Interpretation charts: how to read them

Quick refresher

A chart is an equation solved in advance: a model of the tool or the formation, calculated on a grid of parameters. Its assumptions are written in its heading: the tool and the processing, the borehole, the mud, the tool position and the units.
We apply corrections in order: first for the borehole, then for the enclosing rocks (the beds above and below) and the bed thickness, then for invasion. We feed each chart with readings already corrected by the previous steps. We do not repeat corrections that the tool applied during recording.
The invasion chart (the “tornado” chart) solves the step-invasion model backwards. In this model, the mud filtrate fills a sharp-edged cylinder around the borehole. From two ratios of the readings of three sondes (the measuring parts of the tools), we find $d_i$ , $R_{xo}/R_t$ and the correction $R_t/R_{LLD}$ (the factor between the true resistivity and the reading of the deep laterolog). The laterolog (a resistivity tool that sends focused current into the rock) and the induction log (a resistivity tool that works with magnetic fields) mix the zones in different ways. So their charts are different too.
The accuracy of the answer depends on the place on the chart. Where the lines come together, near the origin and at very deep invasion, a small error in a reading gives a large error in the answer. A point outside the grid means that the model does not fit.
The main mistakes are a chart for the wrong tool or mud, mixed-up units (°F and °C, inches and millimetres, limestone and sandstone porosity units), a correction applied twice, and reading a logarithmic scale as if it were linear.

Key relationships

R = J\,R_{xo} + (1 - J)\,R_t

reading of an electrode sonde with step invasion

\dfrac{1}{R} = \dfrac{G}{R_{xo}} + \dfrac{1 - G}{R_t}

the same for an induction sonde

\dfrac{R_t}{R_a} = \dfrac{1 - G_m}{1 - G_m R_a/R_m}

borehole correction factor of an induction sonde

\mathrm{SSP} = -K\log_{10}\dfrac{R_{mfe}}{R_{we}},\quad K = 65 + 0.24\,T

equation of the SP chart: SSP is the static SP amplitude, $T$ in °C

Common pitfalls

Taking a chart for another tool, another generation of tools or another way of processing.
Plotting readings on the invasion chart without corrections for the borehole and the enclosing rocks.
Entering a temperature in °C on a scale in °F, or a diameter in millimetres on a scale in inches.
Interpolating on a logarithmic scale as if it were linear.
Applying a correction that the tool has already applied during recording.

After this lesson you can

explain that a chart is the graphical solution of a model, and name the main kinds of charts: correction charts, invasion charts, conversion charts, porosity and lithology crossplots, charts for the spontaneous potential (SP, the natural voltage in the borehole) and for saturation, and nomograms
build a simple correction chart and an invasion chart from the equations, and use the invasion chart to find $R_t$ (the resistivity of the uninvaded formation), $R_{xo}$ (the resistivity of the flushed zone, the zone next to the borehole where the filtrate has pushed out the original fluids) and the invasion diameter
read logarithmic scales and interpolate between curves (estimate values between two curves), estimate how a reading error passes into the result, and see where a chart is unreliable
choose a chart by the tool, the mud, the borehole diameter and the units, and recognise the typical mistakes in using it

A chart is an equation solved in advance

Every step of quantitative interpretation is an equation. It is either a model of a tool (how the reading depends on the formation and the borehole) or a model of the formation (how resistivity depends on porosity and saturation). Before there were computers at the wellsite, people solved such equations in advance for a grid of values and drew the result. This gave a chart (interpretation chart). The interpreter plots readings on it and reads off the answer without solving the equation.

Charts differ in how many unknowns they solve and in how they are drawn.

The main kinds of charts

Kind: Correction curves
What it gives: a factor or an addition to a reading
The model behind the chart: the response of the tool to the borehole, the mudcake (the layer of mud solids on the borehole wall) and the enclosing rocks
More detail: this lesson; neutron log

Kind: Invasion charts (“tornado”)
What it gives: $R_t$ , $R_{xo}/R_t$ , $d_i$ from three curves
The model behind the chart: step invasion (a sharp boundary between the invaded zone and the rest of the formation) and the radial characteristics of the sondes (how the readings depend on the distance from the borehole)
More detail: this lesson

Kind: Conversion charts
What it gives: the formation temperature, the water resistivity, the porosity for another matrix
The model behind the chart: the geothermal gradient (the rise of temperature with depth), the Arps formula (how water resistivity changes with temperature), the response of the neutron tool
More detail: Formation water and formation temperature, matrix conversion

Kind: Porosity and lithology crossplots (graphs of one log against another)
What it gives: porosity and mineral composition
The model behind the chart: the responses of the porosity methods to minerals and fluid
More detail: Lithology and porosity from crossplots, Lithology from the photoelectric factor and spectral gamma ray, M–N and MID plots for complex lithologies

Kind: SP charts
What it gives: $R_{we}$ , $R_w$ , the bed thickness correction
The model behind the chart: the static SP amplitude (the largest SP deflection, as in a thick clean bed) and the properties of the solutions
More detail: Determining formation water resistivity, Spontaneous potential (SP) log

Kind: Saturation charts
What it gives: $S_w$ from Archie's equation (which links resistivity, porosity and water saturation) or from the resistivity ratio
The model behind the chart: Archie's equation, the ratio of the resistivities of the zones
More detail: Resistivity–porosity crossplots, Resistivity ratio methods and movable oil

Kind	What it gives	The model behind the chart	More detail
Correction curves	a factor or an addition to a reading	the response of the tool to the borehole, the mudcake (the layer of mud solids on the borehole wall) and the enclosing rocks	this lesson; neutron log
Invasion charts (“tornado”)	$R_t$ , $R_{xo}/R_t$ , $d_i$ from three curves	step invasion (a sharp boundary between the invaded zone and the rest of the formation) and the radial characteristics of the sondes (how the readings depend on the distance from the borehole)	this lesson
Conversion charts	the formation temperature, the water resistivity, the porosity for another matrix	the geothermal gradient (the rise of temperature with depth), the Arps formula (how water resistivity changes with temperature), the response of the neutron tool	Formation water and formation temperature, matrix conversion
Porosity and lithology crossplots (graphs of one log against another)	porosity and mineral composition	the responses of the porosity methods to minerals and fluid	Lithology and porosity from crossplots, Lithology from the photoelectric factor and spectral gamma ray, M–N and MID plots for complex lithologies
SP charts	$R_{we}$ , $R_w$ , the bed thickness correction	the static SP amplitude (the largest SP deflection, as in a thick clean bed) and the properties of the solutions	Determining formation water resistivity, Spontaneous potential (SP) log
Saturation charts	$S_w$ from Archie's equation (which links resistivity, porosity and water saturation) or from the resistivity ratio	Archie's equation, the ratio of the resistivities of the zones	Resistivity–porosity crossplots, Resistivity ratio methods and movable oil

Each chart is calculated for certain conditions. They are printed in its heading or in a corner of the sheet. They include the tool and its generation, the way of processing, the borehole diameter (for example, 8 in, about 203 mm), whether the tool is centred or pressed against the wall, the type and resistivity of the mud, and the units of the scales. These are the assumptions of the model. If the well is different, we first apply corrections that bring the reading to the conditions of the chart, or we take another chart. Processing software does the same on a computer, and it also uses the manufacturer's models. So understanding charts helps us to check the results of the software.

How a correction chart is built

Take a simple case: the borehole correction of an induction sonde. An induction sonde adds up the conductivities of all the zones around it (conductivity is 1/resistivity), each with a weight called a geometric factor (more). The column of mud in the borehole is one more such zone. Its share in the signal, $G_m$ , depends on the borehole diameter, the position of the tool and the design of the sonde, but not on the resistivities. If the formation is uniform (there is no invasion), the reading $R_a$ is linked to $R_t$ like this:

Induction sonde reading with the mud in the borehole

\frac{1}{R_a} = \frac{G_m}{R_m} + \frac{1 - G_m}{R_t}

where:

$R_a$ reading of the sonde (apparent resistivity: the resistivity that the tool shows before any correction), Ω·m
$R_m$ mud resistivity at the temperature in the well, Ω·m
$R_t$ formation resistivity, Ω·m
$G_m$ geometric factor of the borehole: the share of the mud in the signal, —

Units: Ω·m; $G_m$ has no units

Holds when

This is the geometric-factor theory: the skin effect (a weakening of the signal in conductive rock) is negligible or already corrected.
The bed is thicker than the vertical resolution of the sonde (the thinnest bed it can read correctly), and there is no invasion. These are handled by the later corrections.
$G_m$ belongs to a particular tool, borehole diameter and tool position. The values in the example are assumed.

The conductivity of the borehole, $G_m/R_m$ , is an addition to the signal. It is small with a fresh mud (a mud with little salt, so a high resistivity) and in a narrow borehole. It is large with a salty mud, a large borehole diameter and a tool pressed against the wall.

We solve the equation for $R_t$ and divide by $R_a$ . This gives a correction factor that depends on only two numbers with no units, $R_a/R_m$ and $G_m$ :

Borehole correction factor of an induction sonde

\frac{R_t}{R_a} = \frac{1 - G_m}{1 - G_m\,R_a/R_m}

where:

$R_t/R_a$ correction factor for the reading, —
$R_a/R_m$ ratio of the reading to the mud resistivity, —
$G_m$ geometric factor of the borehole, —

Units: dimensionless

Holds when

The same as for the reading with the borehole. The formula makes sense only for $R_a/R_m < 1/G_m$ .

This is the chart itself. The horizontal axis is $R_a/R_m$ , the vertical axis is the factor, and the curves are for different $G_m$ (on manufacturers' charts they are labelled with borehole diameters and stand-offs, the gaps between tool and wall). When $R_a/R_m$ approaches $1/G_m$ , almost all of the signal comes from the mud, the factor goes to infinity, and the chart gives no answer.

CalculatedA borehole correction chart for an induction sonde, calculated with the formula above. The numbers at the curves are the share of the borehole in the signal,

G_m

(assumed values; it grows with the borehole diameter and when the tool moves towards the wall). Points A and B are the example below: a fresh and a salty mud at the same reading.

Worked exampleBorehole correction of an induction sonde with fresh and salty mud

Given (assumed data)

Reading of a deep induction sonde in a thick bed: $R_a = 14$ Ω·m
Mud at formation temperature: A: fresh, $R_m = 0.85$ Ω·m; B and C: salty, $R_m = 0.09$ Ω·m
Share of the borehole in the signal (assumed): $G_m = 0.003$ (216 mm borehole, tool centred) in cases A and B; $G_m = 0.01$ (a wider borehole or the tool at the wall) in case C

Solution

Conductivity from the reading: $\sigma_a = 1000/14 = 71.43$ mS/m.
The contribution of the borehole, $G_m\cdot 1000/R_m$ : A: $0.003\cdot 1000/0.85 = 3.53$ mS/m. B: $0.003\cdot 1000/0.09 = 33.33$ mS/m. C: $0.01\cdot 1000/0.09 = 111.11$ mS/m.
The conductivity of the formation is $\sigma_t = (\sigma_a - G_m\sigma_m)/(1 - G_m)$ . A: $(71.43 - 3.53)/0.997 = 68.10$ mS/m, $R_t = 1000/68.10 = 14.68$ Ω·m, which is 4.9% more than the reading. B: $(71.43 - 33.33)/0.997 = 38.21$ mS/m, $R_t = 26.17$ Ω·m, which is 1.87 times the reading.
C: the contribution of the borehole, 111.11 mS/m, is larger than the whole signal, 71.43 mS/m, so there is no solution. On the chart, $R_a/R_m = 14/0.09 \approx 156$ lies to the right of the asymptote (the line that the curve approaches but never reaches) $1/G_m = 100$ .

Answer: With a fresh mud the correction is 4.9%. With a salty mud the factor is 1.87. In a wide borehole with a salty mud, the reading of the induction sonde is useless for finding

R_t

What it means. The same reading means very different beds, depending on the mud and the borehole. That is why a chart needs its conditions. Cases B and C show why, with a salty mud, we find $R_t$ with a laterolog (choice of method). A correction that nearly doubles the reading is no longer a refinement but a recalculation, sensitive to $G_m$ and $R_m$ . And when the contribution of the borehole is larger than the signal itself, there is no answer. The values of $G_m$ are assumed: for a real tool, we take them from its charts. The data in this example are assumed for illustration.

All correction charts are built in this way: a model of the response, inputs with no units, and a family of curves. For pad tools (tools pressed against the borehole wall), the mudcake and the gap between the pad and the wall play the role of the borehole (mudcake correction of the microresistivity tool, Density log and photoelectric factor (Pe)). For the neutron tool, there is a whole set of conditions, each with its own chart (more). For the gamma ray log, the conditions are the borehole diameter, the mud density and the potassium chloride in the mud (Gamma ray log and natural gamma ray spectrometry). The order is always the same: first the borehole, then the enclosing rocks and the bed thickness, then invasion (more).

The invasion chart: construction and reading

The invasion correction chart solves a problem with three unknowns, $R_t$ , $R_{xo}$ and the invasion diameter $d_i$ , from three readings with different depths of investigation. In the lesson on resistivity logging, this problem was solved by trial: for trial diameters we calculated $R_t$ from the deep and shallow sondes, and looked for the diameter at which the estimates agree. The chart does the same trial in advance, for all possible beds at once.

We build the chart like this. First we set a model, step invasion, and the radial characteristics of the sondes: the share of the flushed zone in the reading, $J(d_i)$ , for the deep and shallow laterologs (pseudo-geometric factor). A microresistivity tool shows $R_{xo}$ itself when the invasion front is farther than about 10 cm from the wall (Microresistivity tools and flushed-zone resistivity). Then we try many imaginary “beds”. For each pair of $d_i$ and $c = R_{xo}/R_t$ , we calculate the readings divided by $R_t$ and plot a point with the coordinates:

Coordinates of a point on the invasion chart for a pair of laterologs

x = \frac{R_{LLD}}{R_{LLS}} = \frac{J_d c + 1 - J_d}{J_s c + 1 - J_s},\qquad y = \frac{R_{LLD}}{R_{xo}} = \frac{J_d c + 1 - J_d}{c}

where:

$R_{LLD},\ R_{LLS}$ readings of the deep and shallow laterologs after corrections for the borehole and the enclosing rocks, Ω·m
$R_{xo}$ reading of the microresistivity tool, corrected for the mudcake, Ω·m
$J_d,\ J_s$ pseudo-geometric factors of the deep and shallow sondes at the given $d_i$ , —
$c$ the ratio $R_{xo}/R_t$ , —

Units: dimensionless

Holds when

A step profile. The bed is thicker than the vertical resolution of the sondes. The corrections for the borehole and the enclosing rocks have been applied.
Laterologs: the zones add up in series, like resistances. For induction sondes the conductivities add up, and the chart is different.
The microresistivity tool shows $R_{xo}$ . The factors $J$ do not depend on the resistivity contrast (for real sondes they depend on it only weakly).

Each imaginary bed gives a point $(x, y)$ . Points with the same $d_i$ are joined into one line, and points with the same $R_{xo}/R_t$ into another. The grid is also labelled with the correction $R_t/R_{LLD} = 1/(J_d c + 1 - J_d)$ . We plot the point of our bed and read $d_i$ , $R_{xo}/R_t$ and $R_t$ without any trial.

CalculatedAn invasion chart for a pair of laterologs and a microresistivity tool, calculated with the step-invasion model and assumed radial characteristics, the same as in the lesson on resistivity logging. Solid lines are equal invasion diameters, dashed lines are equal

R_{xo}/R_t

, and the dotted diagonal is the limit of very deep invasion. Point A is the example below, and B is the bed with very deep invasion from the example on accuracy.

The shape of the grid explains its nickname. While the invasion is so shallow that the laterologs do not see it, both show $R_t$ , so $x = 1$ . The microresistivity tool shows $R_{xo}$ in the chart's model, so $y = R_t/R_{xo}$ . This is why the lines of $R_{xo}/R_t$ start on the vertical axis. As $d_i$ grows, the shallow sonde sees more and more of the flushed zone, the points move to the right, and the grid widens like a funnel. At very deep invasion, the shallow sonde shows practically $R_{xo}$ , $x$ approaches $y$ , and the points press against the diagonal, where different beds are almost indistinguishable.

How to read an invasion chart

1
Readings
Take the readings of the three sondes in the middle of a thick bed and apply the corrections for the borehole and the enclosing rocks. Check that the chart is for these tools, this borehole diameter and this mud. Often it is given for a particular ratio $R_{xo}/R_m$ .
2
Ratios
Calculate $x = R_{LLD}/R_{LLS}$ and $y = R_{LLD}/R_{xo}$ and find the point on the logarithmic scales.
3
Interpolation
Estimate $d_i$ between the nearest lines of equal diameter, and $R_{xo}/R_t$ between the nearest dashed lines. On a logarithmic grid we interpolate by the distance on the sheet, not by the numbers on the labels.
4
Answer
Read the correction $R_t/R_{LLD}$ , find $R_t$ and $R_{xo}$ , and compare $R_{xo}$ with the reading of the microresistivity tool.
5
Check
A point outside the grid, at its edge or near the diagonal is a reason to doubt the model. The cause may be a transition profile (a gradual change of resistivity instead of a step), an annulus (a ring of lower resistivity around the invaded zone), a thin bed, wrong corrections or the wrong chart.

Worked exampleTrue resistivity from the invasion chart

Given (assumed data)

Bed: a thick oil-bearing sandstone, salty mud
Readings after corrections for the borehole and the enclosing rocks: $R_{LLD} = 34.2$ Ω·m, $R_{LLS} = 8.5$ Ω·m, microresistivity tool $R_{xo} = 3.1$ Ω·m
Chart: the one calculated above (the radial characteristics are assumed)

Solution

Ratios: $x = 34.2/8.5 = 4.02$ , $y = 34.2/3.1 = 11.03$ .
On the chart, the point lies between the lines $d_i$ = 1.0 and 1.5 m, closer to the first, and between the dashed lines $R_{xo}/R_t$ = 0.05 and 0.1, closer to 0.05.
We can also solve the equations of the chart, which is what the chart does graphically: $d_i = 1.10$ m, $J_d$ = 0.364, $J_s$ = 0.889, $R_{xo}/R_t$ = 0.0597.
The correction is $R_t/R_{LLD} = 1/(0.364\cdot 0.0597 + 0.636) = 1.52$ , so $R_t = 34.2\cdot 1.52 = 52.0$ Ω·m. Check: $3.1/52.0 = 0.0596$ .
Water saturation from Archie's equation is inversely proportional to $\sqrt{R_t}$ . A calculation from $R_{LLD}$ without the correction would make $S_w$ too high by the factor $\sqrt{1.52} = 1.23$ .

Answer: The invasion diameter is about 1.1 m,

R_{xo}/R_t \approx

0.06, and

R_t \approx

52 Ω·m, which is 1.52 times the reading of the deep sonde.

What it means. Even the deep laterolog gave $R_t$ too low by a third: the low-resistivity flushed zone “mixes into” its reading. The chart gave the answer without any trial. It also showed at once how reliable the answer is: the point lies inside the grid, in an area where the lines of different diameters are well separated (the next example). For real tools, we take $J$ from their characteristics, and the chart will be different in numbers. The data in this example are assumed for illustration.

Worked exampleHow accurate the answer is: two points on one chart

Given (assumed data)

Point A: the bed from the previous example: 34.2 / 8.5 / 3.1 Ω·m
Point B: a bed with very deep invasion: $R_{LLD} = 15.7$ , $R_{LLS} = 2.27$ , $R_{xo} = 2.0$ Ω·m
Reading errors (assumed): deep and shallow sondes ±3%, microresistivity tool ±5%

Solution

Solution for B: $d_i = 2.2$ m, $R_t = 40$ Ω·m, and the correction $R_t/R_{LLD}$ = 2.53. The point is near the diagonal.
The sensitivity of the answer is the number of per cent by which $R_t$ changes for a 1% error in a reading (by the chart's model). A: deep sonde 1.30, shallow sonde −0.38, microresistivity tool 0.08. B: 1.37, −2.07 and 1.70.
The error of $R_t$ from each reading error, in per cent: A: 3.9, 1.1 and 0.4. B: 4.1, 6.2 and 8.5.
Independent errors are added in quadrature (as the square root of the sum of squares). A: $\sqrt{3.9^2 + 1.1^2 + 0.4^2} = 4.1$ %. B: $\sqrt{4.1^2 + 6.2^2 + 8.5^2} = 11$ %.

Answer: With the same reading errors,

R_t

is known to an accuracy of about ±4.1% at point A and ±11% at point B.

What it means. With moderate invasion, $R_t$ comes mainly from the deep sonde, and the chart only slightly increases its error. With very deep invasion, all three readings enter the answer. The errors of the shallow sonde and of the microresistivity tool carry more weight than the error of the deep one, and the correction to the deep sonde is large. There it is better to look for other data: a multi-sonde tool with a greater depth of investigation, a repeat log, or a comparison with water-bearing beds. The reading errors are assumed. The data in this example are assumed for illustration.

Porosity, lithology, SP and saturation charts

The other charts are built on the same principle, and many are covered in their own lessons. Here is what matters when you work with a chart.

Porosity and lithology. The crossplots of neutron against density, and those with the sonic log and with the photoelectric factor (a density-tool curve that is sensitive to the minerals), are calculated from the responses of the tools to minerals and fluid (Lithology and porosity from crossplots, Lithology from the photoelectric factor and spectral gamma ray, M–N and MID plots for complex lithologies). We choose the chart by the tool and the processing of the neutron log, and by the filtrate, fresh or salty, because they have different densities and hydrogen indices (the amount of hydrogen in a unit volume, which the neutron log responds to). We put the readings on the chart with corrections for the measurement conditions (environmental corrections), and the neutron curve in the units for which the chart was built (more).
SP and water resistivity. First, a chart gives a correction for the bed thickness, and with it we get the static amplitude. The chart that turns it into a ratio of equivalent resistivities (the resistivities that the SP equation uses for the mud filtrate and the formation water) is a nomogram of the equation for the static amplitude, with a coefficient $K$ that depends on temperature. One more chart turns the equivalent resistivities into true ones. The whole calculation, with the approximations of these charts, is in the lesson on finding water resistivity.
Saturation. The nomograms of Archie's equation solve $R_0 = F R_w$ and $S_w = (R_0/R_t)^{1/n}$ with straight lines (why this works is in the advanced section). Here $R_0$ is the resistivity of the rock when it is fully filled with water, and $F$ is the formation resistivity factor. The “resistivity against porosity” grids are built so that water-bearing beds lie on a straight line (Resistivity–porosity crossplots). The resistivity-ratio charts give $S_w$ from $R_{xo}/R_t$ and $R_{mf}/R_w$ (Resistivity ratio methods and movable oil).

Worked exampleAn SP chart and a mixed-up temperature scale

Given (assumed data)

Static SP amplitude: −60 mV (after the bed thickness correction)
Formation temperature: 88 °C
Equivalent filtrate resistivity at 88 °C: $R_{mfe} = 0.50$ Ω·m
Temperature scale of the chart: double: °F and °C

Solution

Correct: $K = 65 + 0.24\cdot 88 = 86.12$ mV; $R_{mfe}/R_{we} = 10^{60/86.12} = 4.97$ ; $R_{we} = 0.50/4.97 = 0.1006$ Ω·m.
Mistake: 88 was read on the °F scale (that is about 31 °C): $K = 61 + 0.133\cdot 88 = 72.70$ mV; $10^{60/72.70} = 6.69$ ; $R_{we} = 0.50/6.69 = 0.0747$ Ω·m, which is 25.7% less.
At 24 °C the correct value is $0.1006\cdot 109.5/45.5 = 0.242$ Ω·m. This is more than 0.12, so the water has a moderate salinity, and $R_w \approx R_{we}$ (more).
Water saturation from Archie's equation is proportional to $\sqrt{R_w}$ . The mistake multiplies it by $\sqrt{0.0747/0.1006} = 0.862$ , that is, it makes it too low by 13.8%.

Answer: The mixed-up scale made

R_{we}

too low by 25.7% (0.0747 instead of 0.1006 Ω·m), and the water saturation of all the beds of the interval too low by 13.8%.

What it means. An error in units is systematic: it shifts all the beds in the same way, and inside the interval it cannot be seen. We catch it by comparing with independent data: a water sample, or the apparent water resistivity of water-bearing beds (more). The opposite mistake, a temperature in °F read on the °C scale, makes $K$ and $R_{we}$ too high. The data in this example are assumed for illustration.

Reading accuracy and typical mistakes

Most charts are logarithmic, because resistivities and their ratios change by orders of magnitude (factors of ten). On a logarithmic scale, equal distances mean equal ratios: the step from 1 to 2 is the same as the step from 10 to 20. So the middle between the marks 10 and 20 is not 15 but $\sqrt{10\cdot 20} \approx 14.1$ , and we interpolate by the distance on the sheet. We can read the position of a point on paper to within about 0.5–1 mm. With a decade (one factor of ten) 5–6 cm long, this is 2–5% of the value on each axis.

How a reading error passes into the answer depends on the place on the chart. We estimate it by shifting the reading by the possible error and reading the answer again (the example above). A calculation with the chart's model shows the general picture: the deeper the invasion, the more the answer depends not only on the deep sonde but also on the shallow sonde and the microresistivity tool.

CalculatedThe change in the

R_t

found from the invasion chart above, in per cent, for an error of 1% in each reading, against the invasion diameter (a bed with

R_{xo}/R_t

= 0.06). With shallow invasion the answer depends almost only on the deep sonde. With very deep invasion it depends strongly on all three readings, and on the shallow sonde and the microresistivity tool even more than on the deep one.

Typical mistakes in working with charts

Mistake: A chart for the wrong tool
Example: a laterolog chart used for induction readings; a chart for a tool of another generation or processing
What it risks: $R_t$ is wrong by tens of per cent (example below)

Mistake: Wrong conditions
Example: a chart for an 8 in borehole used in a 12¼ in borehole; a chart for a fresh filtrate used with a salty one
What it risks: an unaccounted effect of the borehole; wrong porosity and lithology

Mistake: Units
Example: °C on the °F scale; millimetres instead of inches; limestone units instead of sandstone units; g/L instead of ppm
What it risks: systematic errors, from a few per cent to several times over

Mistake: Order of corrections
Example: the invasion chart used with readings that have no borehole correction; a correction that the tool already applied is applied again
What it risks: the error is mistaken for invasion, or doubled

Mistake: Temperature
Example: $R_{mf}$ from the heading used without conversion to the formation temperature; the constant of the Arps formula for °F used with a temperature in °C
What it risks: water resistivities are wrong by tens of per cent and by factors of several

Mistake: Extrapolation
Example: a point outside the grid is “read” by extending the curves
What it risks: an answer with no model behind it

Mistake: Reading a logarithmic scale as linear
Example: the middle between 10 and 20 is taken as 15
What it risks: a few per cent on every reading

Mistake	Example	What it risks
A chart for the wrong tool	a laterolog chart used for induction readings; a chart for a tool of another generation or processing	$R_t$ is wrong by tens of per cent (example below)
Wrong conditions	a chart for an 8 in borehole used in a 12¼ in borehole; a chart for a fresh filtrate used with a salty one	an unaccounted effect of the borehole; wrong porosity and lithology
Units	°C on the °F scale; millimetres instead of inches; limestone units instead of sandstone units; g/L instead of ppm	systematic errors, from a few per cent to several times over
Order of corrections	the invasion chart used with readings that have no borehole correction; a correction that the tool already applied is applied again	the error is mistaken for invasion, or doubled
Temperature	$R_{mf}$ from the heading used without conversion to the formation temperature; the constant of the Arps formula for °F used with a temperature in °C	water resistivities are wrong by tens of per cent and by factors of several
Extrapolation	a point outside the grid is “read” by extending the curves	an answer with no model behind it
Reading a logarithmic scale as linear	the middle between 10 and 20 is taken as 15	a few per cent on every reading

Worked exampleInduction readings on a laterolog chart

Given (assumed data)

Bed: thick, fresh mud: a high-resistivity invaded zone
Readings after the borehole corrections: deep induction 7.71 Ω·m, medium induction 15.6 Ω·m, microresistivity tool $R_{xo}$ = 24 Ω·m
Radial characteristics: for clarity, the same as on the chart above; only the model of how the zones mix is different

Solution

The right chart is the induction one, where the zones add up as conductivities: $d_i = 0.90$ m, $R_t = 5.99$ Ω·m.
The laterolog chart, where the zones add up as resistances, gives $d_i = 0.55$ m, $R_t = 4.67$ Ω·m.
The error of $R_t$ is −22%. The water saturation from Archie's equation comes out too high by the factor $\sqrt{5.99/4.67} = 1.13$ .

Answer: The wrong chart gave

R_t

too low, 4.67 instead of 5.99 Ω·m (−22%), and “showed” an invasion of 0.55 m instead of 0.90 m.

What it means. For a laterolog, even a small share of a high-resistivity flushed zone raises the reading strongly. So its chart explains the moderate induction readings by a smaller $R_t$ and a shallow invasion. An induction sonde hardly notices a high-resistivity zone. The two mixing models respond to it in different ways, and they must not be confused. Real tools also differ in their radial characteristics, so the real error would be different in numbers. The data in this example are assumed for illustration.

Terms and symbols

Chart: A graph or nomogram with the solution of a model of the tool or the formation, calculated in advance for a range of parameter values (interpretation chart).
Conditions of a chart: The tool, processing, borehole diameter, tool position, mud and units for which it was calculated.
Correction chart: A factor or an addition to a reading, for the borehole, the mudcake and the enclosing rocks, as a function of inputs with no units.
Invasion chart (“tornado” chart): A grid of lines of equal $d_i$ and $R_{xo}/R_t$ in the coordinates of the ratios of three readings. It gives $R_t$ , $R_{xo}$ and $d_i$ .
Pseudo-geometric factor $J$: The share of the flushed zone in the reading of an electrode sonde with step invasion.
Geometric factor of the borehole $G_m$: The share of the mud in the borehole in the signal of an induction sonde.
Nomogram: A chart made of scales, on which a straight line drawn through the marks of the known quantities gives the answer.

Common mistakes and misconceptions

Wrong

“Charts are the same for all the tools of one method.”

In fact

Each chart is calculated for its own tool, processing and conditions. For the laterolog and the induction log, even the models of zone mixing are different.

Wrong

“Corrections can be applied in any order.”

In fact

The invasion chart is built for readings already corrected for the borehole and the enclosing rocks. Otherwise their distortion will be taken for invasion.

Wrong

“The middle between the marks 10 and 20 is 15.”

In fact

On a logarithmic scale, equal distances are equal ratios: the middle is $\sqrt{10\cdot 20} \approx 14.1$ . We interpolate by the distance on the sheet.

Wrong

“The point is just outside the edge of the chart, so let us extend the curves.”

In fact

Beyond the grid, the model has not been checked. A point outside the grid is a sign that the assumptions are broken: a different invasion profile, a thin bed, errors in corrections or the wrong chart.

Wrong

“The software has calculated it, so we do not need charts.”

In fact

The software solves the same equations with the same models. If we understand the chart, we can check whether the model and the conditions fit, and we notice answers that cannot be right.

In depth: advanced detail

Why a straight line on a nomogram solves an equation

A nomogram with parallel scales solves equations of the form $f_1(u) + f_2(v) = f_3(w)$ . Archie's equation in logarithms is exactly of this kind: $\log_{10} R_0 = \log_{10} F + \log_{10} R_w$ , and then, for $n = 2$ , $\log_{10} S_w = \tfrac12(\log_{10} R_0 - \log_{10} R_t)$ . Here $R_0$ is the resistivity of the rock when it is fully filled with water.

Place two vertical scales at a distance $L$ from each other. On the left scale, we draw the mark $u$ at the height $y_1 = \mu_1 f_1(u)$ , and on the right scale, we draw the mark $v$ at the height $y_2 = \mu_2 f_2(v)$ . Here $\mu_1$ and $\mu_2$ are the scale factors (moduli) of the scales. A straight line through these two marks, at a distance $x$ from the left scale, passes at the height $y_1 + (y_2 - y_1)\,x/L$ . Let us choose the position and the scale of the middle scale like this:

Position and marking of the middle scale of a nomogram with parallel scales

x_3 = \frac{\mu_1}{\mu_1 + \mu_2}\,L,\qquad y_3 = \frac{\mu_1\mu_2}{\mu_1 + \mu_2}\bigl(f_1(u) + f_2(v)\bigr)

where:

$x_3$ distance of the middle scale from the left one, mm
$y_3$ height at which the straight line crosses the middle scale, mm
$\mu_1,\ \mu_2$ moduli of the outer scales: the length of one unit of $f_1$ and $f_2$ on paper, mm
$L$ distance between the outer scales, mm

Units: any units of length on paper

Holds when

The scales are parallel, and heights are measured from a common horizontal line. $f_1$ , $f_2$ and $f_3$ are any monotonic functions (functions that only rise or only fall).

If we put $x_3$ into the equation of the line, we get exactly $y_3$ . A straight line through the marks $u$ and $v$ always crosses the middle scale at a height proportional to $f_1(u) + f_2(v)$ . If we mark this scale by $f_3(w)$ with the modulus $\mu_1\mu_2/(\mu_1 + \mu_2)$ , the mark at the crossing point is the answer $w$ . With equal moduli, the middle scale stands halfway between the outer ones and is squeezed to half the size.

Worked exampleMarking a nomogram for

R_0 = F R_w

Given (assumed data)

Scales: the left one is $\log_{10} R_w$ and the right one is $\log_{10} F$ , both with a modulus of 40 mm per decade; the middle one is halfway between them, with a modulus of 20 mm
Bed (assumed): $R_w = 0.065$ Ω·m, $F = 18$

Solution

Marks on the outer scales: $y_1 = 40\log_{10} 0.065 = -47.5$ mm, $y_2 = 40\log_{10} 18 = 50.2$ mm.
The straight line crosses the middle scale at a height of $(-47.5 + 50.2)/2 = 1.35$ mm.
The middle scale is marked with a modulus of 20 mm: $\log_{10} R_0 = 1.35/20 = 0.0675$ , $R_0 = 1.17$ Ω·m. This is the same as $0.065\cdot 18 = 1.17$ Ω·m.

Answer: A straight line through 0.065 and 18 crosses the middle scale at the mark 1.17 Ω·m.

What it means. A nomogram replaces multiplication by adding lengths, like a slide rule. The second step, water saturation from $R_0$ and $R_t$ , is done in the same way on the next set of three scales. So the nomograms of Archie's equation have two stages with an intermediate scale $R_0$ . The data in this example are assumed for illustration.

Connections to other topics

Log interpretation workflow — where in the order of interpretation charts are needed.
Borehole environment: invaded zone and mudcake — the zones around the borehole and how the sondes mix them.
Resistivity logs: electrical and induction tools — finding $R_t$ from three curves by trial, and the order of corrections.
Microresistivity tools and flushed-zone resistivity — the correction of microresistivity tools for mudcake.
Porosity from the neutron log — environmental corrections and matrix conversion.
Lithology and porosity from crossplots — porosity and lithology crossplots.
Determining formation water resistivity — the SP charts and the step from equivalent to true resistivity.
Resistivity–porosity crossplots — “resistivity against porosity” grids for saturation.
Resistivity ratio methods and movable oil — saturation and mobility from the resistivity ratio.
Formation water and formation temperature — converting resistivities to formation temperature.

Check yourself

1
An induction sonde read 20 Ω·m in a borehole with a salty mud of 0.05 Ω·m. The share of the borehole in the signal is 0.002. What can we say?
1. A.There is no answer: $R_a/R_m$ is larger than $1/G_m$
2. B. $R_t$ is less than 20 Ω·m
3. C.About 80% of the signal comes from the mud; $R_t$ is about 100 Ω·m, but it is better to replace such a correction with a laterolog
4. D.The correction is small: an induction sonde does not feel the mud
Show the answer
Correct answer: C. About 80% of the signal comes from the mud; $R_t$ is about 100 Ω·m, but it is better to replace such a correction with a laterolog
The conductivity from the reading is $1000/20 = 50$ mS/m, and the contribution of the borehole is $0.002\cdot 1000/0.05 = 40$ mS/m, which is 80% of the signal. $\sigma_t = (50 - 40)/0.998 \approx 10$ mS/m, so $R_t \approx 100$ Ω·m, a factor of about 5. Formally there is an answer ( $R_a/R_m = 400 < 500$ ), but a 10% error in $G_m$ or $R_m$ changes it by a factor of 1.4–1.8. So with a salty mud we find $R_t$ with a laterolog.
2
On the invasion chart, the point of a bed lies near the diagonal $x = y$ . What does this mean, and what can we trust?
Show the answer
Near the diagonal, the shallow sonde shows almost $R_{xo}$ : the invasion is very deep. The correction to the deep sonde is large, and the answer is sensitive to all three readings, so $R_t$ is known only roughly. We can trust it only with a wide margin. It is better to bring in sondes with a greater depth of investigation, a repeat log, or a comparison with water-bearing beds of the same section.
3
Why must the invasion chart not be used with readings that have no borehole correction?
1. A.The borehole correction is applied after the invasion correction
2. B.It can be used: the chart accounts for everything
3. C.The borehole affects only the microresistivity tool
4. D.The chart is calculated for readings already corrected for the borehole and the enclosing rocks; otherwise their effect will be taken for invasion
Show the answer
Correct answer: D. The chart is calculated for readings already corrected for the borehole and the enclosing rocks; otherwise their effect will be taken for invasion
The model of the chart is step invasion in a uniform thick bed with no effect of the borehole. If we put readings on it that include the effect of the mud or the enclosing rocks, the chart will read the difference between the sondes as another depth of invasion or another $R_{xo}/R_t$ . Then $R_t$ will come out wrong.
4
On a logarithmic scale, a point lies at 0.3 of the distance between the marks 2 and 5. What is its value?
Show the answer
On a logarithmic scale, fractions of the distance add up in logarithms: $2\cdot(5/2)^{0.3} \approx 2.63$ . Linear interpolation would give $2 + 0.3\cdot 3 = 2.9$ , which is 10% too high.
5
A formation temperature of 150 °F was read on an SP chart on the °C scale. How will $R_{we}$ be wrong?
1. A.It will not change
2. B.Too low
3. C.It depends on the sign of the SP
4. D.Too high: the coefficient $K$ will be larger, and the ratio $R_{mfe}/R_{we}$ smaller
Show the answer
Correct answer: D. Too high: the coefficient $K$ will be larger, and the ratio $R_{mfe}/R_{we}$ smaller
150 °F is about 66 °C, but on the °C scale the chart will “see” 150 °C: $K = 65 + 0.24\cdot 150 = 101$ mV instead of about 81 mV. For the same amplitude, $10^{|\mathrm{SSP}|/K}$ is smaller, $R_{we} = R_{mfe}/10^{|\mathrm{SSP}|/K}$ is larger, and the water saturation of all the beds comes out too high.
6
What do we need to check about a chart before using it?
Show the answer
The tool, its generation and the way of processing; the borehole diameter and the position of the tool (centred, pressed against the wall, stand-off); the type of mud and its resistivity or density, and sometimes the ratio $R_{xo}/R_m$ ; the fluid and the matrix for which the lines were calculated (fresh or salty filtrate, limestone units); the units of all the scales; which corrections have already been applied to the curves during recording. All of this is the assumptions of the model on which the chart is built.

Updated: 17 October 2026