Deriving the equations of motion

The flight dynamics of an aircraft are described by its equations of motion (EOM). We are going to
derive those equations in this chapter.

1 Forces
1.1 The basic force equation
To derive the equations of motion of an aircraft, we start by examining forces. Our starting point in
this is Newton’s second law. However, Newton’s second law only holds in an inertial reference system.
Luckily, the assumptions we have made earlier imply that the Earth-fixed reference frame FE is inertial.
(However, Fb is not an inertial reference frame.) So we will derive the equations of motion with respect
to FE .
Let’s examine an aircraft. Newton’s second law states that
Z Z 
d
F = dF = Vp dm , (1.1)
dt
where we integrate over the entire body. It can be shown that the right part of this equation equals
d
dt (VG m), where VG is the velocity of the center of gravity of the aircraft. If the aircraft has a constant
mass, we can rewrite the above equation into
dVG
F=m = mAG . (1.2)
dt
This relation looks very familiar. But it does imply something very important. The acceleration of the
CG of the aircraft does not depend on how the forces are distributed along the aircraft. It only depends
on the magnitude and direction of the forces.

1.2 Converting the force equation

There is one slight problem. The above equation holds for the FE reference frame. But we usually work
in the Fb reference frame. So we need to convert it. To do this, we can use the relation
 
dVG  dVG 
AG = = + ΩbE × VG . (1.3)
dt E dt b

Inserting this into the above equation will give

 
 u̇ + qw − rv
dVG 
F=m + m ΩbE × VG = m v̇ + ru − pw . (1.4)
 
dt b
ẇ + pv − qu

By the way, in the above equation, we have used that

   
u p
VG =  v  and ΩbE = q  . (1.5)
   

w r

Here, u, v and w denote the velocity components in X, Y and Z direction, respectively. Similarly, p, q
and r denote rotation components about the X, Y and Z axis, respectively.

1
1.3 External forces
Let’s take a look at the forces F our aircraft is subject to. There are two important kinds of forces:
gravity and aerodynamic forces. The gravitational force Fgravity is, in fact, quite simple. It is given by
h iT
FE
gravity = 0 0 mg , (1.6)

where g is the gravitational acceleration. The superscript E indicates that the force is given in
the FE reference frame. However, we want the force in the Fb reference frame. Luckily, we know the
transformation matrix TbE . We can thus find that
 
− sin θ
Fb E
gravity = TbE Fgravity = mg  sin ψ cos θ  . (1.7)
 

cos ψ cos θ

The aerodynamic forces Faero are, however, a lot more difficult. For now, we won’t examine them in
depth. Instead, we simply say that
h iT
Fb
aero = X b
Y b
Z b . (1.8)

By combining this knowledge with the equation of motion for forces, we find that
     
u̇ + qw − rv − sin θ Xb
  b
m v̇ + ru − pw = mg  sin ψ cos θ  +  Y  . (1.9)
  

ẇ + pv − qu cos ψ cos θ Zb

2 Moments
2.1 Angular momentum
Before we’re going to look at moments, we will first examine angular momentum. The angular mo-
mentum of an aircraft BG (with respect to the CG) is defined as
Z
BG = dBG = r × VP dm, (2.1)

where we integrate over every point P in the aircraft. We can substitute


dr 
V P = VG + + ΩbE × r. (2.2)
dt b
If we insert this, and do a lot of working out, we can eventually find that

BG = IG ΩbE . (2.3)

The parameter IG is the inertia tensor, with respect to the CG. It is defined as
  R 
(ry2 + rz2 )dm − (rx ry )dm − (rx rz )dm
R R
Ix −Jxy −Jxz
R 2
(rx + rz2 )dm − (ry rz )dm  .
  R R
IG = −Jxy Iy −Jyz  =  − (rx ry )dm (2.4)
 
R 2
(rx + ry2 )dm
R R
−Jxz −Jyz Iz − (rx rz )dm − (ry rz )dm

We have assumed that the XZ-plane of the aircraft is a plane of symmetry. For this reason, Jxy = Jyz = 0.
This simplifies the inertia tensor a bit.

2
2.2 The moment equation
It is now time to look at moments. We again do this from the inertial reference frame FE . The moment
acting on our aircraft, with respect to its CG, is given by
Z Z Z
d (Vp dm)
MG = dMG = r × dF = r × , (2.5)
dt
where we integrate over the entire body. Luckily, we can simplify the above relation to

dBG 
MG = . (2.6)
dt E
The above relation only holds for inertial reference frames, such as FE . However, we want to have the
above relation in Fb . So we rewrite it to

dBG 
MG = + ΩbE × BG . (2.7)
dt b
By using BG = IG ΩbE , we can continue to rewrite the above equation. We eventually wind up with

dΩbE 
MG = IG + ΩbE × IG ΩbE . (2.8)
dt b
In matrix-form, this equation can be written as
 
Ix ṗ + (Iz − Iy )qr − Jxz (pq + ṙ)
MG = Iy q̇ + (Ix − Iz )pr + Jxz (p2 − r2 ) . (2.9)
 

Iz ṙ + (Iy − Ix )pq + Jxz (qr − ṗ)

Note that we have used the fact that Jxy = Jyz = 0.

2.3 External moments

Let’s take a closer look at MG . Again, we can distinguish two types of moments, acting on our aircraft.
There are moments caused by gravity, and moments caused by aerodynamic forces. Luckily, the moments
caused by gravity are zero. (The resultant gravitational force acts in the CG.) So we only need to consider
the moments caused by aerodynamic forces. We denote those as
h iT
Mb G,aero = L M N . (2.10)

This turns the moment equation into

   
Ix ṗ + (Iz − Iy )qr − Jxz (pq + ṙ) L
2 2 =
Iy q̇ + (Ix − Iz )pr + Jxz (p − r ) M  . (2.11)
 

Iz ṙ + (Iy − Ix )pq + Jxz (qr − ṗ) N

3 Kinematic relations
3.1 Translational kinematics
Now that we have the force and moment equations, we only need to find the kinematic relations for our
aircraft. First, we examine translational kinematics. This concerns the velocity of the CG of the aircraft
with respect to the ground.

3
The velocity of the CG, with respect to the ground, is called the kinematic velocity Vk . In the FE
reference system, it is described by
h iT
Vk = VN VE −VZ . (3.1)

In this equation, VN is the velocity component in the Northward direction, VE is the velocity component
in the eastward direction, and −VZ is the vertical velocity component. (The minus sign is present because,
in the Earth-fixed reference system, VZ is defined to be positive downward.)
However, in the Fb reference system, the velocity of the CG, with respect to the ground, is given by
h iT
VG = u v w . (3.2)

To relate those two vectors to each other, we need a transformation matrix. This gives us

Vk = TEb VG = TTbE VG . (3.3)

This is the translational kinematic relation. We can use it to derive the change of the aircraft position.
To do that, we simply have to integrate the velocities. We thus have
Z t Z t Z t
x(t) = VN dt, y(t) = VE dt and h(t) = −VZ dt. (3.4)
0 0 0

3.2 Rotational kinematics

Now let’s examine rotational kinematics. This concerns the rotation of the aircraft. In the FE reference
system, the rotational velocity is described by the variables ϕ̇, θ̇ and ψ̇. However, in the Fb reference
system, the rotational velocity is described by p, q and r. The relation between these two triples can be
shown to be     
p 1 0 − sin θ ϕ̇
q  = 0 cos ϕ cos θ sin ϕ   θ̇  . (3.5)
    

r 0 − sin ϕ cos θ cos ϕ ψ̇

This is the rotational kinematic relation. It is interesting to note that, if ϕ = θ = ψ = 0, then p = ϕ̇,
q = θ̇ and r = ψ̇. By the way, we can also invert the above relation. We would then get
    
ϕ̇ 1 sin ϕ tan θ cos ϕ tan θ p
 θ̇  = 0 cos ϕ − sin ϕ  q  . (3.6)
    

ψ̇ 0 sin ϕ/ cos θ cos ϕ/ cos θ r