User:Egm6341.s10.team3.sa/Mtg41

Mtg 41: Tue, 13 Apr 10

 Typeset of transparencies, not lecture transcript. Subramanian Annamalai 18:13, 12 August 2010 (UTC)

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HW:
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The given equations of motion of aircraft are


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\mathbf{\dot{z}} = \begin{Bmatrix} \dot{\gamma} \\ \dot{v} \\ \dot{x} \\ \dot{y} \end{Bmatrix} = \begin{Bmatrix} \frac{T-D}{mv}\sin\alpha + \frac{L}{mv}\cos\alpha - \frac{g}{v}\cos\gamma \\ \frac{T-D}{m}\cos\alpha + \frac{L}{m}\sin\alpha - g\sin\gamma \\ v\cos\gamma \\ v\sin\gamma \end{Bmatrix} = \begin{Bmatrix} f_{1}(\mathbf{z}) \\ f_{2}(\mathbf{z}) \\ f_{3}(\mathbf{z}) \\ f_{4}(\mathbf{z}) \end{Bmatrix} = \mathbf{f}(\mathbf{z}) $$ $$
 * $$\displaystyle \color{blue}
 * $$\displaystyle \color{blue}
 * $$\displaystyle (Eq. 1)
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Here, $$\displaystyle \color{blue} D$$ and $$\displaystyle \color{blue} L$$ are
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\begin{cases} D(y,v,\alpha) = \frac{1}{2}C_{d}\rho v^{2}S_{ref} \\ L(y,v,\alpha) = \frac{1}{2}C_{l}\rho v^{2}S_{ref} \end{cases} $$
 * $$\displaystyle \color{blue}
 * $$\displaystyle \color{blue}
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where
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\begin{cases} C_{d}(\alpha) = A_{1}\alpha^2 + A_{2}\alpha + A_{3} \\ C_{l}(\alpha) = B_{1}\alpha + B_{2} \\ \rho(y) = C_{1}y^2 + C_{2}y + C_{3} \end{cases} $$
 * $$\displaystyle \color{blue}
 * $$\displaystyle \color{blue}
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Solve the above equations of motion of aircraft when the below control inputs $$\displaystyle T(t)$$ and $$\displaystyle \color{blue} \alpha(t)$$ are given.


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T(t) = \begin{cases} 6000 \; N, & \mbox{for }0 \leq t < 27 \; \mbox{and} \; 33 \leq t \leq 40 \\ 1000 \; N, & \mbox{for }27 \leq t < 33 \end{cases} $$ $$
 * $$\displaystyle \color{blue}
 * $$\displaystyle \color{blue}
 * $$\displaystyle \color{blue}(Eq. 2)
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\alpha(t) = \begin{cases} 0.03 \; rad, & \mbox{for }0 \leq t < 21 \\ \frac{0.13-0.09}{27-21}(t-21) + 0.09 \; rad, & \mbox{for }21 \leq t < 27 \\ \frac{-0.2+0.13}{33-21}(t-27) - 0.13 \; rad, & \mbox{for }27 \leq t < 33 \\ \frac{-0.13+0.2}{40-33}(t-33) - 0.2\; rad, & \mbox{for }33 \leq t \leq 40 \end{cases} $$ $$
 * $$\displaystyle \color{blue}
 * $$\displaystyle \color{blue}
 * $$\displaystyle \color{blue} (Eq. 3)
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The solution should be obtained using Hermite-Simpson (HS) algorithm and Newton-Raphson (NR) method.

Solve the same equations of motion using "ode45" command in Matlab to verify the solution by HS and NR is correct.




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Question: Why did Simpson's rule pop-out in HS algorithm? (Why not some other rules) Ans: Recall that Simpson's rule can integrate exactly polynomial less than or equal to 3. Refer the theorem on [[media: Egm6341.s10.mtg14.djvu | Page 14-1]] and it's geometric interpretation on [[media: Egm6341.s10.mtg13.djvu | Page 13-3]].

HS Algorithm: Collocation(enforcing compliance with the differential equation at 2 end-points and mid-point);Use cubic Hermitian interpolation Upon enforcing compliance with the differential equation at mid-point (closing gap), then Simpson's rule naturally pops out since it can exactly integrate a cubic polynomial with 3 points.

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HW:
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$$ \displaystyle \color{blue} i)\, Solve\, logistic\, equation\, using\, the\, inconsistent\, trap.\, simpson\, algorithm\, $$

$$ \displaystyle \color{blue} z_{i+1} = z_i + \frac{(\frac{h}{2})}{3}*[f_i + 4f_{i+\frac{1}{2}}+f_{i+1}]\, $$

$$ \displaystyle \color{blue} with\,\, z_{i+1} = \frac{1}{2}*(z_i + z_{i+1})\, $$

$$ \displaystyle \color{blue} ii)\, Compare\, with\, the\, results\, of\, the\, Hermite-Simpson\, Algorithm\, using\, same\, values\, of\, h\, $$


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