Project Bellerophon 644 A.6.0 Trajectory A.6.1 Introduction The trajectory of the launch vehicle is critical to mission success. Without a sub-optimal trajectory, the amount of ∆V required is very large and increases the cost of the vehicle. The first thing that we do to understand the problem is to define the assumptions of our model. The trajectory group is mostly concerned with the translational degrees of freedom of the vehicle and leaves the attitude of the launch vehicle to the D&C group. Therefore, in all of our analysis, we assume that the launch vehicle is a point mass. Our second main assumption is that all of the forces acting on the vehicle are in the same plane, effectively reducing the translational degrees of freedom of the vehicle to two. Once the assumptions are defined, we derive the force model and the equations of motion of the vehicle for each stage. With the equations of motion for the vehicle defined, the design moved to the creation of the steering laws. At this point, the design split into two distinct sections: one for ground and balloon launch and another for aircraft launch. This separation is due to the different initial conditions for each launch type. Once the steering laws are formed, the angles at the end of each stage are optimized to produce the best nominal trajectory for each launch type (ground, balloon and aircraft). Due to the nature of trajectory optimization and design, all members of the trajectory group worked on all aspects of the problem. Even though each member contributed to every section, the sections are written by the trajectory group member that worked the most on that aspect of the design. The following sections go into the details of the trajectory group’s work. Author: Elizabeth Harkness Project Bellerophon 645 A.6.2 Design Methods 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). Author: Brad Ferris Project Bellerophon 646 ⎡ax ⎤ ⎡ cos(θ ) sin(θ ) 0⎤⎡ex ⎤ ⎢a ⎥ = ⎢− sin(θ ) cos(θ ) 0⎥⎢e ⎥ (A.6.2.1.1) ⎢ y ⎥ ⎢ ⎥⎢ y ⎥ ⎣⎢az ⎦⎥ ⎣⎢ 0 0 1⎦⎥⎣⎢ez ⎦⎥ ⎡bφ ⎤ ⎡cos(φ) 0 − sin(φ)⎤⎡ax ⎤ ⎢b ⎥ = ⎢ 0 1 0 ⎥⎢a ⎥ (A.6.2.1.2) ⎢ θ ⎥ ⎢ ⎥⎢ y ⎥ ⎣⎢br ⎦⎥ ⎣⎢sin(φ) 0 cos(φ) ⎦⎥⎣⎢az ⎦⎥ We combine Eqs. (A.6.2.1.1) and (A.6.2.1.2) to get: ⎡ex ⎤ ⎡cos(φ)cos(θ ) − sin(θ ) sin(φ)cos(θ )⎤⎡bφ ⎤ ⎢e ⎥ = ⎢cos(φ)sin(θ ) cos(θ ) sin(φ)sin(θ )⎥⎢b ⎥ (A.6.2.1.3) ⎢ y ⎥ ⎢ ⎥⎢ θ ⎥ ⎣⎢ez ⎦⎥ ⎣⎢ − sin(φ) 0 cos(φ) ⎦⎥⎣⎢br ⎦⎥ 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 Author: Brad Ferris Project Bellerophon 647 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. 0 (A.6.2.1.5) Author: Brad Ferris Project Bellerophon 648 cos sin (A.6.2.1.6) cos′ sin ′ (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) cos sin (A.6.2.1.12) (A.6.2.1.13) sin (A.6.2.1.14) sin 2 cos sin 2 sin sin 2 cos (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 Project Bellerophon 649 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): ′ : sin (A.6.2.1.16) : 2 cos sin (A.6.2.1.17) ′ : 2 sin sin 2 cos (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 Project Bellerophon 650 A.6.2.1.1 Aircraft launch For an aircraft launch, the equations of motion of the launch vehicle are the same as the equations of motion given in Eqs. (A.6.2.1.16) – (A.6.2.1.18) in Section A.6.2.1. For the numerical integration used in the trajectory codes, it is necessary to put the equations of motion into state space. Therefore, in state space form we have the following: ⎡x1 ⎤ ⎡r⎤ ⎢x ⎥ ⎢r⎥ ⎢ 2 ⎥ ⎢ &⎥ ⎢x3 ⎥ ⎢θ ⎥ ⎢ ⎥ = ⎢ ⎥ (A.6.2.1.1.1) ⎢x4 ⎥ ⎢θ&⎥ ⎢x ⎥ ⎢φ⎥ ⎢ 5 ⎥ ⎢ ⎥ ⎣⎢x6 ⎦⎥ ⎣⎢φ&⎦⎥ (A.6.2.1.1.2) ′ sin ′ 1 2 sin 2 cos 1 2 cos sin The equations in Eqs. (A.6.2.1.1.1) – (A.6.2.1.1.2) are subject to the following initial condition: ⎡r(0)⎤ ⎡6,391,000[meters]⎤ ⎢ ⎥ ⎢ ⎥ ⎢r&(0)⎥ ⎢ 0[m / s] ⎥ ⎢θ (0)⎥ ⎢ (−80)[deg.] ⎥ ⎢ ⎥ = ⎢ ⎥ ⎢θ&(0)⎥ ⎢ 0[rad / s] ⎥ ⎢φ(0)⎥ ⎢ (90 − 28)[deg.] ⎥ ⎢ ⎥ ⎢ ⎥ ⎣⎢φ&(0)⎦⎥ ⎣⎢ 0[rad / s] ⎦⎥ Author: Brad Ferris Project Bellerophon 651 A.6.2.1.2 Balloon Launch For a balloon launch, the equations of motion of the launch vehicle are the same as the equations of motion given in Eqs.