Gross rock volume from maps

Quick refresher

For a layer thicker than the accumulation, the rock volume above the contact is the area under the area–depth curve of the top, from the crest to the contact. For a thin layer, we subtract from it the same area for the base.
From contour areas we find the integral slice by slice. The trapezoidal rule is exact if the area grows linearly with depth (a paraboloid dome). The pyramidal rule is exact if it grows as the square of depth (a cone or a pyramid). Simpson's rule gives good accuracy for any smooth curve. On maps their results differ by fractions of a per cent, but the quick “triangle” estimate $\tfrac12 A H$ is wrong by tens of per cent.
The outer pool outline is the line on the map where the contact meets the top. The inner outline is where it meets the base. Between them is the edge-water zone (where there is water under the oil), and inside the inner outline there is oil over the full thickness. The “slab” is the area inside the outer outline times the vertical thickness, $A_\text{OWC}h_\text{TVT}$ (the area is taken at the oil–water contact, OWC). The ratio of the rock volume to the slab is the geometric correction factor. It is less than one.
The closure is the height from the crest to the spill point (the lowest point at which the trap still holds hydrocarbons). Below the spill point there is no accumulation. Until the contact has been penetrated, its depth is a range: not above the deepest oil (“oil down to”), not below the shallowest water (“water up to”), and not below the spill point.
Each metre of error in the contact costs as many cubic metres as the area at the contact (for a thin layer, the area of the edge-water zone). A shift of the map in depth acts as a contact error of the opposite sign, and the method of building the map changes the volume by tens of per cent.

Key relationships

\mathrm{GRV} = \int_{z_\text{crest}}^{z_\text{OWC}} A_\text{top}(z)\,dz

a layer thicker than the accumulation (bottom water everywhere)

\mathrm{GRV} = \int_{z_\text{crest}}^{z_\text{OWC}} \left[A_\text{top}(z) - A_\text{base}(z)\right] dz

a thin layer: the curves of the top and the base

\Delta V = \tfrac{\Delta z}{2}\,(A_i + A_{i+1}),\qquad \Delta V = \tfrac{\Delta z}{3}\left(A_i + A_{i+1} + \sqrt{A_i A_{i+1}}\right)

a slice between neighbouring contours: the trapezoidal rule and the pyramidal rule

\mathrm{GCF} = \frac{\mathrm{GRV}}{A_\text{OWC}\,h_\text{TVT}}

geometric correction factor of a thin layer

\frac{d\,\mathrm{GRV}}{d z_\text{OWC}} = A_\text{top}(z_\text{OWC}) - A_\text{base}(z_\text{OWC})

how much volume one metre of contact depth adds

Common pitfalls

Calculating the volume as the area of the accumulation times the thickness of the layer. For a thin layer this overstates it (in the lesson's example, by 45%).
Using the quick “triangle” estimate without checking the shape of the area–depth curve.
Taking the deepest oil in a well for the contact. It is only a level above which there is no contact.
Getting the curve of the base by shifting the curve of the top, when the thickness of the layer changes over the area.
Calculating the volume from a map built from wells alone, without estimating how much it understates the crest.

After this lesson you can

build an area–depth curve from a structure map and explain why the area under it is the rock volume above the contact
calculate volume from contour areas by the trapezoidal and pyramidal rules, Simpson's rule and the quick “triangle” estimate, and know for which trap shape each rule is exact and in which direction it errs
tell apart a thick layer with bottom water from a thin one with an edge-water zone, draw the outer and inner pool outlines, and calculate the volume from the curves of the top and the base, and the geometric correction factor
find the crest, the spill point and the closure on a map, and estimate the volume for different positions of the contact: “oil down to”, “water up to”, from pressures, and filled to spill
estimate how the volume changes with the contact, with the depths of the map, with the thickness of the layer and with the method used to build the map

The area–depth curve

Imagine cutting the trap with horizontal planes. At depth $z$ , the plane cuts the rock wherever the top of the layer lies above $z$ , that is, inside the contour $z$ . The area of this section, $A_\text{top}(z)$ , is the area bounded by the contour. A thin horizontal slice of thickness $dz$ holds $A_\text{top}(z)\,dz$ of rock. The sum of the slices from the crest to the contact is the whole volume of rock below the top and above the contact.

Rock volume above the contact from the area–depth curve

\mathrm{GRV} = \int_{z_\text{crest}}^{z_\text{OWC}} A_\text{top}(z)\,dz

where:

$\mathrm{GRV}$ gross rock volume between the top and the contact, m³
$A_\text{top}(z)$ area inside the contour of the top at depth $z$ , within the closure of the trap, m²
$z_\text{crest}$ depth of the crest, the highest point of the top, m
$z_\text{OWC}$ depth of the contact (the oil–water contact, OWC, or the gas–water or gas–oil contact), not deeper than the spill point, m

Units: SI: area in m², depth in m, volume in m³; convenient: km² × m = million m³

Holds when

The layer is thicker than the accumulation: under the whole accumulation the base is below the contact, and all the rock between the top and the contact is the layer (bottom water everywhere). The thin layer is in the next section.
The contact is horizontal. Areas are taken only inside the closing contour: parts of the map beyond the spill point do not belong to the trap.
The volume includes all the rock of the layer, non-reservoir rock too. The share of reservoir is taken into account separately (gross and net thickness).

The curve starts from zero area at the crest, and its shape is the shape of the trap. Where the area grows quickly with depth, the limbs are gentle, and each metre of contact adds a lot of volume. Programs calculate the same integral on the grid. In each cell where the top is above the contact, they take the difference between the depth of the contact and the depth of the top, and multiply it by the area of the cell. Then they add up the cells.

CalculatedA map of the top of the same dome as in the topic on structure maps, continued to the north as far as the spill point. The crest is at 2384.6 m. The closing contour (dashed), which is the last contour that still closes around the dome, passes through the saddle (the low point between the dome and the next rise) at 2471.7 m, and the closure is 87 m. The outer pool outline is the 2450 m contour (the oil–water contact, OWC, on the top), and the inner one is the 2420 m contour: there the base of the layer, 30 m thick, meets the OWC. The oil-only zone and the edge-water zone are shaded. Filled circles are wells with oil at the top.

In the past, contour areas were measured with a planimeter (an instrument that gives the area when its pointer traces the contour) or by counting the squares of a grid laid over the map. Now the contour is digitised (traced into the computer as a list of points), and the area of the polygon is found from the coordinates of its vertices:

Area of a polygon from the coordinates of its vertices

A = \frac12\left|\sum_{i=1}^{n} \left(x_i\,y_{i+1} - x_{i+1}\,y_i\right)\right|,\qquad (x_{n+1}, y_{n+1}) = (x_1, y_1)

where:

$x_i,\ y_i$ coordinates of the $i$ -th vertex of the contour on the map, in metres on the ground, m
$n$ number of vertices; we go round them in order, clockwise or anticlockwise, —
$A$ area bounded by the contour, m²

Units: coordinates in metres, area in m²

Holds when

The contour is closed and does not cross itself. The vertices are frequent enough that the straight lines between them follow the contour.
If there is a “hole” (a closed depression) inside the contour, its area is subtracted.

Each term is twice the signed area of the triangle between the origin and a side of the contour. When we go round the contour, the extra triangles cancel each other, and the area inside remains. This is how programs and spreadsheets calculate areas. Units: 1 km² = 247.1 acres, and 1 acre-foot = 1233.5 m³. In oilfield units, the rock volume is expressed in acre-feet.

CalculatedArea–depth curves of the top of the dome (the dots are the areas of the contours every 10 m) and of the base of a layer 30 m thick, which is the same curve lowered by 30 m. The dark fill between them above the oil–water contact (OWC) at 2450 m is the rock volume of the thin layer. The light fill to the left of the base curve would be added if the layer were thicker than the accumulation. The horizontal lines are the contact positions discussed in the lesson: the OWC; “oil down to” (ODT), the deepest oil found in a well; “water up to” (WUT), the shallowest water found in a well; and the spill point.

The trapezoidal and pyramidal rules, and Simpson's rule

On a map, the area is known only at the contours, which are $\Delta z$ apart. Between them we have to assume some shape for the curve. So each rule of calculation is an assumption about the shape of the trap between two neighbouring contours.

Volume of a slice between neighbouring contours: the trapezoidal and pyramidal rules

\Delta V_\text{trap} = \frac{\Delta z}{2}\,(A_i + A_{i+1}),\qquad \Delta V_\text{pyr} = \frac{\Delta z}{3}\left(A_i + A_{i+1} + \sqrt{A_i\,A_{i+1}}\right)

where:

$A_i,\ A_{i+1}$ areas inside neighbouring contours (the smaller one belongs to the upper contour), m² or km²
$\Delta z$ contour interval, the thickness of the slice, m
$\Delta V$ volume of rock in the slice, m³ (million m³ for km²)

Units: areas and interval in consistent units

Holds when

The trapezoidal rule is exact if the area in the slice grows linearly with depth. A dome close to a paraboloid of revolution behaves this way.
The pyramidal rule (the formula for a truncated pyramid) is exact if the horizontal sections are similar and the area grows as the square of depth below the apex: a cone or a pyramid.
A common practical rule: use pyramids when the ratio of the smaller area to the larger one is less than 0.5 (usually near the crest), and trapezoids in the other slices.

If the area grows with depth faster than linearly, the trapezoids slightly overstate the volume of a slice. Pyramids are exact when the width of the trap (the square root of the area) grows linearly with depth. If it grows more and more slowly, pyramids understate the volume of a slice. If it grows faster and faster, they overstate it. For the lesson's dome, trapezoids overstate every slice. Pyramids understate the upper slices and slightly overstate the two lower ones, where the limbs flatten out. In the lower slices, where the areas of neighbouring contours are close, both rules give almost the same result.

Two things remain. One is the cap, the volume above the first contour, from the crest to it. The other is, if we need accuracy, Simpson's rule, which replaces the curve with parabolas through every three neighbouring points.

The cap above the first contour, and Simpson's rule

V_\text{cap} = \frac{A_0 h_\text{cap}}{3}\ \text{(cone)}\ \ldots\ \frac{A_0 h_\text{cap}}{2}\ \text{(paraboloid)},\qquad V = \frac{\Delta z}{3}\left(A_0 + 4A_1 + 2A_2 + \ldots + 4A_{n-1} + A_n\right)

where:

$A_0,\ h_\text{cap}$ area of the first contour below the crest, and the height of the crest above it, m², m
$A_0 \ldots A_n$ areas of the contours at an equal interval $\Delta z$ , from the first to the last, m²
$n$ number of slices; it must be even, —

Units: consistent units

Holds when

A cone cap is a sharp crest (the area grows as the square of depth), and a paraboloid cap is a smooth crest (the area grows linearly). A real cap is usually between them.
Simpson's rule is exact for curves up to a cubic. It needs a smooth curve and an equal interval.

The cap is usually small, and Simpson's rule is almost always more accurate than trapezoids and pyramids. On smooth domes the difference between all the rules is a fraction of a per cent.

Worked exampleRock volume of a dome from the areas of contours

Given (assumed data)

Map: the dome from the topic on structure maps; crest 2384.6 m, contour interval 10 m
Areas of the contours 2390, 2400 … 2450 m, km²: 0.1935; 0.5803; 1.0096; 1.4924; 2.0447; 2.6919; 3.4785
Contact: OWC at 2450 m; the layer is thicker than the accumulation (bottom water everywhere)

Solution

The cap from the crest to the 2390 m contour: $h_\text{cap} = 2390 - 2384.6 = 5.4$ m. As a cone, $0.1935\cdot 5.4/3 = 0.35$ million m³. As a paraboloid, $0.1935\cdot 5.4/2 = 0.52$ million m³.
Trapezoids over six slices: $5\cdot(A_0 + 2A_1 + 2A_2 + 2A_3 + 2A_4 + 2A_5 + A_6) = 5\cdot 19.3098 = 96.55$ million m³. With a cone cap, 96.90 million m³.
Pyramids over the same slices: 95.97 million m³. With the cap, 96.32 million m³. In the top slice, the ratio of areas is $0.1935/0.5803 = 0.333 < 0.5$ . The pyramid gives 3.70 million m³ against 3.87 by trapezoids. If we take the pyramid only for this slice, by the practical rule, we get 96.73 million m³.
Simpson over six slices: $\tfrac{10}{3}(A_0 + 4A_1 + 2A_2 + 4A_3 + 2A_4 + 4A_5 + A_6) = 96.13$ million m³. With a cone cap it is 96.48, and with a paraboloid cap 96.65 million m³.
A calculation from the grid of the model (exact integration of the surface) gives 96.64 million m³. Deviations: trapezoids +0.27%, pyramids −0.33%, the mixed rule +0.09%, and Simpson with a cone −0.17%.
The quick “triangle” estimate: the height of the accumulation is $H = 2450 - 2384.6 = 65.4$ m, and $\tfrac12\cdot 3.4785\cdot 65.4 = 113.7$ million m³, which is 17.7% more.

Answer: The rock volume above the OWC is about 96.6 million m³. All the rules by slices give it to within a fraction of a per cent, and the “triangle” overstates it by 18%.

What it means. The choice of rule changes almost nothing. The error from it is tens of times smaller than the error from an inexact contact or map (last section). The “triangle” assumes that the area grows linearly with depth over the whole height of the accumulation. But on the lesson's dome the limbs flatten, and the area grows faster. For a sharper structure, where the area grows even faster, it would overstate the volume more. For a structure with a flat top and steep limbs, where the area hardly grows with depth, it would understate it. The cap of a smooth crest is better calculated as a paraboloid: the exact value is 0.52 million m³. For the same reason, in the top slice the trapezoid is more accurate, although the ratio of areas is less than 0.5. The exact volume of the slice is 3.84 million m³. The trapezoid errs by +1%, and the pyramid by −4%. Of the cases with a cone cap, the mixed rule is the closest in total only because the shortfall of the pyramid and of the cap itself cancelled the excess of the trapezoids in the other slices. The data in this example are assumed for illustration.

Thick and thin layers: bottom water and the edge-water zone

If the layer is thicker than the accumulation, there is water of the same layer under all the oil. This is bottom water, and the volume is calculated from one curve, the curve of the top. If the layer is thinner, then nearer the crest the base of the layer lies above the contact. There the oil fills the layer over its whole thickness, and water comes only from the edges of the accumulation. This is edge water. On a map there are two outlines (pool outlines). The outer one is the line where the contact crosses the top. The inner one is where it crosses the base. Between them is the edge-water zone, where there is water under the oil. Inside the inner outline is the oil-only zone. We draw the inner outline on the map of the base, which we get from the map of the top and the isochore map (thickness maps).

Rock volume of a thin layer from the curves of the top and the base

\mathrm{GRV} = \int_{z_\text{crest}}^{z_\text{OWC}} \left[A_\text{top}(z) - A_\text{base}(z)\right] dz

where:

$A_\text{top}(z),\ A_\text{base}(z)$ areas inside the contour $z$ on the maps of the top and the base, m²
$z_\text{crest},\ z_\text{OWC}$ depths of the crest and the contact, m

Units: m², m, m³

Holds when

The area at the base is counted from the same crest. Above the base, its curve is zero.
If the vertical thickness of the layer is constant and equal to $h_\text{TVT}$ , the base is the top lowered by $h_\text{TVT}$ , and $A_\text{base}(z) = A_\text{top}(z - h_\text{TVT})$ . If the thickness changes, we need a separate map of the base.

In a horizontal section at depth $z$ , the rock of the layer fills the ring between the contours of the top and the base. Inside the contour of the base at this depth, the rock is already below the layer. For a constant thickness, the volume of a thin layer is the difference of two volumes of a “thick” layer: down to the contact, and down to a level $h_\text{TVT}$ above it.

We compare the volume of a thin layer with a “slab”: the area of the outer outline times the thickness. The slab overstates the volume, because in the edge-water zone the oil fills only part of the thickness. Their ratio is called the geometric correction factor (GCF).

Geometric correction factor

\mathrm{GCF} = \frac{\mathrm{GRV}}{A_\text{OWC}\,h_\text{TVT}};\qquad \text{paraboloid: } \mathrm{GCF} = 1 - \frac{h_\text{TVT}}{2H}\ \ (h_\text{TVT} \le H),\qquad \frac{H}{2h_\text{TVT}}\ \ (h_\text{TVT} \ge H)

where:

$A_\text{OWC}$ area inside the outer pool outline, m²
$h_\text{TVT}$ vertical thickness of the layer, m
$H$ height of the accumulation, from the crest to the contact, m

Units: dimensionless

Holds when

The formulas on the right are for a paraboloid dome, whose area grows linearly with depth. For a cone, $\mathrm{GCF} = \dfrac{1 - (1 - h_\text{TVT}/H)^3}{3h_\text{TVT}/H}$ for $h_\text{TVT} \le H$ , and $H/(3h_\text{TVT})$ for $h_\text{TVT} \ge H$ .
The thickness is constant and the contact is horizontal.

The factor depends on the ratio of the thickness to the height of the accumulation, and on the shape of the trap. A very thin layer ( $h_\text{TVT} \ll H$ ) is almost all in the oil-only zone, and the GCF is close to 1. For a layer thicker than the accumulation, the GCF falls in inverse proportion to the thickness: the slab counts the whole thickness, but the oil fills only the height of the accumulation. The faster the area grows with depth, the lower the GCF at the same thickness. In Russian-language practice, the volume of oil-saturated rock is usually found by integrating a map of oil-saturated thickness over the area. In the oil-only zone the thickness equals the thickness of the layer (for the volume of reservoir, its effective thickness). In the edge-water zone it falls to zero at the outer outline.

Worked exampleA layer 30 m thick: volume and correction factor

Given (assumed data)

Map, contour areas and OWC: as in the previous example; OWC 2450 m
Vertical thickness of the layer: 30 m over the whole trap (the base is the top lowered by 30 m)

Solution

Inner pool outline: the base meets the OWC where the top is at $2450 - 30 = 2420$ m. The area is 1.4924 km². The outer outline is 3.4785 km². The edge-water zone is 1.9861 km², which is 57% of the area of the accumulation.
Volume of a “thick” layer down to the level of 2420 m by trapezoids with a cone cap: $5\cdot(0.1935 + 2\cdot 0.5803 + 2\cdot 1.0096 + 1.4924) + 0.35 = 24.68$ million m³.
Volume of the thin layer: $96.90 - 24.68 = 72.22$ million m³ by trapezoids. The calculation from the grid of the model gives 71.92 million m³.
The slab: $3.4785\cdot 30 = 104.36$ million m³, an overstatement by 45%. $\mathrm{GCF} = 71.92/104.36 = 0.689$ .
For comparison, at $H = 65.4$ m and $h_\text{TVT}/H = 0.459$ , a paraboloid would give $1 - 30/(2\cdot 65.4) = 0.771$ , and a cone 0.611.
The oil-only zone holds $1.4924\cdot 30 = 44.77$ million m³ of rock, and the edge-water zone holds $71.92 - 44.77 = 27.15$ million m³, which is 38% of the volume.

Answer: The rock volume of a layer 30 m thick above the OWC is about 72 million m³. The geometric correction factor is 0.689.

What it means. The edge-water zone takes 57% of the area, but only 38% of the volume of rock: the oil there lies on water, and its thickness falls towards the outer outline. This is where wells water out sooner, so such zones are often drilled and evaluated separately. The factor for the lesson's dome lies between those of the paraboloid and the cone: the area grows with depth faster than linearly. So the formulas for ideal shapes are good only for a first estimate. For an estimate of volume, we calculate it from the maps of the top and the base. The data in this example are assumed for illustration.

CalculatedThe geometric correction factor against the ratio of the layer thickness to the height of the accumulation: the lesson's dome (calculated from the map at an OWC of 2450 m,

H

= 65.4 m), a paraboloid and a cone. The dot is the 30 m layer of the example.

Closure, spill point and the position of the contact

A trap holds an accumulation down to the spill point at most. This is the deepest point down to which it still holds hydrocarbons. If it is overfilled, they leave through this point (crest, closure, spill point). On a map we look for it as a saddle on the way to a neighbouring structure, or to the place where the layers rise up-dip. It is the deepest contour that still closes around the crest. For the lesson's dome, the saddle lies to the north, outside the part of the block that is shown in the topic on structure maps. To calculate the volume, we had to continue the map from seismic data. The closure is 87 m, and the area inside the closing contour is about 6.6 km². In a trap at a fault, the spill point may be the end of the fault, or a window where reservoir rock meets reservoir rock (throw profile).

Until the contact has been penetrated, its depth is bounded from two sides (“oil down to” and “water up to”). A well that found oil down to the base of the layer gives “oil down to” (ODT): the contact is not above this depth. A well with water at the top of the layer gives “water up to” (WUT): the contact is not below it. If no water has been seen yet, the lower limit is the spill point. The contact stands there only if the trap is filled to spill and the seal holds such an accumulation. Formation pressure measurements help to narrow the range. The pressure lines of oil and water cross at the free water level. The contact lies a little above it, by the height of the capillary rise (the height to which capillary forces lift water into the rock; Pressure gradients, fluid contacts and capillary equilibrium, transition zone). In Russian-language reserve estimates, a contact that no well has penetrated is taken conventionally. This is the assumed OWC: for example, at the base of the lowest oil-saturated interval (at the “oil down to” level), or halfway between “oil down to” and “water up to”.

Worked exampleThe contact from wells: from “oil down to” to the spill point

Given (assumed data)

Layer: vertical thickness 30 m, the map of the top is as above
Well 1 (the discovery well): top at 2400.9 m; oil down to the base at 2430.9 m
Well 5 (an appraisal well on the northern limb): top at 2451.8 m; the layer holds water
Spill point: 2471.7 m
Later, wells 3 and 11: penetrated the OWC at 2450 m

Solution

After well 1: the contact is between “oil down to” at 2430.9 m and the spill point at 2471.7 m. The rock volume is from 39.3 to 128.3 million m³, a range of 3.3 times.
After well 5: the lower limit is “water up to” at 2451.8 m. The volume is from 39.3 to 75.6 million m³, a range of 1.9 times.
After wells 3 and 11: the OWC is at 2450 m, and the volume is 71.9 million m³.
A metre of contact at this depth is worth the area of the edge-water zone: $3.4785 - 1.4924 = 1.9861$ km², which is about 1.99 million m³ per metre. An OWC error of ±5 m gives a volume from 62.4 to 82.3 million m³ (−13.2 and +14.5%). It is not symmetric, because the area grows with depth.

Answer: Before the contact is penetrated, the rock volume is known only within 39.3–128.3 million m³. After well 5 it is 39.3–75.6 million m³. With the OWC at 2450 m it is 71.9 million m³.

What it means. “Oil down to” is not a contact but only a limit. A discovery well on the crest often sees oil over the whole thickness of the layer, and a volume calculated “down to the base in the well” is the minimum case. The spill point is the upper limit of the volume, and it is reached only if the trap is filled to the end and the seal holds such an accumulation. So we give the volume of a prospect (a possible accumulation that has not yet been drilled) as a range, and refine it with each well and each pressure measurement. The range passes into a probabilistic estimate of volumes (Probabilistic volumetrics by Monte Carlo simulation) and into reserve categories (Classification of reserves and resources). The data in this example are assumed for illustration.

Uncertainty of the volume: the contact, the depths of the map, the thickness and the method

The rock volume inherits the errors of everything it was calculated from. The simplest way to compare them is to change one assumption at a time and see how much the volume changes. This is how “tornado” diagrams are built: they are charts that rank the assumptions by how much each one changes the result (Sensitivity analysis and scenarios). Four sources matter most.

The contact. A metre of error costs the area at the contact (for a thin layer, the area of the edge-water zone). For gentle structures it is large, and the contact is often the main source of uncertainty.
The depths of the map. A seismic map is converted from time to depth with a velocity model (Seismic interpretation) and fitted to wells. An error that is common to the whole map shifts it in depth and acts as a contact error of the opposite sign: a map 5 m deeper is the same as an OWC 5 m higher. A velocity error between the crest and the limbs stretches or squeezes the relief. The closure and the areas change, although the depth is right at the tied well.
The thickness of the layer and its change over the area, through the isochore map and the inner outline.
The method of building the map. Between wells, the shape of the crest comes from seismic data and the interpolation method (computer contouring). Different methods have different systematic errors.

CalculatedThe outer and inner pool outlines (OWC 2450 m, layer 30 m) from three maps of the same 11 picks: from the model, by minimum curvature, and by the triangle method. The crest of the triangles coincides with well 1, because this method cannot rise above the highest well, and its outlines are clearly smaller. Minimum curvature spreads the outer outline wider, but it lowers the crest.

Worked exampleWhat changes the volume of a thin layer most

Given (assumed data)

Base case: the model map, OWC 2450 m, layer 30 m: 71.92 million m³
Cases: OWC ±5 m; the whole map 5 m deeper or shallower; the relief steeper or gentler by 5% at the same depth at well 1; thickness ±10%; a map from the same picks by minimum curvature and by triangles

Solution

OWC 5 m higher: 62.41 million m³ (−13.2%). 5 m lower: 82.34 million m³ (+14.5%).
Map 5 m deeper: 62.41 million m³, exactly the same as an OWC 5 m higher. 5 m shallower: 82.34 million m³.
Relief 5% steeper (depths measured from well 1, 2400.9 m): 68.62 million m³ (−4.6%). 5% gentler: 75.70 million m³ (+5.3%).
Thickness 27 m: 67.20 million m³ (−6.6%). 33 m: 76.16 million m³ (+5.9%).
Map by minimum curvature: 77.18 million m³ (+7.3%). Map by triangles: 51.34 million m³ (−29%).

Answer: The volume changes most with the method of building the map (triangles, −29%) and with the contact together with a shift of the map in depth (about ±14% per 5 m). Thickness and stretching of the relief give about ±5–7%.

What it means. The sign of the error of a method depends not only on the map but also on the thickness of the layer. For a layer thicker than the accumulation, both maps built from wells give less than the model: minimum curvature gives 91.2 million m³, and triangles give 55.3, against 96.6 million m³. This is because both lower the crest. For a layer 30 m thick, minimum curvature gives more than the model: its outer outline is wider, and the volume of a thin layer is set mostly by the area at the contact. So the statement “a map from wells always understates the volume” is wrong, and we must check each time. The data in this example are assumed for illustration.

Sensitivity of the rock volume of a thin layer (base case: the model map, OWC 2450 m, layer 30 m)

Assumption: Method of building the map
Variant: triangles / minimum curvature
Volume, million m³: 51.34 / 77.18
Change: −28.6% / +7.3%

Assumption: Contact
Variant: OWC −5 / +5 m
Volume, million m³: 62.41 / 82.34
Change: −13.2% / +14.5%

Assumption: Shift of the map
Variant: 5 m deeper / shallower
Volume, million m³: 62.41 / 82.34
Change: −13.2% / +14.5%

Assumption: Thickness of the layer
Variant: 27 / 33 m
Volume, million m³: 67.20 / 76.16
Change: −6.6% / +5.9%

Assumption: Relief
Variant: steeper / gentler by 5%
Volume, million m³: 68.62 / 75.70
Change: −4.6% / +5.3%

Assumption	Variant	Volume, million m³	Change
Method of building the map	triangles / minimum curvature	51.34 / 77.18	−28.6% / +7.3%
Contact	OWC −5 / +5 m	62.41 / 82.34	−13.2% / +14.5%
Shift of the map	5 m deeper / shallower	62.41 / 82.34	−13.2% / +14.5%
Thickness of the layer	27 / 33 m	67.20 / 76.16	−6.6% / +5.9%
Relief	steeper / gentler by 5%	68.62 / 75.70	−4.6% / +5.3%

The base volume is 71.92 million m³. “OWC −5 m” means that the contact is 5 m higher. The rows are ordered by the size of the change. The calculation is made from the lesson's maps; the data are assumed for illustration.

Terms and symbols

GRV, gross rock volume: The volume of rock between the top of the layer and the contact, including non-reservoir rock. For a thin layer, the rock must also lie above the base of the layer.
Area–depth curve: The area inside the contour of the top (or base) against its depth. The area under it from the crest to the contact is the rock volume.
Trapezoidal and pyramidal rules, Simpson's rule: Ways to add the volumes of slices between contours. They are exact for linear, quadratic and up to cubic growth of area with depth.
Outer and inner pool outlines: The lines where the contact meets the top and the base of the layer.
Edge-water zone and oil-only zone: The part of the accumulation between the outlines, where there is water under the oil, and the part inside the inner outline, where the oil fills the whole thickness.
Bottom water and edge water: Water under the whole accumulation (a layer thicker than the accumulation) and water only at the edges (a layer thinner than the accumulation).
Geometric correction factor: The ratio of the rock volume to the area of the outer outline times the thickness of the layer (GCF).
“Oil down to”, “water up to”: The deepest oil and the shallowest water penetrated by wells (ODT, WUT). They are the limits of the depth of the contact.
Closure and spill point: The height of the trap from the crest to the spill point, and the point itself, which is the lower limit of a possible accumulation.

Common mistakes and misconceptions

Wrong

“Rock volume is the area of the accumulation times the thickness of the layer.”

In fact

This holds only for a layer that is very thin compared with the height of the accumulation. Otherwise a correction factor is needed: in the lesson's example, the slab overstates the volume by 45%.

Wrong

“Half the area at the contact times the height of the accumulation is enough.”

In fact

This is the “triangle”, an assumption of linear growth of area. For the lesson's dome it overstates the volume by 18%, and for structures of another shape it may understate it. For an estimate, we calculate from all the contours.

Wrong

“The deepest oil in a well is the OWC.”

In fact

This is “oil down to”, a limit above which there is no contact. Until the contact is penetrated, we give its depth as a range, down to “water up to” or to the spill point, and we narrow it with pressure measurements.

Wrong

“The trapezoidal rule always overstates the volume.”

In fact

It overstates it if the area grows with depth faster than linearly (a convex curve, gentle limbs), and it understates it for the opposite shape. On smooth domes the error is a fraction of a per cent.

Wrong

“A map built from wells alone always understates the volume.”

In fact

The triangle method really cannot raise the crest above the wells, and in the lesson's example it gives 29% less. But minimum curvature for a thin layer gave more than the model. The sign of the error depends on the method, the shape and the thickness.

In depth: advanced detail

Why each rule is exact for its own shape

Let the area in the slice between depths $z_i$ and $z_{i+1} = z_i + \Delta z$ be linear: $A(z) = A_i + (A_{i+1} - A_i)(z - z_i)/\Delta z$ . The integral of a linear function is the mean value times the length of the interval, $\Delta z\,(A_i + A_{i+1})/2$ . This is the trapezoidal rule. The area is linear in depth for a paraboloid of revolution (a bowl shape made by turning a parabola around its vertical axis) $z = z_\text{crest} + k r^2$ : $A = \pi r^2 = \pi (z - z_\text{crest})/k$ (for an elliptical paraboloid, the same with another factor).

If the sections are similar and converge to an apex at depth $z_0$ , the area grows as a square: $A(z) = c\,(z - z_0)^2$ . Then $\int_{z_i}^{z_{i+1}} A\,dz = \tfrac{c}{3}\left[(z_{i+1} - z_0)^3 - (z_i - z_0)^3\right]$ . If we factor the difference of cubes, with $a = \sqrt{A_i/c}$ and $b = \sqrt{A_{i+1}/c}$ , we get $\tfrac{\Delta z}{3}\,(A_i + A_{i+1} + \sqrt{A_i A_{i+1}})$ , because $b - a = \Delta z$ and $a^2 + ab + b^2 = (A_i + A_{i+1} + \sqrt{A_i A_{i+1}})/c$ . This is the formula for a truncated pyramid. Simpson's rule draws a parabola through three neighbouring points and integrates it exactly. The error of this replacement is proportional to the fourth derivative of the curve, so the rule is exact for a cubic dependence too.

The correction factor of a paraboloid: for $A(z) = A_\text{OWC}(z - z_\text{crest})/H$ , the volume under the top down to a depth $d$ below the crest is $A_\text{OWC}d^2/(2H)$ . For a thin layer, $\mathrm{GRV} = \tfrac{A_\text{OWC}}{2H}\left[H^2 - (H - h_\text{TVT})^2\right] = A_\text{OWC}h_\text{TVT}\left(1 - \tfrac{h_\text{TVT}}{2H}\right)$ . This is the formula of the section. For a cone, the same calculation with cubes instead of squares gives the formula in the assumptions.

Volume from a grid, and checking a program

A program builds grids of the top, the base and the surface of the contact. In each cell it takes the thickness of the layer above the contact, $\max\left(0,\ \min(z_\text{OWC}, z_\text{base}) - z_\text{top}\right)$ . The volume is the sum of these thicknesses times the area of the cell. This is the same integral, only taken over area instead of depth: each cell is a column of rock between the top and the contact or the base. A tilted contact (tilted contacts) is set by a surface instead of a horizontal plane. This is the advantage of a grid calculation over an area–depth curve.

A check against the curve. The area inside the contour of the contact on the grid must agree with the area of the contour on the map. The volume must agree with the integral of the area–depth curve to within a fraction of a per cent. A disagreement points to an error in the outline or in the limits of the trap.
Cell size. The cell must be clearly smaller than the distance over which the oil thickness changes in the edge-water zone. For a thin layer with a narrow zone, a large cell distorts its volume.
Limits. The volume is counted only inside the closure and inside its own block. Parts of the grid beyond the spill point, or on the other side of a sealing fault, do not belong to the accumulation.

Connections to other topics

Structure maps and contouring — how the lesson's maps were built, and how interpolation methods differ.
Cross-sections from maps, map combination and faults on maps — the map of the base from the top and isochores, and fault polygons at traps with faults.
Volumetric estimation of hydrocarbons in place — how rock volume turns into hydrocarbons in place.
Traps and timing — the crest, closure and spill point, and what limits the height of an accumulation.
Pressure gradients, fluid contacts and capillary equilibrium — contacts from formation pressure lines.
Probabilistic volumetrics by Monte Carlo simulation — a probabilistic estimate from ranges of rock volume and other parameters.
Geological and technical uncertainty of a project — how the uncertainty of volumes passes into the evaluation of a project.
Classification of reserves and resources — categories of reserves and resources by degree of study.
Seismic interpretation — converting seismic time to depth, and its errors.

Check yourself

1
The areas of the 2140 and 2150 m contours are 2.1 and 3.4 km². Find the volume of the slice between them by the trapezoidal and pyramidal rules. Which rule is more suitable?
Show the answer
Trapezoids: $5\cdot(2.1 + 3.4) = 27.5$ million m³. Pyramids: $\tfrac{10}{3}\,(2.1 + 3.4 + \sqrt{2.1\cdot 3.4}) = \tfrac{10}{3}\cdot(5.5 + 2.672) \approx 27.24$ million m³. The ratio of areas is $2.1/3.4 \approx 0.62 > 0.5$ , so by the practical rule we take trapezoids. The difference is only 1%.
2
A dome is close to a paraboloid, the height of the accumulation is 60 m, and the thickness of the layer is 12 m. What is the geometric correction factor?
1. A.exactly 1
2. B.about 0.9
3. C.about 0.5
4. D.about 0.2
Show the answer
Correct answer: B. about 0.9
For a paraboloid, $\mathrm{GCF} = 1 - h_\text{TVT}/(2H) = 1 - 12/120 = 0.9$ . The layer is thin compared with the accumulation, the edge-water zone is narrow, and the slab overstates the volume by only 11%.
3
A discovery well found oil down to the base of the layer at 1836 m, and an appraisal well found water at the top of the layer at 1862 m. Where is the oil–water contact?
1. A.Somewhere between 1836 and 1862 m
2. B.At 1836 m
3. C.At 1862 m
4. D.At the level of the spill point
Show the answer
Correct answer: A. Somewhere between 1836 and 1862 m
1836 m is “oil down to”, and the contact is not above it. 1862 m is “water up to”, and the contact is not below it. Inside this range, we refine its position by pressure measurements (the crossing of the oil and water pressure lines) or by a new well. We calculate the volume for several cases.
4
The layer is thicker than the accumulation, and the area inside the contour of the contact is 4.2 km². The depth of the contact is known to ±8 m. What is the uncertainty of the rock volume?
Show the answer
A metre of contact costs the area at the contact: about $4.2\cdot 8 \approx 34$ million m³ in each direction. More exactly, it is not symmetric. For a deeper contact the area is larger, and the gain is greater than the loss for a shallower contact.
5
Which map, built from sparse wells without seismic data, will most likely understate the volume of an anticlinal trap more than the others?
1. A.One built by the triangle method
2. B.An interpretive one, with seismic data taken into account
3. C.All the same
4. D.One built by minimum curvature
Show the answer
Correct answer: A. One built by the triangle method
The flat faces of triangles do not raise the crest above the highest well, and they cut off the bulge of the dome between wells. In the lesson's example, the volume is 29% less than the model (and 43% less for a layer thicker than the accumulation). Minimum curvature continues the slopes and errs less, sometimes even in the other direction.
6
The contour of the contact was digitised as a polygon with vertices (0; 0), (1200; 0), (1500; 900) and (300; 1100) m. What is the area inside it?
Show the answer
The terms $x_i y_{i+1} - x_{i+1} y_i$ over the sides of the contour are: $0\cdot0 - 1200\cdot0 = 0$ ; $1200\cdot900 - 1500\cdot0 = 1{,}080{,}000$ ; $1500\cdot1100 - 300\cdot900 = 1{,}380{,}000$ ; $300\cdot0 - 0\cdot1100 = 0$ . The sum is 2,460,000 m², and half of it is 1,230,000 m², which is 1.23 km².

Updated: 16 October 2026