Resistivity logs: electrical and induction tools

Quick refresher

Current in a rock is carried by the ions (charged particles) of the water in the pores. Mineral grains, oil and gas are insulators. So the resistivity of a rock depends on how salty and how warm the water is, on how much water there is, and on how the pore space is built.
An electrode sonde measures the potential difference for a known current. We multiply $\Delta U/I$ by the sonde constant $K$ and get the apparent resistivity $R_a$ : the resistivity of a uniform medium that would give the same reading.
The guard electrodes of a laterolog sonde force the current into the formation in a thin sheet. The laterolog is accurate in salty muds and in high-resistivity and thin beds, but it does not work in oil-based muds (muds whose liquid is oil, which does not conduct current).
An induction sonde measures the conductivity of the rock and does not need a conductive mud. It is the best choice with fresh mud and with oil-based mud. Its accuracy falls in very high-resistivity rocks, with very salty mud and, because of the skin effect, in very conductive rocks.
No sonde shows $R_t$ directly. We take three curves with different depths of investigation and use the step-invasion model (the invaded zone is assumed to have a sharp edge). With it we find $R_t$ , $R_{xo}$ and the invasion diameter $d_i$ .
Make the corrections in this order: first for the borehole, then for the enclosing rocks and the bed thickness, then for invasion.

Key relationships

R_a = K\,\dfrac{\Delta U}{I},\qquad K = 4\pi\,\dfrac{AM\cdot AN}{MN}

apparent resistivity of an electrode sonde, and the sonde constant for point electrodes

C\,[\text{mS/m}] = \dfrac{1000}{R\,[\text{Ω·m}]}

conductivity and resistivity

\delta \approx 503\sqrt{R/f}

skin depth in metres ( $R$ in Ω·m, $f$ in Hz)

R_k = J_k\,R_{xo} + (1 - J_k)\,R_t

reading of a laterolog sonde in the step-invasion model

Common pitfalls

Taking the reading of the deep sonde as $R_t$ without corrections for invasion, the borehole and the enclosing rocks.
Using induction logging in high-resistivity carbonates or with very salty mud, where its error is large.
Relying on a laterolog in a hole drilled with oil-based mud.
Confusing apparent resistivity with the resistivity of any one zone.

After this lesson you can

explain why the resistivity of a rock is set by the water in its pores, and convert resistivity to conductivity and back
describe how potential and gradient sondes, the laterolog and induction logging work, and what apparent resistivity and the sonde constant are
choose a method for measuring $R_t$ from the type of mud, the contrast $R_{mf}/R_w$ and the expected resistivity of the rocks
estimate the skin depth and the error of an induction sonde, and explain why it works worse in very conductive and in very high-resistivity sections
find $R_t$ , $R_{xo}$ and the invasion diameter from three curves with different depths of investigation, and list the corrections in the right order

Resistivity of rocks and fluids

Electrical resistivity ( $R$ ) is how strongly a material resists electric current. Take a sample of length $L$ and cross section $A$ , put a voltage $U$ across it and measure the current $I$ that passes. Its resistivity is $R = (U/I)(A/L)$ , and the unit is the ohm-metre (Ω·m). The reverse of resistivity is the electrical conductivity ( $C$ ). In logging it is given in millisiemens per metre (mS/m; old English-language materials write mmho/m). In Russian-language literature resistivity is written $\rho$ , and the value that a sonde measures is called the apparent resistivity ( $\rho_\text{a}$ ).

Resistivity and conductivity

R = \frac{U}{I}\cdot\frac{A}{L},\qquad C = \frac{1}{R}

where:

$R$ electrical resistivity, Ω·m
$U$ voltage across the sample, V
$I$ current through the sample, A
$A$ cross-sectional area of the sample, perpendicular to the current, m²
$L$ length of the sample along the current, m
$C$ electrical conductivity, S/m

Units: SI. In logging, conductivity is usually given in mS/m: $C = 1000/R$ with $R$ in Ω·m

Holds when

The sample is uniform, and the current flows along it evenly.
Resistivity and conductivity refer to one temperature: in solutions and rocks they change a lot when the temperature rises.

The ratio $U/I$ is the electrical resistance of the sample, in ohms. The factor $A/L$ turns it into a property of the material that does not depend on the size of the sample. A sonde in a borehole does the same: it divides the voltage by the current and multiplies by its own geometric factor. Conductivity is convenient where the contributions add up in parallel (side by side, like separate paths for the current), as in induction logging.

The solid part of most rocks, the mineral matrix (quartz, feldspars, calcite, dolomite), carries almost no current. The exceptions are rare minerals that conduct electrons, such as pyrite and graphite. Oil and gas are insulators too. The current is carried by the ions of the salts dissolved in the pore water, and by positive ions (cations) held on the surface of clay particles. This is why shaly (clay-bearing) beds conduct better than clean ones; see Water saturation in shaly sands. So three things set the resistivity of a rock. The first is the resistivity of the water itself: it is lower when the salinity and the temperature are higher. The second is the amount of water (the porosity and the water saturation). The third is the structure of the pore space, meaning how narrow and winding the paths of the current are. Archie's equation links them in numbers (Archie's equation and saturation of clean reservoirs).

Approximate resistivities of rocks and waters at reservoir conditions

Medium: Formation brines
Resistivity, Ω·m: 0.01–0.1
Why: high salinity and temperature

Medium: Seawater (about 20 °C)
Resistivity, Ω·m: about 0.2
Why: about 35 g/L of dissolved salts

Medium: Fresh water
Resistivity, Ω·m: 10–100 and more
Why: few dissolved salts

Medium: Shales, mudstones
Resistivity, Ω·m: 0.5–10
Why: a lot of bound water, and the clay minerals conduct too

Medium: Water-bearing sandstones and limestones
Resistivity, Ω·m: 0.2–10
Why: depend on $R_w$ (the resistivity of the formation water) and on porosity

Medium: Oil- and gas-bearing reservoirs
Resistivity, Ω·m: from 5–10 up to hundreds and thousands
Why: water fills only a small part of the pores

Medium: Tight carbonates, anhydrite, rock salt
Resistivity, Ω·m: $10^3$ and above
Why: almost no water

Medium	Resistivity, Ω·m	Why
Formation brines	0.01–0.1	high salinity and temperature
Seawater (about 20 °C)	about 0.2	about 35 g/L of dissolved salts
Fresh water	10–100 and more	few dissolved salts
Shales, mudstones	0.5–10	a lot of bound water, and the clay minerals conduct too
Water-bearing sandstones and limestones	0.2–10	depend on $R_w$ (the resistivity of the formation water) and on porosity
Oil- and gas-bearing reservoirs	from 5–10 up to hundreds and thousands	water fills only a small part of the pores
Tight carbonates, anhydrite, rock salt	$10^3$ and above	almost no water

These are orders of magnitude, for guidance. The actual values depend on salinity, temperature, porosity and shale volume. The range is so wide that resistivity curves are drawn on a logarithmic scale (each step is a multiple of the one before), usually from 0.2 to 2000 Ω·m.

Electrode sondes and apparent resistivity

The simplest electrode sonde is a few metal rings (electrodes) on an insulated body. The current electrode $A$ sends a current $I$ into the mud and the rock, and the current returns to a distant electrode $B$ . We record the potential difference $\Delta U$ between the measuring electrodes $M$ and $N$ . The higher the resistivity of the medium, the larger $\Delta U$ is at the same current.

Imagine the sonde in a uniform medium of resistivity $R$ that goes on without end. The current would spread out from the electrode equally in all directions, and the potential at a distance $r$ from it would be $RI/(4\pi r)$ (the derivation is in the advanced section). From this we get the link between the potential difference and the resistivity:

Apparent resistivity of an electrode sonde

R_a = K\,\frac{\Delta U}{I},\qquad K = 4\pi\,\frac{AM\cdot AN}{MN}

where:

$R_a$ apparent resistivity, Ω·m
$\Delta U$ potential difference between the measuring electrodes $M$ and $N$ , V
$I$ current through the current electrode $A$ , A
$K$ sonde constant: a geometric factor that depends only on the distances between the electrodes, m
$AM,\ AN,\ MN$ distances between the electrodes ( $N$ is farther from $A$ than $M$ ), m

Units: SI

Holds when

The electrodes are points, and the return electrode $B$ is infinitely far away.
The constant $K$ is derived for a uniform medium. A borehole is not uniform, so here the formula only defines what we call apparent resistivity.
By the reciprocity principle, the current and measuring electrodes can swap places, and the reading does not change.

In a uniform medium $R_a$ equals the resistivity of that medium for any arrangement of the electrodes; this is what the constant $K$ is for. In a borehole the reading mixes the effect of the mud, the invaded zone, the uninvaded formation and the enclosing rocks (the beds above and below). So $R_a$ is only “apparent”: it is the resistivity of the uniform medium that would give the same $\Delta U$ .

The two classical types of sonde differ in how the electrodes are arranged.

Potential (normal) sonde: electrode $M$ is close to $A$ , and $N$ is very far away. The sonde measures the potential of the point $M$ , and $K \approx 4\pi\,AM$ . The length of the sonde is $AM$ , and the recording point is the middle of $AM$ . Opposite a thick bed the curve is symmetric about the middle of the bed.
Gradient (lateral) sonde: the pair $MN$ sits close together and far from $A$ . The sonde measures the potential gradient (how fast the potential changes with distance). The length of the sonde is the distance from $A$ to the middle of $MN$ (the recording point $O$ ). The curves are not symmetric: a sharp extremum (peak or trough) marks one of the boundaries of a high-resistivity bed, the bottom or the top, depending on which side of $A$ the pair $MN$ is on.

A general rule: the longer the sonde, the deeper it “sees” into the formation. But the bed must also be thicker, so that the neighbouring, enclosing rocks (shoulder beds) do not distort the reading. In beds thinner than the length of the sonde, the curves of potential sondes are distorted so much that they can turn upside down: a thin high-resistivity bed is marked by a minimum instead of a maximum.

Worked exampleApparent resistivity from a gradient sonde

Given (assumed data)

Sonde: gradient sonde A2.0M0.5N: $AM = 2.0$ m, $MN = 0.5$ m
Supply current $I$: 0.10 A
Potential difference $\Delta U$: 8.0 mV

Solution

The distance is $AN = AM + MN = 2.5$ m. The sonde constant is $K = 4\pi\cdot\dfrac{2.0\cdot 2.5}{0.5} = 125.7$ m.
The apparent resistivity is $R_a = K\,\Delta U/I = 125.7\cdot 0.0080/0.10 = 10.1$ Ω·m.
The recording point is the middle of $MN$ . The length of the sonde is $AO = 2.0 + 0.25 = 2.25$ m.

Answer: The apparent resistivity is about 10.1 Ω·m, recorded at the depth of the middle of the pair

MN

What it means. This is the resistivity of the uniform medium in which the sonde would give the same $\Delta U$ at the same current. In a borehole the reading mixes the mud, the invaded zone, the formation itself and the enclosing rocks, so 10.1 Ω·m is not yet $R_t$ . To separate these contributions, we record a set of sondes of different lengths and compare their readings. The data in this example are assumed for illustration.

Laterolog: focusing the current

In an unfocused sonde the current takes the easiest path. Suppose the mud is salty (low mud resistivity $R_m$ ) and the formation has high resistivity. Then most of the current flows along the borehole through the mud and into the conductive enclosing shales. The reading tells more about the mud and the neighbouring beds than about the formation. The larger the ratio $R_t/R_m$ and the contrast between the bed and the enclosing rocks, the stronger the distortion.

The laterolog solves this problem by focusing the current. Above and below the central current electrode $A_0$ there are guard (focusing) electrodes. They send out current of the same polarity. An automatic control sets the guard current so that there is no potential difference between the monitor electrodes (electrodes that only sense the voltage). This means the measuring current cannot spread along the borehole. It is squeezed from both sides, so it goes into the formation as an almost horizontal sheet (a beam), and the reading depends mainly on the rock opposite the sonde.

SchematicThree ways to measure resistivity, drawn as a cross section on one side of the borehole axis. (a) The current of an unfocused electrode in salty mud flows along the mud and into the conductive shales. (b) The guard electrodes of a laterolog sonde force the measuring current into the formation in a thin sheet. (c) An induction sonde drives a ring-shaped eddy current in the rock around the borehole axis. The drawing is not to scale.

The first laterolog sondes were the seven-electrode and the three-electrode types. They differed in the thickness of the current sheet (about 0.8 and 0.3 m): the thinner the sheet, the better the thin beds stand out. These sondes are obsolete now, but their logs remain in the records of old wells. For many years the main tool has been the dual laterolog. One set of electrodes makes two measurements with the same beam thickness, about 0.6 m, but different depths of investigation. The deep sonde (LLD) uses long guard electrodes, about 8.5 m from end to end, and a distant return electrode, so the beam keeps its shape for a long way. In the shallow sonde (LLS) the focusing current returns to neighbouring electrodes, so the beam spreads quickly and the sonde is more sensitive to the invaded zone. The tool is usually run together with a pad microresistivity tool (a small sensor pressed against the borehole wall) that measures $R_{xo}$ (Microresistivity tools and flushed-zone resistivity). Newer multi-sonde laterolog tools give several curves with different depths of investigation at once, which makes it easier to allow for invasion.

The dual laterolog covers roughly 0.2 to 40,000 Ω·m. To stay accurate at both ends of this range, the tool keeps the power constant rather than the current. Power is the product of current and voltage. In a high-resistivity rock the current falls and the voltage rises; in a conductive rock it is the other way round.

Shallow electrode measurements are also made with spherical focusing (spherically focused log, SFL). The guard currents keep two surfaces of equal potential (equipotential surfaces) around the sonde. These surfaces are close to spheres, at roughly 0.2 and 1.3 m from the current electrode, and the potential difference between them is constant. So the current through the layer of rock between them is proportional to the conductivity of that rock. Such a sonde is not very sensitive to the borehole, even with a conductive mud. It serves as a shallow addition to the deep sondes.

How the invaded zone enters the reading depends on the path of the current. The current of a laterolog sonde crosses the mud, the mudcake (the layer of mud solids on the borehole wall), the flushed zone and the formation one after another, in series. So their resistances add up. A high-resistivity invaded zone makes the reading much too high, while a low-resistivity one hardly interferes (worked example).

Induction logging

Induction logging was invented for wells where electrode sondes cannot pass current into the rock: wells drilled with oil-based mud or with air. It soon turned out that in fresh water-based muds too it is often more accurate than the electrode methods.

An alternating current with constant strength and frequency flows through the transmitter coil. The frequency is usually tens of kilohertz; for a long time the standard was about 20 kHz. The alternating magnetic field of the coil drives eddy currents in the surrounding rock. These are rings of current around the borehole axis (coaxial with the borehole). Their strength is proportional to the conductivity of the rock. The eddy currents create their own magnetic field, and this field induces a voltage in the receiver coil. So the signal is proportional to conductivity: an induction sonde measures conductivity, and we get the resistivity as its reciprocal.

The transmitter coil also induces a voltage in the receiver coil directly, even in a vacuum. This direct signal is many times larger than the useful signal, and it is a quarter of a period out of phase with it. Two things remove it: bucking coils wound in the opposite direction, and phase-sensitive detection, which keeps only the part of the signal that is in phase (rises and falls together) with the transmitter current.

Real sondes have several transmitter and receiver coils. The signal of a multi-coil sonde is the sum of the signals of all transmitter–receiver pairs. Each pair has a weight proportional to the number of turns and the areas of its coils, and the sign is taken into account. Choosing the coils well suppresses the contribution of the borehole and the near zone (so the sonde “looks” deeper) and the contribution of the enclosing rocks (so the vertical resolution, the ability to separate thin beds, improves). Such sondes are called focused.

For an induction sonde the zones around the borehole work in parallel: their conductivities add up in the reading, each with a weight called its geometric factor. So a low-resistivity invaded zone makes the reading much too low, while a high-resistivity one hardly interferes. This is the mirror image of the laterolog (comparison in an example).

In a highly conductive rock the induction method meets a difficulty of its own. The eddy currents of the near zone create a field that opposes the transmitter field and weakens it on its way deeper into the formation. This is the skin effect: the higher the conductivity and the frequency, the shallower the depth that the alternating field reaches. For the sonde this means that in low-resistivity rocks the signal grows more slowly than the conductivity. The tool then shows a conductivity that is too low, that is, a resistivity that is too high.

Skin depth

\delta = \sqrt{\frac{2R}{\omega\,\mu_0}} \approx 503\,\sqrt{\frac{R}{f}}

where:

$\delta$ skin depth: the distance over which the amplitude of the alternating field falls by a factor of $e \approx 2.72$ , m
$R$ resistivity of the rock, Ω·m
$f,\ \omega = 2\pi f$ frequency of the current in the transmitter coil, and the angular frequency, Hz, rad/s
$\mu_0$ magnetic constant $4\pi\cdot10^{-7}$ H/m; the magnetic permeability of most rocks is practically equal to it, H/m

Units: SI ( $R$ in Ω·m, $f$ in Hz, $\delta$ in metres)

Holds when

A uniform non-magnetic medium.
It is an estimate for a plane wave. The field of a real sonde is more complicated, but the order of magnitude is the same.
Displacement currents (the part of the current that comes from a changing electric field) are neglected; at frequencies of tens of kilohertz this is justified.

The factor $503 \approx 1/\sqrt{\pi\mu_0}$ . The skin depth grows as the square root of the resistivity. In a high-resistivity rock the field reaches tens of metres, and the skin effect is not noticeable. When $\delta$ becomes comparable to the radius of investigation of the sonde (the distance it “sees”, about 1–2 m), you cannot do without a correction.

Worked exampleSkin depth of an induction sonde in rocks of different resistivity

Given (assumed data)

Frequency $f$: 20 kHz
Resistivities of the rocks: 20, 1 and 0.2 Ω·m
Radius of investigation of the deep sonde: about 1.5 m (assumed)

Solution

At 20 Ω·m: $\delta = 503\sqrt{20/20{,}000} = 15.9$ m.
At 1 Ω·m: $\delta = 3.56$ m; at 0.2 Ω·m: $\delta = 1.59$ m.
If we lower the frequency to 10 kHz, then at 0.2 Ω·m $\delta = 2.25$ m, which is $\sqrt 2$ times more.

Answer: The skin depth is about 15.9 m in a rock of 20 Ω·m, but only 1.59 m in a rock of 0.2 Ω·m. That is about the same as the radius of investigation of the sonde.

What it means. In high-resistivity rocks the skin effect is negligible. In very conductive rocks the eddy currents of the near zone shield the far zone, and without a correction the tool reads a resistivity that is too high. (A very salty mud interferes in a different way: it adds a large signal from the borehole, and without a correction the tool usually reads a resistivity that is too low.) Modern tools apply the correction automatically. Lowering the frequency weakens the effect but also reduces the useful signal. The plane-wave formula gives only an order of magnitude. The data in this example are assumed for illustration.

Generations of induction tools

The first focused six-coil sonde appeared in the late 1950s and was the standard in the 1960s. Then came dual induction tools, with a deep sonde (ILD), a medium sonde (ILM) and a shallow electrode measurement. In the 1980s digital tools appeared. They work at frequencies of 10, 20 and 40 kHz and measure, besides the main signal, a quadrature signal (the part that is a quarter of a period out of phase with it). This gives an accurate skin-effect correction and better resolution. Since the 1990s multi-sonde (array) tools have been common. From the signals of many coil pairs they compute a series of curves with depths of investigation from 25 cm to 2.3 m and a chosen vertical resolution of 0.3, 0.6 or 1.2 m. They also give a two-dimensional picture of the resistivities around the borehole. In Russian-language practice a similar job is done by high-frequency induction isoparametric logging sounding (VIKIZ).

Laterolog or induction: choosing the method

The choice comes down to two questions. Does the mud conduct current? And which zone conducts better, the flushed zone or the uninvaded one? An induction sonde “notices” a high-resistivity invaded zone less ( $R_{xo} > R_t$ , the usual picture with fresh mud). A laterolog notices a low-resistivity one less ( $R_{xo} < R_t$ , typical of salty mud in a productive formation). There is one more reason. The error of an induction sonde is fixed in units of conductivity, so in high-resistivity rocks it becomes large compared with the signal itself.

Which method to choose to find the true resistivity

Conditions: Oil-based mud, air, foam
Method: induction (there is no other choice)
Why: electrode current cannot pass through a mud that does not conduct

Conditions: Fresh mud ( $R_{mf}/R_w$ greater than about 2.5), $R_t$ up to tens or hundreds of Ω·m
Method: induction
Why: a high-resistivity invaded zone has little effect on the sum of conductivities

Conditions: Salty mud ( $R_{mf}/R_w$ less than about 2.5)
Method: laterolog
Why: a conductive mud gives an induction sonde a large signal from the borehole; a low-resistivity invaded zone has little effect on the sum of resistances

Conditions: High-resistivity rocks: carbonates, anhydrites, tight sandstones
Method: laterolog
Why: the conductivity is small, and the error of an induction sonde is large

Conditions: Thin beds with a large resistivity contrast
Method: laterolog for high-resistivity beds, induction for conductive ones
Why: an induction sonde is “pulled” toward conductive layers, a laterolog toward high-resistivity ones

Conditions	Method	Why
Oil-based mud, air, foam	induction (there is no other choice)	electrode current cannot pass through a mud that does not conduct
Fresh mud ( $R_{mf}/R_w$ greater than about 2.5), $R_t$ up to tens or hundreds of Ω·m	induction	a high-resistivity invaded zone has little effect on the sum of conductivities
Salty mud ( $R_{mf}/R_w$ less than about 2.5)	laterolog	a conductive mud gives an induction sonde a large signal from the borehole; a low-resistivity invaded zone has little effect on the sum of resistances
High-resistivity rocks: carbonates, anhydrites, tight sandstones	laterolog	the conductivity is small, and the error of an induction sonde is large
Thin beds with a large resistivity contrast	laterolog for high-resistivity beds, induction for conductive ones	an induction sonde is “pulled” toward conductive layers, a laterolog toward high-resistivity ones

These are rules of thumb for ordinary conditions: moderate porosity and depth of invasion. The boundary $R_{mf}/R_w \approx 2.5$ comes from experience and is not sharp. Near it both methods are suitable. With fresh formation water ( $R_w$ greater than about 1 Ω·m) and high rock resistivities, induction logging loses its advantage. In doubtful cases, record both methods and interpret them together.

Worked exampleError of an induction sonde in a high-resistivity bed

Given (assumed data)

Absolute error in conductivity: ±2 mS/m (assumed)
Resistivities of the beds: 2, 50 and 200 Ω·m

Solution

The conductivities, from $C = 1000/R$ , are 500, 20 and 5 mS/m.
The relative error is $2/C$ : 0.4%, 10% and 40%.
For the 200 Ω·m bed, $C = 5 \pm 2$ mS/m. So $R$ lies between $1000/7 = 143$ and $1000/3 = 333$ Ω·m.

Answer: The same absolute error gives 0.4% in a 2 Ω·m bed and 40% in a 200 Ω·m bed. In the second bed the resistivity is known only to lie between 143 and 333 Ω·m.

What it means. An induction sonde measures conductivity, so it is accurate in conductive sections and loses accuracy in high-resistivity ones. For carbonates, anhydrites and tight gas-bearing rocks with resistivities of hundreds and thousands of Ω·m, we choose a laterolog: its relative error stays roughly constant over a wide range of resistivities. The data in this example are assumed for illustration.

How to find the true resistivity from three curves

No sonde, not even the deepest, removes the effect of the invaded zone completely. The way out is to measure the resistivity with several sondes that look to different depths, and to rebuild the profile from the differences between their readings. This works when three conditions are met. All the sondes see the same interval of the section (their vertical resolutions are close). The effect of the borehole is small or can be corrected reliably. The depths of investigation differ clearly: a shallow sonde sees almost only the flushed zone, a deep one reaches as far as possible into the uninvaded formation, and an intermediate one responds more than the others to the invasion diameter. Typical sets are a dual laterolog with a microresistivity tool, or deep and medium induction sondes with a shallow electrode sonde.

Readings of sondes with different depths of investigation in the step-invasion model

R_k = J_k(d_i)\,R_{xo} + \bigl[1 - J_k(d_i)\bigr]\,R_t,\qquad k = 1, 2, 3

where:

$R_k$ reading of the $k$ -th sonde after corrections for the borehole and the enclosing rocks, Ω·m
$J_k(d_i)$ pseudo-geometric factor: the share of the invaded zone in the reading of the $k$ -th sonde; it grows with $d_i$ , faster for shallow sondes, —
$d_i$ diameter of the invaded zone, m
$R_{xo},\ R_t$ resistivity of the flushed zone and of the uninvaded formation, Ω·m

Units: any consistent units of resistivity, the same for every term

Holds when

The profile is a step: a flushed zone with $R_{xo}$ out to the diameter $d_i$ , and beyond it the formation with $R_t$ . There is no transition zone and no annulus.
This form is for the laterolog, where resistances add up. For induction logging the conductivities $1/R$ are added in the same way, with geometric factors.
The bed is thicker than the vertical resolution of the sondes. Corrections for the borehole and the enclosing rocks have already been made.

There are three unknowns, $R_t$ , $R_{xo}$ and $d_i$ , so we need three readings with different depths of investigation. A microresistivity tool with $J \approx 1$ gives $R_{xo}$ almost directly. The other two equations give $d_i$ and $R_t$ .

For each trial invasion diameter we calculate “its own” value of $R_t$ from the deep sonde and from the shallow sonde. The correct diameter is the one at which the two values agree. Invasion correction charts (tornado charts, named for their shape) do this selection graphically. The ratios of the readings, for example $R_{LLD}/R_{xo}$ and $R_{LLD}/R_{LLS}$ , are plotted on the axes, and the grid of curves gives $R_t/R_{LLD}$ , $d_i$ and $R_{xo}/R_t$ at once. Today the same job is done by inversion software (programs that work backwards from the readings to the formation), especially for multi-sonde tools (Interpretation charts: how to read them).

CalculatedTwo estimates of

R_t

, one from the deep and one from the shallow laterolog sonde, plotted against the assumed invasion diameter. The readings are those of the example below, and the radial responses of the sondes are assumed. The correct solution is the point where the curves cross. The dashed line is the reading of the deep sonde.

Worked exampleTrue resistivity, flushed-zone resistivity and invasion diameter

Given (assumed data)

Readings after corrections for the borehole and the enclosing rocks: deep laterolog $R_{LLD} = 23.3$ Ω·m, shallow $R_{LLS} = 9.3$ Ω·m, microresistivity tool $R_{xo} = 2.0$ Ω·m
Conditions: salty mud, oil-bearing formation: we expect $R_{xo} < R_t$
$J$ at $d_i = 0.50$ m: 0.14 (deep), 0.51 (shallow)
$J$ at $d_i = 0.75$ m: 0.24 and 0.74
$J$ at $d_i = 1.00$ m: 0.33 and 0.86

Solution

The model $R = J R_{xo} + (1 - J) R_t$ gives $R_t = \dfrac{R - J R_{xo}}{1 - J}$ : one estimate from each sonde.
Trial diameter 0.50 m: the deep sonde gives $R_t = (23.3 - 0.14\cdot 2.0)/0.86 = 26.8$ Ω·m and the shallow one gives $R_t = (9.3 - 0.51\cdot 2.0)/0.49 = 16.9$ Ω·m. They do not agree.
Trial diameter 1.00 m: 33.8 and 54.1 Ω·m. Now the shallow sonde gives more than the deep one, so the solution lies between 0.5 and 1.0 m.
Trial diameter 0.75 m: 30.0 and 30.1 Ω·m. The estimates agree: $d_i \approx 0.75$ m, $R_t \approx 30$ Ω·m.
The correction to the deep sonde is $R_t/R_{LLD} = 1.29$ . In Archie's equation with $n = 2$ , the water saturation is inversely proportional to $\sqrt{R_t}$ . So a calculation from the uncorrected $R_{LLD}$ would overstate it by a factor of $\sqrt{1.29} = 1.14$ .

Answer: The invasion diameter is about 0.75 m,

R_t \approx 30

Ω·m (1.29 times the reading of the deep sonde), and

R_{xo} = 2.0

Ω·m.

What it means. Even the deepest sonde read $R_t$ too low, by 22%: the low-resistivity invaded zone is “mixed into” its reading. The water saturation from the uncorrected curve would have been 14% higher than the true one, and the formation would have looked worse than it is. Invasion correction charts make the same selection graphically, and inversion software makes it numerically. The values of $J$ here are assumed; real ones come from the specifications of the particular tool. The data in this example are assumed for illustration.

Even without calculations, how the curves compare tells us what is happening at the borehole:

The curves with different depths of investigation agree: there is no resistivity contrast between the zones. Either there is no invasion (a tight bed), or $R_{xo} \approx R_t$ , and then the deep sonde reads close to $R_t$ . More rarely, very deep invasion looks the same, when all the sondes see only the flushed zone.
Shallow sondes read higher than deep ones: the invaded zone has high resistivity. This is the usual picture of fresh mud in a water-bearing bed.
Shallow sondes read lower than deep ones: the invaded zone has low resistivity. This happens with salty mud. With fresh mud it is a sign of moveable hydrocarbons (hydrocarbons that can flow), if the water-bearing beds of the same section show the opposite picture (Borehole environment: invaded zone and mudcake).

Corrections: borehole, enclosing rocks, thin beds

We correct the reading of a sonde step by step, and the order matters: each correction assumes that the earlier ones have already been made. The manufacturer of the tool supplies the correction charts and algorithms for it.

Order of steps for finding the true resistivity

1
Borehole correction
Take into account the resistivity of the mud at borehole temperature, the borehole diameter from the caliper log (a record of the hole diameter) and the position of the tool in the hole. For an induction sonde the effect of the borehole is large with salty mud and in wide holes. For a laterolog it is large when the ratio of the reading to $R_m$ is very large.
2
Correction for the enclosing rocks and bed thickness
If the bed is thinner than the vertical resolution of the sonde, the neighbouring rocks enter the reading. An induction sonde adds conductivities, so in a thin high-resistivity bed among conductive shales it reads the resistivity too low. A laterolog is more accurate here. The resolution of the dual laterolog is about 0.6 m, and of classical deep induction sondes about 2 m. For multi-sonde induction tools it is chosen during processing.
3
Invasion correction
The last step is to find $R_t$ , $R_{xo}$ and $d_i$ from three curves, as in the example above.
4
Check
Compare $R_t$ with the resistivity of known water-bearing beds of the same section, with core data and with test results. Convert the resistivities of the mud, the filtrate and the formation water to formation temperature (Formation water and formation temperature).

Terms and symbols

Resistivity $R$ and conductivity $C$: Resistivity shows how strongly a material resists current (Ω·m). Conductivity is its reciprocal (S/m; in logging, mS/m): $C = 1000/R$ .
Apparent resistivity $R_a$: The resistivity of a uniform medium that would give the same sonde reading. In a borehole it mixes the effects of the mud, the zones of the formation and the enclosing rocks.
Sonde constant $K$: A geometric factor that turns $\Delta U/I$ into resistivity. For point electrodes $K = 4\pi\,AM\cdot AN/MN$ .
Laterolog: An electrode method in which guard electrodes focus the measuring current into the formation.
Induction log: A method that measures the conductivity of rocks from the eddy currents that the alternating magnetic field of coils induces.
Skin effect: The weakening of an alternating field in a conductive medium. Without a correction, an induction sonde reads too high a resistivity in low-resistivity rocks.
Enclosing rocks: The beds above and below the one being studied (shoulder beds). They distort the readings in beds thinner than the vertical resolution of the sonde.
Invasion correction chart (“tornado” chart): A nomogram (a chart for reading off results) for finding $R_t$ , $R_{xo}$ and $d_i$ from three curves with different depths of investigation.

Common mistakes and misconceptions

Wrong

“The laterolog is more accurate than induction logging because its sonde is focused.”

In fact

The more accurate method is the one that is less affected by the invaded zone and the borehole in the given conditions. With fresh mud and moderate resistivities this is usually induction. With salty mud and in high-resistivity rocks it is the laterolog.

Wrong

“Induction logging works with any mud and in any rocks.”

In fact

It does not need a conductive mud, but its accuracy falls with very salty mud, in wide holes, in very high-resistivity rocks and, because of the skin effect, in very conductive ones.

Wrong

“Apparent resistivity is the resistivity of the formation.”

In fact

It is the resistivity of a uniform medium that would give the same reading. In a borehole it mixes the effects of the mud, the zones of the formation and the enclosing rocks. We get $R_t$ only after corrections.

Wrong

“The longer the sonde, the better.”

In fact

A long sonde sees deeper into the formation, but it separates thin beds worse and is affected more by the enclosing rocks. That is why we record a set of sondes with matched vertical resolution.

Wrong

“The curves with different depths of investigation agree, so the bed is water-bearing.”

In fact

The curves agree only when there is no contrast in what the sondes see. There may be no invasion, or $R_{xo} \approx R_t$ , or (more rarely) the invasion is deeper than the deepest sonde sees. We judge saturation from $R_t$ itself, through Archie's equation, and by comparison with water-bearing beds.

In depth: advanced detail

Where the sonde constant comes from

Let a point electrode in a uniform medium of resistivity $R$ send out a current $I$ . The current spreads out evenly in all directions. All of it passes through a sphere of radius $r$ , so the current density (current per unit area) is $j = I/(4\pi r^2)$ . By Ohm's law the electric field strength is $E = R\,j$ . The potential relative to a point infinitely far away is:

Potential of a point current source in a uniform medium

U(r) = \int_r^{\infty} \frac{R\,I}{4\pi r'^2}\dd r' = \frac{R\,I}{4\pi r}

where:

$U(r)$ potential at a distance $r$ from the electrode relative to a point infinitely far away, V
$R$ resistivity of the medium, Ω·m
$I$ electrode current, A
$r$ distance from the electrode, m

Units: SI

Holds when

The medium is uniform, isotropic (the same in all directions) and infinite.
The electrode is a point, and the return electrode is infinitely far away.

The potential falls in inverse proportion to the distance. From this we get the potential difference between the measuring electrodes:

Potential difference between the electrodes M and N

\Delta U = \frac{R\,I}{4\pi}\left(\frac{1}{AM} - \frac{1}{AN}\right) = \frac{R\,I}{4\pi}\cdot\frac{MN}{AM\cdot AN}

where:

$\Delta U$ potential difference between $M$ and $N$ , V
$AM,\ AN,\ MN$ distances between the electrodes; $M$ and $N$ are on one side of $A$ , so that $MN = AN - AM$ , m
$R,\ I$ resistivity of the medium and the current of the electrode $A$ , Ω·m, A

Units: SI

Holds when

A uniform medium, point electrodes, the return electrode infinitely far away.

Solving this for $R$ , we get $R = K\,\Delta U/I$ with the constant $K = 4\pi\,AM\cdot AN/MN$ .

There are two limiting cases. For $AN \to \infty$ (a potential sonde) $K = 4\pi\,AM$ . For $MN \ll AM$ (an ideal gradient sonde) $K \approx 4\pi\,AO^2/MN$ , where $AO$ is the distance to the middle of $MN$ , because $AM\cdot AN = AO^2 - MN^2/4$ . By the reciprocity principle the current and measuring electrodes can swap places without changing the reading, so in practice the arrangement is chosen for design reasons.

The induction sonde in more detail: geometric factor, phase, anisotropy

Suppose we neglect the skin effect. Then each ring of rock around the axis adds to the signal an amount proportional to its conductivity and to its geometric factor, a multiplier that depends only on the position of the ring relative to the coils. The geometric factors of all space add up to one, so the apparent conductivity is a weighted average: $C_a = \sum_i G_i\,C_i$ . This is why the induction method is linear in conductivity, and why it is convenient to make corrections to it in conductivities.

The skin effect breaks this linearity. Besides the main signal, which is in phase with the transmitter current, the eddy currents create a quadrature signal, shifted by a quarter of a period. It grows together with the skin effect, and measuring it lets us recover the true conductivity more accurately. The useful signal grows roughly as the square of the frequency. So a higher frequency improves accuracy in high-resistivity rocks, but it strengthens the skin effect in conductive ones.

In a vertical well that crosses horizontal beds, the eddy currents of an ordinary induction sonde flow along the bedding, so the sonde measures the resistivity along the layers, $R_h$ . In a thinly laminated formation, where sandy and shaly laminae (thin layers) alternate, the resistivity across the layers, $R_v$ , can be several times larger. The laminae behave like the layers we meet when averaging permeability (Rock permeability): along the layers the conductivities add up, and across them the resistances add up.

Worked exampleA thinly laminated oil-bearing formation as an induction sonde sees it

Given (assumed data)

Structure of the formation: thin alternating laminae: half of the volume is oil-bearing sandstone of 20 Ω·m, half is shale of 1 Ω·m
Borehole: vertical, beds horizontal

Solution

Along the layers the conductivities add up: $1/R_h = 0.5/20 + 0.5/1$ , so $R_h = 1.90$ Ω·m.
Across the layers the resistances add up: $R_v = 0.5\cdot 20 + 0.5\cdot 1 = 10.5$ Ω·m. The anisotropy is $R_v/R_h = 5.5$ .

Answer: The induction sonde will show about 1.90 Ω·m, almost the same as the shale, although half of the formation is oil-bearing sandstone.

What it means. The horizontal eddy currents flow through the conductive shaly laminae and bypass the sandy ones. Such formations are easy to miss. They are called low-resistivity pay: productive zones that show a low resistivity. Anisotropy (the difference between $R_h$ and $R_v$ ) reveals them. Tools with coils of different orientation measure both $R_h$ and $R_v$ , and a model of a laminated formation gives the saturation of the sandy laminae. The data in this example are assumed for illustration.

Resistivity while drilling

Logging-while-drilling (LWD) tools sit in the drill string and measure while the hole is being drilled. They measure resistivity minutes or hours after the bit opens the formation, when the invaded zone is still small (Measurement while drilling). Most often these are electromagnetic wave propagation sondes. A transmitter sends out a wave with a frequency of about 2 MHz or 400 kHz, and two receivers record how its phase and amplitude change between them. Both changes are converted to resistivity. The curve from the phase shift is shallower and has the better vertical resolution. The curve from the attenuation (the drop in amplitude) is deeper.

When these curves differ, as with wireline sondes (those run on a cable), it points to invasion. In deviated and horizontal wells it also points to a nearby bed boundary and to anisotropy. In very high-resistivity rocks the accuracy of wave sondes falls. There, as with salty muds, electrode (galvanic) measurements are used, at the bit and on ring electrodes of the bottomhole assembly.

Connections to other topics

Borehole environment: invaded zone and mudcake — the zones around the borehole, and how their effects mix in the readings of sondes.
Archie's equation and saturation of clean reservoirs — how to calculate water saturation from $R_t$ and $R_{xo}$ .
Microresistivity tools and flushed-zone resistivity — how pad microresistivity tools measure $R_{xo}$ and detect the mudcake.
Interpretation charts: how to read them — how to use correction charts and invasion correction charts.
Resistivity ratio methods and movable oil — saturation and moveable oil, found from the ratio of the resistivities of the zones.
Formation water and formation temperature — how to bring $R_w$ and $R_{mf}$ to formation temperature.
Water saturation in shaly sands — why shaly formations conduct better than clean ones, and how this is allowed for.

Check yourself

1
A well was drilled with an oil-based mud. Which method should we use to measure $R_t$ ?
1. A.Microlaterolog sonde
2. B.Potential sonde
3. C.Dual laterolog
4. D.Induction logging
Show the answer
Correct answer: D. Induction logging
Oil-based mud does not conduct current, so the electrode methods cannot pass current into the rock. These are the laterolog, potential and gradient sondes, and most microresistivity tools. An induction sonde uses a magnetic field to drive currents in the rock and needs no contact with the mud.
2
The section is tight carbonates with an expected $R_t$ of hundreds of Ω·m, and the mud is salty. Which method should we choose to find $R_t$ ?
1. A.Only microresistivity tools
2. B.Dual laterolog
3. C.Any of them: the results will agree
4. D.Induction logging
Show the answer
Correct answer: B. Dual laterolog
Salty mud gives an induction sonde a large signal from the borehole, while the low conductivity of the rock gives only a small useful signal. So the relative error is large. A focused laterolog is designed for exactly this case: a high ratio $R_t/R_m$ and high-resistivity beds.
3
A gradient sonde with $AM = 1.0$ m and $MN = 0.1$ m, at a current of 0.05 A, recorded $\Delta U = 5$ mV. What is the apparent resistivity? Is it equal to $R_t$ ?
Show the answer
$K = 4\pi\cdot 1.0\cdot 1.1/0.1 \approx 138$ m, and $R_a = 138\cdot 0.005/0.05 \approx 13.8$ Ω·m. This is the apparent resistivity. It mixes the effects of the mud, the invaded zone and the enclosing rocks, so it is not $R_t$ . We get $R_t$ only after corrections.
4
Estimate the skin depth for an induction sonde with a frequency of 20 kHz in a rock of resistivity 0.5 Ω·m. Is the skin effect important?
Show the answer
$\delta \approx 503\sqrt{0.5/20{,}000} \approx 2.5$ m. This is about the same size as the radius of investigation of the deep sonde (1–2 m), so the effect is important. Without a correction the sonde would show a conductivity that is too low, that is, a resistivity that is too high. Modern tools correct for it automatically.
5
In a thick sandstone the readings of the deep and shallow laterologs and of the microresistivity tool almost agree. What does this tell us?
1. A.There is no resistivity contrast between the zones: there is no invasion, or $R_{xo} \approx R_t$ , and the deep sonde is close to $R_t$
2. B.The bed is necessarily water-bearing
3. C.The bed is necessarily oil-bearing
4. D.The tool is faulty
Show the answer
Correct answer: A. There is no resistivity contrast between the zones: there is no invasion, or $R_{xo} \approx R_t$ , and the deep sonde is close to $R_t$
When the curves agree, this says only that there is no contrast between the flushed zone and the formation. It happens in tight beds without invasion, or when $R_{xo} \approx R_t$ . More rarely, very deep invasion gives the same picture: then all the sondes, including the deep one, see only the flushed zone, and we cannot find $R_t$ from them. We judge saturation from $R_t$ itself, through Archie's equation, and by comparison with water-bearing beds.
6
Why do we make the invasion correction last, after the corrections for the borehole and the enclosing rocks?
Show the answer
Invasion charts and models are built for readings that are free of the effect of the borehole and of neighbouring beds. If we feed them readings that the mud or the enclosing shales have distorted, the model takes this distortion for invasion, and $R_t$ , $R_{xo}$ and $d_i$ come out wrong.

Updated: 19 October 2026