Chapter 2: Problem 28
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.
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