the sizes must be converted to sieve sizes

The results of different methods differ systematically. So apply criteria and correlations only to data from the same method that produced them.

Any method gives the shares of mass (or volume) in each fraction. From these you draw a histogram and a cumulative curve, which shows what share of the sample is coarser than each size. To select gravel from such a curve, you read the sizes $D_{10}$ , $D_{50}$ , $D_{90}$ and the uniformity coefficient (sieve analysis of formation sand). A geologist describes the same curve with different statistics, in φ units (next section).

Sieves do not work for fractions finer than 0.063 mm. Openings that small clog quickly, and clay particles stick together. So these fractions are separated by settling. You shake the sample up in a tall cylinder of water with a dispersant (a chemical that keeps particles apart). At calculated moments, you draw a little suspension from a set depth with a pipette, or you measure the density of the suspension with a hydrometer. The sampling times follow from how fast fine particles settle.

Settling velocity of a fine particle (Stokes' law)

w = \frac{(\rho_s - \rho_f)\,g\,d^2}{18\,\mu}

where:

$w$ steady settling velocity of the particle, m/s
$\rho_s$ density of the particle; for quartz 2650, kg/m³
$\rho_f$ density of water; about 998 at 20 °C, kg/m³
$g$ acceleration due to gravity, 9.81, m/s²
$d$ particle diameter, m
$\mu$ dynamic viscosity of water (its resistance to flow); about $1.00\cdot10^{-3}$ at 20 °C, Pa·s

Units: SI

Holds when

A single smooth spherical particle in still water, far from walls and other particles.
The water flows slowly around the particle: the particle Reynolds number $\rho_f w d/\mu$ (a number that compares inertia with viscosity) is below about 0.5. For quartz in water at 20 °C this means particles finer than about 0.08 mm.
The particles do not stick together, because a dispersant is added to the sample.

Settling velocity grows as the square of the diameter: a particle half the size settles four times more slowly. So after a known time, at a given depth, only particles finer than a certain size remain in suspension. The same law explains why fine mud settles only in quiet water (transport and deposition).

Worked exampleHow long to wait in a pipette analysis

Given (assumed data)

Particles: quartz, 2650 kg/m³
Water at 20 °C: 998 kg/m³; $1.00\cdot10^{-3}$ Pa·s
Sampling depth: 10 cm
Class boundaries: sand–silt 0.0625 mm; silt–clay 0.0039 mm; for comparison, the 0.01 mm boundary of decimal scales

Solution

Sand–silt boundary: $w = \dfrac{1652\cdot 9.81\cdot(62.5\cdot10^{-6})^2}{18\cdot10^{-3}} = 0.00352$ m/s, that is, 3.52 mm/s. The particle falls the 100 mm in $100/3.52 \approx 28$ s.
Check of the Stokes condition: $\text{Re} = 998\cdot 0.00352\cdot 62.5\cdot10^{-6}/10^{-3} = 0.22$ , which is below 0.5, so the law applies.
Silt–clay boundary ( $d = 3.9$ µm): $w = 0.0137$ mm/s, and the time is $100/0.0137/3600 \approx 2.0$ h.
The 0.01 mm boundary: $w = 0.0900$ mm/s, and the time is $100/0.0900/60 \approx 18.5$ min.

Answer: After 28 s, at a depth of 10 cm, only particles finer than 0.0625 mm remain in suspension. After 18.5 min only those finer than 0.01 mm remain, and after 2.0 h only clay particles finer than 0.0039 mm.

What it means. You dry and weigh the sample drawn with the pipette at each of these moments. This gives the mass of particles finer than each size. The size you find is an equivalent size: clay particles are thin plates, and they settle more slowly than a sphere of the same volume. So they look finer by settling than by laser analysis. At a temperature 5 °C lower, the viscosity of water is about 14% higher, so lengthen the sampling times by the same amount. The data in this example are assumed for illustration.

In thin section, the cutting plane passes through the grains at random and usually cuts them off-centre. So the visible sections are on average smaller than the grains: for spheres the average section is $\pi/4 \approx 0.79$ of the diameter (advanced section). Also, in thin section you count grains, not mass. So sizes from thin sections are converted to sieve sizes with empirical relations, and the sorting in thin section looks worse than in the same rock broken down into grains.

When describing core, you estimate size and sorting with a hand lens, comparing the grains with a reference card (a row of grains of known size and drawings of different sorting). Such an estimate is quick and good enough to name a rock. But calculations, such as selecting gravel or correlating permeability, need laboratory analysis (Core analysis).

The size distribution and the Folk–Ward statistics

The grain sizes of a natural sediment follow approximately a log-normal distribution: it is not the diameters themselves but their logarithms that follow the normal (bell-shaped) distribution (log-normal distribution). So in φ units the distribution is close to normal. On a probability scale (a vertical axis stretched so that a normal distribution plots as a straight line), the cumulative curve is then close to a straight line (probability plot). Bends and tails on such a curve point to a mixture of several groups of grains, for example a sand with some mud that settled at another time.

To describe the curve with a few numbers, we read percentiles $\varphi_p$ from it. The percentile $\varphi_p$ is the size such that $p$ % of the mass of the sample is coarser. These are the same sizes $D_p$ as in gravel selection, only in φ units: $\varphi_p = -\log_2 D_p$ . Linear interpolation between sieves in φ units is the same as interpolation in the logarithms of size.

The Folk–Ward statistics (graphic measures) use percentiles chosen to cover 90% of the sample, from the 5th to the 95th. For a distribution that is normal in φ units, they equal its mean, standard deviation and skewness. They are called graphic because people used to read them straight off the graph of the cumulative curve.

Mean size (graphic mean)

M_z = \frac{\varphi_{16} + \varphi_{50} + \varphi_{84}}{3}

where:

$M_z$ mean grain size, φ units
$\varphi_p$ the size such that $p$ % of the mass of the sample is coarser, φ units

Units: φ units

Holds when

The percentiles are read from the cumulative mass curve (the share coarser than a given size).
For a distribution that is normal in φ units, $M_z$ equals the mean and also the median $\varphi_{50}$ (the middle size).

The median uses only the middle of the curve; $M_z$ also uses its “shoulders”. In millimetres, the mean size $2^{-M_z}$ is the geometric mean of the grain diameters (a mean found by multiplying the diameters, not adding them).

Sorting (graphic standard deviation)

\sigma_I = \frac{\varphi_{84} - \varphi_{16}}{4} + \frac{\varphi_{95} - \varphi_{5}}{6.6}

where:

$\sigma_I$ sorting: the standard deviation of sizes (the typical spread around the mean; inclusive graphic standard deviation), φ units
$\varphi_5,\ \varphi_{16},\ \varphi_{84},\ \varphi_{95}$ percentiles of size, φ units

Units: φ units

Holds when

The formula is applied only to sizes in φ units.
You must measure the tails of the curve. If $\varphi_{95}$ falls on the pan (the fraction that passed through the last sieve), it cannot be found by interpolation, and you measure the fine fraction further by settling or by laser.

For a normal distribution $\varphi_{84} - \varphi_{16} = 2\sigma$ and $\varphi_{95} - \varphi_5 = 3.29\sigma$ , so each term gives about $\sigma/2$ , and the sum is the standard deviation. The second term watches the tails: some clay or gravel mixed in raises $\sigma_I$ a lot, even if the middle of the curve is narrow.

Skewness (graphic skewness)

Sk_I = \frac{\varphi_{16} + \varphi_{84} - 2\varphi_{50}}{2(\varphi_{84} - \varphi_{16})} + \frac{\varphi_{5} + \varphi_{95} - 2\varphi_{50}}{2(\varphi_{95} - \varphi_{5})}

where:

$Sk_I$ skewness of the distribution (inclusive graphic skewness), —
$\varphi_{50}$ median size, φ units

Units: dimensionless; percentiles in φ units

Holds when

The sign depends on the units: in millimetres it is the opposite, so a report must say which units were used.
For a curve that is symmetric in φ units, both terms are zero.

Skewness (how lopsided the distribution is) lies between −1 and +1. Positive means that the curve has a long tail of fine grains (large φ), that is, an excess of fine material. Negative means an excess of coarse material. The first term looks at the central part of the curve, the second at the tails.

Kurtosis (graphic kurtosis)

K_G = \frac{\varphi_{95} - \varphi_{5}}{2.44\,(\varphi_{75} - \varphi_{25})}

where:

$K_G$ kurtosis: how peaked the distribution is (graphic kurtosis), —
$\varphi_{25},\ \varphi_{75}$ quartiles of size (the 25% and 75% percentiles), φ units

Units: dimensionless

Holds when

For a normal distribution $K_G = 1$ . The constant 2.44 is the ratio of the spread between the 5% and 95% percentiles to the spread between the 25% and 75% percentiles of the normal curve.

A $K_G$ greater than 1 (a peaked, or leptokurtic, distribution) means that the middle of the sample is better sorted than its tails, as in a good sand with a little foreign material mixed in. A $K_G$ less than 1 (a flat-topped, or platykurtic, distribution) means a mixture of groups of grains in similar amounts. The classes are: below 0.67, very platykurtic; 0.67–0.90, platykurtic; 0.90–1.11, normal (mesokurtic); 1.11–1.50, leptokurtic; 1.50–3.00, very leptokurtic; above 3.00, extremely leptokurtic.

Descriptive terms for sorting

$\sigma_I$ , φ units: less than 0.35
The sand is…: very well sorted

$\sigma_I$ , φ units: 0.35–0.50
The sand is…: well sorted

$\sigma_I$ , φ units: 0.50–0.71
The sand is…: moderately well sorted

$\sigma_I$ , φ units: 0.71–1.00
The sand is…: moderately sorted

$\sigma_I$ , φ units: 1.00–2.00
The sand is…: poorly sorted

$\sigma_I$ , φ units: 2.00–4.00
The sand is…: very poorly sorted

$\sigma_I$ , φ units: more than 4.00
The sand is…: extremely poorly sorted

$\sigma_I$ , φ units	The sand is…
less than 0.35	very well sorted
0.35–0.50	well sorted
0.50–0.71	moderately well sorted
0.71–1.00	moderately sorted
1.00–2.00	poorly sorted
2.00–4.00	very poorly sorted
more than 4.00	extremely poorly sorted

The boundaries are Folk's. As a rough guide, the sands of desert dunes and beaches are usually well or very well sorted, river sands moderately sorted, and the deposits of mudflows and glaciers poorly or very poorly sorted.

Descriptive terms for skewness

$Sk_I$: +0.30 to +1.00
Distribution: strongly fine-skewed

$Sk_I$: +0.10 to +0.30
Distribution: fine-skewed

$Sk_I$: −0.10 to +0.10
Distribution: nearly symmetrical

$Sk_I$: −0.30 to −0.10
Distribution: coarse-skewed

$Sk_I$: −1.00 to −0.30
Distribution: strongly coarse-skewed

$Sk_I$	Distribution
+0.30 to +1.00	strongly fine-skewed
+0.10 to +0.30	fine-skewed
−0.10 to +0.10	nearly symmetrical
−0.30 to −0.10	coarse-skewed
−1.00 to −0.30	strongly coarse-skewed

The signs apply when you calculate in φ units.

Cumulative curve of sample A in φ units

Sieve opening, mm: 0.595
φ: 0.749
Cumulative coarser, %: 0.3

Sieve opening, mm: 0.420
φ: 1.252
Cumulative coarser, %: 1.8

Sieve opening, mm: 0.297
φ: 1.751
Cumulative coarser, %: 6.0

Sieve opening, mm: 0.250
φ: 2.000
Cumulative coarser, %: 11.0

Sieve opening, mm: 0.210
φ: 2.252
Cumulative coarser, %: 18.0

Sieve opening, mm: 0.177
φ: 2.498
Cumulative coarser, %: 28.0

Sieve opening, mm: 0.149
φ: 2.747
Cumulative coarser, %: 41.0

Sieve opening, mm: 0.125
φ: 3.000
Cumulative coarser, %: 55.0

Sieve opening, mm: 0.105
φ: 3.252
Cumulative coarser, %: 68.0

Sieve opening, mm: 0.088
φ: 3.506
Cumulative coarser, %: 78.0

Sieve opening, mm: 0.074
φ: 3.756
Cumulative coarser, %: 86.0

Sieve opening, mm: 0.053
φ: 4.238
Cumulative coarser, %: 93.0

Sieve opening, mm: 0.044
φ: 4.506
Cumulative coarser, %: 96.0

Sieve opening, mm	φ	Cumulative coarser, %
0.595	0.749	0.3
0.420	1.252	1.8
0.297	1.751	6.0
0.250	2.000	11.0
0.210	2.252	18.0
0.177	2.498	28.0
0.149	2.747	41.0
0.125	3.000	55.0
0.105	3.252	68.0
0.088	3.506	78.0
0.074	3.756	86.0
0.053	4.238	93.0
0.044	4.506	96.0

This is the sand from the example in the lesson on sand control (the same sieves and masses); 4% of the sample passed through the finest sieve. The φ values are given to three decimals so that you can reproduce the interpolation.

Worked exampleFolk–Ward statistics of two samples

Given (assumed data)

Sample A: formation sand; the cumulative curve is in the table above
Sample B: a well-sorted sand, 100 g, sieved on sieves spaced 0.25 φ apart; percentiles $\varphi_5,\ \varphi_{16},\ \varphi_{25},\ \varphi_{50},\ \varphi_{75},\ \varphi_{84},\ \varphi_{95}$ : 0.77; 1.19; 1.34; 1.63; 1.89; 1.99; 2.23

Solution

Sample A, $\varphi_{16}$ : the 16% level lies between the sieves of 0.250 mm (2.000 φ; 11.0% coarser) and 0.210 mm (2.252 φ; 18.0% coarser). Interpolating gives $2.000 + \dfrac{16 - 11}{18 - 11}\cdot 0.252 = 2.18$ .
In the same way we get $\varphi_5 = 1.63$ , $\varphi_{25} = 2.42$ , $\varphi_{50} = 2.91$ , $\varphi_{75} = 3.43$ , $\varphi_{84} = 3.69$ , $\varphi_{95} = 4.42$ .
Mean size: $M_z = (2.18 + 2.91 + 3.69)/3 = 2.93$ φ, that is, $2^{-2.93} = 0.131$ mm, a fine sand.
Sorting: $\sigma_I = (3.69 - 2.18)/4 + (4.42 - 1.63)/6.6 = 0.80$ , which lies between 0.71 and 1.00: moderately sorted.
Skewness: the first term is $(2.18 + 3.69 - 2\cdot 2.91)/(2\cdot 1.51) = 0.017$ and the second is $(1.63 + 4.42 - 2\cdot 2.91)/(2\cdot 2.79) = 0.041$ . Their sum is $Sk_I = 0.06$ , a nearly symmetrical distribution.
Kurtosis: $K_G = (4.42 - 1.63)/(2.44\cdot(3.43 - 2.42)) = 1.13$ , a leptokurtic (peaked) distribution.
Sample B, by the same formulas: $M_z = 1.60$ φ (0.330 mm, a medium sand); $\sigma_I = 0.42$ , well sorted; $Sk_I = -0.14$ , skewed towards coarse grains; $K_G = 1.09$ , normal (mesokurtic) kurtosis.

Answer: Sample A is a fine sand (

M_z

= 2.93 φ), moderately sorted (

\sigma_I

= 0.80), with a nearly symmetrical distribution (

Sk_I

= 0.06); sample B is a medium sand, well sorted (

\sigma_I

= 0.42), with an excess of coarse grains (

Sk_I

= −0.14).

What it means. When screens are selected, sample A counts as well sorted by its uniformity coefficient $C = D_{40}/D_{90} \approx 2.5$ , yet on the geological scale it is only moderately sorted. There is no contradiction: the two scales were built for different tasks, so compare numbers, not words (advanced section). The negative skewness of sample B is the trace of a tail of coarse grains and a shortage of fine ones. This happens when waves or a current keep washing out the fine fraction. But skewness alone does not identify an environment: you also need structures and the sequence of beds (Primary sedimentary structures). The data in this example are assumed for illustration.

CalculatedCumulative curves of samples A and B from the example on a probability scale; the dots are the sieves. The steeper the curve, the better the sorting; a straight line would mean an exactly log-normal distribution. The 16%, 50% and 84% percentiles are marked for sample A.

Grain shape: sphericity and roundness

We describe the shape of a grain at three levels, and each answers its own question. The overall shape is the ratio of the long ( $L$ ), intermediate ( $I$ ) and short ( $S$ ) axes. It depends first of all on the mineral and on how the parent rock broke up. Roundness is how smooth the corners and edges are. It builds up during transport. The surface appearance (shiny, matt, with marks of impacts, or with crystals grown on it) reflects the last stages of transport and the changes after burial.

Zingg shape classes

Class: Equant (about equal in all directions)
$I/L$: more than 2/3
$S/I$: more than 2/3
Example: most quartz grains

Class: Oblate (disc-shaped)
$I/L$: more than 2/3
$S/I$: less than 2/3
Example: mica flakes, flat pebbles

Class: Prolate (rod-shaped)
$I/L$: less than 2/3
$S/I$: more than 2/3
Example: elongated fragments

Class: Bladed (flat and elongated)
$I/L$: less than 2/3
$S/I$: less than 2/3
Example: flat elongated slate fragments

Class	$I/L$	$S/I$	Example
Equant (about equal in all directions)	more than 2/3	more than 2/3	most quartz grains
Oblate (disc-shaped)	more than 2/3	less than 2/3	mica flakes, flat pebbles
Prolate (rod-shaped)	less than 2/3	more than 2/3	elongated fragments
Bladed (flat and elongated)	less than 2/3	less than 2/3	flat elongated slate fragments

The axes of pebbles are measured with callipers, the axes of sand grains under a microscope.

The Zingg classes sort grains into groups. Sphericity instead says in a single number from 0 to 1 how close a grain is to a sphere. Two measures are in common use. Both use the axes of the grain, treating it as an ellipsoid (a squashed or stretched sphere):

Krumbein sphericity

\psi_K = \sqrt[3]{\frac{I\,S}{L^2}}

where:

$\psi_K$ sphericity: the ratio of the diameter of a sphere of the same volume as the grain to the diameter of the sphere that just encloses the grain (the circumscribed sphere), —
$L,\ I,\ S$ long, intermediate and short axes of the grain, mm

Units: any identical units of length

Holds when

The grain is treated as an ellipsoid with axes $L \ge I \ge S$ . For a sphere $\psi_K = 1$ .

The volume of an ellipsoid is proportional to $LIS$ , and the volume of the circumscribed sphere to $L^3$ . The ratio of the volumes is $IS/L^2$ , and the cube root turns it into a ratio of diameters.

Maximum-projection sphericity (Sneed and Folk)

\psi_P = \sqrt[3]{\frac{S^2}{L\,I}}

where:

$\psi_P$ the ratio of the projected area (the area of the shadow) of a sphere of the same volume to the largest projected area of the grain, —
$L,\ I,\ S$ axes of the grain, mm

Units: any identical units of length

Holds when

The grain is treated as an ellipsoid with axes $L \ge I \ge S$ . For a sphere $\psi_P = 1$ .

A settling grain turns its largest projection, $L \times I$ , to the flow. So $\psi_P$ describes how the grain behaves in water better: the smaller $\psi_P$ , the more slowly a grain of the same volume settles.

Worked exampleShape class and sphericity of three grains

Given (assumed data)

Grain 1 (quartz): $L$ = 0.60 mm, $I$ = 0.48 mm, $S$ = 0.40 mm
Grain 2 (a flat siltstone fragment): $L$ = 0.80 mm, $I$ = 0.60 mm, $S$ = 0.20 mm
Grain 3 (an elongated fragment): $L$ = 1.00 mm, $I$ = 0.45 mm, $S$ = 0.36 mm

Solution

Grain 1: $I/L = 0.80$ and $S/I = 0.83$ . Both are greater than 2/3, so the grain is equant. $\psi_K = \sqrt[3]{0.48\cdot 0.40/0.60^2} = 0.81$ ; $\psi_P = \sqrt[3]{0.40^2/(0.60\cdot 0.48)} = 0.82$ .
Grain 2: $I/L = 0.75$ and $S/I = 0.33$ , so the grain is oblate. $\psi_K = 0.57$ ; $\psi_P = 0.44$ .
Grain 3: $I/L = 0.45$ and $S/I = 0.80$ , so the grain is prolate. $\psi_K = 0.55$ ; $\psi_P = 0.66$ .

Answer: The equant grain 1 is the most spherical by both measures (0.81 and 0.82). The disc and the rod are almost equal by Krumbein (0.57 and 0.55), but by maximum projection the disc is markedly farther from a sphere (0.44 compared with 0.66).

What it means. A disc turns flat as it settles and is slowed more than a rod of the same volume. So flat grains, such as mica flakes, are carried farther and deposited together with finer quartz, in quiet water, in thin micaceous laminae (thin layers rich in mica). Lying flat, they make the rock anisotropic (its properties differ with direction): permeability along the bedding is higher than across it. You cannot find roundness from the axes: each of the three grains may be either angular or rounded. The data in this example are assumed for illustration.

CalculatedZingg shape classes (boundaries at 2/3) and lines of equal maximum-projection sphericity

\psi_P

; the dots are the grains of the example.

Roundness, as Wadell defined it, is the ratio of the average radius of curvature of the corners of a grain (how tight the bends of its corners are) to the radius of the largest circle that fits inside its outline. For a perfectly rounded grain it is 1; for a sharp-cornered one it is about 0.1–0.2. In practice we do not measure it but estimate it by comparing the grains with reference drawings.

Roundness classes

Class: Very angular
Roundness: 0.12–0.17
Features: sharp corners and edges, fresh fractures

Class: Angular
Roundness: 0.17–0.25
Features: corners sharp, edges slightly blunted

Class: Subangular
Roundness: 0.25–0.35
Features: corners blunted, faces clearly visible

Class: Subrounded
Roundness: 0.35–0.49
Features: corners rounded, faces still distinguishable

Class: Rounded
Roundness: 0.49–0.70
Features: faces almost gone, outline smooth

Class: Well rounded
Roundness: 0.70–1.0
Features: no corners and no faces

Class	Roundness	Features
Very angular	0.12–0.17	sharp corners and edges, fresh fractures
Angular	0.17–0.25	corners sharp, edges slightly blunted
Subangular	0.25–0.35	corners blunted, faces clearly visible
Subrounded	0.35–0.49	corners rounded, faces still distinguishable
Rounded	0.49–0.70	faces almost gone, outline smooth
Well rounded	0.70–1.0	no corners and no faces

The boundaries are those of the Powers scale; each is about $\sqrt2$ times the previous one. Russian-language descriptions use similar classes, from sharp-edged to well-rounded grains.

What shape and roundness depend on:

Mineral. Micas split into thin flakes, and feldspars split along cleavage planes (planes of easy breaking) into bars. Quartz has no cleavage and gives equant grains. Soft fragments of clayey rocks are rounded and worn away quickly, quartz slowly.
Size. Pebbles become noticeably rounded within the first kilometres to tens of kilometres of transport. Sand grains in water are rounded very slowly: a layer of water cushions the impacts, and the impacts carry little energy.
Transporting medium. Wind rounds sand grains and frosts them (gives them a matt surface) much faster than water does, because impacts in air are stronger.
History. A sand that has been through several cycles of erosion and deposition inherits the roundness from the earlier cycles. So well-rounded quartz points to a long history or to work by wind, and angular grains point to a nearby source and rapid burial. A matt surface can also be a trace of dissolution after burial, and crystal faces grown on a grain are a trace of quartz cement (mineral glue that grew between the grains).

Packing, framework and orientation of grains

How tightly the grains lie decides porosity (the share of the rock volume that is pore space). The ideal packings of equal spheres, cubic (47.6%) and closest (26.0%), show that porosity depends on the arrangement of the grains, not on their size (calculation of packings). Equal spheres poured in at random give a porosity between these limits: about 40% when loose and about 36% after shaking. A freshly deposited well-sorted sand usually has 40–45%, because the grains are not spheres, there is friction between them, and they have no time to settle tightly.

Natural sand is a mixture of sizes, and the key question is where the fine grains end up. While there are few of them, they lie in the pores between the coarse ones. The coarse grains still touch one another and carry the load. This is a rock with a grain framework (grain-supported). When the pores of the framework are full, the extra fine material has nowhere to go except between the coarse grains, and it pushes them apart. The coarse grains then begin to “float” in the filling, and the filling now carries the load (matrix-supported). This is the case in mudflow deposits and in wackes (sandstones with 15–75% fine filling between the grains). In Russian-language petrography, a similar distinction is drawn for cement (the mineral material that glues the grains together). With basal cement, the grains do not touch one another. With pore cement, the cement fills the spaces between grains that touch.

The simplest model of such a mixture has two fractions, so different in size that the fine grains pass freely into the pores of a framework of coarse ones.

Porosity of an ideal mixture of coarse and fine grains

\phi(x) = \begin{cases} 1 - \dfrac{1 - \phi_c}{1 - x}, & x \le x^* \\[8pt] 1 - \dfrac{1}{x/(1 - \phi_f) + 1 - x}, & x \ge x^* \end{cases}

where:

$\phi$ porosity of the mixture, fraction
$x$ share of fine grains in the volume of the solid part (the grains, without the pores), fraction
$\phi_c$ packing porosity of the coarse grains alone, fraction
$\phi_f$ packing porosity of the fine grains alone, fraction
$x^*$ share of fine grains at which they exactly fill the pores of the framework (formula below), fraction

Units: fraction

Holds when

The fine grains are so much smaller than the coarse ones that they pass into the pores of the framework through the narrow gaps between the grains. For close packing, this means finer than about 0.15 of the diameter of the coarse grains (size of the constrictions between grains).
The fine grains pack in the pores as tightly as they would on their own, and the coarse grains as tightly as they would without the fine ones.
Within each fraction, the grains are of one size and shape.

Left branch: the fine grains fill the pores of the framework without pushing it apart, and porosity falls. Right branch: the coarse grains “float” in the fine ones, and they simply replace part of the porous fine material. In real sands the two sizes do not differ so sharply, the fine grains push the framework apart earlier, and porosity falls less. So the model gives a lower bound for the porosity of the mixture.

Composition and porosity of the densest mixture

x^* = \frac{\phi_c\,(1 - \phi_f)}{1 - \phi_c\,\phi_f},\qquad \phi_{min} = \phi_c\,\phi_f

where:

$x^*$ share of fine grains in the volume of the solid part in the densest mixture, fraction
$\phi_{min}$ the lowest porosity of the mixture, fraction

Units: fraction

Holds when

The same assumptions as in the formula for the porosity of the mixture.

In the densest mixture, the only pores left are those inside the fine fraction that has filled the pores of the framework. So the porosity equals the product of the porosities of the two fractions.

Worked exampleHow much silt it takes to “choke” a sand

Given (assumed data)

A framework of coarse sand: packing porosity $\phi_c$ = 0.38
The fine fraction (silt): packing porosity $\phi_f$ = 0.44; the grains are much smaller than the narrow gaps between the grains of the framework

Solution

The share of silt that exactly fills the pores of the framework: $x^* = \dfrac{0.38\cdot(1 - 0.44)}{1 - 0.38\cdot 0.44} = 0.256$ , that is, 25.6% of the grain volume.
The lowest porosity: $\phi_{min} = 0.38\cdot 0.44 = 0.167$ , that is, 16.7%.
With 10% silt the framework is still intact: $\phi = 1 - (1 - 0.38)/(1 - 0.10) = 0.311$ , that is, 31.1%.
With 50% silt the coarse grains already “float”: $\phi = 1 - 1/(0.50/0.56 + 0.50) = 0.282$ , that is, 28.2%.

Answer: Porosity falls from 38% in clean sand to 16.7% at 25.6% silt and rises again to 28.2% at 50%.

What it means. The first 10% of silt takes almost 7 percentage points of porosity from the sand, and it takes far more of the permeability. Once fines fill the pores of the framework, the fluid has to flow through the pores of the silt, and these are narrower by as many times as the silt is finer than the sand. So even a moderate amount of clayey–silty filling (matrix) turns a good sand into a poor reservoir. Rocks in which the coarse grains “float” in the filling, such as wackes and mudflow deposits, are usually not reservoirs. Real mixtures with less contrasting sizes lose less porosity than in the model. The data in this example are assumed for illustration.

CalculatedPorosity of the ideal mixture of coarse sand and silt from the example against the silt fraction. To the left of the minimum, the silt fills the pores of the framework; to the right, the coarse grains “float” in the silt.

More detail:Where the mixture formula comes from

Take a mixture with a unit volume of solid grains: $1 - x$ of coarse grains and $x$ of fine grains. As long as the fine grains fit into the pores of the framework, the framework sets the volume of the rock. Coarse grains of volume $1 - x$ together with their pores occupy a volume of $(1 - x)/(1 - \phi_c)$ . Porosity is the share of the volume not occupied by grains: $\phi = 1 - 1/V = 1 - (1 - \phi_c)/(1 - x)$ .

The fine grains together with their pores occupy a volume $x/(1 - \phi_f)$ . The pores of the framework are full when this volume equals the volume of the pores of the framework, $(1 - x)\,\phi_c/(1 - \phi_c)$ ; solving this gives $x^*$ . Beyond that point the fine grains set the volume of the rock, and the coarse grains simply add their own volume, which has no pores: $V = x/(1 - \phi_f) + (1 - x)$ . At $x = x^*$ both branches give $\phi_c\,\phi_f$ , so the curve is continuous.

Orientation. Elongated and flat grains lie flat. The long axes of sand grains more often point along the current, while pebbles that rolled along the bottom are often turned across it. Flat pebbles in a river tilt against the current and overlap one another like roof tiles. This is called imbrication, and it is one of the signs of the direction of ancient currents (palaeocurrents). Orientation makes permeability anisotropic (different in different directions): it is higher along the bedding and along the current than across them.

Grain contacts. Immediately after deposition, the grains touch at points. On burial, the load presses them into one another: point contacts become linear (straight-line), then concave–convex (one grain pressed into another), and, where grains dissolve under pressure, sutured (jagged and interlocking). The type of contacts seen in thin section shows how far the rock has been compacted. But that already belongs to the history after deposition (Diagenesis: compaction, cementation and dissolution).

Texture and the initial porosity and permeability

As long as a sediment is not compacted and cemented, its capacity and conductivity (how much fluid the rock holds and how easily it flows) are set by its texture. Experiments with artificially packed wet sands showed that porosity hardly depends on grain size, but falls as sorting gets worse: from about 42% in very well sorted sands to 28% in very poorly sorted ones. Permeability depends on both properties. It grows as the square of the grain size (Kozeny–Carman model), and it falls when fine grains narrow the pore channels.

Permeability of a loose sand from size and sorting (Krumbein–Monk)

k = 760\,d_g^{\,2}\,e^{-1.31\,\sigma}

where:

$k$ permeability, D
$d_g$ geometric mean diameter of the grains; from the Folk–Ward statistics, $d_g \approx 2^{-M_z}$ , mm
$\sigma$ sorting: the standard deviation of sizes; in practice, use $\sigma_I$ , φ units

Units: empirical: $k$ in darcies, $d_g$ in millimetres, $\sigma$ in φ units

Holds when

Derived for loose uncemented sands with an approximately log-normal size distribution, packed to a porosity of about 40%. The further the porosity is from this value, the cruder the estimate.
For compacted and cemented rocks the formula greatly overestimates the permeability.
At $\sigma = 0$ it is close to the Kozeny–Carman model for equal spheres at a porosity of 40% (that model gives about 30% more).

The factor $d_g^2$ reflects the width of the pore channels, and $e^{-1.31\sigma}$ reflects how fine grains narrow them. Sorting that is worse by one φ unit cuts permeability by a factor of $e^{1.31} \approx 3.7$ .

In other unit systems (1)

k\,[\text{m}^2] = 7.5\cdot10^{-4}\,d_g^{\,2}\,[\text{m}^2]\;e^{-1.31\,\sigma}

$k$ and $d_g^2$ in square metres, $\sigma$ in φ units.

$760$ D $= 7.5\cdot10^{-10}$ m² per square millimetre of $d_g^2$ , that is, per $10^{-6}$ m²: $7.5\cdot10^{-10}/10^{-6} = 7.5\cdot10^{-4}$ .

Worked exampleInitial permeability of two sands

Given (assumed data)

Sample A (the example above): $M_z$ = 2.93 φ, $\sigma_I$ = 0.80
Sample B: $M_z$ = 1.60 φ, $\sigma_I$ = 0.42

Solution

Mean diameters: $d_g = 2^{-2.93} = 0.131$ mm for A and $2^{-1.60} = 0.330$ mm for B.
Sample A: $k = 760\cdot {0.131}^2\cdot e^{-1.31\cdot 0.80} = 4.6$ D.
Sample B: $k = 760\cdot {0.330}^2\cdot e^{-1.31\cdot 0.42} = 48$ D.
Sand B is 10 times more permeable. Grain size accounts for a factor of $(0.330/0.131)^2 = 6.3$ , and sorting for a factor of $e^{1.31\cdot(0.80 - 0.42)} = 1.6$ .

Answer: Loose sand A would have a permeability of about 4.6 D (darcies) and sand B about 48 D, roughly 10 times more. Grain size accounts for most of the difference.

What it means. This is the permeability of a loose sand immediately after deposition, at a porosity of about 40%. In the formation, these sands are compacted and partly cemented, and their permeability is usually one to three orders of magnitude lower (10 to 1000 times). But the rule “coarser and better sorted means more permeable” is largely preserved. The topics on reservoir quality and on diagenesis explain how compaction, cement and clay minerals change this initial picture, and how reservoir quality is predicted from them. The data in this example are assumed for illustration.

At the scale of a core, texture changes from lamina to lamina. Neighbouring laminae (thin layers), only millimetres thick, differ in grain size and sorting, and so in permeability and capillary pressure (the extra pressure needed for one fluid to push into pores filled with another). Flow along the laminae goes through the best of them, and flow across them runs into the worst. In two-phase flow (two fluids moving together, such as oil and water), one of the phases is held back at the boundaries between laminae (Capillary pressure and the transition zone). The lesson on permeability and the topic on heterogeneity show how to average the permeability of such layers. The lesson on structures shows how laminae stack into cross-bedded sets.

Terms and symbols

Sediment texture: The size, sorting, shape and roundness of the grains and their packing; in Russian-language petrography it is called struktura (“structure”), not tekstura.
Udden–Wentworth scale: A scale of fragment sizes whose boundaries double at each step: sand 1/16–2 mm, silt 1/256–1/16 mm, clay finer than 1/256 mm.
φ units: $\varphi = -\log_2 d$ , with $d$ in millimetres; one Wentworth class is one φ unit wide.
Percentile $\varphi_p$: The size, in φ units, such that $p$ % of the mass of the sample is coarser.
Sorting $\sigma_I$: The standard deviation of sizes in φ units (the spread of sizes); below 0.5 is good, above 1 is poor.
Skewness $Sk_I$: An excess of fine grains (plus) or coarse grains (minus) compared with a symmetric distribution.
Sphericity: How close the shape of a grain is to a sphere; by Krumbein's measure $\sqrt[3]{IS/L^2}$ , by the maximum-projection measure $\sqrt[3]{S^2/(LI)}$ .
Roundness: The smoothness of the corners and edges of a grain; from very angular to well rounded.
Textural maturity: How far a sediment has been cleaned of clay, sorted and rounded.
Grain framework and filling: Coarse grains touch one another and carry the load, or they “float” in fine material, which then carries the load itself.

Common mistakes and misconceptions

Wrong

“Well sorted means coarse-grained.”

In fact

Sorting is the spread of sizes, not how big they are. Desert dust and fine mud can be very well sorted, and a mixture of coarse pebbles and sand is very poorly sorted.

Wrong

“Sorting can be calculated in millimetres by the same formula.”

In fact

The Folk–Ward formulas are designed for φ units. In millimetres the spread depends on the size itself, so sands of different coarseness cannot be compared. Convert the percentiles to φ first.

Wrong

“The laser and sieve analyses of one sample must agree.”

In fact

They measure different diameters: the laser measures an equivalent sphere that scatters light, and the sieve measures the opening that a grain passes through. The laser sees flat particles as larger and finds less clay than settling does. Compare the methods on the same samples, and do not mix their results in one data set.

Wrong

“A rounded grain is round, like a ball.”

In fact

Roundness is how smooth the corners are; sphericity is how close the grain is to a sphere. A rounded flat pebble stays flat, and an angular fragment can be almost equant.

Wrong

“Since porosity hardly depends on grain size, neither does permeability.”

In fact

Permeability grows roughly as the square of the grain size: at the same porosity, a fine sand has a permeability several times lower.

In depth: advanced detail

The engineer's and the geologist's sorting

When screens are selected, sorting is assessed by the uniformity coefficient $C = D_{40}/D_{90}$ . Older reports use Trask's coefficient $S_o = \sqrt{D_{25}/D_{75}}$ , and geologists use $\sigma_I$ . For a log-normal sand, all three depend only on the same standard deviation in φ units, because the distance between any two percentiles is proportional to $\sigma$ .

Uniformity and Trask coefficients for a log-normal sand

C = \frac{D_{40}}{D_{90}} = 2^{\,(z_{0.9} - z_{0.4})\,\sigma} \approx 2^{1.535\,\sigma},\qquad S_o = \sqrt{\frac{D_{25}}{D_{75}}} \approx 2^{0.674\,\sigma}

where:

$C$ uniformity coefficient; $D_p$ is the size such that $p$ % of the mass is coarser, —
$S_o$ Trask sorting coefficient, —
$z_{0.9},\ z_{0.4}$ quantiles of the standard normal distribution (its 90% and 40% points): 1.282 and −0.253, —
$\sigma$ standard deviation of sizes, φ units

Units: dimensionless

Holds when

The size distribution is log-normal (normal in φ units). For real samples the formulas are only approximate.

The threshold of good sorting in screen selection, $C = 3$ , corresponds to $\sigma \approx 1.03$ , and the threshold of poor sorting, $C = 5$ , to $\sigma \approx 1.51$ . An engineer's “well sorted” sand is a geologist's “moderately sorted or better”.

Worked exampleSample A on three scales

Given (assumed data)

Sorting of sample A: $\sigma_I$ = 0.80 (the Folk–Ward example)
Uniformity coefficient from the sieves: about 2.5 (the lesson on sand control)

Solution

For a log-normal sand with the same sorting, $C = 2^{1.535\cdot 0.80} = 2.34$ , which is close to the measured 2.5.
The thresholds of the screen-selection scale, in φ units: $C = 3$ corresponds to $\sigma = \log_2 3/1.535 = 1.03$ , and $C = 5$ corresponds to $\sigma = 1.51$ .
Trask coefficient: $S_o = 2^{0.674\cdot 0.80} = 1.45$ .

Answer: The same sample has

\sigma_I

= 0.80 (moderately sorted for a geologist),

C

≈ 2.5 (well sorted for screen selection) and

S_o

≈ 1.45.

What it means. The calculated $C$ is slightly less than the measured one, because sample A has a heavier tail of fine particles than a log-normal curve. The conversion lets you compare numbers from reports of different years and departments, but the descriptive terms of different scales cannot be compared. The data in this example are assumed for illustration.

Why grains look smaller in thin section

Suppose the plane of the thin section cuts a sphere of radius $R$ at a distance $h$ from its centre, and $h$ is equally likely to have any value from 0 to $R$ . The radius of the section is $\rho = \sqrt{R^2 - h^2}$ . Its mean is the integral $\tfrac{1}{R}\int_0^R \sqrt{R^2 - h^2}\,\dd h = \tfrac{\pi}{4}R$ (the area of a quarter of a circle divided by $R$ ).

Worked exampleAverage visible size of a grain

Given (assumed data)

Grains: spheres 0.25 mm in diameter, cut at random

Solution

Average diameter of the section: $\dfrac{\pi}{4}\cdot 0.25 = 0.196$ mm, which is 21% less than the true diameter.

Answer: In thin section, such grains look 21% smaller on average, and their size distribution is wider, because cuts near the edge of a grain give very small sections.

What it means. So the mean size from a thin section is too small, and the sorting looks worse than it really is. For non-spherical grains the correction is different. In practice, sizes from thin sections are converted to sieve sizes with empirical relations obtained on the same rocks. The data in this example are assumed for illustration.

Connections to other topics

Sedimentary rocks and their classification — names of clastic rocks by grain size, and the framework, matrix and cement.
Primary sedimentary structures — bedding and cross-bedding: structures built of laminae of different texture.
Sand control methods — sieve analysis of formation sand, $D_{10}$ , $D_{50}$ , the uniformity coefficient and gravel selection.
Porosity — ideal packings of spheres and what else determines porosity.
Rock permeability — how permeability depends on grain size, and how to average it over layers.
Distributions and sample size — the log-normal distribution and the probability plot.
Reservoir quality: geological controls on porosity and permeability — how compaction, cement and clay change the initial porosity and permeability.
Reservoir and seal — what makes a rock a reservoir; the two “maturities” of a sandstone.

Check yourself

1
The intermediate axis of a grain is 0.18 mm. What is this in φ units, and to which Wentworth class does the grain belong?
1. A.About 2.5 φ, fine sand
2. B.About −2.5 φ, pebble
3. C.About 0.18 φ, coarse sand
4. D.About 1.8 φ, medium sand
Show the answer
Correct answer: A. About 2.5 φ, fine sand
$\varphi = -\log_2 0.18 = -\ln 0.18/\ln 2 \approx 2.47$ , which is between 2 and 3 φ, that is, 0.125–0.25 mm: a fine sand. Negative φ values belong to gravel and coarser fragments.
2
From the cumulative curve: $\varphi_5$ = 1.2; $\varphi_{16}$ = 1.5; $\varphi_{84}$ = 2.3; $\varphi_{95}$ = 2.6. What is the sorting of the sand?
1. A.Poorly sorted
2. B.Very well sorted
3. C.Well sorted
4. D.Moderately sorted
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Correct answer: C. Well sorted
$\sigma_I = (2.3 - 1.5)/4 + (2.6 - 1.2)/6.6 = 0.20 + 0.21 \approx 0.41$ , which is between 0.35 and 0.50: the sand is well sorted.
3
Two loose sands: the first has a mean size of 0.25 mm and $\sigma_I$ = 0.4; the second 0.5 mm and $\sigma_I$ = 1.2. Which of them is more permeable by the Krumbein–Monk formula, and by how much?
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First sand: $760\cdot 0.25^2\cdot e^{-1.31\cdot 0.4} \approx 28$ D. Second sand: $760\cdot 0.5^2\cdot e^{-1.31\cdot 1.2} \approx 39$ D, which is about 1.4 times more. Grains twice as coarse give a gain of a factor of 4, and the worse sorting takes away a factor of about 2.9. Porosity, however, is higher in the first, well-sorted sand, because porosity hardly depends on grain size but does depend on sorting.
4
A flat slate pebble has smoothed corners and edges. What can be said about its roundness and sphericity?
1. A.Low roundness, high sphericity
2. B.High roundness and sphericity
3. C.Neither can be judged from the shape
4. D.High roundness, low sphericity
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Correct answer: D. High roundness, low sphericity
Smoothed corners show high roundness, and the flat shape (the short axis much smaller than the other two) shows low sphericity. These are independent properties: slate splits into plates, and transport rounds their edges but does not turn them into spheres.
5
A sand with a packing porosity of 40% has had 15% (of the grain volume) of a very fine silt added; the silt has a packing porosity of 45%. What will the porosity be in the ideal mixture model?
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First check whether the silt fits into the pores of the framework: $x^* = 0.40\cdot 0.55/(1 - 0.40\cdot 0.45) \approx 0.27$ . This is greater than 0.15, so the framework is not pushed apart and the left branch applies: $\phi = 1 - (1 - 0.40)/(1 - 0.15) = 1 - 0.60/0.85 \approx 0.29$ . Porosity has fallen from 40% to 29%, and permeability has fallen far more, because the pores of the framework are already about half filled with silt.
6
Which method will give the largest share of particles finer than 4 µm for one and the same sample?
1. A.Sieving
2. B.Settling (pipette)
3. C.Laser diffraction
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Correct answer: B. Settling (pipette)
Sieves do not reach such sizes at all. The laser treats flat clay particles as equivalent spheres based on their large projection, so it “sees” them as larger. In settling, plates settle more slowly than spheres of the same volume, so they are recorded in finer fractions. The share of “clay” comes out largest.

Updated: 12 October 2026