Porosity from the sonic log

Quick refresher

The link between transit time and porosity is empirical, not a mass balance. The time-average equation describes clean compacted rocks whose intergranular pores (the spaces between the grains) are filled with liquid.
The parameters are 182 µs/m for the sandstone matrix (167 for a well-cemented one), 156 for limestone, 143 for dolomite and 620 µs/m for the filtrate (the liquid that enters the rock from the drilling mud). The choice of matrix shifts the result by 2–3 porosity units. So we check the parameters against core (rock samples taken from the well) or another method.
In undercompacted sands the equation gives porosity that is too high. We divide the result by a factor $C_p$ of 1 to 1.6. Roughly, it is the transit time of the neighbouring shales divided by 328 µs/m. More accurately, we find it from a reference water-bearing bed.
The empirical transform is a curve instead of a straight line. It has no fluid transit time, and it needs no correction for undercompaction. At a porosity of 5–20% it gives 1–2.5 units more than the time-average equation.
Shale makes sonic porosity too high: the effective porosity (the volume of the pores outside the shale material) is $\phi_S - V_{sh}\phi_{S,sh}$ . Oil and gas show only in high-porosity sands with shallow invasion of the filtrate.
Vugs hardly show in $\Delta t$ . So sonic porosity is smaller than total porosity (all the pore space, vugs included), and the difference estimates the secondary porosity, if the parameters have been checked in a bed without vugs.

Key relationships

\phi_S = \dfrac{\Delta t - \Delta t_{ma}}{\Delta t_f - \Delta t_{ma}}

time-average equation for clean compacted rocks

\phi_S = \dfrac{\Delta t - \Delta t_{ma}}{\Delta t_f - \Delta t_{ma}}\cdot\dfrac{1}{C_p}

the same with a correction for undercompaction; $C_p \approx \Delta t_{sh}/328$ with $\Delta t_{sh}$ in µs/m

\phi_S \approx C\,\dfrac{\Delta t - \Delta t_{ma}}{\Delta t}

empirical transform, $C \approx 0.67$

\phi_e = \phi_S - V_{sh}\,\phi_{S,sh}

effective porosity of a shaly bed

\phi_2 = \phi_t - \phi_S

estimate of secondary porosity: total minus sonic

Common pitfalls

Using 182 and 620 µs/m for any rock without checking them against core or density porosity.
Applying the time-average equation to loose sands without a correction for undercompaction.
Calculating porosity at points where the transit time is distorted by cycle skipping (the receiver misses the first arrival of the wave and times a later cycle, so the transit time jumps up).
Taking a difference from density porosity for vugs without checking the parameters in a bed without vugs.

After this lesson you can

write the time-average equation, list its assumptions and calculate porosity with it
choose the interval transit times of the matrix and of the fluid, and estimate how the choice affects the result
recognise an undercompacted section (one where the sands are not yet squeezed tight by the weight of the rocks above), apply a correction for undercompaction, and find its factor from a reference bed (a clean water-bearing bed where the porosity is known from another method)
apply the empirical transform and explain how it differs from the time-average equation
correct sonic porosity for shale and estimate the effect of oil and gas
estimate the secondary porosity by comparison with the density and neutron logs, and calibrate the calculation against core

Travel time as a sum of times

The sonic log records the interval transit time $\Delta t$ (Sonic log). This is the time that the compressional wave needs to cross one metre of rock. The rock matrix is fast. The fluid in the pores is three to four times slower. So the more pores there are, the larger $\Delta t$ is. The simplest model says that the wave crosses the matrix and the fluid one after the other. It spends on each a time that is proportional to its share of the volume. The times add up. That is why it is called the time-average equation. M. R. J. Wyllie and co-workers proposed it in 1956, from laboratory measurements on rock samples. Russian-language literature also spells his name as it sounds, “Villi” (in Russian letters).

Time-average equation (Wyllie)

\Delta t = \phi\,\Delta t_f + (1 - \phi)\,\Delta t_{ma} \quad\Rightarrow\quad \phi_S = \frac{\Delta t - \Delta t_{ma}}{\Delta t_f - \Delta t_{ma}}

where:

$\phi_S$ sonic porosity, fraction
$\phi$ porosity: the volume fraction of the pores that are filled with fluid, fraction
$\Delta t$ interval transit time from the log, µs/m
$\Delta t_{ma}$ interval transit time of the matrix, that is, of a tight rock with no pores (table below), µs/m
$\Delta t_f$ interval transit time of the fluid in the pores of the zone of investigation (the part of the rock that the tool reads), which is the mud filtrate, µs/m

Units: the interval transit times in the same units, µs/m or µs/ft; porosity as a fraction

Holds when

A clean rock without shale, with a known mineral composition.
The rock is compacted and cemented. In loose sands the equation gives porosity that is too high.
The pores are small and evenly spread, either intergranular or intercrystalline (between grains or between crystals). Vugs and fractures hardly enter the result.
The pores in the zone of investigation are filled with liquid: the mud filtrate.
The equation is empirical. It is not derived from the mechanics of elastic waves. It describes experimental data for such rocks.

At $\Delta t = \Delta t_{ma}$ the porosity is zero, and at $\Delta t = \Delta t_f$ it is one. For sandstone the denominator is 438 µs/m. So every 4–5 µs/m of interval transit time is one porosity unit (0.01, or one percentage point).

Interval transit times of the matrix and of the fluid for the porosity calculation

Rock or fluid: Sandstone
$\Delta t$ , µs/m: 182; cemented 167
$\Delta t$ , µs/ft: 55.5; 51.0
For the empirical transform: 184 µs/m (56)

Rock or fluid: Limestone
$\Delta t$ , µs/m: 156 (from 156 to 143)
$\Delta t$ , µs/ft: 47.5 (47.6–43.5)
For the empirical transform: 161 µs/m (49)

Rock or fluid: Dolomite
$\Delta t$ , µs/m: 143
$\Delta t$ , µs/ft: 43.5
For the empirical transform: 144 µs/m (44)

Rock or fluid: Anhydrite
$\Delta t$ , µs/m: 164
$\Delta t$ , µs/ft: 50.0
For the empirical transform: —

Rock or fluid: Filtrate of a fresh mud
$\Delta t$ , µs/m: 620
$\Delta t$ , µs/ft: 189
For the empirical transform: not needed

Rock or fluid: Filtrate of a salty mud
$\Delta t$ , µs/m: about 607
$\Delta t$ , µs/ft: 185
For the empirical transform: not needed

Rock or fluid	$\Delta t$ , µs/m	$\Delta t$ , µs/ft	For the empirical transform
Sandstone	182; cemented 167	55.5; 51.0	184 µs/m (56)
Limestone	156 (from 156 to 143)	47.5 (47.6–43.5)	161 µs/m (49)
Dolomite	143	43.5	144 µs/m (44)
Anhydrite	164	50.0	—
Filtrate of a fresh mud	620	189	not needed
Filtrate of a salty mud	about 607	185	not needed

The matrix values are for tight rock without pores. In µs/m they are rounded to one unit. For a mixture of minerals we take an intermediate value from their volume fractions in the solid part. All these numbers are parameters of an empirical relationship, not exact properties of the minerals. So we check them against core.

Worked examplePorosity of a compacted sandstone and the cost of each parameter

Given (assumed data)

Reading of the sonic log: $\Delta t = 265$ µs/m
Rock: clean compacted quartz sandstone; about 280 µs/m in the neighbouring shales
Parameters: $\Delta t_{ma} = 182$ µs/m (55.5 µs/ft), $\Delta t_f = 620$ µs/m (189 µs/ft)

Solution

In metric units: $\phi_S = \dfrac{265 - 182}{620 - 182} = \dfrac{83}{438} = 0.189$ .
In oilfield units: $265\cdot 0.3048 = 80.77$ µs/ft; $\phi_S = \dfrac{80.77 - 55.5}{189 - 55.5} = 0.189$ . The result is the same.
If we take a faster matrix, 167 µs/m: $\dfrac{265 - 167}{620 - 167} = 0.216$ , which is 2.7 porosity units more.
If the mud is salty and $\Delta t_f = 607$ µs/m: $\dfrac{265 - 182}{607 - 182} = 0.195$ , which is 0.6 porosity units more.
If the reading is 6 µs/m too high (about 2 µs/ft): $\dfrac{271 - 182}{620 - 182} = 0.203$ , which is 1.4 porosity units more.

Answer: The sonic porosity is about 19%.

What it means. The system of units does not affect the result. All that matters is that the three interval transit times are in the same units. The result depends most on the choice of the matrix: 2.7 porosity units, against 1.4 for the error of the reading and 0.6 for the fluid. So we do not take the matrix transit time from a table blindly. We check it against core, or against density porosity in water-bearing beds that we know are clean. The data in this example are assumed for illustration.

Undercompacted sands

The time-average equation is “tuned” to rocks with a rigid framework. While the sand grains only touch each other, the framework is soft. The same porosity then gives a noticeably larger $\Delta t$ : the straight line of porosity against $\Delta t$ is steeper. The measure of compaction is the effective stress. This is the difference between the overburden pressure (the weight of the rocks above) and the pore pressure (the pressure of the fluid in the pores). Young sands stay undercompacted (not yet squeezed tight) while this difference is less than about 28–35 MPa (4000–5000 psi). With normal pore pressure that means depths down to about 2–3 km. With abnormally high pore pressure it means greater depths too. The simplest remedy is to divide the result of the equation by an empirical correction factor.

Correction for undercompaction in the time-average equation

\phi_S = \frac{\Delta t - \Delta t_{ma}}{\Delta t_f - \Delta t_{ma}}\cdot\frac{1}{C_p}, \qquad C_p \approx \frac{\Delta t_{sh}}{328}

where:

$\phi_S$ sonic porosity with a correction for undercompaction, fraction
$C_p$ compaction correction factor (on charts, $B_{cp}$ ); not less than 1, —
$\Delta t_{sh}$ interval transit time in the neighbouring shales, µs/m
$\Delta t,\ \Delta t_{ma},\ \Delta t_f$ the same quantities as in the time-average equation, µs/m

Units: $\Delta t_{sh}$ in µs/m; if $\Delta t_{sh}$ is in µs/ft, the divisor is 100

Holds when

Estimating $C_p$ from the neighbouring shales is a rough rule of thumb. Shales compact together with sands. A $\Delta t_{sh}$ above 328 µs/m (100 µs/ft) is taken as a sign of undercompaction. At smaller values we apply no correction ( $C_p = 1$ ).
It is more reliable to find $C_p$ in a reference bed where the porosity is known independently (see the calibration section below).
The bed is clean, and its pores are filled with liquid.

The factor shows how many times the time-average equation overestimates the porosity in a given interval. Usual values are from 1 to 1.6.

Worked exampleYoung loose sand: with and without the correction

Given (assumed data)

Reading in clean water-bearing sand A: $\Delta t = 380$ µs/m (116 µs/ft)
Neighbouring shales: $\Delta t_{sh} = 410$ µs/m (125 µs/ft)
Parameters: $\Delta t_{ma} = 182$ µs/m, $\Delta t_f = 620$ µs/m

Solution

Without the correction: $\phi_S = \dfrac{380 - 182}{620 - 182} = 0.452$ . For a sand this is implausibly high.
In the shales $\Delta t$ is above 328 µs/m, so the section is undercompacted. The correction factor is $C_p \approx 410/328 = 1.25$ .
With the correction: $\phi_S = 0.452/1.25 = 0.362$ .

Answer: The porosity of the sand is about 36%, not 45%.

What it means. Without the correction the pore volume would be overestimated by a quarter. The rule for shales is only a first approximation. The check against a reference bed in the calibration section shows how accurate it is in this section. The data in this example are assumed for illustration.

Shales also obey compaction: their interval transit time falls smoothly with depth. An interval where it departs from this trend towards larger values is a sign of undercompaction and of abnormally high pore pressure (Abnormal formation pressures: origin, prediction and recognition). Sands in such an interval are less compacted than the depth suggests, and their correction is larger.

The empirical transform

The second way is to give up the straight line. People compared $\Delta t$ with independently measured porosity in a large number of beds. The comparisons showed that the real relationship is not a straight line. At low and medium porosity the rock is faster than the time-average equation predicts, while high-porosity sands are much slower. This relationship is linked with the names of Raymer, Hunt and Gardner (1980). Over the usual range of porosity, a simple formula fits it. This formula is the empirical transform.

Empirical transform of interval transit time to porosity

\phi_S = C\,\frac{\Delta t - \Delta t_{ma}}{\Delta t}

where:

$\phi_S$ sonic porosity from the empirical transform, fraction
$C$ empirical constant: from 0.625 to 0.7 for different authors; later comparisons point to 0.67, and old charts use 0.7, —
$\Delta t_{ma}$ interval transit time of the matrix for this transform: 184 µs/m (56 µs/ft) for quartz sandstone, 161 (49) for limestone, 144 (44) for dolomite, µs/m
$\Delta t$ interval transit time from the log, µs/m

Units: the interval transit times in the same units; porosity as a fraction

Holds when

A clean rock, with the pores filled with liquid.
The formula approximates an experimental relationship in the range of ordinary porosity, up to about 35–37%. In looser rocks the grains no longer make a rigid framework, and the relationship is different.
For gas-bearing beds we reduce the constant (see the section on oil and gas).

The formula has no fluid transit time, and it needs no correction for undercompaction. The curvature of the relationship already accounts for the fast growth of $\Delta t$ in high-porosity sands. At a porosity of 5 to 20% the transform gives 1–2.5 porosity units more than the time-average equation with $\Delta t_{ma} = 182$ µs/m. Near 24% the results coincide, and above that the time-average equation gives higher and higher values.

CalculatedCalculated interval transit time of a clean water-bearing sandstone against porosity for three models: the time-average equation (

\Delta t_{ma} = 182

\Delta t_f = 620

µs/m); the same equation for an undercompacted sand with

C_p = 1.25

; and the empirical transform (

C = 0.67

\Delta t_{ma} = 184

µs/m; shown up to a porosity of 37%). The points are the readings of the two examples: 265 and 380 µs/m.

The curve also explains why sandstone has two “table” matrix values. A straight line with $\Delta t_{ma} = 167$ µs/m differs from the empirical curve by about one porosity unit or less at a porosity of 10–20%. The line with 182 µs/m meets the curve at about 24%. When we choose the faster matrix for tight sandstones, we are in effect fitting the straight line to the observations in the range of porosity where we need it.

Worked exampleTwo sandstones by the empirical transform

Given (assumed data)

Compacted sandstone: $\Delta t = 265$ µs/m; 0.189 by the time-average equation
Young sand A: $\Delta t = 380$ µs/m; 0.452 by the time-average equation, 0.362 with the correction from the shales
Parameters of the transform: $C = 0.67$ , $\Delta t_{ma} = 184$ µs/m

Solution

Compacted sandstone: $0.67\cdot(265 - 184)/265 = 0.205$ . This is about one and a half porosity units more than by the time-average equation.
Young sand: $0.67\cdot(380 - 184)/380 = 0.346$ . This is 11 porosity units less than by the equation without the correction, and 1.6 units less than with the correction.

Answer: The transform gives 20% for the compacted sandstone and 35% for the young sand.

What it means. For the compacted rock the two ways differ by one and a half porosity units. That is less than the difference between the two matrix values in the first example. For the loose sand the transform returns a plausible value by itself, without any information about the shales, and it is close to the corrected value. Which way is closer to the truth in a given section is decided by comparison with core or with another porosity method. The data in this example are assumed for illustration.

Calibration: the reference bed and core

The tool measures neither the interval transit time of the matrix, nor the factor $C_p$ , nor the constant $C$ . The interpreter chooses them. We can check the choice only where the porosity is known independently. There are four sources. In an undercompacted section each of them also gives the factor $C_p$ : the ratio of the uncorrected sonic porosity to the independent porosity.

Density log in a clean bed with liquid in the pores and a smooth hole (Porosity from the density log): $C_p = \phi_S/\phi_D$ .
Resistivity of a water-bearing bed. In a clean bed fully saturated with water, the ratio of its resistivity $R_0$ to the resistivity of the water $R_w$ is the formation resistivity factor. We find the porosity from it (Archie's equation and saturation of clean reservoirs).
Neutron log in water-bearing intervals (Porosity from the neutron log). We turn to it when the sands are shaly and the first two ways are unreliable. Shales make both the neutron and the sonic porosity too high. So we treat the difference between the two estimates in water-bearing sands as an effect of undercompaction. This is an approximation, not an exact rule.
Core is a direct comparison of $\Delta t$ with the porosity of the samples. It replaces the table parameters with a local relationship (Core analysis).

Worked exampleCorrection factor from a reference water-bearing bed

Given (assumed data)

Section: young sands from the example on undercompaction; in sand A ( $\Delta t = 380$ µs/m) the density log is unreliable, because the hole is washed out
Reference bed B: clean water-bearing sand, smooth hole: $\Delta t = 372$ µs/m, $\rho_b = 2.11$ g/cm³
Parameters: matrix 182 µs/m and 2.65 g/cm³; filtrate 620 µs/m and 1.00 g/cm³

Solution

Density porosity of bed B: $\phi_D = \dfrac{2.65 - 2.11}{2.65 - 1.00} = 0.327$ .
Sonic porosity of bed B without the correction: $\dfrac{372 - 182}{620 - 182} = 0.434$ .
The correction factor: $C_p = 0.434/0.327 = 1.33$ .
Sand A: $\phi_S = 0.452/1.33 = 0.340$ . The rule for shales ( $C_p = 1.25$ ) gave 0.362.

Answer: From the reference bed,

C_p \approx 1.33

, and the porosity of sand A is about 34%.

What it means. The rule for shales gave a porosity 2.2 porosity units higher. We trust the reference bed more. It belongs to the sands themselves, not to the shales next to them. The conditions are that the bed is really clean, that its pores hold liquid, and that the density log in it is reliable. We carry the factor over only to sands of the same age and the same depth. Compaction grows with depth, and $C_p$ becomes smaller. The data in this example are assumed for illustration.

Core gives more than one factor: from it we build a local relationship of interval transit time and porosity. The order of calibration is the same as for the density log (calibration against core). The graph and the meaning of the parameters are specific to this method.

Calibration against core

1
Depth matching
We match the depths of the core and the log. We use the core gamma log (a gamma ray record made on the core itself) or characteristic tight and porous layers.
2
Reservoir conditions
We bring the core porosity to the effective stress of the reservoir (Rock compressibility and the effect of stress on core properties).
3
The same scale
We average the core values in a window equal to the span of the tool (the distance between the receivers), about 0.6 m. We exclude points with cycle skipping and intervals where the hole is badly washed out.
4
The graph
We plot $\Delta t$ from the log against the core porosity. For rocks of one type the points lie along a straight line $\Delta t = A + B\,\phi$ .
5
Parameters
The intercept $A$ is the interval transit time at zero porosity, that is, the apparent transit time of the matrix. The slope $B$ plays the role of the difference $\Delta t_f - \Delta t_{ma}$ . The ratio $B/(\Delta t_f - A)$ is the correction factor $C_p$ . If it is more than one, the rock behaves as undercompacted.
6
Check
We calculate porosity as $(\Delta t - A)/B$ and compare it with core. The mean difference should be close to zero, and the scatter should be explainable.

Worked exampleA local relationship from seven core–log points

Given (assumed data)

Core porosity (at reservoir conditions, averaged over the window of the tool): 0.084; 0.117; 0.142; 0.168; 0.195; 0.221; 0.246
$\Delta t$ from the log at the same depths, µs/m: 217; 231; 247; 258; 275; 286; 301
“Default” parameters: 182 and 620 µs/m

Solution

The least-squares line (the straight line with the smallest sum of squared deviations from the points): $A = 171.9$ µs/m, $B = 521.6$ µs/m.
The intercept of 171.9 µs/m (52.4 µs/ft) is the apparent interval transit time of the matrix. It lies between 167 and 182 µs/m, as for a cemented sandstone.
The slope is larger than $620 - 171.9 = 448.1$ µs/m. The rock behaves as slightly undercompacted: $C_p = 521.6/448.1 = 1.16$ .
For a reading of 300 µs/m, the “default” parameters give $\phi_S = 0.269$ , and the local line gives 0.246.
On average over the seven points, the “default” calculation is 0.9 porosity units above core. At the most porous point it is 2.6 units above.

Answer: The local relationship is

\Delta t = 171.9 + 521.6\,\phi

. The table parameters overestimate the porosity, and the higher the porosity, the more.

What it means. The mean error of the table parameters is small, but it is not constant. At low porosity the calculation underestimates by about half a unit. At high porosity it overestimates by 1.6–2.6 units. A single shift cannot remove such an error. This is typical of sonic logging: not only the intercept is wrong, the slope is wrong too. The parameters of the line are fitting numbers. The sum $A + B$ = 693 µs/m formally plays the role of the interval transit time of the fluid, but formation water never has such a value. The line is valid for this bed, and for the porosity range of 8–25% from which it was built. How to draw the line, and what to check when core and log disagree, is covered in the topic on density porosity. The data in this example are assumed for illustration.

CalculatedInterval transit time from the log against core porosity for the data of the example: the points, the least-squares line (solid), and the time-average equation with the “default” parameters of 182 and 620 µs/m (dashed). The lines cross near a porosity of 12%. Above that the table parameters overestimate the porosity.

A shaly bed

Shales are usually slower than the sandstone matrix: their interval transit time is from 230 to 500 µs/m. The time-average equation gives them an apparent “porosity” that has nothing to do with the effective pore volume. At $\Delta t_{sh} = 300$ µs/m it is about 0.27. Shale layers bring this “porosity” into the reading in proportion to their share of the volume of the bed. We subtract this contribution, as we do in the density log (shaly bed) and in the neutron log.

Effective porosity of a shaly bed from the sonic log

\phi_e = \phi_S - V_{sh}\,\phi_{S,sh}, \qquad \phi_{S,sh} = \frac{\Delta t_{sh} - \Delta t_{ma}}{\Delta t_f - \Delta t_{ma}}

where:

$\phi_e$ effective porosity: the volume of the pores outside the shale material, as a fraction of the volume of the whole rock, fraction
$\phi_S$ porosity calculated by the time-average equation as for a clean rock, fraction
$V_{sh}$ shale volume, fraction
$\phi_{S,sh}$ the “sonic porosity of shale”: what the same equation shows in a clean shale, fraction
$\Delta t_{sh}$ interval transit time of the shales; we take it from the log in the neighbouring shales, µs/m
$\Delta t_{ma},\ \Delta t_f$ interval transit times of the matrix and of the fluid, the same as in the calculation of $\phi_S$ , µs/m

Units: the interval transit times in the same units; porosity and shale volume as fractions

Holds when

Laminated shale: the wave crosses the sand and shale layers one after the other, and the travel times add up. For thin layers this is an approximation.
The shale layers are the same as the neighbouring shales from which $\Delta t_{sh}$ was taken.
The shale volume $V_{sh}$ was found independently (Determining shale volume).
The rock is compacted, with liquid in the pores.
Shale material dispersed in the pores of a compacted sandstone changes $\Delta t$ little. For it the correction is too large.

This correction is larger than the one in the density log and smaller than the one in the neutron log. The “porosity of shale” is usually 0.2–0.4 for the sonic method, about 0.1 for the density method and 0.30–0.45 for the neutron method.

In other unit systems (1)

Laminated bed: the interval transit time of the sand layers

\Delta t_{sd} = \frac{\Delta t - V_{sh}\,\Delta t_{sh}}{1 - V_{sh}}

$\Delta t_{sd}$ is the interval transit time of the sand layers themselves; $V_{sh}$ is the share of the shale layers in the volume of the bed.

There is no constant: the contribution of the shale layers is removed from the reading, and the remainder is assigned to the sand layers. We calculate the porosity of the layers $\phi_{sd}$ from $\Delta t_{sd}$ with the usual equation, and $\phi_e = (1 - V_{sh})\,\phi_{sd}$ .

Worked exampleShaly sandstone: a third estimate for the same bed

Given (assumed data)

Bed: the shaly water-bearing sandstone from the examples of the topics Porosity from the density log and Porosity from the neutron log: $V_{sh} = 0.25$
Interval transit time: $\Delta t = 282$ µs/m; in the neighbouring shales $\Delta t_{sh} = 300$ µs/m
Parameters: 182 and 620 µs/m; the section is compacted, $C_p = 1$

Solution

As for a clean rock: $\phi_S = \dfrac{282 - 182}{620 - 182} = 0.228$ .
The “porosity of shale”: $\phi_{S,sh} = \dfrac{300 - 182}{620 - 182} = 0.269$ .
The effective porosity: $\phi_e = 0.228 - 0.25\cdot 0.269 = 0.161$ .
If the shale is laminated: $\Delta t_{sd} = \dfrac{282 - 0.25\cdot 300}{0.75} = 276$ µs/m, and the porosity of the layers is 0.215. Check: $0.75\cdot 0.215 = 0.161$ .

Answer: The effective porosity is about 16%, and the porosity of the sand layers is 21.5%.

What it means. The density and neutron logs gave 0.162 for this bed, from the raw readings of 0.188 and 0.257. The sonic log gives 0.161, from the raw 0.228. In this example three methods with three different corrections for shale agree on 16%. This is what a consistent model of the bed looks like. The correction of the sonic method, 6.7 porosity units, is about 2.6 times larger than that of the density method and almost a third smaller than that of the neutron method. The data in this example are assumed for illustration.

Oil and gas in the zone of investigation

How much the fluid affects $\Delta t$ depends on the stiffness of the framework. In a compacted sandstone with its usual porosity of 15–25%, the framework carries the wave. Beds with water, oil and gas give practically the same reading. The looser and more porous the sand, the more the fluid contributes. But the sonde sees only the rock right at the borehole wall (Borehole environment: invaded zone and mudcake), and the hydrocarbons must remain there after the filtrate has invaded. Both conditions are met in high-porosity sands (30% and more) with a high oil or gas saturation and shallow invasion. In them $\Delta t$ is larger than in the same sand with water, and the porosity is overestimated.

Reducing correction of sonic porosity for oil and gas

\phi = k_{\text{hc}}\,\phi_S

where:

$\phi$ porosity with a correction for the fluid, fraction
$\phi_S$ porosity by the time-average equation, calculated as for a water-bearing rock, fraction
$k_{\text{hc}}$ empirical reducing multiplier for hydrocarbons (hc): about 0.9 for oil-bearing beds and about 0.7 for gas-bearing beds, —

Units: fraction

Holds when

A rough rule of thumb for high-porosity sands with shallow invasion. The multiplier depends on the saturation of the zone of investigation and on the stiffness of the rock, so it cannot be exact.
We apply the correction only when the overestimate is obvious, for example by comparison with the water-bearing part of the same bed or with density porosity.
In the empirical transform, for gas-bearing beds we take a smaller constant, $C = 0.6$ , instead of a multiplier.

The multipliers show the order of the effect: gas can raise the sonic porosity by almost one and a half times, and oil by a tenth. In compacted rocks no correction is needed at all.

Worked exampleOne sand, two parts: with water and with gas

Given (assumed data)

Bed: clean high-porosity sand; 310 µs/m in the neighbouring shales, so the section is compacted
Water-bearing part: $\Delta t = 325$ µs/m
Gas-bearing part: $\Delta t = 386$ µs/m; the invasion of the filtrate is shallow

Solution

Water-bearing part: $\phi_S = \dfrac{325 - 182}{620 - 182} = 0.326$ .
Gas-bearing part without the correction: $\dfrac{386 - 182}{620 - 182} = 0.466$ . This is 1.4 times more, although the sand is the same.
With the correction for gas: $0.7\cdot 0.466 = 0.326$ .
The empirical transform with the constant for gas: $0.6\cdot(386 - 184)/386 = 0.314$ .

Answer: The porosity of the sand is about 33%. The reading of 47% in the gas-bearing part is the effect of the gas, not a property of the rock.

What it means. The correction brought the porosity of the gas-bearing part back to the value of the water-bearing part almost exactly. That is because the data of the example were chosen to fit the multiplier 0.7. In reality the multiplier only gives the order of magnitude. For the gas-bearing part, the empirical transform with the gas constant $C = 0.6$ gives 0.314. For the water-bearing part with $C = 0.67$ it gives 0.291, against 0.326 from the time-average equation. So the constant for gas removes the effect of gas only approximately, and the two ways of calculating differ even without gas. That is why we compare values obtained in one way. If density and neutron logs are available, we find the porosity of a gas-bearing bed from their combination (gas in the readings of the neutron log), and the sonic method serves as a check. The data in this example are assumed for illustration.

Vugs: comparison with the density and neutron logs

Dissolution vugs are rounded voids from millimetres to centimetres across. They weaken the rock framework much less than the same volume of intergranular pores does. The wave goes around them through solid rock. So in a vuggy carbonate $\Delta t$ reflects mainly the intergranular porosity, for which the time-average equation remains valid, and the vugs are almost invisible in it. The density and neutron logs, by contrast, count all the voids filled with fluid. As a result the sonic porosity of a vuggy rock is smaller than the total porosity. The difference estimates the secondary porosity (porosity that formed after the rock was deposited, by dissolution or fracturing).

Estimating secondary porosity by comparing methods

\phi_2 = \phi_t - \phi_S

where:

$\phi_2$ estimate of the secondary (vug) porosity: the secondary porosity index, fraction
$\phi_t$ total porosity from the density or neutron log, or from their combination, fraction
$\phi_S$ sonic porosity by the time-average equation, fraction

Units: fraction

Holds when

A clean rock of known mineral composition, with liquid in the pores. Gas and shales distort the three methods in different ways.
The parameters for calculating $\phi_S$ have been checked in an interval without vugs. There $\phi_S$ must agree with $\phi_t$ . Otherwise the error of the equation itself enters the difference.
Vugs are not quite invisible to the wave, and they increase $\Delta t$ a little. So the difference is usually somewhat smaller than their true volume. We exclude from the calculation the points with cycle skipping, which fractures often cause.
This is an index, not a measurement: the errors of the two methods add up.

The ratio $\phi_2/\phi_t$ shows what share of the voids is vugs. A positive difference in a clean carbonate points to vugs. We confirm it with core, borehole image logs and testing. Open fractures usually take only a fraction of a per cent of the volume and hardly affect the difference. A weakened signal and cycle skipping can point to them.

Worked exampleVugs in limestone: what the density and sonic logs see

Given (assumed data)

Rock and fluid: clean limestone: matrix 2.71 g/cm³ and 156 µs/m; fresh filtrate: 1.00 g/cm³ and 620 µs/m
Bed without vugs (from core): $\rho_b = 2.505$ g/cm³, $\Delta t = 211$ µs/m
Bed under study: $\rho_b = 2.40$ g/cm³, $\Delta t = 220$ µs/m

Solution

Check of the parameters in the bed without vugs: the density porosity is $\dfrac{2.71 - 2.505}{2.71 - 1.00} = 0.120$ , and the sonic porosity is $\dfrac{211 - 156}{620 - 156} = 0.119$ . The estimates practically coincide, so the parameters are suitable.
Bed under study, density porosity: $\phi_t = \dfrac{2.71 - 2.40}{2.71 - 1.00} = 0.181$ .
Sonic porosity: $\phi_S = \dfrac{220 - 156}{620 - 156} = 0.138$ .
Secondary porosity: $\phi_2 = 0.181 - 0.138 = 0.043$ . Its share of the total porosity is 24%.

Answer: The total porosity is about 18%, and about 4 porosity units of it are vugs.

What it means. The density fell markedly, but the interval transit time rose by only 9 µs/m. The wave hardly noticed the vugs. If we had calculated porosity from the sonic log alone, the pore volume would have been underestimated by a quarter. The opposite mistake is to take a difference between methods for vugs without checking the parameters. In a bed without vugs the difference must be close to zero. The data in this example are assumed for illustration.

Three porosity methods in different conditions

Condition: Clean bed with liquid
Density: correct
Neutron: correct
Sonic: correct in a compacted rock

Condition: Residual gas
Density: overestimates
Neutron: underestimates
Sonic: hardly changes in a compacted rock

Condition: Shale
Density: depends on the density of the shale
Neutron: strongly overestimates
Sonic: overestimates

Condition: Vugs in the rock
Density: counts them
Neutron: counts them
Sonic: hardly notices them

Condition: Loose uncompacted bed
Density: correct
Neutron: correct
Sonic: overestimates without a correction

Condition: Washout of the hole
Density: strongly distorted
Neutron: distorted
Sonic: compensated

Condition	Density	Neutron	Sonic
Clean bed with liquid	correct	correct	correct in a compacted rock
Residual gas	overestimates	underestimates	hardly changes in a compacted rock
Shale	depends on the density of the shale	strongly overestimates	overestimates
Vugs in the rock	counts them	counts them	hardly notices them
Loose uncompacted bed	correct	correct	overestimates without a correction
Washout of the hole	strongly distorted	distorted	compensated

“Correct” means: with a known mineral matrix. In loose gas-bearing sands the sonic porosity is overestimated, and cycle skipping is possible in gas and in washouts. In Russian-language logging practice one word is used for two different things: a dissolution void in the rock (a vug) and an enlarged section of the borehole (a washout). The first is a property of the bed. The second is a nuisance for the measurement. Using the three methods together on crossplots is the subject of the topic Lithology and porosity from crossplots. Counting gas and shale quantitatively from several methods at once is the subject of Gas, shale, carbonate and pressure effects on porosity logs.

Order of calculation and checks

1
Record quality
Check the units of the scale. Exclude points with cycle skipping, noise spikes and the arrival of the wave through the mud (signs).
2
Zones
Divide the section into beds and rock types. From the interval transit time of the shales, judge whether the section is compacted. Above 328 µs/m a correction for undercompaction is needed.
3
Parameters
Choose the interval transit times of the matrix and of the fluid from the lithology and the type of mud. For a mixture of minerals take an intermediate value.
4
Transform
In compacted rocks use the time-average equation. In undercompacted ones use the same equation with the factor $C_p$ , or the empirical transform. Where there is core or a reference bed, replace the table parameters with local ones.
5
Shale and hydrocarbons
In shaly beds go over to the effective porosity. In high-porosity sands with oil or gas, check whether the result is overestimated.
6
Check
Compare with the density and neutron porosity and with core. In a clean water-bearing bed of known lithology the estimates must agree. In carbonates, a positive difference between the total and the sonic porosity points to vugs.

Terms and symbols

Sonic porosity $\phi_S$: Porosity calculated from the interval transit time of the compressional wave. It reflects mainly the intergranular pores.
Matrix transit time $\Delta t_{ma}$: The interval transit time of a tight rock with no pores. In the calculation it is a parameter of the empirical relationship, which we check against core.
Fluid transit time $\Delta t_f$: The interval transit time of what fills the pores in the zone of investigation, usually the mud filtrate.
Compaction correction factor $C_p$: A number not less than one, by which we divide the result of the time-average equation in undercompacted sands.
Empirical transform: An experimental curve of “interval transit time against porosity” with a constant $C$ . It needs no fluid transit time and no correction for undercompaction.
“Sonic porosity of shale” $\phi_{S,sh}$: The value that the time-average equation gives in a clean shale. We multiply it by the shale volume and subtract the product from $\phi_S$ .
Secondary porosity index $\phi_2$: The difference between the total porosity from the density and neutron logs and the sonic porosity. An estimate of the volume of vugs.

Common mistakes and misconceptions

Wrong

“For sandstone we take 182 and 620 µs/m: these are constants.”

In fact

They are parameters of an empirical relationship. Real sandstones usually give an apparent transit time of the matrix from 167 to 182 µs/m, and a different slope. Without a check against core or density porosity, the error reaches 2–3 porosity units.

Wrong

“The time-average equation is good for any sand.”

In fact

It describes compacted, cemented rocks. In loose sands it overestimates the porosity by tens of per cent of its value. We need a correction for undercompaction, or the empirical transform.

Wrong

“The correction for shale is to subtract $V_{sh}$ .”

In fact

We subtract $V_{sh}\phi_{S,sh}$ : the shale volume multiplied by the “porosity” that the equation gives in the clean shales of the same section. We take the interval transit time of the shales from the log, not from a handbook.

Wrong

“In a gas-bearing bed the sonic porosity is always overestimated.”

In fact

Only in high-porosity sands with shallow invasion. In compacted rocks gas hardly changes $\Delta t$ . But it can cause cycle skipping, which is easily taken for high porosity.

Wrong

“Sonic porosity is lower than density porosity, so one of the curves is wrong.”

In fact

In a clean carbonate this is a usual sign of vugs: the sonic log hardly sees them. The difference is an estimate of the secondary porosity, if the parameters of the calculation have been checked in a bed without vugs.

In depth: advanced detail

Where the constant C in the empirical transform comes from

For porosity up to about 37%, we can write the experimental relationship behind the transform with velocities. The velocity in the rock is the sum of the matrix velocity with the weight $(1 - \phi)^2$ and the fluid velocity with the weight $\phi$ . In interval transit times it looks like this.

Raymer–Hunt–Gardner relationship

\frac{1}{\Delta t} = \frac{(1 - \phi)^2}{\Delta t_{ma}} + \frac{\phi}{\Delta t_f}

where:

$\Delta t$ interval transit time of the rock, µs/m
$\Delta t_{ma},\ \Delta t_f$ interval transit times of the matrix and of the fluid, µs/m
$\phi$ porosity, fraction

Units: the interval transit times in the same units

Holds when

An empirical relationship for clean water-bearing rocks.
The porosity is less than about 37%. In looser rocks the grains no longer make a rigid framework, and the relationship is different.

Multiply both sides by $\Delta t_{ma}$ , and we get $\Delta t_{ma}/\Delta t = (1 - \phi)^2 + \phi\,r$ , where $r = \Delta t_{ma}/\Delta t_f$ . Then $(\Delta t - \Delta t_{ma})/\Delta t = \phi\,(2 - r - \phi)$ . So the constant of the working formula is $C = 1/(2 - r - \phi)$ , and it depends only weakly on porosity.

Worked exampleThe constant C at different porosities of a sandstone

Given (assumed data)

Matrix and fluid: $\Delta t_{ma} = 184$ µs/m, $\Delta t_f = 620$ µs/m
Porosity: 10, 20 and 30%

Solution

The ratio is $r = 184/620 = 0.297$ .
$C = 1/(2 - r - \phi)$ : at 10% it is 0.624, at 20% 0.665, and at 30% 0.713.
Check of the working formula with $C = 0.67$ . The full relationship gives interval transit times of 219, 263 and 318 µs/m at porosities of 10, 20 and 30%. The working formula returns 0.107, 0.201 and 0.282 from them.

Answer: In the porosity range of 10–30% the constant changes from 0.624 to 0.713.

What it means. This also explains the spread of values from 0.625 to 0.7 among different authors: a constant fitted to beds of different porosity comes out different. The value 0.67 is almost exact at a porosity of about 20%. At 10% it overestimates the porosity by 0.7 porosity units, and at 30% it underestimates it by 1.8. The data in this example are assumed for illustration.

Porosity from the shear-wave interval transit time

Tools with an array of receivers, and dipole tools, also measure the interval transit time $\Delta t_s$ of the shear wave (how exactly). In a shear wave the rock moves sideways, across the direction of travel. Experience shows that it grows with porosity in much the same way as that of the compressional wave. Both ways of calculating, the time-average equation and the empirical transform, apply to it with their own matrix parameters.

Approximate parameters for the shear wave

Rock: Sandstone
$\Delta t_s$ , µs/m: 282
$\Delta t_s$ , µs/ft: 86

Rock: Limestone
$\Delta t_s$ , µs/m: 295
$\Delta t_s$ , µs/ft: 90

Rock: Dolomite
$\Delta t_s$ , µs/m: 249
$\Delta t_s$ , µs/ft: 76

Rock: Anhydrite
$\Delta t_s$ , µs/m: 328
$\Delta t_s$ , µs/ft: 100

Rock: “Fluid” (a conventional value)
$\Delta t_s$ , µs/m: 1,148
$\Delta t_s$ , µs/ft: 350

Rock	$\Delta t_s$ , µs/m	$\Delta t_s$ , µs/ft
Sandstone	282	86
Limestone	295	90
Dolomite	249	76
Anhydrite	328	100
“Fluid” (a conventional value)	1,148	350

The values are approximate, and we refine them against core.

The “fluid” row is a fitting parameter, not a property of the liquid: there are no shear waves in a liquid. Its value is chosen so that the time-average equation reproduces the observed relationship of $\Delta t_s$ with porosity. The advantage of the shear wave is that it depends only weakly on gas. Gas hardly changes the resistance of the rock to shear. So the porosity from $\Delta t_s$ in a gas-bearing bed is less distorted than the porosity from the interval transit time of the compressional wave.

Connections to other topics

Sonic log — how the tool measures the interval transit time, what the time depends on, and when it cannot be trusted.
Porosity from the density log — density porosity is a mass balance. We compare sonic porosity with it, and we find the correction factor from it.
Porosity from the neutron log — neutron porosity: a third estimate for the same beds, and its corrections for shale and gas.
Porosity — the kinds of porosity, and why different methods see secondary porosity differently.
Determining shale volume — where the shale volume for the correction comes from.
Core analysis — the core porosity against which we calibrate the calculation.
Abnormal formation pressures: origin, prediction and recognition — undercompaction of shales and sands, and abnormally high pore pressure.
Gas, shale, carbonate and pressure effects on porosity logs — gas, shale and lithology across several porosity methods at once.
Lithology and porosity from crossplots — crossplots with the sonic log, and secondary porosity on them.

Check yourself

1
In which bed will the time-average equation with table parameters give the most reliable porosity?
1. A.Vuggy limestone
2. B.Shaly sandstone with gas
3. C.Clean compacted sandstone with water in the pores
4. D.Young loose sand at a depth of 800 m
Show the answer
Correct answer: C. Clean compacted sandstone with water in the pores
The equation describes clean compacted rocks with intergranular pores filled with liquid. In a loose sand it overestimates the porosity. In a vuggy limestone it does not count the vugs. In a shaly bed the shale makes the result too high. In a gas-bearing bed both an effect of the fluid and cycle skipping are possible.
2
On a log with an oilfield scale, in a clean compacted sandstone $\Delta t = 85$ µs/ft. Find the porosity by the time-average equation in metric and in oilfield units.
Show the answer
In metric units: $85/0.3048 \approx 279$ µs/m. The numerator is $279 - 182 = 97$ , the denominator is $620 - 182 = 438$ µs/m, so $\phi_S \approx 0.22$ . In oilfield units: the numerator is $85 - 55.5 = 29.5$ , the denominator is $189 - 55.5 = 133.5$ µs/ft, and the ratio is the same, 0.22. The main thing is not to mix units within one formula.
3
Young sands: in the neighbouring shales $\Delta t_{sh} = 394$ µs/m (120 µs/ft), and in a clean water-bearing sand $\Delta t = 361$ µs/m. Estimate the porosity of the sand ( $\Delta t_{ma} = 182$ , $\Delta t_f = 620$ µs/m).
Show the answer
Without the correction: the numerator is $361 - 182 = 179$ , the denominator is 438 µs/m, and the ratio is 0.41. In the shales $\Delta t$ is above 328 µs/m, so the section is undercompacted: $C_p \approx 394/328 = 1.2$ . The porosity with the correction is $0.41/1.2 \approx 0.34$ . The empirical transform gives a close value: $0.67\cdot(361 - 184)/361 \approx 0.33$ .
4
Clean limestone, with the pores filled with liquid. The total porosity from the density and neutron logs is 0.16, and the sonic porosity is 0.11. In a neighbouring bed without vugs both estimates agree. What does this mean?
1. A.There is gas in the bed
2. B.The sonic sonde is faulty
3. C.The bed is shaly
4. D.About 5 porosity units are vugs, which the sonic log hardly notices
Show the answer
Correct answer: D. About 5 porosity units are vugs, which the sonic log hardly notices
The density and neutron logs count all the voids, and the sonic log counts mainly the intergranular pores. The difference $0.16 - 0.11 = 0.05$ is an estimate of the secondary porosity, about 30% of all the voids. Gas would raise the density porosity and lower the neutron porosity, not move both in the same way. Shale would raise the sonic porosity, not lower it. The agreement of the estimates in the neighbouring bed shows that the parameters of the calculation are right.
5
Shaly compacted sandstone: $\Delta t = 292$ µs/m, $V_{sh} = 0.20$ , 310 µs/m in the neighbouring shales; the parameters are 182 and 620 µs/m. Find the effective porosity, and the porosity of the sand layers if the shale is laminated.
Show the answer
As for a clean rock: $110/438 = 0.251$ . The “porosity of shale” is $128/438 = 0.292$ . The effective porosity is $0.251 - 0.20\cdot 0.292 = 0.193$ . For the layers, we subtract the contribution of the shales from the reading, $0.20\cdot 310 = 62$ , and divide by 0.80. This gives 287.5 µs/m, so $\phi_{sd} = 105.5/438 = 0.241$ . Check: $0.80\cdot 0.241 = 0.193$ .
6
From core and log we get the line $\Delta t = 175 + 500\,\phi$ (µs/m). What do its parameters mean, and what porosity does it give at 265 µs/m?
Show the answer
The intercept of 175 µs/m is the apparent interval transit time of the matrix. It is faster than the “table” value (182 µs/m). The slope of 500 µs/m is larger than the difference $620 - 175 = 445$ , so the rock behaves as slightly undercompacted: $C_p = 500/445 \approx 1.12$ . At 265 µs/m the porosity is $(265 - 175)/500 = 0.18$ . The table parameters would give 0.19. The line is valid only for its own bed and for the porosity interval from which it was built.

Updated: 3 October 2026