# University of Florida/Egm6321/f09.team1.gzc/Mtg34

Mtg 34: Wed, 23 Mar 11

 1) Supersonic intercaptor: Find angle of attack and thrust to reach a given altitude inmin time Reference: Bryson & Denham 1962 2) Bunt maneuver of aircraft: min altitude to reach target with a bunt maneuver (inverted loop) ${\displaystyle {\color {blue}V}\ =\ velocity\ of\ P\ (airplane\ =\ pt)}$ ${\displaystyle {\color {blue}T}\ =\ thrust\ //\ airplace\ axis}$ ${\displaystyle \alpha \ =\ \angle \ ({\color {blue}V,}\ {\color {red}T})\ =\ angle\ of\ attack}$ ${\displaystyle \gamma \ =\ \angle \ (x,\ {color{blue}V})\ =\ angle\ between\ honz\ x\ axis\ and\ veloctity}$ ${\displaystyle D\ =\ axial\ aero\ force\ //\ {\color {red}T}}$ ${\displaystyle (parall.\ airplane\ axis)}$ ${\displaystyle L\ =\ Tramsv.\ aero.\ force\ \bot \ {\color {red}T}}$ ${\displaystyle W\ =\ mg\ =\ airplane\ weight.}$ ${\displaystyle Equation\ of\ motion\ (EOM)}$ ${\displaystyle {color{blue}Control\ input:}\ {\color {blue}{\underset {time\ dependent}{\underbrace {{\color {red}T}(t)} \ ,\ \alpha (t)}}}}$ ${\displaystyle {\color {blue}state\ var,:}\ x(t),\ y(t)\ v(t),\ r(t)}$ ${\displaystyle {\color {blue}{\begin{cases}&\ {\color {black}D={\frac {L}{2}}\ C_{d}\ \rho \ V^{2}\ S_{reft}}\\&\ {\color {black}L={\frac {L}{2}}\ C_{l}\ \rho \ V^{2}\ S_{reft}}\end{cases}}}}$ ${\displaystyle C_{d}=drg\ coefficient\ {\color {green}|}\ \rho =air\ density}$ ${\displaystyle C_{l}=lift\ coefficient\ {\color {green}|}\ \S _{ref}=ref.\ area\ of\ airplane}$ ${\displaystyle C_{d}=A_{1}\alpha ^{2}+A_{2}\alpha +A_{3}\ {\color {red}(1)}}$ ${\displaystyle (A_{1},\ A_{2},\ A_{3})=curve\ fitting\ coefficient}$ ${\displaystyle C_{l}=B_{1}\alpha +B_{2}\ {\color {red}(2)}}$ ${\displaystyle (B_{1},\ B_{2})=curve\ fitting\ coefficient}$ ${\displaystyle \rho =C_{1}h^{2}+C_{2}h+C_{3}\ {\color {red}(3)}}$ ${\displaystyle (C_{1},\ C_{2},\ C_{3})=curve\ fitting\ coefficient}$ Reference: S & Z 2007 ${\displaystyle {\underline {\color {blue}{Kinematics:}}}\ {\frac {dx}{dt}}=V_{x}=Vcosr\ {\color {red}(4)}}$ ${\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\frac {dy}{dt}}=V_{y}=Vsinr\ {\color {red}(5)}}$ ${\displaystyle {\underline {\color {blue}{Kinematics:}}}{\color {blue}Euler\ equations:}\ {\frac {d}{dt}}{\underline {P}}=\sum _{\color {red}i}{\underline {F}}_{\color {red}i}\ {\color {red}(6)}}$ Note: Parameteriztion of ellipse ${\displaystyle {\color {blue}{\underset {cos^{2}t}{\underbrace {\color {black}\left({\frac {x}{a}}\right)^{2}} }}}+{\color {blue}{\underset {sin^{2}t}{\underbrace {\color {black}\left({\frac {y}{b}}\right)^{2}} }}}=1\ {\color {blue}ellipse}}$ ${\displaystyle <{\color {blue}dl={\color {red}[{\color {blue}dx^{2}+dy^{2}}]^{1/2}}}/math>[itex]Eccentricity:\ e=\left(1-{\frac {b^{2}}{a^{2}}}\right)^{1/2}}$ ${\displaystyle C=\int dl=a\int _{t=0}^{2\pi }[1-e^{2}cos^{2}t]^{1/2}dt}$ ${\displaystyle =41\int _{\alpha =0}^{\frac {\pi }{2}}[1-e^{2}sin^{2}\alpha ]^{\frac {1}{2}}d\alpha =4aE(e)}$