CN102616386B - Uniaxial quick maneuverable spacecraft flywheel configuration and optimization method thereof - Google Patents

Abstract

A method for optimizing the flywheel configuration of a single-axis fast maneuvering spacecraft relates to a spacecraft flywheel configuration and an optimization method for the flywheel configuration. It is designed to solve the problem of underutilization of the flywheel capabilities of fast maneuvering spacecraft. Configuration: There are five flywheels in this configuration, the axis of one flywheel coincides with the axis of the motor shaft, and the other four flywheels are slanted flywheels; the rotation axes of two adjacent slanted flywheels are on the non-motor plane of the spacecraft The included angle of the projection on is 90°, and the included angle between the projection of each obliquely mounted flywheel on the non-maneuvering plane of the spacecraft and the pitch axis of the spacecraft and the included angle of the yaw axis of the spacecraft are both 45°. Optimization: Write the installation matrix of the flywheel with the installation angle as the optimized amount; calculate the distribution matrix of the flywheel with the power consumption of the flywheel system as the index. Determine the maximum torque required by the non-motorized shaft, and finally obtain the optimal installation angle of the motorized shaft torque and adjust it to achieve optimization. The invention is suitable for spacecraft flywheel configuration and its optimization.

Description

The optimization method of single shaft fast reserve spacecraft flywheel configuration

Technical field

The present invention relates to the optimization method of spacecraft flywheel configuration and described flywheel configuration.

Background technology

Current many spacecrafts need to be carried out the task of fast reserve and motor-driven rear fast and stable, conventionally there is to strict restriction the time of the motor-driven specified angle of satellite.In the situation that satellite attitude control algorithm is certain, the ability of satellite executing mechanism has determined the motor-driven required time of satellite, yet what the ability of actuating unit, the increase of quantity were corresponding is the increase of satellite quality, it is very disadvantageous that this attitude to satellite is controlled.In order to complete faster motor-driven task under the prerequisite not increasing actuating unit, conventionally need to reasonably configure the actuating unit of satellite.

At present, do not need the satellite of fast reserve to be conventionally equipped with by four flywheels, be mainly divided into two kinds of three quadrature+mono-angle mount configurations and four angle mount configurations.The flywheel work that the quadrature of three quadrature+mono-angle mount configurations is installed, angle mount flywheel is not worked as backup.Four angle mount configurations be take pitch axis conventionally as installation shaft, and stagger angle is the mounting means of quadratic form optimum, and four flywheels are worked and backuped each other simultaneously.For fast reserve satellite, conventionally carry five flywheels, with the master that is configured as of two flywheel+mono-angle mounts of three quadratures+motorized shaft direction, this kind of configuration mode do not make full use of the ability of flywheel, is therefore necessary its configuration to improve.

Summary of the invention

The present invention utilizes inadequate problem in order to solve fast reserve spacecraft flywheel ability, thereby a kind of optimization method of single shaft fast reserve spacecraft flywheel configuration is provided.

Single shaft fast reserve spacecraft flywheel configuration, includes five flywheels in this configuration, the axis of one of them flywheel and the axis of motorized shaft coincide, and other four flywheels are angle mount flywheel;

The angle of the rotating shaft of each angle mount flywheel and spacecraft maneuver axle is β, and β is real number; The angle of the projection of the rotating shaft of adjacent two angle mount flywheels on the non-plane of maneuver of spacecraft is 90 °, and the angle that is positioned at the projection of each the angle mount flywheel on the non-plane of maneuver of described spacecraft and the angle of spacecraft pitch axis and spacecraft yaw axis is 45 °.

The span of the angle β of the rotating shaft of each angle mount flywheel and spacecraft maneuver axle is:

The optimization method of single shaft fast reserve spacecraft flywheel configuration, it is realized by following steps:

Step 1, according to the angle β of the rotating shaft of each angle mount flywheel and spacecraft maneuver axle, structure flywheel installation matrix U:

$U=\left[\begin{array}{ccccc}1& \cos \mathrm{\β}& \cos \mathrm{\β}& \cos \mathrm{\β}& \cos \mathrm{\β}\\ 0& \frac{\sqrt{2}}{2}\sin \mathrm{\β}& -\frac{\sqrt{2}}{2}\sin \mathrm{\β}& -\frac{\sqrt{2}}{2}\sin \mathrm{\β}& \frac{\sqrt{2}}{2}\sin \mathrm{\β}\\ 0& \frac{\sqrt{2}}{2}\sin \mathrm{\β}& \frac{\sqrt{2}}{2}\sin \mathrm{\β}& -\frac{\sqrt{2}}{2}\sin \mathrm{\β}& -\frac{\sqrt{2}}{2}\sin \mathrm{\β}\end{array}\right]$

Step 2, the installation matrix U obtaining according to the minimum power consumption of spacecraft and step 1, according to formula:

D=U ^T(UU ^T) ^-1

The allocation matrix D of structure flywheel:

$D=\left[\begin{array}{ccc}\frac{1}{1+4{\cos }^2\mathrm{\β}}& 0& 0\\ \frac{\cos \mathrm{\β}}{1+4{\cos }^2\mathrm{\β}}& \frac{\sqrt{2}}{4\sin \mathrm{\β}}& \frac{\sqrt{2}}{4\sin \mathrm{\β}}\\ \frac{\cos \mathrm{\β}}{1+4{\cos }^2\mathrm{\β}}& \frac{\sqrt{2}}{4\sin \mathrm{\β}}& \frac{\sqrt{2}}{4\sin \mathrm{\β}}\\ \frac{\cos \mathrm{\β}}{1+4{\cos }^2\mathrm{\β}}& -\frac{\sqrt{2}}{4\sin \mathrm{\β}}& \frac{\sqrt{2}}{4\sin \mathrm{\β}}\\ \frac{\cos \mathrm{\β}}{1+4{\cos }^2\mathrm{\β}}& \frac{\sqrt{2}}{4\sin \mathrm{\β}}& -\frac{\sqrt{2}}{4\sin \mathrm{\β}}\end{array}\right];$

Step 3, the maximum torque of the non-motorized shaft of spacecraft is estimated, obtained the required matrix-vector T of the non-motorized shaft of spacecraft _nmax:

$T_{N\max }=\left[\begin{array}{c}{T}_{y\max }\\ T_{z\max }\end{array}\right]=\left[\begin{array}{c}{K}_{\mathrm{py}}\left|{\mathrm{\θ}}_{\max }\right|+K_{\mathrm{dy}}\left|{\mathrm{\ω}}_{y\max }\right|\\ K_{\mathrm{pz}}\left|{\mathrm{\ψ}}_{\max }\right|+K_{\mathrm{dz}}\left|{\mathrm{\ω}}_{z\max }\right|\end{array}\right];$

In formula: T _nmaxfor the required matrix-vector of the non-motorized shaft of spacecraft; T _ymaxfor the required maximum torque of pitch axis in axis of rolling mobile process, T _zmaxfor the required maximum torque of yaw axis in axis of rolling mobile process; K _pyproportionality coefficient for the PD controller of pitch axis; K _dydifferential coefficient for the PD controller of pitch axis; K _pzproportionality coefficient for the PD controller of yaw axis; K _dzdifferential coefficient for the PD controller of yaw axis; θ _maxmaxim (conventionally getting the departure higher limit of allowance) for pitch angle in mobile process; ω _ymaxmaxim (conventionally getting the departure higher limit of allowance) for pitch axis cireular frequency in mobile process; ψ _maxmaxim (conventionally getting the departure higher limit of allowance) for yaw angle in mobile process; ω _zmaxmaxim (conventionally getting the departure higher limit of allowance) for yaw axis cireular frequency in mobile process;

Step 4, the required matrix-vector T of the non-motorized shaft of spacecraft obtaining according to the allocation matrix D of the flywheel obtaining in step 2, step 3 _nmaxthe maximum torque that can provide with each flywheel is T _wmaxobtain the optimum stagger angle β of motorized shaft moment _t;

The optimum stagger angle β of step 5, the motorized shaft moment that obtains according to step 4 _t, according to described optimum stagger angle β _tsingle shaft fast reserve spacecraft flywheel configuration is adjusted, thereby realized the optimization of single shaft fast reserve spacecraft flywheel configuration.

The required matrix-vector T of the non-motorized shaft of spacecraft obtaining according to the allocation matrix D of the flywheel obtaining in step 2, step 3 described in step 4 _nmaxthe maximum torque that can provide with each flywheel is T _wmaxobtain the optimum stagger angle β of motorized shaft moment _tconcrete grammar be:

The moment providing according to each flywheel is less than or equal to the principle of the moment upper limit of flywheel, obtains constraint inequality:

DT _cmax≤T _wmax

In formula: T _cmax=[T _xmaxt _ymaxt _zmax] ^t, be the maximum instruction moment that five divided flywheels fit over satellite three axles;

In formula: T _wmaxthe array that the maximum torque value that can provide for each flywheel forms; On the same satellite of normal conditions, the maximum torque value of each flywheel is identical, its size T _wmaxrepresent.T _xmaxthe maximum torque that can provide at the axis of rolling for flywheel;

In above formula: $T_{c\max }=\left[\begin{array}{c}{T}_{x\max }\\ T_{N\max }\end{array}\right]=\left[\begin{array}{ccc}{T}_{x\max }& T_{y\max }& T_{z\max }\end{array}\right];$

Make T _ymax+ T _zmax=T _y+z, by DT _cmax≤ T _wmaxlaunch, obtain:

$\left\{\begin{array}{c}\max \left\{\right|T_{w2}|...|T_{w5}\left|\right\}\≥\frac{\cos \mathrm{\β}}{1+4{\cos }^2\mathrm{\β}}{T}_{x\max }+\frac{\sqrt{2}(T_{z\max }+T_{y\max })}{\left|4\sin \mathrm{\β}\right|}\\ T_{w\max }\≥\frac{1}{1+4{\cos }^2\mathrm{\β}}{T}_{x\max }\end{array}\right.$

Be reduced to:

$\left\{\begin{array}{c}{T}_{x\max }\≤T_{w\max }(1+4{\cos }^2\mathrm{\β})\\ T_{x\max }\≤(T_{w\max }-\frac{\sqrt{2}{T}_{y+z}}{4\sin \mathrm{\β}})(\frac{1}{\cos \mathrm{\β}}+4\cos \mathrm{\β})\\ \frac{\sqrt{2}{T}_{y+z}}{4\sin \mathrm{\β}}\≤T_{w\max }\\ 0\≤\mathrm{\β}\≤\frac{\mathrm{\π}}{2}\end{array}\right.$

According to:

T _xmax=T _wmax(1+4cos ²β) exist inside the subtraction function about β, $T_{x\max }=(T_{w\max }-\frac{\sqrt{2}{T}_{y+z}}{4\sin \mathrm{\β}})(\frac{1}{\cos \mathrm{\β}}+4\cos \mathrm{\β})$ ?

inside the increasing function about β, and

it is inside subtraction function;

The optimum stagger angle β of motorized shaft moment _tvalue be below equation about the solution of β:

$\frac{\sqrt{2}{T}_{y+z}}{4T_{w\max }}=(1-\cos \mathrm{\β})\sin \mathrm{\β}.$

Beneficial effect: the present invention can make full use of fast reserve spacecraft flywheel ability, and flywheel configuration of the present invention has larger moment space in motorized shaft, is applicable to having the spacecraft of single shaft fast reserve ability; Meanwhile, optimization method of the present invention makes flywheel can provide maximum moment in motorized shaft direction, thereby improves the acceleration/accel of satellite; Optimization method of the present invention, is guaranteeing can to control the while to non-motorized shaft, makes the maneuverability of motorized shaft reach optimum, does not ignore the control of non-motorized shaft, is therefore more suitable for real satellite attitude control system.

Accompanying drawing explanation

Fig. 1 is the structural representation of flywheel configuration of the present invention; Fig. 2 is the mapping result schematic diagram of Fig. 1 under O-YZ plane; Fig. 3 is the optimum stagger angle β of motorized shaft moment in optimization method of the present invention _tvalue principle schematic, the part representative function value that is wherein filled with oblique line can be got to obtain scope.

The specific embodiment

The specific embodiment one, single shaft fast reserve spacecraft flywheel configuration, include five flywheels in this configuration, the axis of one of them flywheel and the axis of motorized shaft coincide, and other four flywheels are angle mount flywheel;

The angle of the rotating shaft of each angle mount flywheel and spacecraft maneuver axle is β, and β is real number; The angle of the projection of the rotating shaft of adjacent two angle mount flywheels on the non-plane of maneuver of spacecraft is 90 °, and the angle that is positioned at the projection of each the angle mount flywheel on the non-plane of maneuver of described spacecraft and the angle of spacecraft pitch axis and spacecraft yaw axis is 45 °.

The span of the angle β of the rotating shaft of each angle mount flywheel and spacecraft maneuver axle is:

As depicted in figs. 1 and 2, in figure, RW is flywheel to configuration result; Present embodiment actv. has utilized the ability of flywheel, can guarantee that non-motorized shaft is controlled to the while, makes the maneuverability of motorized shaft reach optimum, is not adding under the prerequisite of other actuating units, has shortened spacecraft maneuver required time.And flywheel configuration of the present invention has larger moment space in motorized shaft, be applicable to having the spacecraft of single shaft fast reserve ability.

The optimization method of the specific embodiment two, single shaft fast reserve spacecraft flywheel configuration, it is realized by following steps:

Step 1, according to the angle β of the rotating shaft of each angle mount flywheel and spacecraft maneuver axle, structure flywheel installation matrix U:

$U=\left[\begin{array}{ccccc}1& \cos \mathrm{\β}& \cos \mathrm{\β}& \cos \mathrm{\β}& \cos \mathrm{\β}\\ 0& \frac{\sqrt{2}}{2}\sin \mathrm{\β}& -\frac{\sqrt{2}}{2}\sin \mathrm{\β}& -\frac{\sqrt{2}}{2}\sin \mathrm{\β}& \frac{\sqrt{2}}{2}\sin \mathrm{\β}\\ 0& \frac{\sqrt{2}}{2}\sin \mathrm{\β}& \frac{\sqrt{2}}{2}\sin \mathrm{\β}& -\frac{\sqrt{2}}{2}\sin \mathrm{\β}& -\frac{\sqrt{2}}{2}\sin \mathrm{\β}\end{array}\right]$

Step 2, the installation matrix U obtaining according to the minimum power consumption of spacecraft and step 1, according to formula:

D=U ^T(UU ^T) ^-1

The allocation matrix D of structure flywheel:

$D=\left[\begin{array}{ccc}\frac{1}{1+4{\cos }^2\mathrm{\β}}& 0& 0\\ \frac{\cos \mathrm{\β}}{1+4{\cos }^2\mathrm{\β}}& \frac{\sqrt{2}}{4\sin \mathrm{\β}}& \frac{\sqrt{2}}{4\sin \mathrm{\β}}\\ \frac{\cos \mathrm{\β}}{1+4{\cos }^2\mathrm{\β}}& \frac{\sqrt{2}}{4\sin \mathrm{\β}}& \frac{\sqrt{2}}{4\sin \mathrm{\β}}\\ \frac{\cos \mathrm{\β}}{1+4{\cos }^2\mathrm{\β}}& -\frac{\sqrt{2}}{4\sin \mathrm{\β}}& \frac{\sqrt{2}}{4\sin \mathrm{\β}}\\ \frac{\cos \mathrm{\β}}{1+4{\cos }^2\mathrm{\β}}& \frac{\sqrt{2}}{4\sin \mathrm{\β}}& -\frac{\sqrt{2}}{4\sin \mathrm{\β}}\end{array}\right];$

Step 3, the maximum torque of the non-motorized shaft of spacecraft is estimated, obtained the maximum torque value T of the non-motorized shaft of spacecraft _nmax:

$T_{N\max }=\left[\begin{array}{c}{T}_{y\max }\\ T_{z\max }\end{array}\right]=\left[\begin{array}{c}{K}_{\mathrm{py}}\left|{\mathrm{\θ}}_{\max }\right|+K_{\mathrm{dy}}\left|{\mathrm{\ω}}_{y\max }\right|\\ K_{\mathrm{pz}}\left|{\mathrm{\ψ}}_{\max }\right|+K_{\mathrm{dz}}\left|{\mathrm{\ω}}_{z\max }\right|\end{array}\right];$

In formula: T _nmaxfor the required matrix-vector of the non-motorized shaft of spacecraft; T _ymaxfor the required maximum torque of pitch axis in axis of rolling mobile process, T _zmaxfor the required maximum torque of yaw axis in axis of rolling mobile process; K _pyproportionality coefficient for the PD controller of pitch axis; K _dydifferential coefficient for the PD controller of pitch axis; K _pzproportionality coefficient for the PD controller of yaw axis; K _dzdifferential coefficient for the PD controller of yaw axis; θ _maxmaxim (conventionally getting the departure higher limit of allowance) for pitch angle in mobile process; ω _ymaxmaxim (conventionally getting the departure higher limit of allowance) for pitch axis cireular frequency in mobile process; ψ _maxmaxim (conventionally getting the departure higher limit of allowance) for yaw angle in mobile process; ω _zmaxmaxim (conventionally getting the departure higher limit of allowance) for yaw axis cireular frequency in mobile process;

The maximum torque value T of step 4, the non-motorized shaft of spacecraft that obtains according to the allocation matrix D of the flywheel obtaining in step 2, step 3 _nmaxthe maximum torque that can provide with each flywheel is T _wmaxobtain the optimum stagger angle β of motorized shaft moment _t;

The optimum stagger angle β of step 5, the motorized shaft moment that obtains according to step 4 _t, according to described optimum stagger angle β _tsingle shaft fast reserve spacecraft flywheel configuration is adjusted, thereby realized the optimization of single shaft fast reserve spacecraft flywheel configuration.

The maximum torque value T of the non-motorized shaft of spacecraft obtaining according to the allocation matrix D of the flywheel obtaining in step 2, step 3 described in step 4 _nmaxthe maximum torque that can provide with each flywheel is T _wmaxobtain the optimum stagger angle β of motorized shaft moment _tconcrete grammar be:

The moment providing according to each flywheel is less than or equal to the principle of the moment upper limit of flywheel, obtains constraint inequality:

DT _cmax≤T _wmax

In formula: T _cmax=[T _xmaxt _ymaxt _zmax] ^t, be the maximum instruction moment that five divided flywheels fit over satellite three axles;

In formula: T _wmaxthe array that the maximum torque value that can provide for each flywheel forms; On the same satellite of normal conditions, the maximum torque value of each flywheel is identical, its size T _wmaxrepresent.T _xmaxthe maximum torque that can provide at the axis of rolling for flywheel;

In above formula, $T_{c\max }=\left[\begin{array}{c}{T}_{x\max }\\ T_{N\max }\end{array}\right]=\left[\begin{array}{ccc}{T}_{x\max }& T_{y\max }& T_{z\max }\end{array}\right];$

Make T _ymax+ T _zmax=T _y+z, by DT _cmax≤ T _wmaxlaunch, obtain:

$\left\{\begin{array}{c}\max \left\{\right|T_{w2}|...|T_{w5}\left|\right\}\≥\frac{\cos \mathrm{\β}}{1+4{\cos }^2\mathrm{\β}}{T}_{x\max }+\frac{\sqrt{2}(T_{z\max }+T_{y\max })}{\left|4\sin \mathrm{\β}\right|}\\ T_{w\max }\≥\frac{1}{1+4{\cos }^2\mathrm{\β}}{T}_{x\max }\end{array}\right.$

In formula: T _x+yit is an intermediate variable;

Be reduced to:

$\left\{\begin{array}{c}{T}_{x\max }\≤T_{w\max }(1+4{\cos }^2\mathrm{\β})\\ T_{x\max }\≤(T_{w\max }-\frac{\sqrt{2}{T}_{y+z}}{4\sin \mathrm{\β}})(\frac{1}{\cos \mathrm{\β}}+4\cos \mathrm{\β})\\ \frac{\sqrt{2}{T}_{y+z}}{4\sin \mathrm{\β}}\≤T_{w\max }\\ 0\≤\mathrm{\β}\≤\frac{\mathrm{\π}}{2}\end{array}\right.$

According to:

T _xmax=T _wmax(1+4cos ²β) exist $\mathrm{\β}\∈\left[\begin{array}{cc}0& \frac{\mathrm{\π}}{2}\end{array}\right]$ Inside the subtraction function about β, $T_{x\max }=(T_{w\max }-\frac{\sqrt{2}{T}_{y+z}}{4\sin \mathrm{\β}})(\frac{1}{\cos \mathrm{\β}}+4\cos \mathrm{\β})$ ? $\mathrm{\β}\∈\left[\begin{array}{cc}0& \frac{\mathrm{\π}}{3}\end{array}\right]$ Inside the increasing function about β, and $\mathrm{\β}\∈\left[\begin{array}{cc}\frac{\mathrm{\π}}{3}& \frac{\mathrm{\π}}{2}\end{array}\right]$ It is inside subtraction function;

The optimum stagger angle β of motorized shaft moment _tvalue be below equation about the solution of β:

$\frac{\sqrt{2}{T}_{y+z}}{4T_{w\max }}=(1-\cos \mathrm{\β})\sin \mathrm{\β}.$

The minimum power consumption of spacecraft described in step 2 is by formula:

$J=\frac{1}{2}{\left(J_w{\stackrel{\·}{\mathrm{\Ω}}}_w\right)}^T\left(J_w{\stackrel{\·}{\mathrm{\Ω}}}_w\right)$

Get minimum acquisition.

In formula, J _wfor the diagonal matrix of each Rotary Inertia of Flywheel composition,

it is the array that each flywheel angular acceleration forms.

In step 3, the maximum torque of the non-motorized shaft of spacecraft is carried out in estimation process, non-arbor is PD controller, and the form of estimation is:

$T_{N\max }=\left[\begin{array}{c}{T}_{y\max }\\ T_{z\max }\end{array}\right]=\left[\begin{array}{c}{K}_{\mathrm{py}}\left|{\mathrm{\θ}}_{\max }\right|+K_{\mathrm{dy}}\left|{\mathrm{\ω}}_{y\max }\right|\\ K_{\mathrm{pz}}\left|{\mathrm{\ψ}}_{\max }\right|+K_{\mathrm{dz}}\left|{\mathrm{\ω}}_{z\max }\right|\end{array}\right]$

Below adopt concrete parameter, implement configuration of the present invention and optimization method:

Steps A, determine that the moment of single flywheel maximum is T _wmax=0.2Nm;

Step B, write out the allocation matrix of satellite;

Step C, according to the value of control algorithm, obtain T _ymaxwith T _zmax: conventionally get:

K _py=K _pz=57.3

θ _max=ψ _max=0.01°=1.7452×10 ^-4rad

K _dy=K _dz=8×57.3=458.4

ω _ymax=ω _zmax=0.005°/s=8.726×10 ^-5rad/s

According to value above and:

$T_{N\max }=\left[\begin{array}{c}{T}_{y\max }\\ T_{z\max }\end{array}\right]=\left[\begin{array}{c}{K}_{\mathrm{py}}\left|{\mathrm{\θ}}_{\max }\right|+K_{\mathrm{dy}}\left|{\mathrm{\ω}}_{y\max }\right|\\ K_{\mathrm{pz}}\left|{\mathrm{\ψ}}_{\max }\right|+K_{\mathrm{dz}}\left|{\mathrm{\ω}}_{z\max }\right|\end{array}\right]$

Obtain T _ymaxwith T _zmaxvalue;

Step D, determine the required maximum torque T of the non-motorized shaft of satellite _ymax=T _zmax=0.05Nm;

Step e, determine the optimum stagger angle of moment, for different configuration modes, resulting result is not identical, and particular case is as follows:

Flywheel configuration of four angle mounts+motorized shaft: this configuration mode need to solve take the equation that β is equation:

$\frac{\sqrt{2}(T_{y\max }+T_{z\max })}{4T_{w\max }}=(1-\cos \mathrm{\β})\sin \mathrm{\β}$

Substitution T _ymax=T _zmax=0.05Nm, T _wmax=0.2Nm, can be in the hope of β _t=42.4 °, T now _xmaxmaxim be 0.635Nm, matrix is installed:

$D=\left[\begin{array}{ccc}0.3146& 0& 0\\ 0.2322& 0.5240& 0.5239\\ 0.2322& -0.5239& 0.5240\\ 0.2322& -0.5239& -0.5239\\ 0.2322& 0.5240& -0.5240\end{array}\right].$

Optimization method in present embodiment makes flywheel can provide maximum moment in motorized shaft direction, thereby improves the acceleration/accel of satellite; The optimization method of present embodiment, is guaranteeing can to control the while to non-motorized shaft, makes the maneuverability of motorized shaft reach optimum, does not ignore the control of non-motorized shaft, is therefore more suitable for real satellite attitude control system.

Claims (2)

1. the optimization method of single shaft fast reserve spacecraft flywheel configuration, the single shaft fast reserve spacecraft flywheel configuration in the method includes five flywheels, and the axis of one of them flywheel and the axis of motorized shaft coincide, and other four flywheels are angle mount flywheel;

The angle of the rotating shaft of each angle mount flywheel and spacecraft maneuver axle is β, and β is real number; The angle of the projection of the rotating shaft of adjacent two angle mount flywheels on the non-plane of maneuver of spacecraft is 90 °, and the angle that is positioned at the projection of each the angle mount flywheel on the non-plane of maneuver of described spacecraft and the angle of spacecraft pitch axis and spacecraft yaw axis is 45 °,

The span of the angle β of the rotating shaft of each angle mount flywheel and spacecraft maneuver axle is:

It is characterized in that: it is realized by following steps:

Step 1, according to the angle β of the rotating shaft of each angle mount flywheel and spacecraft maneuver axle, structure flywheel installation matrix U:

$U=\left[\begin{array}{ccccc}1& \cos \mathrm{\β}& \cos \mathrm{\β}& \cos \mathrm{\β}& \cos \mathrm{\β}\\ 0& \frac{\sqrt{2}}{2}\sin \mathrm{\β}& -\frac{\sqrt{2}}{2}\sin \mathrm{\β}& -\frac{\sqrt{2}}{2}\sin \mathrm{\β}& \frac{\sqrt{2}}{2}\sin \mathrm{\β}\\ 0& \frac{\sqrt{2}}{2}\sin \mathrm{\β}& \frac{\sqrt{2}}{2}\sin \mathrm{\β}& -\frac{\sqrt{2}}{2}\sin \mathrm{\β}& -\frac{\sqrt{2}}{2}\sin \mathrm{\β}\end{array}\right]$

Step 2, the installation matrix U obtaining according to the minimum power consumption of spacecraft and step 1, according to formula:

D=U ^T(UU ^T) ^-1

The allocation matrix D of structure flywheel:

$D=\left[\begin{array}{ccc}\frac{1}{1+4{\cos }^2\mathrm{\β}}& 0& 0\\ \frac{\cos \mathrm{\β}}{1+4{\cos }^2\mathrm{\β}}& \frac{\sqrt{2}}{4\sin \mathrm{\β}}& \frac{\sqrt{2}}{4\sin \mathrm{\β}}\\ \frac{\cos \mathrm{\β}}{1+4{\cos }^2\mathrm{\β}}& \frac{\sqrt{2}}{4\sin \mathrm{\β}}& \frac{\sqrt{2}}{4\sin \mathrm{\β}}\\ \frac{\cos \mathrm{\β}}{1+4{\cos }^2\mathrm{\β}}& -\frac{\sqrt{2}}{4\sin \mathrm{\β}}& \frac{\sqrt{2}}{4\sin \mathrm{\β}}\\ \frac{\cos \mathrm{\β}}{1+4{\cos }^2\mathrm{\β}}& \frac{\sqrt{2}}{4\sin \mathrm{\β}}& -\frac{\sqrt{2}}{4\sin \mathrm{\β}}\end{array}\right];$

Step 3, the maximum torque of the non-motorized shaft of spacecraft is estimated, obtained the required matrix-vector T of the non-motorized shaft of spacecraft _nmax:

$T_{N\max }=\left[\begin{array}{c}{T}_{y\max }\\ T_{z\max }\end{array}\right]=\left[\begin{array}{c}{K}_{\mathrm{py}}\left|{\mathrm{\θ}}_{\max }\right|+K_{\mathrm{dy}}\left|{\mathrm{\ω}}_{y\max }\right|\\ K_{\mathrm{pz}}\left|{\mathrm{\ψ}}_{\max }\right|+K_{\mathrm{dz}}\left|{\mathrm{\ω}}_{z\max }\right|\end{array}\right];$

In formula: T _nmaxfor the required matrix-vector of the non-motorized shaft of spacecraft; T _ymaxfor the required maximum torque of pitch axis in axis of rolling mobile process, T _zmaxfor the required maximum torque of yaw axis in axis of rolling mobile process; K _pyproportionality coefficient for the PD controller of pitch axis; K _dydifferential coefficient for the PD controller of pitch axis; K _pzproportionality coefficient for the PD controller of yaw axis; K _dzdifferential coefficient for the PD controller of yaw axis; θ _maxmaxim for pitch angle in mobile process; ω _ymaxmaxim for pitch axis cireular frequency in mobile process; ψ _maxmaxim for yaw angle in mobile process; ω _zmaxmaxim for yaw axis cireular frequency in mobile process;

Step 4, the required matrix-vector T of the non-motorized shaft of spacecraft obtaining according to the allocation matrix D of the flywheel obtaining in step 2, step 3 _nmaxthe maximum torque that can provide with each flywheel is T _wmaxobtain the optimum stagger angle β of motorized shaft moment _t;

The optimum stagger angle β of step 5, the motorized shaft moment that obtains according to step 4 _t, according to described optimum stagger angle β _tsingle shaft fast reserve spacecraft flywheel configuration is adjusted, thereby realized the optimization of single shaft fast reserve spacecraft flywheel configuration.

2. the optimization method of single shaft fast reserve spacecraft flywheel configuration according to claim 1, is characterized in that the required matrix-vector T of the non-motorized shaft of spacecraft obtaining according to the allocation matrix D of the flywheel obtaining in step 2, step 3 described in step 4 _nmaxthe maximum torque that can provide with each flywheel is T _wmaxobtain the optimum stagger angle β of motorized shaft moment _tconcrete grammar be:

The moment providing according to each flywheel is less than or equal to the principle of the moment upper limit of flywheel, obtains constraint inequality:

DT _cmax≤T _wmax

In formula: T _cmax=[T _xmaxt _ymaxt _zmax] ^t, be the maximum instruction moment that five divided flywheels fit over satellite three axles;

In formula: T _wmaxthe array that the maximum torque value that can provide for each flywheel forms; T _xmaxthe maximum torque that can provide at the axis of rolling for flywheel;

Make T _ymax+ T _zmax=T _y+z, by DT _cmax≤ T _wmaxlaunch, obtain:

$\left\{\begin{array}{c}\max \left\{\right|T_{w2}|...|T_{w5}\left|\right\}\≥\frac{\cos \mathrm{\β}}{1+4{\cos }^2\mathrm{\β}}{T}_{x\max }+\frac{\sqrt{2}(T_{z\max }+T_{y\max })}{\left|4\sin \mathrm{\β}\right|}\\ T_{w\max }\≥\frac{1}{1+4{\cos }^2\mathrm{\β}}{T}_{x\max }\end{array}\right.$

Be reduced to:

$\left\{\begin{array}{c}{T}_{x\max }\≤T_{w\max }(1+4{\cos }^2\mathrm{\β})\\ T_{x\max }\≤(T_{w\max }-\frac{\sqrt{2}{T}_{y+z}}{4\sin \mathrm{\β}})(\frac{1}{\cos \mathrm{\β}}+4\cos \mathrm{\β})\\ \frac{\sqrt{2}{T}_{y+z}}{4\sin \mathrm{\β}}\≤T_{w\max }\\ 0\≤\mathrm{\β}\≤\frac{\mathrm{\π}}{2}\end{array}\right.$

According to:

T _xmax=T _wmax(1+4cos ²β) exist

inside the subtraction function about β, $T_{x\max }=(T_{w\max }-\frac{\sqrt{2}{T}_{y+z}}{4\sin \mathrm{\β}})(\frac{1}{\cos \mathrm{\β}}+4\cos \mathrm{\β})$ ?

inside the increasing function about β, and

it is inside subtraction function;

The optimum stagger angle β of motorized shaft moment _tvalue be below equation about the solution of β:

$\frac{\sqrt{2}{T}_{y+z}}{4T_{w\max }}=(1-\cos \mathrm{\β})\sin \mathrm{\β}.$

Images