Resistivity–porosity crossplots

Quick refresher

Archie's equation in logarithms is $\lg R_t = \lg(aR_w) - m\lg\phi - n\lg S_w$ . Beds with one water and the same water saturation fall on a straight line in the coordinates $\lg R_t$ and $\lg\phi$ .
On the Pickett plot the line of water-bearing beds is set by the exponent $m$ (the slope) and by $aR_w$ , the resistivity on this line at $\phi = 1$ . The lines $S_w = 0.5$ , 0.3 and 0.2 are parallel to it and shifted to the right by the factor $S_w^{-n}$ : at $n = 2$ that is 4, 11.1 and 25 times.
The plot gives $m$ and $aR_w$ by itself, if the water-bearing beds cover a wide range of porosity. It does not give $n$ . We draw the lines of saturation with the chosen $n$ , and at $n = 2.5$ instead of 2 the same point shows more water.
On the Hingle plot the reading of the porosity method is plotted against $R_t^{-1/m}$ . The water line meets the axis of infinite resistivity at the matrix point, and the lines of equal saturation spread out from this point like a fan. Here we need the chosen $m$ , but we do not need the matrix parameter.
We draw the water line through beds that are certainly clean and water-bearing, along the south-west (lower left) edge of the cloud of points. Points to the left of it are a sign of shale, thin beds or a porosity error. They do not mean “more water than pores”.
Hydrocarbons shift the point to the right. Invasion of fresh filtrate that the deep sonde (the tool that reads the resistivity far from the borehole) did not “see through” also shifts it to the right (false oil), and salty filtrate shifts it to the left. Conducting shale shifts it to the left when the water is not very salty. Gas shifts the point down on the neutron porosity (the water is overestimated), and up on the density porosity (it is underestimated).

Key relationships

\lg R_t = \lg(aR_w) - m\lg\phi - n\lg S_w

Archie's equation in logarithms: the basis of the Pickett plot

R_t = \dfrac{aR_w}{\phi^m}\,S_w^{-n}

a line of water saturation: the water line shifted to the right by the factor $S_w^{-n}$

m = \lg\dfrac{R_0(\phi_1)}{R_0(\phi_2)}\Big/\lg\dfrac{\phi_2}{\phi_1}

the cementation exponent from two points of the water line

\rho_b = \rho_{ma} - (\rho_{ma} - \rho_f)\,(aR_w)^{1/m}\,S_w^{-n/m}\,R_t^{-1/m}

the Hingle plot for the density log

Common pitfalls

Drawing the water line through the leftmost points of the cloud without checking that they are clean water-bearing beds.
Putting on one plot intervals with different water or different lithology.
Reading the exponent $m$ from water-bearing beds of a narrow range of porosity.
Forgetting that the lines of saturation were built for the chosen $n$ .
Taking the reading of the deep sonde as $R_t$ when the filtrate has invaded deeply.

After this lesson you can

explain why Archie's equation on logarithmic axes gives a family of parallel straight lines, and what the slope and the position of the lines determine
build a Pickett plot: choose the water-bearing points, draw the line of 100% water saturation, find $m$ and $aR_w$ from it, and draw the lines of water saturation for the chosen $n$
build a Hingle plot from the reading of the density log, and find the matrix density and the water resistivity from it
explain how hydrocarbons, shale, gas, filtrate invasion, and a change of water or lithology shift the points, and not mistake these shifts for saturation
estimate how much the water saturation read from the plot depends on the chosen $n$ and on the choice of the water-bearing points

Archie's equation as a family of straight lines

Archie's equation for a clean rock (Archie's equation and saturation of clean reservoirs) links four quantities of a bed: the resistivity $R_t$ , the porosity $\phi$ , the water saturation $S_w$ and the resistivity of the water $R_w$ . It does so through the coefficients $a$ , $m$ and $n$ . For one bed this is one equation with two unknowns. Even if we know the coefficients, we cannot find the saturation without $R_w$ . But an interval usually has many beds with the same water and the same rock. If we put them all on one plot with suitable coordinates, we can find the parameters that they share from the beds themselves.

Archie's equation in logarithmic coordinates

\lg R_t = \lg(aR_w) - m\lg\phi - n\lg S_w

where:

$R_t$ resistivity of the uninvaded part of the formation (with corrections), Ω·m
$\phi$ porosity, fraction
$S_w$ water saturation, fraction
$aR_w$ the product of the tortuosity factor and the resistivity of the formation water, Ω·m
$m,\ n$ the cementation and saturation exponents, —

Units: resistivities in Ω·m; $\lg$ is the base-10 logarithm

Holds when

Clean beds of one lithology ( $a$ and $m$ are the same) with the same formation water at a close temperature.
$R_t$ is corrected for the borehole, the enclosing rocks and invasion, and the porosity is corrected for environmental conditions and lithology.
The exponent $n$ is the same for all the beds of the interval.

At $S_w = 1$ the last term vanishes. Water-bearing beds fall on a straight line in the coordinates $\lg\phi$ and $\lg R_t$ . Its slope is set by the exponent $m$ , and its position by $aR_w$ . Beds with another, but equal, water saturation fall on a parallel line, shifted along the $\lg R_t$ axis by $n\lg(1/S_w)$ . The same technique, applied to three water-bearing beds, is shown in the topic on finding the water resistivity.

The Pickett plot: construction and reading

On the Pickett plot, the resistivity $R_t$ is plotted along the horizontal axis and the porosity $\phi$ along the vertical axis. Both axes are logarithmic. Usually the decades on both axes are made the same length. (A decade is the span from one power of ten to the next, a factor of ten.) Then we can read the slope of the line directly. When we move one decade along the resistivity axis, the line goes down by $1/m$ of a decade of porosity. When the porosity falls ten times, the resistivity of a water-bearing rock grows by a factor of $10^m$ .

Lines of water saturation on the Pickett plot

R_t = \frac{aR_w}{\phi^m}\,S_w^{-n} = R_0(\phi)\,S_w^{-n}

where:

$R_0(\phi)$ resistivity of a water-bearing rock of the same porosity: a point on the water line, Ω·m
$S_w$ water saturation of the line, fraction
$n$ the chosen saturation exponent, —

Units: Ω·m

Holds when

The same as for Archie's equation in logarithms. $n$ is the chosen value (from core or by analogy).

Each line is a copy of the water line, shifted to the right by the factor $S_w^{-n}$ . At $n = 2$ this is 4 times for $S_w = 0.5$ , 11.1 times for 0.3, 25 times for 0.2 and 100 times for 0.1. We read the water saturation of a bed from the lines between which its point lies, or we calculate it: $S_w = (R_0/R_t)^{1/n}$ .

Order of construction

1
Interval
Choose an interval with one water and one lithology: one horizon, with no change of salinity or of rock type.
2
Points
Plot all the permeable beds. Take the porosity from the density log (in gas-bearing beds, corrected for gas), and take $R_t$ from the deep sonde with corrections (Resistivity logs: electrical and induction tools). Mark the shaly beds by the gamma ray log.
3
Water-bearing beds
Mark the beds that are certainly water-bearing and clean: below the oil–water contact, from tests, from the resistivity profile, or from core.
4
Water line
Draw a straight line through the water-bearing beds, along the south-west edge of the cloud of points. Do not draw it through single extreme points.
5
Parameters
Read $aR_w$ , the resistivity on the line at $\phi = 1$ , and $R_0$ at $\phi = 0.1$ . Then $m = \lg[R_0(0.1)/aR_w]$ , because between $\phi = 0.1$ and 1 there is exactly one decade.
6
Lines of saturation
Draw the lines $S_w = 0.5$ , 0.3 and 0.2 parallel to the water line, shifting it to the right by the factor $S_w^{-n}$ .
7
Saturation and a check
Read the $S_w$ of every bed. Compare $R_w$ and $m$ with water samples, the spontaneous potential (SP, the natural voltage in the borehole) and core (Determining formation water resistivity).

CalculatedPickett plot for the interval of the example: the water line through five water-bearing beds (8–12) and the lines of water saturation at

n = 2

. The productive beds 1–6 lie to the right of the water line, and the shaly bed 7 lies to the left of it. The dotted line is the line of constant bulk volume water 0.055 (advanced section). The porosity scale is stretched: its decade is longer than the resistivity decade, so the lines here are steeper than with equal decades.

Worked exampleWater resistivity, cementation exponent and saturation from a Pickett plot

Given (assumed data)

Interval: sandstones of one horizon; porosity from the density log, $R_t$ from the deep induction sonde with corrections; $a = 1$ , $n = 2$ (assumed)
Productive beds 1–6, $\phi$ and $R_t$ , Ω·m: 0.248 and 17.5; 0.219 and 17.2; 0.193 and 7.75; 0.162 and 5.67; 0.236 and 1.66; 0.105 and 16.1
Bed 7, shaly by the gamma ray log: 0.185 and 0.918
Water-bearing beds 8–12 (below the contact): 0.262 and 0.795; 0.231 and 0.942; 0.204 and 1.25; 0.176 and 1.56; 0.148 and 2.33

Solution

The least-squares line through the five water-bearing beds (the straight line with the smallest sum of squared distances from the points, in the coordinates $\lg\phi$ and $\lg R_t$ ) crosses $\phi = 1$ at $R_t = aR_w = 0.0617$ Ω·m and $\phi = 0.1$ at $R_0 = 4.725$ Ω·m.
The cementation exponent: $m = \lg(4.725/0.0617) = 1.884$ . At $a = 1$ the resistivity of the formation water is $R_w = 0.0617$ Ω·m.
The lines at $n = 2$ : $S_w = 0.5$ is 4 times to the right of the water line, 0.3 is 11.1 times, and 0.2 is 25 times.
Bed 1: on the water line, at the same porosity, $R_0 = 0.0617/0.248^{1.884} = 0.853$ Ω·m, and $S_w = \sqrt{0.853/17.5} = 0.221$ .
The same for the other beds: bed 2 gives 0.250, bed 3 gives 0.420, bed 4 gives 0.579, bed 5 gives 0.751 and bed 6 gives 0.517.
Bed 7: $R_0 = 1.482$ Ω·m is larger than the measured 0.918 Ω·m. The “water saturation” is $\sqrt{1.482/0.918} = 1.27$ , and the point lies to the left of the water line.

Answer:

m \approx 1.884

and

R_w \approx 0.0617

Ω·m. Beds 1 and 2 are productive (

S_w

is 0.221 and 0.250). Beds 3, 4 and the low-porosity bed 6 have a raised water saturation (0.420, 0.579 and 0.517). Bed 5 is in the transition zone (the zone just above the water contact, where the water saturation falls gradually with height), with 0.751. Bed 7 does not fit the model of a clean rock.

What it means. The plot found both $R_w$ and $m$ without water samples or core. The points of the example were calculated with $m = 1.9$ and $R_w = 0.06$ Ω·m, with a scatter of a few per cent, and the plot recovered these parameters to within about 1–3%. A water saturation above one is impossible, so bed 7 does not fit the model of a clean rock: in this rock sequence the shale conducts better than a water-bearing sand of the same porosity. For such a bed we need models of shaly reservoirs (Water saturation in shaly sands). We check the $R_w$ and $m$ that we found against the SP, water samples and core. The data in this example are assumed for illustration.

How to choose the water-bearing points

All the accuracy of the plot rests on the water line, and we draw it through only a few points. So we check each of them: the bed is really water-bearing, clean and thick enough, and its resistivity and porosity were measured reliably. Points with different porosity are useful: the wider the range of porosity, the more accurate the slope.

Water-bearing. The bed lies below the oil–water contact, by tests or sampling, or by the resistivity profile (the change of resistivity from the shallow to the deep sondes; with a fresh mud, the shallow sondes read higher than the deep ones). There are no signs of oil in core or cuttings.
Cleanliness. A low gamma ray reading and a full SP amplitude. Shaly beds do not suit: with moderately salty water they fall to the left of the line, and with strong brine they can fall to the right.
Reliable resistivity. The bed is thicker than the resolution of the sonde, with no effect from the enclosing rocks, and the invasion is shallow or corrected.
Reliable porosity. No washouts in the hole, no gas, and a correct matrix parameter.
One water and one rock. The beds belong to one horizon and one rock type. If the salinity or the lithology changes, we build a separate plot.

If the water-bearing beds cover a narrow range of porosity, we cannot find the slope well. Pairs of “ $m$ and $aR_w$ ” with different slopes describe the points equally well inside the range, and they diverge outside it. Then we take $m$ from core, and we find from the plot only the position of the line. A wrong line makes the biggest errors on beds whose porosity is far from the porosity of the water-bearing points.

Worked exampleWhat happens if a shaly bed gets into the water line

Given (assumed data)

Interval: the same as in the previous example
Error: the water line was drawn through the six leftmost points: the water-bearing beds 8–12 and the shaly bed 7

Solution

The new line: $m = 1.732$ instead of 1.884, and $aR_w = 0.0729$ Ω·m instead of 0.0617.
Bed 1 ( $\phi = 0.248$ ): $S_w = 0.216$ instead of 0.221.
Bed 6 ( $\phi = 0.105$ , far from the water-bearing points): $S_w = 0.474$ instead of 0.517.
The water resistivity from the new line (at $a = 1$ ) is 18% larger.

Answer: The wrong line hardly changes the saturation of beds with a porosity of about 0.2–0.25. But for bed 6 it gives 0.474 instead of 0.517, and it makes

R_w

18% too large.

What it means. The shaly point turned the line: the exponent $m$ became smaller and $aR_w$ larger, and inside the range of porosity of the water-bearing beds these two errors almost cancelled each other. So we cannot see the error from the plot itself. It shows up in a too large $R_w$ when we compare with samples or the SP, and in a too small $m$ when we compare with core. With the wrong line, low-porosity beds are calculated more optimistically than they are. The data in this example are assumed for illustration.

Lines of saturation and the exponent n

The water line does not depend on $n$ : at $S_w = 1$ the last term is zero for any exponent. But the position of the lines of saturation is set by exactly the chosen $n$ . So the Pickett plot finds $R_w$ and $m$ accurately, and it gives the water saturation of productive beds only as accurately as $n$ is known. For water-wet sandstones (where water is the liquid that clings to the grains), usually $n \approx 2$ . In oil-wet rocks (where oil clings to the grains) it is higher (Wettability and interfacial phenomena), and in shaly beds the “apparent” $n$ is lower (Water saturation in shaly sands).

Worked exampleOne point, three saturation exponents

Given (assumed data)

Bed 3 of the example: $\phi = 0.193$ , $R_t = 7.75$ Ω·m; on the water line $R_0 = 1.369$ Ω·m
Variants: $n = 1.8$ ; 2.0; 2.5

Solution

The resistivity index (how many times the resistivity exceeds that of the same rock full of water) $I = R_t/R_0 = 7.75/1.369 = 5.66$ does not depend on $n$ .
$S_w = I^{-1/n}$ : at $n = 1.8$ it is 0.38; at $n = 2$ it is 0.42; at $n = 2.5$ it is 0.50.

Answer: Depending on the chosen

n

, the water saturation of bed 3 is from 0.38 to 0.50.

What it means. A difference of about 0.1 in water saturation, with quite plausible values of $n$ , can decide whether a bed is considered promising. The higher the resistivity index, the more $S_w$ depends on $n$ : the relative error is $\ln I/n^2$ for each unit of error in $n$ . So for productive beds we find $n$ on core with its wettability restored. The data in this example are assumed for illustration.

The Hingle plot

The Pickett plot needs the porosity, and so it needs the matrix parameter: the mineral density for the density log, or the interval transit time (the time a sound wave needs to cross one metre of rock) for the sonic log. In carbonates (limestones and dolomites) of varying composition we often do not know it. The Hingle plot avoids this difficulty. The reading of a porosity method is linear in porosity, so we can plot it directly, without a conversion, if the other axis carries a quantity that is proportional to the porosity of the water-bearing rock. From Archie's equation, $\phi = (aR_w)^{1/m} S_w^{-n/m} R_t^{-1/m}$ . At a constant saturation, porosity is proportional to $R_t^{-1/m}$ .

The Hingle plot for the density log

\rho_b = \rho_{ma} - (\rho_{ma} - \rho_f)\,(aR_w)^{1/m}\,S_w^{-n/m}\,R_t^{-1/m}

where:

$\rho_b$ bulk density from the density log, g/cm³
$\rho_{ma}$ matrix density, g/cm³
$\rho_f$ density of the fluid in the zone of investigation (usually the filtrate), g/cm³
$R_t^{-1/m}$ the horizontal coordinate of the plot (the abscissa); at $m = 2$ it is $1/\sqrt{R_t}$ , that is, the square root of the conductivity, $(\text{Ω·m})^{-1/m}$

Units: densities in g/cm³, resistivities in Ω·m

Holds when

The exponent $m$ is chosen in advance: the horizontal scale is built for it.
The matrix, the fluid and the water are constant through the interval. The reading of the porosity method is linear in porosity: exactly for the density log, for the sonic log with the time-average equation, and approximately for the neutron log.
The beds are clean, and $R_t$ is corrected.

As $R_t \to \infty$ the horizontal coordinate goes to zero, and all the lines of equal saturation meet at the point $\rho_b = \rho_{ma}$ . So the crossing of the water line with the vertical axis gives the matrix density. The slope of the water line is, in absolute value, $(\rho_{ma} - \rho_f)(aR_w)^{1/m}$ . The line of $S_w$ is steeper by the factor $S_w^{-n/m}$ . The plots for the sonic log (the crossing gives the interval transit time of the matrix) and for the neutron log are built in the same way.

CalculatedHingle plot for the carbonate interval of the example: bulk density against

1/\sqrt{R_t}

(the scale is labelled with the values of

R_t

). The water line through the water-bearing beds W1–W6 meets the axis

R_t = \infty

at a matrix density of 2.76 g/cm³. From this point the lines

S_w = 0.5

, 0.3 and 0.2 spread out. On the right is the porosity scale, built from the matrix density that we found.

Worked exampleMatrix density and saturation from a Hingle plot

Given (assumed data)

Interval: limestone with a varying share of dolomite; fresh filtrate, $\rho_f = 1.00$ g/cm³; $a = 1$ , $m = n = 2$ (assumed)
Water-bearing beds W1–W6, $\rho_b$ , g/cm³ and $R_t$ , Ω·m: 2.637 and 6.93; 2.593 and 3.99; 2.549 and 2.38; 2.505 and 1.70; 2.461 and 1.22; 2.417 and 0.911
Productive beds P1–P3: 2.478 and 21.9; 2.566 and 20.0; 2.434 and 11.4

Solution

The values of $1/\sqrt{R_t}$ (the horizontal coordinates) of the water-bearing beds run from 0.38 to 1.05. The line through them (by least squares) crosses the axis $R_t = \infty$ at $\rho_b = 2.76$ g/cm³. It passes through $\rho_b = 2.596$ at the horizontal coordinate 0.5 (that is, at $R_t = 4$ Ω·m).
The matrix density is $\rho_{ma} = 2.76$ g/cm³. The slope is $(2.596 - 2.76)/0.5 = -0.328$ , so $\sqrt{aR_w} = 0.328/(2.76 - 1.00) = 0.1864$ and $aR_w = 0.0347$ Ω·m.
Bed P1: on the water line, a density of 2.478 corresponds to the horizontal coordinate $(2.76 - 2.478)/0.328 = 0.8598$ . The bed itself has $1/\sqrt{21.9} = 0.2137$ . The water saturation is their ratio: $S_w = 0.2137/0.8598 = 0.249$ . The porosity is $(2.76 - 2.478)/(2.76 - 1.00) = 0.160$ .
The same for beds P2 and P3: $S_w$ is 0.378 and 0.298, and the porosity is 0.110 and 0.185.
Suppose we had taken the matrix as limestone (2.71 g/cm³) and calculated with Archie's equation and the same $R_w$ . For bed P1 the porosity would be 0.136 and $S_w = 0.293$ , instead of 0.249.

Answer: The matrix density is 2.76 g/cm³,

R_w \approx 0.0347

Ω·m, and the water saturation of beds P1–P3 is 0.249, 0.378 and 0.298.

What it means. A density of 2.76 g/cm³ lies between calcite (2.71) and dolomite (2.87). About a third of the solid phase is dolomite, which we check on crossplots of the porosity methods (Lithology and porosity from crossplots). A limestone matrix assumed out of habit would have underestimated the porosity and overestimated the water saturation. The points of the example were calculated at $\rho_{ma} = 2.76$ g/cm³ and $R_w = 0.035$ Ω·m, and the plot recovered them to within about 1%. The data in this example are assumed for illustration.

Pickett and Hingle: what we need and what we get

: Axes
Pickett plot: $\lg R_t$ and $\lg\phi$
Hingle plot: the reading of the porosity method and $R_t^{-1/m}$

: We must know
Pickett plot: the matrix parameter (for the porosity)
Hingle plot: the exponent $m$ (for the scale)

: We find
Pickett plot: $m$ and $aR_w$
Hingle plot: the matrix parameter and $aR_w$

: Lines of $S_w$
Pickett plot: parallel straight lines
Hingle plot: a fan from the matrix point

: Convenient for
Pickett plot: sandstones with a known matrix and an unknown $m$
Hingle plot: carbonates of varying composition with a known $m$

	Pickett plot	Hingle plot
Axes	$\lg R_t$ and $\lg\phi$	the reading of the porosity method and $R_t^{-1/m}$
We must know	the matrix parameter (for the porosity)	the exponent $m$ (for the scale)
We find	$m$ and $aR_w$	the matrix parameter and $aR_w$
Lines of $S_w$	parallel straight lines	a fan from the matrix point
Convenient for	sandstones with a known matrix and an unknown $m$	carbonates of varying composition with a known $m$

What shifts the points

The position of a point on the plot reflects not only the saturation. It also reflects any error in the resistivity or the porosity, and any difference of the bed from the model of a clean rock. An experienced interpreter reads the cloud of points as a diagnosis: each cause shifts the point in its own direction.

Directions of the shifts on the Pickett plot

Cause: Oil or gas in the bed
Shift: to the right
Why: higher $R_t$ at the same porosity
How to recognise it: the lines of saturation, the resistivity profile, tests

Cause: Deep invasion of fresh filtrate
Shift: to the right
Why: the high-resistivity invaded zone raises the reading of the deep sonde
How to recognise it: the curves of different depths of investigation disagree (corrections)

Cause: Invasion of salty filtrate
Shift: to the left
Why: the low-resistivity invaded zone lowers the reading
How to recognise it: the shallow sondes read less than the deep ones

Cause: Shale, with water that is not very salty
Shift: to the left
Why: shale conducts in parallel with the water
How to recognise it: gamma ray log, SP

Cause: A thin bed among shales
Shift: to the left
Why: the neighbouring shales lower $R_t$
How to recognise it: thickness less than the resolution of the sonde

Cause: Gas, neutron porosity
Shift: down
Why: the neutron porosity is too low
How to recognise it: neutron–density crossover, gas shows (gas detected in the drilling mud)

Cause: Gas, density porosity
Shift: up
Why: the density porosity is too high
How to recognise it: the same

Cause: Washouts in the hole
Shift: up
Why: the density log overestimates the porosity
How to recognise it: caliper log, density correction

Cause: A different water or rock
Shift: a separate line
Why: different $aR_w$ or $m$
How to recognise it: SP, samples, a change of lithology

Cause	Shift	Why	How to recognise it
Oil or gas in the bed	to the right	higher $R_t$ at the same porosity	the lines of saturation, the resistivity profile, tests
Deep invasion of fresh filtrate	to the right	the high-resistivity invaded zone raises the reading of the deep sonde	the curves of different depths of investigation disagree (corrections)
Invasion of salty filtrate	to the left	the low-resistivity invaded zone lowers the reading	the shallow sondes read less than the deep ones
Shale, with water that is not very salty	to the left	shale conducts in parallel with the water	gamma ray log, SP
A thin bed among shales	to the left	the neighbouring shales lower $R_t$	thickness less than the resolution of the sonde
Gas, neutron porosity	down	the neutron porosity is too low	neutron–density crossover, gas shows (gas detected in the drilling mud)
Gas, density porosity	up	the density porosity is too high	the same
Washouts in the hole	up	the density log overestimates the porosity	caliper log, density correction
A different water or rock	a separate line	different $aR_w$ or $m$	SP, samples, a change of lithology

“To the right” means towards larger resistivities, and “up” means towards larger porosity, on the logarithmic axes as in the lesson's figures. Vugs in carbonates (not in the hole) change $m$ , and they also shift the points to the right (more).

SchematicThe directions in which different causes shift a point on the Pickett plot, from a water-bearing bed W and from a gas-bearing bed G (a diagram; the lengths of the arrows are conventional, except for the gas ones, which are calculated in the example below).

Worked exampleA gas-bearing bed on the plot: what porosity

Given (assumed data)

Bed of the same interval: gas-bearing sandstone with a true porosity of 0.20 and $S_w = 0.25$ ; $R_t = 20.4$ Ω·m
Porosity from the logs: density 0.249, neutron 0.125 (residual gas in the flushed zone, the rock next to the hole that the filtrate has invaded)
Water line: from the first example: $aR_w = 0.0617$ Ω·m, $m = 1.884$ ; $n = 2$

Solution

With the neutron porosity: $R_0 = 0.0617/0.125^{1.884} = 3.102$ Ω·m, and $S_w = \sqrt{3.102/20.4} = 0.39$ .
With the density porosity: $R_0 = 0.847$ Ω·m, and $S_w = 0.20$ .
With the porosity corrected for gas from two methods (more), which is 0.20: $R_0 = 1.280$ Ω·m, and $S_w = 0.25$ .

Answer: By the neutron porosity the bed looks markedly more water-bearing (0.39). By the density porosity it looks a little better than it is (0.20). The true water saturation is 0.25.

What it means. On a plot with the neutron porosity, a gas-bearing bed sinks down towards the water line, and it can even lie on it: the gas will be missed. In gas the density porosity is overestimated by less than the neutron porosity is underestimated, so the water saturation from it is underestimated only a little. But the volume of gas in the pores, $\phi(1 - S_w)$ , is overestimated by about a third with this porosity: $0.249\cdot(1 - 0.20)$ against $0.20\cdot(1 - 0.25)$ . For gas-bearing beds we plot the porosity corrected for gas. The data in this example are assumed for illustration.

It is useful to plot on the same graph the readings of the shallow sonde, which measures the resistivity of the flushed zone $R_{xo}$ (Borehole environment: invaded zone and mudcake). In water-bearing beds the flushed zone is saturated with filtrate, and the points of the shallow sonde fall on their own line, the line $S_{xo} = 1$ . It is parallel to the water line and shifted by the factor $R_{mf}/R_w$ . We use it to check the resistivity of the filtrate $R_{mf}$ . In productive beds, the distance between the points of the deep and shallow sondes tells us whether the filtrate pushed out part of the hydrocarbons. This is covered quantitatively in the topic Resistivity ratio methods and movable oil. Points that lie correctly on one line and incorrectly on the other point to measurement problems: deep invasion, the effect of enclosing rocks, or shallow invasion, in which the shallow sonde does not see the flushed zone.

Order of work and checks

1
Data preparation
Corrections of the resistivity and the porosity, and marks on shaly, thin and gas-bearing beds and on intervals with washouts.
2
Zoning
A separate plot for each interval with one water and one lithology. A recalculation to one temperature is not needed if the interval is short.
3
Water line
Draw it through checked, clean water-bearing beds. With a narrow range of porosity, take $m$ from core.
4
Checking the parameters
Check $R_w$ against samples, the SP and the apparent water resistivity (the water resistivity that each bed would show if it were water-bearing; Determining formation water resistivity). Check $m$ against core. Check the matrix density against crossplots of the porosity methods.
5
Saturation
Draw the lines with $n$ from core. For beds that do not fit the model of a clean rock, use models of shaly reservoirs (Water saturation in shaly sands).
6
Diagnosis
Give every point outside the picture an explanation: shale, gas, invasion, a thin bed, a different water. Unexplained points are a reason to recheck the original curves.

Terms and symbols

Pickett plot: The resistivity and porosity of beds on logarithmic axes: the water line and the parallel lines of water saturation.
Hingle plot: The reading of a porosity method against $R_t^{-1/m}$ : the water line and a fan of lines of saturation from the matrix point.
Water line: The straight line on which water-bearing beds fall ( $S_w = 1$ ). On the Pickett plot it is set by the exponent $m$ and by $aR_w$ .
Line of water saturation: A line of equal $S_w$ , built with the chosen exponent $n$ .
$aR_w$: The resistivity on the water line at a porosity of one. At $a = 1$ it is the resistivity of the formation water.
Bulk volume water $\phi S_w$: The share of the rock volume occupied by water.

Common mistakes and misconceptions

Wrong

“We draw the water line through the leftmost points.”

In fact

The leftmost ones are often shaly and thin beds. We draw the line through checked clean water-bearing beds, and we explain the other points.

Wrong

“The slope gave $m = 1.6$ , so the rock is fractured.”

In fact

The slope is distorted by a narrow range of porosity, shaly points, invasion and a change of lithology. Before we draw a conclusion about the structure of the pores, we check these causes and compare with core.

Wrong

“The point lies on the 0.3 line, so the water saturation is 0.3.”

In fact

Only if $n$ is right, and only for a clean bed with a reliable resistivity and porosity. The lines were built with the chosen $n$ , and for shaly beds we need other models.

Wrong

“Let us put the whole section on one plot.”

In fact

Beds with different water or lithology fall on different lines, and a common straight line through them gives false $m$ and $R_w$ . We build the plot by intervals.

Wrong

“Neutron porosity suits the plot in any bed.”

In fact

In gas-bearing beds it is too low, and the gas can “lie down” on the water line. For gas we take the porosity corrected from two methods.

In depth: advanced detail

Lines of constant bulk volume water

The product $\phi S_w$ is the bulk volume water. In the zone of irreducible water saturation (water held in the pores that cannot be produced) of one rock type it is almost constant (more). On the Pickett plot such beds fall on a line of their own.

A line of constant bulk volume water on the Pickett plot

R_t = aR_w\,(\phi S_w)^{-n}\,\phi^{\,n - m}

where:

$\phi S_w$ bulk volume water: the share of the rock volume occupied by water, fraction
$aR_w,\ m,\ n$ the parameters of the water line and the saturation exponent, Ω·m; —

Units: Ω·m

Holds when

The same as for the lines of water saturation.

At $m = n$ the factor $\phi^{n - m}$ equals one, and the lines of constant bulk volume water on the Pickett plot are vertical. All the beds with the same $\phi S_w$ have the same resistivity: $\phi S_w = \sqrt{aR_w/R_t}$ at $n = 2$ . At $m < n$ the lines slope slightly to the upper right.

Worked exampleWhich beds of the interval are at irreducible water saturation

Given (assumed data)

Beds 1–6 of the first example: the water saturation from the plot: 0.221; 0.250; 0.420; 0.579; 0.751; 0.517

Solution

The bulk volume water $\phi S_w$ : bed 1 is 0.0548; bed 2 is 0.0548; bed 6 is 0.0543.
Beds 3, 4 and 5: 0.0811; 0.0938; 0.177.
The line $\phi S_w = 0.055$ on the plot: at a porosity of 0.25 it passes through $R_t = 17.4$ Ω·m, and at 0.105 through 15.7 Ω·m, that is, almost vertically.

Answer: Beds 1, 2 and 6, at different porosities, have the same bulk volume water of about 0.055, and they lie almost on one vertical line. In beds 3–5 it is larger, and it differs from bed to bed.

What it means. The same $\phi S_w$ in beds of different porosity is a sign of irreducible water saturation (water held in the pores that cannot be produced). Such beds most likely give oil with no water, although the water saturation of the low-porosity bed 6 is above 0.5. Beds 3–5 are in the transition zone, where water inflow is possible. We check the conclusion with capillary pressure curves (core measurements of how much water stays in the rock at a given pressure) and tests. The data in this example are assumed for illustration.

Connections to other topics

Archie's equation and saturation of clean reservoirs — the equation that the plot turns into a family of straight lines, and the meaning of $a$ , $m$ and $n$ .
Determining formation water resistivity — the apparent water resistivity and other ways to find $R_w$ , against which we check the plot.
Water saturation in shaly sands — why shaly beds do not fall on the water line, and how to calculate their saturation.
Resistivity logs: electrical and induction tools — how to get $R_t$ and correct the readings for invasion and enclosing rocks.
Borehole environment: invaded zone and mudcake — the flushed zone and the invasion of filtrate, which shift the points.
Gas, shale, carbonate and pressure effects on porosity logs — porosity in gas-bearing and shaly beds.
Lithology and porosity from crossplots — the matrix parameter that the Hingle plot finds, against the crossplots of porosity methods.
Fluid saturation — bulk volume water and irreducible water.
Wettability and interfacial phenomena — what the saturation exponent $n$ depends on.
Resistivity ratio methods and movable oil — saturation from the ratio of the resistivities of the flushed zone and the formation.

Check yourself

1
On the water line of a Pickett plot the resistivity is 8.0 Ω·m at a porosity of 0.10, and 2.3 Ω·m at a porosity of 0.20. What is the cementation exponent?
1. A.1.5
2. B.2.0
3. C.1.8
4. D.2.3
Show the answer
Correct answer: C. 1.8
$m = \lg(8.0/2.3)/\lg(0.20/0.10) = 0.541/0.301 \approx 1.8$ . When the porosity doubles, the resistivity of the water-bearing rock falls by a factor of $2^m \approx 3.5$ .
2
The point of a bed lies to the left of the water line. What does this mean?
1. A.The bed is saturated with hydrocarbons
2. B.The bed does not fit the model of a clean rock, or was measured wrongly: shale, a thin bed, overestimated porosity, salty filtrate
3. C.The bed is gas-bearing
4. D.The water saturation of the bed is more than one: it is “oversaturated” with water
Show the answer
Correct answer: B. The bed does not fit the model of a clean rock, or was measured wrongly: shale, a thin bed, overestimated porosity, salty filtrate
To the left of the water line means a lower resistivity than that of a completely water-saturated clean rock of the same porosity, that is, $S_w > 1$ , which is physically impossible. So the model or the measurements are wrong. Shale conducts current, a thin bed is lowered by neighbouring shales, the porosity is too high, or the zone of invasion of salty filtrate lowers the reading. Hydrocarbons and gas shift the point to the right.
3
By what factor, at the same porosity, is the resistivity on the line $S_w = 0.25$ larger than on the water line, at $n = 2$ and at $n = 2.5$ ?
Show the answer
The factor is $S_w^{-n}$ . At $n = 2$ it is $0.25^{-2} = 16$ , and at $n = 2.5$ it is $0.25^{-2.5} = 32$ . With a larger $n$ the lines spread farther apart. A bed that lay on the 0.25 line at $n = 2$ will be near the 0.33 line at $n = 2.5$ ( $16^{-1/2.5} \approx 0.33$ ).
4
What must we set in advance to build a Hingle plot?
1. A.The density or interval transit time of the matrix
2. B.The resistivity of the formation water
3. C.The saturation exponent $n$
4. D.The cementation exponent $m$
Show the answer
Correct answer: D. The cementation exponent $m$
The horizontal scale $R_t^{-1/m}$ is built for the chosen $m$ . The plot finds the matrix parameter and $aR_w$ by itself: the first from the crossing of the water line with the axis $R_t = \infty$ , the second from its slope. The exponent $n$ is needed only for the lines of saturation.
5
In the water-bearing beds of an interval there is deep invasion of fresh filtrate, and the deep sonde reads a resistivity 40% above the true one. What will happen if we draw the water line through these beds, and how will it affect the productive beds?
Show the answer
The points of the water-bearing beds are shifted to the right, so the water line will run to the right of the true one: $aR_w$ will come out about 40% too large. The water saturation of the productive beds, calculated from such a line, will be overestimated by a factor of $\sqrt{1.4} \approx 1.18$ , if their readings are not distorted in the same way. If, on the other hand, we draw the line correctly, the water-bearing beds with deep invasion will lie to the right of it, and they can be taken for weakly oil-bearing. So we correct the resistivities for invasion before we build the plot.
6
Why does a Pickett plot with the neutron porosity give a pessimistic estimate in gas-bearing beds, and one with the density porosity a slightly optimistic one?
Show the answer
Gas is poor in hydrogen, and the neutron porosity in it is too low. The point goes down, towards the water line, where for a lower porosity the resistivity of the water-bearing rock is higher. The calculated water saturation is overestimated, and the gas can be missed. In gas the density porosity is too high (gas is light), the point goes up, and the water saturation is underestimated a little. In the lesson's example these are 0.39 and 0.20, against the true 0.25.

Updated: 8 October 2026