# Concise Form of Dimensional EoMs (Undergraduate Version)#

$\newcommand{\pd}[2]{\frac{\partial#1}{\partial#2}}$
$\newcommand{\ppd}[2]{\frac{\partial^2#1}{\partial#2^2}}$
$\newcommand{\pppd}[2]{\frac{\partial^3#1}{\partial#2^3}}$

As demonstrated above, the dimensional EoMs are either symmetric (a.k.a. longitudinal), where $$u,w,q$$ are controlled by $$\theta$$ and $$\delta_e$$, or asymmetric (a.k.a. lateral/directional) where $$v,p,r$$ are controlled by $$\delta_a$$ and $$\delta_r$$. 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:

(68)#$\color{red}{m\dot{u}= \left.\pd{X}{u}\right|_0u + \left.\pd{X}{w}\right|_0w - mg\cdot\cos\theta_0\cdot\theta^\prime}$
(69)#$\color{red}{m\dot{w}= \left.\pd{Z}{u}\right|_0u + \left.\pd{Z}{w}\right|_0w + mU_0q - mg\cdot\sin\Theta_0\cdot\theta^\prime + \left.\pd{Z}{\delta_e}\right|_0\delta_e}$
(70)#$\color{red}{I_{yy}\dot{q} = \left.\pd{M}{u}\right|_0u+\left.\pd{M}{w}\right|_0w+\left.\pd{M}{\dot{w}}\right|_0\dot{w}+\left.\pd{M}{q}\right|_0q+\left.\pd{M}{\delta_e}\right|_0\delta_e}$

## Longitudinal EoMs#

1. The X-Force equation (forward speed equation):

Taking Equation (68) and dividing by $$m$$:

(71)#$\dot{u} = X_uu + X_ww - g\cdot\cos\theta_0\theta^\prime$

where

$X_u\triangleq \frac{1}{m}\left.\pd{X}{u}\right|_0, X_w\triangleq \frac{1}{m}\left.\pd{X}{w}\right|_0$
2. The Z-Force equation (heave equation):

Taking Equation (69) and dividing by $$m$$:

(72)#$\dot{w} = Z_uu + Z_ww + U_eq - g\cdot\sin\theta_0\cdot\theta^\prime + Z_{\delta_e}\delta_e$

where

$Z_u\triangleq \frac{1}{m}\left.\pd{Z}{u}\right|_0, Z_w\triangleq \frac{1}{m}\left.\pd{Z}{w}\right|_0, Z_{\delta_e}\triangleq \frac{1}{m}\left.\pd{Z}{\delta_e}\right|_0$
3. The M-Moment equation (pitching moment equation):

If we take Equation (70) and divide by $$I_{yy}$$:

$\dot{q} = M_uu + M_ww + M_{\dot{w}}\dot{w} + M_qq + M_{\delta_e}\delta_e$

where

$M_u\triangleq \frac{1}{I_{yy}}\left.\pd{M}{u}\right|_0, \text{and the rest of the pattern should be clear}$

There is already an expression for $$\dot{w}$$, Equation (72), which may be substituted in:

$\dot{q} = M_uu + M_ww + M_{\dot{w}}\left(Z_uu + Z_ww + U_eq - g\cdot\sin\theta_0\cdot\theta^\prime + Z_{\delta_e}\delta_e \right) + M_qq + M_{\delta_e}\delta_e$

and can be simplified4 to:

(73)#$\dot{q} = M_u^*u + M_w^*w +M_\omega^*\omega + M_q^*q + M_{\delta_e}^*\delta_e$

where

(74)#$\begin{split}\begin{split}&M_u^*\triangleq M_u + M_{\dot{w}}Z_u\hphantom{======} M_w^*\triangleq M_w+M_{\dot{w}}Z_w\hphantom{======}M_q^*\triangleq M_q+M_{\dot{w}}U_0\\ &M_{\delta_e}^*\triangleq M_{\delta_e}+M_{\dot{w}}Z_{\delta_e}\hphantom{======}M_{\theta}^*\triangleq-M_{\dot{w}}g\sin\theta_0 \end{split}\end{split}$
4. The linearised Euler pitch equation (pitch rate kinematic equation):

$$q$$ is pitch rate, from Eq (48):

$\dot{\theta}^\prime = q$

Now Equations (71), (72), (73), and (48) can be written in matrix form to give the linearised equation of longitudinal motion in concise form:

(75)#\begin{split}\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}\end{split}
$\dot{\vec{x}} = A\vec{x} + B\vec{u}$

Which is in state space form. Equation (75) is a series of simultaneous 1st order ODEs with constant coefficients which may solved to find the longitudinal response ($$u,v,q,\theta$$) of an aircraft due to elevator deflection ($$\delta_e$$).

## Lateral/Directional EoMs#

The asymmetric or lateral/directional EoMs may be similarly manipulated.

(76)#$\color{darkgreen}{m\dot{v}=\left.\pd{Y}{v}\right|_0v - mU_0r+mg\cdot cos\theta_0\cdot\phi^\prime\left.\pd{Y}{\delta_r}\right|_0\delta_r}$
(77)#$\color{darkgreen}{I_{xx}\dot{p} - I_{xz}\dot{r} = \left.\pd{L}{v}\right|_0v + \left.\pd{L}{p}\right|_0p+\left.\pd{L}{r}\right|_0r+\left.\pd{L}{\delta_r}\right|_0\delta_r+\left.\pd{L}{\delta_a}\right|_0\delta_a}$
(78)#$\color{darkgreen}{I_{zz}\dot{r}-I_{xz}\dot{p} = \left.\pd{N}{v}\right|_0v+ \left.\pd{N}{p}\right|_0p + \left.\pd{N}{r}\right|_0r + \left.\pd{N}{\delta_r}\right|_0\delta_r+ \left.\pd{N}{\delta_a}\right|_0\delta_a}$
1. The Y-Force equation (sideslip equation): Taking Equation (76) and dividing by $$m$$:

$\dot{v} = Y_vv-U_er + g\cdot\cos\theta_0\phi + Y_{\delta_r}{\delta_r}$

where

$Y_v\triangleq\frac{1}{m}\pd{Y}{v},\,Y_{\delta_r}\triangleq\frac{1}{m}\pd{Y}{\delta_r}$
2. The two moment equations are coupled, so need to be decoupled. Dividing Equation (77) by the rolling moment of inertia, $$I_xx$$:

(79)#$\dot{p} = \frac{I_{xz}}{I_{xx}}\dot{r} + L_vv + L_pp + L_rr + L_{\delta_r}{\delta_r} + L_{\delta_a}\delta_a$

where

$L_v\triangleq\frac{1}{I_{xx}}\pd{L}{v},\,L_{p}\triangleq\frac{1}{I_{xx}}\pd{L}{p},\,\text{etc.}$

the same can be performed for (78), dividing by the yawing moment of inertia $$I_{zz}$$:

(80)#$\dot{r} = \frac{I_{xz}}{I_{zz}}\dot{p} + N_vv+N_pp + N_rr + N_{\delta_r}\delta_r+N_{\delta_a}\delta_a$

Clearly Equations (79) and (80) are coupled, and must be decoupled if we want to use them in standard form. Substituting equation (80) into (79), after a little manipulation (if you do this yourself, I promise you’ll understand this subject better), the following is achieved:

$\begin{split}\begin{split}\dot{p}\left(1-\frac{I_{xz}^2}{I_{xx}I_{zz}}\right) &= \left(L_v + \frac{I_{xz}}{I_{xx}}N_v\right)v + \left(L_p + \frac{I_{xz}}{I_{xx}}N_p\right)p+\left(L_r+\frac{I_{xz}}{I_{xx}}N_r\right)r +\ldots\\ &\left(L_{\delta_r} + \frac{I_{xz}}{I_{xx}N_{\delta_r}}\right)\delta_r + \left(L_{\delta_a}+\frac{I_{xz}}{I_{xx}}N_{\delta_a}\right)\delta_a\end{split}\end{split}$

Noting that:

$1-\frac{I_{xz}^2}{I_{xx}I_{zz}} = \frac{I_{xx}I_{zz}-I_{zz}^2}{I_{xx}I_{zz}}$

hence, the rolling moment equation may be weitten as:

$\dot{p} = L_v^*v + L_p^*p + L_r^*r+L_{\delta_r}^*\delta_r+L_{\delta_a}^*\delta_a$

where:

$\begin{split}\begin{gathered} L_v^* = \frac{I_{xx}I_{zz}}{I_{xx}I_{zz}-I_{xz}^2}\left(L_v+\frac{I_{xz}}{I_{xx}}N_v\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, L_p^*=\frac{I_{xx}I_{zz}}{I_{xx}I_{zz}-I_{xz}^2}\left(L_p+\frac{I_{xz}}{I_{xx}}N_p\right)\\ L_r^* = \frac{I_{xx}I_{zz}}{I_{xx}I_{zz}-I_{xz}^2}\left(L_r+\frac{I_{xz}}{I_{xx}}N_r\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, L_{\delta_r}^*=\frac{I_{xx}I_{zz}}{I_{xx}I_{zz}-I_{xz}^2}\left(L_{\delta_r}+\frac{I_{xz}}{I_{xx}}N_{\delta_r}\right)\\ L_{\delta_a}^*=\frac{I_{xx}I_{zz}}{I_{xx}I_{zz}-I_{xz}^2}\left(L_{\delta_a}+\frac{I_{xz}}{I_{xx}}N_{\delta_a}\right) \end{gathered}\end{split}$

Similarly, by substituting (79) into (80), the yawing moment equation is yielded:

$\dot{r} = N_v^*v + N_p^*p + N_r^*r + N_{\delta_r}^*\delta_r + N_{\delta_a}^*\delta_a$

where:

$\begin{split}\begin{gathered} N_v^* = \frac{I_{xx}I_{zz}}{I_{xx}I_{zz}-I_{xz}^2}\left(N_v+\frac{I_{xz}}{I_{zz}}L_v\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, N_p^*=\frac{I_{xx}I_{zz}}{I_{xx}I_{zz}-I_{xz}^2}\left(N_p+\frac{I_{xz}}{newI_{zz}}L_p\right)\\ N_r^* = \frac{I_{xx}I_{zz}}{I_{xx}I_{zz}-I_{xz}^2}\left(N_r+\frac{I_{xz}}{I_{zz}}L_r\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, N_{\delta_r}^*=\frac{I_{xx}I_{zz}}{I_{xx}I_{zz}-I_{xz}^2}\left(N_{\delta_r}+\frac{I_{xz}}{I_{zz}}L_{\delta_r}\right)\\ N_{\delta_a}^*=\frac{I_{xx}I_{zz}}{I_{xx}I_{zz}-I_{xz}^2}\left(N_{\delta_a}+\frac{I_{xz}}{I_{zz}}L_{\delta_a}\right) \end{gathered}\end{split}$

often $$I_{xz}$$ is small compared to $$I_{xx}$$ and $$I_{zz}$$, so for these circumstances $$N_v^*\simeq N_v$$, $$L_v^*\simeq L_v$$

3. Lastly, the yaw rate and body rate kinematic equations give:

\begin{split}\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}\end{split}

So finally, the linearised lateral/directional equations of motion may be expressed in state-space form:

(81)#$\begin{split}\begin{bmatrix}\dot{v}\\\dot{p}\\\dot{r}\\\dot{\phi}\\\dot{\psi}\end{bmatrix}=\begin{bmatrix}Y_v & 0 & -U_0 & g\cdot\cos\theta_0 & 0\\L_v^* & L_p^* & L_r^* & 0 & 0\\N_v^* & N_p^* & N_r^* & 0 & 0\\0 & 1 & \tan\theta_0 & 0 & 0 \\ 0 & 0 & \sec\theta_0 & 0 & 0\end{bmatrix}\begin{bmatrix}v\\p\\r\\\phi\\\psi\end{bmatrix}+\begin{bmatrix}Y_{\delta_r} & 0\\L_{\delta_r}^* & L_{\delta_a}^*\\N_{\delta_r}^* & N_{\delta_a}^*\\0 & 0\\0 & 0\end{bmatrix}\begin{bmatrix}\delta_r\\\delta_a\end{bmatrix}\end{split}$

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 $$\psi$$, and it may solved in isolation.

The two sets of matrix equations, (75) and (81), 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

$\dot{\vec{x}} = \boldsymbol{A}\vec{x} + \boldsymbol{B}\vec{u}$

where

$\vec{x} = \text{the state vector (n)}$
$\vec{u} = \text{the control matrix (m)}$
$\boldsymbol{A} = \text{the system matrix (n by n)}$
$\boldsymbol{B} = \text{the control matrix (n by m)}$

for a system with $$n$$ states and $$m$$ 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, $$\boldsymbol{A}$$ - this will be explored in the final module.

• In studies of aircraft control systems, the state space form is easiest to analyse.

• Since $$\boldsymbol{A}$$ and $$\boldsymbol{B}$$ 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 $$u,w,q,\theta$$) to a step change in elevator, $$\delta_e$$.

For the longitudinal equations in state space form, we have:

$\begin{split}\boldsymbol{A}=\begin{bmatrix} X_u & X_w & 0 & -g\cdot\cos\theta_0\\ Z_u & Z_w & U_0 & -g\cdot\sin\theta_0\\ M_u^* & M_w^* & M_q^* & M_\theta^*\\ 0 & 0 & 1 & 0 \end{bmatrix}, \vec{x}=\begin{bmatrix} {u}\\{w}\\{q}\\{\theta} \end{bmatrix}, \boldsymbol{B}= \begin{bmatrix} 0\\Z_{\delta_e}\\M_{\delta_e}^*\\0 \end{bmatrix}, \vec{u}=\left[\delta_e\right]\end{split}$

For the lateral/directional equations in state space form:

$\begin{split}\boldsymbol{A}=\begin{bmatrix}Y_v & 0 & -U_0 & g\cdot\cos\theta_0 & 0\\L_v^* & L_p^* & L_r^* & 0 & 0\\N_v^* & N_p^* & N_r^* & 0 & 0\\0 & 1 & \tan\theta_0 & 0 & 0 \\ 0 & 0 & \sec\theta_0 & 0 & 0\end{bmatrix}, \vec{x} = \begin{bmatrix}v\\p\\r\\\phi\\\psi\end{bmatrix}, \boldsymbol{B}=\begin{bmatrix}Y_{\delta_r} & Y_{\delta_a}\\L_{\delta_r}^* & L_{\delta_a}^*\\N_{\delta_r}^* & N_{\delta_a}^*\\0 & 0\\0 & 0\end{bmatrix}, \vec{u}=\begin{bmatrix}\delta_r\\\delta_a\end{bmatrix}\end{split}$

Some aircraft may have derivatives listed that have been neglected in the preceding - for example, you may find an aircraft with a nonzero $$C_{D_q}$$, which is the nondimensional form of $$X_q$$. 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:

$\begin{split}\boldsymbol{A}=\begin{bmatrix} X_u & X_w & \color{red}{X_q} & -g\cdot\cos\theta_0\\ Z_u & Z_w & U_0 & -g\cdot\sin\theta_0\\ M_u^* & M_w^* & M_q^* & M_\theta^*\\ 0 & 0 & 1 & 0 \end{bmatrix}\end{split}$

Similarly, if you have an aircraft with a $$C_{X_{\delta_e}}$$ and hence $$\cdot X_{\delta_e}$$ term, then this would be inserted into the control matrix

$\begin{split}\boldsymbol{B}= \begin{bmatrix} \color{red}{X_{\delta_e}}\\Z_{\delta_e}\\M_{\delta_e}^*\\0 \end{bmatrix}\end{split}$

# 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 $$X_u$$ or $$Z_w$$. These all have units of $$\left[\text{T}^{-1}\right]$$ or $$s^{-1}$$

4

recalling that simplify often means make look nicer, but complicate the matter by introducing new terminology