Dipmeter Calculations With Stereonets
These tasks include finding the projection of a plane, direction of a line normal to a plane, the line of intersection of two planes, angles between two planes, true dip from two apparent dips and vice versa, and regional dip removal. Some of these functions have been described earlier, using the calculator, SCAT, or tangent diagrams. Some people still prefer the stereonet, but the calculator is easier.
instructions are paraphrased from "Schlumberger Dipmeter
Fundamentals 1981", and the stereonets are copied from the
previous edition dated 1970. For working through stereographic
problems you should have a stereonet such as the one shown below, plus pieces of tracing paper large enough to cover it,
or a plastic overlay, made from a xerographic reproduction of
the next illustration. These two illustrations are used for high angle
dips. Enlarged versions of the central
portions of the illustrations are used for low angle
dips (see below).
The data for each problem are plotted on the tracing paper or overlay, and the stereonet is rotated to suit the differing orientations met with in each case. Although it is usually more convenient to lay the stereonet down and keep it fixed, while rotating the tracing paper over it, keep in mind that it is the tracing paper overlay, and not the net, that represents the fixed Earth.
If you use tracing paper, trace the outer circle of the stereonet on it and mark a "north" point with an N on the circle at some arbitrary point. Tracing the outer circle is necessary so that the two diagrams - overlay and stereonet - can be kept concentric in all orientations. You could achieve the same result by pinning the two layers together so that the tracing paper rotates about the center point of the stereonet. No matter how the overlay is rotated, the N point should be regarded as always pointing north.
If you use a transparent copy of the overlay, the circle and north point (0/360 degrees) are already marked. Use a grease pencil to mark points and lines, so it can be wiped off before the next example.
understand how a stereogram is constructed, imagine standing on
level ground and looking down into a hemisphere contoured at our
feet and extending down into the ground, as if the ground were
transparent. Any plane that passes through the center of a sphere
cuts the spherical surface in an arc called a great circle. If
we stand on an outcrop of a bed dipping down into the ground,
we can imagine that the bed cuts the underground hemisphere with
an arc of a great circle, below, top right.
To project that circle up to the horizontal surface at ground level, we connect every point on the great circle to the zenith point of the sphere, above our head. The intersection of the lines with the horizontal plane form a new circle; many such circles form the north south grid lines of the stereonet, see above, middle left.
The intersections of vertical planes that do not pass through the center of the sphere intersect the hemisphere surface as small circles and can be projected up to the stereogram surface, via the zenith point, exactly as before, see above, lower right. These form the circles that are centered on the north and south poles, forming the east west grid on the stereonet. Superposition of the two sets of circles creates the final stereonet presentation, see above, lower left.
A straight line passing downward at a slant through the point at which we are standing cuts the hemisphere at a point that can be projected onto the stereogram by the same technique. Again, the zenith point provides the reference for the projection, see below, top right. Both lines and planes can be plotted on the same diagram, see below, middle right. Horizontal and vertical planes are special cases; the projection of a horizontal plane is the outer edge of the stereonet, a vertical plane passing through the center is a straight line, below, lower right.
Illustrated below at upper left is how to plot the projection of a plane
dipping 20 degrees in a N 40 degrees E direction:
Lower right shows how to plot the direction of the line
normal to the surface of the plane in example 1.
that it doesn't matter in which direction you count along the
diameter; if you should choose the direction that brings you to
the edge of the net before reaching 90 degrees, jump to the other
end of the diameter and finish counting from there. Check that
both directions bring you to point P.
image below, upper left, shows how to find the line of intersection
of two planes: Given: plane A dips 20 degrees toward N 40 degrees
E (the plane in example 1). plane B dips 30 degrees towards N
20 degrees W.
above, shows how to find the angle between the two
planes in the previous example. Given: plane A dips 20 degrees
toward N 40 degrees E (the plane in example 1). plane B dips 30
degrees towards N 20 degrees W.
Both methods should give the same answers, of course. Notice, however, that with the first method the angle measured directly between PA and PB is 26 degrees, while the angle between the great-circle arcs is 154 degrees. Because 26 degrees + 154 degrees = 180 degrees, we know that 26 degrees is the acute dihedral angle and 154 degrees is the obtuse dihedral angle between the given planes.
The bottom left
illustration shows how to find true dip from dip measured
in two different vertical planes: Given: dip A is 25 degrees,
in a plane N 30 degrees E and dip B is 20 degrees, in a plane
N 40 degrees W
Notice that this procedure can be worked backwards, to find the slope of a bed on any azimuthal direction if the true dip is known. First trace in the great circle for the bedding plane, knowing its dip; then find where this arc cuts a radial line drawn with the desired azimuth. You would need to do this twice to find transverse and longitudinal dip components.
If an inclined formation contains smaller bedded units within it, the computed dips of the subunits need to be corrected, by subtraction of the dip of the major system, to find their dips at the time of deposition. For the stereonet, the problem is that of rotating one plane by an amount, and in a direction, given by the dip of the other.
image at the right shows how to eliminate structural dip from
computed dip. Given: formation dip of 30 degrees, azimuth N 20
degrees E structural dip of 15 degrees, S 40 degrees W
Page Views ---- Since 01 Jan 2015
Copyright 1978 - 2018 E. R. Crain, P.Eng. All Rights Reserved