---
title: "Structural Analysis"
book: "University Earth and Space Sciences — Year 3"
subject: earth
language: en
chapter: 8
exercises: 12
source: https://one-course.com/books/earth/5/en/chapter/8-structural-analysis
license: CC-BY-NC-SA-4.0
credit: "One Earth and Space Sciences Book, One Course (one-course.com)"
---

# Chapter 8 — Structural Analysis

A geologist has measured two hundred bedding planes on the limbs of a fold too large to be seen whole: its hinge lies under a forest, its limbs run for kilometres. Each reading is a pair of numbers. Plotted as points on a circular chart, the [poles](#def-b3-structural-analysis-stereo) of the planes fall along a single curve, a great circle, and the [pole](#def-b3-structural-analysis-stereo) of that great circle is the axis of the fold, a line that nobody has ever seen. This chapter builds the chart, the [stereographic projection](#def-b3-structural-analysis-stereo), and uses it, with the vector tools behind it, to find fold axes, to measure the strain recorded by deformed objects, to recover the stress that drove a set of faults, and to restore a deformed cross-section to its state before the deformation.

**You already know.**

Great and small circles of a sphere are in the Year 1 volume, in its chapter on spherical astronomy; strike, dip, dip direction, trend and plunge in its chapters on geological maps and cross-sections; the upward unit normal of a plane of dip direction $\alpha$ and dip $\delta$, $\vec n = (\sin\delta
\sin\alpha, \sin\delta\cos\alpha, \cos\delta)$ in east, north and up components, and the vector mean of a set of readings in its first chapter on field and laboratory methods. Fold elements and classes, the hinge line of a fold parallel to $\vec n_1\times\vec n_2$ and the interlimb angle $180{}^{\circ} - \arccos(\vec n_1\cdot\vec n_2)$, as well as Anderson’s theory of faulting, are in its chapter on faults, folds and fabrics; stress, the principal stresses, the strain ellipse and simple shear in its chapter on stress, strain and rheology; the rake of a slip vector (from the strike, $-90{}^{\circ}$ for a normal fault, $90{}^{\circ}$ for a reverse fault, 0 sinistral and $180{}^{\circ}$ dextral) and the P and T axes of a focal mechanism in its chapter on the earthquake source. The balanced cross-section and the décollement are in the Year 2 volume, in its chapter on the Alps. Least squares are in [Section 1.4](https://one-course.com/books/earth/5/en/chapter/1-seismology-and-earth-imaging#sec-b3-seismology-imaging-inverse); eigenvalues and eigenvectors of a symmetric matrix are in the mathematics books.

## 8.1 The stereographic projection

A geologist’s data are orientations: the attitude of a bedding plane, a fault, a foliation; the trend and plunge of a lineation, a fold hinge, a slip direction. Each one is a direction through a point, with no position. To compare hundreds of them, imagine them all moved to the centre of one sphere: a line meets the lower half of the sphere at one point, a plane cuts it along a half great circle. A chart of the lower hemisphere then shows lines as points and planes as curves.

**Definition 8.1 (Stereographic projection, equal-angle net, equal-area projection, pole).**

Let a line through the centre $O$ of a sphere of radius $R$ meet the lower hemisphere at $P$, at the angle $\theta = 90{}^{\circ} - p$ from the nadir, where $p$ is the plunge of the line and $t$ its trend. The *stereographic projection* maps $P$ to the point where the straight line from the zenith to $P$ crosses the horizontal plane through $O$; that point lies at the distance $r$ from $O$ in the azimuth $t$. The chart of the lower hemisphere drawn this way, with its grid of great and small circles, is the *equal-angle net* (or Wulff net). The *equal-area projection (of the sphere)* maps $P$ to the point at the distance $r =
R\sqrt2\sin(\theta/2)$ in the azimuth $t$; its net is the Schmidt net. The *pole* of a plane is its normal line, plotted as a point: for a plane of dip direction $\alpha$ and dip $\delta$, the pole trends $\alpha + 180{}^{\circ}$ and plunges $90{}^{\circ} - \delta$.

Both projections put a vertical line at the centre and a horizontal line on the circle of radius $R$, the primitive. They differ in what they keep.

**Theorem 8.2 (The two projections).**

1. The [stereographic projection](#def-b3-structural-analysis-stereo) places $P$ at $r = R\tan(\theta/2)$ . It is conformal: at every point it stretches all directions by the same factor, $1/(2\cos^2(\theta/2))$ , so that it keeps the angles between curves.
2. The equal-area projection multiplies every area by the same factor, $1/2$ : the area of a region of the net is half the area of the region of the hemisphere that it represents.

**Proof.** (a) In the vertical plane through the zenith $Z$, the nadir and $P$ (first figure below), the arc from the nadir to $P$ subtends the angle $\theta$ at the centre, hence the inscribed angle $\theta/2$ at $Z$, which is at the distance $R$ above $O$: $r =
R\tan(\theta/2)$. A small step $R\,\mathrm d\theta$ on the sphere towards the nadir becomes $\mathrm dr = R\,\mathrm d\theta/(2\cos^2(\theta/2))$ on the chart. The circle of points at the angle $\theta$ from the nadir has the circumference $2\pi R\sin\theta$ and becomes a circle of circumference $2\pi
r$; the ratio is $\tan(\theta/2)/\sin\theta = 1/(2\cos^2(\theta/2))$, the same as along the radius. A map that stretches equally in two perpendicular directions stretches equally in all directions and keeps angles.

(b) The cap of the sphere within the angle $\theta$ of the nadir has the area $2\pi R^2(1 - \cos\theta) = 4\pi R^2\sin^2(\theta/2)$; its image is the disc of area $\pi r^2 = 2\pi R^2\sin^2(\theta/2)$, half of it. Both areas are shared out evenly in azimuth, so the sector of the cap between two azimuths and its image keep the ratio $1/2$, and so does the ring between two values of $\theta$, the difference of two caps. Every region is a union of such small ring sectors. ∎

The [stereographic projection](#def-b3-structural-analysis-stereo) also maps every circle of the sphere, great or small, to a circle or a straight line on the chart; we admit this. It makes the [equal-angle net](#def-b3-structural-analysis-stereo) easy to draw with compasses, and a small circle, the cone of the lines at a fixed angle from an axis, keeps its shape. The equal-area projection distorts shapes: a small circle becomes an oval, stretched around the primitive. It is the projection on which a cloud of [poles](#def-b3-structural-analysis-stereo) can be counted, since an equal area of the net holds an equal share of the hemisphere’s directions; structural geologists, and seismologists for their focal mechanisms, use it for their data.

![The two projections, in the vertical plane through the line OP. (a) The stereographic projection sends P to P', where the line from the zenith crosses the horizontal plane: the angle at Z is /2, so r = R ( /2). (b) The equal-area projection carries P, around the nadir, onto the tangent plane at the distance 2R ( /2) of the chord; the chart is then scaled by 1/√2 so that the horizontal lines fall on a primitive of radius R.](https://one-course.com/images/onecourse/chapters/earth-5/b3-structural-analysis/fig-f75a9b4cd820.svg)

*The two projections, in the vertical plane through the line $OP$. (a) The [stereographic projection](#def-b3-structural-analysis-stereo) sends $P$ to $P'$, where the line from the zenith crosses the horizontal plane: the angle at $Z$ is $\theta/2$, so $r = R\tan(\theta/2)$. (b) The equal-area projection carries $P$, around the nadir, onto the tangent plane at the distance $2R\sin(\theta/2)$ of the chord; the chart is then scaled by $1/\sqrt2$ so that the horizontal lines fall on a primitive of radius $R$.*

![The two nets, drawn from the formulas of the theorem above: great circles of the planes striking north–south and dipping every 10 east and west, and small circles every 10 about the north–south axis. On the equal-angle net the small circles are circles; on the equal-area net they are ovals, but equal areas hold equal shares of the directions.](https://one-course.com/images/onecourse/chapters/earth-5/b3-structural-analysis/fig-e79909d180ae.svg)

*The two nets, drawn from the formulas of the theorem above: great circles of the planes striking north–south and dipping every $10{}^{\circ}$ east and west, and small circles every $10{}^{\circ}$ about the north–south axis. On the [equal-angle net](#def-b3-structural-analysis-stereo) the small circles are circles; on the equal-area net they are ovals, but equal areas hold equal shares of the directions.*

**Planes, lines and angles.** A plane of dip direction $\alpha$ and dip $\delta$ cuts the lower hemisphere along a half great circle. Its ends lie on the primitive at the two ends of the strike; its middle is the line of steepest dip, which plunges $\delta$ towards $\alpha$; and its [pole](#def-b3-structural-analysis-stereo) lies $90{}^{\circ}$ from every point of it, in the azimuth $\alpha + 180{}^{\circ}$. A line in the plane is a point on the great circle. To plot data, a transparent overlay is laid on the net and turned about the centre until the data’s strike lies along the net’s north–south diameter; the great circle of the right dip is then traced, and the overlay is turned back. The angle between two lines is measured along the great circle through both points, counting the small-circle divisions it crosses, after turning the overlay until both points lie on one great circle of the net. The angle between two planes is the angle between their [poles](#def-b3-structural-analysis-stereo), or its supplement. A computer does the same with vectors: the angle between two lines of unit vectors $\vec u$ and $\vec v$ is $\arccos|\vec u\cdot\vec v|$.

**Method 8.3 (Rotating data about an axis).**

To undo a tilt, for instance to bring a bed back to the horizontal with the lineations that it carries, or to restore a palaeomagnetic direction measured in tilted strata ([Chapter 4](https://one-course.com/books/earth/5/en/chapter/4-geomagnetism-and-palaeomagnetism#ch-b3-palaeomagnetism)):

1. Choose the axis of rotation: for a tilted bed, its strike line, horizontal; the angle is the dip, in the sense that brings the bed up.
2. On the net, turn the overlay until the axis lies along the north–south diameter, where the small circles are the cones about it.
3. Move every point along its small circle by the angle of rotation, counted on the great circles. A point that would leave the lower hemisphere reappears at the opposite end of its small circle, in the opposite azimuth.
4. Turn the overlay back. With vectors, the rotation of $\vec v$ by the angle $\omega$ about the unit axis $\vec k$ is $\vec v' = \vec v  \cos\omega + (\vec k\times\vec v)\sin\omega + \vec k(\vec k\cdot\vec  v)(1 - \cos\omega)$ .

## 8.2 Fold analysis

A cylindrical fold is generated by a straight line, the fold axis, moved parallel to itself. Every plane tangent to the folded surface contains the axis, so every normal to the surface is perpendicular to the axis: all the [poles](#def-b3-structural-analysis-stereo) lie on the great circle of the plane perpendicular to the axis.

**Definition 8.4 (Pi-diagram, beta-axis).**

The *pi-diagram* of a fold is the plot of the [poles](#def-b3-structural-analysis-stereo) of the planes measured on its folded surface. For a cylindrical fold they fall on a great circle, the $\pi$-girdle; the [pole](#def-b3-structural-analysis-stereo) of the girdle is the *beta-axis*, the estimate of the fold axis given by the measurements.

With two limbs only, the [beta-axis](#def-b3-structural-analysis-pi) is the cross product of their normals, as the Year 1 volume showed for the hinge line. With two hundred scattered readings, a best fit is needed. The vector mean, which served for the readings of one plane, will not do: the [poles](#def-b3-structural-analysis-stereo) of a girdle point every way along it, and their mean says little about the circle they lie on.

**Definition 8.5 (Orientation tensor).**

The *orientation tensor* of $N$ unit vectors $\vec p_1, \dots, \vec p_N$ is the symmetric matrix

$$
T = \frac1N\sum_{i=1}^N \vec p_i\,\vec p_i^{\mathsf T},
  \qquad T_{jk} = \frac1N\sum_i p_{ij}\,p_{ik} .
$$

It does not change when a vector is replaced by its opposite, so it suits axes and [poles](#def-b3-structural-analysis-stereo), which have no sense. Its eigenvalues $\lambda_1 \ge \lambda_2
\ge \lambda_3 \ge 0$ add up to its trace, $1$.

**Theorem 8.6 (Cluster and girdle from the orientation tensor).**

For any unit vector $\vec u$,

$$
\frac1N\sum_{i=1}^N (\vec p_i\cdot\vec u)^2 = \vec u^{\mathsf T}T\,\vec u ,
$$

which takes its smallest value, $\lambda_3$, when $\vec u$ is the eigenvector $\vec e_3$ of the smallest eigenvalue, and its largest, $\lambda_1$, along the eigenvector $\vec e_1$. Since $\vec p_i\cdot\vec u$ is the sine of the angle between $\vec p_i$ and the plane perpendicular to $\vec u$, the great circle perpendicular to $\vec e_3$ is the girdle that minimises the sum of the squared sines of the distances of the [poles](#def-b3-structural-analysis-stereo) to it, and $\sqrt{\lambda_3}$ is their root-mean-square sine. The axis $\vec e_1$ is the line closest to the vectors, the mean axis of a cluster.

**Proof.** $(\vec p_i\cdot\vec u)^2 = \vec u^{\mathsf T}\vec p_i\,\vec p_i^{\mathsf T}\vec
u$; average over $i$. The matrix $T$ is symmetric, so it has an orthonormal basis of eigenvectors, in which $\vec u = \sum_k u_k\vec e_k$ with $\sum_k
u_k^2 = 1$ and $\vec u^{\mathsf T}T\vec u = \sum_k\lambda_k u_k^2$. This is a weighted mean of the eigenvalues, with weights $u_k^2$ that add up to 1: it lies between $\lambda_3$ and $\lambda_1$ and reaches them for $\vec u = \vec
e_3$ and $\vec e_1$. ∎

The eigenvalues describe the shape of the data. A cluster has $\lambda_1$ near 1 and the other two small; a girdle has $\lambda_3$ near 0, the [poles](#def-b3-structural-analysis-stereo) being spread along the circle so that $\lambda_1$ and $\lambda_2$ share the rest; uniformly scattered vectors give three eigenvalues near $1/3$. A fold is cylindrical, within the precision of the readings, when $\sqrt{\lambda_3}$ is no larger than the scatter of repeated readings of one plane. A conical fold, or two generations of folds, spread the [poles](#def-b3-structural-analysis-stereo) away from any one great circle and raise $\lambda_3$.

The figure below plots twelve readings from the two limbs of the Varden anticline, the series’ invented fold ([Problem 8.1](#pb-b3-structural-analysis-1) computes its axis). The [poles](#def-b3-structural-analysis-stereo) of the gentle limbs cluster near the centre of the net in two groups, which together define a girdle running almost east–west; the [beta-axis](#def-b3-structural-analysis-pi) lies on the primitive, almost due north. The mean planes of the two limbs, drawn as great circles, cross at the same point: the [beta-axis](#def-b3-structural-analysis-pi) is the line common to all the limbs.

![Pi-diagram of the Varden anticline on the equal-area net, lower hemisphere: the poles of twelve bedding readings (dots), the best-fit girdle from the orientation tensor (blue), the great circles of the two mean limbs (dashed) and the beta-axis (triangle), where all three great circles meet. A bed dipping east has its pole to the west of the centre.](https://one-course.com/images/onecourse/chapters/earth-5/b3-structural-analysis/fig-ba2875214877.svg)

*[Pi-diagram](#def-b3-structural-analysis-pi) of the Varden anticline on the equal-area net, lower hemisphere: the [poles](#def-b3-structural-analysis-stereo) of twelve bedding readings (dots), the best-fit girdle from the [orientation tensor](#def-b3-structural-analysis-orientation-tensor) (blue), the great circles of the two mean limbs (dashed) and the [beta-axis](#def-b3-structural-analysis-pi) (triangle), where all three great circles meet. A bed dipping east has its [pole](#def-b3-structural-analysis-stereo) to the west of the centre.*

## 8.3 Finite strain analysis

A deformed rock records its strain when it contains objects whose original shape, or whose original arrangement, is known: round ooids and reduction spots, fossils of known symmetry, pebbles, or simply the centres of grains that were once evenly spaced.

**Definition 8.7 (Finite strain analysis, Fry method).**

*Finite strain analysis* is the measurement of the strain ellipse, or ellipsoid, that a rock has undergone, from the shapes, orientations or spacings of objects whose state before the deformation is known. In the *Fry method*, the vector from the centre of every object to the centre of every other is plotted from one common origin; when the centres were spread with a minimum spacing and no preferred direction, the empty region around the origin is an ellipse of the shape and orientation of the strain ellipse.

**Proposition 8.8 (The Fry void).**

If a homogeneous strain maps every vector $\vec d$ to $F\vec d$, and no two centres were closer than $d_0$ before the strain, then after it the region around the origin of the Fry plot that is free of points contains the ellipse $\{F\vec d : |\vec d| < d_0\}$, of semi-axes $S_1d_0$ and $S_2d_0$ along the principal directions of the strain.

**Proof.** A homogeneous strain moves every point by the same linear map, plus a translation, so the difference of two centres becomes $F\vec d$. Before the strain no difference lay in the disc of radius $d_0$; afterwards none lies in its image, which is the strain ellipse of the Year 1 volume scaled by $d_0$. ∎

In practice the centres are not all at the minimum spacing, and the void is read as the inner edge of the densest ring of points; its ratio estimates the strain ratio $R_s = S_1/S_2$ slightly low. The figure below makes the test on an invented thin section, strained to $R_s = 2$ with the long axis at $30{}^{\circ}$ from the horizontal: the ellipse fitted to the edge of the void gives $R_s = 1.9$ and $30{}^{\circ}$. When the objects themselves are measured, initially round ones become ellipses of the axial ratio $R_s$ exactly; initially elliptical ones, with random orientations, end with ratios $R_f$ scattered about $R_s$, the largest when their long axes started parallel to the stretching direction, and their final orientations $\varphi$ gather around it. The plot of $R_f$ against $\varphi$, read with tables or a computer, separates the strain from the original shapes.

![The Fry method on an invented population of grain centres, laid with a minimum spacing of 1 and then strained to R_s = 2, long axis at 30 anticlockwise from the horizontal, area kept. (a) The centres, in which the eye hardly sees the strain. (b) The vectors between all pairs of centres closer than 2.6: the empty centre is the strain ellipse. The red ellipse is fitted to the nearest point in each of 36 sectors: ratio 1.9, long axis at 30.](https://one-course.com/images/onecourse/chapters/earth-5/b3-structural-analysis/fig-4faf354a0fe1.svg)

*The [Fry method](#def-b3-structural-analysis-strain-analysis) on an invented population of grain centres, laid with a minimum spacing of 1 and then strained to $R_s = 2$, long axis at $30{}^{\circ}$ anticlockwise from the horizontal, area kept. (a) The centres, in which the eye hardly sees the strain. (b) The vectors between all pairs of centres closer than 2.6: the empty centre is the strain ellipse. The red ellipse is fitted to the nearest point in each of 36 sectors: ratio 1.9, long axis at $30{}^{\circ}$.*

In three dimensions the strain ellipsoid has the principal stretches $X \ge Y
\ge Z$. Its shape is plotted on the Flinn diagram, with $a = X/Y$ against $b =
Y/Z$ and the parameter

$$
k = \frac{a - 1}{b - 1} .
$$

On the line $k = 1$, which holds plane strain at constant volume ($Y = 1$, $X = 1/Z$, so $a = b$), the ellipsoid is neither cigar nor pancake. Above it, $k > 1$, the ellipsoid is prolate, stretched in one direction by a constriction: rocks deformed this way grow a strong lineation. Below, $k < 1$, it is oblate, flattened, and the rock grows a strong foliation.

![The Flinn diagram of strain ellipsoids. The undeformed state is the origin, a = b = 1; the cigar-shaped ellipsoids of constriction lie towards the vertical axis, the pancakes of flattening towards the horizontal one.](https://one-course.com/images/onecourse/chapters/earth-5/b3-structural-analysis/fig-9c1534db2b6a.svg)

*The Flinn diagram of strain ellipsoids. The undeformed state is the origin, $a = b = 1$; the cigar-shaped ellipsoids of constriction lie towards the vertical axis, the pancakes of flattening towards the horizontal one.*

## 8.4 Palaeostress

Anderson’s theory says on which planes new faults form, and that they slip perpendicular to $\sigma_2$. Most faults, though, are older weaknesses reactivated in a stress field that they did not choose, and they slip obliquely. Their slip directions, read on the striations of the fault surfaces, still record the stress, if each fault slips the way the stress pushes it.

**Definition 8.9 (Palaeostress, Wallace–Bott hypothesis, stress ratio, fault-slip inversion).**

*Palaeostress* is the state of stress that acted in a rock in the past, as far as it can be deduced from the structures it left: in practice, the directions of the principal stresses and the *stress ratio*

$$
\Phi = \frac{\sigma_2 - \sigma_3}{\sigma_1 - \sigma_3}, \qquad 0 \le \Phi
  \le 1 .
$$

The *Wallace–Bott hypothesis* states that a fault slips parallel to the shear stress resolved on its plane by a uniform stress, each fault independently. A *fault-slip inversion* is the search for the principal directions and the stress ratio whose predicted slip directions best fit a set of measured faults and striations.

**Theorem 8.10 (Slip direction on a plane).**

Let $S$ be the stress tensor, compressions counted positive, and $\vec n$ the upward unit normal of a fault. Write

$$
\vec\tau = S\vec n - (\vec n\cdot S\vec n)\,\vec n .
$$

Under the [Wallace–Bott hypothesis](#def-b3-structural-analysis-palaeostress) the hanging wall slips relative to the footwall along $-\vec\tau$, the shear traction that it exerts on the footwall. Moreover:

1. the direction of $\vec\tau$ is unchanged when $S$ is replaced by $aS +  bI$ with $a > 0$ ; it therefore depends only on the principal directions and on $\Phi$ , and the reduced tensor $S' =  \mathrm{diag}(1, \Phi, 0)$ in the principal frame $(\vec e_1, \vec  e_2, \vec e_3)$ predicts the same slip as $S$ ;
2. if $\vec n$ is perpendicular to one principal direction, the slip is perpendicular to it too;
3. with $n_k = \vec n\cdot\vec e_k$ , the direction is that of $\vec\tau'  = \vec\tau_0 + \Phi\,\vec\tau_1$ , linear in $\Phi$ , where $\vec\tau_0  = n_1(\vec e_1 - n_1\vec n)$ and $\vec\tau_1 = n_2(\vec e_2 - n_2\vec  n)$ .

**Proof.** With compressions positive, the hanging wall pushes on the footwall, across the plane of normal $\vec n$ pointing into the hanging wall, with the traction $-S\vec n$; its part along the plane is $-\vec\tau$. When the fault slips, the hanging wall moves the way it drags the footwall, along $-\vec\tau$. (Check: with $\sigma_1$ vertical, $S = \sigma_1\vec e_z\vec
e_z^{\mathsf T}$ and $-\vec\tau = -\sigma_1\cos\delta\,(\vec e_z -
\cos\delta\,\vec n)$ points down the dip: a normal fault.) (a) $(aS + bI)\vec n - (\vec n\cdot(aS + bI)\vec n)\vec n = a\vec\tau + b(\vec
n - \vec n) = a\vec\tau$. Writing $\sigma_k = \sigma_3 + (\sigma_1 -
\sigma_3)s_k$ with $s = (1, \Phi, 0)$ gives $S = \sigma_3 I + (\sigma_1 -
\sigma_3)S'$. (b) $S\vec n = \sum_k\sigma_kn_k\vec e_k$; if $n_j = 0$, neither $S\vec n$ nor $\vec n$ has a component along $\vec e_j$, and neither has $\vec\tau$. (c) $S'\vec n = n_1\vec e_1 + \Phi n_2\vec e_2$ and $\vec n\cdot S'\vec n =
n_1^2 + \Phi n_2^2$; collect the terms with and without $\Phi$. ∎

Part (b) is Anderson’s case: a fault that contains $\sigma_2$ slips perpendicular to it, whatever $\Phi$, so that it says nothing of the [stress ratio](#def-b3-structural-analysis-palaeostress). Part (c) turns a measured slip into $\Phi$. Let $\vec u$ be the unit slip vector of the hanging wall, from the rake, and $\vec w = \vec n\times\vec
u$ the direction in the plane perpendicular to it. The predicted slip is parallel to $\vec u$ when $(\vec\tau_0 + \Phi\vec\tau_1)\cdot\vec w = 0$, and since $\vec n\cdot\vec w = 0$,

$$
\Phi = -\frac{n_1\,(\vec e_1\cdot\vec w)}{n_2\,(\vec e_2\cdot\vec w)} ,
$$

provided the predicted slip then points along $+\vec u$ and not $-\vec u$.

When the principal directions are not known, each fault gives one equation, the angle of its slip within its plane, for four unknowns: three angles that orient the principal axes, and $\Phi$. Four faults of different attitudes are the minimum, and a few tens are usual. The inversion minimises the sum of the squared angles between the measured and the predicted slips, by least squares on a grid or by iteration as in [Section 1.4](https://one-course.com/books/earth/5/en/chapter/1-seismology-and-earth-imaging#sec-b3-seismology-imaging-inverse). Faults that do not fit form a second population, perhaps of another age.

The figure below shows five faults of the Varden Hills: the main normal fault and four smaller ones, all with normal slip. A grid search over $\Phi$ and the trend of $\sigma_3$, with $\sigma_1$ vertical, fits all five within a quarter of a degree, with $\sigma_3$ horizontal towards $120{}^{\circ}$.

![Fault-slip data of the Varden Hills on the equal-area net, lower hemisphere: five fault planes (great circles) with the slip direction of each hanging wall (dots), and the principal stresses of the best-fitting stress: _1 vertical (square), _2 and _3 horizontal (diamonds and triangles, at both ends of each axis). The main fault, dipping 60 towards 120, contains _2 and slips straight down its dip; the others slip obliquely.](https://one-course.com/images/onecourse/chapters/earth-5/b3-structural-analysis/fig-4995cee04a61.svg)

*Fault-slip data of the Varden Hills on the equal-area net, lower hemisphere: five fault planes (great circles) with the slip direction of each hanging wall (dots), and the principal stresses of the best-fitting stress: $\sigma_1$ vertical (square), $\sigma_2$ and $\sigma_3$ horizontal (diamonds and triangles, at both ends of each axis). The main fault, dipping $60{}^{\circ}$ towards $120{}^{\circ}$, contains $\sigma_2$ and slips straight down its dip; the others slip obliquely.*

The P and T axes of the Year 1 volume are not the principal stresses. They are built from the fault plane and the slip alone, at $45{}^{\circ}$ from the plane in the plane that contains the slip and the normal; a new fault, in rock with an angle of friction near $30{}^{\circ}$, makes about $30{}^{\circ}$ with $\sigma_1$, and a reactivated one any angle. One mechanism gives the P and T axes; only many faults of different attitudes, inverted together, give the stress. The hypothesis itself has limits: the stress may vary from fault to fault, blocks may rotate, and faults that meet hinder each other’s slip.

## 8.5 Restoring a cross-section

A cross-section is an interpretation: between the surface and the boreholes, the geologist draws what is likely. The Year 2 volume, in its chapter on the Alps, called a section balanced when it can be returned to an undeformed state with the lengths of the beds and the areas of the units conserved, folds and thrusts rooting on a décollement. Restoring is the test.

**Definition 8.11 (Section restoration, pin line, kink-band method).**

*Section restoration* undoes, on a cross-section drawn parallel to the direction of tectonic transport, the deformation it shows, event by event in the reverse order, keeping the lengths of the beds and the areas of the units in the plane of the section. A *pin line* is a line across the beds, placed where they did not slip along each other (in the undeformed foreland, or on the axial surface of a symmetric fold), which is taken to be perpendicular to the beds in the restored section; lengths are measured from it. In the *kink-band method* the folds are drawn as straight limbs of constant bed thickness joined at sharp hinges, whose axial surfaces bisect the angle between the limbs; each bed is then a broken line, whose length is the sum of its straight segments.

A bed folded into straight limbs dipping $\delta$ either way, over the horizontal distance $L$, has the length $L/\cos\delta$: its shortening is $S = L(1/\cos\delta - 1)$, the relative change of length $\cos\delta - 1$. Areas give the other half of the balance.

**Proposition 8.12 (Depth to detachment).**

Let a layer at the height $h$ above a horizontal detachment be shortened by $S$ in the plane of a section, and let the rock between it and the detachment keep its area, with no material crossing the ends of the section. Then the area $A$ by which the folded layer stands above its original level satisfies

$$
A = S\,h , \qquad\text{so that}\qquad h = \frac AS .
$$

**Proof.** Before the deformation the rock under the layer occupied $L_0 h$ over the length $L_0$. After it, the same rock lies over the length $L_0 - S$, under the layer, which stands at the original level plus the uplift: its area is $(L_0 - S)h + A$. Equating the two areas gives $A = Sh$. ∎

Two layers give both unknowns: their excess areas differ by $S$ times their vertical separation. The figure below sets out the construction.

![Restoring a kink-band fold, drawn to balance with limbs at 50. (a) Two competent layers folded into straight limbs of constant thickness over a weak layer that thickens into the core, between two pin lines on the axial surfaces of the synclines, which stayed at the original level. (b) The restored section: each bed laid straight with its length measured along the limbs, L_0, longer than the deformed section by the shortening S. The hatched area A, by which the top layer stands above its original level, equals S times its height h above the detachment; the weak layer has the same area in both panels.](https://one-course.com/images/onecourse/chapters/earth-5/b3-structural-analysis/fig-08d8cad150ee.svg)

*Restoring a kink-band fold, drawn to balance with limbs at $50{}^{\circ}$. (a) Two competent layers folded into straight limbs of constant thickness over a weak layer that thickens into the core, between two [pin lines](#def-b3-structural-analysis-restoration) on the axial surfaces of the synclines, which stayed at the original level. (b) The restored section: each bed laid straight with its length measured along the limbs, $L_0$, longer than the deformed section by the shortening $S$. The hatched area $A$, by which the top layer stands above its original level, equals $S$ times its height $h$ above the detachment; the weak layer has the same area in both panels.*

A section crossed by a fault is restored in the reverse order of the events: the youngest structure first. A normal fault of throw $t_h$ and dip $\delta$ has the heave $t_h/\tan\delta$ and the slip $t_h/\sin\delta$ on a section parallel to its dip; the hanging wall is moved back up the fault by the slip. On a section oblique to the slip, both the dip of the fault and its displacement are apparent, and a section that seems to balance may not: this is why sections are drawn parallel to the transport direction.

**In the field — Measuring an attitude.**

![](https://one-course.com/images/onecourse/chapters/earth-5/b3-structural-analysis/img-cd31dfccab61.jpg)

*A compass-clinometer laid against an inclined rock surface, with a field notebook (AI-generated illustration).*

To measure a bedding plane, the geologist lays the open lid of a compass-clinometer flat against the surface, chosen smooth and representative, and turns the body until its bubble shows it level: the needle then gives the dip direction, and the clinometer, with the instrument turned on its edge along the steepest line, the dip. The reading is noted as dip direction and dip, for instance $075/13$, with the place and the kind of surface. Azimuths read with a magnetic needle are corrected for the local declination. A smartphone with a geology application measures the same thing in one step: held flat against the plane, its accelerometers give the direction of gravity and hence the dip, and its magnetometer the azimuth. Field geologists compare it regularly with a compass, since nearby iron and the phone’s own circuits can disturb the magnetometer; they repeat each reading and average the normals with the vector mean.

## 8.6 Exercises

**Exercise 8.1 ★.**

Give the trend and plunge of the [pole](#def-b3-structural-analysis-stereo) of the plane $040/30$ (dip direction, dip). Give the attitude of the plane whose [pole](#def-b3-structural-analysis-stereo) trends $315{}^{\circ}$ and plunges $20{}^{\circ}$.

**Solution of Exercise 8.1.**

The [pole](#def-b3-structural-analysis-stereo) trends $40 + 180 = 220{}^{\circ}$ and plunges $90 - 30 = 60{}^{\circ}$. The plane dips $90 - 20 = 70{}^{\circ}$ towards $315 - 180 = 135{}^{\circ}$: $135/70$.

**Exercise 8.2 ★.**

A line plunges $30{}^{\circ}$. At what distance from the centre does it plot on an [equal-angle net](#def-b3-structural-analysis-stereo) and on an equal-area net, both of radius $10\,\mathrm{cm}$?

**Solution of Exercise 8.2.**

$\theta = 60{}^{\circ}$. Equal-angle: $10\tan30{}^{\circ} = 5.77\,\mathrm{cm}$; equal-area: $10\sqrt2\sin30{}^{\circ} = 7.07\,\mathrm{cm}$.

**Exercise 8.3 ★.**

The limbs of a fold have the attitudes $090/30$ and $270/30$. Compute the angle between their [poles](#def-b3-structural-analysis-stereo) with the dot product of the normals, the interlimb angle, and the class of the fold.

**Solution of Exercise 8.3.**

$\vec n_1 = (\sin30{}^{\circ}, 0, \cos30{}^{\circ})$, $\vec n_2 = (-\sin30{}^{\circ}, 0,
\cos30{}^{\circ})$, $\vec n_1\cdot\vec n_2 = -0.25 + 0.75 = 0.5$: the [poles](#def-b3-structural-analysis-stereo) are $60{}^{\circ}$ apart, the interlimb angle is $120{}^{\circ}$, an open fold.

**Exercise 8.4 ★.**

A fault $090/60$ carries striations of rake $-45{}^{\circ}$. With the strike vector $\vec s = (0, 1, 0)$ and the down-dip vector $\vec d$ of the plane, the slip of the hanging wall is $\vec u = \cos r\,\vec s - \sin r\,\vec d$. Compute $\vec d$, $\vec u$, and the trend and plunge of the slip. Name the fault.

**Solution of Exercise 8.4.**

$\vec d = (\cos60{}^{\circ}, 0, -\sin60{}^{\circ}) = (0.5, 0, -0.866)$ and $\vec u =
0.707\,\vec s + 0.707\,\vec d = (0.354, 0.707, -0.612)$: trend $\operatorname{atan2}(0.354, 0.707) = 26.6{}^{\circ}$, plunge $\arcsin0.612 =
37.8{}^{\circ}$. The rake lies between $-90{}^{\circ}$ (normal) and 0 (sinistral): an oblique normal fault with a sinistral component.

**Exercise 8.5 ★★.**

The limbs of a fold have the attitudes $060/40$ and $250/30$. Compute the trend and plunge of its axis, and its interlimb angle.

**Solution of Exercise 8.5.**

$$
\begin{gather*}
  \vec n_1 = (0.5567, 0.3214, 0.7660),\qquad \vec n_2 = (-0.4698, -0.1710, 0.8660),\\
  \vec n_1\times\vec n_2 = (0.4093, -0.8420, 0.0558),
\end{gather*}
$$

of length $0.9379$. Its opposite, $-\vec n_1\times\vec n_2$, points down: trend $334.1{}^{\circ}$, plunge $\arcsin(0.0558/0.9379) = 3.4{}^{\circ}$. $\vec n_1\cdot\vec
n_2 = 0.3469$, $69.7{}^{\circ}$ between the normals, interlimb angle $110.3{}^{\circ}$: an open fold.

**Exercise 8.6 ★★.**

Compute the fraction of the area of the net that lies inside the circle of lines plunging more than $60{}^{\circ}$, on the equal-area and on the [equal-angle net](#def-b3-structural-analysis-stereo), and the fraction of the directions of the lower hemisphere that plunge more than $60{}^{\circ}$. Which net should be used to count [poles](#def-b3-structural-analysis-stereo)?

**Solution of Exercise 8.6.**

Plunge above $60{}^{\circ}$ means $\theta < 30{}^{\circ}$. Equal-area: $r^2 =
2\sin^215{}^{\circ} = 0.134$, $13.4\,\%$ of the net; equal-angle: $\tan^215{}^{\circ} = 0.072$, $7.2\,\%$. On the sphere the cap holds $1 - \cos30{}^{\circ} = 0.134$ of the hemisphere: only the equal-area net shows it with its true share, so [poles](#def-b3-structural-analysis-stereo) are counted on it.

**Exercise 8.7 ★★.**

Three [poles](#def-b3-structural-analysis-stereo) lie in the vertical east–west plane, at $-20{}^{\circ}$, $0$ and $20{}^{\circ}$ from the vertical: $\vec p = (\sin a, 0, -\cos a)$. Compute their [orientation tensor](#def-b3-structural-analysis-orientation-tensor), its eigenvalues and eigenvectors, and the [beta-axis](#def-b3-structural-analysis-pi). Why is the smallest eigenvalue exactly zero?

**Solution of Exercise 8.7.**

$$
\begin{gather*}
  T_{xx} = \frac{2\sin^220{}^{\circ}}3 = 0.078,\qquad
  T_{zz} = \frac{2\cos^220{}^{\circ} + 1}3 = 0.922,\\
  T_{xz} = -\frac{\sin20{}^{\circ}\cos20{}^{\circ} - \sin20{}^{\circ}\cos20{}^{\circ}}3 = 0,
\end{gather*}
$$

and every term with $y$ is 0. The eigenvalues are $0.922$ (vertical), $0.078$ (east) and $0$ (north): the [beta-axis](#def-b3-structural-analysis-pi) is horizontal, north–south. The three [poles](#def-b3-structural-analysis-stereo) lie in one plane, so none has a component along its normal, and $\vec u^{\mathsf T}T\vec u = 0$ for $\vec u$ along it.

**Exercise 8.8 ★★.**

Place on the Flinn diagram the ellipsoid $X : Y : Z = 3 : 1.5 : 1$, then the ellipsoid of constant volume with $X = 1.8$ and $Y = 1.1$. Compute $k$ for each and name the strain.

**Solution of Exercise 8.8.**

$a = 2$, $b = 1.5$, $k = 1/0.5 = 2$: constriction. $Z = 1/(1.8\times1.1) =
0.505$, $a = 1.636$, $b = 2.178$, $k = 0.636/1.178 = 0.54$: flattening.

**Exercise 8.9 ★★.**

In a section cut parallel to the stretching, initially round ooids are ellipses of axial ratio $2.25$, all parallel, and the area of each is unchanged. Compute the principal stretches and the extension along the long axis. Before the strain, the ooid centres were never closer than $0.4\,\mathrm{mm}$: give the semi-axes of the void of the Fry plot.

**Solution of Exercise 8.9.**

$S_1/S_2 = 2.25$ and $S_1S_2 = 1$: $S_1 = 1.5$, $S_2 = 0.667$; the long axis was extended by $50\,\%$. The void has the semi-axes $1.5\times0.4 =
0.60\,\mathrm{mm}$ and $0.667\times0.4 = 0.27\,\mathrm{mm}$.

**Exercise 8.10 ★★★.**

At $\theta = 60{}^{\circ}$ from the nadir, compute the stretch along the radius, $\mathrm dr/(R\,\mathrm d\theta)$, and around the circle, $r/(R\sin\theta)$, for both projections. Check the theorem on the projections, and give the ratio of the axes of the oval into which the equal-area net turns a very small circle there.

**Solution of Exercise 8.10.**

Equal-angle: $1/(2\cos^230{}^{\circ}) = 0.667$ along the radius and $\tan30{}^{\circ}/\sin60{}^{\circ} = 0.577/0.866 = 0.667$ around: equal, conformal. Equal-area: $\mathrm dr/(R\,\mathrm d\theta) = (\sqrt2/2)\cos30{}^{\circ} = 0.612$ and $\sqrt2\sin30{}^{\circ}/\sin60{}^{\circ} = 0.707/0.866 = 0.816$; the product is $0.500$, the area factor of the theorem. A small circle becomes an oval of axes in the ratio $0.816/0.612 = 1.33$, long around the centre.

**Exercise 8.11 ★★★.**

In a strike-slip regime, $\sigma_1$ is horizontal towards $000{}^{\circ}$, $\sigma_3$ horizontal towards $090{}^{\circ}$ and $\sigma_2$ vertical. A vertical fault strikes $020{}^{\circ}$. Without computing $\vec\tau$, show that its slip is horizontal for every $\Phi$. Then compute $-\vec\tau$ for $\Phi = 0.4$ with the plane’s normal $\vec n = (\cos20{}^{\circ}, -\sin20{}^{\circ}, 0)$ and give the sense of slip.

**Solution of Exercise 8.11.**

The fault is vertical, so its normal is horizontal and perpendicular to the vertical $\sigma_2$; by part (b) of the theorem the slip is perpendicular to $\sigma_2$, horizontal, for every $\Phi$. With $\sigma_1 = 1$ along north, $\sigma_3 = 0$ along east: $S\vec n = (0, -0.342, 0)$, $\vec n\cdot S\vec n =
0.117$, $\vec\tau = (0, -0.342, 0) - 0.117\,(0.940, -0.342, 0) = (-0.110,
-0.302, 0)$ and $-\vec\tau = (0.110, 0.302, 0)$, towards $020{}^{\circ}$. The block on the side of $\vec n$, the east side, moves north-north-east along the strike: seen from the west side, it moves to the left, a sinistral fault (rake 0). $\sigma_2$ did not enter the computation.

**Exercise 8.12 ★★★.**

A section $20\,\mathrm{km}$ long crosses a train of detachment folds. Two horizons $1.0\,\mathrm{km}$ apart stand above their original levels by the areas $6.0\,\mathrm{km}^{2}$ (lower) and $7.5\,\mathrm{km}^{2}$ (upper). Compute the shortening, the depth of the detachment below the lower horizon, the original length of the section and its relative change of length.

**Solution of Exercise 8.12.**

$S = (7.5 - 6.0)/1.0 = 1.5\,\mathrm{km}$; $h = 6.0/1.5 = 4.0\,\mathrm{km}$ below the lower horizon. The section was $20 + 1.5 = 21.5\,\mathrm{km}$ long, and changed by $-1.5/21.5 = -7.0\,\%$.

## 8.7 Problem: Folds and Faults of the Varden Hills

**Problem 8.1.**

Weekend problem — the axis of a fold, the stress of a fault set and the restoration of a section, in an invented region

In the Varden Hills, the series’ invented region, a Mesozoic cover rests on a basement of schist and granite. The cover is folded into an upright anticline and syncline with straight limbs, then cut by a normal fault. A geologist measured twelve bedding planes on the two limbs of the anticline (dip direction/dip, in degrees):

| east limb | 075/13 | 084/11 | 089/13 | 079/14 | 086/12 | 072/11 |
| --- | --- | --- | --- | --- | --- | --- |
| west limb | 283/11 | 272/13 | 281/14 | 288/10 | 276/13 | 274/12 |

The sums of the upward unit normals of the six readings of each limb are $\vec S_E = (1.2583, 0.2017, 5.8604)$ and $\vec S_W = (-1.2445, 0.1886,
5.8636)$ (east, north, up). Five faults carry striations (dip direction/dip, rake of the hanging wall’s slip): $120/60/{-90}$ (the main fault), $150/55/{-76}$, $090/65/{-108}$, $200/70/{-78}$ and $060/50/{-105}$. The main fault has a throw of $200\,\mathrm{m}$. The anticline and syncline axes are $4\,\mathrm{km}$ apart, so that the folds repeat every $8\,\mathrm{km}$ from east to west.

**Part I — Poles and the beta-axis.**

1. Give the trend and plunge of the [pole](#def-b3-structural-analysis-stereo) of the reading $075/13$ , and its distance from the centre of an equal-area net of radius $10\,\mathrm{cm}$ .
2. Compute the upward unit normal of the reading $075/13$ ; check it against $\vec S_E$ in size.
3. Compute the mean attitude of each limb from $\vec S_E$ and $\vec S_W$ , the mean resultant lengths, and the root-mean-square scatter of the readings about their means.
4. Compute $\vec S_E\times\vec S_W$ , and the trend and plunge of the [beta-axis](#def-b3-structural-analysis-pi) .
5. Compute the angle between the two mean normals, the interlimb angle, and the class of the fold.
6. The axial plane of this symmetric fold contains the [beta-axis](#def-b3-structural-analysis-pi) and is perpendicular to $\vec m_E - \vec m_W$ , the difference of the mean unit normals. Give its attitude; is the fold upright?
7. Give the attitude of the $\pi$ -girdle, and say why the [poles](#def-b3-structural-analysis-stereo) lie on it.

**Part II — The orientation tensor.** The [orientation tensor](#def-b3-structural-analysis-orientation-tensor) of the twelve [poles](#def-b3-structural-analysis-stereo) is

$$
T = \begin{pmatrix} 0.04400 & 0.00035 & 0.00109\\ 0.00035 & 0.00144 & 0.03179\\
  0.00109 & 0.03179 & 0.95456 \end{pmatrix},
$$

with the eigenvalues $0.95562$, $0.04400$ and $0.00038$.

8. Check that the trace of $T$ is 1. With $\vec b = (0.0053, -0.9994,  0.0333)$ , the unit vector along $\vec S_E\times\vec S_W$ , compute $\vec b^{\mathsf T}T\vec b$ . What does the result say about $\vec b$ ?
9. Compute the root-mean-square angle between the [poles](#def-b3-structural-analysis-stereo) and the girdle.
10. Explain the value of $\lambda_2$ from the dips of the limbs.
11. Is the fold cylindrical within the precision of the readings?
12. The plunge of the [beta-axis](#def-b3-structural-analysis-pi) is set by the north–south tilt of the girdle, to which each [pole](#def-b3-structural-analysis-stereo) contributes with the scatter found in question 9. Estimate the standard error of the plunge.

**Part III — The palaeostress.**

13. What kind of fault is the main fault? Following Anderson, give the directions of $\sigma_1$ , $\sigma_2$ and $\sigma_3$ .
14. Show, with the theorem on the slip direction, that the main fault slips straight down its dip for every value of $\Phi$ . What does it tell of $\Phi$ ?
15. For the fault $150/55$ , compute $\vec n$ and its components $n_1 =  \vec n\cdot\vec e_1$ and $n_2 = \vec n\cdot\vec e_2$ , with $\vec e_1$ vertical and $\vec e_2$ horizontal towards $030{}^{\circ}$ .
16. Its slip gives $\vec w = \vec n\times\vec u = (0.771, 0.605, 0.198)$ . Compute $\Phi$ .
17. The other three small faults give $\Phi = 0.294$ , $0.301$ and $0.306$ . Give the mean, and say where the spread comes from.
18. What does $\Phi = 0.3$ say of $\sigma_2$ ?
19. Compute the P and T axes of the main fault (at $45{}^{\circ}$ from the plane, in its dip plane), and compare them with $\sigma_1$ and $\sigma_3$ .

**Part IV — Restoring an east–west section.**

20. Compute the apparent dip of the east limb, $080.9/12.3$ , on an east–west section.
21. Taking the limbs at $12{}^{\circ}$ on the section, compute the restored length of one fold wavelength, the shortening and the relative change of length.
22. Compute the height of the anticline crest above the synclines and the excess area of one wavelength, assuming that the synclines stayed at the original level.
23. Compute the depth of the detachment beneath the folded bed. Can the folds be detached at the base of the cover, a few hundred metres down?
24. Compute the heave and the slip of the main fault on a section along $120{}^{\circ}$ .
25. On the east–west section, compute the apparent dip of the fault and its apparent heave. Which section should be restored?
26. In what order are the fault and the folds restored, and where are the [pin lines](#def-b3-structural-analysis-restoration) placed?
27. What is the plunge of the Varden fold axis, recovered from the twelve readings?

**Solution of Problem 8.1.**

**1.** The [pole](#def-b3-structural-analysis-stereo) trends $255{}^{\circ}$ and plunges $77{}^{\circ}$; $\theta =
13{}^{\circ}$ and $r = 10\sqrt2\sin6.5{}^{\circ} = 1.60\,\mathrm{cm}$ from the centre, towards $255{}^{\circ}$.

**2.** $\vec n = (\sin13{}^{\circ}\sin75{}^{\circ}, \sin13{}^{\circ}\cos75{}^{\circ},
\cos13{}^{\circ}) = (0.2173, 0.0582, 0.9744)$. Six nearly parallel unit vectors add up to nearly 6: $|\vec S_E| = 5.9974$.

**3.** East limb: $\bar\alpha = \operatorname{atan2}(1.2583, 0.2017) =
80.9{}^{\circ}$ and $\bar\delta = \arccos(5.8604/5.9974) = 12.3{}^{\circ}$; $\bar R
= 5.9974/6 = 0.99956$ and the scatter is $\sqrt{2\times0.00044} =
0.0297\,\mathrm{rad}$, or $1.7{}^{\circ}$. West limb: $\bar\alpha =
\operatorname{atan2}(-1.2445, 0.1886) = 278.6{}^{\circ}$ and $\bar\delta =
\arccos(5.8636/5.9972) = 12.1{}^{\circ}$; $\bar R = 0.99953$, scatter $1.8{}^{\circ}$.

**4.**

$$
\vec S_E\times\vec S_W = (0.0774, -14.6714, 0.4883),
$$

of length $14.680$. Its opposite points down: the trend is $\operatorname{atan2}(-0.0774, 14.6714) = 359.7{}^{\circ}$ and the plunge $\arcsin(0.4883/14.680) = 1.9{}^{\circ}$.

**5.** The cosine of the angle between the mean normals is $32.835/35.968 = 0.9129$, the dot product of the sums divided by their lengths: $24.1{}^{\circ}$ between the normals, an interlimb angle of $155.9{}^{\circ}$: an open fold, close to gentle.

**6.**

$$
\begin{gather*}
  \vec m_E = (0.2098, 0.0336, 0.9772),\qquad \vec m_W = (-0.2075, 0.0314, 0.9777),\\
  \vec m_E - \vec m_W = (0.4173, 0.0022, -0.0005):
\end{gather*}
$$

a horizontal normal towards $089.7{}^{\circ}$. The axial plane is vertical, striking $359.7{}^{\circ}$, practically north–south: the fold is upright.

**7.** The girdle is the plane whose [pole](#def-b3-structural-analysis-stereo) is the [beta-axis](#def-b3-structural-analysis-pi): dip direction $179.7{}^{\circ}$, dip $90 - 1.9 = 88.1{}^{\circ}$, almost the east–west vertical plane. Every limb contains the fold axis, so every normal is perpendicular to it and lies in the plane perpendicular to it.

**8.** $0.04400 + 0.00144 + 0.95456 = 1.00000$. $\vec b^{\mathsf T}T\vec
b = 0.00038$, the smallest eigenvalue: $\vec b$ is, to the rounding, the eigenvector $\vec e_3$, and the girdle through the two mean [poles](#def-b3-structural-analysis-stereo) is the best-fit girdle of all twelve.

**9.** $\arcsin\sqrt{0.00038} = \arcsin0.0195 = 1.1{}^{\circ}$.

**10.** Along the girdle the [poles](#def-b3-structural-analysis-stereo) sit about $12.2{}^{\circ}$ from the vertical, on either side; $\sin^212.2{}^{\circ} = 0.045$, the mean square of their east–west components, which is $\lambda_2$.

**11.** Yes: the [poles](#def-b3-structural-analysis-stereo) stray from the girdle by $1.1{}^{\circ}$ on average, no more than the $1.7{}^{\circ}$ scatter of the readings of one limb about their mean.

**12.** $1.1{}^{\circ}/\sqrt{12} = 0.3{}^{\circ}$: the plunge is $1.9{}^{\circ}$ $\pm$ $0.3{}^{\circ}$.

**13.** A rake of $-90{}^{\circ}$: a pure normal fault, dip-slip. Anderson: $\sigma_1$ vertical, $\sigma_3$ horizontal along the dip direction, $120{}^{\circ}$–$300{}^{\circ}$, $\sigma_2$ horizontal along the strike, $030{}^{\circ}$–$210{}^{\circ}$.

**14.** The normal of $120/60$ lies in the vertical plane trending $120{}^{\circ}$, which holds $\vec e_1$ and $\vec e_3$: $n_2 = 0$. By part (b) the slip is perpendicular to $\sigma_2$, the strike, for every $\Phi$: it is dip-slip. The fault tells nothing of $\Phi$; in the formula both $n_2$ and $\vec e_2\cdot\vec w$, hence numerator and denominator, vanish.

**15.** $\vec n = (\sin55{}^{\circ}\sin150{}^{\circ}, \sin55{}^{\circ}\cos150{}^{\circ},
\cos55{}^{\circ}) = (0.4096, -0.7094, 0.5736)$; $n_1 = 0.574$; $n_2 = 0.5\times
0.4096 + 0.866\times(-0.7094) = -0.410$.

**16.** $\vec e_1\cdot\vec w = 0.198$ and $\vec e_2\cdot\vec w = 0.5\times
0.771 + 0.866\times0.605 = 0.909$, so $\Phi = -(0.574\times0.198)/(-0.410\times
0.909) = 0.1137/0.3727 = 0.305$.

**17.** $(0.305 + 0.294 + 0.301 + 0.306)/4 = 0.30$. The rakes were rounded to the degree; one degree of rake moves $\Phi$ by about 0.02 on these faults, so the half-degree of rounding explains the spread.

**18.** $\sigma_2 - \sigma_3 = 0.3(\sigma_1 - \sigma_3)$: $\sigma_2$ is closer to $\sigma_3$ than to $\sigma_1$. Both horizontal stresses are well below the vertical one; the region was extended mostly towards $120{}^{\circ}$, a little along $030{}^{\circ}$.

**19.** With $\vec n = (0.750, -0.433, 0.500)$ and the slip $\vec u =
(0.433, -0.250, -0.866)$, $(\vec n - \vec u)/\sqrt2$ plunges $75{}^{\circ}$ towards $300{}^{\circ}$: P; $(\vec n + \vec u)/\sqrt2$ plunges $15{}^{\circ}$ towards $120{}^{\circ}$: T. They lie in the same vertical plane as $\sigma_1$ and $\sigma_3$ but $15{}^{\circ}$ away from them, because the fault makes $30{}^{\circ}$, not $45{}^{\circ}$, with $\sigma_1$.

**20.** $\tan a = \tan12.3{}^{\circ}\times\cos(90{}^{\circ} - 80.9{}^{\circ}) =
0.2180\times0.9874 = 0.2153$: $a = 12.1{}^{\circ}$.

**21.** $L_0 = 8000/\cos12{}^{\circ} = 8178.7\,\mathrm{m}$, $S =
178.7\,\mathrm{m}$, a relative change of $\cos12{}^{\circ} - 1 =
-2.2\,\%$.

**22.** $4000\tan12{}^{\circ} = 850.2\,\mathrm{m}$; $A = \tfrac12\times8000\times
850.2 = 3.40 \times 10^{6}\,\mathrm{m}^{2}$.

**23.** $h = 3.40 \times 10^{6}/178.7 = 19.0\,\mathrm{km}$. A detachment at the base of the cover would give $h$ of a few hundred metres and an excess area some fifty times smaller: the folds are not detached there. Either the basement folded with the cover, or the synclines sank below the original level, which makes $A$ smaller; the section alone cannot tell.

**24.** Heave $200/\tan60{}^{\circ} = 115.5\,\mathrm{m}$, slip $200/\sin60{}^{\circ}
= 230.9\,\mathrm{m}$.

**25.** $\tan a = \tan60{}^{\circ}\cos30{}^{\circ} = 1.500$: $a = 56.3{}^{\circ}$, and the apparent heave is $200/1.5 = 133.3\,\mathrm{m}$. The slip of the fault is down the dip, towards $120{}^{\circ}$: the section along $120{}^{\circ}$ shows true lengths and should be the one restored.

**26.** The fault cuts the folds, so it is younger and is restored first, moving the hanging wall back up the fault by $230.9\,\mathrm{m}$; the folds are then unfolded. The [pin lines](#def-b3-structural-analysis-restoration) go on the axial surfaces of the synclines, in the footwall, where the symmetric folds had no slip between the beds.

**27.** The Varden fold axis plunges **$1.9{}^{\circ}$** towards the north (trend $359.7{}^{\circ}$): the $2{}^{\circ}$ built into the region, recovered to the $0.3{}^{\circ}$ that twelve readings allow.