M–N and MID plots for complex lithologies

Quick refresher

A rock with liquid in the pores is a mixture of matrix and fluid by volume. So all beds with one matrix lie on a straight line that runs from the fluid point (100% pores) to the matrix point. The slope of this line does not depend on porosity. On the sonic–density plane it is $M$ , and on the neutron–density plane it is $N$ .
On the M–N plot the minerals are points. In the lesson's model they are quartz (0.62–0.65; 0.81), calcite (0.585; 0.83), dolomite (0.49–0.52; 0.78), anhydrite (0.505; 0.70), gypsum (0.37; 1.01) and salt (0.97; 1.18), with $N$ given first. The points of quartz and dolomite shift slightly with porosity, because the neutron response of these rocks is not linear.
A mixture of three minerals lies inside the triangle of their points. But the position in it gives shares weighted by $\rho_{ma} - \rho_f$ , not volumes. One point usually lies in several triangles. We choose by the geology, by $P_e$ (the photoelectric factor, a log curve that responds to the minerals), and by cuttings (rock chips brought up by the mud) and core (rock samples taken from the well).
Vugs raise the point on the M–N plot vertically, because the sonic log does not see them. Gas shifts it up and to the right, and shale shifts it down and to the left, towards the shale point.
MID plot: we calculate $\rho_{maa}$ from the apparent porosity of the neutron–density crossplot, and $\Delta t_{maa}$ from the sonic–neutron crossplot. (The apparent porosity is the porosity that a crossplot shows if we assume that the bed holds only liquid.) We use the same transform that was used to build the mineral points. Gas moves the point towards quartz and beyond. Vugs move it to the left, shale down, and gypsum far down and to the left.
An error of one porosity unit in the apparent porosity shifts $\rho_{maa}$ by about 0.02 g/cm³ and $\Delta t_{maa}$ by 5–6 µs/m. That is comparable to the difference between quartz, calcite and dolomite. So both plots are more reliable for choosing the minerals than for finding their shares.

Key relationships

M = 0.003048\,\dfrac{\Delta t_f - \Delta t}{\rho_b - \rho_f}

$\Delta t$ in µs/m; with $\Delta t$ in µs/ft the multiplier is 0.01

N = \dfrac{\phi_{Nf} - \phi_N}{\rho_b - \rho_f}

neutron porosity in limestone units (the tool is scaled to read the true porosity in limestone)

\rho_{maa} = \dfrac{\rho_b - \phi_{ta}\rho_f}{1 - \phi_{ta}}

apparent matrix density

\Delta t_{maa} = \dfrac{\Delta t - \phi_{ta}\Delta t_f}{1 - \phi_{ta}}

apparent matrix transit time (time-average equation)

\phi_2 = (1 - \phi_t)\left(\dfrac{M}{M_{ma}} - 1\right)

secondary porosity from the rise of the point above the mineral point

Common pitfalls

Reading the position of the point in the triangle of the M–N plot as volume fractions of the minerals.
Taking the mineral points from a chart for another filtrate, tool or transit time transform.
Reading the lithology from the point of a gas-bearing, shaly or vuggy bed without corrections.
Trusting the mineral shares from the MID plot as much as the choice of minerals.

After this lesson you can

explain why the parameters $M$ and $N$ hardly depend on porosity, and calculate them from log readings
find the mineral points on the M–N plot, read a mixture of three minerals, and explain why the answer is not unique and how we narrow it down
estimate the secondary porosity (the pore space that formed after the rock was deposited, such as vugs) from the shift of the point on the M–N plot, and recognise gas and shale on it
calculate the apparent matrix density and transit time from the apparent porosity, and read the MID plot
estimate how an error in the apparent porosity and the choice of the transit time transform (the formula that turns transit time into porosity) shift the points, and understand where the plots give the composition only approximately

How to take porosity out of the coordinates

On a crossplot of two porosity methods (Lithology and porosity from crossplots) the point of a bed depends on both the minerals and the porosity. So the mineral lines carry porosity marks. To choose minerals, it is convenient to move to coordinates that have no porosity in them at all. The key is the balance of volumes. In a clean bed with liquid, a reading that is linear in volumes is the average of the readings of the matrix and of the fluid, weighted by volume: $L = (1 - \phi)L_{ma} + \phi L_f$ . So on the plane of two such readings, the bed point lies on the segment between the matrix point and the fluid point. Porosity is the fraction of the way from the matrix to the fluid.

All beds with one matrix lie on one straight line that starts at the fluid point, whatever their porosity. Only the matrix sets its slope. This is the quantity that does not depend on porosity. There is also a second option. We can estimate the porosity, and extend the line from the fluid point through the bed point to zero porosity. This gives the apparent parameters of the matrix. The M–N plot is built on the first idea, and the MID plot on the second.

CalculatedNeutron–density crossplot extended to the fluid point (100% pores: neutron porosity 1, density 1.00 g/cm³). The dashed straight lines are rays from the fluid point to the mineral points. The solid lines are beds of quartz, calcite and dolomite with liquid in the pores, from 0 to 40%, from the lesson's model. The calcite line coincides with its ray. The quartz and dolomite lines depart slightly from theirs, because the neutron response is not linear.

N

is the ratio of the horizontal step of the ray to the vertical one.

The parameters M and N

Parameter

M

: the slope on the sonic–density plane

M = 0.003048\,\frac{\Delta t_f - \Delta t}{\rho_b - \rho_f}

where:

$\Delta t$ interval transit time from the sonic log, µs/m
$\Delta t_f$ interval transit time of the fluid in the zone of investigation (the part of the rock that the tool measures): 620 µs/m (189 µs/ft) for fresh filtrate (the liquid that enters the rock from the drilling mud), about 607 µs/m (185 µs/ft) for salty filtrate, µs/m
$\rho_b$ bulk density from the log, g/cm³
$\rho_f$ fluid density: 1.00 for fresh filtrate, about 1.1 for salty filtrate, g/cm³

Units: $\Delta t$ in µs/m, density in g/cm³; $M$ is a conventional quantity (µs/ft per 1 g/cm³, divided by 100)

Holds when

A clean bed with liquid in the pores, and the readings corrected for environmental conditions.
The interval transit time is linear in volumes (the time-average equation). With the empirical transform, $M$ changes slightly with porosity.
The multiplier is a conventional scale: 0.01 with $\Delta t$ in µs/ft, that is, $0.01\cdot 0.3048 = 0.003048$ with µs/m. The calculation and the chart must use the same multiplier.

How much interval transit time the rock “loses” for every g/cm³ of increase in density, as we go from the fluid point to the bed point. The multiplier brings $M$ to values near one, like $N$ .

In other unit systems (1)

Oilfield units

M = 0.01\,\frac{\Delta t_f - \Delta t}{\rho_b - \rho_f}

$\Delta t$ and $\Delta t_f$ in µs/ft, densities in g/cm³

0.01 is a conventional scale. 0.003048 in the metric form is the same multiplier, with µs/ft converted to µs/m.

Parameter

N

: the slope on the neutron–density plane

N = \frac{\phi_{Nf} - \phi_N}{\rho_b - \rho_f}

where:

$\phi_N$ neutron porosity in limestone units (the tool is scaled to read the true porosity in limestone), after corrections for environmental conditions, fraction
$\phi_{Nf}$ neutron reading of the fluid: 1 for fresh filtrate, fraction
$\rho_b,\ \rho_f$ bulk density and fluid density, g/cm³

Units: neutron porosity as a fraction, density in g/cm³; $N$ in cm³/g

Holds when

A clean bed with liquid. The neutron curve is in limestone units of the same tool for which the mineral points were built.
$N$ does not depend on porosity only if the neutron reading is linear in volumes. For sandstone and dolomite it is not linear, and their $N$ changes by 0.02–0.04 between 0 and 30% pores.

The slope of the ray from the fluid point to the bed point: how much neutron porosity the rock loses for each unit of increase in density. Of the main minerals, quartz has the largest value, and dolomite has the smallest.

If we put into the formulas the parameters of a pure mineral with no pores (its interval transit time, density and neutron reading), we get the point of that mineral. A bed of the same rock with liquid in the pores would give the same point at any porosity, if the responses were linear.

M

and

N

of the minerals (the lesson's model, fresh filtrate)

Mineral: Quartz
$M$: 0.809
$N$: 0.618 → 0.645
Note: $N$ grows with porosity from 0 to 30%

Mineral: Calcite
$M$: 0.827
$N$: 0.585
Note: does not depend on porosity

Mineral: Dolomite
$M$: 0.777
$N$: 0.524 → 0.488
Note: $N$ falls with porosity from 0 to 30%

Mineral: Anhydrite
$M$: 0.702
$N$: 0.505
Note: usually tight

Mineral: Gypsum
$M$: 1.011
$N$: 0.370
Note: water of crystallisation (water locked in the crystal)

Mineral: Rock salt
$M$: 1.184
$N$: 0.971
Note: usually tight

Mineral: The lesson's shale
$M$: 0.659
$N$: 0.419
Note: depends on the shales and the tool

Mineral	$M$	$N$	Note
Quartz	0.809	0.618 → 0.645	$N$ grows with porosity from 0 to 30%
Calcite	0.827	0.585	does not depend on porosity
Dolomite	0.777	0.524 → 0.488	$N$ falls with porosity from 0 to 30%
Anhydrite	0.702	0.505	usually tight
Gypsum	1.011	0.370	water of crystallisation (water locked in the crystal)
Rock salt	1.184	0.971	usually tight
The lesson's shale	0.659	0.419	depends on the shales and the tool

Parameters of the model (density, g/cm³; interval transit time by the time-average equation, µs/m; neutron reading with no pores): quartz 2.65, 182, −0.02; calcite 2.71, 156, 0; dolomite 2.87, 143, +0.02; anhydrite 2.98, 164, 0; gypsum 2.35, 172, 0.5; salt 2.03, 220, 0; shale 2.48, 300, 0.38. For salty filtrate (about 1.1 g/cm³ and 607 µs/m, neutron reading of the fluid 1), all the points shift up and to the right. The calcite point goes to (0.621; 0.854). On charts, dolomite is sometimes given by several points for different porosity ranges, for the same reason as in the model. The neutron readings of evaporites in the model are approximate. Real tools read a few units below zero in salt, and 0.5–0.6 in gypsum, and on the charts of particular tools their points are shifted.

For quartz, calcite and dolomite, $M$ is almost the same (0.78–0.83). For the density and sonic logs these minerals differ from the fluid in almost the same proportion. This is exactly why the sonic–density crossplot separates them so poorly. $N$ separates them better (0.49–0.65), and the evaporites differ strongly in $M$ . Together, $M$ and $N$ put all six substances in different places on the plane.

CalculatedM–N plot from the lesson's model (fresh filtrate): the mineral points (for quartz and dolomite, segments from 0 to 30% porosity), the quartz–calcite–dolomite triangle and the sides to anhydrite, and the shale point. The square is the bed of the example below. The arrows show the shift of the point by vugs and gas for limestones, and by shale for sandstone.

Mixtures of three minerals and ambiguity

For a mixture of minerals, $M$ and $N$ are the averages of the mineral points, taken with the same weights. So the point of a mixture of three minerals lies inside the triangle of their points, and the point of a mixture of two lies on the segment between them. The weights are not proportional to the volumes of the minerals. They are proportional to the product of the volume and $\rho_{ma} - \rho_f$ (why). For quartz, calcite, dolomite and anhydrite this shifts the shares by a few per cent. For gypsum and salt it shifts them noticeably more.

The main difficulty is ambiguity. Six mineral points make many triangles, and the point of a bed usually lies in several at once. Which one to choose is decided by the geology (which minerals are possible in this sequence and at this depth), the photoelectric factor and spectrometry (Lithology from the photoelectric factor and spectral gamma ray), cuttings and core.

Worked exampleDolomite with anhydrite, or limestone with dolomite and anhydrite?

Given (assumed data)

Bed: a clean carbonate bed at a depth of about 2500 m, with the pores filled with fresh filtrate
Readings: $\rho_b = 2.673$ g/cm³; $\phi_N = 0.127$ in limestone units; $\Delta t = 200$ µs/m; $P_e = 3.23$ barn/electron
Mineral points at a porosity of about 10% ( $N$ ; $M$ ): quartz (0.635; 0.809), calcite (0.585; 0.827), dolomite (0.504; 0.777), anhydrite (0.505; 0.702), gypsum (0.370; 1.011)
Volumetric photoelectric cross sections of the minerals (how strongly 1 cm³ of the mineral absorbs gamma rays), barn/cm³: quartz 4.80; calcite 13.76; dolomite 8.97; anhydrite 14.95

Solution

$M = 0.003048\cdot\dfrac{620 - 200}{2.673 - 1} = 0.765$ , $N = \dfrac{1 - 0.127}{1.673} = 0.522$ .
The point lies in four triangles: dolomite–anhydrite–quartz, calcite–dolomite–anhydrite, and two with gypsum. As gypsum is buried deeper, it loses its water of crystallisation and becomes anhydrite. So at a depth of 2500 m there is usually no gypsum, and two hypotheses remain.
Hypothesis 1, dolomite–anhydrite–quartz: the weights (the solution of three linear equations) are 0.646, 0.218 and 0.136. We divide them by $\rho_{ma} - 1$ (1.87; 1.98; 1.65) and scale the results so that they add up to 1. The volume fractions are 0.642, 0.205 and 0.153.
Hypothesis 2, calcite–dolomite–anhydrite: the weights are 0.218, 0.476 and 0.306, and the volume fractions are 0.238, 0.474 and 0.288.
Porosity from the density, with the matrix of each hypothesis: $\rho_{ma}$ = 2.859 and 2.864 g/cm³, and $\phi = \dfrac{\rho_{ma} - 2.673}{\rho_{ma} - 1}$ = 0.100 and 0.102. The two porosities are almost the same.
The expected $P_e$ : the volumetric cross section of the matrix is 9.56 and 11.83 barn/cm³, and that of the rock is $(1 - \phi)U_{ma} + 0.397\phi$ . We divide it by the electron density index $\dfrac{2.673 + 0.1883}{1.0704} = 2.673$ g/cm³ and get 3.23 and 3.99 barn/electron. The recorded value is 3.23.

Answer: The bed is dolomite with anhydrite and quartz: about 64% dolomite, 21% anhydrite and 15% quartz, with a porosity of about 10%.

What it means. The M–N plot narrowed the choice to two plausible compositions, but it could not choose between them itself. Both hypotheses explain the three porosity methods equally well and give almost the same porosity. The photoelectric factor decided. The second hypothesis has a lot of calcite and anhydrite, with $P_e$ about 5, and it would give 3.99 instead of the recorded 3.23. The readings were calculated for a bed of 65% dolomite, 20% anhydrite and 15% quartz, with a porosity of 10%. The data in this example are assumed for illustration.

Vugs, gas and shale on the M–N plot

Vugs. The sonic log sees mainly the intergranular pores, while the density and neutron logs see all the voids (secondary porosity). In a vuggy bed the interval transit time is smaller than in a rock of the same density without vugs. So $M$ grows, and $N$ does not change: the point rises vertically above the point of its mineral. The height of the rise measures the share of vugs.

Secondary porosity from the rise of the point on the M–N plot

\phi_2 = (1 - \phi_t)\left(\frac{M}{M_{ma}} - 1\right)

where:

$\phi_2$ secondary porosity: vugs and fractures, fraction
$\phi_t$ total porosity from the density and neutron logs, fraction
$M,\ M_{ma}$ $M$ of the bed and $M$ of its mineral; only their ratio enters the formula, conventional units of $M$

Units: fraction

Holds when

A clean bed with liquid, made of one known mineral.
The sonic log sees only the intergranular porosity $\phi_1$ and follows the time-average equation. The density and neutron logs see the total porosity $\phi_t = \phi_1 + \phi_2$ .

The numerator of $M$ is proportional to $1 - \phi_1$ , and the denominator is proportional to $1 - \phi_t$ . So $M = M_{ma}\dfrac{1 - \phi_1}{1 - \phi_t}$ . Solving this for $\phi_1$ gives the formula. It is the same secondary porosity index $\phi_t - \phi_S$ , read from the M–N plot.

Worked exampleVuggy limestone above the calcite point

Given (assumed data)

Bed: clean limestone with vugs (the same bed as in the topic on crossplots)
Readings: $\rho_b = 2.402$ g/cm³; $\phi_N = 0.180$ in limestone units; $\Delta t = 212$ µs/m
Calcite: $M_{ma} = 0.827$ , $N = 0.585$ ; interval transit time 156 µs/m

Solution

$M = 0.003048\cdot\dfrac{620 - 212}{2.402 - 1} = 0.887$ , $N = \dfrac{1 - 0.180}{1.402} = 0.585$ . The point is exactly above calcite, higher by 0.060.
The total porosity from the neutron and density logs is 0.180 (the point of the neutron–density crossplot lies on the limestone line).
$\phi_2 = (1 - 0.180)\cdot(0.887/0.827 - 1) = 0.059$ .
Check: the intergranular porosity is $0.180 - 0.059 = 0.121$ , and the sonic porosity is $\dfrac{212 - 156}{464} = 0.121$ .

Answer: Of the total porosity of 18%, about 6 porosity units are vugs.

What it means. The rise of the point on the M–N plot and the secondary porosity index are the same thing, written in two ways. If we ignored the vugs, we could take the point above calcite for a mixture of calcite with gypsum or with salt: minerals with a large $M$ . The readings were calculated for a limestone with a total porosity of 18%, of which 6 units are vugs. The data in this example are assumed for illustration.

Gas in the flushed zone (the rock next to the hole, which the mud filtrate has invaded) lowers both the density and the neutron porosity, and it changes the interval transit time of a compacted rock little. The denominator of both parameters, $\rho_b - \rho_f$ , becomes smaller, and the numerator of $N$ grows. So the point goes up and to the right. Take the gas-bearing limestone from the topic on crossplots ( $\rho_b = 2.371$ g/cm³, $\phi_N = 0.102$ , and $\Delta t = 230$ µs/m, the same as in that limestone with liquid). It gives $M = 0.867$ and $N = 0.655$ , to the right of and above the quartz point (0.64; 0.81). Without a correction for gas, we would read it as a quartz rock.

Shale shifts the point towards the shale point. In the lesson's model, this point lies lower and to the left of all the main minerals (0.419; 0.659). Take the shaly sandstone from the same topic ( $V_{sh} = 0.3$ ; $\rho_b = 2.345$ g/cm³, $\phi_N = 0.237$ ; interval transit time 285 µs/m by the same model). It gives the point (0.567; 0.759), inside the triangles with anhydrite. It is easily taken for a mixture of quartz, dolomite and anhydrite. The position of the shale point depends on the clay minerals and on the tool. So we take it from the neighbouring shale beds.

The MID plot: apparent matrix parameters

The second way to get rid of porosity is to estimate it and to recalculate the readings to zero porosity. That is, we extend the line from the fluid point through the bed point to the matrix point. We need a porosity that hardly depends on the minerals: the apparent total porosity $\phi_{ta}$ . We read it from a crossplot of two methods, assuming a bed with liquid.

Apparent matrix density

\rho_{maa} = \frac{\rho_b - \phi_{ta}\,\rho_f}{1 - \phi_{ta}}

where:

$\rho_{maa}$ apparent matrix density, g/cm³
$\phi_{ta}$ apparent total porosity from the neutron–density crossplot, fraction
$\rho_b,\ \rho_f$ bulk density and fluid density, g/cm³

Units: g/cm³

Holds when

A clean bed with liquid of known density; the apparent porosity is correct.

The mass balance, solved for the matrix density.

Apparent matrix transit time

\Delta t_{maa} = \frac{\Delta t - \phi_{ta}\,\Delta t_f}{1 - \phi_{ta}}

where:

$\Delta t_{maa}$ apparent matrix transit time, µs/m
$\phi_{ta}$ apparent total porosity from the sonic–neutron crossplot, built with the same transform, fraction
$\Delta t,\ \Delta t_f$ interval transit time from the log and of the fluid (620 µs/m for fresh filtrate), µs/m

Units: µs/m (or µs/ft: the same for all quantities)

Holds when

This form is for the time-average equation. The empirical transform has its own form (below).
The mineral points on the chart were built with the same transform as the calculation.

The time-average equation, solved for the matrix transit time.

In other unit systems (1)

For the empirical transform

\Delta t_{maa} = \Delta t\left(1 - \frac{\phi_{ta}}{C}\right)

$C \approx 0.67$ ; the interval transit times in the same units

from $\phi = C\dfrac{\Delta t - \Delta t_{ma}}{\Delta t}$ , solved for $\Delta t_{ma}$ (empirical transform).

Which porosity to take. For $\rho_{maa}$ we take it from the neutron–density crossplot, and for $\Delta t_{maa}$ from the sonic–neutron crossplot (crossplots with the sonic log). In both cases we read the porosity from the isoporosity lines (lines of equal porosity) of the two minerals whose lines the point lies between. For points above the sandstone line, we take the porosity of the sandstone with the same neutron reading. If the bed is clean, has liquid, and its minerals are the same as on the lines, both porosities are right, and the apparent parameters equal the matrix parameters. Both coordinates mix by volume. So a mixture of three minerals lies inside a triangle, and, unlike on the M–N plot, the position in it gives volume fractions.

Mineral points on the MID plot (the lesson's model)

Mineral: Quartz
$\rho_{maa}$ , g/cm³: 2.65
$\Delta t_{maa}$ , µs/m: 182
By the empirical transform, µs/m: 184

Mineral: Calcite
$\rho_{maa}$ , g/cm³: 2.71
$\Delta t_{maa}$ , µs/m: 156
By the empirical transform, µs/m: 161

Mineral: Dolomite
$\rho_{maa}$ , g/cm³: 2.87
$\Delta t_{maa}$ , µs/m: 143
By the empirical transform, µs/m: 144

Mineral: Anhydrite
$\rho_{maa}$ , g/cm³: 2.98
$\Delta t_{maa}$ , µs/m: 164
By the empirical transform, µs/m: —

Mineral: Rock salt
$\rho_{maa}$ , g/cm³: 2.03
$\Delta t_{maa}$ , µs/m: 220
By the empirical transform, µs/m: —

Mineral: Gypsum
$\rho_{maa}$ , g/cm³: 2.35
$\Delta t_{maa}$ , µs/m: 172
By the empirical transform, µs/m: —

Mineral	$\rho_{maa}$ , g/cm³	$\Delta t_{maa}$ , µs/m	By the empirical transform, µs/m
Quartz	2.65	182	184
Calcite	2.71	156	161
Dolomite	2.87	143	144
Anhydrite	2.98	164	—
Rock salt	2.03	220	—
Gypsum	2.35	172	—

This is how the parameters of the minerals themselves fall. Anhydrite and salt, read by the rules of the MID plot, shift by 0.02–0.03 g/cm³ and by 2–9 µs/m. Gypsum goes far down and to the left: its water of crystallisation makes the apparent porosity too high (in the lesson's model, to 82 µs/m and 3.09 g/cm³).

CalculatedMID plot from the lesson's model (time-average equation, fresh filtrate): the points of quartz, calcite, dolomite and anhydrite, the triangle of the main minerals, the lesson's shale point, and the directions to salt and to gypsum. Letters mark the beds of the example below: dolomitised limestone A, gas-bearing limestone B and vuggy limestone C. The arrows show the shift from the calcite point by gas and by vugs.

Worked exampleThree limestones on the MID plot

Given (assumed data)

Bed A: $\rho_b = 2.508$ g/cm³; $\phi_N = 0.173$ ; $\Delta t = 221$ µs/m
Bed B: $\rho_b = 2.371$ g/cm³; $\phi_N = 0.102$ ; $\Delta t = 230$ µs/m; high resistivity, gas shows (gas detected in the drilling mud)
Bed C: $\rho_b = 2.402$ g/cm³; $\phi_N = 0.180$ ; $\Delta t = 212$ µs/m
Conditions: fresh filtrate (1.00 g/cm³; 620 µs/m); the crossplots and the mineral points are those of the lesson's model, with the time-average equation

Solution

Bed A: $\phi_{D,ls} = 0.118$ , and the point is below the limestone line. From the limestone and dolomite lines, $\phi_{ta} = 0.150$ on the neutron–density crossplot and 0.150 on the sonic–neutron crossplot. $\rho_{maa} = \dfrac{2.508 - 0.150}{1 - 0.150} = 2.774$ g/cm³, $\Delta t_{maa} = \dfrac{221 - 0.150\cdot 620}{1 - 0.150} = 150.6$ µs/m.
Point A lies on the calcite–dolomite side. From the density, the share of dolomite is $\dfrac{2.774 - 2.71}{0.16} = 0.40$ , and from the interval transit time it is $\dfrac{156 - 150.6}{13} = 0.42$ .
Bed B: $\phi_{D,ls} = 0.198$ , with a neutron porosity of 0.102. This is above the sandstone line, so $\phi_{ta} = 0.147$ (the sandstone with the same neutron reading). From the sonic–neutron crossplot it is 0.130. Then $\rho_{maa} = 2.607$ g/cm³ and $\Delta t_{maa} = 171.7$ µs/m. This is near the quartz point, and the density is even lower than that of quartz.
Bed C: $\phi_{D,ls} = 0.180$ , a point on the limestone line, and $\phi_{ta} = 0.180$ . From the sonic–neutron crossplot it is 0.139. Then $\rho_{maa} = 2.710$ g/cm³, which is calcite. But $\Delta t_{maa} = 146.1$ µs/m, which is to the left of calcite by 10 µs/m.

Answer: A is a limestone with 40% dolomite in the matrix. B is “quartz” by the MID plot, but in fact a limestone with gas. C is a limestone with vugs, which shifted the point to the left.

What it means. The readings were calculated for these beds. A is 60% calcite and 40% dolomite at a porosity of 15%. B is a limestone with a porosity of 16% and gas in 40% of the pores of the flushed zone. C is a limestone with a total porosity of 18%, of which 6 units are vugs. The MID plot showed all three cases. But B and C can be read correctly only with other data. We find gas from the resistivity, the gas shows and the neutron–density crossover (the places where the neutron and density curves cross over each other). We find vugs from the apparent density, which equals that of calcite (for dolomite it would be about 2.87). The data in this example are assumed for illustration.

Gas, vugs, shale and evaporites on the MID plot

Gas lowers the density almost without changing the neutron–density apparent porosity, so $\rho_{maa}$ falls. The neutron reading in gas is too low, so the apparent porosity from the sonic–neutron crossplot comes out smaller than the true one, and $\Delta t_{maa}$ grows. The point goes up and to the right, towards quartz and beyond, like bed B. Gas is shown by the neutron–density crossover, the resistivity and the gas shows. $P_e$ helps to recognise the mineral: on the $U_{maa}$ – $\rho_{maa}$ crossplot the same bed stays above the calcite point (gas on this crossplot).
Vugs reduce $\Delta t_{maa}$ with no change in $\rho_{maa}$ , so the point goes to the left, like bed C. By the interval transit time we could take it for dolomite, but the apparent density of dolomite would be about 2.87.
Shale plots low: the lesson's shale gives $\rho_{maa} \approx 3.01$ g/cm³ and $\Delta t_{maa} \approx 145$ µs/m, lower than dolomite. Its neutron porosity includes bound water, so the apparent porosity is too high. Where exactly the shale point will lie depends on the clay minerals, the bound water and the neutron tool.
Evaporites. Rock salt is far up and to the right (220 µs/m; 2.03 g/cm³), and anhydrite is low down (164; 2.98). Gypsum shows a neutron porosity of about 0.5, so the apparent porosity is huge, and the point goes far down and to the left.

How accurate are the apparent parameters? An error in the apparent porosity shifts the point along the straight line that goes out from the fluid point. If the porosity is too high, the point moves farther from the fluid, to larger densities and smaller interval transit times.

Sensitivity of the apparent parameters to the apparent porosity

\frac{\partial\rho_{maa}}{\partial\phi_{ta}} = \frac{\rho_{maa} - \rho_f}{1 - \phi_{ta}}, \qquad \frac{\partial\Delta t_{maa}}{\partial\phi_{ta}} = -\frac{\Delta t_f - \Delta t_{maa}}{1 - \phi_{ta}}

where:

$\rho_{maa},\ \Delta t_{maa}$ apparent matrix density and transit time, g/cm³; µs/m
$\phi_{ta}$ apparent total porosity, fraction
$\rho_f,\ \Delta t_f$ density and interval transit time of the fluid, g/cm³; µs/m

Units: g/cm³ and µs/m per unit of porosity (as a fraction)

Holds when

Small porosity errors; the derivatives were obtained by differentiating the formulas of the apparent parameters.

For a limestone with a porosity of 15%, they are 2.0 g/cm³ and −546 µs/m per unit of porosity, that is, 0.02 g/cm³ and 5.5 µs/m for every 0.01. This is comparable to the size of the triangle: quartz and calcite differ by 0.06 g/cm³ and 26 µs/m, and calcite and dolomite by 0.16 g/cm³ and 13 µs/m.

An apparent porosity read from the lines of the wrong pair of minerals is wrong by half a unit, a unit or more. An example is a sandy dolomite or a carbonate with anhydrite. So the MID plot shows with confidence which minerals are present, and where gas and vugs shift the point, but it gives the mineral shares only approximately (example). The M–N plot does not need the apparent porosity, but the position in its triangle is not the volume fractions.

How to work with the M–N and MID plots

1
Data
Depth-match the curves and correct them for environmental conditions. Edit out cycle skipping on the sonic curve. Exclude intervals with washouts in the hole and with a large density correction $\Delta\rho$ . Check that the filtrate (fresh or salty) is the one for which the mineral points were built.
2
Cloud of points
We plot many depths at once. Clean beds group near the mineral points and along the sides of the triangles. Shaly beds stretch towards the shale point. Gas-bearing and vuggy beds move off in their own directions. The cloud shows the section. A single point does not.
3
Reference points
We take the shale point from the neighbouring shale packages. If the section has tight anhydrite and salt, we use them to check the chart and the filtrate: their points must fall in their places.
4
Gas and vugs
We confirm gas by the neutron–density crossover, the resistivity and the gas shows. We confirm vugs by the secondary porosity index. We correct such points, or do not use them to choose minerals.
5
Choice of minerals
We compare the triangles that hold the cloud with the geology, $P_e$ and spectrometry (Lithology from the photoelectric factor and spectral gamma ray), cuttings and core.
6
Calculation
We calculate the final porosity and the shares of the chosen minerals. We do it from a pair of curves, from the density and $P_e$ , or with a multi-mineral program, and we check them against core.

Terms and symbols

Fluid point: The readings of a pure fluid (100% pores) on the plane of two methods. The straight lines of beds with one matrix go out from it.
Parameters $M$ and $N$: The slopes of the “fluid–bed” lines on the sonic–density and neutron–density planes. They hardly depend on porosity.
M–N plot: A crossplot of $N$ against $M$ with mineral points. Mixtures of three minerals lie inside triangles, and the shares are weighted by $\rho_{ma} - \rho_f$ .
Apparent total porosity $\phi_{ta}$: Porosity read from a crossplot of two methods, assuming a bed with liquid.
Apparent matrix parameters $\rho_{maa}$ and $\Delta t_{maa}$: The density and interval transit time recalculated to zero porosity with $\phi_{ta}$ .
MID plot: A crossplot of $\Delta t_{maa}$ against $\rho_{maa}$ (or $U_{maa}$ against $\rho_{maa}$ ) with mineral points. Mixtures are given in volume fractions.

Common mistakes and misconceptions

Wrong

“The position of the point in the triangle of the M–N plot gives the volume fractions of the minerals.”

In fact

These are shares weighted by $\rho_{ma} - \rho_f$ . For a mixture of quartz and anhydrite, half and half by volume, the point goes 55% of the way from quartz to anhydrite. We get the volumes by dividing the weights by $\rho_{ma} - \rho_f$ .

Wrong

“ $M$ and $N$ do not depend on porosity at all.”

In fact

Only when the responses are linear. The neutron response of sandstone and dolomite is not linear, so their points shift with porosity by 0.02–0.04 in $N$ . The empirical transform of the interval transit time makes $M$ depend on porosity.

Wrong

“A point above calcite on the M–N plot is an unknown mineral.”

In fact

More often these are vugs: the sonic log does not see them, and $M$ grows at the same $N$ . The rise of the point gives the secondary porosity.

Wrong

“The MID plot gives exact mineral shares.”

In fact

It is reliable for choosing the minerals. The shares depend on the apparent porosity, and an error of one unit in it shifts $\rho_{maa}$ by 0.02 g/cm³ and $\Delta t_{maa}$ by 5–6 µs/m. We calculate the final shares from the chosen minerals.

Wrong

“We can calculate $\Delta t_{maa}$ with any transform.”

In fact

Only with the one by which the mineral points on the chart were built. Mixing the time-average equation with the empirical one shifts the point by tens of µs/m.

In depth: advanced detail

Why the position in the M–N triangle is not volume fractions

Take a clean bed with liquid whose matrix is a mixture of minerals with volume fractions $f_i$ (shares of the matrix), and take linear responses. The numerators and denominators of $M$ and $N$ are differences between the readings of the bed and of the fluid. Each of them is a sum of the contributions of the minerals. The contributions are proportional to the volumes, and they are multiplied by the share of the matrix $1 - \phi$ . This multiplier cancels, which is why there is no dependence on porosity. The ratio of the two sums turns out to be an average of the mineral parameters, with weights that are proportional to the terms of the denominator:

M and N of a mixture of minerals

M = \sum_i w_i M_i, \qquad N = \sum_i w_i N_i, \qquad w_i = \frac{f_i\,(\rho_{ma,i} - \rho_f)}{\sum_j f_j\,(\rho_{ma,j} - \rho_f)}

where:

$M_i,\ N_i$ parameters of the $i$ -th mineral, in units of $M$ and $N$
$f_i$ volume fraction of the mineral in the matrix, fraction
$w_i$ weight of the mineral in the point of the mixture, fraction
$\rho_{ma,i},\ \rho_f$ densities of the mineral and of the fluid, g/cm³

Units: $w_i$ and $f_i$ are fractions; $M$ and $N$ are in their own units (see above)

Holds when

The responses of all three methods are linear in volumes. For sandstone and dolomite, which have a non-linear neutron response, we take the mineral points at the porosity of the bed.

The point of a mixture is inside the triangle, but it divides the triangle in the proportion of the weights. A mineral with a larger difference $\rho_{ma} - \rho_f$ “pulls” the point harder than its volume would. The reverse conversion is $f_i \propto w_i/(\rho_{ma,i} - \rho_f)$ .

For the main minerals the difference $\rho_{ma} - \rho_f$ lies between 1.65 (quartz) and 1.98 (anhydrite). So the weights and the volumes differ by a few per cent. In the example with anhydritic dolomite, the weights 0.646, 0.218 and 0.136 gave the volumes 0.642, 0.205 and 0.153. For a mixture of quartz and anhydrite, half and half by volume, the weight of anhydrite is already 0.55. For gypsum and salt the differences are 1.35 and 1.03, and with them the discrepancy is larger still. On the MID plot and on the $U_{maa}$ – $\rho_{maa}$ crossplot there is no such problem: their coordinates mix by volume themselves.

Sandy dolomite: why the MID plot gives the composition only approximately

Worked exampleSandy dolomite on the MID plot and on the

U_{maa}

–

\rho_{maa}

crossplot

Given (assumed data)

Readings: $\rho_b = 2.497$ g/cm³; $\phi_N = 0.197$ ; $\Delta t = 234$ µs/m; $P_e = 2.58$ barn/electron; fresh filtrate
Bed (known in advance): dolomite with an admixture of quartz sand

Solution

$\phi_{D,ls} = 0.125$ , and the point is below the limestone line. From the limestone and dolomite lines, the apparent porosity is 0.166 on the neutron–density crossplot and 0.176 on the sonic–neutron crossplot.
$\rho_{maa} = \dfrac{2.497 - 0.166}{1 - 0.166} = 2.795$ g/cm³, $\Delta t_{maa} = \dfrac{234 - 0.176\cdot 620}{1 - 0.176} = 151.6$ µs/m.
At this density the dolomite–quartz side passes through 156.3 µs/m, and the dolomite–calcite side passes through 149.1. The point is closer to the second. The calcite–dolomite–quartz triangle gives 31% calcite, 58% dolomite and 12% quartz.
The $U_{maa}$ – $\rho_{maa}$ crossplot with the same porosity: $U_{maa} = 7.68$ barn/cm³, and the same triangle gives 2% calcite, 65% dolomite and 33% quartz.
Finally, the “sandstone–dolomite” pair on the neutron–density crossplot: the porosity is 0.170 and the dolomite is 70% of the matrix.

Answer: The MID plot took a sandy dolomite for a dolomite with calcite (31% calcite). The crossplot with

U_{maa}

found an almost right composition. The “sandstone–dolomite” pair gives a porosity of 17% and 70% dolomite.

What it means. The readings were calculated for a bed of 70% dolomite and 30% quartz, with a porosity of 17%. The apparent porosities were read from the limestone and dolomite lines, and there is no quartz in that pair. They came out wrong by −0.4 and +0.6 porosity units. For the MID plot this is enough: quartz and calcite are close on it, and the point slid to the calcite–dolomite side. On the crossplot with $U_{maa}$ , the same porosity error is almost harmless: quartz and calcite differ in $U$ by almost a factor of three. So the working order is that the plots choose the minerals, and we calculate the shares from the chosen pair or triple. The data in this example are assumed for illustration.

The same mechanism makes the MID plot underestimate the anhydrite in carbonates. Anhydrite shifts the point on the neutron–density crossplot towards the dolomite line (for a tight dolomite, beyond it). The apparent porosity read from the limestone and dolomite lines comes out too low. In the lesson's model it is low by about half a unit for every 10% of anhydrite, and $\rho_{maa}$ falls with it. The M–N plot does not use apparent porosity. So for such a bed it gives the composition more accurately, provided that the triangle is chosen correctly.

Connections to other topics

Lithology and porosity from crossplots — crossplots of two methods, from which we read the apparent porosity.
Lithology from the photoelectric factor and spectral gamma ray — the photoelectric factor and the $U_{maa}$ – $\rho_{maa}$ crossplot, which remove the ambiguity.
Porosity from the sonic log — the time-average equation, the empirical transform and the secondary porosity index.
Porosity from the neutron log — limestone units and the non-linear response of sandstone and dolomite.
Porosity from the density log — matrix and fluid density in the porosity calculation.
Gas, shale, carbonate and pressure effects on porosity logs — gas, shale and carbonates in the readings of the three methods.
Sonic log — interval transit time, cycle skipping and the quality of the sonic log.
Interpretation charts: how to read them — charts for particular tools and filtrates.

Check yourself

1
Why do $M$ and $N$ hardly depend on porosity?
1. A.They are calculated from a porosity that has been corrected
2. B.A bed with liquid is a mixture of matrix and fluid by volume, so beds with one matrix lie on one straight line out of the fluid point, and $M$ and $N$ are the slopes of this line
3. C.The multiplier 0.003048 compensates for porosity
4. D.Density and interval transit time do not depend on porosity
Show the answer
Correct answer: B. A bed with liquid is a mixture of matrix and fluid by volume, so beds with one matrix lie on one straight line out of the fluid point, and $M$ and $N$ are the slopes of this line
In the numerator and denominator of $M$ and $N$ there are differences with the readings of the fluid, and both are proportional to the matrix share $1 - \phi$ . The multiplier cancels. So the bed point on the M–N plot coincides with the point of its matrix. This is exact for linear responses, and approximate if the response is not linear, as for the neutron log in sandstone and dolomite.
2
A clean bed with fresh filtrate: $\rho_b = 2.525$ g/cm³, $\phi_N = 0.192$ , $\Delta t = 228$ µs/m. Find $M$ and $N$ and estimate the lithology from the points of the lesson's model.
Show the answer
$M = 0.003048\cdot\dfrac{620 - 228}{1.525} \approx 0.783$ , $N = \dfrac{0.808}{1.525} \approx 0.530$ . The point lies near the dolomite (about 0.50; 0.78) – quartz (about 0.64; 0.81) side, about a quarter of the way from dolomite along $N$ . The weight of quartz is about 0.23. With the correction for $\rho_{ma} - \rho_f$ , this is about a quarter of the volume of the matrix. The bed is a sandy dolomite. We then calculate the porosity from the “sandstone–dolomite” pair.
3
The point of a limestone on the M–N plot lies straight above the calcite point: the same $N$ , a larger $M$ . What is this most likely?
1. A.An error of the neutron log
2. B.An admixture of dolomite
3. C.Shale
4. D.Vugs or fractures, which the sonic log does not see
Show the answer
Correct answer: D. Vugs or fractures, which the sonic log does not see
The density and neutron logs see vugs, so $N$ does not change. But the interval transit time is smaller than in a rock of the same density without vugs, and $M$ grows. Dolomite would shift the point down and to the left, and shale down and to the left, towards the shale point. The secondary porosity is $(1 - \phi_t)(M/M_{ma} - 1)$ .
4
Where does gas in the flushed zone shift the point of a limestone on the MID plot?
1. A.To the left, to smaller apparent interval transit times
2. B.Nowhere: gas is not seen on the MID plot
3. C.Up and to the right, towards quartz and beyond
4. D.Down, to larger apparent densities
Show the answer
Correct answer: C. Up and to the right, towards quartz and beyond
Gas lowers the density at almost the same neutron–density apparent porosity, so $\rho_{maa}$ falls. The neutron reading is also too low, so the apparent porosity from the sonic–neutron crossplot is smaller than the true one, and $\Delta t_{maa}$ grows. A gas-bearing limestone comes to the quartz point. We confirm gas by the resistivity, the gas shows and the crossover of the curves, and we recognise the mineral by $P_e$ .
5
The apparent porosity of a limestone (true value 15%) is read with an error of +0.01. By how much will $\rho_{maa}$ and $\Delta t_{maa}$ shift, and is this dangerous for the choice between calcite and dolomite?
Show the answer
$\partial\rho_{maa}/\partial\phi_{ta} = \dfrac{2.71 - 1}{0.85} \approx 2.0$ g/cm³, $\partial\Delta t_{maa}/\partial\phi_{ta} = -\dfrac{620 - 156}{0.85} \approx -546$ µs/m. An error of 0.01 gives +0.02 g/cm³ and −5.5 µs/m. The point shifts towards dolomite by an eighth of the distance in density (0.16 g/cm³) and by almost a half in interval transit time (13 µs/m). This does not change the choice of mineral, but it distorts the dolomite share markedly.
6
The matrix is half quartz and half anhydrite by volume. What share of anhydrite will the position of the point on the M–N plot show, if we read it as a volume fraction?
Show the answer
The weights are proportional to $f_i(\rho_{ma,i} - \rho_f)$ : for quartz $0.5\cdot 1.65 = 0.825$ , and for anhydrite $0.5\cdot 1.98 = 0.99$ . The weight of anhydrite is $0.99/1.815 \approx 0.55$ . The point goes 55% of the way from quartz to anhydrite, and without a conversion the anhydrite will be overestimated by 5 percentage points. We get the volumes by dividing the weights by $\rho_{ma} - \rho_f$ .

Updated: 28 October 2026