PlanetPhysics/2D With Drag Airship Optimal Control

2D with Drag and No Wind

{\mathbf Disclaimer}: This is a [[../Work/|work]] in progress... \\

The [[../Thrust/|force]] due to drag in 2D is given by

$F_d=\frac{1}{2}{C}_d\rho A{v}^2$

Unlike the 1D case, which uses the absolute value, the direction of the [[../CoriolisEffect/|drag force]] in 2D is taken into account by the [[../AbsoluteMagnitude/|magnitude]] of the [[../Velocity/|velocity]] times the component velocities such that the x and y components of the drag force are

$F_d^x=-\frac{1}{2}{C}_d\rho A\dot{x}\sqrt{{\dot{x}}^2+{\dot{y}}^2}$ $F_d^y=-\frac{1}{2}{C}_d\rho A\dot{y}\sqrt{{\dot{x}}^2+{\dot{y}}^2}$

The equations of [[../CosmologicalConstant/|motion]] for this problem are

$\dot{x}=\dot{x}$ $\dot{y}=\dot{y}$ $\ddot{x}=F_d^x+\frac{T}{m}\cos \alpha$ $\ddot{y}=F_d^y+\frac{T}{m}\sin \alpha$

where T is the applied Torque in Newtons, $\alpha$ is the angle of applied torque counter clockwise from the negative x-axis and $m$ is the [[../CosmologicalConstant/|mass]] of the airship.

The performance measure for the minimum [[../CosmologicalConstant/|energy]] problem assuming there is no effort required to change the angle of attack

$J=\underset{0}{\overset{t_f}{\int }}\frac{T^2}{m^2}dt$

The [[../Hamiltonian2/|Hamiltonian]] for this optimal control problem

$\mathscr{H}=\frac{T^2}{m^2}+p_1\dot{x}+p_2\dot{y}+p_3(F_d^x+\frac{T}{m}\cos \alpha )+p_4(F_d^y+\frac{T}{m}\sin \alpha )$

The neccessary conditions for the unconstrained control inputs

$\frac{\partial \mathscr{H}}{\partial \alpha }=0$ $\frac{\partial \mathscr{H}}{\partial T}=0$ \\ $\frac{\partial \mathscr{H}}{\partial \alpha }=-p_3\frac{T}{m}\sin \alpha +p_4\frac{T}{m}\cos \alpha =0$ $\frac{p_{4}T}{m}\cos \alpha =\frac{p_{3}T}{m}\sin \alpha$ $\frac{p_4}{p_3}=\frac{\sin \alpha }{\cos \alpha }$

$\tan \alpha =\frac{p_4}{p_3}$

$\frac{\partial \mathscr{H}}{\partial T}=\frac{2T}{m^2}+\frac{p_3}{m}\cos \alpha +\frac{p_4}{m}sin\alpha =0$ $\frac{2T}{m}+p_3\cos \alpha +p_{4}sin\alpha =0$ $\frac{2T}{m}=-(p_3\cos \alpha +p_{4}sin\alpha )$

$T=-\frac{m}{2}(p_3\cos \alpha +p_{4}sin\alpha )$

The neccessary conditions for the costates are

$\dot{p_1}=-\frac{\partial \mathscr{H}}{\partial x}=0$

$\dot{p_2}=-\frac{\partial \mathscr{H}}{\partial y}=0$

$\dot{p_3}=-\frac{\partial \mathscr{H}}{\partial \dot{x}}=-p_1+\frac{1}{2}{p}_3{C}_d\rho A\sqrt{{\dot{x}}^2+{\dot{y}}^2}+\frac{p_3{C}_d\rho A{\dot{x}}^2}{\sqrt{{\dot{x}}^2+{\dot{y}}^2}}+\frac{p_4{C}_d\rho A\dot{x}\dot{y}}{\sqrt{{\dot{x}}^2+{\dot{y}}^2}}$

$\dot{p_4}=-\frac{\partial \mathscr{H}}{\partial \dot{y}}=-p_2+\frac{1}{2}{p}_4{C}_d\rho A\sqrt{{\dot{x}}^2+{\dot{y}}^2}+\frac{p_4{C}_d\rho A{\dot{y}}^2}{\sqrt{{\dot{x}}^2+{\dot{y}}^2}}+\frac{p_3{C}_d\rho A\dot{x}\dot{y}}{\sqrt{{\dot{x}}^2+{\dot{y}}^2}}$