Exploring the Multidimensional Steady Flight Envelope of a Business Jet Aircraft Silvia Ferrari Princeton University FAA/NASA Joint University Program on Air Transportation, FAA Tech Center, Atlantic City, NJ October 26-27, 2000

Table of Contents Introduction and motivation Proportional-integral neural network controller Nonlinear business jet aircraft model Trim map definition Exploring aircraft multidimensional flight envelope Computation and storage of trim data Trim control surfaces Conclusions and future work At the last JUP meeting, the role and pre-training of the forward neural network in the overall control system, was discussed and demonstrated by modelling the longitudinal trim data of a transport aircraft reduced model. Here, a procedure for mapping non-symmetric trim for a full, 6 d.o.f. model of a business jet, is proposed. This procedure uses all of the aircraft’s equations of motion to determine the trim control settings in the case of coupled longitudinal and lateral-directional dynamics. Therefore, the command input vector includes both longitudinal and lateral-directional command inputs. The aircraft angles and coordinate frames that are involved in the solution are presented here. In particular, for non-level flight, 3-dimensional (spherical) trigonometric relationships between Euler, airflow, and flight path angles are needed. This procedure allows one to determine the control settings for fully-coupled coordinated maneuvers, such as coordinated climbing or descending turns.

Introduction Stringent operational requirements introduce: Complexity Nonlinearity Uncertainty Gain scheduling: Design a set of “local” linear controllers Interpolate local operating point designs Gain scheduling limitations: Nonlinearities dominate Fast transitions between operating points Nonlinear control design takes advantage of knowledge gained from linear controllers Mention references (i.e. prior approaches) and underline the difference between this method and the old ones

Neural Networks for Control: Copying with Complexity Applicability to nonlinear systems Applicability to multi-variable systems Parallel distributed processing and hardware implementation Learning Flexible logic INTRODUCE PI STRUCTURE AND IDEA FOR PINN MOTIVATION… 1- .. by replacing the gains with neural networks 2- Explain well what is covered here and what is instead the ultimate goal - this is why the training is called "pre-training" - and that performance baselines for this phase are established by an equivalent linear model.. 3- The overall procedure provides improved globala control w.r.t. gain scheduling, because ... 4- This phase already defines a global nonlinear neural network controller with less effort than.. (also compare to previous such methods) and is demonstrated here

Motivation for Neural Network-Based Controller Network of networks motivated by linear control structure Multiphase learning: Pre-training On-line training during piloted simulations or testing Improved global control Pre-training phase alone provides: A global neural network controller An excellent initialization point for on-line learning INTRODUCE PI STRUCTURE AND IDEA FOR PINN MOTIVATION… 1- .. by replacing the gains with neural networks 2- Explain well what is covered here and what is instead the ultimate goal - this is why the training is called "pre-training" - and that performance baselines for this phase are established by an equivalent linear model.. 3- The overall procedure provides improved globala control w.r.t. gain scheduling, because ... 4- This phase already defines a global nonlinear neural network controller with less effort than.. (also compare to previous such methods) and is demonstrated here

Proportional-Integral Controller Closed-loop stability: yc x(t) ys(t) CI CF CB Hu Hx + - u(t) PLANT Delta’s omitted for simplicity. *Study the formulas and notation from the paper in case of questions.. yc = desired output, (xc,uc) = set point.

Proportional-Integral Neural Network Controller ys(t) + uI(t) NNI - yc + uc u(t) x(t) NNF PLANT + + e + uB(t) Note the total control is the sum of uc and the NNs contributions.. a NNB SVG - xc CSG Where:

Nonlinear Business Jet Aircraft Model Aircraft Equations of Motion: Body axis: XB, YB, ZB XB ZB Thrust Drag Lift V mg p q r a b YB State vector: Control vector: Parameters: Output vector: The aircraft nonlinear model is in terms of the given state and control vectors. In this and the following slides, the body frame, the inertial frame and all of the significant aircraft angle are introduced to illustrate the complexity of the problem at hand. For the model:

Aircraft Body Frame and Flight Path Angles Flight path angles: g, x, m XE . YE x g V Command input: m not shown on this figure

Aircraft Angles Airflow angles: a, b Euler angles: f, q, y YB XB ZB V u w Kinematic equations: : non-orthogonal transformation matrix

Neural Network Modelling of Aircraft Trim Maps The trim equations, xc yc uc CSG are solved for the control settings, uc, over initial estimate of operational domain, (X0, P0). p(a,yc) Command State Generator (CSG): employs kinematic equations to produce xc yc uc NNF The trim equations are obtained by setting the left-hand side of part of the equations of motion to zero, as this is a quasi-static equilibrium. These nonlinear equations are usually solved for the control settings over that which is an initial estimate of the aircraft operational domain. Once the region of operation over which trim can be achieved by the available controls (i.e., where a solution u* exists) is numerically determined, one can define this to be the effective aircraft flight envelope. This “envelope” will be multi-dimensional in nature, because of the dimensionality of x and p. The cartoons show how the results obtained by the non-symmetrical trim mapping, UT, provide the (final) data to train the forward neural network. In essence, the forward neural network is modelling the inverse of the nonlinear map provided by the aircraft equations of motions that are contained in fT. The aircraft flight envelope consists of all (x,p), for which a solution uc, s.t.: a a = [ h ]

Example of Coordinated Maneuver YB XB ZB q mg f acent V Steady Climbing Turn: acent = centripetal acceleration = rate of turn Command input: [Etkin, “Dynamics of Flight”]

Estimated Business Jet Aircraft Steady Flight Envelope Based on simplified aircraft model, and aerodynamics and thrust characteristics Stall speed: CL = CLmax Thrust-limited minimum speed h (m) Maximum allowable dynamic pressure The aircraft nonlinear model is in terms of the given state and control vectors. In this and the following slides, the body frame, the inertial frame and all of the significant aircraft angle are introduced to illustrate the complexity of the problem at hand. Compressible Mach number limit V (m/s) [Stengel, “Flight Dynamics and Control”]

Spherical Mapping of Aircraft Angles Spherical trigonometric relations: x a Angles defined on a unit sphere: Redundant set of angles: f, a The equations of motion and the command input are formulated in terms of all three sets of aircraft angles commonly known as: Euler, airflow, and flight path angles. As exaplined in the reference [Kalviste, J. “Spherical Mapping and Analysis of Aircraft Angles for Maneuvering Flight”, AIAA Journal of Aircraft, V. 24, no. 8, August 1987], these three sets of angles are related by spherical trigonometric nonlinear equations, in the general form shown (“fn” stands for “nonlinear function of”). In particular, at least two of the angles, in this case f and a, are usually redundant. That is, once the remaining angles are specified, these two are completely determined by one of the above relationships. assuming x = y. [Kalviste, 1987]

Non-Symmetrical Trim Solution at One Operating Condition Given the command input, , and the scheduling variable, h, In a coordinated maneuver: Solve the trim equations, , for Subject to, Aircraft’s kinematic equations: In a coordinated maneuver, such as the one illustrated in the previous slide, two of the Euler angular rates are zero and the rate of turn, dy/dt, is constant, representing the steady angular rate of the turning velocity vector. With these conditions, the kinematic equations simplify to the form shown providing an expression for each of the body angular rates in terms of q, f, dy/dt, all of which are constant during the motion. The command input also provides constant known values of the elements shown. Therefore, if we set 6 of the equations of motion to zero, as shown, we can solve for the 4 elements of the control vector, as well as for the pitch angle and the rate of turn. This is possible because the remaining angles, f and a, are known through the trigonometric relations. Spherical trigonometric relations: Airflow angles definitions:

Computed Three-Dimensional Flight Envelope h (m) g (deg) V (m/s) h (m) The aircraft nonlinear model is in terms of the given state and control vectors. In this and the following slides, the body frame, the inertial frame and all of the significant aircraft angle are introduced to illustrate the complexity of the problem at hand. b = m = 0 g (deg) V (m/s)

Computed Multi-Dimensional Flight Envelope h (m) g (deg) V (m/s) b = m =0 b = 5, m = -20 h (m) The aircraft nonlinear model is in terms of the given state and control vectors. In this and the following slides, the body frame, the inertial frame and all of the significant aircraft angle are introduced to illustrate the complexity of the problem at hand. Similarly for all: V (m/s) g (deg)

Rotations of Three-Dimensional Envelope View b = 5, m = -20 Rotations of Three-Dimensional Envelope View h (m) h (m) g (deg) g (deg) V (m/s) V (m/s) h (m) h (m) The aircraft nonlinear model is in terms of the given state and control vectors. In this and the following slides, the body frame, the inertial frame and all of the significant aircraft angle are introduced to illustrate the complexity of the problem at hand. g (deg) V (m/s) g (deg) V (m/s)

Multidimensional Flight Envelope Storage Using Cell Arrays Cell arrays are containers that can hold other MATLAB arrays Multidimensional envelope V-H envelope (nested cell) b m g In a coordinated maneuver, such as the one illustrated in the previous slide, two of the Euler angular rates are zero and the rate of turn, dy/dt, is constant, representing the steady angular rate of the turning velocity vector. With these conditions, the kinematic equations simplify to the form shown providing an expression for each of the body angular rates in terms of q, f, dy/dt, all of which are constant during the motion. The command input also provides constant known values of the elements shown. Therefore, if we set 6 of the equations of motion to zero, as shown, we can solve for the 4 elements of the control vector, as well as for the pitch angle and the rate of turn. This is possible because the remaining angles, f and a, are known through the trigonometric relations.

Optimal Parameters Storage Using Cell Arrays To each operating point, there corresponds a vector of optimal parameters Optimal parameter cell Multidimensional cell V h b m g In a coordinated maneuver, such as the one illustrated in the previous slide, two of the Euler angular rates are zero and the rate of turn, dy/dt, is constant, representing the steady angular rate of the turning velocity vector. With these conditions, the kinematic equations simplify to the form shown providing an expression for each of the body angular rates in terms of q, f, dy/dt, all of which are constant during the motion. The command input also provides constant known values of the elements shown. Therefore, if we set 6 of the equations of motion to zero, as shown, we can solve for the 4 elements of the control vector, as well as for the pitch angle and the rate of turn. This is possible because the remaining angles, f and a, are known through the trigonometric relations.

Computation of Trim Optimal Parameters Throughout Flight Envelope Algorithm Search for Vmin Search for Vmax Form a vector, v(h) = Vmin : DV : Vmax Store v(h), Compute: ua*[g, m, b, v(h)] h h (m) In a coordinated maneuver, such as the one illustrated in the previous slide, two of the Euler angular rates are zero and the rate of turn, dy/dt, is constant, representing the steady angular rate of the turning velocity vector. With these conditions, the kinematic equations simplify to the form shown providing an expression for each of the body angular rates in terms of q, f, dy/dt, all of which are constant during the motion. The command input also provides constant known values of the elements shown. Therefore, if we set 6 of the equations of motion to zero, as shown, we can solve for the 4 elements of the control vector, as well as for the pitch angle and the rate of turn. This is possible because the remaining angles, f and a, are known through the trigonometric relations. V (m/s) Vmin(h) Vmax(h) g = 0, b = 0, m = -50

Example: Business Jet Aircraft Trim Surfaces g = 3, b = 5, m = -10 dR (deg) dA (deg) h (m) V (m/s) h (m) V (m/s) dT (%) dS (deg) In a coordinated maneuver, such as the one illustrated in the previous slide, two of the Euler angular rates are zero and the rate of turn, dy/dt, is constant, representing the steady angular rate of the turning velocity vector. With these conditions, the kinematic equations simplify to the form shown providing an expression for each of the body angular rates in terms of q, f, dy/dt, all of which are constant during the motion. The command input also provides constant known values of the elements shown. Therefore, if we set 6 of the equations of motion to zero, as shown, we can solve for the 4 elements of the control vector, as well as for the pitch angle and the rate of turn. This is possible because the remaining angles, f and a, are known through the trigonometric relations. V (m/s) h (m) V (m/s) h (m)

Summary and Conclusions Non-symmetrical trim study: Full, nonlinear, coupled aircraft dynamics Trim solution at one operating point Multidimensional flight envelope computation Multidimensional envelope and trim data storage Results: trim control surfaces