Commit e6255e67 authored by Maximilian Vitz's avatar Maximilian Vitz
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Updated Solution 3

parent c5ba8877
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### Error Propagation
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1)
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$a = 6.1 \pm 0.1$ cm
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$b = 2.8 \pm 0.1$ cm
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With 100% correlation one ends up with $\rho = 1$.
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$\Rightarrow \text{cov}(a,b) = \rho \sigma_{a} \sigma_{b} = \sigma_{a} \sigma_{b}$
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$\sigma_{f}^{2} = \left( \frac{\delta f}{\delta a} \right)^{2} \sigma_{a}^{2} + \left( \frac{\delta f}{\delta b} \right)^{2} \sigma_{b}^{2} + 2 \left( \frac{\delta A}{\delta a} \right) \left( \frac{\delta f}{\delta b} \right) \text{cov}(a,b) = \left( \frac{\delta f}{\delta a} \right)^{2} \sigma_{a}^{2} + \left( \frac{\delta f}{\delta b} \right)^{2} \sigma_{b}^{2} + 2 \left( \frac{\delta f}{\delta a} \right) \left( \frac{\delta f}{\delta b} \right) \sigma_{a} \sigma_{b} = \left( \frac{\delta f}{\delta a} \sigma_{a} + \frac{\delta f}{\delta b} \sigma_{b} \right)^{2}$
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$f = A(a,b) = a \cdot b$
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$\Rightarrow \sigma_{A} = \frac{\delta A}{\delta a} \sigma_{a} + \frac{\delta A}{\delta b} \sigma_{b} = b \sigma_{a} + a \sigma_{b} = 0.89 \text{cm}^{2}$
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$f = C(a,b) = 2a + 2b$
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$\Rightarrow \sigma_{C} = \frac{\delta C}{\delta a} \sigma_{a} + \frac{\delta C}{\delta b} \sigma_{b} = 2 \sigma_{a} + 2 \sigma_{b} = 0.4 \text{cm}$
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2)
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With 0% correlation one ends up with $\rho = 0$.
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$\Rightarrow \text{cov}(a,b) = \rho \sigma_{a} \sigma_{b} = 0$
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$\sigma_{f}^{2} = \left( \frac{\delta f}{\delta a} \right)^{2} \sigma_{a}^{2} + \left( \frac{\delta f}{\delta b} \right)^{2} \sigma_{b}^{2} + 2 \left( \frac{\delta A}{\delta a} \right) \left( \frac{\delta f}{\delta b} \right) \text{cov}(a,b) = \left( \frac{\delta f}{\delta a} \right)^{2} \sigma_{a}^{2} + \left( \frac{\delta f}{\delta b} \right)^{2} \sigma_{b}^{2} $
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$f = A(a,b) = a \cdot b$
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$\Rightarrow \sigma_{A} = \sqrt{\left( \frac{\delta A}{\delta a} \right)^{2} \sigma_{a}^{2} + \left( \frac{\delta A}{\delta b} \right)^{2} \sigma_{b}^{2}} = \sqrt{b^{2} \sigma_{a}^{2} + a^{2} \sigma_{b}^{2}} = 0.67 \text{cm}^{2}$
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$f = C(a,b) = 2a + 2b$
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$\Rightarrow \sigma_{C} = \sqrt{\left( \frac{\delta C}{\delta a} \right)^{2} \sigma_{a}^{2} + \left( \frac{\delta C}{\delta b} \right)^{2} \sigma_{b}^{2}} = \sqrt{4 \sigma_{a}^{2} + 4 \sigma_{b}^{2}} = 0.28 \text{cm}$
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3)
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$\left(\begin{array}{cc}
\sigma_{A}^{2} & \text{cov}(A,C)\\
\text{cov}(A,C) & \sigma_{C}^{2}
\end{array}\right) = \left(\begin{array}{cc}
\frac{\delta A}{\delta a} & \frac{\delta A}{\delta b}\\
\frac{\delta C}{\delta a} & \frac{\delta C}{\delta b}
\end{array}\right) \cdot \left(\begin{array}{cc}
\sigma_{a}^{2} & \text{cov}(a,b)\\
\text{cov}(a,b) & \sigma_{b}^{2}
\end{array}\right) \cdot \left(\begin{array}{cc}
\frac{\delta A}{\delta a} & \frac{\delta C}{\delta a}\\
\frac{\delta A}{\delta b} & \frac{\delta C}{\delta b}
\end{array}\right) = \left(\begin{array}{cc}
b & a\\
2 & 2
\end{array}\right) \cdot \left(\begin{array}{cc}
\sigma_{a}^{2} & \text{cov}(a,b)\\
\text{cov}(a,b) & \sigma_{b}^{2}
\end{array}\right) \cdot \left(\begin{array}{cc}
b & 2\\
a & 2
\end{array}\right)$
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$\Rightarrow \left(\begin{array}{cc}
\sigma_{A}^{2} & \text{cov}(A,C)\\
\text{cov}(A,C) & \sigma_{C}^{2}
\end{array}\right) = \left(\begin{array}{cc}
\sigma_{a}^{2} b^{2} + \sigma_{b}^{2} a^{2} + 2 a b \text{cov} (a,b) & 2 (b \sigma_{2}^2 + a \sigma_{b}^2 + \text{cov}(a,b)(a+b))\\
2 (b \sigma_{2}^2 + a \sigma_{b}^2 + \text{cov}(a,b)(a+b)) & 4 \sigma_{a}^2 + 4 \sigma_{b}^2 + 8 \text{cov}(a,b)
\end{array}\right)$
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$\Rightarrow \text{cov}(A,C) = 2 (b \sigma_{2}^2 + a \sigma_{b}^2 + \text{cov}(a,b)(a+b))$
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$\rho_{A,C} = \frac{\text{cov}(A,C)}{\sigma_{A}\sigma_{C}}$
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With 100% correlation one ends up with $\rho_{a,b} = 1$, $\sigma_{a}= \sigma_{b}$.
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$\rho_{A,C} = \frac{\text{cov}(A,C)}{\sigma_{A}\sigma_{C}} = \frac{4(a+b)}{4(a+b)} = 1$
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With 0% correlation one ends up with $\rho_{a,b} = 0$, $\sigma_{a}= \sigma_{b}$.
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$\rho_{A,C} = \frac{\text{cov}(A,C)}{\sigma_{A}\sigma_{C}} = \frac{2(a+b)}{\sqrt{8(a^{2}+b^{2})}} = 0.938$
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### Triangulation
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1)
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$l = 101.2$ m, $\sigma_{l} = 0.1$ m
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$\alpha = 87.4^{\circ}$, $\sigma_{\alpha} = 0.1^{\circ}$
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$\beta = 87.8^{\circ}$, $\sigma_{\beta} = 0.1^{\circ}$
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$d = \frac{l\sin{(\alpha)} \sin{(\beta)}}{\sin{(\alpha +\beta)}}$
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$\sigma_{d}^{2} = \sigma_{\alpha}^{2} \left( - \frac{l\sin{(\alpha)} \sin{(\beta)} \cos{(\alpha +\beta)}}{\sin^{2}{(\alpha +\beta)}} + \frac{l\cos{(\alpha)} \sin{(\beta)}}{\sin{(\alpha +\beta)}} \right)^{2} + \sigma_{\beta}^{2} \left( - \frac{l\sin{(\alpha)} \sin{(\beta)} \cos{(\alpha +\beta)}}{\sin^{2}{(\alpha +\beta)}} + \frac{l\sin{(\alpha)} \cos{(\beta)}}{\sin{(\alpha +\beta)}} \right)^{2} + \sigma_{l}^{2} \left( \frac{\sin{(\alpha)} \sin{(\beta)}}{\sin{(\alpha +\beta)}} \right)^{2}$
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$\Rightarrow d =1207.36$ m, $\sigma_{d} = 35.63$ m
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2)
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Assuming $\sigma_{l} = 0$.
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$\Rightarrow \sigma_{d}^{2} = \sigma_{\alpha}^{2} \left( - \frac{l\sin{(\alpha)} \sin{(\beta)} \cos{(\alpha +\beta)}}{\sin^{2}{(\alpha +\beta)}} + \frac{l\cos{(\alpha)} \sin{(\beta)}}{\sin{(\alpha +\beta)}} \right)^{2} + \sigma_{\beta}^{2} \left( - \frac{l\sin{(\alpha)} \sin{(\beta)} \cos{(\alpha +\beta)}}{\sin^{2}{(\alpha +\beta)}} + \frac{l\sin{(\alpha)} \cos{(\beta)}}{\sin{(\alpha +\beta)}} \right)^{2}$
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$\Rightarrow \sigma_{d} = 36.61$ m
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The error on the distance is dominated by the error on the angles.
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3)
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$\alpha = \beta$, $\sigma_{\alpha} = \sigma_{\beta} = \sigma$
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$\sigma_{d}^{2} = \sigma_{\alpha}^{2} \left( - \frac{l\sin{(\alpha)} \sin{(\beta)} \cos{(\alpha +\beta)}}{\sin^{2}{(\alpha +\beta)}} + \frac{l\cos{(\alpha)} \sin{(\beta)}}{\sin{(\alpha +\beta)}} \right)^{2} + \sigma_{\beta}^{2} \left( - \frac{l\sin{(\alpha)} \sin{(\beta)} \cos{(\alpha +\beta)}}{\sin^{2}{(\alpha +\beta)}} + \frac{l\sin{(\alpha)} \cos{(\beta)}}{\sin{(\alpha +\beta)}} \right)^{2}$
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$\Rightarrow \sigma_{d}^{2} = 2 \sigma^{2} l^{2} \left( \frac{-\sin{(\alpha)} \sin{(\alpha)} \cos{(2\alpha)}}{\sin^{2}{(2\alpha)}} + \frac{\cos{(\alpha)} \sin{(\alpha)}}{\sin{(2 \alpha)}} \right)^{2} = 2 \sigma^{2} l^{2} \left( \frac{-\sin^{2}{(\alpha)} \cos{(2\alpha)} + \cos{(\alpha)} \sin{(\alpha)}\sin{(2\alpha)}}{\sin^{2}{(2\alpha)}} \right)^{2}$
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Use: $\hspace{0.5cm}$ $\cos(2 \alpha) = \cos^{2}{(\alpha)} - \sin^{2}{(\alpha)}$ $\hspace{0.5cm}$ and $\hspace{0.5cm}$ $\sin(2 \alpha) = 2\sin{(\alpha)}\cos{(\alpha)}$
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$\Rightarrow \sigma_{d}^{2} = 2 \sigma^{2} l^{2} \left( \frac{-\sin^{2}{(\alpha)} \cos^{2}{(\alpha)} + \sin^{2}{(\alpha)} \sin^{2}{(\alpha)} + 2 \cos^{2}{(\alpha)} \sin^{2}{(\alpha)}}{\sin^{2}{(2\alpha)}} \right)^{2} = 2 \sigma^{2} l^{2} \left( \frac{ \sin^{2}{(\alpha)} \cdot [ - \cos^{2}{(\alpha)} + \sin^{2}{(\alpha)} + 2 \cos^{2}{(\alpha)} ]}{\sin^{2}{(2\alpha)}} \right)^{2}= 2 \sigma^{2} l^{2} \left( \frac{ \sin^{2}{(\alpha)} \cdot [ \cos^{2}{(\alpha)} + \sin^{2}{(\alpha)]} }{\sin^{2}{(2\alpha)}} \right)^{2}$
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Use: $\hspace{0.5cm}$ $\sin(2 \alpha) = 2\sin{(\alpha)}\cos{(\alpha)}$
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$\Rightarrow \sigma_{d}^{2} = 2 \sigma^{2} l^{2} \left( \frac{ \sin^{2}{(\alpha)}}{\sin^{2}{(2\alpha)}} \right)^{2} = 2 \sigma^{2} l^{2} \left( \frac{ \sin^{2}{(\alpha)}}{4 \sin^{2}{(\alpha)}\cos^{2}{(\alpha)}} \right)^{2} = \frac{ \sigma^{2} l^{2}}{8 \cos^{4}{(\alpha)}}$
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$\Rightarrow \sigma_{d} \approx \frac{ \sigma l}{\sqrt{8} } \frac{(l/2)^{2} + d}{(l/2)^{2}} \approx \frac{ \sigma l}{\sqrt{8} } \frac{4 d^{2}}{l^{2}} = \sqrt{2} \sigma \frac{d^{2}}{l}$
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``` python
```
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