Cross-sections from maps, map combination and faults on maps

Quick refresher

A section is built from a map by carrying over the points where the section line crosses the contours. The true dip is seen only on a section across the strike and without scale distortion. On other sections the dip is apparent (smaller than the true dip). With the vertical scale enlarged $V$ times, all slopes look steeper. Both scales are written on every section.
An isochore map is the difference of two structure surfaces: the thickness in the vertical direction. Isopachs (the true thickness, measured at right angles to the layer) are smaller than it by a factor of $1/\cos\delta_b$ . The map of the base is made by adding isochores, not isopachs, to the map of the top. Thicknesses from deviated wells are first reduced to the vertical thickness.
A normal fault on a map is a polygon between the lines where the layer meets the fault plane in the footwall (the block below the fault plane) and in the hanging wall (the block above it). The width of the strip, measured at right angles to the fault, is the heave $w$ . The difference between the depths of its edges is the throw $t$ . Their ratio is the tangent of the dip of the fault plane.
A contour of the layer ends at a fault where it is crossed by the contour of the fault plane with the same value, in both walls. So each horizon has its own polygon. The polygon of the base is shifted down the dip of the fault plane.
The throw along a fault changes smoothly, without a change of sign, and falls towards the tips. Where faults merge, the throws add up. A well that crosses a fault inside the layer penetrates an incomplete thickness. We do not use it for the isochore map.

Key relationships

\mathrm{tan}\,\delta' = \mathrm{tan}\,\delta_b\sin\beta,\qquad \mathrm{tan}\,\delta_V = V\,\mathrm{tan}\,\delta'

apparent dip on a section at angle $\beta$ to the strike, and the angle that is drawn with scale distortion $V$ (topic Stress, strain and structural analysis)

h_\text{TVT} = z_\text{base} - z_\text{top},\qquad h = h_\text{TVT}\cos\delta_b

isochore as the difference of the depths of the base and the top, and the isopach

\mathrm{tan}\,\delta = \frac{t}{w}

dip of the fault plane from the polygon: the throw $t$ is the difference of the depths of the edges, and the heave $w$ is the width normal to the fault

\Delta q = \frac{h_\text{TVT}}{\mathrm{tan}\,\delta \pm \mathrm{tan}\,\delta_b}

shift of the polygon of the base relative to the polygon of the top, down the dip of the fault plane; “+” if the layers dip towards the fault plane

Common pitfalls

Reading dip angles from a section with a distorted vertical scale, or from a section that is oblique to the strike.
Building the base by adding isopachs, or depth differences from deviated wells without recalculation, to the top.
Drawing a fault as a single line and gridding over it: the step turns into a steep bend.
Taking one fault polygon for all horizons.
Plotting on the isochore map a thickness met by a well that crossed the fault inside the layer.

After this lesson you can

build a section from a structure map: choose the line and the scales, carry over the crossings with contours, complete the base and the faults, and explain how the direction of the section and the vertical exaggeration change the visible angles
get isochore and isopach maps as the difference of two surfaces, build a base map from a top map and a thickness map, and understand which thickness to take from deviated wells and from wells near faults
draw faults as polygons by the lines where the layer meets the fault plane, and find from a polygon the heave, the throw and the dip of the fault plane
draw fault plane contours from the fault points in wells, and use them to find where the contours of the layer end in both walls and where the polygon of the base is displaced
check the consistency of maps, sections and wells: the throw profile along a fault, the sum of throws where faults merge, and the missing interval in wells

A section from a structure map

A map and a section describe one three-dimensional surface. We build a section from a map when there are few wells on the line, or none at all. The exact depth of the layer on the line is known only where the line crosses the contours. Between the crossings we restore the surface in the same way as on the map: smoothly, continuing the slopes of the neighbouring parts. The topic structure contours and sections explains what an apparent dip is and what a section with a distorted vertical scale is. Here we show how a section is built from a map and what is checked.

Building a section from a structure map

1
Choose the line
For true angles, choose a line across the strike of the layers or of the fault. For a tie to wells, choose a line through the wells (the line becomes broken, but it keeps the general direction). Label the ends (A–A′). Use one orientation rule for all the sections of a project, for example west or south on the left.
2
Set the scales
The horizontal scale is the scale of the map. The vertical scale is first taken the same ( $V = 1$ ): only on such a section are the angles and shapes correct. If we need to show thin layers, we enlarge the vertical scale and write $V$ .
3
Carry over the crossings
Along the line we mark the distances from its start to each contour, and the values of the contours. We use a strip of paper, a ruler or a calculation from the grid. Then we plot the points on the profile.
4
Draw the top
We join the points with a smooth line. Between two crossings of the same contour, on a crest or in a syncline, the surface goes past its value but stays within one contour interval. If it rose higher than the next value (or sank lower), the line would cross that contour too.
5
Add the faults
The edges of the fault polygon on the section line are the points where the top meets the fault plane in the two walls. Through them we draw the fault plane at the angle that was found from its contours. The layer is not continued across the polygon.
6
Add the base and the wells
We lay off the base from the top, vertically, by the thickness from the isochore map (next section). At wells on the line we plot the picks and the points where they crossed faults.
7
Check
Do the throws and angles of the faults, the depths in wells and the shape of folds agree with the map? A contradiction means an error in the map or in the section. We correct both.

Vertical exaggeration does not change depths, but it changes angles: the tangent of each slope is multiplied by $V$ . Gentle limbs of folds become visible, and faults become almost vertical. At $V = 5$ a normal fault dipping at 52° is drawn at 81°, and by eye it is hard to tell a steep fault from a gentle one. Vertical thicknesses are stretched $V$ times, and distances at right angles to the layers are distorted. So we do not take true thicknesses from such a section.

CalculatedSection A–A′ built from the map of the top and the isochore map (the map is in the section on faults): without scale distortion, and with the vertical scale enlarged 5 times. On the left is the hanging wall, with a crest of its own. On the right is the footwall, whose highest point rests against the fault plane. On this section the limbs dip at no more than 6°, and at

V = 5

they look as if they dip up to 28°. A fault with a true dip of 52° looks almost vertical.

The direction of the section matters both for layers and for faults. Section A–A′ was drawn across the strike of the fault, so the fault plane is seen on it at its true angle. The layers on the limbs of the fold strike in another direction, and on the same section their dip is apparent, smaller than the true one (apparent dip).

Worked exampleSection A–A′: angles on the section and on the map

Given (assumed data)

Section line: A–A′, azimuth 60°, across the strike of the fault (150°); length 2.8 km
Crossings with the contours of the top in the footwall (from point A): 1910 m at 1934.6 m, 1920 m at 2027.8 m
Dip of the top at the middle of this stretch, from the map: 7.84° towards an azimuth of 21.3° (at right angles to the contours)
Crossings with the 1940 m contour in the hanging wall: at 1148.3 and 1208.7 m; the 1930 m contour does not reach the line
Vertical scale for display: 5 times larger than the horizontal one

Solution

Apparent angle on the section between the contours 1910 and 1920 m: $\mathrm{tan}\,\delta' = 10/(2027.8 - 1934.6) = 10/93.2 = 0.1073$ , so $\delta' = 6.12°$ .
Check from the map: the angle between the section line and the dip direction is $\omega = 60° - 21.3° = 38.7°$ , and $\mathrm{tan}\,\delta' = \mathrm{tan}\,7.84°\cdot\cos 38.7° = 0.1075$ . This is almost the same as from the crossings. Over the stretch of 93.2 m the slope changes a little.
Fault plane: the section is perpendicular to its strike, so on a section without distortion the fault plane is seen at its true angle of 52°. The strike of the layer here is 111.3°, which differs by 39° from the strike of the fault. So the dip of the layer is apparent.
At $V = 5$ the limb is drawn at $\arctan(5\cdot 0.1073) = 28.2°$ , and the fault plane at $\arctan(5\,\mathrm{tan}\,52°) = 81.1°$ .
Crest of the hanging wall: between the crossings of the 1940 m contour (1148.3 and 1208.7 m), the top rises above 1940 m, but not to 1930 m. Otherwise the 1930 m contour would cross the line. In the model, the crest on the line is 1939.9 m.

Answer: On the section without distortion, the limb of the footwall dips at 6.12°, while its true dip is 7.84°. The fault plane dips at 52°. With a five-fold exaggeration they look like 28.2° and 81.1°.

What it means. A section across the strike of a fault does not have to be across the strike of the layers. The same section shows the true angle of the fault plane and the apparent angle of the layer. From a section with a distorted scale we can read an angle only by dividing the tangent by $V$ . The rule “not higher than the next contour” limits the crest from above, but it does not give its height. Between the crossings the surface is restored from the slopes, and the shape of the crest is refined by seismic data. The data in this example are assumed for illustration.

Thickness maps as the difference of surfaces

The thickness of a layer is shown on two kinds of maps (kinds of maps). An isochore map shows the thickness in the vertical direction, $h_\text{TVT}$ (true vertical thickness, TVT): the difference between the depths of the base and of the top at each point. An isopach map (isopach map) shows the true thickness $h$ , measured at right angles to the bedding (true stratigraphic thickness, TST). If grids of the top and the base have been built, the isochore map is their difference at each node, and programs make it this way. In this lesson the dip angle of the layer is $\delta_b$ , and the dip angle of the fault plane is $\delta$ , as in the topic Faults.

Isochore as the difference of surfaces, and isopach

h_\text{TVT}(x, y) = z_\text{base}(x, y) - z_\text{top}(x, y),\qquad h = h_\text{TVT}\cos\delta_b

where:

$h_\text{TVT}$ thickness of the layer in the vertical direction (isochore) at a point of the map, m
$z_\text{top},\ z_\text{base}$ depths of the top and the base below sea level at the same point of the map, m
$h$ true thickness (isopach), at right angles to the bedding, m
$\delta_b$ dip angle of the layer at this point, °

Units: lengths in the same units; angles in degrees

Holds when

The top and the base at the point belong to one block: depths are not subtracted across a fault (section on fault polygons).
For $h$ , the top and the base are nearly parallel over a distance of the order of the layer thickness. Otherwise the thickness is measured at right angles to the mean surface.
Depths are measured from one level and are positive downwards.

The isochore is needed for volume: the rock volume between two structure surfaces is found by adding up (integrating) the isochore over the area (Gross rock volume from maps). The isopach is needed for geology: it shows where the layer was deposited thicker, and it does not depend on a tilt that came later. At a dip of up to 9°, as on the limbs of the fold in the lesson, they differ by no more than about 1%. At 30° they differ by 13%.

Often the top is clearly seen on seismic data, but the base is not. Then the isochore map is built from wells (with a geological idea: thinning on the crest of a growing structure, the direction of sediment supply), and the maps are combined: base = top + isochores, node by node. The errors add up: the error of the base is the error of the top plus the error of the thickness. Other maps are combined in the same way. Isopachs are divided by $\cos\delta_b$ to get isochores. Maps of effective thickness, porosity and oil saturation are multiplied into a map of hydrocarbon pore thickness, which is the net thickness times porosity times oil saturation (averaging and hydrocarbon pore thickness). A contact is overlaid on the map of the top to find the outline of the accumulation.

Wells of the lesson: what logging gave

Well: 1
Wall: footwall
Top, m: 1874.1
Base, m: 1911.8
Thickness, m: 37.7

Well: 2
Wall: hanging wall
Top, m: 1941.7
Base, m: 1979.0
Thickness, m: 37.3

Well: 3
Wall: footwall
Top, m: 1948.1
Base, m: 1991.4
Thickness, m: 43.3

Well: 4
Wall: hanging wall
Top, m: 1984.7
Base, m: 2027.1
Thickness, m: 42.4

Well: 5
Wall: footwall
Top, m: 1883.3
Base, m: 1906.3
Thickness, m: 23.0

Well: 6
Wall: hanging wall
Top, m: 1976.4
Base, m: 2018.6
Thickness, m: 42.2

Well: 7
Wall: footwall
Top, m: 1913.2
Base, m: 1954.9
Thickness, m: 41.6

Well	Wall	Top, m	Base, m	Thickness, m
1	footwall	1874.1	1911.8	37.7
2	hanging wall	1941.7	1979.0	37.3
3	footwall	1948.1	1991.4	43.3
4	hanging wall	1984.7	2027.1	42.4
5	footwall	1883.3	1906.3	23.0
6	hanging wall	1976.4	2018.6	42.2
7	footwall	1913.2	1954.9	41.6

Depths are below sea level. The thickness is the one penetrated vertically (the wells are vertical). The wells crossed the fault at these depths: well 1 at 1738.6 m; well 2 at 2368.5 m; well 5 at 1883.3 m; well 6 at 2106.8 m; well 7 at 1609.9 m (from the missing interval of the section). Well 5 penetrated the layer right below the fault. The data are assumed for illustration.

Worked exampleThe base from the top and isochores: the relief of the base is larger

Given (assumed data)

Well 1 (crest of the footwall): top 1874.1 m, base 1911.8 m
Well 3 (eastern limb): top 1948.1 m, base 1991.4 m; dip of the layer from the map 6.9°

Solution

Isochores: in well 1, $h_\text{TVT} = 1911.8 - 1874.1 = 37.7$ m. In well 3, $1991.4 - 1948.1 = 43.3$ m. On the crest the layer is thinner by 5.6 m.
Difference of depths between the wells: for the top, $1948.1 - 1874.1 = 74.0$ m, and for the base, $1991.4 - 1911.8 = 79.6$ m. The base difference is larger by the same 5.6 m: the base is bent more strongly than the top.
Isopach in well 3: $h = 43.3\cdot\cos 6.9° = 43.0$ m, which is 0.3 m less than the isochore.
If the base were built by adding a constant 40 m to the top, in well 1 it would lie 2.3 m deeper than the real one, and in well 3 it would lie 3.3 m higher.

Answer: The isochore is 37.7 m on the crest and 43.3 m on the limb. The relief of the base between the wells is 5.6 m larger than the relief of the top.

What it means. Thinning of the layer over a crest is a sign that the fold was growing while the layer was being deposited: the lower layers had time to bend more (growth of a structure on a flattened diagram). For volume, this means that the base cannot be found by shifting the top by a constant thickness. For a thin layer, an error of a few metres in the base changes the area where there is no water under the oil (Gross rock volume from maps). The difference between the isochore and the isopach is small at a gentle dip, and it is often neglected. On steep limbs it cannot be. The data in this example are assumed for illustration.

CalculatedIsochore map of the layer (every 2 m): over the crest the layer is thinner, and towards the limbs the thickness grows to 46 m. At the wells, the penetrated thickness is shown. In well 5, 23.0 m was penetrated against a block isochore of 37.3 m: the well crossed the fault inside the layer. The grey strip along the fault runs from the line where the top meets the fault plane in the footwall to the line where the base meets it in the hanging wall. Here a vertical well will meet an incomplete layer. In the middle part of the strip, where the throw is greater than the thickness of the layer, it will not meet the layer at all.

A fault on the map: cutoff lines and the polygon

The topic faults on maps and seismic data explains how a fault looks on a structure map. A normal fault is a strip with no layer, a reverse fault is an overlap, and contours break off and shift at the fault. Here we show how such a strip is built and what can be calculated from it.

The edges of the strip are the fault cutoffs, the lines where the layer meets the fault plane. In the footwall (the block below the fault plane), the layer is cut off by the fault plane from above. In the hanging wall (the block above it), the layer is cut off from below. Before the displacement, the two lines were one. Each point of the cutoff in the hanging wall is a point that moved down the fault plane, together with the hanging wall. So for a displacement along the dip, two points on opposite edges that face each other across the fault (along the line at right angles to it) differ exactly by the displacement vector. Horizontally they differ by the horizontal amplitude of the fault, the heave $w$ , and vertically by the vertical amplitude, the throw $t$ . The area between the lines is the fault polygon.

Throw, heave and the dip of the fault plane from a fault polygon

w = \Delta_\perp,\qquad t = z_\text{hw} - z_\text{fw},\qquad \mathrm{tan}\,\delta = \frac{t}{w}

where:

$\Delta_\perp$ width of the polygon, measured at right angles to the strike of the fault, m
$w$ heave (horizontal amplitude) of the fault at this place, m
$z_\text{fw},\ z_\text{hw}$ depths of the layer at the edges of the polygon in the footwall and the hanging wall, at points that lie opposite each other across the fault, along the line at right angles to it, m
$t$ throw (vertical amplitude), m
$\delta$ dip angle of the fault plane, °

Units: lengths in metres (on the ground, not on the map); angles in degrees

Holds when

The displacement is along the dip of the fault plane. For an oblique displacement, the “paired” points on the two edges are offset along the fault by the strike-slip component, and these paired points are the ones to compare (components of displacement).
The fault plane is planar (not curved) over the width of the strip.
The dip of the layers does not affect the relations: at the edges of the strip we compare one and the same point of the layer before and after the displacement (the derivation is in the advanced section).

A polygon is a ready measurement of the fault. It is narrow near the tips and wide in the middle, and it shows how the throw changes. The difference of the depths of the edges divided by the width gives the dip of the fault plane, even if it is not seen on seismic data. The depths of the edges are not the same as the difference of contours carried across the fault. That difference gives the vertical separation of the layer $s$ , which differs from $t$ for dipping layers (vertical separation of a layer).

CalculatedStructure map of the top of the layer (contours every 10 m), cut by a normal fault. The grey polygon is where the top is absent. The dashed outline is the polygon of the base for the same fault, shifted down the dip of the fault plane (to the south-west). The hanging wall west of the fault has dropped by 75 m in the middle of the fault. Towards the tips the throw falls to zero and the contours close. Dots are wells, the dotted line is the section line A–A′, and the frame is the part that is shown in the figure with the fault plane contours.

Worked exampleThrow, strip width and fault dip from the map

Given (assumed data)

Top on the line A–A′ (normal to the fault): the footwall ends at 1428.2 m from point A at a depth of 1868.9 m, and the hanging wall ends at 1369.6 m at a depth of 1943.9 m
Base at the same place: footwall at 1400.0 m, at a depth of 1905.0 m; hanging wall at 1341.4 m, at a depth of 1980.0 m
Isochore near the fault and the slope of the top across it: 37.3 m; the top rises in the direction of the dip of the fault plane with a slope of 0.041 (the layer dips towards the fault plane)

Solution

The heave is the width of the strip: $w = 1428.2 - 1369.6 = 58.6$ m.
The throw is the difference of the depths of the edges: $t = 1943.9 - 1868.9 = 75.0$ m. The hanging wall is lowered, so the fault is a normal fault.
Dip of the fault plane: $\mathrm{tan}\,\delta = 75.0/58.6 = 1.2799$ , so $\delta = 52.0°$ .
Base: $w = 1400.0 - 1341.4 = 58.6$ m and $t = 1980.0 - 1905.0 = 75.0$ m. These are the same, and the polygon is shifted by $1428.2 - 1400.0 = 28.2$ m down the dip of the fault plane.
Check of the shift by the formula: $\Delta q = 37.3/(1.2799 + 0.041) = 28.2$ m.

Answer: On the line A–A′ the fault lowers the hanging wall by 75.0 m, with a strip width of 58.6 m. The fault plane dips at 52.0°, and the polygon of the base is shifted by 28.2 m towards the hanging wall.

What it means. The throw and the dip of the fault plane were taken from one map, without seismic data and wells. The angle agrees with the one that fault points in wells will give (next section): two independent estimates agree. The shift of the polygon of the base is a little less than $h_\text{TVT}/\mathrm{tan}\,\delta$ , because the layer rises in the direction of the dip of the fault plane and meets it earlier. The data in this example are assumed for illustration.

Each horizon has its own polygon. The base meets the fault plane deeper than the top, which means farther down its dip. A simple relation shows by how much. It follows from the same section across the strike of the fault (the derivation is in the advanced section):

Shift of the polygon of the base relative to the polygon of the top

\Delta q = \frac{h_\text{TVT}}{\mathrm{tan}\,\delta \pm \mathrm{tan}\,\delta_b}

where:

$\Delta q$ horizontal shift of the cutoff of the base relative to the cutoff of the top, in the direction of the dip of the fault plane, m
$h_\text{TVT}$ vertical thickness of the layer at the fault, m
$\delta$ dip angle of the fault plane, °
$\delta_b$ dip angle of the layer in a section across the strike of the fault, °

Units: m; angles in degrees

Holds when

“+” if the layer dips towards the fault plane, and “−” if it dips the same way, as in the formula for the vertical separation of a layer in the topic on faults.
The layer and the fault plane are planar (not curved) over the distance $\Delta q$ , and the thickness is constant. The same holds for both walls.

For a thickness of 37 m, a fault plane dipping at 52° and an almost horizontal layer, the base lies 29 m away from the top, which is almost 3 mm on a map at 1:10,000. Between the cutoff lines of the top and of the base, the layer is incomplete in the vertical direction: in the footwall it is cut off by a wedge from above, and in the hanging wall from below. This matters for the rock volume near the fault, and for wells that are sited close to it.

In mapping programs, polygons are set as boundaries that interpolation does not cross. Each block is built from its own data, and inside the polygon there are no values. It is better to set the fault plane itself as a surface. Then the program finds the cutoff lines for each horizon itself, and the polygons of all the horizons agree. If the fault is not set at all, the grid will pass through it continuously, and a step of tens of metres will turn into a steep bend a few cells wide (computer contouring).

Fault plane contours: how the cutoff lines are found

A fault plane is a surface like the top of a layer, and it is drawn with contours too. This is the fault surface map. Its data are the depths at which wells crossed the fault (found from the missing or repeated interval, which is a part of the section that the well misses or meets twice because of the fault; a fault in a well) and the fault lines on seismic sections, converted to depth. The contours of a planar (not curved) fault plane are parallel straight lines spaced $\Delta z/\mathrm{tan}\,\delta$ apart. At a dip of 52° and an interval of 10 m, they are 7.8 m apart. So they are drawn on large-scale fragments or with a large interval.

The main rule of construction: a contour of the layer ends at the fault where it is crossed by the fault plane contour of the same value. At every point of the cutoff line, the layer and the fault plane are at one depth. So the point lies on the contours of both surfaces with the same value. The rule holds for both walls. The 1960 m contour of the footwall and the 1960 m contour of the hanging wall end on the same fault plane contour, but at different points of it, some distance apart along the fault. The distance between them is the familiar offset of contours at a fault.

Offset of like-valued contours along a fault

\Delta s = \frac{t}{\mathrm{tan}\,\delta_\parallel}

where:

$\Delta s$ distance along the fault between the points where the same contour of the layer ends in the footwall and in the hanging wall, m
$t$ throw at the place where the contour ends in the hanging wall, m
$\mathrm{tan}\,\delta_\parallel$ average slope of the cutoff line in the footwall along the strike of the fault over the stretch $\Delta s$ ; in practice, the apparent slope of the layer along the fault, —

Units: m

Holds when

The displacement is along the dip: a point of the hanging wall and its “pair” in the footwall lie opposite each other across the fault, along the line at right angles to it, and differ in depth by $t$ .
The layer is inclined along the fault. If it strikes parallel to the fault ( $\delta_\parallel = 0$ ), the contours are not offset but run along the strip, and we read the throw from the number of contours that are “missed” across it.

The larger the throw and the flatter the layer along the fault, the larger the offset. From the end of a contour in the hanging wall, we go across the strip to its “pair” in the footwall, which is $t$ shallower. Then we go along the cutoff line of the footwall until it sinks by these $t$ metres. In the lesson's example, the 1960 m contour ends in the footwall 646 m to the north-west of where it ends in the hanging wall. Where it leaves the hanging wall, the throw is 68.38 m, and the cutoff line of the footwall sinks towards the northern tip of the fault by 0.1058 m per metre on average: 68.38/0.1058 ≈ 646 m. The reverse problem, the throw from the offset of contours, is $t = \Delta s\,\mathrm{tan}\,\delta_\parallel$ .

Building a polygon from fault plane contours

1
Build the fault surface map
Use the fault points in wells and on seismic data, with the same contour interval as the map of the layer.
2
Continue the contours of the layer
In each block the contours are continued past the fault along the slopes of their own block, as if there were no fault.
3
Mark the crossing points
Mark the places where the continued contour of a block meets the fault plane contour of the same value. For the footwall we take its contours, and for the hanging wall, its own.
4
Join the points
The points of the footwall are joined into its cutoff line, and the points of the hanging wall into theirs. Between them is the polygon. The continuations of contours inside the polygon are erased.
5
Repeat for each horizon
The fault plane is the same, but the cutoff lines of the base and of other layers are their own.

CalculatedA close-up of the map near the northern part of the fault (the frame on the map above): contours of the top every 10 m and contours of the fault plane every 20 m. The 1960 m contour of the top in the footwall ends at the northern edge of the close-up, and in the hanging wall at the southern edge. Both ends lie on the 1960 m contour of the fault plane (bold dash). Inside the polygon, this contour joins them. The width of the strip, which is the heave, and the depths of its edges, whose difference is the throw, are shown across the strip.

Worked exampleA fault plane from three wells, and the edge of the polygon on a section

Given (assumed data)

Fault in wells (from the missing interval): well 1 (x = 2312.6 m; y = 1565.0 m) at 1738.6 m; well 6 (x = 1660 m; y = 2120 m) at 2106.8 m; well 7 (x = 2830 m; y = 870 m) at 1609.9 m
Points of line A–A′ at 1430 and 1420 m from A: x = 2226.0 and 2217.3 m, y = 1515.0 and 1510.0 m
Top of the footwall from the map at these points: 1868.95 and 1868.54 m

Solution

The three-point problem for the elevation of the fault plane $e = -z$ (three-point problem): the slopes of the elevation to the east and to the north are $g_x = 1.1086$ and $g_y = 0.6401$ .
$\mathrm{tan}\,\delta = \sqrt{g_x^2 + g_y^2} = 1.2801$ , so $\delta = 52.0°$ . The fault plane dips against the vector $(g_x;\ g_y)$ , towards an azimuth of 240°, and its strike is 150°. The contours of the fault plane every 10 m are $10/1.2801 = 7.8$ m apart.
Depth of the fault plane at the points of line A–A′: $z = 1738.6 - g_x(x - 2312.6) - g_y(y - 1565.0)$ . At 1430 m from A it is 1866.61 m, and at 1420 m it is 1879.46 m.
The footwall exists as long as its top is below the fault plane. The difference “top minus fault plane” is $1868.95 - 1866.61 = 2.34$ m and $1868.54 - 1879.46 = -10.92$ m. The sign has changed. By linear interpolation, the edge is at $1430 - 10\cdot 2.34/(2.34 + 10.92) = 1428.2$ m from A.

Answer: The fault plane dips at 52.0° towards an azimuth of 240°. The top of the footwall meets it on line A–A′ at 1428.2 m from point A, which is the same place where it ends on the map.

What it means. The angle of the fault plane from three wells agrees with the angle taken from the polygon in the previous example. And the edge built from the fault plane contours agrees with the edge on the map. So the fault surface map and the maps of the walls are consistent. Three points define a plane, but they do not check it. The check is a fourth well or seismic data. Real fault planes are curved: listric normal faults flatten with depth, and segments are joined by steps. From many points the fault plane is mapped in the same way as a layer. The data in this example are assumed for illustration.

Throw along a fault, and matching maps, sections and wells

We check a fault on a map with the throw profile. We read the throw off the polygon at equal distances along the fault and draw a graph. For a single fault it is smooth, with the largest value near the middle and zeros at the tips (displacement and throws), and the sign of the throw does not change.

CalculatedThrow along the lesson's fault: the points were taken from the map every 200 m as the difference of the depths of the edges of the polygon of the top. The line is the model (the base has the same values). The dashed line is the thickness of the layer at the fault. Farther than 0.8 km from the middle, the throw is smaller than the thickness. There the layer is partly matched with itself: “sandstone-to-sandstone” windows.

A jump in throw is an error in construction, a junction of two fault segments, or a crossing with another fault.
A change of sign along one fault almost always means an error. Most likely there are two different faults.
Different horizons have a similar throw if the fault formed after them. If it grew while sediments were being deposited (a growth fault), the throw grows with depth, and the thicknesses in the hanging wall are greater (growth faults).
Merging faults: the throw of the common continuation is about equal to the sum of the throws of the branches at the junction.

The throw profile at once shows where a layer is matched with itself across the fault. Where the throw is smaller than the thickness of the layer (in the lesson's example, farther than 0.8 km from the middle of the fault), “sandstone-to-sandstone” windows remain. That is where the fault is checked for sealing (juxtaposition of layers and SGR). In the middle of the fault the throw of 75 m is twice the thickness. The layer of the walls rests entirely against impermeable rock, if there is any there.

What is checked when maps, sections and wells are matched

What: Section and map
Condition: the same crossings with the contours, the same position and throws of faults, and angles corrected for $V$ and the direction
How: read them off the map and off the section

What: Polygons of horizons
Condition: one fault plane; deeper horizons are shifted down the dip by $\Delta q$ ; the throw is almost the same unless the fault grew during deposition
How: overlay the maps

What: Throw profile
Condition: smooth, without a change of sign, zero at the tips; at junctions, a sum
How: a graph of $t$ along the fault

What: Isochore map
Condition: $h_\text{TVT} \ge 0$ ; the zero isochore runs between wells with the layer and without it; wells near faults do not distort the map
How: the difference of the grids of the base and the top

What: Wells
Condition: the picks are on the maps; the depth of the fault is on the fault surface map; the missing interval equals the separation $s$
How: a table of discrepancies

What	Condition	How
Section and map	the same crossings with the contours, the same position and throws of faults, and angles corrected for $V$ and the direction	read them off the map and off the section
Polygons of horizons	one fault plane; deeper horizons are shifted down the dip by $\Delta q$ ; the throw is almost the same unless the fault grew during deposition	overlay the maps
Throw profile	smooth, without a change of sign, zero at the tips; at junctions, a sum	a graph of $t$ along the fault
Isochore map	$h_\text{TVT} \ge 0$ ; the zero isochore runs between wells with the layer and without it; wells near faults do not distort the map	the difference of the grids of the base and the top
Wells	the picks are on the maps; the depth of the fault is on the fault surface map; the missing interval equals the separation $s$	a table of discrepancies

Fluid contacts in one block are usually horizontal, and in different blocks they may differ if the fault separates them. A tilted contact needs an explanation (tilted contacts).

Worked exampleWell 5 near the fault: a thin layer or a cut one?

Given (assumed data)

Well 5 (x = 2204.7 m; y = 1525.8 m), footwall right next to the fault: the layer from 1883.3 to 1906.3 m; above the top, correlation shows that part of the seal is missing
Neighbouring wells 1 and 2: isochores 37.7 and 37.3 m
Maps at the point of well 5: the top of the footwall, continued to the well, is 1869.1 m; the top of the hanging wall, continued to the same place, is 1946.8 m; the fault plane is at 1883.3 m; the isochore of the footwall is 37.3 m
The fault here, from the throw profile: $t = 75.0$ m, $w = 58.6$ m; over a stretch of length $w$ in front of the well, the top dips towards the fault plane at 2.75°

Solution

Penetrated thickness: $1906.3 - 1883.3 = 23.0$ m, which is 14.3 m less than the isochore of 37.3 m.
The top of the layer in well 5 coincides with the depth of the fault plane: the well entered the layer straight from the fault. The cut-off part of the layer is $1883.3 - 1869.1 = 14.2$ m. Together with the penetrated part, $14.2 + 23.0 = 37.2$ m, which is the isochore to within rounding.
How much of the section should be missing: the vertical separation of the layer is $s = 1946.8 - 1869.1 = 77.7$ m. By the formula $s = t + w\,\mathrm{tan}\,\delta_b$ (vertical separation of a layer) it is practically the same: $75.0 + 58.6\cdot\mathrm{tan}\,2.75° = 77.8$ m (the difference comes from rounding).
Correlation with well 1 shows that about 78 m of the section is missing at a depth of 1883.3 m. The maps and the well agree.

Answer: The 23.0 m in well 5 is not thinning of the layer but a cut by the fault: the isochore there is 37.3 m, and about 78 m of the section is missing at the fault.

What it means. For the isochore map we take 37.3 m (or we do not use well 5). Otherwise a false “bull's eye” of a thin layer will appear along the fault. For rock volume, on the other hand, the cut-off wedge matters. Vertical wells in a strip about 87 m wide along the middle of the fault penetrate an incomplete layer, and in its middle part, where the throw is greater than the thickness, they do not penetrate it at all (the grey strip on the isochore map). The agreement of the missing interval with the vertical separation from the maps is one of the best checks that the fault surface map and the maps of the walls are consistent. The data in this example are assumed for illustration.

Terms and symbols

Map-derived cross-section, section line: A vertical section along a line (A–A′), built by carrying over the crossings of the line with contours. Angles on it are true only across the strike and without scale distortion.
Vertical exaggeration $V$: How many times the vertical scale of a section is larger than the horizontal one. The tangents of all slopes are multiplied by $V$ .
Isochore map and isopach map: Maps of vertical thickness ( $h_\text{TVT}$ , the difference between the depths of the base and the top) and of true thickness ( $h = h_\text{TVT}\cos\delta_b$ ).
Map combination: Getting a map from other maps. Examples: base = top + isochores; isochores = isopachs / $\cos\delta_b$ ; hydrocarbon pore thickness as the product of maps.
Fault cutoff: The line along which a layer meets the fault plane in one wall. It is the edge of the fault polygon.
Fault polygon: The area between the cutoff lines of a horizon in the two walls. For a normal fault the horizon is absent there.
Fault surface map: A structure map of the fault plane, made from fault points in wells and from seismic data.
Throw profile: A graph of the throw along the fault.

Common mistakes and misconceptions

Wrong

“The dip angle can be read from any section.”

In fact

The true angle is seen only on a section across the strike and without scale distortion. On an oblique section the angle is smaller, and on a section with exaggeration $V$ the tangent is larger by a factor of $V$ .

Wrong

“The map of the base is the map of the top plus the isopach map.”

In fact

We must add isochores, which are vertical thicknesses. Isopachs are smaller by a factor of $1/\cos\delta_b$ , and depth differences from deviated wells are distorted by the direction of the hole.

Wrong

“A fault on a map is a line.”

In fact

A normal fault on a structure map is a strip between the cutoff lines of the layer, and a reverse fault is an overlap. The width of the strip is the heave. If the fault is not set, the grid will smear the step into a bend.

Wrong

“One fault polygon is good for all horizons.”

In fact

Each horizon has its own cutoff lines. A deeper horizon meets the fault plane farther down its dip, and its polygon is shifted by $h_\text{TVT}/(\mathrm{tan}\,\delta \pm \mathrm{tan}\,\delta_b)$ .

Wrong

“A thin layer in a well near a fault is a thinning of facies.”

In fact

First we check whether the well crossed the fault inside the layer: the depth of the fault from the fault surface map, the missing interval, a comparison with the neighbours. A cut-off thickness is not plotted on the isochore map.

In depth: advanced detail

Why the strip width equals the heave at any dip of the layers

Take a point $P$ on the cutoff line of the footwall. Before the displacement, the layer of both walls and the fault plane met at it. When the displacement happened along the dip, the hanging wall moved along the fault plane by a vector with a horizontal component $w$ (at right angles to the strike, in the direction of the dip) and a vertical component $t$ . The piece of the layer of the hanging wall that was at $P$ ended up at point $P'$ . It is still on the layer of the hanging wall and still on the fault plane, because the displacement vector lies in its plane. So $P'$ is a point of the cutoff line in the hanging wall. The distance from $P$ to $P'$ is $w$ horizontally and $t$ vertically, whatever happens to the dip of the layer.

The difference of depths of the two walls at one point of the map behaves differently. This is the vertical separation of the layer $s$ . It compares different pieces of the layer, and so it depends on its dip: $s = t \pm w\,\mathrm{tan}\,\delta_b$ (on faults). On a vertical section at an angle $\beta_F$ to the strike of the fault, the strip looks wider, $w/\sin\beta_F$ , and the fault plane looks flatter: $\mathrm{tan}\,\delta' = \mathrm{tan}\,\delta\sin\beta_F$ . So we read the throw and the dip of the fault plane along the line at right angles to the fault, comparing the “paired” points.

Where the shift of the polygon of the base comes from

Let us draw a section across the strike of the fault. The axis $q$ is horizontal, in the direction of the dip of the fault plane, and depths are down. The fault plane is $z = z_0 + q\,\mathrm{tan}\,\delta$ . The top of the layer is a straight line $z = a + g\,q$ , where $g$ is its slope along $q$ : $g = \mathrm{tan}\,\delta_b$ if the layer sinks the same way as the fault plane dips, and $g = -\mathrm{tan}\,\delta_b$ if it dips towards it. The base is the same line, lower by $h_\text{TVT}$ .

The top meets the fault plane at $a + g\,q_\text{top} = z_0 + q_\text{top}\,\mathrm{tan}\,\delta$ , and the base at $a + h_\text{TVT} + g\,q_\text{base} = z_0 + q_\text{base}\,\mathrm{tan}\,\delta$ . Subtracting, we get $q_\text{base} - q_\text{top} = h_\text{TVT}/(\mathrm{tan}\,\delta - g)$ , that is, $\Delta q = h_\text{TVT}/(\mathrm{tan}\,\delta \pm \mathrm{tan}\,\delta_b)$ , with “+” for layers that dip towards the fault plane. The hanging wall is the same layer moved by $(w, t)$ , so its shift is the same. When the layer dips the same way as the fault plane, and almost as steeply, the denominator goes to zero. The layer runs along the fault plane, and the cutoff lines are far apart. If the layer is steeper than the fault plane, the order of the lines changes, as does the rule of the missing section in the topic Faults.

Connections to other topics

Structure maps and contouring — data for structure maps, contouring and gridding, kinds of maps.
Faults — types of faults, throws, the missing interval in a well, matching of layers and sealing.
Stress, strain and structural analysis — apparent dip, a section with a distorted scale, the three-point problem.
Well correlation — consistent picks and thicknesses in deviated wells.
Gross rock volume from maps — rock volume from maps of the top and the base, outlines of the oil-bearing area.
Traps and timing — traps at faults, closure and tilted contacts.
Seismic interpretation — following horizons and faults on seismic sections and converting them to depth.
Practical structural geology — recognising structures and balancing sections.

Check yourself

1
On a section across the strike of a fault, drawn with the vertical scale enlarged 4 times, the fault plane is drawn at an angle of 78°. What is its true dip?
1. A.78°
2. B.about 85°
3. C.about 50°
4. D.about 20°
Show the answer
Correct answer: C. about 50°
The tangent of the drawn angle is $V$ times the true one: $\mathrm{tan}\,\delta = \mathrm{tan}\,78°/4 = 4.705/4 \approx 1.18$ , so $\delta \approx 49.6°$ . We must divide the tangents, not the angles: 78°/4 = 19.5° is a gross error.
2
The polygon of a normal fault on a structure map is 40 m wide, measured at right angles to the fault. The top at its edges, opposite each other, is at depths of 2236 and 2296 m. Find the throw, the heave and the dip of the fault plane. Which wall has dropped?
Show the answer
The heave is the width of the strip, $w = 40$ m. The throw is the difference of the depths of the edges, $t = 2296 - 2236 = 60$ m. Then $\mathrm{tan}\,\delta = 60/40 = 1.5$ , and $\delta \approx 56.3°$ . The edge at a depth of 2296 m belongs to the hanging wall. It has dropped, so this is a normal fault. The dip of the layers does not affect the answer.
3
The top of a layer is reliably mapped from seismic data, and the base is not. How do we get a map of the base?
1. A.Add an isopach map to the map of the top
2. B.Add to the map of the top an isochore map, built from the vertical thicknesses in wells
3. C.Add the vertical depth differences from all wells, including deviated ones, without recalculation
4. D.Shift the map of the top by the average thickness of the layer
Show the answer
Correct answer: B. Add to the map of the top an isochore map, built from the vertical thicknesses in wells
The base lies below the top by the vertical thickness, which is the isochore. Isopachs are smaller by a factor of $1/\cos\delta_b$ . Depth differences in deviated wells are distorted by the direction of the hole. A constant shift does not take account of the change of thickness: in the lesson's example, the relief of the base is several metres larger than the relief of the top.
4
A normal fault dips at 60°, the vertical thickness of the layer is 40 m, and the layer is almost horizontal. By how much, and in which direction, is the polygon of the base shifted relative to the polygon of the top?
Show the answer
$\Delta q = h_\text{TVT}/\mathrm{tan}\,\delta = 40/\mathrm{tan}\,60° = 40/1.732 \approx 23$ m. The shift is horizontal, in the direction of the dip of the fault plane, which is towards the hanging wall. Between the cutoff lines of the top and the base the layer is incomplete in the vertical direction.
5
Two normal faults merge into one. Just before the junction their throws are 25 and 40 m. What throw should we expect on the common fault just beyond the junction?
1. A.40 m
2. B.about 65 m
3. C.32.5 m
4. D.about 15 m
Show the answer
Correct answer: B. about 65 m
The displacements of the branches add up. The common fault carries what two branches carried before, about 25 + 40 = 65 m. If the map shows something clearly different, there is an error in the construction of one of the branches or at the junction.
6
In a well near a fault the layer was penetrated with a thickness of 19 m, and in the neighbouring wells it is 44–46 m. About 50 m of section is missing above the top of the layer. What do we do with this thickness when we build the isochore map?
Show the answer
This is a cut by the fault: the well entered the layer straight from the fault, and the upper part of the layer stayed on the other side of the fault plane. We do not plot 19 m on the isochore map (a false “bull's eye” would appear). We take the isochore of the block from the neighbours. We check the depth of the fault against the fault surface map, and the missing interval against the vertical separation of the layer from the maps. For the volume, the strip of incomplete layer along the fault is counted separately.

Updated: 16 October 2026