# Patent

## Title
Heading reference command and control algorithm systems and methods for aircraft turn-to-target maneuvers

## Abstract

Systems and methods are provided for determining a final heading of a turning vehicle, such as a rotorcraft. The system may include an algorithm that calculates an advance prediction of a final heading that will be achieved after control input is terminated.

## Background
This invention was made with Government support under contract number DAAH 10-00-C-0052 awarded by the United States Army. The Government has certain rights in this invention.


BACKGROUND OF THE INVENTION
1. Field of the Invention
The invention is related to vehicle control, and more specifically, to systems and methods for precisely and quickly pointing a vehicle in a desired direction.
2. Description of the Related Art
Turning a machine, such as a vehicle, or a portion thereof, quickly to point in a desired direction can be difficult. Turning precisely and quickly can be a challenge for flying and/or hovering vehicles, such as, for example, a helicopter or other rotorcraft (e.g., a tilt-rotor aircraft, a vertical/short takeoff and landing (VSTOL) aircraft), whether it is piloted or an unmanned aerial vehicle (UAV). Typically, internal and external effects, such as moments of inertia, angular momentum, structural limits, aerodynamic loads, and control surface actuator authority and rate limits must be dealt with.
For military aircraft, it may be especially important to have a capability to point a vehicle quickly in a desired direction, for example, to acquire, track, and fire upon an enemy target in advance of the enemy firing upon the aircraft. In this regard, the United States Army has established a performance specification entitled “AERONAUTICAL DESIGN STANDARD (ADS-33E-PRF) HANDLING QUALITIES REQUIREMENTS FOR MILITARY ROTORCRAFT,” that includes the following requirement for flight in Good Visual Environment (GVE) conditions:
“3.11.17 Turn to Target (Handling Qualities Requirement)
From a stable hover at an altitude of less than 20 ft complete a 180 deg turn. Turns must be completed in both directions. Final rotorcraft heading must be achieved within 5 seconds of initiating the turn within +/−3 degrees of the target.”
This task may be difficult with state-of-the-art technology because it is typically difficult for a pilot to time his or her directional control inputs precisely enough to get the aircraft to stop at the intended terminal heading without significant heading overshoot and time consuming heading backup toward the target heading. The heading overshoot and resulting heading backup problems are a result of aircraft yaw acceleration and jerk (time derivative of acceleration) limitations that prevent the pilot from instantaneously arresting the yaw rate developed during the turn.
In single rotor helicopters, these yaw acceleration and jerk limitations are imposed by the combined effect of tail rotor gearbox torque limits, tailrotor collective pitch actuator authority and rate limits, and limitations on allowable yaw acceleration due to inertial structural load limitations on attachment points for external stores.
Tandem rotor helicopters and tiltrotor aircraft face similar limits on allowable yaw acceleration and jerk due to rotor flapping induced blade and hub loads and differential cyclic pitch actuator authority and rate limits. Thus, the rapid turn to target task is difficult in virtually all vehicles with Vertical Takeoff Or Landing (VTOL) capabilities.
Meeting this requirement with existing rotorcraft control systems and configurations would rely heavily on pilot skill and training. For example, a pilot would need to anticipate, while turning a rotorcraft, the heading (azimuth) that the rotorcraft would achieve at the end of the turn once input force is released from the controls (e.g., rudder pedals), largely relying on experience and “feel” of the rotorcraft's handling characteristics, as well as mentally factoring in environmental conditions such as wind, altitude, etc. This results in an undesirably high pilot workload, and meeting the requirement using existing rotorcraft control systems and configurations could be impossible, or very difficult. In addition, there is a possibility of a pilot inadvertently exceeding the mechanical limits of a rotorcraft tail rotor gearbox and/or airframe while attempting to turn to a target quickly.
One possible solution is to build a rotorcraft with a large tail rotor, such as was proposed to be included in the Comanche helicopter program. However, such a solution may require an excessive cost, as well as design tradeoffs, such as increased weight and reduced agility.
This disclosure is directed toward overcoming one or more problems or disadvantages associated with the prior art.
SUMMARY OF THE INVENTION
According to one aspect of the invention, a computerized system predicts a final attitude of a maneuvering aircraft. The system includes a computer processor and an algorithm programmed into the computer processor. The algorithm is adapted to calculate, during a maneuver, a predicted final aircraft attitude based on aircraft parameters. The algorithm may include calculations based upon model following control laws.
According to another aspect of the invention, the algorithm may further include a plant canceller element. The final attitude may include a final heading and/or a final pitch attitude.
In accordance with a further aspect of the invention, the model following control laws may be tailored to maintain aircraft angular acceleration within specified limits, such as, for example, limits of angular acceleration and/or angular jerk about an aircraft yaw axis.
According to other aspects of the invention, the predicted final aircraft attitude is calculated using one of an exact solution algorithm, an iterative solution algorithm, and an approximate solution algorithm.
In accordance with a still further aspect of the invention, a method of predicting a final attitude of a maneuvering aircraft is provided. The method includes providing a computer processor, programming an algorithm into the computer processor to calculate a predicted final aircraft attitude based on aircraft parameters, and calculating, during a maneuver, the predicted final aircraft attitude.
The features, functions, and advantages can be achieved independently in various embodiments of the present invention or may be combined in yet other embodiments.



BRIEF DESCRIPTION OF THE DRAWINGS
FIG. 1 is diagrammatic plan view of a rotorcraft that may incorporate the invention, shown in three phases of an exemplary turning maneuver;
FIG. 2 is a block diagram illustrating a system and method of turning an aircraft according to one aspect of the invention;
FIG. 3 is a block diagram illustrating a generic architecture that may be used for explicit model following control laws;
FIG. 4 is a block diagram illustrating the architecture of FIG. 3 in further detail;
FIG. 5 is a block diagram of a system and method that includes a yaw axis explicit model following control law architecture as modified to perform yaw acceleration and jerk limiting;
FIG. 6 is a plot that shows a time history of a final heading prediction cue curve overlaying a time history of commanded heading curve for a 180 degree turn of an aircraft implementing the invention;
FIG. 7 is an illustration of an exemplary cockpit display that includes a final heading prediction icon according to the invention;
FIG. 8 is a plot that shows piloted simulation predictions for a maneuver in which data are presented for the maneuver conducted with and without using the invention;
FIG. 9 is bar graph that shows an average agility factor for three differently configured aircraft.



DETAILED DESCRIPTION
FIG. 1 illustrates the operation of a Heading Reference Command and Control Algorithm according to one aspect of the invention. With reference to FIG. 1, a pilot 30 is faced with an example of a maneuver in which he or she has to execute a maximum rate hover turn of approximately 90 degrees to point a helicopter 32 at an intended target 34, for example, a tank, within weapons firing constraints. The maneuver may be broken down into three phases in FIG. 1: a first phase, generally indicated at 36 in which the pilot 30 turns the helicopter 32 at a maximum yaw rate toward the target 34 in order to place a cursor 38 displayed along with the target 34 on a visual display device such as a helmet-mounted monocle 40; a second phase, generally indicated at 42, in which the pilot 30 relaxes directional control input force when the cursor 38 is displayed on the monocle 40 over the target 34; and a third phase, generally indicated at 44, in which the helicopter 32 automatically aligns with a heading pointing toward the target 34, within firing constraints.
A Heading Reference Command & Control Algorithm, described in further detail below, makes the rapid turn to target task illustrated in FIG. 1 possible with a reasonable level of pilot workload by providing a “Heading Reference Cue” in the pilot's monocle 40. The monocle 40 may be an Integrated Helmet and Display Sighting System (IHADSS) monocle. The IHADSS monocle-may be part of a synthetic vision system, and may project a synthetic image of the outside world, such as, for example, an image generated by a Forward Looking Infrared (FLIR) sensor (not shown). The FLIR image may be a virtual image of the outside world that is enhanced by the FLIR sensor to provide improved vision at night or in other degraded visual environments.
The field of view of the IHADSS monocle may be slaved, or head-tracked, to the pilot's head, thus the pilot 30 may simply turn his or her head to view objects to the left and right of the helicopter 32 through the enhanced vision capabilities of the FLIR sensor. Flight display symbology may be superimposed over the FLIR image in the IHADSS monocle, allowing the pilot 30 to monitor critical flight display parameters while his or her vision is directed outside the cockpit.
The cursor 38 may be a “Heading Reference Cue,” that indicates the closest possible heading at which the helicopter 32 can be stopped with no heading overshoot. The “Heading Reference Cue” may be fixed with respect to the virtual image of the outside world depicted by the FLIR imagery in the IHADSS monocle.
When the pilot 30 uses the “Heading Reference Cue,” he or she may apply full directional control input and easily command maximum yaw rate during the turn to minimize a time line to a firing solution. As shown in FIG. 1, as the helicopter 32 turns toward the target 34, the “Heading Reference Cue” may move toward the target 34 in the pilot's IHADSS display. When the “Heading Reference Cue” is superimposed over the image of the target 34, the pilot 30 may simply relax directional control input force to command the helicopter 32 to stop at the target heading with no heading overshoot. High fidelity simulations conducted under the Helicopter Active Control Technology (HACT) program predict that the Heading Reference Command & Control Algorithm will decrease time lines to target engagement in the scenario depicted by FIG. 1 by as much as 60% in critical nighttime operational scenarios or in other degraded visual environments.
While the preceding discussion focuses on providing a reference cue for a directional axis targeting task, the concept of providing an attitude reference cue and integrated control laws to lower pilot workload and shorten timelines required to achieve the target attitude may also be applied to the other axes of control. For example, there are mission requirements for pitch axis pointing tasks and limitations on aircraft pitch axis response capabilities that are analogous to those discussed in association with the directional axis task depicted in FIG. 1.
It is straightforward to extend the functionality of the cue depicted in FIG. 1 to serve as an attitude reference for both yaw and pitch axis pointing tasks. For example, the cue in FIG. 1 may move horizontally and vertically with respect to the virtual FLIR image to cue the pilot to both directional and longitudinal axis pointing requirements. When the heading/pitch attitude reference cue is superimposed over the intended target in the FLIR image, the pilot could simultaneously relax input forces in both the longitudinal and directional axis controls to bring the aircraft to the required final heading and pitch attitudes with no overshoot and in the minimum amount of time.
It should be noted that the symbology for the Heading Reference Cue could take many different forms than the symbol shown in FIG. 1. The specific symbol used to depict the Heading Reference Cue in FIG. 1 is presented solely for the purpose of illustrating the concept through a single simplified example. The specific size, shape, color, and geometry of the cue would vary depending upon the specific application. Similarly, multiple sources besides the line of sight from the pilot's eye to the target depicted in FIG. 1 could be used to drive the Heading Reference Cue. For example, lines of sight from Target Acquisition and Designation System (TADS) sensors, which may be located at various points on the aircraft, may be used for targeting tasks.
It should also be noted that the Heading Reference Command and Control Algorithm is equally applicable to aircraft equipped with conventional center stick and pedal cockpit controls or aircraft equipped with sidestick controllers. In an aircraft equipped with conventional directional pedals, the pilot may relax directional control input force by relaxing his leg muscles and allowing the directional pedals to center. In an aircraft equipped with a twist grip sidestick controller to control the yaw axis, the pilot may relax directional control input force by relaxing his wrist muscles and allowing the torsional axis of the sidestick controller to center. The pilot may relax directional control input force at an appropriate time indicated by the cue to command the aircraft to stop at the target heading with equivalent proficiency whether the aircraft is equipped with directional pedals or sidestick controllers.
FIG. 2 is a block diagram illustrating one example of system level operation of the Heading Reference Command & Control Algorithm. An overall system 46 may include an Aircraft Response Element 48, a Model Following Control Laws Module 50, Final Heading Prediction Algorithm Module 52, a Final Heading Cue Display Module 54, and/or a Pilot response element 56. Referring to FIG. 2, an input to the overall system may be a “Desired Final Heading” and an output of the overall system may be an actual aircraft heading. The Heading Reference Command & Control Algorithm effectively minimizes the error between the actual aircraft heading and the desired final heading by providing the pilot with a display cue that allows the pilot to close the loop on the final heading with a low workload and minimal training.
The Aircraft Response Element 48 represents the basic heading and yaw rate response of the aircraft to a directional actuator command input (δRACT). For most VTOL aircraft, the basic heading response of the aircraft to actuator commands is too sluggish to provide precise pointing capabilities with a reasonable level of pilot workload. The Model Following Control Laws Module 50 may shape the pilot's directional-control inputs to provide crisp heading and yaw rate response characteristics that lower pilot workload required to control the directional axis. The Model Following Control Laws Module 50 may also provide feedback of sensed aircraft heading and yaw rate for disturbance rejection and stability enhancement. The Model Following Control Laws Module 50 may replace the sluggish inherent dynamics of the aircraft, defined approximately by the bare airframe yaw rate damping (NR) and yaw acceleration control sensitivity (NδR) derivatives, with the desired crisp and predictable yaw axis handling qualities of a first order system with the desired bandwidth (K1) and maximum achievable yaw rate specified by military handling qualities specifications such as Aeronautical Design Standard (ADS) 33.
The Heading Reference Command & Control Algorithm system 46 advances the state of the art by introducing aircraft structural limits on yaw acceleration and jerk into the model following control laws. By limiting actual aircraft yaw acceleration and jerk to the values used as inputs to the Final Heading Prediction Algorithm Module 52, the Model Following Control Laws Module 50 helps to ensure accuracy of the Final Heading Prediction Algorithm Module 52 by rejecting errors between actual and assumed maximum yaw acceleration and jerk capabilities, for example, due to unmodelled aircraft dynamics or atmospheric turbulence. In addition to improving the accuracy of the Final Heading Prediction Algorithm Module 52, the Model Following Control Laws Module 50 may also actively control the helicopter 32 to ensure that the desired final heading is attained with no residual heading error or overshoot when the pilot 30 reacts to the display cue 38 converging with the target 34 by relaxing directional control input force.
While the block diagram shown in FIG. 2 is drawn for the yaw axis, a similar block diagram could be drawn for the pitch axis. The input to the pitch axis block diagram analogous to FIG. 2 would be “Desired Final Pitch Attitude” and its output would be “Aircraft Pitch Attitude.” Referring to FIG. 2, the “Final Heading Prediction Algorithm” Module 52 may be replaced by a “Final Pitch Attitude Prediction Algorithm” Module, and the “Final Heading Cue Display” Module 54 may be replaced by a “Final Pitch Attitude Cue Display” Module.
FIG. 3 shows a generic architecture that maybe used for explicit model following control laws. The generic model following control law architecture provides an aircraft response to pilot commands that is equal to a desired response command model “D(s)” 58. A Plant Canceller Module 60 cancels the inherent aircraft dynamics, denoted by P(s) in FIG. 3, while substituting the desired response of the command model, denoted by D(s) in FIG. 3. Any errors between the desired response command model and sensed aircraft response data may be passed through a Stability Compensation Module 62 and fedback through a Limited Authority Port 64 to actively mitigate model following errors. A Sensor Module 66 may be used to provide the sensed aircraft response data.
Errors between the command model response and actual sensed aircraft response, referred to as model following errors, may be caused by external disturbances such as wind gusts as well as inexact cancellation of the aircraft dynamics by the Plant Canceller Module 60. Since the explicit model following control law architecture minimizes errors between sensed and commanded aircraft response, the output of the Stability Compensation Module 62 may be limited to a relatively small magnitude by the limited authority port without affecting performance of the system. Limiting the port authority provides robustness to sensor failures by allowing the pilot enough authority to easily override any erroneous stability compensation inputs caused by sensor failures. The aircraft module 68 represents the system being controlled by the model following control laws, but of course the system being controlled could be another type of machine, such as a robot arm or a crane, that may need to be precisely pointed in a desired direction.
FIG. 4 shows how the generic explicit model following control law architecture may be applied to the yaw axis of the aircraft.

    D(s)=Transfer Function of Desired Response Command Model
    H(s)=Transfer Function of Control Law Stability Compensation Element
    Kr=Yaw Rate Error Feedback Gain (in/(rad/sec))
    KΨ=Heading Error Feedback Gain (in/rad)
    NR=Yaw rate damping derivative (rad/sec)
    NδR=Yaw rate control response derivative (rad/sec2/in)
    P(s)=Transfer Function of Unaugmented Aircraft Response
    s=Laplace Variable (rad/sec)
    δR=Directional Control Input (%)
    {dot over (Ψ)}MAX=Maximum Yaw Rate (Units=Radians/sec)


The input may be a pilot directional control input in units of percent and the output may be aircraft heading. A small deadzone function 70, on the order of +/−5% of full control authority, may be applied to the pilot directional control input so that the pilot can easily command zero yaw rate by relaxing directional controller input force even if the controller does not return to an exactly centered position. A nonlinear shaping function 72 may be applied to the output of the deadzone so that control sensitivity is lower for small control inputs and higher for larger control inputs. The nonlinear shaping provides precise control for small heading changes while allowing the pilot to exploit full yaw rate capability for large control deflections. The output of the nonlinear shaping may be divided by a factor of 100% and multiplied by the maximum allowable yaw rate to scale the yaw rate command so that a full authority directional control input commands the maximum allowable yaw rate.
The Plant Canceller Module 60 may be designed based on the assumption that the aircraft directional axis dynamics can be represented by a first order response characterized by yaw rate damping and directional control response sensitivity derivatives. The first order response assumed for the aircraft dynamics is an approximation. For example, additional unmodelled dynamics may be present in the aircraft response and the yaw rate damping and directional control response sensitivity derivatives may differ from the assumed values as gross weight, center of gravity, and atmospheric conditions vary.
The explicit model following control law architecture is robust to reasonable discrepancies in assumed and actual aircraft dynamics, therefore adequate plant cancellation is achieved as long as the assumed aircraft dynamics roughly match the actual aircraft dynamics. In general, adequate model following performance is obtained as long as the yaw rate damping and directional control response sensitivity derivatives assumed in the aircraft model are within +/−50% of the values that describe an equivalent first order response model of the actual aircraft dynamics.
As shown in FIG. 4, the Plant Canceller Module 60 may be a function of desired yaw axis control response bandwidth, bare airframe yaw damping, and bare airframe directional control response sensitivity. The yaw rate desired response command model 58 may be formed by cascading a first order lag filter with the desired yaw axis control response bandwidth (K1) with models of actuator dynamics and system digital and transport delays. Commanded yaw rate may be integrated to calculate the commanded heading angle. Yaw rate error may be calculated by subtracting sensed yaw rate from commanded yaw rate. Heading error may be calculated by subtracting sensed heading angle from commanded heading angle. Stability compensation may be calculated by multiplying the yaw rate and heading errors by appropriate feedback gains and structural filters. The stability compensation command may be limited by the port before being fedback to the directional control actuator to provide robustness to sensor failures.
The generic explicit model following control law architecture may also be applied to the pitch axis of the aircraft using the same or similar architecture shown in FIG. 4. Pitch rate damping, pitch control response sensitivity, pitch axis desired bandwidth, maximum pitch rate, pitch attitude feedback gain, and pitch rate feedback gain would replace their yaw axis counterparts in the explicit model following control law architecture block diagram shown in FIG. 4.
FIG. 5 is a block diagram of a system and method that includes a yaw axis explicit model following control law architecture as modified to perform yaw acceleration and jerk limiting. The block diagram of FIG. 5 differs from that of FIG. 4 in that the dynamics of the Plant Canceller Module 60 and desired response command model 58 are combined into an integrated yaw rate command model 74. It is possible to combine the Plant Canceller Module 60 and the desired response command model 58 because the denominators of the transfer functions of both elements contain the same poles. The integrated yaw rate command model calculates commanded yaw rate as the output of an integrator, referred to as the yaw rate command integrator. Authority limiting the input to the yaw rate command integrator limits commanded yaw acceleration, while rate limiting the input to the yaw rate command integrator limits commanded yaw jerk. The explicit model following control law architecture shown in FIG. 5 provides equivalent model following performance as the architecture shown in FIG. 4, while adding the additional capability to limit yaw acceleration and jerk precisely within explicitly specified values.
A pitch axis explicit model following control law architecture that performs acceleration and jerk limiting may also be implemented using architecture analogous to that shown in FIG. 5 with the appropriate substitution of pitch axis parameters for the corresponding yaw axis parameters in FIG. 5.
FIG. 6 shows a time history of the final heading prediction cue curve 82 (solid curve) overlaying a time history of commanded heading curve 84 (dashed curve) for a case where the aircraft performs a 180 degree right turn followed by a 180 degree left turn. The final heading prediction cue calculates the final heading associated with the release of directional control input force. Referring to FIG. 6, at time equal to 76 seconds the time rate of change of heading angle is virtually zero, thus the aircraft can obviously be stopped at the currently commanded heading. Therefore, at time equal to 76 seconds, the final heading prediction is identical to the currently commanded heading angle. Between time equal to 76 and 78 seconds, the time rate of change of heading increases to a large value as the pilot conducts the right turn.
As the time rate of change of heading increases, the final heading prediction cue curve 82 diverges from the commanded heading curve 84, indicating that the final heading that will result from releasing directional control input force will be slewed significantly to the right of the currently commanded heading angle. For example, as illustrated in FIG. 6, at time roughly equal to 78 seconds, the commanded heading is roughly 250 degrees while the predicted final stopping heading is roughly 300 degrees. This means that if the pilot were to release directional control input force at time equal to 78 seconds that the aircraft would continue-to yaw an additional 50 degrees before it comes to a stop.
As the heading rate slows between time equal to 80 and 82 seconds, the final heading prediction cue curve 82 once again converges to the commanded heading curve 84. The final heading prediction cue diverges again from the commanded heading curve 84 as the aircraft initiates the left turn between time equal to 82 and 84 seconds, indicating that the final stopping heading will be slewed significantly to the left of the currently commanded heading angle when the leftward time rate of change of heading angle is large.
FIG. 7 shows a final heading prediction cue display 86 that can be used as an alternative to the one illustrated in FIG. 1 in aircraft that are not equipped with the synthetic visionics capability and/or head-up flight display capability of the Integrated Helmet and Display Sighting System (IHADSS). The final heading prediction cue display shown in FIG. 7 is a low cost option that indicates the predicted final stopping heading by a diamond shaped symbol 88 displayed on a heading tape 90. The current heading angle is the heading shown in the center of the heading tape 90. The current heading angle may also be boxed to make it easier for the pilot to recognize the current heading angle. FIG. 7 depicts a high yaw rate right hover turn, wherein the current heading angle is 360 degrees and the predicted final stopping heading is 70 degrees.
The simplified heading tape based final heading prediction cue display shown in FIG. 7 can be used in aircraft equipped with night vision goggles as well as aircraft equipped only with conventional head-down flight displays. High fidelity piloted simulations using the final heading prediction cue display of FIG. 7 indicate that this simplified display provides significant benefit for demanding turn to target maneuvers such as those defined in Aeronautical Design Standard 33 (ADS-33).
The specific size, shape, color, geometry, and location on the flight display symbology of the cue may vary from that shown in FIG. 7 depending upon the specific application. The specific symbol used to depict the Heading Reference Cue in FIG. 7 is presented solely for the purpose of illustrating through a single simplified example how the concept can be applied in aircraft without synthetic visionics capabilities.
The U.S. Army developed the ADS-33E Turn to Target Mission Task Element (MTE) to measure directional axis handling qualities, maneuverability, and agility. The Turn to Target Mission Task Element (MTE) defines quantitative adequate and desired performance criteria for heading control and position maintenance for an aggressive 180 degree heading turn. For Level 1 handling qualities in Attack helicopters, ADS-33E requires that the 180 degree Turn to Target MTE be completed in less than 5.0 seconds in daytime conditions and less than 10 seconds in nighttime conditions where the degraded visual environment makes the aircraft harder to fly. Given the current emphasis on nighttime tactics, it would obviously be desirable to provide equivalent agility in the Turn to Target MTE in both day and night visual conditions.
FIG. 8 shows piloted simulation predictions for the Turn to Target MTE conducted with an Apache helicopter in nighttime conditions. Data are presented for the Turn to Target maneuver conducted with and without the Final Heading Prediction cue generated by the Heading Reference Command and Control Algorithm. As shown in FIG. 8, it takes 7.6 seconds for the pilot to complete the Turn to Target maneuver without the Final Heading Prediction cue, as indicated by a solid curve 92. When the Final Heading Prediction cue is provided, the pilot is able to reduce the time it takes to perform the maneuver to 4.1 seconds, as indicated by a dotted curve 94. Further, the Final Heading Prediction cue allows the maneuver to be conducted with no heading overshoot, providing an absolutely deadbeat and predictable yaw rate response during the deceleration to the target heading.
The time it takes to complete the Turn to Target maneuver is a measure of maneuverability and agility. Handling qualities rating is another distinct metric that measures pilot workload required to achieve a given level of maneuverability and agility. In other words, for a given helicopter and flight control system, it may be possible for a pilot with exceptional skill to perform the Turn to Target MTE in 5 seconds, whereas a pilot with average skill may not be able to perform the maneuver in 5 seconds. Also, a pilot may be able to improve performance by devoting more attention, or increasing workload, to perform the task within the desired tolerance and allotted time. The Cooper Harper Point Rating (CHPR) system provides quantitative handling qualities ratings that measure the ability of service pilots to perform Mission Task Elements (MTEs) such as the Turn to Target maneuver in operational conditions. On the Cooper-Harper scale, lower numerical ratings indicate better handling qualities, with a CHPR of 1 indicating “best” handling qualities and a CHPR of 10 indicating worst handling qualities.
Piloted simulation predicts that handling qualities ratings for the Turn to Target MTE improve from CHPR 7 for current fleet Apaches to CHPR 3 when the current fleet Apaches are equipped with a modified flight control system that includes the Heading Reference Command and Control Algorithm. A Cooper-Harper handling qualities rating of 7 indicates that the pilot could not perform the task to even adequate performance standards with the current Apache flight control system even with a high level of pilot workload. A Cooper-Harper handling qualities rating of 3 indicates that the modified flight control system allows the pilot to perform the task to desired performance standards with a low level of pilot workload. Thus the Heading Reference Command and Control Algorithm significantly improves handling qualities while it increases usable maneuverability and agility.
FIG. 9 is a bar graph that quantifies the progress toward achievement of the U.S. Department of Defense's (DoD's) Rotary Wing Vehicle Technology Development Approach (RWV TDA) Subarea Goals for maneuverability and agility resulting from using flight control systems that incorporate the Heading Reference Command and Control Algorithm according to one aspect of the invention. The data shown in FIG. 9 quantify improvements in Usable Agility/Maneuverability that were measured within the context of the U.S. Army HACT Science and Technology program. The metric for this Subarea Goal is average usable agility factor in aggressive maneuvers. Agility factor is defined as the ratio of usable performance to inherent performance that is attainable by service pilots in combat conditions. Inherent performance is defined as the aircraft performance limit with “ideal” pilot compensation. Usable agility factor is averaged over the Turn to Target maneuver as well as Bob-up/Bob-down, Pirouette; Slalom, and Pull-up/Pushover maneuvers performed in both day and Degraded Visual Environment (DVE) conditions. For the Turn to Target maneuver, the usable agility metric is the ratio of the time required to complete the Turn to Target maneuver in an AH-64 Apache helicopter with “ideal” pilot compensation, calculated as 4.1 seconds by the DoD, to the time it takes an actual pilot to complete the maneuver while maintaining at least Level 2 handling qualities. Note that Level 2 handling qualities correspond to Cooper-Harper handling qualities ratings of 4 through 6.
The current fleet Apache AH-64A -flight control system allows service pilots to exploit only 30% of the agility inherent in the Apache airframe as indicated by a left-hand bar 96 in FIG. 9. Implementation of a Partial Authority HACT Flight Control System (PAHFCS) that includes the Heading Reference Command & Control Algorithm into a force modernized Apache helicopter is predicted to increase average usable agility factor by. 66%, allowing service pilots to exploit 50% of the agility inherent in the Apache airframe as indicated by a center bar 98 in FIG. 9. Note that the PAHFCS utilizes the mechanical flight control system implemented in current fleet Apache helicopters. Current fleet Apache helicopters augment the pilot's direct mechanical control inputs with inputs from limited authority stability augmentation actuators whose control authority is limited to only roughly +/−10% of the pilot's total control authority. Thus the current fleet Apache flight control actuator hardware is referred to as a “Partial Authority” flight control augmentation system because the actuators used for stability and control augmentation have only “part” of the authority available to the pilot. An Apache upgrade with the PAHFCS is a relatively low cost option for increasing usable agility in the Apache, requiring extensive flight control computer software changes but minimal flight control hardware changes. A partial authority HFCS upgrade including the Heading Reference Command & Control Algorithm is predicted to achieve the 2005 RWV TDA Maneuverability/Agility Subarea Goal.
A more capable but mote expensive design option would be to replace the mechanical flight control system currently used in the Apache with a Fly-By-Wire flight control system that allows the HFCS to command full actuator authority. The full authority HFCS upgrade would require extensive flight control hardware changes and extensive flight control computer processor and software changes. As shown by the right bar 100 in FIG. 9, the full authority HFCS implementation would allow pilots to exploit roughly 65% of the agility inherent in the Apache airframe, achieving the year 2010 RWV TDA Agility/Maneuverability Subarea Goal.
The DoD defines connectivity between the “Increase in Usable Agility/Maneuverability Subarea Goal” and System Level Payoffs in the RWV TDA. For example, attaining the 2005 (Phase 2) Agility/Maneuverability Subarea Goal is predicted to increase mission capability by 65%, reduce major accident rate by 10%, increase probability of survival by 4.5%, and increase mission reliability by 20% in comparison to current fleet Apache helicopters. Thus the Heading Reference Command & Control Algorithm is predicted to have significant benefits in expanding mission capability, increasing flight safety, improving survivability, and increasing mission reliability.
In accordance with one aspect of the invention, a final heading prediction algorithm may be derived and based upon the solution to the continuous time domain control problem. A general derivation of a heading reference command and control algorithm is set forth below.
General Derivation of Heading Reference Command & Control Algorithm
There exists a state: X.
Its derivatives exist: {dot over (X)}, {umlaut over (X)}, , . . .
Limits on the derivatives are defined: {dot over (X)}MAX, {umlaut over (X)}MAX, and MAX such that |{dot over (X)}|≦{dot over (X)}MAX etc.
The basic relationships are defined as exponential: {dot over (X)}=−K X
Thus: {umlaut over (X)}=−K {dot over (X)}, =−K {umlaut over (X)}, {umlaut over (X)}=K2 X, =−K3 X etc.
Given an arbitrary initial condition, an optimal “at rest” condition may be achieved as follows:
Initial condition defined as: {dot over (X)}={dot over (X)}0, {umlaut over (X)}={umlaut over (X)}0, =undefined
“At rest” is defined as all derivative states=0.
There are two basic conditions that may exist.

    
    
        1. The acceleration (deceleration) profile must be “captured”.
        2. The acceleration profile is being tracked.
        
            a. Constant (maximum) deceleration is active.
            b. Exponential curve is being tracked.
        
        
    
    


Starting from an arbitrary initial condition the capture may be to either the exponential or the constant (maximum) acceleration portions of the profile.
To the Exponential:



  
    
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          )
        
      
    
  




  
    
      
        t
        
          1
          exp
        
        2
      
      -
      
        
          
            2
            ⁢
            
              (
              
                
                  K
                  ⁡
                  
                    (
                    
                      
                        X
                         ̈
                      
                      0
                    
                    )
                  
                
                -
                
                  
                    
                      X
                      ⃛
                    
                    MAX
                  
                  ⁢
                  
                    
                  
                  ⁢
                  sign
                  ⁢
                  
                    
                  
                  ⁢
                  
                    (
                    
                      
                        X
                        .
                      
                      0
                    
                    )
                  
                
              
              )
            
          
          
            K
            ⁡
            
              (
              
                
                  
                    X
                    ⃛
                  
                  MAX
                
                ⁢
                
                  
                
                ⁢
                sign
                ⁢
                
                  
                
                ⁢
                
                  (
                  
                    
                      X
                      .
                    
                    0
                  
                  )
                
              
              )
            
          
        
        ⁢
        
          t
          
            1
            exp
          
        
      
      -
      
        
          2
          ⁢
          
            (
            
              
                K
                ⁡
                
                  (
                  
                    
                      X
                      .
                    
                    0
                  
                  )
                
              
              +
              
                
                  X
                   ̈
                
                0
              
            
            )
          
        
        
          K
          ⁡
          
            (
            
              
                
                  X
                  ⃛
                
                MAX
              
              ⁢
              
                
              
              ⁢
              sign
              ⁢
              
                
              
              ⁢
              
                (
                
                  
                    X
                    .
                  
                  0
                
                )
              
            
            )
          
        
      
    
    =
    0
  




  
    
      t
      
        1
        exp
      
    
    =
    
      
        
          
            K
            ⁡
            
              (
              
                
                  X
                   ̈
                
                0
              
              )
            
          
          -
          
            
              
                X
                ⃛
              
              MAX
            
            ⁢
            
              
            
            ⁢
            sign
            ⁢
            
              
            
            ⁢
            
              (
              
                
                  X
                  .
                
                0
              
              )
            
          
        
        
          K
          ⁡
          
            (
            
              
                
                  X
                  ⃛
                
                MAX
              
              ⁢
              
                
              
              ⁢
              sign
              ⁢
              
                
              
              ⁢
              
                (
                
                  
                    X
                    .
                  
                  0
                
                )
              
            
            )
          
        
      
      +
      
        
          
            
              
                
                  
                    (
                    
                      
                        
                          K
                          ⁡
                          
                            (
                            
                              
                                X
                                 ̈
                              
                              0
                            
                            )
                          
                        
                        -
                        
                          
                            
                              X
                              ⃛
                            
                            MAX
                          
                          ⁢
                          
                            
                          
                          ⁢
                          sign
                          ⁢
                          
                            
                          
                          ⁢
                          
                            (
                            
                              
                                X
                                .
                              
                              0
                            
                            )
                          
                        
                      
                      
                        K
                        ⁡
                        
                          (
                          
                            
                              
                                X
                                ⃛
                              
                              MAX
                            
                            ⁢
                            
                              
                            
                            ⁢
                            sign
                            ⁢
                            
                              
                            
                            ⁢
                            
                              (
                              
                                
                                  X
                                  .
                                
                                0
                              
                              )
                            
                          
                          )
                        
                      
                    
                    )
                  
                  2
                
                +
              
            
          
          
            
              
                
                  2
                  ⁢
                  
                    (
                    
                      
                        K
                        ⁡
                        
                          (
                          
                            
                              X
                              .
                            
                            0
                          
                          )
                        
                      
                      +
                      
                        
                          X
                           ̈
                        
                        0
                      
                    
                    )
                  
                
                
                  K
                  ⁡
                  
                    (
                    
                      
                        
                          X
                          ⃛
                        
                        MAX
                      
                      ⁢
                      
                        
                      
                      ⁢
                      sign
                      ⁢
                      
                        
                      
                      ⁢
                      
                        (
                        
                          
                            
                              X
                              .
                            
                            ⁢
                            
                              
                            
                          
                          0
                        
                        )
                      
                    
                    )
                  
                
              
            
          
        
      
    
  




  
    
      Δ
      ⁢
      
        
          X
          .
        
        
          1
          ⁢
          
            
          
          ⁢
          exp
        
      
    
    =
    
      
        
          
            X
             ̈
          
          0
        
        ⁢
        
          t
          
            1
            exp
          
        
      
      -
      
        
          
            X
            ⃛
          
          MAX
        
        ⁢
        
          
        
        ⁢
        sign
        ⁢
        
          
        
        ⁢
        
          (
          
            
              X
              .
            
            0
          
          )
        
        ⁢
        
          
            t
            
              1
              exp
            
            2
          
          2
        
      
    
  




  
    
      Δ
      ⁢
      
        
      
      ⁢
      
        X
        
          1
          ⁢
          
            
          
          ⁢
          exp
        
      
    
    =
    
      
        
          
            X
             ̈
          
          0
        
        ⁢
        
          t
          
            1
            exp
          
        
      
      +
      
        
          
            X
             ̈
          
          0
        
        ⁢
        
          
            t
            
              1
              exp
            
            2
          
          2
        
      
      -
      
        
          
            X
            ⃛
          
          MAX
        
        ⁢
        
          
        
        ⁢
        sign
        ⁢
        
          
        
        ⁢
        
          (
          
            
              X
              .
            
            0
          
          )
        
        ⁢
        
          
            t
            
              1
              exp
            
            3
          
          6
        
      
    
  



To the Constant Acceleration:




  
    
      t
      
        1
        MAX
      
    
    =
    
      
        
          
            -
            
              
                X
                 ̈
              
              MAX
            
          
          ⁢
          
            
          
          ⁢
          sign
          ⁢
          
            
          
          ⁢
          
            (
            
              
                X
                .
              
              0
            
            )
          
        
        -
        
          
            X
             ̈
          
          0
        
      
      
        
          -
          
            
              X
              ⃛
            
            MAX
          
        
        ⁢
        sign
        ⁢
        
          
        
        ⁢
        
          (
          
            
              X
              .
            
            0
          
          )
        
      
    
  




  
    
      Δ
      ⁢
      
        
          X
          .
        
        
          1
          MAX
        
      
    
    =
    
      
        
          
            X
             ̈
          
          0
        
        ⁢
        
          t
          
            1
            MAX
          
        
      
      -
      
        
          
            X
            ⃛
          
          MAX
        
        ⁢
        
          
        
        ⁢
        sign
        ⁢
        
          
        
        ⁢
        
          (
          
            
              X
              .
            
            0
          
          )
        
        ⁢
        
          
            t
            
              1
              MAX
            
            2
          
          2
        
      
    
  




  
    
      Δ
      ⁢
      
        
      
      ⁢
      
        X
        
          1
          MAX
        
      
    
    =
    
      
        
          
            X
            .
          
          0
        
        ⁢
        
          t
          
            1
            MAX
          
        
      
      +
      
        
          
            X
             ̈
          
          0
        
        ⁢
        
          
            t
            
              1
              MAX
            
            2
          
          2
        
      
      -
      
        
          
            X
            ⃛
          
          MAX
        
        ⁢
        
          
        
        ⁢
        sign
        ⁢
        
          
        
        ⁢
        
          (
          
            
              X
              .
            
            0
          
          )
        
        ⁢
        
          
            t
            
              1
              MAX
            
            3
          
          6
        
      
    
  



    If t1MAX≦t1exp then:





  
    
      Δ
      ⁢
      
        
      
      ⁢
      
        X
        1
      
    
    =
    
      Δ
      ⁢
      
        
      
      ⁢
      
        X
        
          1
          MAX
        
      
    
  




  
    
      t
      2
    
    =
    
      
        
          
            X
            .
          
          0
        
        +
        
          Δ
          ⁢
          
            
              X
              .
            
            
              1
              MAX
            
          
        
        -
        
          
            
              
                X
                 ̈
              
              MAX
            
            ⁢
            
              
            
            ⁢
            sign
            ⁢
            
              
            
            ⁢
            
              (
              
                
                  X
                  .
                
                0
              
              )
            
          
          K
        
      
      
        
          
            X
             ̈
          
          MAX
        
        ⁢
        
          
        
        ⁢
        sign
        ⁢
        
          
        
        ⁢
        
          (
          
            
              X
              .
            
            0
          
          )
        
      
    
  




  
    
      Δ
      ⁢
      
        
          X
          .
        
        2
      
    
    =
    
      
        -
        
          
            X
             ̈
          
          MAX
        
      
      ⁢
      
        
      
      ⁢
      sign
      ⁢
      
        
      
      ⁢
      
        (
        
          
            X
            .
          
          0
        
        )
      
      ⁢
      
        t
        2
      
    
  




  
    
      Δ
      ⁢
      
        
      
      ⁢
      
        X
        2
      
    
    =
    
      
        
          (
          
            
              
                X
                .
              
              0
            
            +
            
              Δ
              ⁢
              
                
                  X
                  .
                
                
                  1
                  MAX
                
              
            
          
          )
        
        ⁢
        
          t
          2
        
      
      -
      
        
          
            X
             ̈
          
          MAX
        
        ⁢
        
          
        
        ⁢
        sign
        ⁢
        
          
        
        ⁢
        
          (
          
            
              X
              .
            
            0
          
          )
        
        ⁢
        
          
            t
            2
            2
          
          2
        
      
    
  




  
    
      Δ
      ⁢
      
        
      
      ⁢
      
        X
        3
      
    
    =
    
      
        
          
            X
             ̈
          
          MAX
        
        ⁢
        
          
        
        ⁢
        sign
        ⁢
        
          
        
        ⁢
        
          (
          
            
              X
              .
            
            0
          
          )
        
      
      
        K
        2
      
    
  



    Else:

ΔX1=ΔX1exp

ΔX2=0







  
    Δ
    ⁢
    
      
    
    ⁢
    
      X
      3
    
  
  =
  
    
      
        
          X
          .
        
        0
      
      +
      
        Δ
        ⁢
        
          
            X
            .
          
          
            1
            exp
          
        
      
    
    K
  




The algorithm may be based on a set of equations that allow the “final heading reference” to be computed continuously. Calculations may be based on knowing the yaw bandwidth (K), yaw rate command {dot over (Ψ)}c “derived from the stick”, maximum yaw rate {dot over (Ψ)}MAX, maximum yaw acceleration {umlaut over (Ψ)}MAX, and maximum jerk rate {dot over ({umlaut over (Ψ)}MAX of the aircraft. Based on this aircraft information an algorithm was derived that predicts the heading response for a given yaw rate command. The algorithm is set forth below.



  
    IF
    ⁢
    
      
    
    ⁢
    
      (
      
        
          abs
          ⁡
          
            (
            
              
                Ψ
                .
              
              c
            
            )
          
        
        ≤
        
          
            
              Ψ
               ̈
            
            MAX
          
          
            K
            1
          
        
      
      )
    
    ⁢
    
      
    
    ⁢
    THEN
  




  
    
      
    
    ⁢
    
      
        t
        1
      
      =
      
        
          
            
              -
              
                K
                1
              
            
            *
            
              (
              
                
                  
                    Ψ
                    .
                  
                  C
                
                +
                
                  Δ
                  ⁢
                  
                    
                      Ψ
                      .
                    
                    1
                    ′
                  
                
              
              )
            
          
          -
          
            
              Ψ
               ̈
            
            C
          
        
        
          
            -
            
              
                Ψ
                ⃛
              
              MAX
            
          
          *
          
            SIGN
            (
            
              
                Ψ
                .
              
              C
            
            )
          
        
      
    
  




  
    
      
    
    ⁢
    
      
        Δ
        ⁢
        
          
            Ψ
            .
          
          1
          ′
        
      
      =
      
        
          
            
              Ψ
               ̈
            
            C
          
          *
          
            t
            1
          
        
        -
        
          
            
              Ψ
              ⃛
            
            MAX
          
          *
          
            SIGN
            (
            
              
                Ψ
                .
              
              C
            
            )
          
          *
          
            
              t
              1
              2
            
            2.0
          
        
      
    
  




  
    
      
    
    ⁢
    
      
        ΔΨ
        1
      
      =
      
        
          
            
              Ψ
              .
            
            C
          
          *
          
            t
            1
          
        
        +
        
          
            
              Ψ
               ̈
            
            C
          
          *
          
            
              t
              1
              2
            
            2.0
          
        
        -
        
          
            
              Ψ
              ⃛
            
            MAX
          
          *
          
            SIGN
            (
            
              
                Ψ
                .
              
              C
            
            )
          
          *
          
            
              t
              1
              3
            
            6.0
          
        
      
    
  




  
    
      
    
    ⁢
    
      ΔΨ
      =
      
        
          
            
              
                Ψ
                .
              
              C
            
            +
            
              Δ
              ⁢
              
                
                  Ψ
                  .
                
                1
                ′
              
            
          
          
            K
            1
          
        
        +
        
          ΔΨ
          1
        
      
    
  




  ELSE




  
    
      
    
    ⁢
    
      
        t
        1
      
      =
      
        
          
            
              -
              
                
                  Ψ
                   ̈
                
                MAX
              
            
            *
            
              SIGN
              (
              
                
                  Ψ
                  .
                
                C
              
              )
            
          
          -
          
            
              Ψ
               ̈
            
            C
          
        
        
          
            -
            
              
                Ψ
                ⃛
              
              MAX
            
          
          *
          
            SIGN
            (
            
              
                Ψ
                .
              
              C
            
            )
          
        
      
    
  




  
    
      
    
    ⁢
    
      
        Δ
        ⁢
        
          
            Ψ
            .
          
          1
        
      
      =
      
        
          
            
              Ψ
               ̈
            
            C
          
          *
          
            t
            1
          
        
        -
        
          
            
              Ψ
              ⃛
            
            MAX
          
          *
          
            SIGN
            (
            
              
                Ψ
                .
              
              C
            
            )
          
          *
          
            
              t
              1
              2
            
            2.0
          
        
      
    
  




  
    
      
    
    ⁢
    
      
        ΔΨ
        1
      
      =
      
        
          
            
              Ψ
              .
            
            C
          
          *
          
            t
            1
          
        
        +
        
          
            
              Ψ
               ̈
            
            C
          
          *
          
            
              t
              1
              2
            
            2.0
          
        
        -
        
          
            
              Ψ
              ⃛
            
            MAX
          
          *
          
            SIGN
            (
            
              
                Ψ
                .
              
              C
            
            )
          
          *
          
            
              t
              1
              3
            
            6.0
          
        
      
    
  




  
    
      
    
    ⁢
    
      
        t
        2
      
      =
      
        
          
            
              Ψ
              .
            
            C
          
          +
          
            Δ
            ⁢
            
              
                Ψ
                .
              
              1
            
          
          -
          
            
              
                
                  Ψ
                   ̈
                
                MAX
              
              *
              
                SIGN
                (
                
                  
                    Ψ
                    .
                  
                  C
                
                )
              
            
            
              K
              1
            
          
        
        
          
            
              Ψ
               ̈
            
            MAX
          
          *
          
            SIGN
            (
            
              
                Ψ
                .
              
              C
            
            )
          
        
      
    
  




  
    
      
    
    ⁢
    
      
        ΔΨ
        2
      
      =
      
        
          
            (
            
              
                
                  Ψ
                  .
                
                C
              
              +
              
                Δ
                ⁢
                
                  
                    Ψ
                    .
                  
                  1
                
              
            
            )
          
          *
          
            t
            2
          
        
        -
        
          
            
              Ψ
               ̈
            
            MAX
          
          *
          
            SIGN
            (
            
              
                Ψ
                .
              
              C
            
            )
          
          *
          
            
              t
              2
              2
            
            2.0
          
        
      
    
  




  
    
      
    
    ⁢
    
      
        ΔΨ
        3
      
      =
      
        
          
            
              Ψ
               ̈
            
            MAX
          
          *
          
            SIGN
            ⁡
            
              (
              
                
                  Ψ
                  .
                
                C
              
              )
            
          
        
        
          K
          1
          2
        
      
    
  




  
    
      
    
    ⁢
    
      ΔΨ
      =
      
        
          ΔΨ
          1
        
        +
        
          ΔΨ
          2
        
        +
        
          ΔΨ
          3
        
      
    
  




  ENDIF




  
    
      
    
    ⁢
    
      
        Ψ
        PREDICTION
      
      =
      
        
          Ψ
          C
        
        +
        ΔΨ
      
    
  



Where,

    Ψc=Yaw Command (Units =Radians)
    {dot over (Ψ)}c=Yaw Rate Command (Units =Radians/sec)
    {umlaut over (Ψ)}c=Commanded Yaw Acceleration (Units =Radians/sec2)
    {umlaut over (Ψ)}MAX=Yaw Acceleration Limit “Max Acceleration” (Units =Radians/sec2)
    {dot over ({umlaut over (Ψ)}MAX=Yaw Jerk Limit “Maximum Jerk” (Units =Radians/sec3)
    K =Yaw Desired Bandwidth (Units =Radians/sec)


Note the algorithm is broken in two parts, resulting in ΔΨ being calculated differently depending upon the state of




  
    (
    
      
        abs
        ⁡
        
          (
          
            
              Ψ
              .
            
            c
          
          )
        
      
      ≤
      
        
          
            Ψ
             ̈
          
          MAX
        
        K
      
    
    )
  
  .




Both PSI dot and double dot are signals generated within the Yaw Axis Command model. They are labeled as Yaw Rate Command and Commanded Yaw Acceleration respectively. Both the Yaw Rate Command and Commanded Yaw Acceleration result from the pilot's directional input in lieu of limits such as Max Yaw Rate, Desired Yaw Bandwidth, etc. Note the Yaw Command may be derived in the Attitude Command Model as a Time Rate of Change of Aircraft Heading Angle Command, alternatively referred to as an Inertial Yaw Rate Command coming from the Yaw Axis Command Model. Since helicopters often hover at non-zero pitch and roll angles, it is preferable to yaw about earth-fixed, or inertial, axes rather than yaw about the aircraft body axes for hover pointing tasks.
A variant of the foregoing algorithm has been found to perform well in a fully-digital rotorcraft control system, and is set forth below:



  
    IF
    ⁢
    
      
    
    ⁢
    
      (
      
        
          abs
          ⁡
          
            (
            
              
                Ψ
                .
              
              c
            
            )
          
        
        <=
        
          
            
              Ψ
               ̈
            
            MAX
          
          K
        
      
      )
    
    ⁢
    
      
    
    ⁢
    THEN
  




  
    
      
    
    ⁢
    
      
        t
        1
        ′
      
      =
      
        
          
            
              -
              K
            
            *
            
              
                Ψ
                .
              
              C
            
          
          -
          
            
              Ψ
               ̈
            
            C
          
        
        
          
            -
            
              
                Ψ
                ⃛
              
              MAX
            
          
          *
          
            SIGN
            (
            
              
                Ψ
                .
              
              C
            
            )
          
        
      
    
  




  
    
      
    
    ⁢
    
      
        Δ
        ⁢
        
          
            Ψ
            .
          
          1
          ′
        
      
      =
      
        
          
            
              Ψ
               ̈
            
            C
          
          *
          
            t
            1
            ′
          
        
        -
        
          
            
              Ψ
              ⃛
            
            MAX
          
          *
          
            SIGN
            (
            
              
                Ψ
                .
              
              C
            
            )
          
          *
          
            
              
                (
                
                  t
                  1
                  ′
                
                )
              
              2
            
            2.0
          
        
      
    
  




  
    
      
    
    ⁢
    
      
        t
        1
      
      =
      
        
          
            
              -
              K
            
            *
            
              (
              
                
                  
                    Ψ
                    .
                  
                  C
                
                +
                
                  Δ
                  ⁢
                  
                    
                      Ψ
                      .
                    
                    1
                    ′
                  
                
              
              )
            
          
          -
          
            
              Ψ
               ̈
            
            C
          
        
        
          
            -
            
              
                Ψ
                ⃛
              
              MAX
            
          
          *
          
            SIGN
            (
            
              
                Ψ
                .
              
              C
            
            )
          
        
      
    
  




  
    
      
    
    ⁢
    
      
        Δ
        ⁢
        
          
            Ψ
            .
          
          1
        
      
      =
      
        
          
            
              Ψ
               ̈
            
            C
          
          *
          
            t
            1
          
        
        -
        
          
            
              Ψ
              ⃛
            
            MAX
          
          *
          
            SIGN
            (
            
              
                Ψ
                .
              
              C
            
            )
          
          *
          
            
              t
              1
              2
            
            2.0
          
        
      
    
  




  
    
      
    
    ⁢
    
      
        ΔΨ
        1
      
      =
      
        
          
            
              Ψ
              .
            
            C
          
          *
          
            t
            1
          
        
        +
        
          
            
              Ψ
               ̈
            
            C
          
          *
          
            
              t
              1
              2
            
            2.0
          
        
        -
        
          
            
              Ψ
              ⃛
            
            MAX
          
          *
          
            SIGN
            (
            
              
                Ψ
                .
              
              C
            
            )
          
          *
          
            
              t
              1
              3
            
            6.0
          
        
      
    
  




  
    
      
    
    ⁢
    
      ΔΨ
      =
      
        
          
            
              
                Ψ
                .
              
              C
            
            +
            
              Δ
              ⁢
              
                
                  Ψ
                  .
                
                1
              
            
          
          K
        
        +
        
          ΔΨ
          1
        
      
    
  




  ELSE




  
    
      
    
    ⁢
    
      
        t
        1
      
      =
      
        
          
            
              -
              
                
                  Ψ
                   ̈
                
                MAX
              
            
            *
            
              SIGN
              (
              
                
                  Ψ
                  .
                
                C
              
              )
            
          
          -
          
            
              Ψ
               ̈
            
            C
          
        
        
          
            -
            
              
                Ψ
                ⃛
              
              MAX
            
          
          *
          
            SIGN
            (
            
              
                Ψ
                .
              
              C
            
            )
          
        
      
    
  




  
    
      
    
    ⁢
    
      
        Δ
        ⁢
        
          
            Ψ
            .
          
          1
        
      
      =
      
        
          
            
              Ψ
               ̈
            
            C
          
          *
          
            t
            1
          
        
        -
        
          
            
              Ψ
              ⃛
            
            MAX
          
          *
          
            SIGN
            (
            
              
                Ψ
                .
              
              C
            
            )
          
          *
          
            
              t
              1
              2
            
            2.0
          
        
      
    
  




  
    
      
    
    ⁢
    
      
        ΔΨ
        1
      
      =
      
        
          
            
              Ψ
              .
            
            C
          
          *
          
            t
            1
          
        
        +
        
          
            
              Ψ
               ̈
            
            C
          
          *
          
            
              t
              1
              2
            
            2.0
          
        
        -
        
          
            
              Ψ
              ⃛
            
            MAX
          
          *
          
            SIGN
            (
            
              
                Ψ
                .
              
              C
            
            )
          
          *
          
            
              t
              1
              3
            
            6.0
          
        
      
    
  




  
    
      
    
    ⁢
    
      
        t
        2
      
      =
      
        
          
            
              Ψ
              .
            
            C
          
          +
          
            Δ
            ⁢
            
              
                Ψ
                .
              
              1
            
          
          -
          
            
              
                
                  Ψ
                   ̈
                
                MAX
              
              *
              
                SIGN
                (
                
                  
                    Ψ
                    .
                  
                  C
                
                )
              
            
            K
          
        
        
          
            
              Ψ
               ̈
            
            MAX
          
          *
          
            SIGN
            (
            
              
                Ψ
                .
              
              C
            
            )
          
        
      
    
  




  
    
      
    
    ⁢
    
      
        ΔΨ
        2
      
      =
      
        
          
            (
            
              
                
                  Ψ
                  .
                
                C
              
              +
              
                Δ
                ⁢
                
                  
                    Ψ
                    .
                  
                  1
                
              
            
            )
          
          *
          
            t
            2
          
        
        -
        
          
            
              Ψ
               ̈
            
            MAX
          
          *
          
            SIGN
            (
            
              
                Ψ
                .
              
              C
            
            )
          
          *
          
            
              t
              2
              2
            
            2.0
          
        
      
    
  




  
    
      
    
    ⁢
    
      
        ΔΨ
        3
      
      =
      
        
          
            
              Ψ
               ̈
            
            MAX
          
          *
          
            SIGN
            ⁡
            
              (
              
                
                  Ψ
                  .
                
                C
              
              )
            
          
        
        
          K
          2
        
      
    
  




  
    
      
    
    ⁢
    
      ΔΨ
      ⁢
      
        
      
      =
      
        
      
      ⁢
      
        
          ΔΨ
          1
        
        ⁢
        
          
        
        +
        
          
        
        ⁢
        
          ΔΨ
          2
        
        ⁢
        
          
        
        +
        
          
        
        ⁢
        
          ΔΨ
          3
        
      
    
  




  ENDIF




  
    
      
    
    ⁢
    
      
        Ψ
        PREDICTION
      
      ⁢
      
        
      
      =
      
        
      
      ⁢
      
        
          Ψ
          C
        
        ⁢
        
          
        
        +
        
          
        
        ⁢
        ΔΨ
      
    
  



The variant of a heading prediction algorithm set forth above explicitly defines and updates all possible states calculated by the heading prediction algorithm during each digital update cycle of the flight control computer, ensuring constant computational throughput and timing requirements for the heading prediction algorithm and eliminating any possibility of computational overflow or underflow conditions resulting from ill conditioned inputs to the heading prediction algorithm. This is a robust heading prediction algorithm that eliminates the potential for computational singularities that may exist in previous alternative heading prediction algorithms when those algorithms are used to predict the terminal heading of aircraft that are compliant with ADS-33 handling qualities specifications.
As still further alternatives, the following heading prediction implementations have also been developed and are set forth below.
Exact Solution Implementation of Final Heading Reference Algorithm
To the Constant Acceleration:



  
    
      t
      
        1
        CA
      
    
    =
    
      
        
          
            
              ψ
               ̈
            
            MAX
          
          ⁢
          
            sign
            (
            
              
                ψ
                .
              
              C
            
            )
          
        
        +
        
          
            ψ
             ̈
          
          C
        
      
      
        
          
            ψ
            ⃛
          
          MAX
        
        ⁢
        
          sign
          ⁡
          
            (
            
              
                ψ
                .
              
              C
            
            )
          
        
      
    
  




  
    
      
        If
        ⁢
        
          
        
        ⁢
        
          t
          
            1
            CA
          
        
      
      <
      
        0
        ⁢
        
          
        
        ⁢
        then
        ⁢
        
          
        
        ⁢
        
          t
          
            1
            CA
          
        
      
    
    =
    0
  




  
    
      Δ
      ⁢
      
        
          ψ
          .
        
        
          1
          MAX
        
      
    
    =
    
      
        
          
            ψ
             ̈
          
          C
        
        ⁢
        
          t
          
            1
            CA
          
        
      
      -
      
        
          
            ψ
            ⃛
          
          MAX
        
        ⁢
        
          sign
          ⁡
          
            (
            
              
                ψ
                .
              
              C
            
            )
          
        
        ⁢
        
          
            t
            
              1
              CA
            
            2
          
          2
        
      
    
  




  
    
      t
      2
    
    =
    
      
        
          
            
              ψ
              .
            
            C
          
          +
          
            Δ
            ⁢
            
              
                ψ
                .
              
              
                1
                MAX
              
            
          
        
        
          
            
              ψ
               ̈
            
            MAX
          
          ⁢
          
            sign
            ⁡
            
              (
              
                
                  ψ
                  .
                
                C
              
              )
            
          
        
      
      -
      
        1
        K
      
    
  




  
    
      If
      ⁢
      
        
      
      ⁢
      
        t
        2
      
    
    >
    
      0
      ⁢
      
        
      
      ⁢
      Then
      ⁢
      
        :
      
    
  




  
    
      Δψ
      
        1
        MAX
      
    
    =
    
      
        
          
            ψ
            .
          
          C
        
        ⁢
        
          t
          
            1
            CA
          
        
      
      +
      
        
          
            ψ
             ̈
          
          C
        
        ⁢
        
          
            t
            
              1
              CA
            
            2
          
          2
        
      
      -
      
        
          
            ψ
            ⃛
          
          MAX
        
        ⁢
        
          sign
          ⁡
          
            (
            
              
                ψ
                .
              
              C
            
            )
          
        
        ⁢
        
          
            t
            
              1
              CA
            
            3
          
          6
        
      
    
  




  
    
      Δ
      ⁢
      
        
          ψ
          .
        
        2
      
    
    =
    
      
        -
        
          
            ψ
             ̈
          
          MAX
        
      
      ⁢
      
        sign
        ⁡
        
          (
          
            
              ψ
              .
            
            C
          
          )
        
      
      ⁢
      
        t
        2
      
    
  




  
    
      Δψ
      2
    
    =
    
      
        
          (
          
            
              
                ψ
                .
              
              C
            
            +
            
              Δ
              ⁢
              
                
                  ψ
                  .
                
                
                  1
                  MAX
                
              
            
          
          )
        
        ⁢
        
          t
          2
        
      
      -
      
        
          
            ψ
             ̈
          
          MAX
        
        ⁢
        
          sign
          ⁡
          
            (
            
              
                ψ
                .
              
              C
            
            )
          
        
        ⁢
        
          
            t
            2
            2
          
          2
        
      
    
  




  
    
      Δψ
      3
    
    =
    
      
        
          
            ψ
             ̈
          
          MAX
        
        ⁢
        
          sign
          ⁡
          
            (
            
              
                ψ
                .
              
              C
            
            )
          
        
      
      
        K
        2
      
    
  




  
    Δψ
    =
    
      
        Δψ
        
          1
          MAX
        
      
      +
      
        Δψ
        2
      
      +
      
        Δψ
        3
      
    
  



    Else:

To the Exponential:






  
    
      t
      
        1
        exp
      
    
    =
    
      
        
          
            K
            ⁢
            
              
                ψ
                 ̈
              
              C
            
          
          -
          
            
              
                ψ
                ⃛
              
              MAX
            
            ⁢
            
              sign
              ⁡
              
                (
                
                  
                    ψ
                    .
                  
                  C
                
                )
              
            
          
        
        
          K
          ⁢
          
            
              ψ
              ⃛
            
            MAX
          
          ⁢
          
            sign
            ⁡
            
              (
              
                
                  ψ
                  .
                
                C
              
              )
            
          
        
      
      +
      
        
          
            
              
                
                  
                    (
                    
                      
                        
                          K
                          ⁢
                          
                            
                              ψ
                               ̈
                            
                            C
                          
                        
                        -
                        
                          
                            
                              ψ
                              ⃛
                            
                            MAX
                          
                          ⁢
                          
                            sign
                            ⁡
                            
                              (
                              
                                
                                  ψ
                                  .
                                
                                C
                              
                              )
                            
                          
                        
                      
                      
                        K
                        ⁢
                        
                          
                            ψ
                            ⃛
                          
                          MAX
                        
                        ⁢
                        
                          sign
                          ⁡
                          
                            (
                            
                              
                                ψ
                                .
                              
                              C
                            
                            )
                          
                        
                      
                    
                    )
                  
                  2
                
                +
              
            
          
          
            
              
                
                  2
                  ⁢
                  
                    (
                    
                      
                        K
                        ⁢
                        
                          
                            ψ
                            .
                          
                          C
                        
                      
                      +
                      
                        
                          ψ
                           ̈
                        
                        C
                      
                    
                    )
                  
                
                
                  K
                  ⁢
                  
                    
                      ψ
                      ⃛
                    
                    MAX
                  
                  ⁢
                  
                    sign
                    ⁡
                    
                      (
                      
                        
                          ψ
                          .
                        
                        C
                      
                      )
                    
                  
                
              
            
          
        
      
    
  




  
    
      Δ
      ⁢
      
        
          ψ
          .
        
        
          1
          exp
        
      
    
    =
    
      
        
          
            ψ
             ̈
          
          C
        
        ⁢
        
          t
          
            1
            exp
          
        
      
      -
      
        
          
            ψ
            ⃛
          
          MAX
        
        ⁢
        
          sign
          ⁡
          
            (
            
              
                ψ
                .
              
              C
            
            )
          
        
        ⁢
        
          
            t
            
              1
              exp
            
            2
          
          2
        
      
    
  




  
    
      Δψ
      
        1
        exp
      
    
    =
    
      
        
          
            ψ
            .
          
          C
        
        ⁢
        
          t
          
            1
            exp
          
        
      
      +
      
        
          
            ψ
             ̈
          
          C
        
        ⁢
        
          
            t
            
              1
              exp
            
            2
          
          2
        
      
      -
      
        
          
            ψ
            ⃛
          
          MAX
        
        ⁢
        
          sign
          ⁡
          
            (
            
              
                ψ
                .
              
              C
            
            )
          
        
        ⁢
        
          
            t
            
              1
              exp
            
            3
          
          6
        
      
    
  




  
    
      Δψ
      3
    
    =
    
      
        
          
            ψ
            .
          
          C
        
        +
        
          Δ
          ⁢
          
            
              ψ
              .
            
            
              1
              exp
            
          
        
      
      K
    
  




  
    Δψ
    =
    
      
        Δψ
        
          1
          exp
        
      
      +
      
        Δψ
        3
      
    
  




  
    End
    ⁢
    
      
    
    ⁢
    if
  




  
    
      ψ
      PREDICTION
    
    =
    
      
        ψ
        C
      
      +
      Δψ
    
  



Iterative Solution Implementation of Final Heading Reference Algorithm

To the Constant Acceleration:




  
    
      t
      
        1
        CA
      
    
    =
    
      
        
          
            
              ψ
               ̈
            
            MAX
          
          ⁢
          
            sign
            (
            
              
                ψ
                .
              
              C
            
            )
          
        
        +
        
          
            ψ
             ̈
          
          C
        
      
      
        
          
            ψ
            ⃛
          
          MAX
        
        ⁢
        
          sign
          ⁡
          
            (
            
              
                ψ
                .
              
              C
            
            )
          
        
      
    
  




  
    
      
        If
        ⁢
        
          
        
        ⁢
        
          t
          
            1
            CA
          
        
      
      <
      
        0
        ⁢
        
          
        
        ⁢
        then
        ⁢
        
          
        
        ⁢
        
          t
          
            1
            CA
          
        
      
    
    =
    0
  




  
    
      Δ
      ⁢
      
        
          ψ
          .
        
        
          1
          MAX
        
      
    
    =
    
      
        
          
            ψ
             ̈
          
          C
        
        ⁢
        
          t
          
            1
            CA
          
        
      
      -
      
        
          
            ψ
            ⃛
          
          MAX
        
        ⁢
        
          sign
          ⁡
          
            (
            
              
                ψ
                .
              
              C
            
            )
          
        
        ⁢
        
          
            t
            
              1
              CA
            
            2
          
          2
        
      
    
  




  
    
      t
      2
    
    =
    
      
        
          
            
              ψ
              .
            
            C
          
          +
          
            Δ
            ⁢
            
              
                ψ
                .
              
              
                1
                MAX
              
            
          
        
        
          
            
              ψ
               ̈
            
            MAX
          
          ⁢
          
            sign
            ⁡
            
              (
              
                
                  ψ
                  .
                
                C
              
              )
            
          
        
      
      -
      
        1
        K
      
    
  




  
    
      If
      ⁢
      
        
      
      ⁢
      
        t
        2
      
    
    >
    
      0
      ⁢
      
        
      
      ⁢
      Then
      ⁢
      
        :
      
    
  




  
    
      Δψ
      
        1
        MAX
      
    
    =
    
      
        
          
            ψ
            .
          
          C
        
        ⁢
        
          t
          
            1
            CA
          
        
      
      +
      
        
          
            ψ
             ̈
          
          C
        
        ⁢
        
          
            t
            
              1
              CA
            
            2
          
          2
        
      
      -
      
        
          
            ψ
            ⃛
          
          MAX
        
        ⁢
        
          sign
          ⁡
          
            (
            
              
                ψ
                .
              
              C
            
            )
          
        
        ⁢
        
          
            t
            
              1
              CA
            
            3
          
          6
        
      
    
  




  
    
      Δ
      ⁢
      
        
          ψ
          .
        
        2
      
    
    =
    
      
        -
        
          
            ψ
             ̈
          
          MAX
        
      
      ⁢
      
        sign
        ⁡
        
          (
          
            
              ψ
              .
            
            C
          
          )
        
      
      ⁢
      
        t
        2
      
    
  




  
    
      Δψ
      2
    
    =
    
      
        
          (
          
            
              
                ψ
                .
              
              C
            
            +
            
              Δ
              ⁢
              
                
                  ψ
                  .
                
                
                  1
                  MAX
                
              
            
          
          )
        
        ⁢
        
          t
          2
        
      
      -
      
        
          
            ψ
             ̈
          
          MAX
        
        ⁢
        
          sign
          ⁡
          
            (
            
              
                ψ
                .
              
              C
            
            )
          
        
        ⁢
        
          
            t
            2
            2
          
          2
        
      
    
  




  
    
      Δψ
      3
    
    =
    
      
        
          
            ψ
             ̈
          
          MAX
        
        ⁢
        
          sign
          ⁡
          
            (
            
              
                ψ
                .
              
              C
            
            )
          
        
      
      
        K
        2
      
    
  




  
    Δψ
    =
    
      
        Δψ
        
          1
          MAX
        
      
      +
      
        Δψ
        2
      
      +
      
        Δψ
        3
      
    
  



    Else:

To the Exponential:






  
    
      t
      
        1
        exp
      
    
    =
    
      
        
          K
          *
          
            (
            
              
                
                  Ψ
                  .
                
                C
              
              +
              
                Δ
                ⁢
                
                  
                
                ⁢
                
                  
                    Ψ
                    .
                  
                  1
                  ′
                
              
            
            )
          
        
        +
        
          
            Ψ
            ..
          
          C
        
      
      
        
          
            Ψ
            ...
          
          MAX
        
        *
        SIGN
        ⁢
        
          
        
        ⁢
        
          (
          
            
              Ψ
              .
            
            C
          
          )
        
      
    
  




  
    
      Δ
      ⁢
      
        
      
      ⁢
      
        
          Ψ
          .
        
        1
        ′
      
    
    =
    
      
        
          
            Ψ
            ..
          
          C
        
        *
        
          t
          
            1
            exp
          
        
      
      -
      
        
          
            Ψ
            ...
          
          MAX
        
        *
        SIGN
        ⁢
        
          
        
        ⁢
        
          (
          
            
              Ψ
              .
            
            C
          
          )
        
        *
        
          
            t
            
              1
              exp
            
            2
          
          2.0
        
      
    
  




  
    
      Δ
      ⁢
      
        
      
      ⁢
      
        Ψ
        1
      
    
    =
    
      
        
          
            Ψ
            .
          
          C
        
        *
        
          t
          
            1
            exp
          
        
      
      +
      
        
          
            Ψ
            ..
          
          C
        
        *
        
          
            t
            
              1
              exp
            
            2
          
          2.0
        
      
      -
      
        
          
            Ψ
            ...
          
          MAX
        
        *
        SIGN
        ⁢
        
          
        
        ⁢
        
          (
          
            
              Ψ
              .
            
            C
          
          )
        
        *
        
          
            t
            
              1
              exp
            
            3
          
          6.0
        
      
    
  




  
    
      Δ
      ⁢
      
        
      
      ⁢
      
        ψ
        3
      
    
    =
    
      
        
          
            ψ
            .
          
          C
        
        +
        
          Δ
          ⁢
          
            
          
          ⁢
          
            
              Ψ
              .
            
            1
            ′
          
        
      
      K
    
  




  
    
      Δ
      ⁢
      
        
      
      ⁢
      ψ
    
    =
    
      
        Δψ
        1
      
      +
      
        Δψ
        3
      
    
  




  
    
      End
      ⁢
      
        
      
      ⁢
      if
    
    ⁢
    
      
    
  




  
    
      ψ
      PREDICTION
    
    =
    
      
        ψ
        C
      
      +
      Δψ
    
  



Approximate Solution Implementation of Final Heading Reference Algorithm

To the Constant Acceleration:




  
    
      t
      
        1
        CA
      
    
    =
    
      
        
          
            
              ψ
              ..
            
            MAX
          
          ⁢
          sign
          ⁢
          
            
          
          ⁢
          
            (
            
              
                ψ
                .
              
              C
            
            )
          
        
        +
        
          
            ψ
            ..
          
          C
        
      
      
        
          
            ψ
            ...
          
          MAX
        
        ⁢
        sign
        ⁢
        
          
        
        ⁢
        
          (
          
            
              ψ
              .
            
            C
          
          )
        
      
    
  




  
    
      
        If
        ⁢
        
          
        
        ⁢
        
          t
          
            1
            CA
          
        
      
      <
      
        0
        ⁢
        
          
        
        ⁢
        then
        ⁢
        
          
        
        ⁢
        
          t
          
            1
            CA
          
        
      
    
    =
    0
  




  
    
      Δ
      ⁢
      
        
      
      ⁢
      
        
          ψ
          .
        
        
          1
          MAX
        
      
    
    =
    
      
        
          
            ψ
            ..
          
          C
        
        ⁢
        
          t
          
            1
            CA
          
        
      
      -
      
        
          
            ψ
            ...
          
          MAX
        
        ⁢
        sign
        ⁢
        
          
        
        ⁢
        
          (
          
            
              ψ
              .
            
            C
          
          )
        
        ⁢
        
          
            t
            
              1
              CA
            
            2
          
          2
        
      
    
  




  
    
      t
      2
    
    =
    
      
        
          
            
              ψ
              .
            
            C
          
          +
          
            Δ
            ⁢
            
              
            
            ⁢
            
              
                ψ
                .
              
              
                1
                MAX
              
            
          
        
        
          
            
              ψ
              ..
            
            MAX
          
          ⁢
          sign
          ⁢
          
            
          
          ⁢
          
            (
            
              
                ψ
                .
              
              C
            
            )
          
        
      
      -
      
        1
        K
      
    
  




  
    
      If
      ⁢
      
        
      
      ⁢
      
        t
        2
      
    
    >
    
      0
      ⁢
      
        
      
      ⁢
      Then
      ⁢
      
        :
      
    
  




  
    
      Δ
      ⁢
      
        
      
      ⁢
      
        ψ
        
          1
          MAX
        
      
    
    =
    
      
        
          
            ψ
            .
          
          C
        
        ⁢
        
          t
          
            1
            CA
          
        
      
      +
      
        
          
            ψ
            ..
          
          C
        
        ⁢
        
          
            t
            
              1
              CA
            
            2
          
          2
        
      
      -
      
        
          
            ψ
            ...
          
          MAX
        
        ⁢
        sign
        ⁢
        
          
        
        ⁢
        
          (
          
            
              ψ
              .
            
            C
          
          )
        
        ⁢
        
          
            t
            
              1
              CA
            
            3
          
          6
        
      
    
  




  
    
      Δ
      ⁢
      
        
      
      ⁢
      
        
          ψ
          .
        
        2
      
    
    =
    
      
        -
        
          
            ψ
             ̈
          
          MAX
        
      
      ⁢
      sign
      ⁢
      
        
      
      ⁢
      
        (
        
          
            ψ
            .
          
          C
        
        )
      
      ⁢
      
        t
        2
      
    
  




  
    
      Δ
      ⁢
      
        
      
      ⁢
      
        ψ
        2
      
    
    =
    
      
        
          
            (
            
              
                
                  ψ
                  .
                
                C
              
              +
              
                Δ
                ⁢
                
                  
                
                ⁢
                
                  
                    ψ
                    .
                  
                  
                    1
                    MAX
                  
                
              
            
            )
          
          ⁢
          
            t
            2
          
        
        -
        
          
            
              ψ
              ..
            
            MAX
          
          ⁢
          
            sign
            ⁢
            
              
                
              
              ⁢
              
                
              
            
            (
            
              
                ψ
                .
              
              C
            
            )
          
          ⁢
          
            
              t
              2
              2
            
            2
          
          ⁢
          
            

          
          ⁢
          
            Δ
            ⁢
            
              
            
            ⁢
            
              ψ
              3
            
          
        
      
      =
      
        
          
            
              
                
                  ψ
                  ..
                
                MAX
              
              ⁢
              sign
              ⁢
              
                
              
              ⁢
              
                (
                
                  
                    ψ
                    .
                  
                  C
                
                )
              
            
            
              K
              
                2
                ⁢
                
                  
                
              
            
          
          ⁢
          
            

          
          ⁢
          Δ
          ⁢
          
            
          
          ⁢
          ψ
        
        =
        
          
            Δ
            ⁢
            
              
            
            ⁢
            
              ψ
              
                1
                MAX
              
            
          
          +
          
            Δψ
            2
          
          +
          
            Δψ
            3
          
        
      
    
  



    Else:

To the Exponential:






  
    
      t
      
        1
        exp
      
      ′
    
    =
    
      
        
          K
          *
          
            
              Ψ
              .
            
            C
          
        
        +
        
          
            Ψ
            ..
          
          C
        
      
      
        
          
            Ψ
            ...
          
          MAX
        
        *
        SIGN
        ⁢
        
          
        
        ⁢
        
          (
          
            
              Ψ
              .
            
            C
          
          )
        
      
    
  




  
    
      Δ
      ⁢
      
        
      
      ⁢
      
        
          Ψ
          .
        
        1
        ′
      
    
    =
    
      
        
          
            Ψ
            ..
          
          C
        
        *
        
          t
          
            1
            exp
          
          ′
        
      
      -
      
        
          
            Ψ
            ...
          
          MAX
        
        *
        
          SIGN
          ⁢
          
            
              
            
            ⁢
            
              
            
          
          (
          
            
              Ψ
              .
            
            C
          
          )
        
        *
        
          
            t
            
              1
              exp
            
            ′2
          
          2.0
        
      
    
  




  
    
      t
      
        1
        exp
      
    
    =
    
      
        
          K
          *
          
            (
            
              
                
                  Ψ
                  .
                
                C
              
              +
              
                Δ
                ⁢
                
                  
                
                ⁢
                
                  
                    Ψ
                    .
                  
                  1
                  ′
                
              
            
            )
          
        
        +
        
          
            Ψ
            ..
          
          C
        
      
      
        
          
            Ψ
            ...
          
          MAX
        
        *
        SIGN
        ⁢
        
          
        
        ⁢
        
          (
          
            
              Ψ
              .
            
            C
          
          )
        
      
    
  




  
    
      Δ
      ⁢
      
        
      
      ⁢
      
        
          Ψ
          .
        
        1
      
    
    =
    
      
        
          
            Ψ
            ..
          
          C
        
        *
        
          t
          
            1
            exp
          
        
      
      -
      
        
          
            Ψ
            ...
          
          MAX
        
        *
        SIGN
        ⁢
        
          
        
        ⁢
        
          (
          
            
              Ψ
              .
            
            C
          
          )
        
        *
        
          
            t
            
              1
              exp
            
            2
          
          2.0
        
      
    
  




  
    
      Δ
      ⁢
      
        
      
      ⁢
      
        Ψ
        1
      
    
    =
    
      
        
          
            Ψ
            .
          
          C
        
        *
        
          t
          
            1
            exp
          
        
      
      +
      
        
          
            Ψ
            ..
          
          C
        
        *
        
          
            t
            
              1
              exp
            
            2
          
          2.0
        
      
      -
      
        
          
            Ψ
            ...
          
          MAX
        
        *
        
          SIGN
          ⁢
          
            
              
            
            ⁢
            
              
            
          
          (
          
            
              Ψ
              .
            
            C
          
          )
        
        *
        
          
            t
            
              1
              exp
            
            3
          
          6.0
        
      
    
  




  
    
      Δ
      ⁢
      
        
      
      ⁢
      
        ψ
        3
      
    
    =
    
      
        
          
            ψ
            .
          
          C
        
        +
        
          Δ
          ⁢
          
            
          
          ⁢
          
            
              Ψ
              .
            
            1
            ′
          
        
      
      K
    
  




  
    
      Δ
      ⁢
      
        
      
      ⁢
      ψ
    
    =
    
      
        Δψ
        1
      
      +
      
        Δψ
        3
      
    
  




  
    
      End
      ⁢
      
        
      
      ⁢
      if
    
    ⁢
    
      
    
  




  
    
      ψ
      PREDICTION
    
    =
    
      
        ψ
        C
      
      +
      Δψ
    
  



Of the three implementations set forth above, the exact solution is likely to produce the most accurate heading prediction, but it is the most computationally demanding. The iterative solution is the more exact of the other two less computationally demanding solutions, but it may sometimes have convergence problems. The approximate solution will never have convergence problems, but it is the least exact solution.
The invention provides the pilot with a significantly reduced workload and allows the pilot to focus more attention on other tasks, thereby improving mission effectiveness and increasing safety. A system configured in accordance with one aspect of the invention has been demonstrated in an AH-64A helicopter simulator to achieve the aforementioned turn to target performance specification goals for maneuverability and agility of helicopters established by the U.S. Army.
Other aspects and features of the present invention can be obtained from a study of the drawings, the disclosure, and the appended claims. For example, although the foregoing has focused on piloted rotorcraft, the invention could be used on other aircraft, machines such as robot arms and cranes, that need to be precisely pointed in a desired direction, and in connection with flight controls for remotely piloted vehicles, such as, for example, unmanned helicopters as means for a remote pilot and/or an autonomous flight control system to better control such a vehicle. The invention is applicable to both fixed wing and helicopter aircraft in both “coordinated” and “uncoordinated” pointing (or hybrid) tasks. For instance, at a given speed, the heading rate (analogous to yaw rate for a hovering vehicle) is proportional to the bank angle in a coordinated turn. The heading acceleration is proportional to the bank angle rate. The heading “jerk” is proportional to the bank angle acceleration.

## Claims

1. A computerized system that continuously predicts a final aircraft state of a maneuvering aircraft provided a control input of the aircraft is at least one of released and relaxed, the system comprising:
a computer processor;
and an algorithm programmed into said computer processor and adapted to calculate, during a maneuver, the predicted final aircraft state based on a mathematical command model, which comprises a basic heading and yaw rate response, comprising at least one of a desired final heading and a desired final pitch attitude, at least a first three derivatives of at least one of the desired final heading and the desired final pitch attitude, limiting values for each of the at least first three derivatives, closed loop algorithms, a plant canceller element, and a control input for at least one of the desired final heading and the desired final pitch attitude.

2. The computerized system of claim 1, wherein the basic heading and the yaw rate response comprise both the desired final heading and the desired final pitch attitude.

3. The computerized system of claim 1, wherein the mathematical command model is tailored to maintain aircraft angular acceleration within specified limits.

4. The computerized system of claim 3, wherein the specified limits include limits of angular acceleration about an aircraft yaw axis.

5. The computerized system of claim 1, wherein the mathematical command model is tailored to maintain aircraft angular jerk within specified limits.

6. The computerized system of claim 5, wherein the specified limits include limits of angular jerk about an aircraft yaw axis.

7. The computerized system of claim 1, wherein the predicted final aircraft state is calculated using an exact solution algorithm.

8. The computerized system of claim 1, wherein the predicted final aircraft state is calculated using an iterative solution algorithm.

9. The computerized system of claim 1, wherein the predicted final aircraft state is calculated using an approximate solution algorithm.

10. A method of continually predicting a final aircraft state of a maneuvering aircraft provided a control input is at least one of released and relaxed, the method comprising:
providing a computer processor;
programming an algorithm which includes a mathematical command model, comprising a basic heading and yaw rate response, which comprises at least one of a desired final heading and a desired final pitch attitude, at least a first three derivatives of each of at least one of the desired final heading and the desired final pitch attitude limiting values for each of the at least first three derivatives, closed loop algorithms, a plant canceller element, and at least one control input for at least one of the desired final heading and the desired final pitch attitude; and
calculating, during a maneuver, said predicted final aircraft state using said algorithm.

11. The method of claim 10, wherein the basic heading and the yaw rate response comprise both the desired final heading and the desired final pitch attitude.

12. The method of claim 10, wherein the mathematical command model is tailored to maintain aircraft angular acceleration within specified limits.

13. The method of claim 10, wherein calculating the predicted final aircraft state includes using an exact solution algorithm.

14. The method of claim 10, wherein calculating the predicted final aircraft state includes using an iterative solution algorithm.

15. The method of claim 10, wherein calculating the predicted final aircraft state includes using an approximate solution algorithm.