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[UAV] kinematic analysis from single propeller to four rotor UAV

2022-06-30 04:35:00 【Zhao-Jichao】

List of articles

1 Force analysis of single propeller

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When a single propeller rotates , Pictured above M 1 M1 M1 and M 2 M2 M2, Will produce an upward lift T 1 , T 2 T_1, T_2 T1​,T2​ And rotational moment M 1 , M 2 M_1, M_2 M1​,M2​, The lift direction is vertical and the propeller is vertically upward , The direction of the rotational moment is generally assumed to be parallel to the plane of the aircraft , Perpendicular to boom .

2 Coordinate change

Before analyzing the impact of the four propellers on the UAV , First establish two coordinate systems , They are the world coordinate system S g S_g Sg​ And body coordinate system S b S_b Sb​. At the same time, the transformation relationship between the two is given .

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In the world coordinate system S g S_g Sg​ And body coordinate system S b S_b Sb​ The transformation matrix between is ：

M g 2 b = [ cos ⁡ θ cos ⁡ ψ cos ⁡ θ sin ⁡ ψ − sin ⁡ θ sin ⁡ ϕ sin ⁡ θ cos ⁡ ψ − cos ⁡ ϕ sin ⁡ ψ sin ⁡ ϕ sin ⁡ θ sin ⁡ ψ + cos ⁡ ϕ cos ⁡ ψ sin ⁡ ϕ cos ⁡ θ cos ⁡ ϕ sin ⁡ θ cos ⁡ ψ + sin ⁡ ϕ sin ⁡ ψ cos ⁡ ϕ sin ⁡ θ sin ⁡ ψ − sin ⁡ ϕ cos ⁡ ψ cos ⁡ ϕ cos ⁡ θ ] M_{g2b} = \left[\begin{matrix} \cos\theta \cos\psi & \cos\theta \sin\psi & -\sin\theta \\ \sin\phi \sin\theta \cos\psi - \cos\phi \sin\psi & \sin\phi \sin\theta \sin\psi + \cos\phi \cos\psi & \sin\phi \cos\theta \\ \cos\phi \sin\theta \cos\psi + \sin\phi \sin\psi & \cos \phi \sin\theta \sin\psi - \sin\phi \cos\psi & \cos\phi \cos\theta \\ \end{matrix}\right] Mg2b​=⎣⎡​cosθcosψsinϕsinθcosψ−cosϕsinψcosϕsinθcosψ+sinϕsinψ​cosθsinψsinϕsinθsinψ+cosϕcosψcosϕsinθsinψ−sinϕcosψ​−sinθsinϕcosθcosϕcosθ​⎦⎤​

The transformation between the ground coordinate system and the body coordinate system satisfies the equation （ For the replacement of this matrix relation, please refer to ）
S b = M g 2 b ⋅ S g    or    S g = M b 2 g ⋅ S b S_b = M_{g2b} \cdot S_g ~~ \text{ or } ~~ S_g = M_{b2g} \cdot S_b Sb​=Mg2b​⋅Sg​   or   Sg​=Mb2g​⋅Sb​

M b 2 g = [ cos ⁡ θ cos ⁡ ψ sin ⁡ ϕ sin ⁡ θ cos ⁡ ψ − cos ⁡ ϕ sin ⁡ ψ cos ⁡ ϕ sin ⁡ θ cos ⁡ ψ + sin ⁡ ϕ sin ⁡ ψ cos ⁡ θ sin ⁡ ψ sin ⁡ ϕ sin ⁡ θ sin ⁡ ψ + cos ⁡ ϕ cos ⁡ ψ cos ⁡ ϕ sin ⁡ θ sin ⁡ ψ − sin ⁡ ϕ cos ⁡ ψ − sin ⁡ θ sin ⁡ ϕ cos ⁡ θ cos ⁡ ϕ cos ⁡ θ ] M_{b2g}= \left[\begin{matrix} \cos\theta \cos\psi & \sin\phi \sin\theta \cos\psi - \cos\phi \sin\psi & \cos\phi \sin\theta \cos\psi + \sin\phi \sin\psi \\ \cos\theta \sin\psi & \sin\phi \sin\theta \sin\psi + \cos\phi \cos\psi & \cos\phi \sin\theta \sin\psi - \sin\phi \cos\psi \\ -\sin\theta & \sin\phi \cos\theta & \cos\phi \cos\theta \\ \end{matrix}\right] Mb2g​=⎣⎡​cosθcosψcosθsinψ−sinθ​sinϕsinθcosψ−cosϕsinψsinϕsinθsinψ+cosϕcosψsinϕcosθ​cosϕsinθcosψ+sinϕsinψcosϕsinθsinψ−sinϕcosψcosϕcosθ​⎦⎤​

For the detailed derivation process of coordinate system transformation, please refer to ： The first 5 Chapter - Unmanned aerial vehicle (uav) UAV model analysis .

3 The impact of the four propellers on the UAV

According to the difference between the established body coordinates and the UAV arm , UAVs can be divided into “ Ten ” Word flight mode and “X” Type a flight mode , As shown on the left and right of the figure below . About “ Ten ” Word flight mode , More discussion . But in actual flight , Use “X” Type mode more . So here we use “X” Type coordinates to perform the following analysis .

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3.1 Effect of rotor on position

About the plane right Z b Z_b Zb​ The influence of the axis , It's easy to understand , At the same time, the lift of the propeller is increased T i T_i Ti​, The UAV will be lifted vertically , So there is
T Z b = T 1 + T 2 + T 3 + T 4 (1) T_{Z_b} = T_1 + T_2 + T_3 + T_4 \tag{1} TZb​​=T1​+T2​+T3​+T4​(1)

among T Z b T_{Z_b} TZb​​ It is expressed in the body coordinate system Z b Z_b Zb​ The force on an aircraft .

At this time, don't worry about mapping the acceleration to the Newton's law Z ¨ b \ddot{Z}_b Z¨b​ On , Otherwise, it would be difficult to switch down .

The effect of gravitational acceleration is not considered here , Because now Z b Z_b Zb​ Or the value in the body coordinate system . And the body coordinate system will change , The acceleration of gravity is constant relative to the world coordinate system .

But how to change the lift of the four propellers , Will not occur in X b X_{b} Xb​ and Y b Y_{b} Yb​ Change of direction , So there is
T X b = 0 (2) T_{X_b} = 0 \tag{2} TXb​​=0(2)

T Y b = 0 (3) T_{Y_b} = 0 \tag{3} TYb​​=0(3)

3.2 Effect of rotor on attitude angle

At the same time, the UAV also has three attitude angles , They are rolling angles ϕ \phi ϕ, Pitch angle θ \theta θ, Heading angle ψ \psi ψ, Note that these three attitude angles are relative to the body coordinate system S b S_b Sb​ In terms of the . How do they change ？

First, we study the relatively simple heading angle ψ \psi ψ. We know that when the propeller rotates, it will produce a rotating torque , The rotating torque generated by the four propellers is just the opposite direction of the two adjacent propellers , So it just offsets . Otherwise, the UAV will rotate in place . So we want the drone to rotate in place , It can be realized by adjusting the speed of the four propellers . Be careful , At this time, whether it is “ Ten ” Word or “X” The type pattern is the same . For example, we can reduce M 1 ( T 1 ) , M 3 ( T 3 ) M_1(T_1), M_3(T_3) M1​(T1​),M3​(T3​), Improve M 2 ( T 2 ) , M 4 ( T 4 ) M_2(T_2), M_4(T_4) M2​(T2​),M4​(T4​), So the aircraft keeps the total lift constant , Counter clockwise rotation occurs , To change the heading angle .

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When air resistance is not considered , Yes
ψ ¨ = ( − M 1 + M 2 − M 3 + M 4 ) I z (4) \ddot{\psi} = \frac{(-M_1 + M_2 - M_3 + M_4)}{I_z} \tag{4} ψ¨​=Iz​(−M1​+M2​−M3​+M4​)​(4)

When considering air resistance , Yes
ψ ¨ = ( − M 1 + M 2 − M 3 + M 4 − K 6 ψ ˙ ) I z (4) \ddot{\psi} = \frac{(-M_1 + M_2 - M_3 + M_4 - K_6 \dot{\psi})}{I_z} \tag{4} ψ¨​=Iz​(−M1​+M2​−M3​+M4​−K6​ψ˙​)​(4)

among K 6 K_6 K6​ It's the drag coefficient （drag coefficient）, Determined by aerodynamics .

And change the roll angle ϕ \phi ϕ And pitch angle θ \theta θ, Is the lift difference between the four propellers of the aircraft . The first is the roll angle ϕ \phi ϕ, Because the roll angle is around X b X_b Xb​ The shaft rotates , According to the figure , We need to improve T 2 , T 3 T_2, T_3 T2​,T3​, Reduce T 1 , T 4 T_1, T_4 T1​,T4​. Empathy , Change the pitch angle θ \theta θ, The pitch angle is around Y b Y_b Yb​ The shaft rotates , So promote T 1 , T 2 T_1, T_2 T1​,T2​, Reduce T 3 , T 4 T_3, T_4 T3​,T4​.

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When air resistance is not considered , Yes
ϕ ¨ = L ( − T 1 + T 2 + T 3 − T 4 ) I x (5) \ddot{\phi} = \frac{L (- T_1 + T_2 + T_3 - T_4)}{I_x} \tag{5} ϕ¨​=Ix​L(−T1​+T2​+T3​−T4​)​(5)

θ ¨ = L ( T 1 + T 2 − T 3 − T 4 ) I y (6) \ddot{\theta} = \frac{L (T_1 + T_2 - T_3 - T_4)}{I_y} \tag{6} θ¨=Iy​L(T1​+T2​−T3​−T4​)​(6)

When considering air resistance , Yes
ϕ ¨ = L ( − T 1 + T 2 + T 3 − T 4 − K 4 ϕ ˙ ) I x (5) \ddot{\phi} = \frac{L (- T_1 + T_2 + T_3 - T_4 - K_4 \dot{\phi})}{I_x} \tag{5} ϕ¨​=Ix​L(−T1​+T2​+T3​−T4​−K4​ϕ˙​)​(5)

θ ¨ = L ( T 1 + T 2 − T 3 − T 4 − K 5 θ ˙ ) I y (6) \ddot{\theta} = \frac{L (T_1 + T_2 - T_3 - T_4 - K_5 \dot{\theta})}{I_y} \tag{6} θ¨=Iy​L(T1​+T2​−T3​−T4​−K5​θ˙)​(6)

A little bit of the relationship between torque and angular momentum is used here , Please refer to 【 control 】 Attitude angle analysis of four rotor UAV .

4 Output in body coordinates

In this way, we get the expression of six outputs of UAV （ Notice that gravity is not taken into account at this time , And the stress analysis is in the body coordinate system ）.
{ T X b = 0 T Y b = 0 T Z b = T 1 + T 2 + T 3 + T 4 ϕ ¨ = L ( − T 1 + T 2 + T 3 − T 4 − K 4 ϕ ˙ ) I x θ ¨ = L ( T 1 + T 2 − T 3 − T 4 − K 5 θ ˙ ) I y ψ ¨ = ( − M 1 + M 2 − M 3 + M 4 − K 6 ψ ˙ ) I z (7) \left\{\begin{aligned} T_{X_b} &= 0 \\ T_{Y_b} &= 0 \\ T_{Z_b} &= T_1 + T_2 + T_3 + T_4 \\ \ddot{\phi} &= \frac{L (- T_1 + T_2 + T_3 - T_4 - K_4 \dot{\phi})}{I_x} \\ \ddot{\theta} &= \frac{L (T_1 + T_2 - T_3 - T_4 - K_5 \dot{\theta})}{I_y} \\ \ddot{\psi} &= \frac{(-M_1 + M_2 - M_3 + M_4 - K_6 \dot{\psi})}{I_z} \\ \end{aligned}\right. \tag{7} ⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​TXb​​TYb​​TZb​​ϕ¨​θ¨ψ¨​​=0=0=T1​+T2​+T3​+T4​=Ix​L(−T1​+T2​+T3​−T4​−K4​ϕ˙​)​=Iy​L(T1​+T2​−T3​−T4​−K5​θ˙)​=Iz​(−M1​+M2​−M3​+M4​−K6​ψ˙​)​​(7)

When the plane is flying at low speed , The drag coefficient is often negligible , So there can be
{ T X b = 0 T Y b = 0 T Z b = T 1 + T 2 + T 3 + T 4 ϕ ¨ = L ( − T 1 + T 2 + T 3 − T 4 ) I x θ ¨ = L ( T 1 + T 2 − T 3 − T 4 ) I y ψ ¨ = ( − M 1 + M 2 − M 3 + M 4 ) I z (7) \left\{\begin{aligned} T_{X_b} &= 0 \\ T_{Y_b} &= 0 \\ T_{Z_b} &= T_1 + T_2 + T_3 + T_4 \\ \ddot{\phi} &= \frac{L (- T_1 + T_2 + T_3 - T_4)}{I_x} \\ \ddot{\theta} &= \frac{L (T_1 + T_2 - T_3 - T_4)}{I_y} \\ \ddot{\psi} &= \frac{(-M_1 + M_2 - M_3 + M_4)}{I_z} \\ \end{aligned}\right. \tag{7} ⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​TXb​​TYb​​TZb​​ϕ¨​θ¨ψ¨​​=0=0=T1​+T2​+T3​+T4​=Ix​L(−T1​+T2​+T3​−T4​)​=Iy​L(T1​+T2​−T3​−T4​)​=Iz​(−M1​+M2​−M3​+M4​)​​(7)

The position state in the body coordinates X b , Y b , Z b X_b, Y_b, Z_b Xb​,Yb​,Zb​ It is inconvenient for us to study , So we use coordinate transformation , Convert to world coordinates S g S_g Sg​ Next .

5 The position variable is converted to the world coordinate system

Because the influence of gravity was ignored before , Next we need to take into account the acceleration of gravity .

First, the force received by the UAV in the body coordinate system T X b , T Y b , T Z b T_{X_b}, T_{Y_b}, T_{Z_b} TXb​​,TYb​​,TZb​​ Convert to the world coordinate system . At the same time, add Z g Z_g Zg​ The effect of the gravitational acceleration of the axis .

[ T X g T Y g T Z g ] = [ cos ⁡ θ cos ⁡ ψ sin ⁡ ϕ sin ⁡ θ cos ⁡ ψ − cos ⁡ ϕ sin ⁡ ψ cos ⁡ ϕ sin ⁡ θ cos ⁡ ψ + sin ⁡ ϕ sin ⁡ ψ cos ⁡ θ sin ⁡ ψ sin ⁡ ϕ sin ⁡ θ sin ⁡ ψ + cos ⁡ ϕ cos ⁡ ψ cos ⁡ ϕ sin ⁡ θ sin ⁡ ψ − sin ⁡ ϕ cos ⁡ ψ − sin ⁡ θ sin ⁡ ϕ cos ⁡ θ cos ⁡ ϕ cos ⁡ θ ] [ T X b T Y b T Z b ] = [ cos ⁡ ϕ sin ⁡ θ cos ⁡ ψ + sin ⁡ ϕ sin ⁡ ψ cos ⁡ ϕ sin ⁡ θ sin ⁡ ψ − sin ⁡ ϕ cos ⁡ ψ cos ⁡ ϕ cos ⁡ θ ] ( T 1 + T 2 + T 3 + T 4 ) − [ 0 0 m g ] (8) \begin{aligned} \left[\begin{matrix} T_{X_g} \\ T_{Y_g} \\ T_{Z_g} \\ \end{matrix}\right]&= \left[\begin{matrix} \cos\theta \cos\psi & \sin\phi \sin\theta \cos\psi - \cos\phi \sin\psi & \cos\phi \sin\theta \cos\psi + \sin\phi \sin\psi \\ \cos\theta \sin\psi & \sin\phi \sin\theta \sin\psi + \cos\phi \cos\psi & \cos\phi \sin\theta \sin\psi - \sin\phi \cos\psi \\ -\sin\theta & \sin\phi \cos\theta & \cos\phi \cos\theta \\ \end{matrix}\right] \left[\begin{matrix} T_{X_b} \\ T_{Y_b} \\ T_{Z_b} \\ \end{matrix}\right] \\ &= \left[\begin{matrix} \cos\phi \sin\theta \cos\psi + \sin\phi \sin\psi \\ \cos \phi \sin\theta \sin\psi - \sin\phi \cos\psi \\ \cos\phi \cos\theta \\ \end{matrix}\right] (T_1 + T_2 + T_3 + T_4) - \left[\begin{matrix} 0 \\ 0 \\ m g \\ \end{matrix}\right] \end{aligned} \tag{8} ⎣⎡​TXg​​TYg​​TZg​​​⎦⎤​​=⎣⎡​cosθcosψcosθsinψ−sinθ​sinϕsinθcosψ−cosϕsinψsinϕsinθsinψ+cosϕcosψsinϕcosθ​cosϕsinθcosψ+sinϕsinψcosϕsinθsinψ−sinϕcosψcosϕcosθ​⎦⎤​⎣⎡​TXb​​TYb​​TZb​​​⎦⎤​=⎣⎡​cosϕsinθcosψ+sinϕsinψcosϕsinθsinψ−sinϕcosψcosϕcosθ​⎦⎤​(T1​+T2​+T3​+T4​)−⎣⎡​00mg​⎦⎤​​(8)

At this time, consider the use of Newton's law , Switch to acceleration . The acceleration at this time is just in the world coordinate system .

T X g = m X ¨ g = ( cos ⁡ ϕ sin ⁡ θ cos ⁡ ψ + sin ⁡ ϕ sin ⁡ ψ ) ( T 1 + T 2 + T 3 + T 4 ) (9) T_{X_g} = m \ddot{X}_g = (\cos\phi \sin\theta \cos\psi + \sin\phi \sin\psi)(T_1 + T_2 + T_3 + T_4) \tag{9} TXg​​=mX¨g​=(cosϕsinθcosψ+sinϕsinψ)(T1​+T2​+T3​+T4​)(9)

T Y g = m Y ¨ g = ( cos ⁡ ϕ sin ⁡ θ sin ⁡ ψ − sin ⁡ ϕ cos ⁡ ψ ) ( T 1 + T 2 + T 3 + T 4 ) (10) T_{Y_g} = m \ddot{Y}_g = (\cos \phi \sin\theta \sin\psi - \sin\phi \cos\psi)(T_1 + T_2 + T_3 + T_4) \tag{10} TYg​​=mY¨g​=(cosϕsinθsinψ−sinϕcosψ)(T1​+T2​+T3​+T4​)(10)

T Z g = m Z ¨ g = ( cos ⁡ ϕ cos ⁡ θ ) ( T 1 + T 2 + T 3 + T 4 ) − m g (11) T_{Z_g} = m \ddot{Z}_g = (\cos\phi \cos\theta)(T_1 + T_2 + T_3 + T_4) - mg \tag{11} TZg​​=mZ¨g​=(cosϕcosθ)(T1​+T2​+T3​+T4​)−mg(11)

The six outputs of the UAV , Where the acceleration is in the world coordinate system ：
{ X ¨ g = ( cos ⁡ ϕ sin ⁡ θ cos ⁡ ψ + sin ⁡ ϕ sin ⁡ ψ ) ( T 1 + T 2 + T 3 + T 4 ) m Y ¨ g = ( cos ⁡ ϕ sin ⁡ θ sin ⁡ ψ − sin ⁡ ϕ cos ⁡ ψ ) ( T 1 + T 2 + T 3 + T 4 ) m Z ¨ g = ( cos ⁡ ϕ cos ⁡ θ ) ( T 1 + T 2 + T 3 + T 4 ) m − g ϕ ¨ = L ( − T 1 + T 2 + T 3 − T 4 − K 4 ϕ ˙ ) I x θ ¨ = L ( T 1 + T 2 − T 3 − T 4 − K 5 θ ˙ ) I y ψ ¨ = ( − M 1 + M 2 − M 3 + M 4 − K 6 ψ ˙ ) I z (12) \left\{\begin{aligned} \ddot{X}_g &= \frac{(\cos\phi \sin\theta \cos\psi + \sin\phi \sin\psi)(T_1 + T_2 + T_3 + T_4)}{m} \\ \ddot{Y}_g &= \frac{(\cos \phi \sin\theta \sin\psi - \sin\phi \cos\psi)(T_1 + T_2 + T_3 + T_4)}{m} \\ \ddot{Z}_g &= \frac{(\cos\phi \cos\theta)(T_1 + T_2 + T_3 + T_4)}{m} - g \\ \ddot{\phi} &= \frac{L (- T_1 + T_2 + T_3 - T_4 - K_4 \dot{\phi})}{I_x} \\ \ddot{\theta} &= \frac{L (T_1 + T_2 - T_3 - T_4 - K_5 \dot{\theta})}{I_y} \\ \ddot{\psi} &= \frac{(-M_1 + M_2 - M_3 + M_4 - K_6 \dot{\psi})}{I_z} \\ \end{aligned}\right. \tag{12} ⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​X¨g​Y¨g​Z¨g​ϕ¨​θ¨ψ¨​​=m(cosϕsinθcosψ+sinϕsinψ)(T1​+T2​+T3​+T4​)​=m(cosϕsinθsinψ−sinϕcosψ)(T1​+T2​+T3​+T4​)​=m(cosϕcosθ)(T1​+T2​+T3​+T4​)​−g=Ix​L(−T1​+T2​+T3​−T4​−K4​ϕ˙​)​=Iy​L(T1​+T2​−T3​−T4​−K5​θ˙)​=Iz​(−M1​+M2​−M3​+M4​−K6​ψ˙​)​​(12)

Again , When the influence of air resistance is ignored , Yes
{ X ¨ g = ( cos ⁡ ϕ sin ⁡ θ cos ⁡ ψ + sin ⁡ ϕ sin ⁡ ψ ) ( T 1 + T 2 + T 3 + T 4 ) m Y ¨ g = ( cos ⁡ ϕ sin ⁡ θ sin ⁡ ψ − sin ⁡ ϕ cos ⁡ ψ ) ( T 1 + T 2 + T 3 + T 4 ) m Z ¨ g = ( cos ⁡ ϕ cos ⁡ θ ) ( T 1 + T 2 + T 3 + T 4 ) m − g ϕ ¨ = L ( − T 1 + T 2 + T 3 − T 4 ) I x θ ¨ = L ( T 1 + T 2 − T 3 − T 4 ) I y ψ ¨ = ( − M 1 + M 2 − M 3 + M 4 ) I z (12) \left\{\begin{aligned} \ddot{X}_g &= \frac{(\cos\phi \sin\theta \cos\psi + \sin\phi \sin\psi)(T_1 + T_2 + T_3 + T_4)}{m} \\ \ddot{Y}_g &= \frac{(\cos \phi \sin\theta \sin\psi - \sin\phi \cos\psi)(T_1 + T_2 + T_3 + T_4)}{m} \\ \ddot{Z}_g &= \frac{(\cos\phi \cos\theta)(T_1 + T_2 + T_3 + T_4)}{m} - g \\ \ddot{\phi} &= \frac{L (- T_1 + T_2 + T_3 - T_4)}{I_x} \\ \ddot{\theta} &= \frac{L (T_1 + T_2 - T_3 - T_4)}{I_y} \\ \ddot{\psi} &= \frac{(-M_1 + M_2 - M_3 + M_4)}{I_z} \\ \end{aligned}\right. \tag{12} ⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​X¨g​Y¨g​Z¨g​ϕ¨​θ¨ψ¨​​=m(cosϕsinθcosψ+sinϕsinψ)(T1​+T2​+T3​+T4​)​=m(cosϕsinθsinψ−sinϕcosψ)(T1​+T2​+T3​+T4​)​=m(cosϕcosθ)(T1​+T2​+T3​+T4​)​−g=Ix​L(−T1​+T2​+T3​−T4​)​=Iy​L(T1​+T2​−T3​−T4​)​=Iz​(−M1​+M2​−M3​+M4​)​​(12)

Here we control the relationship , It can also be said that the variable relationship is clear . I know six outputs X g , Y g , Z g , ϕ , θ , ψ X_g, Y_g, Z_g, \phi, \theta, \psi Xg​,Yg​,Zg​,ϕ,θ,ψ, The lift of the four propellers can be calculated from the above formula T 1 , T 2 , T 3 , T 4 T_1, T_2, T_3, T_4 T1​,T2​,T3​,T4​ 了 , And torque M 1 , M 2 , M 3 , M 4 M_1, M_2, M_3, M_4 M1​,M2​,M3​,M4​ And lift are both positively related to rotational speed , It can also be understood as knowing the speed that each propeller needs to reach .

6 Simplified analysis

But we don't study the flight of UAV under aerodynamics , Therefore, the influence of air resistance will be ignored in the following discussion . At the same time, for convenience , We assume that M i = C ⋅ T i , i = 1 , 2 , 3 , 4 M_i = C \cdot T_i, i=1,2,3,4 Mi​=C⋅Ti​,i=1,2,3,4. Define four variables that have the following relationship with the propeller lift ,
[ u f u ϕ u θ u ψ ] = [ 1 1 1 1 − L L L − L L L − L − L − C C − C C ] [ T 1 T 2 T 3 T 4 ] = [ T 1 + T 2 + T 3 + T 4 ( − T 1 + T 2 + T 3 − T 4 ) L ( T 1 + T 2 − T 3 − T 4 ) L ( − T 1 + T 2 − T 3 + T 4 ) C ] (13) \begin{aligned} \left[\begin{matrix} u_f \\ u_\phi \\ u_\theta \\ u_\psi \\ \end{matrix}\right]&= \left[\begin{matrix} 1 & 1 & 1 & 1 \\ -L & L & L & -L \\ L & L & -L & -L \\ -C & C & -C & C \\ \end{matrix}\right] \left[\begin{matrix} T_1 \\ T_2 \\ T_3 \\ T_4 \\ \end{matrix}\right] \\ &= \left[\begin{matrix} T_1 + T_2 + T_3 + T_4 \\ (-T_1 + T_2 + T_3 - T_4) L \\ (T_1 + T_2 - T_3 - T_4) L \\ (-T_1 + T_2 - T_3 + T_4) C \\ \end{matrix}\right] \end{aligned} \tag{13} ⎣⎢⎢⎡​uf​uϕ​uθ​uψ​​⎦⎥⎥⎤​​=⎣⎢⎢⎡​1−LL−C​1LLC​1L−L−C​1−L−LC​⎦⎥⎥⎤​⎣⎢⎢⎡​T1​T2​T3​T4​​⎦⎥⎥⎤​=⎣⎢⎢⎡​T1​+T2​+T3​+T4​(−T1​+T2​+T3​−T4​)L(T1​+T2​−T3​−T4​)L(−T1​+T2​−T3​+T4​)C​⎦⎥⎥⎤​​(13)

such ,（12） It can be simplified to
{ X ¨ g = ( cos ⁡ ϕ sin ⁡ θ cos ⁡ ψ + sin ⁡ ϕ sin ⁡ ψ )   u f m Y ¨ g = ( cos ⁡ ϕ sin ⁡ θ sin ⁡ ψ − sin ⁡ ϕ cos ⁡ ψ )   u f m Z ¨ g = ( cos ⁡ ϕ cos ⁡ θ )   u f m − g ϕ ¨ = u ϕ I x θ ¨ = u θ I y ψ ¨ = u ψ I z (14) \left\{\begin{aligned} \ddot{X}_g &= \frac{(\cos\phi \sin\theta \cos\psi + \sin\phi \sin\psi)~ u_f}{m} \\ \ddot{Y}_g &= \frac{(\cos \phi \sin\theta \sin\psi - \sin\phi \cos\psi)~ u_f}{m} \\ \ddot{Z}_g &= \frac{(\cos\phi \cos\theta)~ u_f}{m} - g \\ \ddot{\phi} &= \frac{u_\phi}{I_x} \\ \ddot{\theta} &= \frac{u_\theta}{I_y} \\ \ddot{\psi} &= \frac{u_\psi}{I_z} \\ \end{aligned}\right. \tag{14} ⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧​X¨g​Y¨g​Z¨g​ϕ¨​θ¨ψ¨​​=m(cosϕsinθcosψ+sinϕsinψ) uf​​=m(cosϕsinθsinψ−sinϕcosψ) uf​​=m(cosϕcosθ) uf​​−g=Ix​uϕ​​=Iy​uθ​​=Iz​uψ​​​(14)

Once again,

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