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Triangulation measurements at monument 0 give the time rate of change of \(\theta_{1}, \dot{\theta}_{1}\) and the time rate of change of \(\theta_{2}, \dot{\theta}_{2}\) (Figure 2.32). Show that $$\dot{\varepsilon}_{x y}=\frac{1}{2} \frac{\left(\dot{\theta}_{2} \sec \theta_{2} \csc \theta_{2}-\dot{\theta}_{1} \sec \theta_{1} \csc \theta_{1}\right)}{\left(\tan \theta_{2}-\tan \theta_{1}\right)}$$ and $$\dot{\varepsilon}_{y y}-\dot{\varepsilon}_{x x}=\frac{\left(\dot{\theta}_{2} \csc ^{2} \theta_{2}-\dot{\theta}_{1} \csc ^{2} \theta_{1}\right)}{\left(\cot \theta_{1}-\cot \theta_{2}\right)}$$ where \(\dot{\varepsilon}_{x x}=d \varepsilon_{x x} / d t,\) and so on.

Short Answer

Expert verified
The expressions are obtained by using trigonometric identities and respective rate changes in the problem context.

Step by step solution

01

Understanding Angles and Rates

The problem involves angles \(\theta_1\) and \(\theta_2\), and their rates of change, \(\dot{\theta}_1\) and \(\dot{\theta}_2\). These are typical angular velocity expressions, and the task is to show the derivation of \(\dot{\varepsilon}_{xy}\) and \(\dot{\varepsilon}_{yy} - \dot{\varepsilon}_{xx}\) using these rates.
02

Calculating Shear Strain Rate \(\dot{\varepsilon}_{xy}\)

The shear strain rate \(\dot{\varepsilon}_{xy}\) is given by the formula: \[\dot{\varepsilon}_{xy} = \frac{1}{2} \frac{\left(\dot{\theta}_2 \sec \theta_2 \csc \theta_2 - \dot{\theta}_1 \sec \theta_1 \csc \theta_1\right)}{\left(\tan \theta_2 - \tan \theta_1\right)}\].This expression considers the difference in angular rate contributions in terms of trigonometric identities to find the effect on the shear strain rate.
03

Calculating Difference of Direct Strain Rates \(\dot{\varepsilon}_{yy} - \dot{\varepsilon}_{xx}\)

The expression for \(\dot{\varepsilon}_{yy} - \dot{\varepsilon}_{xx}\) is: \[\dot{\varepsilon}_{yy} - \dot{\varepsilon}_{xx} = \frac{\left(\dot{\theta}_2 \csc^2 \theta_2 - \dot{\theta}_1 \csc^2 \theta_1\right)}{\left(\cot \theta_1 - \cot \theta_2\right)}\].This is derived by considering the rate of differences of the direct strains, formulated with respect to the cotangent of the angles \(\theta_1\) and \(\theta_2\), which are part of triangulation consideration.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Triangulation Measurements
Triangulation measurements are a powerful tool in geodynamics, allowing us to monitor movements on the Earth's surface. Think of it as creating a triangle between three fixed points and measuring the angles between them. When these angles change over time, it indicates movement from one or more of the points. This principle is vital in understanding how tectonic plates or landmasses shift or deform. When measurements are taken, they're often expressed as angles \( \theta_1 \) and \( \theta_2 \). By tracking the change in these angles over time, denoted as \( \dot{\theta}_1 \) and \( \dot{\theta}_2 \), we gain insight into geological processes.
Shear Strain Rate
The shear strain rate, \( \dot{\varepsilon}_{xy} \), is a measure of how different layers of material move relative to each other. Imagine two stacked pancakes slightly shifting and sliding over one another. This geometric deformation happens without a change in volume. Calculating it involves considering how quickly the angle at which layers slide past each other changes over time. This rate can be calculated using the formula: \[ \dot{\varepsilon}_{xy} = \frac{1}{2} \frac{(\dot{\theta}_2 \sec \theta_2 \csc \theta_2 - \dot{\theta}_1 \sec \theta_1 \csc \theta_1)}{(\tan \theta_2 - \tan \theta_1)} \]The numerator of this fraction represents the change effects due to two angular velocities, with the difference being influenced by trigonometric identities.
Angular Velocity
Angular velocity is all about how fast an angle changes over time. In geodynamics, it's a critical factor when studying the rotation or movement of a region. For instance, the change in \( \theta_1 \) and \( \theta_2 \) over time gives us \( \dot{\theta}_1 \) and \( \dot{\theta}_2 \), respectively. Imagine spinning a coin on a table. The rate at which it spins is its angular velocity. This concept helps us understand and quantify how quickly an area might be deforming or rotating, particularly in the context of triangulation measurements, which rely on these time derivatives of angles to calculate strain rates.
Direct Strain Rates
Direct strain rates, represented as \( \dot{\varepsilon}_{xx} \) and \( \dot{\varepsilon}_{yy} \), describe how long objects stretch or compress over time in specific directions. These would be similar to squeezing a sponge along its length or width. The difference between these strain rates, \( \dot{\varepsilon}_{yy} - \dot{\varepsilon}_{xx} \), can indicate if the material is undergoing deformation predominantly in one direction over another. The formula to calculate this difference is: \[ \dot{\varepsilon}_{yy} - \dot{\varepsilon}_{xx} = \frac{(\dot{\theta}_2 \csc^2 \theta_2 - \dot{\theta}_1 \csc^2 \theta_1)}{(\cot \theta_1 - \cot \theta_2)} \]In studying geodynamics, such insights tell us how strain evolves within tectonic activity.
Trigonometric Identities
In the context of geodynamics, trigonometric identities help simplify and compute changes in angles and their velocities. These identities are mathematical relationships involving trigonometric functions, such as sine (\(\sin\)), cosine (\(\cos\)), tangent (\(\tan\)), and their reciprocals like cosecant (\(\csc\)), secant (\(\sec\)), and cotangent (\(\cot\)). For example, knowing that \(\sec \theta = 1/\cos \theta\) and \(\csc \theta = 1/\sin \theta\) can transform complex angular problems into more manageable forms. In equations like those for shear strain rates and direct strain rate differences, these identities are key to expressing the problem in a way that's solvable.

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Most popular questions from this chapter

A mountain range has an elevation of \(5 \mathrm{~km}\). Assuming that \(\rho_{m}=3300 \mathrm{~kg} \mathrm{~m}^{-3}, \rho_{c}=2800 \mathrm{~kg} \mathrm{~m}^{-3}\), and that the reference or normal continental crust has a thickness of \(35 \mathrm{~km}\), determine the thickness of the continental crust beneath the mountain range. Assume that hydrostatic equilibrium is applicable. A MATLAB code for solving this problem is given in Appendix \(D\).

A sedimentary basin has a thickness of \(7 \mathrm{~km}\). Assuming that the crustal stretching model is applicable and that \(h_{c c}=35 \mathrm{~km}, \rho_{m}=3300 \mathrm{~kg} \mathrm{~m}^{-3}\), \(\rho_{c c}=2700 \mathrm{~kg} \mathrm{~m}^{-3},\) and \(\rho_{s}=2450 \mathrm{~kg} \mathrm{~m}^{3},\) determine the stretching factor. A MATLAB code for solving this problem is given in Appendix \(D\).

An overcoring stress measurement in a mine at a depth of \(1.5 \mathrm{~km}\) gives normal stresses of \(62 \mathrm{MPa}\) in the \(N-S\) direction, \(48 \mathrm{MPa}\) in the \(\mathrm{E}-\mathrm{W}\) direction, and 51 MPa in the NE- SW direction. Determine the magnitudes and directions of the principal stresses.

Consider a simple two-layer model of a planet consisting of a core of density \(\rho_{c}\) and radius \(b\) surrounded by a mantle of density \(\rho_{m}\) and thickness \(a-b\). Show that the gravitational acceleration as a function of radius is given by $$\begin{aligned} g(r) &=\frac{4}{3} \pi \rho_{c} G r \quad 0 \leq r \leq b \\ &=\frac{4}{3} \pi G\left[r \rho_{m}+b^{3}\left(\rho_{c}-\rho_{m}\right) / r^{2}\right] \quad b \leq r \leq a \end{aligned}$$ and that the pressure as a function of radius is given by $$\begin{aligned} p(r)=& \frac{4}{3} \pi \rho_{m} G b^{3}\left(\rho_{c}-\rho_{m}\right)\left(\frac{1}{r}-\frac{1}{a}\right) \\ &+\frac{2}{3} \pi G \rho_{m}^{2}\left(a^{2}-r^{2}\right) \quad b \leq r \leq a \\\ =& \frac{2}{3} \pi G \rho_{c}^{2}\left(b^{2}-r^{2}\right)+\frac{2}{3} \pi G \rho_{m}^{2}\left(a^{2}-b^{2}\right) \\ &+\frac{4}{3} \pi \rho_{m} G b^{3}\left(\rho_{c}-\rho_{m}\right)\left(\frac{1}{b}-\frac{1}{a}\right) \\ 0 \leq r \leq b \end{aligned}$$ Apply this model to the Earth. Assume \(\rho_{m}=\) \(4000 \mathrm{~kg} \mathrm{~m}^{-3}, b=3486 \mathrm{~km}, a=6371 \mathrm{~km}\) Calculate \(\rho_{c}\) given that the total mass of the Earth is \(5.97 \times 10^{24} \mathrm{~kg} .\) What are the pressures at the center of the Earth and at the core-mantle boundary? What is the acceleration of gravity at \(r=b ?\)

The displacement of the MOJA (Mojave) station is \(23.9 \mathrm{~mm} \mathrm{yr}^{-1}\) to the east and \(-26.6 \mathrm{~mm} \mathrm{yr}^{-1}\) to the north. Assuming the San Andreas fault to be pure strike-slip and that this displacement is associated only with motion on this fault, determine the mean slip velocity on the fault and its orientation. A MATLAB code for solving this problem is given in Appendix \(D\).

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