diff --git a/mr/Ub4/mr4.pdf b/mr/Ub4/mr4.pdf index aa072d1..9f7ab82 100644 --- a/mr/Ub4/mr4.pdf +++ b/mr/Ub4/mr4.pdf Binary files differ diff --git a/mr/Ub4/mr4.tex b/mr/Ub4/mr4.tex index 6608b98..378914e 100644 --- a/mr/Ub4/mr4.tex +++ b/mr/Ub4/mr4.tex @@ -83,37 +83,36 @@ \Aufgabe{}{8} \begin{enumerate}[(a)] - \item $v_R = sin(\frac{\pi}{2}-\theta_w+\phi)\cdot v_R = v_R \cdot (sin(\frac{\pi}{2}-\theta_w)\cdot cos(\phi) + cos(\frac{\pi}{2}-\theta_w) \cdot sin(\phi)) - $ + \item $v_w = \cos\left(\theta_w-\phi\right)\cdot v_R = v_R \cdot \underbrace{\left(\cos\left(\theta_w\right)\cdot \cos\left(\phi\right) + \sin\left(\theta_w\right) \cdot \sin\left(\phi\right)\right)}_{T}$ \item \begin{align*} - v_1&= v_R \cdot (sin(\frac{\pi}{2}+\frac{\pi}{6})\cdot cos(\phi) + cos(\frac{\pi}{2}+\frac{\pi}{6}) \cdot sin(\phi))-l ^R\omega\\ - &=v_R \cdot (\frac{\sqrt{3}}{2} cos(\phi)- \frac{1}{2} sin(\phi)-l ^R\omega\\ - v_2&= v_R \cdot (sin(\frac{\pi}{2}-\frac{\pi}{2})\cdot cos(\phi) + cos(\frac{\pi}{2}+\frac{\pi}{2}) \cdot sin(\phi))-l ^R\omega\\ - &=v_R \cdot sin(\phi) -l^R\omega\\ - v_3&= v_R \cdot (sin(\frac{\pi}{2}+\frac{5\pi}{6})\cdot cos(\phi) + cos(\frac{\pi}{2}+\frac{5\pi}{6}) \cdot sin(\phi))-l ^R\omega\\ - &=v_R \cdot (-\frac{\sqrt{3}}{2} cos(\phi)- \frac{1}{2} sin(\phi))-l ^R\omega\\ - ^Rv_x &= v_R * cos(\phi)\\ - ^Rv_y &= v_R * sin(\phi)\\ + v_1&= v_R \cdot \left(\sin\left(\frac{\pi}{2}+\frac{\pi}{6}\right)\cdot \cos\left(\phi\right) + \cos\left(\frac{\pi}{2}+\frac{\pi}{6}\right) \cdot \sin\left(\phi\right)\right)-l ^R\omega\\ + &=v_R \cdot \left(\frac{\sqrt{3}}{2} \cos\left(\phi\right)- \frac{1}{2} \sin\left(\phi\right)\right)-l ^R\omega\\ + v_2&= v_R \cdot \left(\sin\left(\frac{\pi}{2}-\frac{\pi}{2}\right)\cdot \cos\left(\phi\right) + \cos\left(\frac{\pi}{2}+\frac{\pi}{2}\right) \cdot \sin\left(\phi\right)\right)-l ^R\omega\\ + &=v_R \cdot \sin\left(\phi\right) -l^R\omega\\ + v_3&= v_R \cdot \left(\sin\left(\frac{\pi}{2}+\frac{5\pi}{6}\right)\cdot \cos\left(\phi\right) + \cos\left(\frac{\pi}{2}+\frac{5\pi}{6}\right) \cdot \sin\left(\phi\right)\right)-l ^R\omega\\ + &=v_R \cdot \left(-\frac{\sqrt{3}}{2} \cos\left(\phi\right)- \frac{1}{2} \sin\left(\phi\right)\right)-l ^R\omega\\ + ^Rv_x &= v_R * \cos\left(\phi\right)\\ + ^Rv_y &= v_R * \sin\left(\phi\right)\\ \\ \Rightarrow \\ \begin{pmatrix} v_{w,1} \\ v_{w,2}\\ v_{w,3}\\ - \end{pmatrix} &= + \end{pmatrix} &= \begin{pmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2} &-l\\ 1 & 0 &-l\\ -\frac{\sqrt{3}}{2} & -\frac{1}{2} &-l\\ - \end{pmatrix} \cdot + \end{pmatrix} \cdot \begin{pmatrix} ^Rv_x \\ ^Rv_y\\ ^R\omega\\ - \end{pmatrix} + \end{pmatrix} \end{align*} - + \end{enumerate} - +\pagebreak \Aufgabe{}{8} \begin{enumerate}[(a)] \item $ ^bg= @@ -121,27 +120,26 @@ 0 \\ 0\\ -g \end{pmatrix} $ - \item $ - ^ba_m = ^bg+^ba_l\\ + \item $^ba_m = ^bg+^ba_l\\ $ if $ ^ba_m=0$ \\ $ ^ba_l = -^bg $ - \item \begin{align*} + \item static $\Rightarrow ^ba_l=0$\begin{align*} ^ba_m &= b_g+^ba_l\\ &= ^b\mathbf{R}_w \cdot \begin{pmatrix} 0 \\ 0\\ -g \end{pmatrix} \\ - &= + &= \begin{pmatrix} - sin(\theta)g \\-cos(\theta)*sin(\phi)g \\ -cos(\theta)*cos(\phi)g + \sin(\theta)g \\-\cos(\theta)*\sin(\phi)g \\ -\cos(\theta)*\cos(\phi)g \end{pmatrix}\\ - ^ba_{m,x}&= sin(\theta)g \\ - \theta &= sin^{-1}(\frac{^ba_{m,x}}{g})\\ - ^ba_{m,y} &= -g\cdot sin(\phi)\cdot \sqrt{1-\left(\frac{^ba_x}{g}\right)^2} \\ + ^ba_{m,x}&= \sin(\theta)g \\ + \theta &= \sin^{-1}(\frac{^ba_{m,x}}{g})\\ + ^ba_{m,y} &= -g\cdot \sin(\phi)\cdot \sqrt{1-\left(\frac{^ba_x}{g}\right)^2} \\ \phi&= \sin^{-1} \left(-\frac{^ba_y}{g*\sqrt{1-\left(\frac{^ba_x}{g}\right)^2}}\right) \end{align*} - \end{enumerate} + \end{enumerate}\pagebreak \Aufgabe{}{8} \begin{enumerate}[(a)] \item \begin{align*} @@ -151,38 +149,34 @@ p_0 &= 101325 Pa\\ p(a)&= p_0 \cdot exp(-\beta a)\\ p(a)_{lin}&= p(\overline{a}) + p'(\overline{a})\cdot (\overline{a}-a)\\ - p'(\overline{a})& = -\beta\cdot p_0 \cdot exp(-\beta\overline{a})\\ - p(a)_{lin}&= p_0 \cdot exp(-\beta\overline{a}) -\beta\cdot p_0 \cdot exp(-\beta\overline{a})\cdot (a-\overline{a})\\ + p'(\overline{a})& = -\beta\cdot p_0 \cdot exp(-\beta\overline{a})\\ + p(a)_{lin}&= p_0 \cdot exp(-\beta\overline{a}) -\beta\cdot p_0 \cdot exp(-\beta\overline{a})\cdot (a-\overline{a})\\ &= p_0 \cdot exp(-\beta\overline{a})(1-\beta(a-\overline{a})) \end{align*} - + \item \begin{align*} \overline{a}_1&= 100m \\ -p(a)_{lin1} &= p_0 \cdot exp(-\beta100m)(1-\beta(a-100m))\\ - p(a)_{lin1} &= 101317.937 - 11.86959596*a \\ +p(a)_{model} &= p_0 \cdot exp(-\beta100m)(1-\beta(a-100m))\\ + p(a)_{model} &= 101317.937 - 11.86959596*a \\ \overline{a_2}&= 10000m\\ -p(a)_{lin1} &= p_0 \cdot exp(-\beta 10000m)(1-\beta(a-10000m))\\ - p(a)_{lin2} &= 67675.77342 - 3.670864739*a\\ +p(a)_{passenger} &= p_0 \cdot exp(-\beta 10000m)(1-\beta(a-10000m))\\ + p(a)_{passenger} &= 67675.77342 - 3.670864739*a\\ \end{align*} - \item - \begin{align*} - \sigma_p= 5Pa\\ - p_{0l} &=101320Pa\\ - p_{0h} &=101330Pa\\ - p(a)_{lin1,p_{0l}} &= 101312.9373 - 11.86901024a\\ - p(a)_{lin1,p_{0h}} &= 101322.9366 - 11.87018168a\\ - \sigma_{a1} &= |\frac{p(a)_{lin1,p_{0l}}-p(a)_{lin1,p_{0h}}}{2}|\\ - &= |4.999651466 - 0.0005857190211a|\\ - p(a)_{lin2,p_{0l}} &= 100661.9642 - 10.66795293a\\ - p(a)_{lin2,p_{0h}} &= 100671.8993 - 10.66900582a\\ - \sigma_{a2} &= |\frac{p(a)_{lin2,p_{0l}}-p(a)_{lin2,p_{0h}}}{2}|\\ - &= |4.967526857 - 0.0005264485258a| - \end{align*} - Habe keinen Beleg gefunden wie man das Sigma wirklich berechnet. Das hier hab ich mir nur aus dem Kopf gezogen - %TODO JP muss seine Lösung coden - \end{enumerate} - - + \item + \begin{math} + \sigma_p= 5Pa\\ + a(p)=\frac{p-p_0e^{-\beta\bar{a}}-p_0\beta\bar{a}e^{-\beta\bar{a}}}{-p_0\beta e^{-\beta\bar{a}}}=-\frac{p}{\beta p_0}\cdot e^{\beta\bar{a}}+\beta^{-1}+\bar{a} + \end{math} + \begin{align*} + a(p\pm\sigma_p)&=-\frac{p\pm\sigma_p}{\beta p_0}\cdot e^{\bar{a}\beta}+\beta^{-1}+\bar{a}\\ + &=-\frac{p}{\beta p_0}\cdot e^{\bar{a}\beta}\pm\frac{\sigma_p}{\beta p_0}\cdot e^{\bar{a}\beta}+\beta^{-1}+\bar{a}\\ + \sigma_a&=\frac{\sigma_p}{\beta p_0}\cdot e^{\bar{a}\beta} + \end{align*} + $\Rightarrow \sigma_{a_{model}}=\frac{5}{\beta p_0}\cdot e^{100\beta}\approx 0.42124,\\ + \sigma_{a_{passenger}}=\frac{5}{\beta p_0}\cdot e^{10000\beta}\approx 1.3621$ + \end{enumerate} + + \end{document}