Aircraft¶

In this section we consider the model of a unmanned vertical take-off aircraft. The aircraft has two permanently mounted thrusters on the wings which can apply the thrust forces $F_1$ and $F_2$ independently of each other. The two engines are inclined by an angle $\alpha$ with respect to the aircraft-fixed axis $\eta_2$ and engage in the points $P_1 = (l, h)$ and $P_2 = (-l,-h)$ . The coordinates of the center of mass $M$ of the aircraft in the inertial system are denoted by $z_1$ and $z_2$ . At the same time, the point is the origin of the plane coordinate system. The aircraft axes are rotated by the angle $\theta$ with respect to the $z_2$ -axis.

Through the establishment of the momentum balances for the model one obtains the equations

$\begin{eqnarray*} m \ddot{z}_1 & = & - \sin(\theta)(F_1 + F_2)\cos(\alpha) + \cos(\theta)(F_1 - F_2)\sin(\alpha) \\ m \ddot{z}_2 & = & \cos(\theta)(F_1 + F_2)\sin(\alpha) + \sin(\theta)(F_1 - F_2)\cos(\alpha) - mg \\ J \ddot{\theta} & = & (F_1 - F_2)(l \cos(\alpha) + h \sin(\alpha)) \end{eqnarray*}$

With the state vector $x = [z_1, \dot{z}_1, z_2, \dot{z}_2, \theta, \dot{\theta}]^T$ and $u = [u_1, u_2]^T = [F_1, F_2]^T$ the state space representation of the system is as follows.

$\begin{eqnarray*} \dot{x}_1 & = & x_2 \\ \dot{x}_2 & = & \frac{1}{m}(-\sin(x_5)(u_1 + u_2)\cos(\alpha) + \cos(x_5)(u_1 - u_2)\sin(\alpha)) \\ \dot{x}_3 & = & x_4 \\ \dot{x}_2 & = & \frac{1}{m}(\cos(x_5)(u_1 + u_2)\cos(\alpha) + \sin(x_5)(u_1 - u_2)\sin(\alpha)) - g \\ \dot{x}_5 & = & x_6 \\ \dot{x}_6 & = & \frac{1}{J}(l \cos(\alpha) + h \sin(\alpha)) \end{eqnarray*}$

For the aircraft, a trajectory should be planned that translates the horizontally aligned flying object from a rest position (hovering) along the $z_1$ and $z_2$ axis back into a hovering position. The hovering is to be realized on the boundary conditions of the input. Therefor the derivatives of the state variables should satisfy the following conditions.

$\dot{z}_1 = \ddot{z}_1 = \dot{z}_2 = \ddot{z_2} = \dot{\theta} = \ddot{\theta} = 0$

For the horizontal position applies $\theta = 0$ . These demands yield the boundary conditions for the inputs.

$F_1(0) = F_1(T) = F_2(0) = F_2(T) = \frac{mg}{2 \cos(\alpha)}$

Source Code¶# vertical take-off aircraft

# import trajectory class and necessary dependencies
from pytrajectory import ControlSystem
from sympy import sin, cos
import numpy as np
from numpy import pi

# define the function that returns the vectorfield
def f(x,u):
    x1, x2, x3, x4, x5, x6 = x  # system state variables
    u1, u2 = u                  # input variables
    
    # coordinates for the points in which the engines engage [m]
    l = 1.0
    h = 0.1

    g = 9.81    # graviational acceleration [m/s^2]
    M = 50.0    # mass of the aircraft [kg]
    J = 25.0    # moment of inertia about M [kg*m^2]

    alpha = 5/360.0*2*pi    # deflection of the engines

    sa = sin(alpha)
    ca = cos(alpha)

    s = sin(x5)
    c = cos(x5)
    
    ff = np.array([             x2,
                    -s/M*(u1+u2) + c/M*(u1-u2)*sa,
                                x4,
                    -g+c/M*(u1+u2) +s/M*(u1-u2)*sa ,
                                x6,
                    1/J*(u1-u2)*(l*ca+h*sa)])
    
    return ff 

# system state boundary values for a = 0.0 [s] and b = 3.0 [s]
xa = [ 0.0, 0.0, 0.0, 0.0, 0.0, 0.0]
xb = [10.0, 0.0, 5.0, 0.0, 0.0, 0.0]

# boundary values for the inputs
ua = [0.5*9.81*50.0/(cos(5/360.0*2*pi)), 0.5*9.81*50.0/(cos(5/360.0*2*pi))]
ub = [0.5*9.81*50.0/(cos(5/360.0*2*pi)), 0.5*9.81*50.0/(cos(5/360.0*2*pi))]

# create trajectory object
S = ControlSystem(f, a=0.0, b=3.0, xa=xa, xb=xb, ua=ua, ub=ub)

# don't take advantage of the system structure (integrator chains)
# (this will result in a faster solution here)
S.set_param('use_chains', False)

# also alter some other method parameters to increase performance
S.set_param('kx', 5)

# run iteration
S.solve()