Determining shale volume

Quick refresher

Shale volume is the share of the rock's volume taken up by shale material, together with its bound water and silt. The share of the clay minerals alone is usually smaller.
The gamma ray index $I_{GR}$ shows where the reading lies between the clean-formation baseline and the shale baseline of the same section. We choose the baselines from the typical beds of each zone, not from the extreme values of the log.
The straight-line (linear) estimate $V_{sh} = I_{GR}$ is an upper limit. Larionov, Stieber and Clavier fitted empirical relationships (formulas based on measurements, not on theory) to core data (measurements on rock samples from the well). For the same index, these give a smaller shale volume.
Each shale indicator can give too high a shale volume under conditions of its own. The GR does so because of uranium and feldspars (common minerals that contain potassium, which is weakly radioactive). The SP does so because of hydrocarbons (oil and gas). The neutron–density pair does so because of a wrongly chosen matrix (the mineral that we assume the rock is made of). So from several checked estimates we usually take the smallest.
The choice of baselines and relationship can change the net thickness twofold. We support the choice with core data and record it in the report.

Key relationships

I_{GR} = \dfrac{GR - GR_{cl}}{GR_{sh} - GR_{cl}}

gamma ray index (in Russian-language practice, the double-difference parameter $\Delta J_\gamma$ )

V_{sh} = \dfrac{2^{\,C I_{GR}} - 1}{2^{\,C} - 1}

Larionov relationship: $C = 3.7$ for Tertiary (younger) rocks, $C = 2$ for older rocks

V_{sh} = \dfrac{I_{GR}}{3 - 2\,I_{GR}}

Stieber relationship

V_{sh} \approx 1 - \dfrac{\mathrm{PSP}}{\mathrm{SSP}}

shale volume from the SP in a water-bearing bed

Common pitfalls

Taking the absolute maximum of the GR (a uranium-rich shale) as the shale baseline, and the absolute minimum (coal, anhydrite, salt) as the clean-formation baseline.
Using one pair of baselines for the whole borehole, although they shift from one formation to the next and with depth.
Using a nonlinear relationship out of habit, without checking it on core.
Comparing the shale volume from logs with the clay mineral content from X-ray analysis of core as if they were the same quantity.

After this lesson you can

tell shale volume (the volume of shale material) apart from clay mineral content, and understand which of the two each method gives
pick the clean-formation and shale baselines on a GR log and calculate the gamma ray index
turn the index into shale volume with the linear relationship and with the nonlinear ones (Larionov, Stieber, Clavier), and explain why you chose one
estimate shale volume from the spontaneous potential (SP, the natural voltage in the borehole) and from the gap between the neutron and density logs (two porosity logs), and choose the final value from several indicators
estimate how the choice of baselines and relationship changes the net thickness of a bed

What we call shale volume

In log analysis the word “shale” means a rock, not a mineral. It is a fine-grained mixture of three things. The first is clay minerals (the main kinds are illite, kaolinite, smectite and chlorite). The second is silt grains of quartz and feldspar (grains between clay and sand in size). The third is water bound to the surface of the clay particles. Shale volume $V_{sh}$ is the share of the bed's volume taken up by such material, with the same properties as the neighbouring shale beds. This is the main idea behind every log estimate. We call a reference shale bed 100% shale. We call a clean sandstone or a clean carbonate rock (limestone or dolomite) 0% shale. Then we find where the bed we study lies between the two.

Shale volume and clay volume ( $V_{cl}$ , the share taken up by the clay minerals themselves) are different quantities. A shale bed may be only half clay minerals. The rest is silt and bound water. So X-ray analysis of core, which gives the mineral composition, usually shows smaller values than the shale volume from logs. In Russian-language practice, people often use core to find the shale content by mass. This is the share of the fine (pelitic) fraction, for example particles smaller than 0.01 mm. They then convert it into a shale volume, written $K_{\text{sh}}$ in Russian-language notation. When we compare logs with core, it is important to compare quantities that are defined in the same way.

Shale material can sit in a bed in different ways. It can lie in thin layers between sandy layers (laminated shale). It can be scattered in the pores and on the surface of the grains (dispersed shale). Or it can form grains and fragments that replace part of the rock's solid frame (structural shale). For the same $V_{sh}$ , these forms affect porosity, permeability (how easily fluid flows through the rock) and resistivity in different ways. Dispersed clay blocks the pore channels and lowers permeability the most. Laminated shale hardly changes the properties of the sandy layers between the shales. Saturation models for shaly sands allow for the way the shale is distributed. See Water saturation in shaly sands.

What shale volume is used for

Task: Picking reservoir rock
How $V_{sh}$ is used: intervals with a shale volume above the cut-off are left out of the net thickness (Volumetric estimation of hydrocarbons in place)

Task: Porosity
How $V_{sh}$ is used: correcting the readings of the density, neutron and sonic logs for shale (Gas, shale, carbonate and pressure effects on porosity logs)

Task: Saturation
How $V_{sh}$ is used: shaly-sand models used instead of Archie's equation (Water saturation in shaly sands)

Task: Reservoir quality
How $V_{sh}$ is used: relationships between shale volume and permeability, rock classification, choosing the intervals to perforate (to shoot holes through the steel casing that lines the well)

Task	How $V_{sh}$ is used
Picking reservoir rock	intervals with a shale volume above the cut-off are left out of the net thickness (Volumetric estimation of hydrocarbons in place)
Porosity	correcting the readings of the density, neutron and sonic logs for shale (Gas, shale, carbonate and pressure effects on porosity logs)
Saturation	shaly-sand models used instead of Archie's equation (Water saturation in shaly sands)
Reservoir quality	relationships between shale volume and permeability, rock classification, choosing the intervals to perforate (to shoot holes through the steel casing that lines the well)

Clean and shale baselines, the gamma ray index

The first step is to pick two reference lines on the gamma ray log. The clean-formation baseline $GR_{cl}$ (the “clean line”) is the typical reading of a clean sandstone or carbonate with no shale. The shale baseline $GR_{sh}$ (the “shale line”) is the typical reading of thick shale beds. With these two lines, we turn the reading of a bed into a number with no units, the gamma ray index.

Gamma ray index

I_{GR} = \frac{GR - GR_{cl}}{GR_{sh} - GR_{cl}}

where:

$I_{GR}$ gamma ray index, fraction
$GR$ GR reading in the interval we study, API
$GR_{cl}$ clean-formation baseline, API
$GR_{sh}$ shale baseline, API

Units: any GR units, the same for all three readings (API, the standard gamma ray unit, or µR/h); the index has no units

Holds when

Only the shale material makes the bed radioactive.
The shale material in the bed is the same as in the reference shale bed, and the clean part is the same as in the reference clean bed.
The bed is thicker than the vertical resolution of the tool (the thinnest bed the tool can read correctly).
We cut off values outside the baselines: we use 0 if $I_{GR} < 0$ , and 1 if $I_{GR} > 1$ .

The index is the fraction of the way from clean rock to shale on the radioactivity scale. It is not yet the shale volume. How to turn it into $V_{sh}$ depends on the rocks (the next section). In Russian-language practice the same quantity is called the double-difference parameter $\Delta J_\gamma$ .

SchematicPicking the baselines on an illustrative GR log. We draw the clean-formation baseline (28 API) and the shale baseline (128 API) through typical beds, not through the extreme values. A coal streak and a uranium-rich shale give the extreme values, but they are not suitable for baselines.

How to pick the baselines

1
Split the section into zones
Draw the baselines separately for each formation (a named rock unit) or group of beds. With depth, shales are squeezed tighter and change in composition, and their readings shift. Using one pair of baselines for the whole borehole is a common mistake.
2
Shale baseline
Take the typical value of thick, uniform shale beds next to the bed you study. For example, take the value that only 5–10% of the readings exceed in the shaly part of the zone. Do not count uranium-rich shales or single peaks.
3
Clean-formation baseline
Take the typical value of beds that are known to be clean and are of the same type as the bed you study: sandstones for sandstones, carbonates for carbonates. Coal, anhydrite and salt give lower values, so they are not suitable for the baseline.
4
Check with other methods
Use the SP, the neutron and density logs and core to make sure that the reference beds really are clean or shaly. They must not be a radioactive sandstone (one that is radioactive but not shaly) or a “cold” shale (a shale with unusually low radioactivity).
5
Record the choice
Write the baselines and the reasons for them in the report. The net thickness depends on them directly, and we revise them when new data arrive.

Worked exampleHow the baselines change the index

Given (assumed data)

GR reading in the bed we study: 73 API
Typical clean sandstones and shales of the zone: 28 API and 128 API
Absolute minimum and maximum of the log: 12 API (coal) and 186 API (uranium-rich shale)

Solution

With the typical baselines: $I_{GR} = (73 - 28)/(128 - 28) = 0.45$ .
With the absolute extremes: $I_{GR} = (73 - 12)/(186 - 12) = 0.35$ .
The difference is 0.10 in index units, so the index is too low by 22%.

Answer: Correctly chosen baselines give an index of 0.45. The extremes of the log give 0.35.

What it means. The uranium-rich shale raised the shale baseline, so the bed looks much cleaner than it is. The coal lowered the clean-formation baseline, which pushes the index the other way, but less strongly. The error in the index passes into shale volume and then into net thickness. If the cut-off lies between the two estimates, the choice of baselines decides whether the bed counts as reservoir rock. The data in this example are assumed for illustration.

From the index to shale volume

The simplest assumption is that shale volume equals the index: $V_{sh} = I_{GR}$ . This is the linear (straight-line) estimate. It is true when two conditions hold. First, the radioactivity of the mixture is the sum of the contributions of clean rock and shale, each in proportion to its volume. Second, the shale material in the bed is the same as in the reference bed. Comparisons with core have shown that in many sections the linear estimate gives too high a shale volume, especially in the middle of the range. So we treat it as an upper limit. To get a better value, we use empirical relationships (relationships based on measurements, not derived from theory) that were fitted to core. All of them give 0 for clean rock and practically 1 for shale, and they lie below the straight line $V_{sh} = I_{GR}$ .

Larionov relationship

V_{sh} = \frac{2^{\,C\,I_{GR}} - 1}{2^{\,C} - 1}

where:

$V_{sh}$ shale volume, fraction
$I_{GR}$ gamma ray index, fraction
$C$ Larionov coefficient, which sets the curvature: 3.7 for Tertiary (Paleogene and Neogene) unconsolidated rocks (loose, poorly cemented), 2 for older consolidated rocks (well cemented), —

Units: dimensionless; $V_{sh}$ and $I_{GR}$ as fractions

Holds when

It is an empirical relationship: the coefficient $C$ reflects the rocks it was fitted to, and we refine it with core.
The index uses correctly chosen clean-formation and shale baselines.
The radioactivity comes from the shale material.

The larger $C$ is, the more the curve bends below the straight line $V_{sh} = I_{GR}$ , and the smaller the shale volume it gives for the same index. As $C \to 0$ , the relationship becomes linear. In Russian-language practice people apply it to the double-difference parameter $\Delta J_\gamma$ and fit $C$ to core. You often see the formula written with the coefficients 0.083 and 0.33. These are special cases of the same formula (see the other forms).

In other unit systems (2)

Tertiary rocks, C = 3.7

V_{sh} = 0.083\left(2^{\,3.7\,I_{GR}} - 1\right)

$V_{sh}$ and $I_{GR}$ as fractions.

$0.083 \approx 1/(2^{3.7} - 1) = 1/12.0$ .

Older rocks, C = 2

V_{sh} = 0.33\left(2^{\,2\,I_{GR}} - 1\right)

$V_{sh}$ and $I_{GR}$ as fractions.

$0.33 \approx 1/(2^{2} - 1) = 1/3$ .

Stieber relationship

V_{sh} = \frac{I_{GR}}{3 - 2\,I_{GR}}

where:

$V_{sh}$ shale volume, fraction
$I_{GR}$ gamma ray index, fraction

Units: dimensionless; fractions

Holds when

It is an empirical relationship fitted to Tertiary sand–shale deposits.
It is a special case of a family of curves with the curvature parameter $A = 3$ . The general form is in the advanced section.

For small indices it gives about a third of the linear estimate. At $I_{GR} = 0.5$ it gives exactly half of it (0.25).

Clavier relationship

V_{sh} = 1.7 - \sqrt{3.38 - \left(I_{GR} + 0.7\right)^2}

where:

$V_{sh}$ shale volume, fraction
$I_{GR}$ gamma ray index, fraction

Units: dimensionless; fractions

Holds when

It is an empirical relationship. Geometrically, it is an arc of a circle with its centre at the point $(-0.7;\ 1.7)$ , passing through $(0;\ 0)$ and $(1;\ 1)$ .

Its curvature lies between the Larionov relationships for older rocks and for Tertiary rocks.

CalculatedShale volume as a function of the gamma ray index for the linear estimate and four empirical relationships. The dots mark the bed from the lesson's examples (

I_{GR} = 0.45

). The horizontal line is the cut-off of 0.35 from the example on net thickness.

Worked exampleThe same index, five relationships

Given (assumed data)

Gamma ray index $I_{GR}$: 0.45 (the bed from the previous example)

Solution

Linear estimate: $V_{sh} = 0.45$ .
Larionov, older rocks: $(2^{0.9} - 1)/3 = 0.289$ .
Clavier: $1.7 - \sqrt{3.38 - 1.15^2} = 0.266$ .
Stieber: $0.45/(3 - 0.9) = 0.214$ .
Larionov, Tertiary rocks: $(2^{1.665} - 1)/(2^{3.7} - 1) = 0.181$ .

Answer: For the same index of 0.45, shale volume comes out anywhere from 0.181 to 0.45. The largest value is 2.5 times the smallest.

What it means. Choosing a relationship is not a mere formality. With a shale-volume cut-off of 0.30, the linear estimate rejects the bed as non-reservoir, and all the nonlinear ones keep it. Only a comparison with core can show which one is right for a given section. The data in this example are assumed for illustration.

Which relationship to choose

Conditions: No core, and we want a cautious estimate
A reasonable choice: linear
Why: it gives the largest shale volume, an upper limit

Conditions: Young (Tertiary) unconsolidated sand–shale sequences
A reasonable choice: Larionov with $C = 3.7$ , or Stieber
Why: they were fitted to such rocks

Conditions: Older consolidated rocks (Mesozoic, Paleozoic)
A reasonable choice: Larionov with $C = 2$ , or Clavier
Why: the first was fitted to such rocks, and the second has a similar moderate curvature

Conditions: Thin alternating layers of sandstone and shale, thinner than the vertical resolution of the GR tool
A reasonable choice: closer to the linear one
Why: the tool averages the layers in proportion to their thickness, so the index is close to the share of shale layers

Conditions: Core with measured shale volume is available
A reasonable choice: a relationship fitted to the core
Why: it accounts for the local shales and clean rocks

Conditions	A reasonable choice	Why
No core, and we want a cautious estimate	linear	it gives the largest shale volume, an upper limit
Young (Tertiary) unconsolidated sand–shale sequences	Larionov with $C = 3.7$ , or Stieber	they were fitted to such rocks
Older consolidated rocks (Mesozoic, Paleozoic)	Larionov with $C = 2$ , or Clavier	the first was fitted to such rocks, and the second has a similar moderate curvature
Thin alternating layers of sandstone and shale, thinner than the vertical resolution of the GR tool	closer to the linear one	the tool averages the layers in proportion to their thickness, so the index is close to the share of shale layers
Core with measured shale volume is available	a relationship fitted to the core	it accounts for the local shales and clean rocks

In the basic split into Tertiary and older rocks, Jurassic and Cretaceous sandstones count as older rocks. They hold many oil and gas accumulations of Mangyshlak and of the South Turgay Basin. But here too the final choice is made with core.

To see at what index a bed passes the shale-volume cut-off, we invert the relationship (turn it around, so that the index comes out of the shale volume). For the Larionov relationship we can write the result as a formula:

Index that corresponds to a given shale volume (the Larionov relationship inverted)

I^{*} = \frac{1}{C}\,\log_2\!\left[1 + V^{*}\left(2^{\,C} - 1\right)\right]

where:

$I^{*}$ gamma ray index at which the shale volume equals $V^{*}$ , fraction
$V^{*}$ the given shale volume, for example the cut-off value, fraction
$C$ Larionov coefficient, —

Units: dimensionless; $\log_2$ is the base-2 logarithm, $\log_2 x = \ln x/\ln 2$

Holds when

The same as for the Larionov relationship.

Beds with an index not higher than $I^{*}$ pass the shale-volume cut-off. The larger $C$ is, the higher this threshold index.

Worked exampleNet thickness with different relationships

Given (assumed data)

Interval: 6 m; the index is averaged over 0.5 m
Indices from top to bottom: 0.10; 0.16; 0.22; 0.28; 0.40; 0.45; 0.50; 0.57; 0.59; 0.68; 0.76; 0.90
Shale-volume cut-off: $V_{sh} \le 0.35$

Solution

Threshold index: for the linear estimate, $I^{*} = 0.35$ . For the Larionov relationship for older rocks, $I^{*} = \log_2(1 + 0.35\cdot 3)/2 = 0.518$ . For Tertiary rocks, $I^{*} = 0.643$ .
Readings at or below the threshold: 4, 7 and 9 out of 12.
Net thickness: 2.0, 3.5 and 4.5 m. The net-to-gross ratio (net thickness divided by the whole thickness of 6 m) is 0.33, 0.58 and 0.75.

Answer: Depending on the relationship chosen, the net thickness of the interval is from 2.0 to 4.5 m.

What it means. The same logs and the same cut-off give net thicknesses where the largest is more than twice the smallest. The volume of hydrocarbons in place changes in the same proportion, because the net-to-gross ratio is one of its multipliers. So we support the choice of relationship with core data and record it in the report, together with the baselines and the cut-off. When the choice is uncertain, we calculate the options. The data in this example are assumed for illustration.

Other shale indicators

The GR is not the only indicator, and not always the best one. Shales differ from clean rocks in other ways too. They weaken the spontaneous potential (SP), they hold a lot of bound water, and they conduct electric current. An estimate is more reliable when methods that rest on different physics confirm it, because each method has its own sources of error.

Shale volume from the SP

V_{sh} \approx 1 - \frac{\mathrm{PSP}}{\mathrm{SSP}}

where:

$\mathrm{PSP}$ pseudo-static SP amplitude in the bed we study: the deflection from the shale baseline, mV
$\mathrm{SSP}$ static SP amplitude: the deflection in a thick, clean, water-bearing bed with the same formation water, mV

Units: mV; the ratio has no units

Holds when

The bed is water-bearing. Hydrocarbons also reduce the SP amplitude, and then the estimate is too high.
The bed is thick enough, or the amplitude is corrected for its thickness and resistivity.
The resistivities of the mud filtrate (the liquid that seeps from the drilling mud into the rock) and of the formation water differ clearly. The water in the bed is the same as in the reference clean bed.
The mud conducts electricity: the SP is not recorded in oil-based muds.

Shale in the bed weakens the electrochemical potential (the natural voltage that arises where waters with different salt content meet). So the SP amplitude falls from the static value in a clean bed to zero on the shale baseline. In Russian-language practice the ratio $\mathrm{PSP}/\mathrm{SSP}$ is called the relative SP amplitude $\alpha_\text{SP}$ . People build the link between it and shale volume from core, area by area. To learn where the SP comes from, see Spontaneous potential (SP) log.

Shale volume from the gap between neutron and density porosity

V_{sh} \approx \frac{\phi_N - \phi_D}{\phi_{N,sh} - \phi_{D,sh}}

where:

$\phi_N,\ \phi_D$ neutron and density porosity of the bed, calculated for the same matrix, fraction
$\phi_{N,sh},\ \phi_{D,sh}$ the same quantities in the reference shale bed, fraction

Units: fractions

Holds when

The pores of the bed are filled with liquid. Gas brings the curves together or swaps them, and then the estimate is too low.
The matrix is chosen correctly. In dolomite and in rocks with heavy minerals, the gap appears even without shale.
The shale in the bed is the same as in the reference bed.
The readings have been corrected for the effect of the borehole.

Shales contain a lot of bound water and hydroxyl groups (OH groups that are part of the clay minerals). The neutron log, which responds to hydrogen, sees a large “porosity” in them. The density log sees a smaller one. The gap between the two curves grows with shale volume and does not depend on radioactivity, so uranium and feldspars cannot fool this indicator.

Other data are useful too. One kind is the uranium-free gamma ray and the thorium content from the spectral gamma ray log (a log that measures potassium, uranium and thorium separately). Neither of these two responds to uranium. Another is the volume of clay-bound water from the nuclear magnetic resonance (NMR) log (Nuclear magnetic resonance (NMR) logging). The last is core: X-ray analysis of the mineral composition, grain-size analysis and thin sections (thin slices of rock studied under a microscope). Core is the standard against which we calibrate (check and adjust) all the log estimates.

What fools the shale indicators

Indicator: GR
Makes shale volume too high: feldspars, micas and glauconite (minerals that contain potassium), uranium, potassium chloride in the mud
Makes it too low, or does not work: “cold” kaolinitic shales (shales rich in kaolinite, a clay mineral with little potassium)

Indicator: Uranium-free GR
Makes shale volume too high: feldspars, micas, heavy minerals
Makes it too low, or does not work: “cold” shales

Indicator: SP
Makes shale volume too high: hydrocarbons in the bed, thin beds, high bed resistivity
Makes it too low, or does not work: oil-based mud; filtrate and formation water with the same salt content

Indicator: Neutron and density
Makes shale volume too high: a wrong matrix (dolomite, heavy minerals)
Makes it too low, or does not work: gas in the bed

Indicator	Makes shale volume too high	Makes it too low, or does not work
GR	feldspars, micas and glauconite (minerals that contain potassium), uranium, potassium chloride in the mud	“cold” kaolinitic shales (shales rich in kaolinite, a clay mineral with little potassium)
Uranium-free GR	feldspars, micas, heavy minerals	“cold” shales
SP	hydrocarbons in the bed, thin beds, high bed resistivity	oil-based mud; filtrate and formation water with the same salt content
Neutron and density	a wrong matrix (dolomite, heavy minerals)	gas in the bed

Each indicator gives too high a value under conditions of its own, so from several estimates we often take the smallest. We do this only after checking that the low value is not caused by gas, a matrix error or another disturbance.

Worked exampleThree indicators in one bed

Given (assumed data)

GR of the bed; clean-formation and shale baselines: 82 API; 30 and 130 API; a zone of older rocks ( $C = 2$ )
SP: static amplitude and amplitude in the bed: −90 mV and −76 mV; the bed is water-bearing
Neutron and density porosity of the bed: 0.25 and 0.21 (sandstone matrix)
The same quantities in the reference shales: 0.40 and 0.12

Solution

GR: $I_{GR} = (82 - 30)/(130 - 30) = 0.52$ . With the Larionov relationship for older rocks, $V_{sh} = 0.35$ .
SP: $V_{sh} = 1 - 76/90 = 0.16$ .
Neutron and density: $V_{sh} = (0.25 - 0.21)/(0.40 - 0.12) = 0.14$ .
Two estimates that rest on different physics are close (0.16 and 0.14), while the GR gives a value 2.5 times larger.

Answer: We take a shale volume of about 0.14, the smallest of the estimates that agree. The high GR reading is most likely caused by potassium feldspars or micas.

What it means. With a cut-off of 0.30, an estimate from the GR alone would reject this bed. We test the idea about feldspars with the spectral gamma ray log and with core. The spectral log should show a lot of potassium and little thorium, compared with the shales of the same section. If the bed held oil, the hydrocarbons would reduce the SP amplitude, and the SP estimate would also come out too high. The data in this example are assumed for illustration.

Finding shale volume, step by step

1
Split the section into zones
By formation and rock type. Each zone has its own baselines and its own relationship.
2
Check the curves
Correct the GR for the borehole and the mud. Match the records by depth. Mark the intervals with washouts (places where the borehole is wider than the bit) and the intervals where potassium chloride in the mud raises the GR.
3
Pick the baselines and calculate the index
Use the reference clean and shale beds of the zone. Cut off values outside the baselines at 0 and 1.
4
Turn the index into shale volume
Use a relationship fitted or checked with core. Without core, use the linear estimate as an upper limit and a nonlinear one as an alternative.
5
Compare with the other indicators
SP, neutron and density, spectral log. If they disagree, find the reason (uranium, feldspars, gas, matrix) and choose the reliable estimate.
6
Estimate the effect
See how net thickness, porosity and saturation change for reasonable choices of baselines and relationships. Record the choice in the report.

Terms and symbols

Shale volume, $V_{sh}$: The share of the rock's volume taken up by shale material, together with its bound water and silt. In Russian-language literature it is written $K_{\text{sh}}$ .
Clay volume, $V_{cl}$: The share of the volume taken up by the clay minerals themselves. It is usually smaller than $V_{sh}$ .
Clean-formation baseline and shale baseline: The typical readings of the log at the reference clean beds and the reference shale beds of the zone.
Gamma ray index, $I_{GR}$: The position of the GR reading between the baselines. In Russian-language practice it is the double-difference parameter $\Delta J_\gamma$ .
Larionov coefficient, $C$: The curvature parameter of the Larionov relationship: 3.7 for Tertiary rocks and 2 for older rocks. We refine it with core.
Relative SP amplitude, $\alpha_\text{SP}$: The ratio of the SP amplitude in a bed to the static amplitude in a clean water-bearing bed.

Common mistakes and misconceptions

Wrong

“ $V_{sh} = I_{GR}$ is the shale volume.”

In fact

The linear index is an upper limit. In many sections core shows a smaller shale volume. Nonlinear relationships allow for this, but we choose and check them against core.

Wrong

“A nonlinear relationship is always more accurate than the linear one.”

In fact

It is more accurate where core has confirmed it. In thin alternating layers of sandstone and shale that the tool averages together, the linear estimate is often closer to the truth.

Wrong

“The GR is high, so the bed is not a reservoir.”

In fact

Feldspars, micas, glauconite and uranium also make the GR high. Before rejecting the bed, we check the SP, the neutron–density log, the spectral log and core.

Wrong

“Shale volume from logs must match the clay content from X-ray analysis of core.”

In fact

X-ray analysis gives the content of clay minerals. The log gives the volume of shale material, including silt and bound water. The first is usually smaller. We compare quantities that are defined in the same way.

Wrong

“The clean-formation and shale baselines are the same for the whole well.”

In fact

With depth, and from one formation to the next, the readings of shales and of clean rocks shift. We draw the baselines zone by zone, close to the interval we study.

In depth: advanced detail

Why the link between the index and shale volume is nonlinear

The GR is proportional to the mass concentration of the radioactive elements (their amount per kilogram of rock), not to their volume concentration (their amount per cubic metre of rock). See Gamma ray log and natural gamma ray spectrometry. If shale is denser than clean porous rock, its share by mass is larger than its share by volume, and the index grows faster than shale volume. Let shale and clean rock be mixed in volume fractions $V_{sh}$ and $1 - V_{sh}$ , and let the index equal the mass fraction of shale. Solving for $V_{sh}$ gives a relationship of the same form as Stieber's:

Mixing by mass, and the general form of the Stieber relationship

V_{sh} = \frac{I_{GR}}{A - (A - 1)\,I_{GR}}, \qquad A = \frac{\rho_{sh}}{\rho_{cl}}

where:

$A$ curvature parameter; for mixing by mass, the ratio of the densities, —
$\rho_{sh}$ bulk density of the shale, g/cm³
$\rho_{cl}$ bulk density of the clean rock together with the fluid in its pores, g/cm³

Units: dimensionless; the densities are in the same units

Holds when

The index equals the mass fraction of shale: the shale in the bed is the same as in the reference bed, and only shale is radioactive.
The densities of the components are constant.

At $A = 1$ the relationship is linear, and at $A = 3$ it is the same as the Stieber formula. In the other direction, from a pair of values “index and shale volume” we find the equivalent parameter as $A = I_{GR}\,(1/V_{sh} - 1)/(1 - I_{GR})$ .

Worked exampleHow much of the nonlinearity the density difference explains

Given (assumed data)

Shale density $\rho_{sh}$: 2.55 g/cm³
Density of a porous clean sandstone $\rho_{cl}$: 2.20 g/cm³
Gamma ray index: 0.45

Solution

Curvature parameter from the density difference: $A = 2.55/2.20 = 1.159$ .
Shale volume: $V_{sh} = 0.45/(1.159 - 0.159\cdot 0.45) = 0.41$ .
For the same index, the empirical relationships give 0.289 (Larionov, $C = 2$ ) and 0.181 (Larionov, $C = 3.7$ ). These correspond to $A \approx 2.0$ and $A \approx 3.7$ .

Answer: The density difference lowers shale volume only from 0.45 to 0.41. The empirical curves need a parameter

A

from 2.0 to 3.7 instead of 1.159.

What it means. The density difference alone cannot explain the nonlinearity. A likely reason is that the shale material inside the reservoir differs from the reference shales in composition and radioactivity. The size of this effect varies from area to area, which is why the relationship is fitted to core. The data in this example are assumed for illustration.

How to fit the Larionov coefficient to core

If core gives the shale volume in intervals with different indices, we choose the coefficient $C$ so that the curve passes through these points with the smallest error. For a first estimate, the point $I_{GR} = 0.5$ is convenient. Put it into the Larionov relationship and write $y = 2^{C/2}$ . Then $V_{sh} = (y - 1)/(y^2 - 1) = 1/(y + 1)$ , and so:

Larionov coefficient from the shale volume at an index of 0.5

C = 2\log_2\!\left(\frac{1}{V_{0.5}} - 1\right)

where:

$V_{0.5}$ shale volume from core in intervals with an index of about 0.5, fraction
$C$ Larionov coefficient, —

Units: dimensionless

Holds when

The shale volume from core is defined in the same way as from logs: the volume of shale material, not only of clay minerals.
The index in the core intervals is calculated with the same baselines as in the whole zone.
One point gives only a first approximation. With all the data, we fit $C$ by the least-squares method (the method that makes the sum of the squared errors as small as possible).

For $V_{0.5} = 1/3$ we get $C = 2$ , and for $V_{0.5} = 0.217$ we get about 3.7. These are the two classic relationships.

Worked exampleCalibration against core

Given (assumed data)

Average shale volume from core in intervals with an index of about 0.5: 0.26
The bed we study: $I_{GR} = 0.45$

Solution

$1/0.26 - 1 = 2.846$ ; $\log_2 2.846 = 1.509$ ; $C = 2\cdot 1.509 = 3.02$ .
Check at the calibration point: $1/(2^{C/2} + 1) = 0.26$ .
Shale volume of the bed with an index of 0.45: $V_{sh} = 0.220$ . This lies between the Larionov estimates for older rocks (0.289) and for Tertiary rocks (0.181).

Answer: The Larionov coefficient for this zone is about 3.02. At an index of 0.45 the shale volume is 0.220.

What it means. One point is not enough. We check the calibration over the whole range of indices and in several wells. The scatter of the core points around the curve shows how accurately shale volume can be estimated from the GR in this zone at all. The data in this example are assumed for illustration.

Connections to other topics

Gamma ray log and natural gamma ray spectrometry — where the radioactivity of rocks comes from and when the GR gives a wrong answer.
Open-hole logging: why, how and when — what logging measures and why thin beds look more shaly.
Water saturation in shaly sands — how shale volume enters the calculation of saturation in shaly sands.
Archie's equation and saturation of clean reservoirs — why Archie's equation gives too high a water saturation in shaly rocks.
Volumetric estimation of hydrocarbons in place — cut-offs, net thickness and the calculation of hydrocarbons in place.
Gas, shale, carbonate and pressure effects on porosity logs — how shale distorts the porosity from the density, neutron and sonic logs.
Spontaneous potential (SP) log — the spontaneous potential and its link with shale volume.
Core analysis — core as the standard for calibrating shale volume.

Check yourself

1
The GR of a bed is 70 API, the clean-formation baseline is 30 API and the shale baseline is 130 API. What is the gamma ray index?
1. A.0.40
2. B.0.30
3. C.0.54
4. D.0.70
Show the answer
Correct answer: A. 0.40
$I_{GR} = (70 - 30)/(130 - 30) = 0.40$ . This is the position of the reading between the baselines, not the shale volume itself. The nonlinear relationships give a smaller shale volume.
2
Which relationship gives the largest shale volume at an index of 0.5?
1. A.Larionov for Tertiary rocks
2. B.Stieber
3. C.Clavier
4. D.Linear
Show the answer
Correct answer: D. Linear
The linear one gives 0.50, Clavier about 0.31, Stieber 0.25 and Larionov for Tertiary rocks about 0.22. All the empirical relationships lie below the straight line, which is why the linear estimate is treated as an upper limit.
3
A water-bearing bed: the static SP amplitude is −80 mV and the amplitude in the bed we study is −60 mV. What is the shale volume from the SP?
1. A.0.33
2. B.0.75
3. C.0.25
4. D.0.60
Show the answer
Correct answer: C. 0.25
$V_{sh} \approx 1 - 60/80 = 0.25$ . The estimate is valid for a water-bearing bed. In an oil-bearing bed the hydrocarbons make the SP amplitude smaller as well, so this formula would give too high a shale volume.
4
Why is it risky to draw the shale baseline through the highest GR value in the well when the section has uranium-rich shales?
Show the answer
The shale baseline will be too high. The denominator of the index (the difference between the two baselines) grows, so the index, and with it the shale volume, comes out too low. Shaly beds then pass the cut-off, and the net thickness and the volume of hydrocarbons in place come out too high.
5
Why do we often take the smallest of several shale-volume estimates (from the GR, the SP and the neutron–density log)?
Show the answer
Each indicator responds to shale, but also to something else that pushes its estimate up: the GR to uranium and feldspars, the SP to hydrocarbons and thin beds, the neutron–density pair to a wrong matrix. The smallest estimate is closest to the truth, provided we have checked that gas or another disturbance did not push it down.
6
Why is the shale volume from the gap between neutron and density porosity too low in a gas-bearing bed?
Show the answer
Gas contains little hydrogen, so the neutron porosity falls. The low density of gas raises the density porosity. The curves move closer together or swap places, and the gap that should have shown the shale disappears.

Updated: 28 October 2026