Linearisation draft
As demonstrated above, the dimensional EoMs are either symmetric (a.k.a. longitudinal), where u , w , q u,w,q u , w , q are controlled by θ \theta θ and δ e \delta_e δ e , or asymmetric (a.k.a. lateral/directional) where v , p , r v,p,r v , p , r are controlled by δ a \delta_a δ a and δ r \delta_r δ r . This is a consequence of our assumptions made to remove stability derivatives.
That is, this was enforced by the assumption of zero cross-coupling rather than being a natural consequence of the physics.
Since these assumptions have been made, the two sets of equations may be treated in isolation from each other and presented in a matrix form. First, look at the symmetric equations:
m u ˙ = ∂ X ∂ u ∣ 0 u + ∂ X ∂ w ∣ 0 w − m g ⋅ cos θ 0 ⋅ θ ′ (1) \def\pd#1#2{\frac{\partial#1}{\partial#2}}
\color{red}{m\dot{u}= \left.\pd{X}{u}\right|_0u + \left.\pd{X}{w}\right|_0w - mg\cdot\cos\theta_0\cdot\theta^\prime}\tag{1} m u ˙ = ∂ u ∂ X 0 u + ∂ w ∂ X 0 w − m g ⋅ c o s θ 0 ⋅ θ ′ ( 1 )
m w ˙ = ∂ Z ∂ u ∣ 0 u + ∂ Z ∂ w ∣ 0 w + m U 0 q − m g ⋅ sin Θ 0 ⋅ θ ′ + ∂ Z ∂ δ e ∣ 0 δ e (2) \def\pd#1#2{\frac{\partial#1}{\partial#2}}
\color{red}{m\dot{w}= \left.\pd{Z}{u}\right|_0u + \left.\pd{Z}{w}\right|_0w + mU_0q - mg\cdot\sin\Theta_0\cdot\theta^\prime + \left.\pd{Z}{\delta_e}\right|_0\delta_e}\tag{2} m w ˙ = ∂ u ∂ Z 0 u + ∂ w ∂ Z 0 w + m U 0 q − m g ⋅ s i n Θ 0 ⋅ θ ′ + ∂ δ e ∂ Z 0 δ e ( 2 )
I y y q ˙ = ∂ M ∂ u ∣ 0 u + ∂ M ∂ w ∣ 0 w + ∂ M ∂ w ˙ ∣ 0 w ˙ + ∂ M ∂ q ∣ 0 q + ∂ M ∂ δ e ∣ 0 δ e (3) \def\pd#1#2{\frac{\partial#1}{\partial#2}}
\color{red}{I_{yy}\dot{q} = \left.\pd{M}{u}\right|_0u+\left.\pd{M}{w}\right|_0w+\left.\pd{M}{\dot{w}}\right|_0\dot{w}+\left.\pd{M}{q}\right|_0q+\left.\pd{M}{\delta_e}\right|_0\delta_e}\tag{3} I y y q ˙ = ∂ u ∂ M 0 u + ∂ w ∂ M 0 w + ∂ w ˙ ∂ M 0 w ˙ + ∂ q ∂ M 0 q + ∂ δ e ∂ M 0 δ e ( 3 )
Longitudinal EoMs
The X-Force equation (forward speed equation ):
Taking Equation (1) and dividing by m m m :
u ˙ = X u u + X w w − g ⋅ cos θ 0 θ ′ (4) \dot{u} = X_uu + X_ww - g\cdot\cos\theta_0\theta^\prime\tag{4} u ˙ = X u u + X w w − g ⋅ cos θ 0 θ ′ ( 4 )
where
X u ≜ 1 m ∂ X ∂ u ∣ 0 , X w ≜ 1 m ∂ X ∂ w ∣ 0 \def\pd#1#2{\frac{\partial#1}{\partial#2}}
X_u\triangleq \frac{1}{m}\left.\pd{X}{u}\right|_0, X_w\triangleq \frac{1}{m}\left.\pd{X}{w}\right|_0 X u ≜ m 1 ∂ u ∂ X 0 , X w ≜ m 1 ∂ w ∂ X 0
The Z-Force equation (heave equation ):
Taking Equation (2) and dividing by m m m :
w ˙ = Z u u + Z w w + U e q − g ⋅ sin θ 0 ⋅ θ ′ + Z δ e δ e (5) \dot{w} = Z_uu + Z_ww + U_eq - g\cdot\sin\theta_0\cdot\theta^\prime + Z_{\delta_e}\delta_e\tag{5} w ˙ = Z u u + Z w w + U e q − g ⋅ sin θ 0 ⋅ θ ′ + Z δ e δ e ( 5 )
where
Z u ≜ 1 m ∂ Z ∂ u ∣ 0 , Z w ≜ 1 m ∂ Z ∂ w ∣ 0 , Z δ e ≜ 1 m ∂ Z ∂ δ e ∣ 0 \def\pd#1#2{\frac{\partial#1}{\partial#2}}
Z_u\triangleq \frac{1}{m}\left.\pd{Z}{u}\right|_0, Z_w\triangleq \frac{1}{m}\left.\pd{Z}{w}\right|_0, Z_{\delta_e}\triangleq \frac{1}{m}\left.\pd{Z}{\delta_e}\right|_0 Z u ≜ m 1 ∂ u ∂ Z 0 , Z w ≜ m 1 ∂ w ∂ Z 0 , Z δ e ≜ m 1 ∂ δ e ∂ Z 0
The M-Moment equation (pitching moment equation ):
If we take Equation (3) and divide by I y y I_{yy} I y y :
q ˙ = M u u + M w w + M w ˙ w ˙ + M q q + M δ e δ e \dot{q} = M_uu + M_ww + M_{\dot{w}}\dot{w} + M_qq + M_{\delta_e}\delta_e q ˙ = M u u + M w w + M w ˙ w ˙ + M q q + M δ e δ e
where
M u ≜ 1 I y y ∂ M ∂ u ∣ 0 , and the rest of the pattern should be clear \def\pd#1#2{\frac{\partial#1}{\partial#2}}
M_u\triangleq \frac{1}{I_{yy}}\left.\pd{M}{u}\right|_0, \text{and the rest of the pattern should be clear} M u ≜ I y y 1 ∂ u ∂ M 0 , and the rest of the pattern should be clear
There is already an expression for w ˙ \dot{w} w ˙ , Equation (5) , which may be substituted in:
q ˙ = M u u + M w w + M w ˙ ( Z u u + Z w w + U e q − g ⋅ sin θ 0 ⋅ θ ′ + Z δ e δ e ) + M q q + M δ e δ e \dot{q} = M_uu + M_ww + M_{\dot{w}}\left(Z_uu + Z_ww + U_eq - g\cdot\sin\theta_0\cdot\theta^\prime + Z_{\delta_e}\delta_e \right) + M_qq + M_{\delta_e}\delta_e q ˙ = M u u + M w w + M w ˙ ( Z u u + Z w w + U e q − g ⋅ sin θ 0 ⋅ θ ′ + Z δ e δ e ) + M q q + M δ e δ e
and can be simplified1 to:
q ˙ = M u ∗ u + M w ∗ w + M ω ∗ ω + M q ∗ q + M δ e ∗ δ e (6) \dot{q} = M_u^*u + M_w^*w +M_\omega^*\omega + M_q^*q + M_{\delta_e}^*\delta_e\tag{6} q ˙ = M u ∗ u + M w ∗ w + M ω ∗ ω + M q ∗ q + M δ e ∗ δ e ( 6 )
where
M u ∗ ≜ M u + M w ˙ Z u = = = = = = M w ∗ ≜ M w + M w ˙ Z w = = = = = = M q ∗ ≜ M q + M w ˙ U 0 M δ e ∗ ≜ M δ e + M w ˙ Z δ e = = = = = = M θ ∗ ≜ − M w ˙ g sin θ 0 (7) \begin{split}&M_u^*\triangleq M_u + M_{\dot{w}}Z_u\hphantom{======} M_w^*\triangleq M_w+M_{\dot{w}}Z_w\hphantom{======}M_q^*\triangleq M_q+M_{\dot{w}}U_0\\
&M_{\delta_e}^*\triangleq M_{\delta_e}+M_{\dot{w}}Z_{\delta_e}\hphantom{======}M_{\theta}^*\triangleq-M_{\dot{w}}g\sin\theta_0
\end{split}\tag{7} M u ∗ ≜ M u + M w ˙ Z u ====== M w ∗ ≜ M w + M w ˙ Z w ====== M q ∗ ≜ M q + M w ˙ U 0 M δ e ∗ ≜ M δ e + M w ˙ Z δ e ====== M θ ∗ ≜ − M w ˙ g sin θ 0 ( 7 )
The linearised Euler pitch equation (pitch rate kinematic equation ):
q q q is pitch rate, from Eq (12) :
θ ˙ ′ = q \dot{\theta}^\prime = q θ ˙ ′ = q
Now Equations (4) , (5) , (6) , and (12) can be written in matrix form to give the linearised equation of longitudinal motion in concise form :
[ u ˙ w ˙ q ˙ θ ˙ ] = [ X u X w 0 − g ⋅ cos θ 0 Z u Z w U 0 − g ⋅ sin θ 0 M u ∗ M w ∗ M q ∗ M θ ∗ 0 0 1 0 ] [ u w q θ ] + [ 0 Z δ e M δ e ∗ 0 ] [ δ e ] (8) \begin{aligned}
\begin{bmatrix} \dot{u}\\\dot{w}\\\dot{q}\\\dot{\theta}\end{bmatrix} &= \begin{bmatrix}
X_u & X_w & 0 & -g\cdot\cos\theta_0\\
Z_u & Z_w & U_0 & -g\cdot\sin\theta_0\\
M_u^* & M_w^* & M_q^* & M_\theta^*\\
0 & 0 & 1 & 0
\end{bmatrix}\begin{bmatrix}
{u}\\{w}\\{q}\\{\theta}
\end{bmatrix} + \begin{bmatrix}
0\\Z_{\delta_e}\\M_{\delta_e}^*\\0
\end{bmatrix}\left[\delta_e\right]\end{aligned}\tag{8} u ˙ w ˙ q ˙ θ ˙ = X u Z u M u ∗ 0 X w Z w M w ∗ 0 0 U 0 M q ∗ 1 − g ⋅ cos θ 0 − g ⋅ sin θ 0 M θ ∗ 0 u w q θ + 0 Z δ e M δ e ∗ 0 [ δ e ] ( 8 )
x ⃗ ˙ = A x ⃗ + B u ⃗ \dot{\vec{x}} = A\vec{x} + B\vec{u} x ˙ = A x + B u
Which is in state space form. Equation (8) is a series of simultaneous 1st order ODEs with constant coefficients which may solved to find the longitudinal response (u , v , q , θ u,v,q,\theta u , v , q , θ ) of an aircraft due to elevator deflection (δ e \delta_e δ e ).
Lateral/Directional EoMs
The asymmetric or lateral/directional EoMs may be similarly manipulated.
m v ˙ = ∂ Y ∂ v ∣ 0 v − m U 0 r + m g ⋅ c o s θ 0 ⋅ ϕ ′ ∂ Y ∂ δ r ∣ 0 δ r (9) \def\pd#1#2{\frac{\partial#1}{\partial#2}}
\color{darkgreen}{m\dot{v}=\left.\pd{Y}{v}\right|_0v - mU_0r+mg\cdot cos\theta_0\cdot\phi^\prime\left.\pd{Y}{\delta_r}\right|_0\delta_r}\tag{9} m v ˙ = ∂ v ∂ Y 0 v − m U 0 r + m g ⋅ cos θ 0 ⋅ ϕ ′ ∂ δ r ∂ Y 0 δ r ( 9 )
I x x p ˙ − I x z r ˙ = ∂ L ∂ v ∣ 0 v + ∂ L ∂ p ∣ 0 p + ∂ L ∂ r ∣ 0 r + ∂ L ∂ δ r ∣ 0 δ r + ∂ L ∂ δ a ∣ 0 δ a (10) \def\pd#1#2{\frac{\partial#1}{\partial#2}}
\color{darkgreen}{I_{xx}\dot{p} - I_{xz}\dot{r} = \left.\pd{L}{v}\right|_0v + \left.\pd{L}{p}\right|_0p+\left.\pd{L}{r}\right|_0r+\left.\pd{L}{\delta_r}\right|_0\delta_r+\left.\pd{L}{\delta_a}\right|_0\delta_a}\tag{10} I xx p ˙ − I x z r ˙ = ∂ v ∂ L 0 v + ∂ p ∂ L 0 p + ∂ r ∂ L 0 r + ∂ δ r ∂ L 0 δ r + ∂ δ a ∂ L 0 δ a ( 10 )
I z z r ˙ − I x z p ˙ = ∂ N ∂ v ∣ 0 v + ∂ N ∂ p ∣ 0 p + ∂ N ∂ r ∣ 0 r + ∂ N ∂ δ r ∣ 0 δ r + ∂ N ∂ δ a ∣ 0 δ a (11) \def\pd#1#2{\frac{\partial#1}{\partial#2}}
\color{darkgreen}{I_{zz}\dot{r}-I_{xz}\dot{p} = \left.\pd{N}{v}\right|_0v+ \left.\pd{N}{p}\right|_0p + \left.\pd{N}{r}\right|_0r + \left.\pd{N}{\delta_r}\right|_0\delta_r+ \left.\pd{N}{\delta_a}\right|_0\delta_a}\tag{11} I z z r ˙ − I x z p ˙ = ∂ v ∂ N 0 v + ∂ p ∂ N 0 p + ∂ r ∂ N 0 r + ∂ δ r ∂ N 0 δ r + ∂ δ a ∂ N 0 δ a ( 11 )
The Y-Force equation (sideslip equation ): Taking Equation (9) and dividing by m m m :
v ˙ = Y v v − U e r + g ⋅ cos θ 0 ϕ + Y δ r δ r \dot{v} = Y_vv-U_er + g\cdot\cos\theta_0\phi + Y_{\delta_r}{\delta_r} v ˙ = Y v v − U e r + g ⋅ cos θ 0 ϕ + Y δ r δ r
where
Y v ≜ 1 m ∂ Y ∂ v , Y δ r ≜ 1 m ∂ Y ∂ δ r \def\pd#1#2{\frac{\partial#1}{\partial#2}}
Y_v\triangleq\frac{1}{m}\pd{Y}{v},\,Y_{\delta_r}\triangleq\frac{1}{m}\pd{Y}{\delta_r} Y v ≜ m 1 ∂ v ∂ Y , Y δ r ≜ m 1 ∂ δ r ∂ Y
The two moment equations are coupled, so need to be decoupled. Dividing Equation (10) by the rolling moment of inertia, I x x I_xx I x x :
p ˙ = I x z I x x r ˙ + L v v + L p p + L r r + L δ r δ r + L δ a δ a (12) \dot{p} = \frac{I_{xz}}{I_{xx}}\dot{r} + L_vv + L_pp + L_rr + L_{\delta_r}{\delta_r} + L_{\delta_a}\delta_a\tag{12} p ˙ = I xx I x z r ˙ + L v v + L p p + L r r + L δ r δ r + L δ a δ a ( 12 )
where
L v ≜ 1 I x x ∂ L ∂ v , L p ≜ 1 I x x ∂ L ∂ p , etc. \def\pd#1#2{\frac{\partial#1}{\partial#2}}
L_v\triangleq\frac{1}{I_{xx}}\pd{L}{v},\,L_{p}\triangleq\frac{1}{I_{xx}}\pd{L}{p},\,\text{etc.} L v ≜ I xx 1 ∂ v ∂ L , L p ≜ I xx 1 ∂ p ∂ L , etc.
the same can be performed for (11) , dividing by the yawing moment of inertia I z z I_{zz} I z z :
r ˙ = I x z I z z p ˙ + N v v + N p p + N r r + N δ r δ r + N δ a δ a (13) \dot{r} = \frac{I_{xz}}{I_{zz}}\dot{p} + N_vv+N_pp + N_rr + N_{\delta_r}\delta_r+N_{\delta_a}\delta_a\tag{13} r ˙ = I z z I x z p ˙ + N v v + N p p + N r r + N δ r δ r + N δ a δ a ( 13 )
Clearly Equations (12) and (13) are coupled, and must be decoupled if we want to use them in standard form. Substituting equation (13) into (12) , after a little manipulation (if you do this yourself, I promise you’ll understand this subject better), the following is achieved:
p ˙ ( 1 − I x z 2 I x x I z z ) = ( L v + I x z I x x N v ) v + ( L p + I x z I x x N p ) p + ( L r + I x z I x x N r ) r + … ( L δ r + I x z I x x N δ r ) δ r + ( L δ a + I x z I x x N δ a ) δ a \begin{split}\dot{p}\left(1-\frac{I_{xz}^2}{I_{xx}I_{zz}}\right) &= \left(L_v + \frac{I_{xz}}{I_{xx}}N_v\right)v + \left(L_p + \frac{I_{xz}}{I_{xx}}N_p\right)p+\left(L_r+\frac{I_{xz}}{I_{xx}}N_r\right)r +\ldots\\ &\left(L_{\delta_r} + \frac{I_{xz}}{I_{xx}N_{\delta_r}}\right)\delta_r + \left(L_{\delta_a}+\frac{I_{xz}}{I_{xx}}N_{\delta_a}\right)\delta_a\end{split} p ˙ ( 1 − I xx I z z I x z 2 ) = ( L v + I xx I x z N v ) v + ( L p + I xx I x z N p ) p + ( L r + I xx I x z N r ) r + … ( L δ r + I xx N δ r I x z ) δ r + ( L δ a + I xx I x z N δ a ) δ a
Noting that:
1 − I x z 2 I x x I z z = I x x I z z − I z z 2 I x x I z z 1-\frac{I_{xz}^2}{I_{xx}I_{zz}} = \frac{I_{xx}I_{zz}-I_{zz}^2}{I_{xx}I_{zz}} 1 − I xx I z z I x z 2 = I xx I z z I xx I z z − I z z 2
hence, the rolling moment equation may be weitten as:
p ˙ = L v ∗ v + L p ∗ p + L r ∗ r + L δ r ∗ δ r + L δ a ∗ δ a \dot{p} = L_v^*v + L_p^*p + L_r^*r+L_{\delta_r}^*\delta_r+L_{\delta_a}^*\delta_a p ˙ = L v ∗ v + L p ∗ p + L r ∗ r + L δ r ∗ δ r + L δ a ∗ δ a
where:
L v ∗ = I x x I z z I x x I z z − I x z 2 ( L v + I x z I x x N v ) L p ∗ = I x x I z z I x x I z z − I x z 2 ( L p + I x z I x x N p ) L r ∗ = I x x I z z I x x I z z − I x z 2 ( L r + I x z I x x N r ) L δ r ∗ = I x x I z z I x x I z z − I x z 2 ( L δ r + I x z I x x N δ r ) L δ a ∗ = I x x I z z I x x I z z − I x z 2 ( L δ a + I x z I x x N δ a ) \begin{gathered}
L_v^* = \frac{I_{xx}I_{zz}}{I_{xx}I_{zz}-I_{xz}^2}\left(L_v+\frac{I_{xz}}{I_{xx}}N_v\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, L_p^*=\frac{I_{xx}I_{zz}}{I_{xx}I_{zz}-I_{xz}^2}\left(L_p+\frac{I_{xz}}{I_{xx}}N_p\right)\\
L_r^* = \frac{I_{xx}I_{zz}}{I_{xx}I_{zz}-I_{xz}^2}\left(L_r+\frac{I_{xz}}{I_{xx}}N_r\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, L_{\delta_r}^*=\frac{I_{xx}I_{zz}}{I_{xx}I_{zz}-I_{xz}^2}\left(L_{\delta_r}+\frac{I_{xz}}{I_{xx}}N_{\delta_r}\right)\\
L_{\delta_a}^*=\frac{I_{xx}I_{zz}}{I_{xx}I_{zz}-I_{xz}^2}\left(L_{\delta_a}+\frac{I_{xz}}{I_{xx}}N_{\delta_a}\right)
\end{gathered} L v ∗ = I xx I z z − I x z 2 I xx I z z ( L v + I xx I x z N v ) L p ∗ = I xx I z z − I x z 2 I xx I z z ( L p + I xx I x z N p ) L r ∗ = I xx I z z − I x z 2 I xx I z z ( L r + I xx I x z N r ) L δ r ∗ = I xx I z z − I x z 2 I xx I z z ( L δ r + I xx I x z N δ r ) L δ a ∗ = I xx I z z − I x z 2 I xx I z z ( L δ a + I xx I x z N δ a )
Similarly, by substituting (12) into (13) , the yawing moment equation is yielded:
r ˙ = N v ∗ v + N p ∗ p + N r ∗ r + N δ r ∗ δ r + N δ a ∗ δ a \dot{r} = N_v^*v + N_p^*p + N_r^*r + N_{\delta_r}^*\delta_r + N_{\delta_a}^*\delta_a r ˙ = N v ∗ v + N p ∗ p + N r ∗ r + N δ r ∗ δ r + N δ a ∗ δ a
where:
N v ∗ = I x x I z z I x x I z z − I x z 2 ( N v + I x z I z z L v ) N p ∗ = I x x I z z I x x I z z − I x z 2 ( N p + I x z n e w I z z L p ) N r ∗ = I x x I z z I x x I z z − I x z 2 ( N r + I x z I z z L r ) N δ r ∗ = I x x I z z I x x I z z − I x z 2 ( N δ r + I x z I z z L δ r ) N δ a ∗ = I x x I z z I x x I z z − I x z 2 ( N δ a + I x z I z z L δ a ) \begin{gathered}
N_v^* = \frac{I_{xx}I_{zz}}{I_{xx}I_{zz}-I_{xz}^2}\left(N_v+\frac{I_{xz}}{I_{zz}}L_v\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, N_p^*=\frac{I_{xx}I_{zz}}{I_{xx}I_{zz}-I_{xz}^2}\left(N_p+\frac{I_{xz}}{newI_{zz}}L_p\right)\\
N_r^* = \frac{I_{xx}I_{zz}}{I_{xx}I_{zz}-I_{xz}^2}\left(N_r+\frac{I_{xz}}{I_{zz}}L_r\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, N_{\delta_r}^*=\frac{I_{xx}I_{zz}}{I_{xx}I_{zz}-I_{xz}^2}\left(N_{\delta_r}+\frac{I_{xz}}{I_{zz}}L_{\delta_r}\right)\\
N_{\delta_a}^*=\frac{I_{xx}I_{zz}}{I_{xx}I_{zz}-I_{xz}^2}\left(N_{\delta_a}+\frac{I_{xz}}{I_{zz}}L_{\delta_a}\right)
\end{gathered} N v ∗ = I xx I z z − I x z 2 I xx I z z ( N v + I z z I x z L v ) N p ∗ = I xx I z z − I x z 2 I xx I z z ( N p + n e w I z z I x z L p ) N r ∗ = I xx I z z − I x z 2 I xx I z z ( N r + I z z I x z L r ) N δ r ∗ = I xx I z z − I x z 2 I xx I z z ( N δ r + I z z I x z L δ r ) N δ a ∗ = I xx I z z − I x z 2 I xx I z z ( N δ a + I z z I x z L δ a )
often I x z I_{xz} I x z is small compared to I x x I_{xx} I xx and I z z I_{zz} I z z , so for these circumstances N v ∗ ≃ N v N_v^*\simeq N_v N v ∗ ≃ N v , L v ∗ ≃ L v L_v^*\simeq L_v L v ∗ ≃ L v
Lastly, the yaw rate and body rate kinematic equations give:
r = ψ ˙ cos θ 0 ⟹ ψ ˙ = r ⋅ sec θ 0 p = ϕ ˙ − ψ ˙ sin θ 0 ⟹ ϕ ˙ = p + r ⋅ tan θ 0 \begin{aligned}
r &= \dot{\psi}\cos\theta_0\\
\implies \dot{\psi} &= r\cdot\sec\theta_0\\
p &= \dot{\phi}-\dot{\psi}\sin\theta_0\\
\implies\dot{\phi} &= p + r\cdot\tan\theta_0
\end{aligned} r ⟹ ψ ˙ p ⟹ ϕ ˙ = ψ ˙ cos θ 0 = r ⋅ sec θ 0 = ϕ ˙ − ψ ˙ sin θ 0 = p + r ⋅ tan θ 0
So finally, the linearised lateral/directional equations of motion may be expressed in state-space form:
[ v ˙ p ˙ r ˙ ϕ ˙ ψ ˙ ] = [ Y v 0 − U 0 g ⋅ cos θ 0 0 L v ∗ L p ∗ L r ∗ 0 0 N v ∗ N p ∗ N r ∗ 0 0 0 1 tan θ 0 0 0 0 0 sec θ 0 0 0 ] [ v p r ϕ ψ ] + [ Y δ r 0 L δ r ∗ L δ a ∗ N δ r ∗ N δ a ∗ 0 0 0 0 ] [ δ r δ a ] (14) \begin{bmatrix}\dot{v}\\\dot{p}\\\dot{r}\\\dot{\phi}\\\dot{\psi}\end{bmatrix}=\begin{bmatrix}Y_v & 0 & -U_0 & g\cdot\cos\theta_0 & 0\\L_v^* & L_p^* & L_r^* & 0 & 0\\N_v^* & N_p^* & N_r^* & 0 & 0\\0 & 1 & \tan\theta_0 & 0 & 0 \\ 0 & 0 & \sec\theta_0 & 0 & 0\end{bmatrix}\begin{bmatrix}v\\p\\r\\\phi\\\psi\end{bmatrix}+\begin{bmatrix}Y_{\delta_r} & 0\\L_{\delta_r}^* & L_{\delta_a}^*\\N_{\delta_r}^* & N_{\delta_a}^*\\0 & 0\\0 & 0\end{bmatrix}\begin{bmatrix}\delta_r\\\delta_a\end{bmatrix}\tag{14} v ˙ p ˙ r ˙ ϕ ˙ ψ ˙ = Y v L v ∗ N v ∗ 0 0 0 L p ∗ N p ∗ 1 0 − U 0 L r ∗ N r ∗ tan θ 0 sec θ 0 g ⋅ cos θ 0 0 0 0 0 0 0 0 0 0 v p r ϕ ψ + Y δ r L δ r ∗ N δ r ∗ 0 0 0 L δ a ∗ N δ a ∗ 0 0 [ δ r δ a ] ( 14 )
These are a series of linear differential equations which may be solved to give the aircraft lateral/directional response to inputs of rudder and aileron. You can observe that only the final equation has terms that involve ψ \psi ψ , and it may solved in isolation.
The two sets of matrix equations, (8) and (14) , are known as the equations of motion in concise form .
As will prove to be integral to Module 5, these are in state space form
x ⃗ ˙ = A x ⃗ + B u ⃗ \dot{\vec{x}} = \boldsymbol{A}\vec{x} + \boldsymbol{B}\vec{u} x ˙ = A x + B u
where
x ⃗ = the state vector (n) \vec{x} = \text{the state vector (n)} x = the state vector (n)
u ⃗ = the control matrix (m) \vec{u} = \text{the control matrix (m)} u = the control matrix (m)
A = the system matrix (n by n) \boldsymbol{A} = \text{the system matrix (n by n)} A = the system matrix (n by n)
B = the control matrix (n by m) \boldsymbol{B} = \text{the control matrix (n by m)} B = the control matrix (n by m)
for a system with n n n states and m m m controls. There are three reasons why we write the equations of motion in state space form:
The aircraft stability characteristics are obtained directly from the system matrix, A \boldsymbol{A} A - this will be explored in the final module.
In studies of aircraft control systems, the state space form is easiest to analyse.
Since A \boldsymbol{A} A and B \boldsymbol{B} B are constant matrices, we can numerically integrate the equations of motion (using a Runge-Kutta method) to obtain the time response of the aircraft to a given control displacement - we may wish to calculate the longitudinal response (time histories of u , w , q , θ u,w,q,\theta u , w , q , θ ) to a step change in elevator, δ e \delta_e δ e .
For the longitudinal equations in state space form, we have:
A = [ X u X w 0 − g ⋅ cos θ 0 Z u Z w U 0 − g ⋅ sin θ 0 M u ∗ M w ∗ M q ∗ M θ ∗ 0 0 1 0 ] , x ⃗ = [ u w q θ ] , B = [ 0 Z δ e M δ e ∗ 0 ] , u ⃗ = [ δ e ] \boldsymbol{A}=\begin{bmatrix}
X_u & X_w & 0 & -g\cdot\cos\theta_0\\
Z_u & Z_w & U_0 & -g\cdot\sin\theta_0\\
M_u^* & M_w^* & M_q^* & M_\theta^*\\
0 & 0 & 1 & 0
\end{bmatrix}, \vec{x}=\begin{bmatrix}
{u}\\{w}\\{q}\\{\theta}
\end{bmatrix}, \boldsymbol{B}= \begin{bmatrix}
0\\Z_{\delta_e}\\M_{\delta_e}^*\\0
\end{bmatrix}, \vec{u}=\left[\delta_e\right] A = X u Z u M u ∗ 0 X w Z w M w ∗ 0 0 U 0 M q ∗ 1 − g ⋅ cos θ 0 − g ⋅ sin θ 0 M θ ∗ 0 , x = u w q θ , B = 0 Z δ e M δ e ∗ 0 , u = [ δ e ]
For the lateral/directional equations in state space form:
A = [ Y v 0 − U 0 g ⋅ cos θ 0 0 L v ∗ L p ∗ L r ∗ 0 0 N v ∗ N p ∗ N r ∗ 0 0 0 1 tan θ 0 0 0 0 0 sec θ 0 0 0 ] , x ⃗ = [ v p r ϕ ψ ] , B = [ Y δ r Y δ a L δ r ∗ L δ a ∗ N δ r ∗ N δ a ∗ 0 0 0 0 ] , u ⃗ = [ δ r δ a ] \boldsymbol{A}=\begin{bmatrix}Y_v & 0 & -U_0 & g\cdot\cos\theta_0 & 0\\L_v^* & L_p^* & L_r^* & 0 & 0\\N_v^* & N_p^* & N_r^* & 0 & 0\\0 & 1 & \tan\theta_0 & 0 & 0 \\ 0 & 0 & \sec\theta_0 & 0 & 0\end{bmatrix}, \vec{x} = \begin{bmatrix}v\\p\\r\\\phi\\\psi\end{bmatrix}, \boldsymbol{B}=\begin{bmatrix}Y_{\delta_r} & Y_{\delta_a}\\L_{\delta_r}^* & L_{\delta_a}^*\\N_{\delta_r}^* & N_{\delta_a}^*\\0 & 0\\0 & 0\end{bmatrix}, \vec{u}=\begin{bmatrix}\delta_r\\\delta_a\end{bmatrix} A = Y v L v ∗ N v ∗ 0 0 0 L p ∗ N p ∗ 1 0 − U 0 L r ∗ N r ∗ tan θ 0 sec θ 0 g ⋅ cos θ 0 0 0 0 0 0 0 0 0 0 , x = v p r ϕ ψ , B = Y δ r L δ r ∗ N δ r ∗ 0 0 Y δ a L δ a ∗ N δ a ∗ 0 0 , u = [ δ r δ a ]
Some aircraft may have derivatives listed that have been neglected in the preceding - for example, you may find an aircraft with a nonzero C D q C_{D_q} C D q , which is the nondimensional form of X q X_q X q . You can hopefully appreciate that it can be readily re-inserted into the longitudinal equations of motion by choosing the logical place to insert into the system matrix:
A = [ X u X w X q − g ⋅ cos θ 0 Z u Z w U 0 − g ⋅ sin θ 0 M u ∗ M w ∗ M q ∗ M θ ∗ 0 0 1 0 ] \boldsymbol{A}=\begin{bmatrix}
X_u & X_w & \color{red}{X_q} & -g\cdot\cos\theta_0\\
Z_u & Z_w & U_0 & -g\cdot\sin\theta_0\\
M_u^* & M_w^* & M_q^* & M_\theta^*\\
0 & 0 & 1 & 0
\end{bmatrix} A = X u Z u M u ∗ 0 X w Z w M w ∗ 0 X q U 0 M q ∗ 1 − g ⋅ cos θ 0 − g ⋅ sin θ 0 M θ ∗ 0
Similarly, if you have an aircraft with a C X δ e C_{X_{\delta_e}} C X δ e and hence ⋅ X δ e \cdot X_{\delta_e} ⋅ X δ e term, then this would be inserted into the control matrix
B = [ X δ e Z δ e M δ e ∗ 0 ] \boldsymbol{B}= \begin{bmatrix}
\color{red}{X_{\delta_e}}\\Z_{\delta_e}\\M_{\delta_e}^*\\0
\end{bmatrix} B = X δ e Z δ e M δ e ∗ 0
Units
As we’ll discover, it’s useful to know the dimensions of the different derivatives. Rather than writing them all out, the following guide will help:
Force Derivative wrt Velocity Perturbation
This is the terms like X u X_u X u or Z w Z_w Z w . These all have units of [ T − 1 ] \left[\text{T}^{-1}\right] [ T − 1 ] or s − 1 s^{-1} s − 1