A.6.2.1 Equations of Motion1

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 frame. The frame is fixed in the Earth, and rotates with the Earth. The unit vector lies along the Earth’s axis of rotation. Next we need an intermediate coordinate frame, the frame. The frame is situated such that lies along ; is offset from by angle θ. Angle θ is analogous to longitude. The final coordinate frame is the frame. This frame is situated such that lies along ; is offset from 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 frame to the frame. To do this, we first define transformation matrices which represent the simple rotation from to (Eq. (A.6.2.1.1)), and from to (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).

(A.6.2.1.1)

(A.6.2.1.2)

We combine Eqs. (A.6.2.1.1) and (A.6.2.1.2) to get:

(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 - 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 ).

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 is the angle at which thrust is offset from the direction of the velocity. For the trajectory model, the thrust is always aligned with the velocity, therefore 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).

(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 frame.

(A.6.2.1.5)

(A.6.2.1.6)

(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.

(A.6.2.1.8)

(A.6.2.1.9)

(A.6.2.1.10)

In order to obtain the kinematic part of the equations of motion, we first establish a position vector (). 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.6.2.1.11)

With the angular velocity and the position vector, we can differentiate to get acceleration.

(A.6.2.1.12)

(A.6.2.1.12)

(A.6.2.1.13)

(A.6.2.1.14)

(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 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):

(A.6.2.1.16)

(A.6.2.1.17)

(A.6.2.1.18)

If we desire the orientation of the launch vehicle with respect to the Earth fixed frame (), 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