A specified time flapping preset performance control method for two-stage wings

Abstract

The application discloses a specified time flapping preset performance control method for a two-section wing, and specifically comprises the following steps: constructing a flapping control model of the two-section wing under different flight modes, and determining model parameters; designing a flapping control performance envelope of the two-section wing for specified time convergence based on the flapping control model; constructing a specified time flapping preset performance controller of the two-section wing, and determining specified time flapping preset performance control parameters. The application can guarantee that transient and steady flapping performances of the two-section wing meet aerodynamic requirements under various flight conditions, and can flexibly specify the convergence time of the flapping frequency of the flapping control system, and is not affected by flapping control parameters and flapping initial states.

Description

Technical Field

[0001] This invention relates to the field of biomimetic flapping-wing aircraft technology, and in particular to a method for controlling the pre-set flapping performance of a two-segment wing at a specified time. Background Technology

[0002] Ornithopter aircraft are aircraft designed based on biomimetic principles, mimicking the flight patterns of flying creatures such as birds or insects. Compared to traditional fixed-wing and rotary-wing aircraft, ornithopter aircraft have advantages such as high maneuverability, high flight efficiency, low energy consumption, and good stealth. Research shows that ornithopter aircraft possess a unique high-lift generation mechanism, exhibiting excellent flight capabilities in small-scale, unsteady flow fields, and low Reynolds numbers, giving them a significant advantage over fixed-wing and rotary-wing aircraft. Furthermore, ornithopter aircraft have enormous development potential and application space in both civilian and military fields.

[0003] Early flapping-wing aircraft mostly simplified the flapping wing into a rigid body model with a single degree of freedom for up-and-down flapping. The advantages of single-stage flapping-wing aircraft are simple structure and stable reliability; the disadvantages are poor aerodynamic performance and low flight efficiency due to the lack of consideration for the bending, deformation, and torsional movements of bird wings. In recent years, with in-depth research, it has been discovered that multi-stage and multi-degree-of-freedom flexible flapping wings have better aerodynamic performance and can better mimic bird flight characteristics. Among these, two-stage flapping-wing aircraft have received the most attention.

[0004] Although two-stage flapping-wing aircraft have the aforementioned advantages in flapping flight, they still face many challenges in practical applications. Currently, the flapping control of two-stage flapping-wing aircraft mainly suffers from the following problems:

[0005] First, the overshoot and steady-state error of the flapping frequency of a two-stage wing are affected by aerodynamic performance. Existing flapping methods do not take these factors into account, which will make it difficult for the transient and steady-state performance of the two-stage wing flapping control system to meet aerodynamic requirements, thereby affecting flight efficiency and even threatening flight safety.

[0006] Secondly, the convergence time of the two-stage flapping control is affected by the control parameters and the initial flapping state, making it difficult to control flexibly. In many application scenarios, the flapping control system needs to respond to the flapping frequency within a specified time. If it cannot converge to the desired flapping frequency within the specified time, its aerodynamic performance will often be greatly reduced. Summary of the Invention

[0007] To address the aforementioned problems in the flapping control of two-segment flapping aircraft, this invention proposes a time-based preset performance control method for two-segment flapping, which can guarantee the transient and steady-state performance of the two-segment flapping control system and allows for flexible specification of the convergence time of the flapping frequency.

[0008] The technical solution to achieve the purpose of this invention is as follows:

[0009] A method for pre-setting performance control of flapping at a specified time for a two-stage wing includes:

[0010] Construct flapping control models for two-segment wings under different flight modes and determine the model parameters;

[0011] Based on the flapping control model, a flapping control performance envelope for a two-segment wing that converges at a specified time is designed. The flapping control performance envelope can be flexibly customized according to the flight scenario to ensure that the transient and steady-state flapping performance of the two-segment wing meets various aerodynamic requirements.

[0012] A two-stage wing flapping preset performance controller is constructed, and the preset performance control parameters for flapping at a specified time are determined. The controller strategy is not affected by the flapping control parameters and the initial flapping state, and can ensure that the system responds to the flapping frequency within the specified time.

[0013] Furthermore, the flapping control models for the two-segment wing under different flight modes are constructed as follows:

[0014] Construct the kinematic model, aerodynamic model, load model, motor drive model, and electronic control model of the two-stage wing flapping mechanism;

[0015] By integrating kinematic models, aerodynamic models, load models, motor drive models, and electronic control models, a flapping control model is constructed.

[0016] Furthermore, the kinematic model is as follows:

[0017] ξ=θ0-∠CDA

[0018]

[0019] β1=∠BCD-∠HEF

[0020] ∠BCD=arc((c 2 +b 2 -(a 2 +d 2 -2ad cos(θ c -θ0))) / 2cb)

[0021] Where ξ is the spanwise flapping angle of the flapping mechanism, θ0 is the frame mounting angle, d is the straight-line distance between the output shaft and the flapping rotation axis of the wing, ∠CDA is the angle between the line connecting the crank shaft and the flapping shaft and the rocker arm, and l BD β1 is the distance between the crank end point B and the rocker arm pivot point D, β1 is the spanwise turning angle of the inner and outer airfoils, ∠BCD is the angle between the connecting rod and the rocker arm, ∠HEF is the fixed angle of the auxiliary mechanism, and θ is the distance between the crank end point B and the rocker arm pivot point D. cLet be the crank rotation angle, and a, b, and c be the lengths of the crank, connecting rod, and rocker arm in the crank-rocker mechanism, respectively.

[0022] Furthermore, the aerodynamic model is as follows:

[0023]

[0024]

[0025] Where l is the total length of the inner and outer wings of the flapping wing, L(t) and D(t) are the total lift and drag of the single wing at time t, respectively, and θ is the velocity tilt angle. parameter Inner wing segment V rx =V inner,rx V ry =V inner,ry ; Outer wing section V rx =V outer,rx V ry =V outer,ry V inner,rx V inner,ry V is the projection of the flapping inner wing velocity onto the X and Y axes. outer,rx V outer,ry The projection of the flapping wing velocity onto the X and Y axes. and Let dF be the component of the aircraft's center of mass velocity along the X and Y axes at time t, and r be the distance of the wing element from the flapping axis. L and dF D Let be the lift and drag forces acting on the wing element at time t, respectively:

[0026]

[0027]

[0028] Among them, C L and C D α represents the dimensionless lift and drag coefficients, respectively, and α is the angle of attack of a single wing element. V is the pitch angle of the aircraft, and V is the velocity of the wing element. ρ is the air density, and c is the chord length at this airfoil element.

[0029] Furthermore, the construction of the load model specifically includes:

[0030] Force analysis is performed based on the structure of the inner and outer wings of the two-section flapping wing and their kinematic relationship:

[0031]

[0032]

[0033] In the formula, F1, F2 and T1, T2 are the constraint forces and moments of the hinges at the flapping axis and the turning axis of the inner and outer wings, respectively; F air1 F air2 With T air1 T air2 These are the aerodynamic forces acting on the inner and outer wings, respectively, and the moments acting on the flapping axis and the turning axis of the inner and outer wings, respectively; F I2 F I2 These are the resultant inertial forces acting on the inner and outer wing surfaces, respectively; G1, G2, and T. G1 T G2 The gravitational force and stress moment acting on the inner and outer wings; r DE Let T be the vector pointing to the wing surface, where i is the unit normal vector perpendicular to the wing surface; wing This refers to the torque at the root of the flapping wing on one side;

[0034] The root load of a single-sided two-section flapping wing is calculated as follows:

[0035] T wing =T air +T G +T I

[0036]

[0037] Among them, T air T G With T I These are the aerodynamic torque, gravitational torque, and inertial torque acting on a single-sided flapping bearing, respectively; inner With l outer These are the lengths of the inner and outer wings, respectively; n1 and n2 are the normal vectors of the inner and outer wing surfaces; r c1 r c2 α1 and α2 are the radius vectors from the center of mass of the inner and outer wings to the flapping axis and the turning axis of the inner and outer wings, respectively; α1 and α2 are the flapping angular velocity of the inner wing and the rotational angular acceleration of the outer wing relative to the inner wing; J inner,D J outer,E These are the moments of inertia of the inner and outer wings relative to the axis of rotation, dF. air (t,r) represents the instantaneous aerodynamic force dF acting on the unit airfoil. air (t,r)=dF L (t,r)+dF D (t,r);

[0038] Determine the motor output shaft load T load,m With reducer output shaft load T gear The torque balance between them is:

[0039]

[0040] In the formula, η g Let i be the efficiency of the reducer, and i be the overall reduction ratio.

[0041] The torque transmitted from the two-stage wing flapping mechanism to the motor shaft via the reducer is determined as follows:

[0042]

[0043] In the formula, η hinge Let ∠CBA be the transmission efficiency of a hinged joint, and ∠CBA be the angle between the crank and the connecting rod.

[0044] Furthermore, the motor drive model specifically includes:

[0045] Voltage balance model:

[0046] E m =R m I m +K e ω m

[0047] Torque balance model:

[0048]

[0049] T m =K m (I m -I0)

[0050] Among them, E m R is the input voltage to the motor. m L is the internal resistance of the motor. m For motor inductance, I m I is the motor current, I0 is the no-load current, and K is the motor current. e ω is the constant of the reverse induced electromotive force. m J is the angular velocity of the motor. m.equ T is the equivalent rotational inertia of the motor shaft. m T is the output torque of the motor. load.m K represents the motor shaft load torque. m is the torque constant.

[0051] Furthermore, the electronic control model is as follows:

[0052] E m =σE b

[0053] Among them, E b σ is the battery output voltage, and σ is the equivalent voltage coefficient.

[0054] Furthermore, the flapping control performance envelope function is:

[0055]

[0056] In the formula, σ0 and σ s Let T be a positive real number, representing the initial boundary and steady-state boundary of the performance envelope function, respectively. s ∈R represents the specified convergence time. χ∈(0.5,1) is the envelope function adjustment parameter;

[0057] The flapping control performance envelope function satisfies: (1) σ(t)>0 and (2) And when t≥T s σ(t) = σ s .

[0058] Furthermore, the two-segment wing flapping preset performance controller at a specified time is as follows:

[0059]

[0060] Where k is the control gain to be adjusted. g = K m / J m,equ J m.equ K is the equivalent rotational inertia of the motor shaft. m Where I is the torque constant and I0 is the no-load current. Here, e(t) represents the normalized error, and e(t) represents the real-time tracking error. This is the desired angular velocity of the reducer output shaft.

[0061] Furthermore, the control gain k to be adjusted is solved using the Lyapunov stability analysis method.

[0062] Compared with the prior art, the main advantages of the present invention are as follows:

[0063] (1) It can flexibly set the flapping control performance envelope according to the flight scenario to ensure that the transient and steady-state flapping performance such as overshoot, convergence rate and steady-state error of the two-stage wing flapping process meets various aerodynamic requirements.

[0064] (2) The convergence time of the two-stage wing flapping control can be freely set. The set flapping frequency convergence time is not affected by the flapping control parameters and the initial flapping state, which can ensure that the flapping control system responds to the flapping frequency within a specified time. Attached Figure Description

[0065] Figure 1 This is a simplified diagram of the flapping drive mechanism (inner wing).

[0066] Figure 2 This is a simplified diagram of the flapping drive mechanism (outer wing).

[0067] Figure 3 This is a simplified model diagram of a two-section wing.

[0068] Figure 4 This is a cross-sectional view of the airfoil of a two-section wing.

[0069] Figure 5 (a) is a schematic diagram of the force analysis of the inner wing; (b) is a schematic diagram of the force analysis of the outer wing.

[0070] Figure 6 This is a schematic diagram of the torque transmission of a two-section wing flapping mechanism on one side.

[0071] Figure 7 This is a curve showing the tracking error during takeoff.

[0072] Figure 8 This is a graph showing the angular velocity tracking during takeoff.

[0073] Figure 9 This is a curve showing the flapping angle of the inner wing during takeoff.

[0074] Figure 10 This is a curve showing the folding angle of the inner and outer wings during takeoff.

[0075] Figure 11 This is a graph showing the motor current curve in takeoff mode.

[0076] Figure 12 This is a graph showing the motor shaft torque in takeoff mode.

[0077] Figure 13 This is a tracking error curve for cruise mode.

[0078] Figure 14 This is a graph showing the angular velocity tracking in cruise mode.

[0079] Figure 15 This is a curve showing the flapping angle of the inner wing in cruise mode.

[0080] Figure 16 This is a curve showing the folding angle of the inner and outer wings in cruise mode.

[0081] Figure 17 This is a graph showing the motor current in cruise mode.

[0082] Figure 18 This is a graph showing the motor shaft torque in cruise mode.

[0083] Figure 19 This is a tracking error curve for landing mode.

[0084] Figure 20 This is a graph showing the angular velocity tracking during landing.

[0085] Figure 21This is a curve showing the flapping angle of the inner wing during landing.

[0086] Figure 22 This is a curve showing the folding angle of the inner and outer wings during landing.

[0087] Figure 23 This is a graph showing the motor current in landing mode.

[0088] Figure 24 This is a graph showing the motor shaft torque in landing mode. Detailed Implementation

[0089] The technical solution of the present invention will now be described in detail with reference to the accompanying drawings.

[0090] Example 1

[0091] This embodiment considers the flapping actuation method of a two-stage wing and designs a flapping control performance envelope with convergence at a specified time. It also proposes a design strategy for a preset performance controller for flapping at a specified time, specifically including:

[0092] First, a flapping control model for the two-segment wing is established, and the model parameters are determined. This flapping control model includes the kinematic model, aerodynamic model, load model, and motor drive model and ESC model related to the flapping mechanism.

[0093] (1) Kinematic model of a two-stage wing flapping mechanism. The inner wing of the two-stage wing uses a crank-rocker structure to convert circular motion into flapping motion. The specific structure can be simplified as follows: Figure 1 The crank's two ends are A and B, the connecting rod's two ends are B and C, and one end of the rocker arm is connected to the connecting rod's end C. The rocker arm's pivot point is marked D. The outer wing is assisted by a two-bar, three-joint mechanism, whose structure can be simplified as follows: Figure 2 The auxiliary rods HE and EF intersect at point E relative to the rocker arm, and the auxiliary rod HG intersects the connecting rod at point G. The fixed angle ∠HEF of the auxiliary mechanism remains constant. Based on geometric relationships, the crank rotation angle θ can be used as a reference. c The spanwise flapping angle ξ of the flapping mechanism is calculated.

[0094] ξ=θ0-∠CDA

[0095] In the formula, θ0 is the frame mounting angle, θ0=arccos(l / d) is the angle between the line connecting the crankshaft and the flapping shaft and the base, d is the straight-line distance between the output shaft and the flapping rotation axis of the wing, and l is the horizontal distance between the shafts. This angle is constant during the flapping process. According to geometric relationships, the angle ∠CDA between the line AD connecting the crankshaft and the flapping shaft and the rocker arm satisfies...

[0096]

[0097] l BD The distance between the crank end point B and the rocker arm pivot point D;

[0098] Similarly, based on the crank rotation angle θ c The angle ∠HEF, which is fixed to the auxiliary mechanism, can be used to calculate the spanwise turning angle β1 of the inner and outer wings.

[0099] β1=∠BCD-∠HEF

[0100] ∠BCD=arc((c 2 +b 2 -(a 2 +d 2 -2ad cos(θ c -θ0))) / 2cb)

[0101] In the formula: ∠BCD is the angle between the connecting rod and the rocker arm, and a, b, and c correspond to the lengths of the crank, connecting rod, and rocker arm in the crank-rocker mechanism, respectively.

[0102] (2) Aerodynamic Model of Two-Segment Wing Flapping Mechanism. To accurately describe the instantaneous motion of the inner and outer wings of a two-segment flapping wing aircraft and simplify the analysis of the forces acting on the wings, the wing surface is divided into n wing elements of length dl along the wingspan direction, as follows: Figure 3 As shown. The total aerodynamic force generated by flapping is obtained by superimposing the aerodynamic forces calculated from each element on the wing along the wingspan. A ground coordinate system OXYZ is established with the aerodynamic center of the airfoil as the origin. Figure 4 This is a cross-sectional view of an airfoil element in a ground coordinate system. V in the figure... x With V y These are the components of the velocity V of the airfoil element along the X and Y axes, respectively, and α is the angle of attack of the airfoil. The pitch angle of the aircraft.

[0103] Assuming the flapping inner wing is at a distance r from the flapping axis inner The wing velocity at point V inner,r The distance between the outer wing and the pivot axis r outer The wing velocity at point V outer,r Based on the kinematic model analysis of the two-stage flapping wing, it can be concluded that...

[0104]

[0105]

[0106] l inner V is the length of the inner wing section. inner,r V outer,r The projections on the X and Y axes are respectively

[0107]

[0108]

[0109] In the formula, ζ represents V outer,r The angle between the aircraft and the horizontal plane. The components of the aircraft's center of mass's velocity along the X and Y axes at time t are respectively... and The absolute velocity and velocity angle of the wing's aerodynamic center are then:

[0110]

[0111]

[0112] In the formula, (inner wing segment V) rx =V inner,rx V ry =V inner,ry ; Outer wing section V rx =V outer,rx V ry =V outer,ry Then the angle of attack of the wing unit at this time is...

[0113]

[0114] The aerodynamic force acting on a unit length of the wing due to the flapping of the wing is:

[0115]

[0116]

[0117] Where dF L and dF D C represents the lift and drag acting on the wing element, respectively. L and C D Let dF be the dimensionless lift and drag coefficients. L dF D By projecting along the X and Y axes respectively and integrating along the span, the total lift and drag of a single wing can be obtained as follows:

[0118]

[0119] (3) Load Model of Two-Segment Flapping Mechanism. To calculate the torque ultimately applied to the motor shaft by the load on the two-segment wing, the instantaneous aerodynamic force on the unit wing surface, the inertial torque on the wing root, and the gravitational torque are analyzed, and the dynamic wing root torque load is calculated. Based on the structure of the inner and outer wings of the two-segment flapping wing and their kinematic relationship, its force analysis is as follows: Figure 5 .

[0120] By dynamic and static method achievable

[0121]

[0122]

[0123] In the formula, F1, F2 and T1, T2 are the constraint forces and moments of the hinges at the flapping axis and the turning axis of the inner and outer wings, respectively; F air1 F air2 With T air1 T air2 These are the aerodynamic forces acting on the inner and outer wings, respectively, and the moments acting on the flapping axis and the turning axis of the inner and outer wings, respectively; F I2 F I2 These are the resultant inertial forces acting on the inner and outer wing surfaces, respectively; G1, G2, and T. G1 T G2 The gravitational force and stress moment acting on the inner and outer wings; r DE Let be the vector pointing from , and ...

[0124] T wing =T air +T G +T I

[0125]

[0126] Where n1 and n2 are the normal vectors of the inner and outer wing surfaces; r c1 r c2 α1 and α2 are the radius vectors from the center of mass of the inner and outer wings to the flapping axis and the turning axis of the inner and outer wings, respectively; α1 and α2 are the flapping angular velocity of the inner wing and the rotational angular acceleration of the outer wing relative to the inner wing; J inner,D J outer,E These are the moments of inertia of the inner and outer wings relative to the axis of rotation, dF. air (t,r) represents the instantaneous aerodynamic force dF acting on the unit airfoil. air (t,r)=dF L (t,r)+dF D (t,r). The root torque of the two-stage wing is applied to the reducer output shaft through the crank-connecting rod, and then finally applied to the motor shaft through the reducer. The load transmission of the flapping mechanism is as follows: Figure 6 As shown, shaft A is the extension shaft of the speed reduction device.

[0127] According to torque balance, we have

[0128]

[0129] If we consider rod BC as a two-force member, then we have

[0130]

[0131] In the formula F 23 F is the force exerted by the connecting rod on the rocker arm. 21 T is the force exerted by the connecting rod on the crank. gear η is the torque output by the reducer to a single flapping wing. hinge Let T be the transmission efficiency of a hinged joint. Considering the transmission efficiency of the reducer, the motor output shaft load T is... load,m With reducer output shaft load T gear The torque balance between them is

[0132]

[0133] In the formula, η g Let be the efficiency of the reducer, and i be the overall reduction ratio. Therefore, the torque transmitted from the two-stage wing flapping mechanism to the motor shaft via the reducer is...

[0134]

[0135] (4) Motor Drive Model. The two-section wing uses a brushless DC motor as the main power source for driving the flapping mechanism. Let E... m R is the input voltage to the motor. m L is the internal resistance of the motor. m For motor inductance, I m K represents the motor current. e ω is the constant of the reverse induced electromotive force. m Let be the angular velocity of the motor. Then, according to the equivalent model of a DC motor, voltage balance is achieved.

[0136]

[0137] The current fluctuations caused by the inductance of a typical brushless DC motor are negligible, so the above equation can be simplified to a first-order form.

[0138] E m =R m I m +K e ω m

[0139] Furthermore, due to torque balance,

[0140]

[0141] T m =K m (I m -I0)

[0142] In the formula, J m.equ T is the equivalent rotational inertia of the motor shaft. m T is the output torque of the motor.load.m K represents the motor shaft load torque. m is the torque constant.

[0143] (5) ESC Model. The speed of a brushless DC motor is regulated by adjusting the duty cycle of the input voltage using an electronic speed controller (ESC). Therefore, the relationship between the duty cycle and the motor's equivalent input voltage needs to be established. Assuming the input power and output power of the ESC are equal, and letting the battery output voltage E... b Equivalent output voltage of the motor m The equivalent voltage coefficient σ has

[0144] E m =σE b

[0145] The equivalent voltage coefficient is directly related to the duty cycle of the electronic speed controller (ESC) during voltage pulse width modulation (PWM). The duty cycle x of the ESC has an exponential relationship with the equivalent voltage coefficient σ, and its empirical function is:

[0146]

[0147] In the formula, a, b, c, and d are all constants.

[0148] (6) Two-stage wing flapping control model. Integrating the kinematic model, aerodynamic model, load model, motor drive model, and ESC model of the flapping mechanism, the two-stage wing power system model is simplified as follows:

[0149]

[0150] In the formula, x = ω m ,u=I m , g = K m / J m,equ .

[0151] Secondly, we design a flapping control performance envelope that converges within a specified time. The flapping control performance envelope function with specified time convergence is defined as follows.

[0152]

[0153] In the formula, σ0 and σ s Let T be a positive real number, representing the initial boundary and steady-state boundary of the performance envelope function, respectively. s ∈R represents the specified convergence time. x∈(0.5,1) is the envelope function adjustment parameter. Differentiating the above performance envelope function with respect to time yields...

[0154]

[0155] Wherein, the parameter α is related to χ and satisfies The performance envelope function designed in this invention satisfies: (1) σ(t)>0 and (2) And when t≥T s σ(t) = σ s .

[0156] Next, a pre-defined performance controller for flapping at a specified time is constructed. To achieve flapping frequency control of the two-stage wing, a pre-defined performance controller for flapping at a specified time is designed based on the flapping control performance envelope function that converges at a specified time. The rotational angular velocity ω of the crank (i.e., the output shaft of the reducer) of the flapping mechanism is defined. m Then the speed tracking error satisfies

[0157]

[0158] Based on the defined flapping control performance envelope function and speed tracking error, the following normalized error is defined.

[0159]

[0160] Based on this normalized error, a time-specified performance controller for flapping is designed, which takes the form of:

[0161]

[0162] In the formula, k is the control gain to be adjusted.

[0163] Finally, the preset performance control parameters for flapping at a specified time are determined. In the two-stage wing preset performance control method for flapping at a specified time of the present invention, the control parameters to be adjusted include the envelope function adjustment parameter χ and the specified convergence time T. s And the control gain k. The envelope function adjustment parameter χ is selected within the range (0.5, 1), and the selection criterion is the desired convergence rate. The larger χ is, the tighter the envelope of the flutter control performance convergence at a specified time; the specified convergence time T s The choice of is determined based on the actual engineering needs; the selection of the control gain k is based on the Lyapunov stability analysis method. Construct the Lyapunov function. Taking the derivative of the function with respect to time, we can obtain

[0164]

[0165] Substituting the designed preset performance controller for the specified time flapping, we can obtain...

[0166]

[0167] Clearly, as long as k > 0, according to Lyapunov's stability theory, the flapping control system of the two-stage wing can achieve the desired flapping effect at a specified time T. sInternal stability. The larger the value of gain k, the faster the convergence within a specified convergence time T. s The internal energy reaches a convergence state more quickly.

[0168] Example 2

[0169] Taking the three typical flight modes of a two-stage flapping wing aircraft—takeoff, cruise, and landing—as examples, the implementation process of a pre-set performance control method for flapping at a specified time for two-stage wings includes the following five specific steps.

[0170] Step 1: Determine the flapping control model and parameters for a two-segment wing under a typical flight model.

[0171] The kinematic model, aerodynamic model, load model, and motor drive model and ESC model related to the flapping mechanism of the two-stage wing flapping mechanism are shown in Example 1. The specific parameters are determined based on the flight parameters under three typical flight modes. The flight parameters under these three modes are shown in Table 1.

[0172] Table 1 Flight parameters under three typical flight modes

[0173]

[0174] Based on the above flight parameters and the overall structure of the two-stage flapping-wing aircraft, the relevant parameters of the flapping control model are determined as follows: Motor drive parameter K m = 0.00348 N·m·A -1 I0 = 0.460 A, J m.equ =7.54×10 -5 kg·m 2 Transmission parameter η hinge =0.98, η g =0.97 3 A, i = 20; Regarding aerodynamic parameters, the empirical formula for the relationship between the lift-drag coefficient and the wing angle of attack of a flapping-wing aircraft is:

[0175]

[0176] Step 2: Design the flapping control performance envelope for a two-segment wing that converges at a specified time.

[0177] The flapping control performance envelope function converges within a specified time.

[0178]

[0179] The relevant parameters of the flapping control performance envelope function are selected in Table 2 for the three typical flight modes.

[0180] Table 2. Flapping control performance envelope parameters under three typical flight modes

[0181]

[0182] Step 3: Construct a preset performance controller for the specified time flapping.

[0183] The specified time flapping preset performance controller is designed as

[0184]

[0185] Step 4: Determine the preset performance control parameters for the specified time flapping.

[0186] The envelope function is adjusted by setting the parameter χ and the convergence time T. s As given in step 2, using a trial-and-error method, the control gain k is selected as 6 in the three typical flight modes.

[0187] Based on the specified-time flapping preset performance control method for two-stage wings proposed in this invention, the simulation results of the two-stage wing flapping control system in takeoff mode are as follows: Figure 7-12 As shown.

[0188] Simulation results of the two-stage wing flapping control system in cruise mode are as follows: Figure 13-18 As shown.

[0189] Simulation results of the two-stage wing flapping control system in landing mode are as follows: Figure 19-24 As shown.

[0190] according to Figure 7-24 The envelope of the flapping control performance that converges at a specified time is in T. s The two-stage wing flapping control system converges to the steady-state value in 5s. The tracking error of the two-stage wing flapping control system can smoothly converge to the steady-state boundary within a specified time in all three modes. The flapping control system will output a smooth tracking signal.

[0191] This invention can ensure that the transient and steady-state flapping performance of the two-stage wing meets the aerodynamic requirements under various flight conditions, while also allowing for flexible specification of the convergence time of the flapping control system's flapping frequency, unaffected by flapping control parameters and the initial flapping state.

Claims

1. A method for controlling the pre-set performance of flapping at a specified time for a two-stage wing, characterized in that, include: Construct flapping control models for two-segment wings under different flight modes and determine the model parameters; Based on the flapping control model, a flapping control performance envelope for a two-segment wing that converges at a specified time is designed. Construct a two-stage wing flapping preset performance controller at a specified time and determine the preset performance control parameters for flapping at a specified time; The flapping control performance envelope function is: In the formula and Let be positive real numbers, and represent the initial boundary and steady-state boundary of the performance envelope function, respectively. The convergence time can be specified. , Adjust the parameters for the envelope function; The flapping control performance envelope function satisfies: (1) and (2) And when hour ; The two-segment wing flapping preset performance controller at a specified time is: in, The control gain to be adjusted. , , This is the equivalent rotational inertia of the motor shaft. The torque constant is This is the no-load current. To normalize the error, To track errors in real time; This is the desired angular velocity of the reducer output shaft.

2. The method for controlling the pre-set flapping performance of a two-segment wing at a specified time according to claim 1, characterized in that, The flapping control models for a two-segment wing under different flight modes include: Construct the kinematic model, aerodynamic model, load model, motor drive model, and electronic control model of the two-stage wing flapping mechanism; By integrating kinematic models, aerodynamic models, load models, motor drive models, and electronic control models, a flapping control model is constructed.

3. The method for controlling the pre-set flapping performance of a two-section wing at a specified time according to claim 2, characterized in that, The kinematic model is as follows: in, For the flapping mechanism to extend to the flapping angle, For rack mounting angle, This is the straight-line distance between the output shaft and the wing flapping rotation axis. The angle between the line connecting the crankshaft and the rocker arm is denoted by . The distance between the crank end point B and the rocker arm pivot point D. For the inward and outward wing span to turn angle, The angle between the connecting rod and the rocker arm. For the auxiliary mechanism to fix the angle, The crank rotation angle, These represent the lengths of the crank, connecting rod, and rocker arm in a crank-rocker mechanism.

4. The method for controlling the pre-set flapping performance of a two-section wing at a specified time according to claim 2, characterized in that, The aerodynamic model is as follows: in, The total length of the flapping wing's inner and outer wings. , Let t represent the total lift and drag of a single wing at time t. For velocity tilt angle, ,parameter , Inner wing section , Outer wing section , , , For the flapping speed of the inner wing at Projection on axis , For flapping of the outer wing, the wing speed is at Projection on axis and For the center of mass of the aircraft Flight speed along the moment , The component of the axis, r, is the distance of the wing element from the flapping axis. and Let be the lift and drag forces acting on the wing element at time t, respectively: in, and These are the dimensionless lift and drag coefficients, respectively. For the single wing angle of attack, , The pitch angle of the aircraft. For the velocity of the airfoil element, , air density, This is the chord length at this wing unit.

5. The method for controlling the pre-set flapping performance of a two-segment wing at a specified time according to claim 2, characterized in that, Constructing the load model specifically includes: Force analysis is performed based on the structure of the inner and outer wings of the two-section flapping wing and their kinematic relationship: In the formula, , and , These are the constraint forces and moments of the hinges at the flapping axis and the inner and outer wing turning axes, respectively; , and , These are the aerodynamic forces acting on the inner and outer wings, and the torques acting on the flapping axis and the turning axis of the inner and outer wings, respectively; , These are the resultant inertial forces acting on the inner and outer wing surfaces, respectively. , and , The weight and stress moment acting on the inner and outer wings; From the vector that points to, It is the unit normal vector perpendicular to the wing surface; This refers to the torque at the root of the flapping wing on one side; The root load of a single-sided two-section flapping wing is calculated as follows: in, , and These are the combined torques of aerodynamic force, gravitational force, and inertial force on a single-sided flapping bearing, respectively. and These are the lengths of the inner and outer wings, respectively; , These are the normal vectors for the inner and outer wing surfaces; , These are the radius vectors from the center of mass of the inner and outer wings to the flapping axis and the turning axis of the inner and outer wings, respectively. , The flapping angular velocity of the inner wing and the rotational angular acceleration of the outer wing relative to the inner wing are given by the inner wing. , These are the moments of inertia of the inner and outer wings relative to the axis of rotation. Instantaneous aerodynamic forces on a single airfoil surface ; Determine the motor output shaft load With the load on the output shaft of the reducer The torque balance between them is: In the formula, For the efficiency of the reducer, This is the overall reduction ratio; The torque transmitted from the two-stage wing flapping mechanism to the motor shaft via the reducer is determined as follows: In the formula, The transmission efficiency of a hinged joint. It is the angle between the crank and the connecting rod.

6. The method for controlling the pre-set flapping performance of a two-segment wing at a specified time according to claim 2, characterized in that, The motor drive model specifically includes: Voltage balance model: Torque balance model: in, This is the input voltage for the motor. The internal resistance of the motor, For motor inductance, This is the motor current. This is the no-load current. This is the constant of the reverse induced electromotive force. The angular velocity of the motor. This is the equivalent rotational inertia of the motor shaft. For the motor output torque, This is the motor shaft load torque. is the torque constant.

7. The method for controlling the pre-set performance of flapping at a specified time for a two-section wing according to claim 2, characterized in that, The electronic control model is as follows: in, This is the battery output voltage. This is the equivalent voltage coefficient.

8. The method for controlling the pre-set flapping performance of a two-segment wing at a specified time according to claim 1, characterized in that, The control gain to be adjusted The Lyapunov stability analysis method is used to solve the problem.

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