Lab 2 - Greek Hills map area

400 m

450m

*
30 0 m

500m

Q.1.

Recognising hills on a topo map The hills are shown by the two areas enclosed or partly-enclosed by contours with elevations that increase inwards (*).

350m

Q.2.
20 0 m

250

150m

100m

3 00m

Identifying unconformities on maps The identification of unconformities (abbreviated U/C) on maps will be discussed using cross-sections as examples to highlight a number of geometric relationships, but these points are just as applicable to their recognition on maps.

The simplest U/C is termed a disconformity, where the discontinuity and bedding both above and below, are parallel. This type cannot be recognised on a cross-section or geological map solely on the basis of it's relationships with adjacent units. The example given below left shows a parallel sequence of rocks representing time periods A to I with the exception of F. In this case therefore, the contact E-G representing the period of missing geological time is a disconformity. Note that this would be impossible to identify if the rocks were not labeled with their true ages, or symbols given for unconformities.

Angular unconformities can however be identified on the basis of the geometric relationships between it and adjacent rock units. This is due to the contrast between the concordant (parallel; GI) and discordant (non-parallel; A-E) orientations of bedding across it (above and below respectively). Furthermore, the relative ages of the two different sets of rocks can be established by looking at their geometry relative to the U/C. The series of rocks A-E are recognised as older, partly because they are structurally below the overlying sequence (but this is not always the case, if later tilting, folding or faulting has occurred), but principally because the layers and its contacts are NOT parallel to the U/C. This is shown by the bedding planes which when traced up-dip (shown with arrows) terminate or are truncated abruptly at the former erosion surface (hollow circles). In contrast, bedding above is parallel to the U/C so that G is the only rock of the younger series adjacent to it. Also note how H and I are a constant stratigraphic distance from U/C.

40 0m

*
500

I H G E D C B A

bedding younger than U/C

I H G E D C B A

}
disconformity

bedding older than U/C

} }

bedding younger than U/C bedding older than U/C

angular unconformity (U/C)

The outcrop of the U/C is identified where a series of stratigraphic contacts are 'lost' or 'disappear' when traced across the map. Following certain bed contacts (highlighted by arrows), these are seen to stop abruptly and can be traced no further (shown by the circles). The line on which these points occur is the U/C shown as thicker black lines. On the other hand, also note how the conglomerate is always in contact with the U/C irrespective of where the latter is found on the map.

Having identified the unconformity and seen the bedding relationships on either side, it should now be easy to determine which beds are above (horizontal ornamentation) and which are below the U/C (diagonal pattern) .

Q.3. 1.)

Attitude of bedding beneath the unconformity

Find a SINGLE stratigraphic contact which a.) crosses the same contour in two or more different geographic locations, and b.) crosses a contour (or contours) of a different elevation (or elevations at least once (but preferably more). Tip: choose a contact where the crossing points between contacts and contours (or contour intersections - C.I.) for certain elevations are as abundant and widely spaced as possible. This will make the answers more accurate.

The example provided uses the contact between a limestone and a shale (thick black line). All possible contour intersections are highlighted as circles and labeled with their respective elevations. Note that no points were taken from the other shale/limestone contact or the top of the shale (marked with Xs) since they are different planes. The reasons for this being chosen are a.) the number of different contour lines that it crosses (3; 300m, 350m & 400m) and b.) its widespread distribution on the map (note spacing of C.I.s for 350m).

500m
400 m
4 50m

X X
350m

350m 300m
250 m

300

X X 300m

200

150m

100m

X
30 0m

40 0m

350m
500 m

X 400m X

500m
400 m
4 50m

2.)

350m
350m

300m
250 m

30 0

2 00

s.

300m
35 l.

150m

0m

100m

30 0m

45E

Strike lines can now be drawn for this surface showing how the elevation of this contact varies across the map. The best estimate is from the 350m C.I.s because of the large distance between them. Note that a 400m s.l. would not be possible to determine by itself since there is no second C.I. for that elevation. LABEL ALL STRIKE LINES with their respective elevations (and if dealing with more than one particular feature, then also labels distinguishing one set from another, especially if they overlap on the map). Only one strike line is needed to determine the trend of the strike. Measure the clockwise angle from N for the azimuth. Strike = 045.
upper s.l.
dip angle

l.

s.

0m

30

350m

40

0m

s.

l.

Dip is analogous to gradient in that it describes the slope of a surface except that the former is simply stated as an angle measured from the horizontal. Like gradient, the graphical determination of dip using descriptive geometry involves the use and proportions of horizontal and vertical distances. In the following procedure, the horizontal distance is given by the spacing or separation of strike lines, while the vertical distances are defined by the differences in elevation also given by the strike lines.
lh s

35 0m

s. l.

s. l.

30 0m

SE

4.)

Like a cross-section across a map, the dip line construction (or profile) has two mutually opposite bearings, NW and SE; and two vertical directions, Up and Down. The two compass bearings are obviously constrained by the trend of the line ON the map. The vertical sense of direction is decided by the individual PROVIDED THAT up and down are ALWAYS CONSISTENT. In this example, 'up' relative to the folding line is towards the top right-hand corner of the page, and vice versa.

40

rh

0m

s. l.

40 0m

3.)
500 m

400m

n p ve a gi n m ce s o an .l. st f s di o al n nt atio o r riz pa ho s e by vertical distance defined by difference in s.l. values

lower s.l.

3.)
N W

U
f.l .( 40

(map of strike lines, with other info removed)

0m )

Draw a line perpendicular to the strike lines. This is a folding line (f.l.) which may be visualized like a cross-section or profile across a map except we are only concerned with strike lines and not topography, etc. The folding line also doubles up as a horizontal datum (or line of constant elevation) as in a x-section. Assign the folding line a suitable elevation. It is generally better (though not essential) to label with the highest elevation in use (f.l. (400m)).

5.)
lh s =

W U

0m

10

f.l

s. l.

50

0m

30

dip angle

rh

SE

6.)

The dip is given by the acute angle between the horizontal, i.e. f.l., and the dip line, = 16E. The dip direction can be determined in one of two ways. Firstly, by finding the direction of the profile or f.l. towards which the dip line is inclined or gets progressively lower. In the above example, the dip line is inclined DOWN TOWARDS the lhs of f.l., that is, to the NW. Alternatively, since down dip results in a decrease in the elevation of the feature in question, the dip can simply be determined from finding the geographic direction in which the strike lines elevation's decrease, i.e. NW. The latter has the advantage of not needing a folding line/dip line construction, simply inspection of the values of the strike lines. Full answer: 16ENW/043. Attitude of bedding above the unconformity The orientation of the younger series is shown by both the unconformity itself, and the contact between the conglomerate and the overlying sandstone. At all places within the map, the unconformity is generally half way between the 400m and 450m contours irrespective of the orientation of the topographic slope. This is also shown by the overlying contact whose elevation appears to be about 475m. Only one orientation is possible where the whole plane is at the same elevation, that is horizontal.

Q.4.

40

0m

s. l.

35

d.l.

0m

s. l.

.( 40

0m

Where the 400m strike line intersects the folding line, it can also be located with regard to the profile at the same point since it is at exactly the same elevation as that designated for the folding line. Other strike line elevations are lower however, so that when they intersect the profile line, they must be marked 'below' f.l. at their appropriate elevation. Therefore, the position of the contact at 350m and 300m elevations have been located 50m and 100m LOWER respectively on the profile (shown as black, open circles). The points on the profile form the dip line (d.l.) of the contact.

Q.6. Recognising oldest and youngest lithologies on map The dip of the older rocks has already been established in Q.3. as being to the NW. From the decreasing contours, the valley however slopes to the ESE. Therefore, the beds dip in the opposite direction to the slope of the valley (case B on p.9). Viewing a profile of the valley and the rocks that strike across it (see right), it should become apparent that the oldest rocks are found in the lowermost reaches of the valley, i.e. in the lower shale. However, even within the shale, it is possible to find the oldest part within the shale by finding the part that is stratigraphically the furthest from the top of the unit (highlighted in black). The increase in age within the shale is shown schematically with arrows pointing away from the upper contact, 'converging' towards 'O'. Since the younger series is still horizontal, it follows from the 'layer-cake' stratigraphy that the youngest rocks will occur at the highest elevations. These are given by the two hills previously identified in Q.1., and are now labeled 'Y'.

WNW

ESE dip direction increase in age slop e of seen at surface valle y

you

nge

st

old

est

cross-section/ profile of valley

increasing age of rocks

500m
4 00 m

Y
450m

35 0m
300 m

250

200

150m
100m

O
40 0m
300m

50 0

Q.6.

Determining thicknesses Thicknesses for the horizontal sequence is simplest to determine since their thickness is measured vertically (perpendicular to horizontal). This can therefore be calculated using the difference in elevations between the top and bottom of each unit. An estimated elevation of 425m for the base of the conglomerate is reasonable since it outcrops about halfway between the 400m and 450m contours. The top, likewise, can be estimated to be at about 475m elevation. Thickness of conglomerate ~50m. The sandstone's base equates with the top of the conglomerate at about 475m. The top of the sandstone does not outcrop since it has been eroded. The map does not provide any information on how much as been eroded away. The elevation for the top is unknown, and so only a minimum thickness can be determined. The map does however show that sandstone is present above 500m giving a 25m thickness as an absolute minimum. A variation is to estimate the highest topographic elevation, e.g. 525m, to provide a more 'realistic' minimum which in this case would be ~75m. This answer although probably closer depends on the reliability of the estimate, whereas the former is more conservative but based on firmer logic.

Folding line/dip line constructions are necessary to determine the thicknesses for the inclined units. To avoid repetition, only a few examples will be worked through. Case 1: layer with one strike of same elevation for top and bottom each, e.g. sandstone The adjacent diagram shows the two contacts highlighted and labeled. Note that they are on opposite sides of the sandstone, and in contact with two different units. 1.) From the map, it is possible to find three points where the top is at 400m (t400), thereby giving a 400m strike line for the top (t 400m s.l.). 2.) Likewise, the lower contact with the shale is also found in three separate place to be at 400m as well (although it can be as low as one).
500m

40 0
top

m
450 m
t400

top
thic kne ss

bottom

b400
350 m

f.l
bo tto

.4

00 m

s. l.

3 00
E
25 0 m

00

16

20 0

t4

t400 t400

150m

s. l.

b400
3 00 m

b400
5 00 m

It is now possible to draw a profile across the sandstone by using a folding line/dip line construction. 3.) Draw the folding line perpendicular to the strike lines. This is the trend of the profile. Label and assign it the same elevation as that of the strike lines - f.l. 400m. 4.) Where the strike lines intersect f.l., measure down the dip angles (16E) so that the dip lines are inclined downwards relative to the NW end of f.l. Note that the dip lines must be parallel. If there was any uncertainty as to which contact was the top and vice versa, there should no longer be if the construction is done correctly (dip lines labeled). 5.) Measure the thickness perpendicular to the dip lines, NOT perpendicular to the folding line.

40

0m

100m

40 0m

Case 2: layer where one side not observed (only minimum thickness), using same elevation strike lines, e.g. upper limestone 1.) The base (= top of previous sandstone) is known to occur at 400m at three places (b400), giving a 400m strike line (b 400m s.l.). 2.) The youngest part of the sandstone is by the NW corner of the map since it is 'furthest' stratigraphically from the base. The elevation at this point is unknown, but there is a point nearby within the sandstone which is known for certain to occur at 400m (t400?). This will be used as a minimum, and the problem is then solved the same way as the previous case, with the exception that the final answer will be a minimum thickness rather than true thickness. Case 3: layer where no pair of strike lines of same elevation for top and bottom, e.g. lower shale. 1.) Like the second example, one contact can be identified at a definite elevation (t250), giving a 250m strike line (t 250m s.l.). 2.) Also like case 2, a point must be found within the unit which is stratigraphically 'furthest' from the opposite contact. This will be 'O' from Q.6. Unfortunately, the elevation near 'O' is ~100m, NOT 250m. Using 'O' as our minimum, it is possible to draw a strike line for this stratigraphic level for 100m (bmin 100m s.l.).

top

tm

t400?

in

40

0m

s. l.

min thic imum kn e ss

bottom
f.l
bo tto

.4

b400

00

NW corner of map
b400

b400

top

16 E top

t250

f.l .2

t2

50

50

m
b100?

thi c

kn e

t250
mi nim um

ss

s. l.

40

0m

16

s. l.

bo tto m?

00

3.) Draw a folding line and assign it the highest elevation in use, i.e. 250m (f.l. 250m). 4.) Where the t 250 s.l. intersects the folding line, measure the appropriate dip (both angle and direction) and draw in the dip line (labeled 'top'). 5.) The dip line for the minimum base cannot be drawn directly from f.l. since this represents the 250m level on our profile. However, since the line represents an elevation that is 150m lower, this can be measured down on the profile and marked at the appropriate depth. 6.) The dip line can be drawn through this point parallel to the other dip line (labeled 'bottom?'). 7.) Measure thickness between dip lines.

m 16 ' E

15

'f. l. 1

0'

SE corner of map

Thicknesses (top to bottom): youngest sandstone conglomerate youngest limestone oldest sandstone youngest shale oldest limestone oldest shale >25m (or 50m*) ~50m >75m (or 112m*) 55m 75m 95m >220m

* depending on the 'minimum' used

m in

10

0m

s. l.

Cross-section X-Y, Greek Hills map area (lab 2)

NW

500m sandstone

unconformity
conglomerate

400m

300m

limestone

200m

sandstone

100m

? ? ? ? ? ?
0m 100m 200m

shale

0m

limestone

shale

-100m

?
l abs

_o_ -A-