A.6.2.1 Equations of Motion
1
A.6.2.1 Equations of Motion
The first step to developing the equations of motion for the launch vehicle is setting up the
appropriate coordinate systems. A complete set of coordinate systems is displayed in Fig.
A.6.2.1.1. We start with the 𝑒̂𝑖 frame. The 𝑒̂𝑖 frame is fixed in the Earth, and rotates with the
Earth. The unit vector 𝑒̂𝑧 lies along the Earth’s axis of rotation. Next we need an intermediate
coordinate frame, the 𝑎̂𝑖 frame. The 𝑎̂𝑖 frame is situated such that 𝑎̂𝑧 lies along 𝑒̂𝑧 ; 𝑎̂𝑥 is offset
from 𝑒̂𝑥 by angle θ. Angle θ is analogous to longitude. The final coordinate frame is the 𝑏̂𝑖
frame. This 𝑏̂𝑖 frame is situated such that 𝑏̂𝜃 lies along 𝑎̂𝑦 ; 𝑏̂𝑟 is offset from 𝑎̂𝑧 by angle Φ.
Angle Φ is analogous to latitude.
Fig.A.6.2.1.1: Coordinate frames used for developing equations of motion.
(Amanda Briden)
We need to derive the transformation matrix to get from the 𝑏̂𝑖 frame to the 𝑒̂𝑖 frame. To do this,
we first define transformation matrices which represent the simple rotation from 𝑒̂𝑖 to 𝑎̂𝑖 (Eq.
(A.6.2.1.1)), and from 𝑎̂𝑖 to 𝑏̂𝑖 (Eq. (A.6.2.1.2)). After we define these two transformation
matrices, we can combine Eqs. (A.6.2.1.1) and (A.6.2.1.2) to get the desired transformation
matrix in Eq. (A.6.2.1.3).
Author: Brad Ferris
A.6.2.1 Equations of Motion
2
 a x   cos( ) sin(  ) 0 ex 
a    sin(  ) cos( ) 0 e 
 y 
 y 
 a z   0
0
1 ez 
(A.6.2.1.1)
b  cos( ) 0  sin(  )  a x 
b    0
1
0  a y 
  
br   sin(  ) 0 cos( )   a z 
(A.6.2.1.2)
We combine Eqs. (A.6.2.1.1) and (A.6.2.1.2) to get:
ex  cos( ) cos( )  sin(  ) sin(  ) cos( ) b 
e    cos( ) sin(  ) cos( ) sin(  ) sin(  )  b 
 y 
  
ez    sin(  )
0
cos( )  br 
(A.6.2.1.3)
Next, we analyze the forces acting on the launch vehicle. The free body diagram of the launch
vehicle is shown in Fig. (A.6.2.1.2). The angle γ is the flight path angle, or the angle between
the direction of velocity and the local horizon.
For the purpose of developing a trajectory, the launch vehicle is treated as a point mass. The
stability of the launch vehicle is left to the D&C group. Therefore, this model of the launch
vehicle does not take into account the launch vehicle’s moments of inertia, the location of the
center of pressure, or any other perturbing forces that affect stability. The forces weight, drag,
and thrust are assumed to act in the 𝑏̂𝑟 - 𝑏̂𝜃 plane. To summarize, the launch vehicle is modeled
as a dot flying through space.
Weight (W) is the force the Earth exerts on the launch vehicle. Weight is a function of the mass
of the launch vehicle. Therefore, weight varies over the flight as the launch vehicle burns fuel
and discards inert mass (stages and payload fairings). The weight of the launch vehicle is always
directed to the center of the Earth (opposite 𝑏̂𝑟 ).
Thrust (T) is the force developed by the engine. A detailed description about modeling thrust is
given in Section A.6.2.1.5. In modeling thrust, the angle 𝛼’ is the angle at which thrust is offset
Author: Brad Ferris
A.6.2.1 Equations of Motion
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from the direction of the velocity. For the trajectory model, the thrust is always aligned with the
velocity, therefore 𝛼’ is always equal to zero.
Drag (D) is the force the atmosphere exerts on the launch vehicle. A detailed description about
modeling drag is given in Section A.6.2.1.6. Drag always acts opposite of the direction of the
velocity.
Fig.A.6.2.1.2: Free-body diagram of launch vehicle used to develop equations of motion.
(Brad Ferris)
In order to develop the equations of motion, we use Newton’s Second Law. Newton’s Second
Law states that force (F) is equal to mass (m) multiplied by acceleration (a).
𝐹̅ = 𝑚𝑎̅
(A.6.2.1.4)
The equations of motion satisfy Newton’s Second Law. The treatment of the forces is called
kinetics, and the treatment of the motion (acceleration) is called kinematics.
In order to analyze the kinetics, it is necessary to break up the forces into their components in the
𝑏̂𝑖 frame.
̅ = 0𝑏̂𝜃 − 𝑚𝑔𝑏̂𝑟
𝑊
Author: Brad Ferris
(A.6.2.1.5)
A.6.2.1 Equations of Motion
4
̅ = −𝐷 cos(𝛾) 𝑏̂𝜃 − 𝐷sin⁡(𝛾)𝑏̂𝑟
𝐷
(A.6.2.1.6)
𝑇̅ = 𝑇 cos(𝛾 + 𝛼′) 𝑏̂𝜃 + 𝑇sin⁡(𝛾 + 𝛼′)𝑏̂𝑟
(A.6.2.1.7)
Next, we sum all of the forces in each direction. We introduce a term for the force of the wind in
each direction (Fwi). A detailed description about modeling wind force is given in Section
A.6.2.1.4.
∑ 𝐹𝑟 = −𝑚𝑔 − 𝐷𝑠𝑖𝑛(𝛾) + 𝑇𝑠𝑖𝑛(𝛾 + 𝛼 ′ ) + 𝐹𝑤𝑟
(A.6.2.1.8)
∑ 𝐹𝜃 = −𝐷𝑐𝑜𝑠(𝛾) + 𝑇𝑐𝑜𝑠(𝛾 + 𝛼 ′ ) + 𝐹𝑤𝜃
(A.6.2.1.9)
∑ 𝐹𝜙 = 𝐹𝑤𝜙
(A.6.2.1.10)
In order to obtain the kinematic part of the equations of motion, we first establish a position
̂𝑟 ).
vector (𝑟̅ = 𝑟𝑏
We need to differentiate the position vector two times in order to get
acceleration. It is necessary to use the Basic Kinematic Equation (BKE) to differentiate, because
the bi frame is not inertially fixed. With the BKE, we introduce angular velocity (ω).
𝑎
𝑑𝑥̅
𝑑𝑡
𝑏
=
𝑑𝑥̅
𝑑𝑡
+ 𝑎𝜔
̅ 𝑏 × 𝑥̅
(A.6.2.1.11)
With the angular velocity and the position vector, we can differentiate to get acceleration.
𝑒
𝜔
̅ 𝑏 = 𝜃̇𝑎̂𝑧 + 𝜙̇𝑏̂𝜃
(A.6.2.1.12)
𝑒
𝜔
̅ 𝑏 = 𝜃̇ cos(𝜃) 𝑏̂𝑟 − 𝜃̇ sin(𝜃) 𝑏̂𝜙 + 𝜙̇𝑏̂𝜃
(A.6.2.1.12)
𝑟̅ = 𝑟𝑏̂𝑟
(A.6.2.1.13)
𝑟̅̇ = 𝑟̇ 𝑏̂𝑟 + 𝑟𝜙̇𝑏̂𝜙 + 𝑟𝜃̇sin⁡(𝜙)𝑏̂𝜃
(A.6.2.1.14)
𝑟̅̈ = [𝑟̈ − 𝑟𝜃̇ 2 sin(𝜙)2 − 𝑟𝜙̇ 2 ]𝑏̂𝑟 + [2𝑟̇ 𝜙̇ + 𝑟𝜙̈ − 𝑟𝜃̇ 2 cos(𝜙) sin(𝜙)]𝑏̂𝜙 + [2𝑟̇ 𝜃̇ sin(𝜙) +
𝑟𝜃̈ sin(𝜙) + 2𝑟𝜃̇𝜙̇ cos(𝜙)]𝑏̂𝜃
(A.6.2.1.15)
Now that we have the kinetics and kinematics describing the motion of the launch vehicle, we
can use Newton’s Second Law, Eq. (A.6.2.1.4), to get the equations of motion. Recall that
kinetics pertains to the treatment of the forces, Eqs. (A.6.2.1.8) – (A.6.2.1.10), and the
Author: Brad Ferris
A.6.2.1 Equations of Motion
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kinematics pertains to the treatment of the motion. The mass required for Newton’s Second Law
is the mass of the launch vehicle (m). Since the launch vehicle is burning fuel and discarding
inert mass, the mass of the launch vehicle is a function of time, m=m(t). Applying Newton’s
Second Law, the final EOMs for the launch vehicle are defined in Eqs, (A.6.2.1.16) –
(A.6.2.1.18):
𝑏̂𝑟 : −𝑚𝑔 − 𝐷𝑠𝑖𝑛(𝛾) + 𝑇𝑠𝑖𝑛(𝛾 + 𝛼 ′ ) + 𝐹𝑤𝑟 = 𝑚[𝑟̈ − 𝑟𝜃̇ 2 sin(𝜙)2 − 𝑟𝜙̇ 2 ] (A.6.2.1.16)
𝑏̂𝜙 : 𝐹𝑤𝜙 = 𝑚[2𝑟̇ 𝜙̇ + 𝑟𝜙̈ − 𝑟𝜃̇ 2 cos(𝜙) sin(𝜙)]
(A.6.2.1.17)
𝑏̂𝜃 : −𝐷𝑐𝑜𝑠(𝛾) + 𝑇𝑐𝑜𝑠(𝛾 + 𝛼 ′ ) + 𝐹𝑤𝜃 = 𝑚[2𝑟̇ 𝜃̇ sin(𝜙) + 𝑟𝜃̈ sin(𝜙) + 2𝑟𝜃̇𝜙̇ cos(𝜙)]
(A.6.2.1.18)
If we desire the orientation of the launch vehicle with respect to the Earth fixed frame (𝑒̂𝑖 ), we
apply the transformation matrix in Eq. (A.6.2.1.3) to Eqs. (A.6.2.1.16) – (A.6.2.1.18).
Author: Brad Ferris