GEOL 4334
Lab 05: Geologic Maps II: Structure Contours & 3-Point Problems
v. 2015
Lab 5: Geologic Maps II: Structure Contours & 3 Point Problems
Objectives:
Ø
Ø
Construct and evaluate structure contours on various geologic features
Calculate the orientation of a planar geologic surface using a variety of 3 point methods
Materials:
tracing paper, pencils, ruler, protractor, divider, trig-function calculator
Structure Contours
A structure contour is an imaginary line that connects points of equal elevation (a contour) on a
structural surface such as a fault, the top of a stratigraphic bed, or a buried erosional surface (unconformity).
Structure contour maps can be thought of as a “topographic” map that contours some structural or
sedimentological surface, just as a real topographic map contours the earth’s surface.
Figure 1 A (above) shows a shaded oblique perspective of an inclined plane with a NW strike and a 75°
dip to the SW. Structure contours are drawn at 100’ intervals (dashed white lines). Figure 1B depicts the map
view of the structure contours. Note how the star symbols on Figure 1B correlate with the stars as projected
from the 400’ contour in Figure 1A.
Structure contour maps are often used in the petroleum industry to describe sub-surface structures
where drill hole data is available, but only limited or no ground surface structural data is available. Examples
might include domal petroleum traps or tabular aquifers.
The data used to construct a structure contour map can be presented in terms of elevations above (or
below) sea level or in terms of depths below the ground surface. You must always specify the datum plane
on every structure contour map. Note that if the ground surface is not horizontal (if you are drilling on a
sloping hill side, for example) and the data are provided as depth below ground surface, a correction must be
applied so that all points represent depth below the same horizontal datum plane. Sea level is a common datum
plane for most contour maps.
Structural Analysis of Hydrocarbon Systems Lab • Page 1 of 5
GEOL 4334
Lab 05: Geologic Maps II: Structure Contours & 3-Point Problems
v. 2015
Study Figure 2 to help visualize
topographic contours, the “rule of V’s”,
and structure contours. In the top block
diagram (Fig. 2A), a transparent layer
about 100’ thick is gently inclined to the
north. Note how the trace of the planar
unit “v’s up the drainage”, indicating that
it dips “up the drainage”. Compare Fig.
2A and 2C to help visualize the shape of
the contact.
The black solid line in Fig. 2A
represents the trace of the “top” of the
planar unit where it crosses the 300’
topo contour. The dashed black line
connects the two localities where the
planar surface intercepts the 400’ topo
contour interval (Fig. 2A). Hence, these
two lines represent structure contours
on the inclined unit. Note that these
lines are parallel to the strike of the
plane – i.e., they are horizontal lines
contained within an inclined plane! Yay!
Using the quadrant convention, what’s the strike of the unit?
Figure 2B displays a cross
section drawn parallel to X-X’. Visualize
the N-dipping plane in 3D. Each black
dot on the topographic profile represents
the point where the topography
intersects the map contour interval.
Figure 2C is a map view of the
trace of the inclined plane on a shadedtopographic relief map (notice the “v”
pattern) and structure contours drawn
every 20 feet. Each white circle
represents a point where the structure
contour intersects the outcrop trace of
the unit. Also shown is a topographic
profile and the inclined bed (Fig. 2D).
3-point problem teaser…
Notice that it’s possible to solve for the
dip, δ, using trigonometry:
Tan δ = Δ elev. (ft)/map distance (ft) –
see if you can “see” this relationship in
Fig. 2C and D.
Cool!
Figure 2. Three views of structure
contours.
Structural Analysis of Hydrocarbon Systems Lab • Page 2 of 5
GEOL 4334
Lab 05: Geologic Maps II: Structure Contours & 3-Point Problems
v. 2015
Can you contour non-planar surfaces?
Fig. 3
Non-planar structures such as folds and domes may also be contoured. In the above diagram (Fig. 3A)
a south-plunging anticline is contoured in feet below the local land surface. (Do you remember what plunging means?) The left
hand diagram shows an oblique perspective; Figure 3B shows the 2-D contour map. Note the plunging anticline
symbol.
Structure Contour – a line of equal elevation on a continuous or projected structural surface.
Structure Datum – Stratigraphic or structural surface (sometimes called a horizon) on which the contours are
drawn (e.g., the top of a reservoir, unconformity, contact, fault, etc.).
Structural Elevation – Elevation on the datum above or below sea level (or depth below Earth’s surface).
NOTES:
Structural Analysis of Hydrocarbon Systems Lab • Page 3 of 5
GEOL 4334
Lab 05: Geologic Maps II: Structure Contours & 3-Point Problems
v. 2015
3 Point Problems in Detail
Fig. 4A depicts three wells drilled at an elevation of 1000’ above sea level. Each well intersects an inclined fault
plane at a different depth below the local ground surface. What is the orientation of the fault plane in the
subsurface? Knowing how to calculate the orientation of the fault
plane will allow you to then predict where the fault plane would
map surface at 1000 ‘ above sea level
project further into the subsurface. Thus, you will have a better
understanding of the 3D geometry of rock bodies, reservoirs and
well 1
other features in the un-exposed subsurface.
well 2
A
2 900’
700’
line 23
3
2
1000’
800’
0
500 ft.
3
V=H
-200’
D
2 900’
1
0’
70 00’
6 00’ ’
5 00
4 00’
3 0’
20 00’
1
ft.
700’
L
eS
ov
ab
REMEMBER!
vertical depth’s to
inclined fault plane
400’
2
L
3
0’
70 00’
6 00’ ’
H
5 00
=
V
4 00’
3 0’
20 00’
1
vertical depth’s to inclined fault plane
Note that the depth to 1’ in
Figure 4C is 700’; in other words a
well drilled at point 1’ on line 23
would intersect the fault plane at an
elevation of 300’ above sea level.
Method 2: calculate the difference between depthto-fault between middle well and shallow well, and,
the deepest well and shallow well. The ratio
represents the distance along line 23 to point 1’:
Intermediate well – shallow well = 700’ – 400’ =
300’
Deep well – shallow well = 900’ – 400’ = 500’
th
Therefore, point 1’ is 3/5 ’s of distance from well 3
along line 23. Use the map scale to measure this
distance.
ft.
eS
ov
b
a
3
500 ft.
Method 1: Draw a profile plane as in
D with the same scale as the map;
draw a line that connects the depth to
fault in wells 2 and 3. Then, locate
the depth to fault in well 1 along that
line (i.e., 700’, or 300’ on the profile).
Project that point up to line 23 (Fig.
D).
200’
SL
0
Step 2 (Fig. 4D): How to find point 1’?
600’
400’
400’
400’
well 3
700’ (Fig. 4C). Once you find point 1’
on line 23, you have established the
strike line fo the plane.
C
1
N
Overlay tracing paper and precisely trace the well locations, map
scale and use a ruler to connect the wells. The triangle 123 in Figure
B essentially represents the fault plane, with each apex of the
triangle at different depths. We need to find a line of equal elevation
– i.e., a structure contour or a strike line - contained in this plane.
Distinguish the shallowest and deepest wells (e.g., well 2 and well 3,
respectively). Somewhere along the line 23 is point 1’ at a depth of
B
900’
700’
Step 1 (Fig. 4B): Visualize and sketch the problem.
Figure 4A-D.
Structural Analysis of Hydrocarbon Systems Lab • Page 4 of 5
GEOL 4334
Lab 05: Geologic Maps II: Structure Contours & 3-Point Problems
Step 3 (Fig. 4E): Draw structure contours.
E
Connect points 1 and 1’ and you’ve just drawn a structure contour at a
depth of 700’. Complete the map by drawing a contour at 100’
intervals.
v. 2015
2
1
700’
800
700
Step 4: Calculate the strike of the fault plane using either azimuthal or
quadrant conventions.
900’
900
’
’
’
400’
’
’
400
3
N
Step 5 (Fig. 4F): Define dip direction & calculate the dip angle, δ.
600
500
0
500 ft.
’
Measure the map distance (MD) between two structure contours using
the map scale. Note the vertical distance between the two structure contours (Δelev.). Using trig we know that
-1
-1
δ = tan (Δelev /MD) = tan (200’/400’) = 27°
ft.
Therefore, the orientation of the fault plane is
N20°E/27°NW. Be able to visualize the fault plane in
the resulting structure contour map (Fig. 4G).
tal
δ
700’
zon
dire
900’
hori
2
1
dip
0
ctio
n
500
F
400’
900
800
’
’
well 2
600
’
700
’
fold
500
well 3
40
27° 0’
Figure 4. E-G.
NOTES:
Structural Analysis of Hydrocarbon Systems Lab • Page 5 of 5
’
N
line
N
3
500 ft.
well 1
400
300 ’
200 ’
100 ’
’
SL
-100
-200 ’
’
0
G
0
500 ft.