i- determination of the true dip and strike...iv- determination of the true dip and strike - a...

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1 Stereographic Projection and its use in solving structural problems Dr. Mohamed Abdel Wahed

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Page 1: I- Determination of the true dip and strike...IV- Determination of the true dip and strike - A bedding plane has the following apparent dips : Given two apparent dips 1- The angle

1

Stereographic Projection

and its use in solving structural problems

Dr. Mohamed Abdel Wahed

Page 2: I- Determination of the true dip and strike...IV- Determination of the true dip and strike - A bedding plane has the following apparent dips : Given two apparent dips 1- The angle

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I- Plotting the projection of a line

Draw the projection of a line that has the following orientation : The direction (= trend = bearing ) of the line = N60W . Angle of plunge = 40

.

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II- Plotting the projection of a plane

Given : strike - true dip - direction of dip

Draw the projection of a plane of the following orientation :

Strike = N40E . Dip = 60 SE

.

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III- Plotting the projection of a plane Given : true dip - direction of dip

Draw the projection of a plane of the following orientation :

Dip = 60 , direction of dip =

S30W

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IV- Determination of the true dip and strike

- A bedding plane has the following apparent dips :

Given two apparent dips

1- The angle of apparent dip = 40 on a direction = N70E. 2- The angle of apparent dip = 60 on a direction = N30W.

Determine the angle of true dip and the direction of strike of the bedding plane.

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V- Determination of the apparent dip in a certain direction

given the true dip and strike A bedding plane has the following attitude : Strike = N60E , dip = 50 SE. Determine the angle of apparent dip in the direction S30W.

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VI- Intersection of two planes The two limbs of a fold have the following attitudes :

The first limb (A) strikes N20E and dips 60 SE. The second limb (B) strikes N60E and dips 50 NW

. Determine the orientation of he fold axis (trend and plunge).

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VII- Rotation of data on the stereographic projection An unconformity surface separates between two groups of beds; Group (A) above the unconformity surface and group (B) below. The attitude of these beds are : Group (A) : strikes N25W and dips 30 SW. Group (B) : strikes N60E and dips 40 SE

. Fined the attitude (strike and dip) of beds (B) at the time beds (A) was horizontal.

Answer: Strike = N31°E , Dip = 51°SE

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Determine the orientation of the stress regime responsible for the development of these structures.

VIII- Determination of the stress regime given the orientation of two conjugate joints or faults

Two conjugate faults have the following orientations : 1- The first faults : strikes N40E and dips 60 SE. 2- The second fault : strikes N80W and dips 50 NE.

1- Draw the projection of the faults as two arcs. The point of intersection of the two planes is the location of (σ 2).

Determine the trend and plunge of (σ 2).

2- a- Rotate (σ 2) to the horizontal axis of

the net.

Determination of (σ 1):

b- Plot the arc of the plane normal to (σ 2).

c- Bisect (نصف) the angle between

the two planes (A and B), i.e. the angle C-D, and locate the position of (σ 1). UDetermine the trend and plunge of (σ 1).

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3- Measure 90° from (σ 1) along the plane X-X , and locate the position of (σ 3).

Determination of (σ 3):

Determine the trend and plunge of (σ 3).

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IX- Data Representation

1- Rose Diagrams The following table represents a set of 64 joint planes. Represent these data in a rose diagram.

Example Data

Strike Dip Strike Dip N65E 50SE N46E 82SE N75E 53S N27E 67NW N53E 85NW N41E 85NW N39E 82SE N58E 87SE N53E 82SE N48E 82SE N58E 66SE N42E 82NW N50E 75SE N10E 82E N51E 85NW N52E 75SE N40E 87NW N49E 68SE N30E 85SE N36E 89SE N59E 82SE N45E 81NW N44E 88SE N41E 87NW N54E 86SE N36E 63NW N71E 82N N14E 86W N53E 83NW N48E 81NW N52E 86SE N27E 81SE N26E 80SE N46E 89NW N40E 78SE N58E 62SE N86E 85N N35E 81SE N67E 89SE N37E 88SE N61E 85SE N55E 70NW N72E 85S N38E 80SE N54E 86SE N47E 77NW N32E 67SE N34E 85SE N58E 87NW N45E 74NW N84E 86S N53E 90SE N50E 81NW N57E 90SE N48E 85NW N66E 90SE N50E 79NW N45E 90SE N51E 86NW N47E 90SE N40E 88SE N54E 90SE N53E 84NW N45E 90SE

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To plot the data as a rose diagram follow the following steps: 1-UConstruct the frequency table for the dataU. (الجدول التكرارى )

Determine number of joints in each trend category. Then calculate the percentage of the number of joints in each category with respect to the total number of joints. Any joint trend that lies at the boundary between two categories must be added to both categories.

Trend N0-10E N10-20E

N20-30E

N30-40E N40-50E N50-60E N60-

70E N70-80E

N80-90E

planes 1 11 1111 111111 111111

11111111 11111111 111111

11111111 11111111 111111

1111 111 11

Number of joints 1 2 4 12 22 22 4 3 2

% 1.56 % 3.13 % 6.25 % 18.8 % 34.4 % 34.4 % 6.25 % 4.8 % 3.2 %

You can choose the categories limits as:

Terend 0-9 10-19 20-29 30-39 40-49 50-59 60-69 70-79 80-89 90-99 so on No. of joints 0 2 3 9 19 22 4 3 2 0

UTrends must be in azimuth Both methods of choosing the categories limits are correct and can be used.

Both are used in the computer software dealing with plotting rose diagrams. You can use the number of joints in each category or the percent of joints in

each category in constructing the rose diagram.

Trend N0-10W

N10-20W

N20-30W

N31-40W

N40-50W

N50-60W

N60-70W

N70-80W

N80-90W

planes -- -- -- -- -- -- -- -- -- Number of joints 0 0 0 0 0 0 0 0 0

% 0 0 0 0 0 0 0 0 0

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Use the following polar net to construct the rose diagram

Polar net

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2- UConstruct the rose diagram as followsU: a- Determine the maximum number of joints in the categories in the previous table

(which is 22 joints in the categories (N41-50E) and (N51-60E). Represent this number (22 joints) on the diagram by a suitable distance along the radius of the circle. If the radius of the circle = 5 cm, then the 22 joints can be represented by 5 cm on the diagram, measured from its center outwards.

b- In the diagram, hachure (ظلل) a triangle with its side = 5 cm in the category N41-50E and also in the category N51-60E.

c- The distance corresponding to the number of joints in the other categories can be calculated as follows:

22 joints ----------- is represented by ---------- 5 cm 11 joints ------- should be represented by ------ X cm X = (5 X 11) / 22 = 2.7 cm 22 joints ----------- is represented by ---------- 5 cm 4 joints ------- should be represented by ------ X cm X = (5 X 4) / 22 = 0.91 cm And so on.

Trend N0-10E

N11-20E

N21-30E

N31-40E

N41-50E

N51-60E

N61-70E

N71-80E

N81-90E

Number of joints 1 2 4 12 22 22 4 3 2

% 1.56 % 3.13 % 6.25 % 18.8 % 34.4 % 34.4 % 6.25 % 4.8 % 3.2 %

Distance on the

diagram

0.22 cm

0.45 cm

0.91 cm

2.7 cm

5 cm

5. cm

0.91 cm

0.68 cm

0.45 cm

d- In the diagram, hachure (ظلل) triangles with their sides = the lengths calculated

for each category in the table. e- The resulting diagram will be as follows:

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If you are constructing rose diagrams for several parts of a study area for the

reason of comparison between these parts, the percentage of the number of joints in each category with respect to the total number of joints must be used in plotting.

This is because the total number of joints in the different subareas within the study area will be different, that will give erroneous impression and interpretation when comparing the intensity of the major trends in the different diagrams.

In this rose diagram, only the trends (strikes) of joints area represented. The angle of dip of these joints are not represented in the diagram. So the joints in the maximum concentration shown in the diagram (N50E) may be of uniform dips and represent a main trend, or may vary greatly in dip and represent more than one main trend. In such case, the point contour diagram is more appropriate for the representation of data. In point contour diagrams. Both strike and dip are represented in the diagram.

The rose diagram of the previous data Showing the main trend of joints = N50E

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• The point contour diagram represents the data on the stereographic projection (stereographic net).

2- Point-contour diagrams

• In the point contour diagrams, planes are not represented as arcs but are projected as poles

• .

The pole is the projection of the line normal to the plane, so its projection on the stereographic net is a point

as follows.

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- The projection of the line normal to the plane (pole) is a point that is characterized by:

1- The pole lies on a line normal to the strike of the plane. 2- The pole 3- The pole

lies in opposite direction to the dip of the plane. lies at a distance = angle of dip of the plane measured from

the center of the net

.

Draw the pole to the bedding plane that strikes

Draw the projection of the pole to a plane as follows:

N30E and dips 60 NW. 1-Draw the projection of the plane as

an arc. 2- Rotate the arc to the vertical axis of

the net. Plot the pole to the arc at an angle = 60 measured from the center of the net and in opposite direction to the dip.

3- Rotate to the North position.

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You can plot the pole to the plane

without drawing the arc as follows : 1- Just mark the position of the strike

of the plane (at N30E). 2- Draw the pole to the plane at 60

measured from the center of the net in opposite direction to the dip.

3- Rotate to the North position.

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• To calculate point densities on a stereogram, count the number of points that are found within a standard sampling area of the net (= 1% of the total area of the net).

The Point-Contour Diagram : To construct the point-contour diagram the density of the poles in the plot must be transformed into numbers (to be contoured) as follow:

• The method for contouring described here makes use of the Kalsbeek counting net (see fig)

1- The data must be plotted as poles or lines (for lineation) using an equal-area net, (fig. a)

2- The Kalsbeek counting net (see Fig. b) has a layout which consists of mutually overlapping hexagons, each with an area equal to 1/100 of the total area of the full stereogram.

3- The Kalsbeek net is placed under the tracing paper of the plotted data. A new clean tracing paper is put on both, and all are fixed together at the center.

4- The number of points occurring in each hexagon is recorded and written at the center of the hexagon (Fig. c) and, in this manner, an array of numbers covering the stereogram (Fig. d) is obtained.

5- The obtained density values can be contoured (Fig. e).

6- For convenient density distribution values, the number of points in each hexagon is transformed as a % of the total number of points in the plot, then the % values are contoured.

Point contour diagrams can be easily constructed by computer software. Many are available for free in the internet, e.g.: Stereonet and GEOrient software

In fact there is no need to follow the manual methodology.

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Example of point contour diagram: Equal area, lower hemisphere projection of poles to joint data, and the determination of the main joint orientations from the contour diagram. N = 128 joints , contours are at 0, 2, 4, 6, and 8 % .

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Rose and point contour diagram for the same data set constructed by stereonet-v 10 software. Compare the results.