CN102616386A - Uniaxial quick maneuverable spacecraft flywheel configuration and optimization method thereof - Google Patents

Abstract

A uniaxial quick manoeuvrable spacecraft flywheel configuration and an optimization method thereof relate to the spacecraft flywheel configuration and the optimization method thereof and aim to solve the problem that the capability of the quick manoeuvrable spacecraft flywheel is not fully utilized. The configuration comprises five flywheels, the axis of one of the flywheels coincides with that of a manoeuvrable shaft, and the other four flywheels are obliquely mounted flywheels; the included angle of the projections of the rotating shafts of two adjacent obliquely mounted flywheels on a non manoeuvrable plane of the spacecraft is 90 degrees; and the included angle between the projection of each flywheel on the non manoeuvrable plane of the spacecraft and pitch axis of the spacecraft, as well as between the projection and the yaw axis of the spacecraft is 45 degrees. According to the optimization method, a mounting angle is a mounting matrix for the flywheels, which is written according to optimization quantities; the power consumption of a flywheel system servers as an indicator to obtain a distribution matrix for the flywheels; and a maximum moment required by a non manoeuvrable shaft is ensured, and the mounting angle of the manoeuvrable shaft with an optimal moment is finally obtained so as to realize optimization through adjustment. The invention is suitable for the spacecraft flywheel configuration and the optimization thereof.

Description

The optimization method of single shaft fast reserve spacecraft flywheel configuration and said flywheel configuration

Technical field

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

Background technology

Present many spacecrafts need be carried out the task of fast reserve and motor-driven back fast and stable, and there is strict restriction the time to the motor-driven specified angle of satellite usually.Under the certain situation of satellite attitude control algorithm; The ability of satellite executing mechanism has determined the motor-driven required time of satellite; Yet the ability of actuating unit, the increase of quantity is corresponding is the increase of satellite quality, and this attitude control to satellite is very disadvantageous.In order under the prerequisite that does not increase actuating unit, to accomplish motor-driven task faster, need carry out reasonable configuration to the actuating unit of satellite usually.

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

Summary of the invention

The present invention utilizes inadequate problem in order to solve fast reserve spacecraft flywheel ability, thereby the optimization method of a kind of single shaft fast reserve spacecraft flywheel configuration and said 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 the angle mount flywheel;

The angle of the rotating shaft of each angle mount flywheel and spacecraft maneuver axle is β, and β is a 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 °, is positioned at the projection of each the angle mount flywheel on the non-plane of maneuver of said spacecraft and the angle of spacecraft pitch axis and the angle of spacecraft yaw axis and 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 rotating shaft of each angle mount flywheel and the angle β of spacecraft maneuver axle, the structure flywheel the 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 that obtains 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 the formula: T _NmaxBe the required matrix-vector of the non-motorized shaft of spacecraft; T _YmaxBe the required maximum torque of pitch axis in axis of rolling mobile process, T _ZmaxBe 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 (getting the departure higher limit of allowance usually) for pitch angle in the mobile process; ω _YmaxMaxim (getting the departure higher limit of allowance usually) for pitch axis cireular frequency in the mobile process; ψ _MaxMaxim (getting the departure higher limit of allowance usually) for yaw angle in the mobile process; ω _ZmaxMaxim (getting the departure higher limit of allowance usually) for yaw axis cireular frequency in the 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 that obtains in the 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 said optimum stagger angle β _TSingle shaft fast reserve spacecraft flywheel configuration is adjusted, thus the optimization of realization single shaft fast reserve spacecraft flywheel configuration.

The maximum torque value T of the non-motorized shaft of spacecraft that obtains according to the allocation matrix D of the flywheel that obtains in the step 2, step 3 described in the 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:

Be less than or equal to the principle of the moment upper limit of flywheel according to the moment that each flywheel provided, obtain the constraint inequality:

DT _cmax≤T _wmax

In the formula: T _Cmax=[T _XmaxT _YmaxT _Zmax] ^T, be the maximum instruction moment that five divided flywheels fit over three on satellite;

In the formula: T _WmaxThe array that the maximum torque value that can provide for each flywheel is formed; The maximum torque value of each flywheel is all identical on the same satellite of normal conditions, and its size is used T _WmaxExpression.T _XmaxThe maximum torque that can provide at the axis of rolling for flywheel;

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

Make T _Ymax+ T _Zmax=T _Y+z, with 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 the formula: T _X+yIt is an intermediate variable;

Be reduced to:

$\left\{\begin{array}{c}{T}_{x\max }\&le;T_{w\max }(1+{4\cos }^2\mathrm{\&beta;})\\ T_{x\max }\&le;(T_{w\max }-\frac{\sqrt{2}{T}_{y+\mathrm{<figure-callout\; id="4"\; label="z"\; filenames="HDA0000148403800000021.png"\; state="\{\{state\}\}">z</figure-callout>}}}{4\sin \mathrm{\&beta;}})(\frac{1}{\cos \mathrm{\&beta;}}+4\cos \mathrm{\&beta;})\\ \frac{\sqrt{2}{T}_{y+\mathrm{<figure-callout\; id="4"\; label="z"\; filenames="HDA0000148403800000021.png"\; state="\{\{state\}\}">z</figure-callout>}}}{4\sin \mathrm{\&beta;}}\&le;T_{w\max }\\ 0\&le;\mathrm{\&beta;}\&le;\frac{\mathrm{\&pi;}}{2}\end{array}\right.$

According to:

T _Xmax=T _Wmax(1+4cos ²β) exist $\mathrm{\&beta;}\&Element;\left[\begin{array}{cc}0& \frac{\mathrm{\&pi;}}{2}\end{array}\right]$ In be subtraction function about β, $T_{x\mathrm{Max}}=(T_{w\mathrm{Max}}-\frac{\sqrt{2}{T}_{y+\mathrm{<figure-callout\; id="4"\; label="z"\; filenames="HDA0000148403800000021.png"\; state="\{\{state\}\}">z</figure-callout>}}}{4\mathrm{Sin}\mathrm{\&beta;}})(\frac{1}{\mathrm{Cos}\mathrm{\&beta;}}+4\mathrm{Cos}\mathrm{\&beta;})$ $\mathrm{\&beta;}\&Element;\left[\begin{array}{cc}0& \frac{\mathrm{\&pi;}}{3}\end{array}\right]$ In be increasing function about β, and $\mathrm{\&beta;}\&Element;\left[\begin{array}{cc}\frac{\mathrm{\&pi;}}{3}& \frac{\mathrm{\&pi;}}{2}\end{array}\right]$ In be subtraction function;

The optimum stagger angle β of motorized shaft moment then _TValue be following equation separating about β:

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

Beneficial effect: the present invention can make full use of fast reserve spacecraft flywheel ability, and flywheel configuration of the present invention has bigger moment space, the spacecraft that is fit to have single shaft fast reserve ability in motorized shaft; Simultaneously, optimization method of the present invention makes flywheel in the motorized shaft direction maximum moment can be provided, 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, therefore more is applicable to the real satellite attitude control system.

Description of drawings

Fig. 1 is the structural representation of flywheel configuration of the present invention; Fig. 2 is the mapping result scheme drawing of Fig. 1 under the O-YZ plane; Fig. 3 is the optimum stagger angle β of motorized shaft moment in the optimization method of the present invention _TThe value principle schematic, the part representative function value that wherein is filled with oblique line can get 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 the angle mount flywheel;

The angle of the rotating shaft of each angle mount flywheel and spacecraft maneuver axle is β, and β is a 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 °, is positioned at the projection of each the angle mount flywheel on the non-plane of maneuver of said spacecraft and the angle of spacecraft pitch axis and the angle of spacecraft yaw axis and is 45 °.

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

The configuration result is as depicted in figs. 1 and 2, and RW is a flywheel among the figure; This embodiment actv. has utilized the ability of flywheel, can guarantee to non-motorized shaft control simultaneously, makes the maneuverability of motorized shaft reach optimum, under the prerequisite of not adding other actuating units, has shortened the spacecraft maneuver required time.And flywheel configuration of the present invention has bigger moment space, the spacecraft that is fit to have single shaft fast reserve ability in motorized shaft.

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

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

Step 2, the installation matrix U that obtains 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:

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

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:

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

In the formula: T _NmaxBe the required matrix-vector of the non-motorized shaft of spacecraft; T _YmaxBe the required maximum torque of pitch axis in axis of rolling mobile process, T _ZmaxBe 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 (getting the departure higher limit of allowance usually) for pitch angle in the mobile process; ω _YmaxMaxim (getting the departure higher limit of allowance usually) for pitch axis cireular frequency in the mobile process; ψ _MaxMaxim (getting the departure higher limit of allowance usually) for yaw angle in the mobile process; ω _ZmaxMaxim (getting the departure higher limit of allowance usually) for yaw axis cireular frequency in the 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 that obtains in the 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 said optimum stagger angle β _TSingle shaft fast reserve spacecraft flywheel configuration is adjusted, thus the optimization of realization single shaft fast reserve spacecraft flywheel configuration.

The maximum torque value T of the non-motorized shaft of spacecraft that obtains according to the allocation matrix D of the flywheel that obtains in the step 2, step 3 described in the 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:

Be less than or equal to the principle of the moment upper limit of flywheel according to the moment that each flywheel provided, obtain the constraint inequality:

DT _cmax≤T _wmax

In the formula: T _Cmax=[T _XmaxT _YmaxT _Zmax] ^T, be the maximum instruction moment that five divided flywheels fit over three on satellite;

In the formula: T _WmaxThe array that the maximum torque value that can provide for each flywheel is formed; The maximum torque value of each flywheel is all identical on the same satellite of normal conditions, and its size is used T _WmaxExpression.T _XmaxThe maximum torque that can provide at the axis of rolling for flywheel;

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

Make T _Ymax+ T _Zmax=T _Y+z, with DT _Cmax≤T _WmaxLaunch, obtain:

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

In the formula: T _X+yIt is an intermediate variable;

Be reduced to:

$\left\{\begin{array}{c}{T}_{x\max }\&le;T_{w\max }(1+{4\cos }^2\mathrm{\&beta;})\\ T_{x\max }\&le;(T_{w\max }-\frac{\sqrt{2}{T}_{y+\mathrm{<figure-callout\; id="4"\; label="z"\; filenames="HDA0000148403800000021.png"\; state="\{\{state\}\}">z</figure-callout>}}}{4\sin \mathrm{\&beta;}})(\frac{1}{\cos \mathrm{\&beta;}}+4\cos \mathrm{\&beta;})\\ \frac{\sqrt{2}{T}_{y+\mathrm{<figure-callout\; id="4"\; label="z"\; filenames="HDA0000148403800000021.png"\; state="\{\{state\}\}">z</figure-callout>}}}{4\sin \mathrm{\&beta;}}\&le;{\mathrm{<figure-callout\; id="0"\; label="T\; w\; max"\; filenames="HDA0000148403800000021.png"\; state="\{\{state\}\}">T</figure-callout>}}_{w\max }\\ 0\&le;\mathrm{\&beta;}\&le;\frac{\mathrm{\&pi;}}{2}\end{array}\right.$

According to:

T _Xmax=T _Wmax(1+4cos ²β) exist $\mathrm{\&beta;}\&Element;\left[\begin{array}{cc}0& \frac{\mathrm{\&pi;}}{2}\end{array}\right]$ In be subtraction function about β, $T_{x\mathrm{Max}}=(T_{w\mathrm{Max}}-\frac{\sqrt{2}{T}_{y+\mathrm{<figure-callout\; id="4"\; label="z"\; filenames="HDA0000148403800000021.png"\; state="\{\{state\}\}">z</figure-callout>}}}{4\mathrm{Sin}\mathrm{\&beta;}})(\frac{1}{\mathrm{Cos}\mathrm{\&beta;}}+4\mathrm{Cos}\mathrm{\&beta;})$ $\mathrm{\&beta;}\&Element;\left[\begin{array}{cc}0& \frac{\mathrm{\&pi;}}{3}\end{array}\right]$ In be increasing function about β, and $\mathrm{\&beta;}\&Element;\left[\begin{array}{cc}\frac{\mathrm{\&pi;}}{3}& \frac{\mathrm{\&pi;}}{2}\end{array}\right]$ In be subtraction function;

The optimum stagger angle β of motorized shaft moment then _TValue be following equation separating about β:

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

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

$J=\frac{1}{2}{\left(J_w{\stackrel{\&CenterDot;}{\mathrm{\&Omega;}}}_w\right)}^T\left(J_w{\stackrel{\&CenterDot;}{\mathrm{\&Omega;}}}_w\right)$

Get minimum acquisition.

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

It is the array that each flywheel angular acceleration is formed.

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

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

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

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

Step B, write out the allocation matrix of satellite;

Step C, obtain T according to the value of control algorithm _YmaxWith T _Zmax: get usually:

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 top win the confidence and:

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

Obtain T _YmaxWith T _ZmaxValue;

Step D, confirm the required maximum torque T of the non-motorized shaft of satellite _Ymax=T _Zmax=0.05Nm;

Step e, confirm the optimum stagger angle of moment, for different configuration modes, resulting result is also inequality, and particular case is following:

Four angle mounts+flywheel configuration of motorized shaft: it is the equality of equation that this configuration mode need be found the solution with β:

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

Substitution T _Ymax=T _Zmax=0.05Nm, T _Wmax=0.2Nm can be in the hope of β _T=42.4 °, the T of this moment _XmaxMaxim be 0.635Nm, matrix then 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 this embodiment makes flywheel in the motorized shaft direction maximum moment can be provided, thereby improves the acceleration/accel of satellite; The optimization method of this 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, therefore more is applicable to the real satellite attitude control system.

Claims (4)

1. single shaft fast reserve spacecraft flywheel configuration, it is characterized in that: 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 the angle mount flywheel;

The angle of the rotating shaft of each angle mount flywheel and spacecraft maneuver axle is β, and β is a 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 °, is positioned at the projection of each the angle mount flywheel on the non-plane of maneuver of said spacecraft and the angle of spacecraft pitch axis and the angle of spacecraft yaw axis and is 45 °.

2. single shaft fast reserve spacecraft flywheel configuration according to claim 1; It is characterized in that the span of the angle β of the rotating shaft of each angle mount flywheel and spacecraft maneuver axle is:

3. the optimization method of the said single shaft fast reserve of claim 1 spacecraft flywheel configuration, it is characterized in that: it is realized by following steps:

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

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

Step 2, the installation matrix U that obtains 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:

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

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:

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

In the formula: T _NmaxBe the required matrix-vector of the non-motorized shaft of spacecraft; T _YmaxBe the required maximum torque of pitch axis in axis of rolling mobile process, T _ZmaxBe 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 the mobile process; ω _YmaxMaxim for pitch axis cireular frequency in the mobile process; ψ _MaxMaxim for yaw angle in the mobile process; ω _ZmaxMaxim for yaw axis cireular frequency in the 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 that obtains in the 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 said optimum stagger angle β _TSingle shaft fast reserve spacecraft flywheel configuration is adjusted, thus the optimization of realization single shaft fast reserve spacecraft flywheel configuration.

4. the optimization method of single shaft fast reserve spacecraft flywheel configuration according to claim 3 is characterized in that described in the step 4 allocation matrix D according to the flywheel that obtains in the step 2, the maximum torque value T of the non-motorized shaft of spacecraft that step 3 obtains _NmaxThe maximum torque that can provide with each flywheel is T _WmaxObtain the optimum stagger angle β of motorized shaft moment _TConcrete grammar be:

Be less than or equal to the principle of the moment upper limit of flywheel according to the moment that each flywheel provided, obtain the constraint inequality:

DT _cmax≤T _wmax

In the formula: T _Cmax=[T _XmaxT _YmaxT _Zmax] ^T, be the maximum instruction moment that five divided flywheels fit over three on satellite;

In the formula: T _WmaxThe array that the maximum torque value that can provide for each flywheel is formed; T _XmaxThe maximum torque that can provide at the axis of rolling for flywheel;

Make T _Ymax+ T _Zmax=T _Y+z, with DT _Cmax≤T _WmaxLaunch, obtain:

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

In the formula: T _X+yIt is an intermediate variable;

Be reduced to:

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

According to:

T _Xmax=T _Wmax(1+4cos ²β) exist $\mathrm{\&beta;}\&Element;\left[\begin{array}{cc}0& \frac{\mathrm{\&pi;}}{2}\end{array}\right]$ In be subtraction function about β, $T_{x\mathrm{Max}}=(T_{w\mathrm{Max}}-\frac{\sqrt{2}{T}_{y+z}}{4\mathrm{Sin}\mathrm{\&beta;}})(\frac{1}{\mathrm{Cos}\mathrm{\&beta;}}+4\mathrm{Cos}\mathrm{\&beta;})$ $\mathrm{\&beta;}\&Element;\left[\begin{array}{cc}0& \frac{\mathrm{\&pi;}}{3}\end{array}\right]$ In be increasing function about β, and $\mathrm{\&beta;}\&Element;\left[\begin{array}{cc}\frac{\mathrm{\&pi;}}{3}& \frac{\mathrm{\&pi;}}{2}\end{array}\right]$ In be subtraction function;

The optimum stagger angle β of motorized shaft moment then _TValue be following equation separating about β:

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

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