Aircraft Equations of Motion draft
Summary of the equations of motion
With the translational (Eq. (2) ) and rotational (Eq. (5) ) defined, and the relationship between the Euler angles and the body rates (this equation and Eq. (4) ), fully unconstrained flight in 6DoF can be described.
In total twelve equations have been derived (body/attitude rates are only three but the matrix is tough to invert, so I’ll include both versions below) that can describe aircraft position and attitude in a Newtonian framework:
Translational Motion:
m [ U ˙ + Q W − R V V ˙ + R U − P W W ˙ + P V − Q U ] = − m g sin θ − D cos α + L sin α + T cos θ T m g sin ϕ cos θ + F A Y + F T Y m g cos ϕ cos θ − D sin α − L cos α − T sin θ T m\begin{bmatrix} \dot{U} + Q\,W - R\,V \\ \dot{V} + R\,U-P\,W\\\dot{W}+P\,V-Q\,U\end{bmatrix}=\begin{matrix} -mg\sin\theta - D\cos\alpha + L\sin\alpha + T\cos\theta_T\\mg\sin\phi\cos\theta + F_{A_Y} + F_{T_Y}\\mg\cos\phi\cos\theta - D\sin\alpha - L\cos\alpha - T\sin\theta_T\end{matrix} m U ˙ + Q W − R V V ˙ + R U − P W W ˙ + P V − Q U = − m g sin θ − D cos α + L sin α + T cos θ T m g sin ϕ cos θ + F A Y + F T Y m g cos ϕ cos θ − D sin α − L cos α − T sin θ T Angular Motion:
[ P ˙ ⋅ I x x Q ˙ ⋅ I y y R ˙ ⋅ I z z + + + Q ⋅ R ( I z z − I y y ) P ⋅ R ( I x x − I z z ) P ⋅ Q ( I y y − I x x ) − + + ( R ˙ + P ⋅ Q ) I x z ( P 2 − R 2 ) I x z ( Q ⋅ R − P ˙ ) I x z ] = [ L M N ] \left[{\begin{matrix} \dot{P}\cdot I_{xx} \\ \dot{Q}\cdot I_{yy} \\\dot{R}\cdot I_{zz}\end{matrix}} \hspace{.5cm} \begin{matrix} + \\+\\+\end{matrix} \hspace{.5cm} {\begin{matrix} Q\cdot R\left(I_{zz} - I_{yy}\right) \\ P\cdot R\left(I_{xx} - I_{zz}\right) \\ P\cdot Q\left(I_{yy} - I_{xx}\right)\end{matrix}} \hspace{.5cm} \begin{matrix} - \\+\\+\end{matrix} \hspace{.5cm} {\begin{matrix}\left(\dot{R} + P\cdot Q\right) I_{xz} \\ \left(P^2 - R^2\right) I_{xz} \\ \left(Q\cdot R - \dot{P}\right) I_{xz}\end{matrix}} \right] = \begin{bmatrix} L \\ M \\ N\end{bmatrix} P ˙ ⋅ I xx Q ˙ ⋅ I y y R ˙ ⋅ I z z + + + Q ⋅ R ( I z z − I y y ) P ⋅ R ( I xx − I z z ) P ⋅ Q ( I y y − I xx ) − + + ( R ˙ + P ⋅ Q ) I x z ( P 2 − R 2 ) I x z ( Q ⋅ R − P ˙ ) I x z = L M N Body angular rate due to an attitude rate:
[ P Q R ] = [ 1 0 − sin θ 0 cos ϕ sin ϕ cos θ 0 − sin ϕ cos ϕ cos θ ] [ ϕ ˙ θ ˙ ψ ˙ ] \begin{aligned}
\begin{bmatrix} P\\Q\\R\end{bmatrix} &= \begin{bmatrix}
1 & 0 & -\sin\theta\\
0 & \cos\phi & \sin\phi\cos\theta\\
0 & -\sin\phi & \cos\phi\cos\theta
\end{bmatrix}\begin{bmatrix} \dot{\phi}\\\dot{\theta}\\\dot{\psi}\end{bmatrix}\end{aligned} P Q R = 1 0 0 0 cos ϕ − sin ϕ − sin θ sin ϕ cos θ cos ϕ cos θ ϕ ˙ θ ˙ ψ ˙ Attitude rate due to a body rate
[ ϕ ˙ θ ˙ ψ ˙ ] = [ 1 sin ϕ tan θ cos ϕ tan θ 0 cos ϕ − sin ϕ 0 sin ϕ cos θ cos ϕ cos θ ] [ P Q R ] \begin{aligned}
\begin{bmatrix} \dot{\phi}\\\dot{\theta}\\\dot{\psi}\end{bmatrix}&= \begin{bmatrix}
1 & \sin\phi\tan\theta & \cos\phi\tan\theta\\
0 & \cos\phi & -\sin\phi \\
0 & \frac{\sin\phi}{\cos\theta} & \frac{\cos\phi}{\cos\theta}
\end{bmatrix}\begin{bmatrix} P\\Q\\R\end{bmatrix}\end{aligned} ϕ ˙ θ ˙ ψ ˙ = 1 0 0 sin ϕ tan θ cos ϕ c o s θ s i n ϕ cos ϕ tan θ − sin ϕ c o s θ c o s ϕ P Q R ← Inertia Tensor Calculation Why Aircraft Need Derivatives →