Skip to content
Brent$117.6Gas TTF$1,192USD453.87 ₸
MunaiHub
Logs and petrophysicsLogging basics Basic 18 min read

Log interpretation workflow

Log interpretation is the sequence of steps that turns checked curves into formation properties: lithology (rock type), shale volume, porosity, saturation and, in the end, net pay (the thickness of reservoir rock that holds hydrocarbons) and the parameters for the volumetric estimate (the calculation of hydrocarbons in place from the rock volume). Each step uses the results of the previous steps, so the order of the steps matters as much as the method of each step. Quick-look interpretation goes through the same steps quickly, with simplified models. It helps us decide at the wellsite (the drilling site) whether to test the formation and whether to run casing.

Why it matters

The other lessons of the library take each method separately: shale volume from the gamma ray log, porosity from density, formation water resistivity, Archie's equation (which gives saturation from resistivity) and the shaly-sand models (which handle sands that contain shale). In a real well we apply them together, and the result of one step becomes the input of the next. An error in water resistivity passes into all the saturations. An error in shale volume passes into porosity, saturation and net thickness. This lesson is a map: it shows in what order to go, what to check at each step, and where to find the details. It also gives a complete quick-look interpretation of an assumed well, from the readings to net thickness and hydrocarbon pore thickness.

Quick refresher

  • The order of the steps: data checks and depth matching → zonation (dividing the section into beds by lithology, reservoirs and fluids) → shale volume → porosity → formation water resistivity → saturation → cut-offs and net thickness → averaging → uncertainty.
  • Each step uses the previous ones: effective porosity needs shale volume, RwR_w from a water-bearing bed needs the porosity of that bed, saturation needs porosity, RwR_w and RtR_t, and net thickness needs all of them together.
  • Formation water resistivity is a single number per interval. We check it in several ways. The model with the adopted parameters must give Sw≈1S_w \approx 1 in water-bearing beds.
  • High resistivity does not by itself mean oil: tight rocks are also highly resistive. A productive bed must pass the cut-offs for shale volume, porosity and saturation at the same time.
  • We average porosity over thickness and water saturation over pore volume. The main result for the estimate of hydrocarbons in place is the hydrocarbon pore thickness H=∑h ϕe (1−Sw)H = \sum h\,\phi_e\,(1 - S_w).
  • Marginal beds (beds close to the cut-off limits) change the result most. The choice of the saturation model, of the shale-volume relationship and of the cut-offs can include them or exclude them entirely.
Key relationships
ϕe=ϕD−Vsh ϕD,sh\phi_e = \phi_D - V_{sh}\,\phi_{D,sh}

effective porosity of a shaly bed from density

Rwa=Rt ϕm/aR_{wa} = R_t\,\phi^m/a

water resistivity from a clean water-bearing bed

Sw=(a Rwϕem Rt)1/nS_w = \left(\dfrac{a\,R_w}{\phi_e^m\,R_t}\right)^{1/n}

Archie's equation for clean beds

H=∑khk ϕe,k (1−Sw,k)H = \sum_k h_k\,\phi_{e,k}\,(1 - S_{w,k})

hydrocarbon pore thickness over the beds that pass the cut-offs

δHH=−Sw1−Sw⋅δSwSw\dfrac{\delta H}{H} = -\dfrac{S_w}{1 - S_w}\cdot\dfrac{\delta S_w}{S_w}

how a relative error in saturation leads to an error in the pore thickness of a bed

Common pitfalls
  • Starting the calculation without checking the data and the depth matching of the runs.
  • Taking a highly resistive tight streak for a productive one without looking at porosity.
  • Taking RwR_w from a shaly water-bearing bed, or not bringing it to formation temperature.
  • Choosing the saturation model or the shale volume to suit the desired result.
  • Averaging water saturation over thickness instead of over pore volume.
  • Reporting a single number for net thickness, with no alternatives, when a marginal bed changes it by a third.

After this lesson you can

  • list the interpretation steps in order and explain which results each step takes from the previous ones
  • name, for each step, the main method, the main check and the lesson where it is covered in detail
  • divide the section into beds by lithology, permeability and fluid, and find the oil–water contact (the surface below which the rock holds water instead of oil)
  • make a quick-look interpretation of several beds: shale volume, porosity, water resistivity, water saturation
  • apply cut-offs (limit values that separate reservoir from non-reservoir rock) and find the net thickness and the net pay, the net-to-gross ratio, the average parameters and the hydrocarbon pore thickness
  • assess which decisions change the result most, and present it together with a range

The interpretation map: the steps and their order

Interpretation answers four questions: what the section is made of, where the reservoirs are, what is in them, and how much there is. We answer them step by step, and the dependencies set the order of the steps. The effective porosity of a shaly bed needs the shale volume. The water resistivity from a water-bearing bed needs the porosity of that bed. The water saturation needs the porosity, the water resistivity and the true resistivity of the formation. So an error at any step flows down the chain, and we place checks between the steps, not only at the end.

Log interpretation steps and their linkscore and tests0. Data and depthscalibrations, repeat section, depth matching1. Zonationlithology, reservoirs, fluids, contacts2. Shale volume VshGR, SP, neutron and density3. Porosity φedensity, neutron, sonic, NMR4. Water resistivity Rwsamples, SP, water-bearing beds5. Saturation SwArchie, shaly-sand models6. Cut-offsreservoir and productive reservoir7. AveragingNTG, mean φ and Sw, pore thickness H8. Uncertainty and conclusionvariants, recommendations for testscheck: in water Sw ≈ 1

A vertical diagram of nine blocks from top to bottom: 0 — data checks and depth matching; 1 — zonation: lithology, reservoirs, fluids; 2 — shale volume; 3 — porosity; 4 — formation water resistivity; 5 — saturation; 6 — cut-offs and net thickness; 7 — averaging and hydrocarbon pore thickness; 8 — uncertainty and conclusion. The blocks are joined by arrows pointing down. To the right of blocks 2 to 5 a feedback arrow runs from saturation to water resistivity and shale volume, labelled “check: water-bearing beds give S_w of about 1”. To the left of blocks 2 to 6 is a band labelled “core and tests”.

SchematicThe interpretation steps and their links. Arrows pointing down pass the results from step to step. The arrow pointing back is the check against water-bearing beds: if the model does not give Sw≈1S_w \approx 1 in them, we review the water resistivity, the coefficients and the shale volume. Core data and tests calibrate steps two to six.
The interpretation steps and where they are covered
Step
0. Data and depths
Question
can the curves be trusted; do the runs agree in depth
Step
1. Zonation
Question
where the shales, reservoirs and non-reservoir rocks are; which fluids; where the contacts are
Step
2. Shale volume
Question
what share of the volume is occupied by shale
Step
3. Porosity
Question
how many pores there are — total and effective porosity
Step
4. Water resistivity
Question
RwR_w at formation temperature
Step
5. Saturation
Question
what share of the pores is occupied by water; whether the oil is moveable
Step
6. Cut-offs
Question
what to count as reservoir and as productive
Step
7. Averaging
Question
net thickness, average parameters, hydrocarbon pore thickness
Details
averaging
Step
8. Uncertainty
Question
how reliable the result is and what changes it

In manual interpretation we do many of the steps with charts: graphs for borehole corrections, porosity, lithology and saturation. The topic Interpretation charts: how to read them shows how to read them and where they go wrong.

We can interpret in two ways. In bed-by-bed interpretation, we divide the section into beds, take the characteristic readings opposite each bed, and calculate the properties of the bed. This is how we do a quick-look. It is also how the traditional report on the results of well logging is organised. For each bed it gives the interval, the thickness, the properties and the fluid type: oil, gas, water or “unclear”. In sample-by-sample interpretation, a program calculates the properties at every recording depth and then sums the intervals that pass the cut-offs. The sample-by-sample approach shows changes within a bed better. The bed-by-bed approach is clearer and easier to check. The steps are the same in both, and in this lesson the quick-look goes bed by bed.

Steps 0–1: data checks and zonation

Before calculating, we check the data: the heading, the calibrations and the repeat section, the logging speed, the quality indicators of the methods, and the depth matching of the runs (checklist). Then we check the borehole: the caliper log against the bit, the type of mud, and the expected invasion profile (first checks). We mark unreliable intervals at once. In washouts the density and the neutron porosity are distorted, so we evaluate such a bed by other methods or exclude it.

Then we divide the section into beds. This step is qualitative (it uses no calculations), but everything that follows depends on it: each bed gets its own readings and its own parameters.

  • Lithology. We recognise shales by a high gamma ray (GR) and by the spontaneous potential (SP) at the shale baseline, the line that the SP follows opposite shales (Gamma ray log and natural gamma ray spectrometry, Spontaneous potential (SP) log). We recognise sandstones, limestones, dolomites and evaporites (such as salt and anhydrite) by density, the photoelectric factor and the crossplots of the porosity methods (Lithology and porosity from crossplots, Lithology from the photoelectric factor and spectral gamma ray). We place bed boundaries at the middle of the transition of the curves.
  • Reservoirs. The direct signs of permeability are mudcake on the caliper log (a diameter smaller than the bit), separation of resistivity curves of different depths of investigation, and positive microlog separation on the microresistivity tools (more). An indirect sign is the deflection of the SP from the shale baseline. High density with a low GR and coincident resistivity curves indicate a tight streak (a thin bed with very little pore space).
  • Fluids. We compare beds of the same lithology and porosity by their deep resistivity: hydrocarbons raise it many times over (Resistivity logs: electrical and induction tools). With fresh mud, the ratio of the flushed-zone resistivity to the true resistivity, Rxo/RtR_{xo}/R_t, is close to Rmf/RwR_{mf}/R_w in a water-bearing bed. In an oil-bearing bed it is much lower (resistivity profile). The crossover of the neutron and density curves shows gas (more). The oil–water contact (OWC) shows up as a drop in resistivity within one clean bed. The drop is abrupt, or gradual if there is a capillary transition zone above the contact (Capillary pressure and the transition zone).

From here on, we carry out all the steps on one assumed vertical well. The lesson Log acquisition, processing and presentation covers its heading and runs. The formation temperature is about 80 °C, the mud is fresh, and the filtrate resistivity at formation temperature is RmfR_{mf} = 0.191 Ω·m. The density run is tied to the resistivity run by the GR, and the neutron curve is recorded in sandstone units.

Quick-look interpretation of the assumed well, bed by bedGR0–150 APIR, Ω·m0.2–200Vsh0–1φe0–0.3Sw0–1FlagsRtRxo0.350.100.50res.payABCDGE1E2F26002605261026152620262526302635OWC

Six narrow vertical tracks and a depth column from 2598 to 2636 metres, with the letters of the beds A, B, C, D, G, E1, E2, F on the left. GR: about 102 API in the shales, 22–29 API in sandstones B, C, E1, E2, 58 API in bed D, 83 API in bed G. Resistivities on a logarithmic scale: the deep tool about 31 Ω·m in B, 56 Ω·m in the tight streak C, about 10 Ω·m in D, 14 Ω·m in E1 and 1.6 Ω·m in the water-bearing E2; the microresistivity tool 4.7–7.9 Ω·m in the permeable sandstones. The result tracks are steps by bed: shale volume with a cut-off line at 0.35 (exceeded by the shales and bed G), effective porosity with a cut-off line at 0.10 (below it are C and G), water saturation with a cut-off line at 0.50 (0.195 in B, 0.44 in D, 0.344 in E1, 1 in E2). On the right are the flags: reservoir — B, D, E1, E2; productive bed — B, D and E1 above the contact at 2621 metres.

CalculatedThe assumed well, 2598–2636 m: the readings and the results of the bed-by-bed quick-look. From left to right: GR; resistivities (deep tool and microresistivity tool); shale volume, effective porosity and water saturation, with the cut-offs dashed; and the reservoir and pay flags. Bed D is marginal.
Readings opposite the beds of the assumed well
Bed
A
Interval, m
2600.0–2606.0
GR, API
102
ρb\rho_b, g/cm³ / ϕN\phi_N
2.550 / 0.330
RtR_t / RxoR_{xo}, Ω·m
3.2 / 3.2
Bed
B
Interval, m
2606.0–2612.0
GR, API
27
ρb\rho_b, g/cm³ / ϕN\phi_N
2.261 / 0.244
RtR_t / RxoR_{xo}, Ω·m
30.6 / 5.5
Bed
C
Interval, m
2612.0–2613.2
GR, API
29
ρb\rho_b, g/cm³ / ϕN\phi_N
2.567 / 0.061
RtR_t / RxoR_{xo}, Ω·m
56.3 / 52
Bed
D
Interval, m
2613.2–2617.0
GR, API
58
ρb\rho_b, g/cm³ / ϕN\phi_N
2.380 / 0.242
RtR_t / RxoR_{xo}, Ω·m
9.58 / 7.2
Bed
G
Interval, m
2617.0–2618.6
GR, API
83
ρb\rho_b, g/cm³ / ϕN\phi_N
2.505 / 0.255
RtR_t / RxoR_{xo}, Ω·m
4.3 / 4.3
Bed
E1
Interval, m
2618.6–2621.0
GR, API
25
ρb\rho_b, g/cm³ / ϕN\phi_N
2.325 / 0.203
RtR_t / RxoR_{xo}, Ω·m
14 / 7.9
Bed
E2
Interval, m
2621.0–2629.0
GR, API
22
ρb\rho_b, g/cm³ / ϕN\phi_N
2.318 / 0.201
RtR_t / RxoR_{xo}, Ω·m
1.58 / 4.7
Bed
F
Interval, m
2629.0–2635.0
GR, API
102
ρb\rho_b, g/cm³ / ϕN\phi_N
2.550 / 0.330
RtR_t / RxoR_{xo}, Ω·m
3.2 / 3.2

Neutron porosity is in sandstone units; RtR_t is the deep tool after corrections; RxoR_{xo} is the microresistivity tool corrected for mudcake; the resistivities are at formation temperature. Caliper: 210–211 mm opposite sandstones B, D, E1 and E2 (mudcake, with a 215.9 mm bit), 216 mm opposite streak C, 228–236 mm opposite the shales (washouts).

Zonation of the section
Bed
A, F
Rock
shale: high GR, washouts
Reservoir
no
Fluid
—
Bed
B
Rock
clean sandstone
Reservoir
yes: mudcake, RxoR_{xo} and RtR_t differ
Fluid
oil: RtR_t is 19 times higher than in E2
Bed
C
Rock
tight (cemented) sandstone: low GR, high density
Reservoir
no: no mudcake, Rxo≈RtR_{xo} \approx R_t
Fluid
—; the high RtR_t is due to low porosity
Bed
D
Rock
shaly sandstone: GR 58 API
Reservoir
yes: mudcake
Fluid
probably oil: RtR_t 9.58 Ω·m
Bed
G
Rock
silty shale
Reservoir
no
Fluid
—
Bed
E1
Rock
clean sandstone
Reservoir
yes
Fluid
oil
Bed
E2
Rock
clean sandstone
Reservoir
yes
Fluid
water: the microresistivity tool reads three times higher than the deep tool

The OWC is at 2621.0 m: in one and the same clean sandstone RtR_t falls from 14.0 to 1.58 Ω·m. There is no gas: on scales compatible with sandstone the neutron and density curves do not cross over in the sandstones.

Steps 2–3: shale volume and porosity

We determine shale volume before porosity. We need it to correct the porosity of shaly beds and to choose the saturation model. The main method is the gamma ray index, which shows how far the GR reading lies between the clean-formation baseline and the shale baseline of the same sequence. We use it with a relationship checked against core (Determining shale volume). We check the result with other indicators: the SP and the difference between neutron and density porosity (more). We draw the shale baseline through thick, uniform shales, not through the peak of a highly radioactive streak.

In a sand–shale (siliciclastic) section we most often take porosity from the density log, using the matrix density (the density of the rock grains) from the lithology and the filtrate density (Porosity from the density log). In a shaly bed we subtract the contribution of the shale. This gives the effective porosity ϕe\phi_e (shale correction). The neutron and sonic logs serve as checks (Porosity from the neutron log, Porosity from the sonic log). In mixed lithology we find porosity and composition from crossplots (Lithology and porosity from crossplots). In gas-bearing beds we use the density–neutron pair with a gas correction (Gas, shale, carbonate and pressure effects on porosity logs). Nuclear magnetic resonance (NMR) logging divides the pore space into bound and free water (Nuclear magnetic resonance (NMR) logging).

Worked exampleShale volume and effective porosity by bed
Given (assumed data)
GR baselines
clean rock 22 API (bed E2), shale 102 API (beds A and F, excluding the highly radioactive streak)
Relationship
Larionov relationship for older rocks (a standard shale-volume relationship), C=2C = 2 (adopted from the core of the field; assumed)
Densities
matrix (quartz) 2.65, filtrate 1.00, shale 2.55 g/cm³; neutron porosity in shale 0.33
Readings
the table above
Solution
  1. Density “porosity” of shale: ϕD,sh=(2.65−2.55)/1.65=0.061\phi_{D,sh} = (2.65 - 2.55)/1.65 = 0.061.
  2. Bed B: IGR=(27−22)/80=0.0625I_{GR} = (27 - 22)/80 = 0.0625, Vsh=(22⋅0.0625−1)/3=0.030V_{sh} = (2^{2\cdot 0.0625} - 1)/3 = 0.030; ϕD=(2.65−2.261)/1.65=0.236\phi_D = (2.65 - 2.261)/1.65 = 0.236; ϕe=0.236−0.030⋅0.061=0.234\phi_e = 0.236 - 0.030\cdot 0.061 = 0.234.
  3. Bed C: Vsh=0.043V_{sh} = 0.043, ϕD=0.050\phi_D = 0.050, ϕe=0.047\phi_e = 0.047 — tight.
  4. Bed D: IGR=(58−22)/80=0.450I_{GR} = (58 - 22)/80 = 0.450, Vsh=(22⋅0.450−1)/3=0.289V_{sh} = (2^{2\cdot 0.450} - 1)/3 = 0.289; ϕD=(2.65−2.380)/1.65=0.164\phi_D = (2.65 - 2.380)/1.65 = 0.164; ϕe=0.164−0.289⋅0.061=0.146\phi_e = 0.164 - 0.289\cdot 0.061 = 0.146. Neutron check: Vsh≈(0.242−0.164)/(0.33−0.061)=0.290V_{sh} \approx (0.242 - 0.164)/(0.33 - 0.061) = 0.290 — it agrees with the GR estimate.
  5. Bed G: Vsh=0.626V_{sh} = 0.626, ϕe=0.050\phi_e = 0.050. Beds E1 and E2: Vsh=0.018V_{sh} = 0.018 and 0, ϕe=0.196\phi_e = 0.196 and 0.201.
Answer: Shale volume: B — 0.030, C — 0.043, D — 0.289, G — 0.626, E1 — 0.018; effective porosity: B — 0.234, C — 0.047, D — 0.146, G — 0.050, E1 — 0.196, E2 — 0.201.

What it means. Two independent indicators of the shale volume of bed D agree (0.289 from the GR and 0.290 from the neutron and density). This means that we chose the baselines and the relationship sensibly. A linear estimate would give a shale volume of 0.450 in D; this will be useful at the uncertainty step. The shale correction reduced the porosity of D by 0.018, and the porosity of the clean beds by only thousandths. Residual oil (oil that the filtrate leaves behind) in the flushed zone lowers the density slightly and overestimates ϕD\phi_D. Here this effect is below 0.01, so we ignore it. The data in this example are assumed for illustration.

Step 4: formation water resistivity

RwR_w is a single number for the whole interval that contains one water, and all the saturations depend on it (more). We take it at formation temperature and check it in several ways. The ways are water samples taken in tests, the SP, the apparent water resistivity RwaR_{wa} (the water resistivity calculated from the resistivity and porosity of a clean water-bearing bed), and the Pickett plot (a log–log plot of resistivity against porosity) (Determining formation water resistivity, from the SP, Resistivity–porosity crossplots). RwaR_{wa} needs the porosity of the water-bearing bed, which is why this step comes after porosity. The bed itself must be clean and thick, because in shaly beds RwaR_{wa} comes out too low.

Worked exampleWater resistivity from bed E2 and a check against a sample
Given (assumed data)
Bed E2
clean, below the contact: ϕe\phi_e = 0.201, RtR_t = 1.58 Ω·m
Coefficients
a=1a = 1, m=2m = 2
Water sample of the same horizon from a neighbouring well
0.146 Ω·m at 25 °C
Formation temperature and filtrate
80.3 °C, RmfR_{mf} = 0.191 Ω·m (lesson Log acquisition, processing and presentation)
Solution
  1. Rwa=Rt ϕem/a=1.58⋅0.2012=0.0638R_{wa} = R_t\,\phi_e^m/a = 1.58\cdot 0.201^2 = 0.0638 Ω·m. We take it as RwR_w.
  2. We bring the sample to formation temperature with the Arps formula: 0.146⋅25+21.580.3+21.5=0.06670.146\cdot\dfrac{25 + 21.5}{80.3 + 21.5} = 0.0667 Ω·m. This is 4.5% higher, and the discrepancy is within the accuracy of the methods.
  3. The contrast between filtrate and water is Rmf/Rw=0.191/0.0638=3.0R_{mf}/R_w = 0.191/0.0638 = 3.0. In bed E2, Rxo/Rt=4.7/1.58=3.0R_{xo}/R_t = 4.7/1.58 = 3.0, as it should be in a water-bearing bed. In bed B, 5.5/30.6=0.185.5/30.6 = 0.18, which is 17 times smaller: a sign of oil.
Answer: RwR_w = 0.0638 Ω·m at 80.3 °C. The water sample confirms it to within 4.5%.

What it means. The check “Sw≈1S_w \approx 1 in a water-bearing bed” tells us nothing for E2 itself, because we took RwR_w from E2. The independent checks are the water sample and the ratio of the flushed-zone and formation resistivities. If the sample had differed from RwaR_{wa} by several times, we would have looked for the cause, not averaged the two values (reconciling the estimates). The data in this example are assumed for illustration.

Step 5: saturation

We calculate saturation from the true formation resistivity RtR_t, the porosity and RwR_w. RtR_t is the deep tool reading after corrections for the borehole, the enclosing rocks (the beds above and below) and invasion (how to find it). In clean beds Archie's equation works, with coefficients from core (Archie's equation and saturation of clean reservoirs). In shaly beds the clays conduct current, and Archie's equation overestimates water saturation. There we need a shaly-sand model. We choose it according to the form of the shale distribution (how the shale is spread in the rock), and we check it on water-bearing shaly beds (Water saturation in shaly sands). We choose the model before the calculation, not by which result we like better.

The flushed-zone resistivity gives a second saturation, SxoS_{xo}, and with it a check of oil moveability. If SxoS_{xo} is noticeably larger than SwS_w, the filtrate has displaced some of the oil, which means that the oil is moveable (more).

Worked exampleWater saturation of the beds
Given (assumed data)
Parameters
RwR_w = 0.0638 Ω·m, a=1a = 1, m=n=2m = n = 2
Shales
RshR_{sh} = 3.2 Ω·m (thick shales A and F)
Model
Archie at Vsh≤0.10V_{sh} \le 0.10; at higher shale volume, Simandoux (a shaly-sand model) in its common form (models)
Porosity and shale volume
from the previous example
Solution
  1. Bed B (Archie): Sw=0.06380.2342⋅30.6=0.195S_w = \sqrt{\dfrac{0.0638}{0.234^2\cdot 30.6}} = 0.195.
  2. Bed E1: Sw=0.06380.1962⋅14.0=0.344S_w = \sqrt{\dfrac{0.0638}{0.196^2\cdot 14.0}} = 0.344. There is more water at the contact than in B.
  3. Bed D (at n=2n = 2, Simandoux gives a quadratic equation, αSw2+γSw=1/Rt\alpha S_w^2 + \gamma S_w = 1/R_t): α=ϕe2/Rw=0.1462/0.0638=0.334\alpha = \phi_e^2/R_w = 0.146^2/0.0638 = 0.334, γ=Vsh/Rsh=0.289/3.2=0.0903\gamma = V_{sh}/R_{sh} = 0.289/3.2 = 0.0903, 1/Rt=0.10441/R_t = 0.1044; Sw=−γ+γ2+4α/Rt2α=0.440S_w = \dfrac{-\gamma + \sqrt{\gamma^2 + 4\alpha/R_t}}{2\alpha} = 0.440.
  4. For comparison, Archie in bed D gives Sw=0.0638/(0.1462⋅9.58)=0.559S_w = \sqrt{0.0638/(0.146^2\cdot 9.58)} = 0.559, which is higher by 0.119. Archie attributes the conductivity of the clays to water.
  5. Bed C by Archie: Sw=0.72S_w = 0.72. A number comes out, but in a tight rock with ϕe\phi_e = 0.047 it means nothing.
  6. Moveable oil in B: Sw/Sxo=(Rxo/Rt)/(Rmf/Rw)=0.18/3.0=0.245S_w/S_{xo} = \sqrt{(R_{xo}/R_t)/(R_{mf}/R_w)} = \sqrt{0.18/3.0} = 0.245, hence Sxo≈0.80S_{xo} \approx 0.80. The filtrate has displaced most of the oil.
Answer: SwS_w: B — 0.195, E1 — 0.344, D — 0.440 by Simandoux (0.559 by Archie). The oil of bed B is moveable, Sxo≈0.80S_{xo} \approx 0.80.

What it means. In nearly clean beds the models differ little. Simandoux would give 0.190 in bed B instead of 0.195, so here Archie overestimates the water saturation only slightly. Bed D is marginal. By the shaly-sand model its water saturation is 0.440, and by Archie it is 0.559. With a cut-off of 0.50, the bed is productive in the first case and not in the second. Core data decide which model is right: the water saturation of samples cut with oil-based mud, and the cation exchange capacity of the clays (their ability to hold exchangeable ions). A test of bed D decides it too. A wish to show more hydrocarbons in place does not. The data in this example are assumed for illustration.

Steps 6–7: cut-offs, net thickness, averaging

Cut-offs (in Russian-language reserves estimation also called “conditional limits”, from the Russian konditsionnye predely) divide the section. Shale volume and porosity cut-offs separate reservoir from non-reservoir rock. A water saturation cut-off separates productive reservoir from water-bearing reservoir (more). We apply the cut-offs in sequence, and each level gives its own thickness. The gross thickness is the whole bed from top to base. The net thickness is the sum of the reservoir intervals. The net pay is the reservoir with hydrocarbons above the contact. The cut-offs themselves are a decision, not a property of the rock. The advanced section covers where they come from.

Then we average the parameters over the same intervals for which we determined the thickness: porosity over thickness, and water saturation over pore volume. The result for the estimate of hydrocarbons in place is the hydrocarbon pore thickness H=∑h ϕe (1−Sw)H = \sum h\,\phi_e\,(1 - S_w). This is the thickness of the layer of hydrocarbons that we would get if we gathered them out of the pores (more). We present the result as a table by bed and a summary for the unit. In the logging report we give each bed a fluid type and recommendations: which intervals to test and perforate.

Worked exampleNet thickness and volumetric parameters of the assumed well
Given (assumed data)
Field cut-offs (assumed)
Vsh≤0.35V_{sh} \le 0.35, ϕe≥0.10\phi_e \ge 0.10, Sw≤0.50S_w \le 0.50
Unit
the sequence from the top of bed B (2606.0 m) to the base of E2 (2629.0 m); OWC 2621.0 m
Bed properties
from the previous examples
Solution
  1. Reservoir (by shale volume and porosity): B, D, E1, E2. C (ϕe\phi_e = 0.047) and G (VshV_{sh} = 0.626) did not pass. The net thickness is 6.0+3.8+2.4+8.0=20.26.0 + 3.8 + 2.4 + 8.0 = 20.2 m, against a gross thickness of 23.0 m. So the net-to-gross ratio is NTG=0.88\mathrm{NTG} = 0.88.
  2. Productive reservoir (above the OWC and Sw≤0.50S_w \le 0.50): B, D, E1. The net pay is 12.2 m.
  3. Average porosity over thickness: pore thickness ∑hϕe=6.0⋅0.234+3.8⋅0.146+2.4⋅0.196=1.404+0.5548+0.4704=2.429\sum h\phi_e = 6.0\cdot 0.234 + 3.8\cdot 0.146 + 2.4\cdot 0.196 = 1.404 + 0.5548 + 0.4704 = 2.429 m, and ϕˉ=2.429/12.2=0.199\bar\phi = 2.429/12.2 = 0.199.
  4. Average water saturation over pore volume: ∑hϕeSw=1.404⋅0.195+0.5548⋅0.440+0.4704⋅0.344=0.6797\sum h\phi_e S_w = 1.404\cdot 0.195 + 0.5548\cdot 0.440 + 0.4704\cdot 0.344 = 0.6797 m, and Sˉw=0.6797/2.429=0.280\bar S_w = 0.6797/2.429 = 0.280.
  5. Hydrocarbon pore thickness by bed: B 1.404⋅(1−0.195)=1.1301.404\cdot(1 - 0.195) = 1.130 m, D 0.311 m, E1 0.309 m. In total H=1.75H = 1.75 m. The same result comes from 12.2⋅0.1991⋅(1−0.2798)12.2\cdot 0.1991\cdot(1 - 0.2798).
Answer: Net thickness 20.2 m (NTG 0.88), net pay 12.2 m, ϕˉ\bar\phi = 0.199, Sˉw\bar S_w = 0.280, hydrocarbon pore thickness 1.75 m.

What it means. Under every square metre of area at the well there are about 1.75 m³ of oil at reservoir conditions. Bed B gives 65% of this pore thickness. The marginal bed D gives 18%, although in thickness it is almost a third of the productive section. Bed E2 is included in the net thickness but not in the net pay, because it is a water-bearing reservoir. That is why the NTG here refers to the whole penetrated interval, including the water-bearing part below the OWC. For the gross rock volume above the contact, we apply a fraction calculated only for the interval above the contact; in this example it is smaller. If the estimate of hydrocarbons in place takes the NTG for the reservoir, we must take the averages for the reservoir too. We must not mix the levels (why). The data in this example are assumed for illustration.

Results of the quick-look by bed
Bed
B
VshV_{sh}
0.030
ϕe\phi_e
0.234
SwS_w
0.195
Result
reservoir, oil
Bed
C
VshV_{sh}
0.043
ϕe\phi_e
0.047
SwS_w
—
Result
non-reservoir: tight
Bed
D
VshV_{sh}
0.289
ϕe\phi_e
0.146
SwS_w
0.440
Result
reservoir, oil (marginal)
Bed
G
VshV_{sh}
0.626
ϕe\phi_e
0.050
SwS_w
—
Result
non-reservoir: shaly
Bed
E1
VshV_{sh}
0.018
ϕe\phi_e
0.196
SwS_w
0.344
Result
reservoir, oil
Bed
E2
VshV_{sh}
0.000
ϕe\phi_e
0.201
SwS_w
1.000
Result
reservoir, water

In the logging report, we mark bed D as productive, with a reservation about the saturation model, and we recommend testing it. Beds B and E1 are productive, and E2 is water-bearing.

Step 8: uncertainty

Each step brings its own uncertainty. It comes from the parameters (the Archie coefficients aa, mm, nn, and RwR_w, RshR_{sh}, the matrix and shale densities), the choice of model and relationship, the GR baselines, the cut-offs, and the data themselves (depth matching, corrections, thin beds). The influence is of two kinds. Continuous parameters shift the result smoothly: an error of 10% in RwR_w changes SwS_w by 5% of its value at n=2n = 2 (more). “Yes or no” decisions can include or exclude a marginal bed entirely, and so change the net thickness in a jump. These decisions are the saturation model, the shale-volume relationship and the cut-offs.

How an error in saturation leads to an error in the hydrocarbon pore thickness of a bed
δHH=−δSw1−Sw=−Sw1−Sw⋅δSwSw\frac{\delta H}{H} = -\frac{\delta S_w}{1 - S_w} = -\frac{S_w}{1 - S_w}\cdot\frac{\delta S_w}{S_w}
where:
  • HHhydrocarbon pore thickness of the bed, h ϕe (1−Sw)h\,\phi_e\,(1 - S_w), m
  • δH\delta Hits change, m
  • SwS_wwater saturation of the bed, fraction
  • δSw\delta S_werror in water saturation (absolute), fraction

Units: dimensionless (relative changes)

Holds when
  • The thickness and porosity of the bed do not change; only SwS_w does.
  • The error is small (linear approximation).
  • The bed stays on the same side of the cut-offs. Otherwise its pore thickness changes in a jump, all at once.

The same relative error in SwS_w costs more the higher the water saturation. With an error of 5% (an error of 10% in RwR_w at n=2n = 2) the pore thickness changes by 1.25% at Sw=0.2S_w = 0.2, by 5% at Sw=0.5S_w = 0.5 and by almost 12% at Sw=0.7S_w = 0.7.

Worked exampleWhat changes the result of the assumed well most
Given (assumed data)
Base case
the previous examples: net pay 12.2 m, HH = 1.75 m
Variants
one change at a time: the saturation model, the shale-volume relationship, RwR_w, mm, nn, the cut-offs (table below)
Solution
  1. RwR_w 10% higher: the SwS_w of bed B rises from 0.195 to 0.205, and by the formula above its pore thickness falls by about 0.1951−0.195⋅5 %=1.2 %\dfrac{0.195}{1 - 0.195}\cdot 5\,\% = 1.2\,\%; the whole HH falls by 1.8%.
  2. m=2.2m = 2.2 instead of 2, with RwR_w recalculated from bed E2 with the same mm (0.0463 Ω·m): HH changes by only 0.1%. The error in mm almost cancels because the porosity of the productive beds is close to that of E2. On the other hand, this RwR_w is 31% lower than the water sample — the discrepancy itself points to a wrong mm.
  3. m=2.2m = 2.2 with RwR_w from the water sample (0.0667 Ω·m): the SwS_w of bed D is 0.516 — above the cut-off; the bed drops out, and HH is lower by 22.5%.
  4. Archie in bed D: SwS_w = 0.559, the bed drops out, net pay is 8.4 m, and HH is lower by 17.8%. Linear shale volume: the VshV_{sh} of bed D is 0.450 — above the cut-off of 0.35, and the result is almost the same: HH is lower by 18.6%.
  5. n=1.8n = 1.8 and 2.22.2: HH is higher by 4.5% and lower by 4.3%; bed D remains productive.
Answer: Continuous parameters change HH by a few per cent, and the decisions about bed D by 17.8–22.5%: the result rests on one marginal bed.

What it means. That is why the report gives not only the base value but also the alternatives. Most importantly, it states what would reduce the uncertainty: a test of bed D, the water saturation from core cut with oil-based mud, and the cation exchange capacity of the clays. If we set all the parameters to their extreme values at once, the range would be too wide and too improbable. Instead, we assess their joint influence with probabilities (Probabilistic volumetrics by Monte Carlo simulation, Geological and technical uncertainty of a project). The data in this example are assumed for illustration.

Variants of the quick-look of the assumed well
Variant
Base case
Bed D
productive, SwS_w 0.440
hpayh_{\text{pay}}, m
12.2
ΔH\Delta H
—
Variant
Bed D saturation by Archie
Bed D
non-productive: SwS_w 0.559
hpayh_{\text{pay}}, m
8.4
ΔH\Delta H
−17.8%
Variant
Linear shale volume, Vsh=IGRV_{sh} = I_{GR}
Bed D
non-reservoir: VshV_{sh} 0.450
hpayh_{\text{pay}}, m
8.4
ΔH\Delta H
−18.6%
Variant
RwR_w 10% higher / lower
Bed D
productive, SwS_w 0.456 / 0.422
hpayh_{\text{pay}}, m
12.2
ΔH\Delta H
−1.8% / +1.8%
Variant
m=1.8m = 1.8 / 2.22.2, RwR_w recalculated from bed E2
Bed D
productive, SwS_w 0.429 / 0.451
hpayh_{\text{pay}}, m
12.2
ΔH\Delta H
+0.1% / −0.1%
Variant
m=2.2m = 2.2, RwR_w from the water sample
Bed D
non-productive: SwS_w 0.516
hpayh_{\text{pay}}, m
8.4
ΔH\Delta H
−22.5%
Variant
n=1.8n = 1.8 / 2.22.2
Bed D
productive, SwS_w 0.411 / 0.466
hpayh_{\text{pay}}, m
12.2
ΔH\Delta H
+4.5% / −4.3%
Variant
Cut-off SwS_w 0.40 instead of 0.50
Bed D
non-productive: SwS_w 0.440
hpayh_{\text{pay}}, m
8.4
ΔH\Delta H
−17.8%
Variant
Cut-off VshV_{sh} 0.25 instead of 0.35
Bed D
non-reservoir: VshV_{sh} 0.289
hpayh_{\text{pay}}, m
8.4
ΔH\Delta H
−17.8%

Each variant changes one decision, and everything else stays as in the base case. The two cut-off rows and the row with the Archie model give the same result: in all three, bed D drops out.

Terms and symbols

Zonation
Division of the section into beds by lithology, permeability and fluid. It is the basis of bed-by-bed interpretation.
Quick-look interpretation
A fast interpretation after logging, for decisions at the wellsite. The steps are the same, but the parameters are preliminary.
Bed-by-bed and sample-by-sample interpretation
A calculation from readings averaged opposite beds, or at every recording depth with the intervals summed afterwards.
Cut-offs
Thresholds of shale volume, porosity and water saturation by which reservoir and productive reservoir are identified.
Net thickness and net pay
The sum of the reservoir intervals, and the sum of the reservoir intervals with hydrocarbons above the contact.
Net-to-gross ratio (NTG)
The ratio of net thickness to gross thickness.
Hydrocarbon pore thickness HH
The volume of hydrocarbons at reservoir conditions per unit area. It equals ∑h ϕe (1−Sw)\sum h\,\phi_e\,(1 - S_w).
Fluid type
The conclusion of the logging report for each bed: oil, gas, water or “unclear”.
Marginal bed
A bed with properties near the cut-offs. Whether it enters the net thickness depends on the model and the parameters.

Common mistakes and misconceptions

Wrong

“The resistivity is high, so it is oil.”

In fact

Resistivity rises both with hydrocarbons and with low porosity. A tight streak can be the most resistive bed in the section. Porosity, shale volume, saturation and the signs of permeability together decide productivity.

Wrong

“The order of the steps does not matter — we will calculate everything anyway.”

In fact

Effective porosity needs shale volume, RwaR_{wa} needs porosity, and saturation needs both. A step made too early rests on assumptions that we later forget and never check.

Wrong

“We will take RwR_w from the neighbouring field.”

In fact

RwR_w is one number per interval, and its error passes into all the saturations. We bring it to formation temperature and check it in several ways in our own well: against samples, the SP and water-bearing beds.

Wrong

“We will choose the saturation model that gives more oil.”

In fact

We choose the model according to the form of the shale distribution, and we check it on water-bearing beds and on core before anyone looks at the result. Otherwise marginal beds become productive at the interpreter's wish.

Wrong

“Net thickness is a single number.”

In fact

It is the result of a chain of decisions. If one marginal bed changes it by a third, the report gives alternatives and recommends a test that will settle the question.

In depth: advanced detail

Where the cut-offs come from

A cut-off separates rock that practically does not yield fluid under reservoir pressure drawdowns (the pressure drop that pushes fluid to the well) from rock that does. That is why we look for it through permeability. From core we build a porosity–permeability relationship (Reservoir quality: geological controls on porosity and permeability), and we find the porosity at which the permeability falls to the limit. For oil the limit is often taken to be of the order of 1 mD (millidarcy, a unit of permeability), and for gas it is lower. We choose the shale-volume cut-off by the same logic or from test results. We choose the water-saturation cut-off from relative permeabilities: above it, the bed yields mainly water (Relative permeability, Fractional flow function and mobility ratio).

Worked examplePorosity cut-off from the relationship with permeability
Given (assumed data)
Relationship from the core of the bed (assumed)
log⁡10k=−2.4+24 ϕ\log_{10} k = -2.4 + 24\,\phi, kk in mD
Permeability limit
1 mD; variants 0.5 and 2 mD
Pore thickness of the beds
hϕeh\phi_e from the examples of the lesson
Solution
  1. At 1 mD: ϕ=(log⁡101+2.4)/24=0.100\phi = (\log_{10} 1 + 2.4)/24 = 0.100 — the adopted cut-off.
  2. At 0.5 and 2 mD: ϕ=(log⁡100.5+2.4)/24=0.087\phi = (\log_{10} 0.5 + 2.4)/24 = 0.087 and ϕ=(log⁡102+2.4)/24=0.113\phi = (\log_{10} 2 + 2.4)/24 = 0.113.
  3. What share of the pore thickness the cut-off removes: streak C has 1.2⋅0.047=0.05641.2\cdot 0.047 = 0.0564 m out of 4.094 m of total pore thickness ∑hϕe\sum h\phi_e in the beds that passed the shale-volume cut-off (B, C, D, E1, E2). That is 1.4%.
Answer: The porosity cut-off is 0.100 (0.087–0.113 at a limit of 0.5–2 mD). In every variant streak C is cut off and bed D passes.

What it means. A cut-off is an agreement with a justification, not a constant of the rock. We check it by how much pore thickness it removes. If the removed streaks contain a noticeable share of the pores, the cut-off needs a special justification, and the result needs alternatives. Here the removed streak contains only 1.4% of the pores. The data in this example are assumed for illustration.

Connections to other topics

Check yourself

  1. 1

    Why is shale volume determined before porosity and saturation?

    1. A.Without it one cannot calculate the effective porosity of a shaly bed or choose the saturation model
    2. B.Shale volume is determined last, from saturation
    3. C.It is the convention in reports
    4. D.Shale volume affects neither porosity nor saturation
    Show the answer

    Correct answer: A. Without it one cannot calculate the effective porosity of a shaly bed or choose the saturation model

    The effective porosity of a shaly bed is the density porosity minus the contribution of the shale, ϕe=ϕD−VshϕD,sh\phi_e = \phi_D - V_{sh}\phi_{D,sh}, and shaly-sand models use VshV_{sh} directly. Fitting the shale volume to the saturation is a vicious circle.

  2. 2

    The most resistive bed of the section, 45 Ω·m, has a density porosity of 0.04, there is no mudcake on the caliper log, and the resistivity curves of different depths of investigation coincide. What is it most likely to be?

    1. A.A tight non-reservoir rock
    2. B.Shale
    3. C.An oil-bearing reservoir
    4. D.A gas-bearing reservoir
    Show the answer

    Correct answer: A. A tight non-reservoir rock

    The small amount of water in the low porosity explains the high resistivity here, not oil. The absence of mudcake, and the coincidence of curves of different depths of investigation, say that there is no invasion, that is, the bed is practically impermeable. It will not pass the porosity cut-off, and a saturation calculation for it is meaningless.

  3. 3

    A clean bed 4 m thick: ϕe=0.25\phi_e = 0.25, Rt=20R_t = 20 Ω·m, Rw=0.05R_w = 0.05 Ω·m, a=1a = 1, m=n=2m = n = 2. Find SwS_w and the hydrocarbon pore thickness of the bed.

    Show the answer

    Sw=0.05/(0.252⋅20)=0.05/1.25=0.04=0.20S_w = \sqrt{0.05/(0.25^2\cdot 20)} = \sqrt{0.05/1.25} = \sqrt{0.04} = 0.20. The pore thickness is H=h ϕe (1−Sw)=4⋅0.25⋅0.80=0.80H = h\,\phi_e\,(1 - S_w) = 4\cdot 0.25\cdot 0.80 = 0.80 m — under every square metre there is 0.8 m³ of oil at reservoir conditions.

  4. 4

    RwR_w is overestimated by 10% (n=2n = 2). In which bed will the hydrocarbon pore thickness change more, in per cent?

    1. A.The same in both
    2. B.In a bed with Sw=0.5S_w = 0.5
    3. C.In a bed with Sw=0.2S_w = 0.2
    4. D.In neither: RwR_w does not affect the pore thickness
    Show the answer

    Correct answer: B. In a bed with Sw=0.5S_w = 0.5

    In both beds SwS_w will rise by 5% of its value, but the pore thickness is proportional to 1−Sw1 - S_w: δH/H=−Sw1−Sw⋅5%\delta H/H = -\dfrac{S_w}{1 - S_w}\cdot 5\%, that is −1.25% at Sw=0.2S_w = 0.2 and −5% at Sw=0.5S_w = 0.5. Errors in the parameters hit beds with a high water saturation hardest.

  5. 5

    A test confirmed that the bed below the contact is water-bearing, but the model with the adopted parameters gives Sw=0.78S_w = 0.78 in it. What should be done?

    Show the answer

    Do not “clip” the result to 1 (100%). Look for what underestimates the water saturation. RwR_w may be too low (taken from a shaly bed or from the water of another horizon, or converted to an overestimated temperature). mm may be too low. Porosity may be too high (a wrong matrix, gas, washouts). RtR_t may be too high (not corrected for invasion by fresh filtrate). In the shaly-sand model, the shale volume may be too high or RshR_{sh} too low. Apply the corrected parameters to the whole interval. Otherwise the same error would overestimate the hydrocarbon saturation of the productive beds.

  6. 6

    By Archie's equation, bed D of the assumed well fails the water-saturation cut-off; by the Simandoux model it passes. How should this be reflected in the report?

    Show the answer

    Give both alternatives: with bed D, a net pay of 12.2 m and a pore thickness of 1.75 m; without it, 8.4 m and 1.44 m (18% less). State that core does not confirm the choice of model, and recommend what will settle the question: a test of bed D, and determination of the water saturation and cation exchange capacity on core. Never choose a model to suit the desired result.

Updated: 16 October 2026