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【Paper】2006_ Time-Optimal Control of a Hovering Quad-Rotor Helicopter

2022-06-30 04:34:00 【Zhao-Jichao】

Lai L C, Yang C C, Wu C J. Time-optimal control of a hovering quad-rotor helicopter[J]. Journal of Intelligent and Robotic Systems, 2006, 45(2): 115-135.

List of articles

1. Introduction
2. Dynamical Equations of a Quad-Rotor Helicopter

1. Introduction

2. Dynamical Equations of a Quad-Rotor Helicopter

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To illustrate the motion of the helicopter , Pictured 1 A schematic diagram is shown in . In the working space of the helicopter , Define the world coordinate system and the body coordinate system . The world coordinate system represents a coordinate system that all discussions can refer to , The body coordinate system is a coordinate system attached to the helicopter . In order to keep the four rotor helicopter hovering , The four driving forces $F_i, i = 1,2,3,4$ The size of the helicopter is adjusted to a quarter of the original helicopter weight . By changing the speed of four rotors at the same time , Can realize the body along $z$ The vertical movement of the axis . By reversing 1 Number and 3 Speed of rotor No , keep 2 Number and 4 Speed of rotor No , Can realize the body along $x$ Forward movement in axis direction . Reverse change 2 Number and 4 Speed of rotor No , keep 1 Number and 3 Speed of rotor No , The utility model can realize the following functions of the machine body $y$ The lateral movement of the shaft . The yaw motion is related to the difference between the moments produced by the rotor . For clockwise rotation , rotor 2 and 4 Speed should be increased , To overcome the rotor 1 and 3 The speed of . On the other hand , To rotate counterclockwise , rotor 3 and 1 The speed should be increased , To overcome the rotor 2 and 4 Speed of . surface 1 The symbol names used in this paper are summarized .

According to the Euler angle, the rotation transformation matrix between the world coordinate system and the body coordinate system can be obtained .

$R_{EB} = \left[\begin{matrix} \cos\theta \cos\psi & \sin\theta \cos\psi \sin\phi - \sin\psi \cos\phi & \sin\theta \cos\psi \cos\phi + \sin\psi \sin\phi \\ \cos\theta \sin\psi & \sin\theta \sin\psi \sin\phi \red{-} \cos\psi \cos\phi & \sin\theta \sin\psi \cos\phi - \cos\psi \sin\phi \\ -\sin\theta & \cos\theta \sin\phi & \cos\theta \cos\phi \\ \end{matrix}\right] \tag{1}$

The original text is the above formula , There should be a sign error （ Red ）, The correct should be as follows
$R_{EB} = \left[\begin{matrix} \cos\theta \cos\psi & \sin\theta \cos\psi \sin\phi - \sin\psi \cos\phi & \sin\theta \cos\psi \cos\phi + \sin\psi \sin\phi \\ \cos\theta \sin\psi & \sin\theta \sin\psi \sin\phi + \cos\psi \cos\phi & \sin\theta \sin\psi \cos\phi - \cos\psi \sin\phi \\ -\sin\theta & \cos\theta \sin\phi & \cos\theta \cos\phi \\ \end{matrix}\right] \tag{1}$

Then the velocity transformation between the body coordinate system and the world coordinate system is obtained
$\left[\begin{matrix} u \\ v \\ w \\ \end{matrix}\right] =R_{EB} \left[\begin{matrix} u_B \\ v_B \\ w_B \\ \end{matrix}\right] \tag{2}$

Similarly , The acceleration 、 Rotation speed 、 Location 、 Both forces and moments can be based on $R_{EB}$ Convert between coordinate systems .

In the body coordinate system , Force is defined as
$F_{B}= \left[\begin{matrix} F_{xB} \\ F_{yB} \\ F_{zB} \\ \end{matrix}\right]= \left[\begin{matrix} 0 \\ 0 \\ \sum_{k=1}^{4} F_k \\ \end{matrix}\right] \tag{3}$

In the world coordinate system , Force can be described as
$\begin{aligned} \left[\begin{matrix} F_{x} \\ F_{y} \\ F_{z} \\ \end{matrix}\right]= R_{EB} \cdot F_B = (\sum_{k=1}^{4} F_k) \left[\begin{matrix} \sin\theta \cos\psi \cos\phi + \sin\psi \sin\phi \\ \sin\theta \sin\psi \cos\phi - \cos\psi \sin\phi \\ \cos\theta \cos\phi \\ \end{matrix}\right] \end{aligned} \tag{4}$

therefore ,UAV The movement equation in the world coordinate system is
$\left[\begin{matrix} \ddot{x} \\ \ddot{y} \\ \ddot{z} \\ \end{matrix}\right]= \left[\begin{matrix} F_{x} - K_1 \cdot \dot{x} \\ F_{y} - K_2 \cdot \dot{y} \\ F_{z} - mg - K_3 \cdot \dot{z} \\ \end{matrix}\right] \tag{5}$

here $K_i, i= 1,2,3$ It's the drag coefficient . Be careful , These coefficients are negligible at low speeds .

therefore , The equation of motion can be derived from the balance of force and moment .

$\ddot{\phi} = \frac{l (F_3 - F_1 - K_4 \dot{\phi})}{I_x} \tag{6}$

$\ddot{\theta} = \frac{l (F_4 - F_2 - K_5 \dot{\theta})}{I_y} \tag{7}$

$\ddot{\psi} = \frac{(M_1 - M_2 + M_3 - M_4 - K_6\dot{\psi})}{I_z} \tag{8}$

here $l$ The length from the center of gravity of the UAV to each rotor , $M_i, i=1,2,3,4$ Is the moment of the rotor , $I_x, I_y, I_z$ Express $x, y, z$ Moment of inertia in direction . use $F_1, F_2, F_3, F_4$ To express ,（8） It can be rewritten as

$\ddot{\psi} = \frac{(F_1 - F_2 + F_3 - F_4 - K_6^{'} \dot{\psi})}{I_z^{'}} \tag{9}$

here $I_z^{'} = I_z /C, K_6^{'}=K_6/C$ , $C$ Is a scaling factor .

For ease of calculation TOMP problem , Define the input as
$\begin{aligned} u_1 =& F_1 + F_2 + F_3 + F_4 \\ u_2 =& - F_2 + F_4 \\ u_3 =& - F_1 + F_3 \\ u_4 =& F_1 - F_2 + F_3 - F_4 \\ \end{aligned} \tag{10}$

The input can be expressed as a matrix
$\left[\begin{matrix} u_1 \\ u_2 \\ u_3 \\ u_4 \\ \end{matrix}\right]= \left[\begin{matrix} 1 & 1 & 1 & 1 \\ 0 & -1 & 0 & 1 \\ -1 & 0 & 1 & 0 \\ 1 & -1 & 1 & -1 \\ \end{matrix}\right] \left[\begin{matrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ \end{matrix}\right] \tag{11}$

So the single force will be
$\left[\begin{matrix} F_1 \\ F_2 \\ F_3 \\ F_4 \\ \end{matrix}\right]= \frac{1}{4} \left[\begin{matrix} 1 & 0 & -2 & 1 \\ 1 & -2 & 0 & -1 \\ 1 & 0 & 2 & 1 \\ 1 & 2 & 0 & -1 \\ \end{matrix}\right] \left[\begin{matrix} u_1 \\ u_2 \\ u_3 \\ u_4 \\ \end{matrix}\right] \tag{12}$

therefore , The dynamic equations of the four rotor helicopter are obtained

$\ddot{x}= \frac{(\sin \psi \sin \phi + \cos \psi \sin \theta \cos \phi) u_1 - K_1 \dot{x}}{m} \tag{13}$

$\ddot{y}= \frac{(\sin \psi \sin \theta \cos \phi - \cos \psi \sin \phi) u_1 - K_2 \dot{y}}{m} \tag{14}$

$\ddot{z}= \frac{(\cos \phi \cos \theta) u_1 - K_3 \dot{z}}{m} -g \tag{15}$

$\ddot{\phi} = \frac{(u_3 - K_4 \dot{\phi})~ l}{I_x} \tag{16}$

$\ddot{\theta} = \frac{(u_2 - K_5 \dot{\theta}) ~l}{I_y} \tag{17}$

$\ddot{\psi} = \frac{(u_4 - K_6^{'} \dot{\psi})}{I_z^{'}} \tag{18}$

For hovering flight , Speed and roll angle 、 The pitch and yaw angles must be zero , And the four driving forces are $F_1(0)=F_2(0)=F_3(0)=F_4(0)=\frac{mg}{4}$ . Substitute these conditions into the formula （13） To formula （18）, It can be found that there will be zero acceleration .