1. 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

2. 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.

3. 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

4. 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

5. 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

6. 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.

7. 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:

8. 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:

9. 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

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

11. 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 ]

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

13. 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”]

14. 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]

15. 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:

16. 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)

17. 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)

18. 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)

19. 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.

20. 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.

21. 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

22. 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)

23. 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