JPH10167197A - Attitude control for three-axes satellite by meand of wheel and unloading wheel and spin satellite attitude restoring device by using wheels - Google Patents

Abstract

PROBLEM TO BE SOLVED: To control the position of a three-axes satellite and to stop the precession of a pin satellite to restore the position thereof and to unload a reaction wheel by using wheels without using jet propulsion so as to keep the function of an artificial satellite for a long time. SOLUTION: The position of a three-axes satellite is controlled by producing positive and negative torque of the same direction by a resultant angular momentum vector produced by changing the directions of a pair of wheels 1a, 1b in the opposite directions and the position of a spin satellite is restored by relatively increasing the resultant angular momentum. The reaction of a reaction wheel when it is unloaded is absorbed and stopped by the pair of wheels 1a, 1b having the same moment of inertia and the rotation of the wheels 1a, 1b is stopped by changing the directions of thereof.

Description

DETAILED DESCRIPTION OF THE INVENTION

[0001]

BACKGROUND OF THE INVENTION 1. Field of the Invention The present invention relates to an attitude control of a three-axis satellite based on a combined angular momentum vector of two wheels, an unloading of a reaction wheel by a combined angular momentum, and an attitude recovery apparatus of a spin satellite.

[0002]

2. Description of the Related Art An artificial satellite cannot keep its attitude stable during flight in space due to disturbance torque caused by solar radiation pressure torque, geomagnetic torque, gravitational gradient torque and the like.

[0003] In the case of a three-axis stabilization type satellite, attitude control is performed by using the angular momentum of the momentum wheel and the auction caused by the increase or decrease in the number of revolutions of the reaction wheel. When the rotation speed of the wheel reaches the upper limit and becomes uncontrollable, a gas jet is injected to unload the wheel. Therefore, the attitude control becomes impossible within a short period of time when the installed fuel runs out.

[0004] In the case of a spin-stabilized satellite, precession of the spin axis occurs mainly due to solar radiation pressure torque.
It is controlled by injecting gas. As in the case of the three-axis stabilization system, if the installed fuel runs out, attitude control becomes impossible within a short period of time.

When attitude control is performed by changing the direction of a rotating wheel (Japanese Patent Publication No. 7-2480), when the direction of a rotating wheel is changed, an angular momentum vector operates at a phase of 90 degrees.
It is not possible to obtain control torque in the same direction for the satellite body. In addition, if another rotating wheel is mounted, the angular momentum vector is also buffered at a phase of 90 degrees on the wheel, and the control is complicated and difficult. To return the direction of the wheel by the magnetic torquer, the angular momentum vector is required. Works great, and the torque of the magnetic torquer is a small torque, which takes a long time and is put to practical use.

[0006]

SUMMARY OF THE INVENTION It is an object of the present invention to change the direction of a rotating wheel and to obtain a control torque in the same direction with respect to a main body of a satellite.

[0007] Unloading a wheel without using gas jet injection

An object of the present invention is to stop the precession motion of the spin satellite without using the gas jet injection and to restore the attitude, and the purpose is to continue the function of the artificial satellite for a long time.

[0009]

A pair of two wheels having the same angular momentum and changing the directions of the two wheels in a symmetrical direction so that the direction of the resultant angular momentum vector is the same axis direction, is the main body of the satellite. , A forward and reverse control torque in the same direction is obtained.

The two wheels having the same moment of inertia and the rotation axis of the reaction wheel are made parallel to each other to absorb the reaction at the time of unloading of the reaction wheel by the two wheels to stop the rotation of the reaction wheel. Change the rotation direction to the opposite direction,
Stop wheel rotation.

A pair of two wheels having the same angular momentum is used as a pair to change the directions of the two wheels in a symmetrical direction so that the angular momentum is relatively increased to restore the attitude of the spin satellite.

[0012]

BEST MODE FOR CARRYING OUT THE INVENTION "Rigid body theory by centrifugal force" proposed by the inventor will be described. Conventionally, the precession of a rotating rigid body has been explained by the following equation using angular momentum and vector representation. IωΩ = WL (I = Mr ² ) where M = inertial mass of rigid body, W = rigid body , R = radius of inertia of rigid body, ω = angular velocity of rigid body, Ω = angular velocity of precession of rigid body, L =
According to this equation, when Iω and W are constant, if L is increased or Ω is increased by applying an external force, the moment of IωΩ acts as large as possible, but it cannot occur in reality. This theory analyzes the centrifugal force and Mr ² ω ² · sin θ = WL · sin α where α = inclination angle of the rotation axis with respect to vertical, θ = deviation angle (pseudonym, explained in this paper) This is for explaining the characteristics. "Introduction""Momentary centerline of rotation during linear movement" Radius r as shown in Fig. 1
When rigid outer circumference velocity v is linearly moved at a speed V ₁ while rotating in the arrow direction, provided that contact point E is the circle of the time line AB and the radius r ₁ of the peripheral velocity V ₁ radius r ₁ instantaneous center point of rotation And the instantaneous center line (vertical direction) of the rotating body. "Instantaneous center line of rotation during precession" when rigid as in FIG. 2 precess at a speed V ₁ while rotating in the arrow direction, provided that the radius r ₁ peripheral speed V ₁ when the radius r ₁ The point E where the circle touches is
As in FIG. 1, the speed is zero at the instantaneous center point of the rotation and the line AE
Is the instantaneous center line of the rotating body. "Theory""Motion direction of mass during precession" As shown in Fig. 2, when a rigid body precesses, each mass acts in the direction of the tangent of the rotation direction of the rigid body and the tangent of the precession direction. Its tangent is perpendicular to the instantaneous center line of rotation. Proof In FIG. 3, E is the instantaneous center point of rotation G is the center of gravity of the rigid body BDC = right angle DBC = d A is the fulcrum of the rotation axis AG is the rotation axis Angle CGD = br ₀ = Distance of GC Angle A'AG = distance perpendicular _{AB = r 0 · cosb AB =} GD AG = BD rotation axis of length L = AG distance line A'B 'and E'D' parallel ray lines E'D 'passes the rotation center of gravity of the rigid body The center line precession speed perpendicular to the axis is V, the angular speed around the rotation axis is ω, the rotation speed of the mass point of radius r ₀ is v _{0, the} deviation angle (intersection angle between the rotation axis and the instantaneous center line, angle EAG pseudonym) is θ When the distance between the instantaneous center point E of rotation and the center point G of rotation is r, the rotational speed of the mass point C in the side view parallel to E′D ′ is v _0.
・ Sinb The precession speed of the mass point C in the plan view is V · secd The precession speed parallel to the rotation axis is V · secd · sind = V · tan
d Front view is a composite diagram of v ₀ · sinb and Vtand.
When the angle GCF is x, v ₀ · sin btan x = V tan ────────────── equation v ₀ = r ₀ · ω ────────────── equation V = rω ────────────── assignment _{tanx = Vtand / v 0 · sinb} , the the equation _{tanx = rωtand / r 0 · ωsinb} = rtand / r 0 · sinb ───── ───────── Equation r = L tan θ を Substituting into the equation tanx = L tan θ tan / r ₀ · sinb tan ─ since an expression Ltand = r ₀ · sinb (see FIG. 3) is therefore tanx = tan .theta, intersection angles between the instantaneous center line AE and rotation mass point of the tangent can be proved is perpendicular亊. "Direction of centrifugal force during precession" The direction of centrifugal force is considered to act in the direction of motion of the mass point from the center of rotation. When the rigid body rotates and precesses, the centrifugal force direction of the mass point acts perpendicular to the instantaneous center line of rotation and can be considered to be inclined by the deviation angle from the rotational direction of the rigid body around the rotation axis. Definition of inertial mass] When the center of a homogeneous disk having a radius R is rotating around a fixed axis, the radius of inertia r is given by πR ² -2πr ² = 0.

(Equation 1) Let dm be the mass of the mass point of the homogeneous disk, and define the inertial mass M as M = {dm} (Definition 2). Definitions 1 and 2, the sum of centrifugal forces F is F = Mrω ² (ω = angular velocity) “Generation of deviation moment” When a rigid body precesses, the torsional force of the rigid body is pseudonymously referred to as the deviation moment. The angle of intersection (deviation angle) between the instantaneous center line of rotation during precession and the rotation axis is θ. The direction of centrifugal force of the rotating body has an inclination of θ with respect to the rotating body.Each mass point exerts a force in the direction of centrifugal force, and a force acts to tilt the rotating body in the direction of centrifugal force. total Mromega ² moment produce and moments lambda a Mrω ² · sinθ multiplied by sin [theta lambda, the center ^{Λ = Mrω 2 · sinθ · r} = Mr 2 ω 2 · multiplied by the distance r from the sin [theta rotational this moment lambda Is the deviation moment. Since the instantaneous center line of the rotation is located directly above the rotation axis, the direction of the deviation moment is in the antigravity direction (negative direction) with respect to the fulcrum. When a rigid body of gravity W precesses at a tilt angle of α with respect to the vertical axis on the rotation axis of length L, the moment of WL · sinα acts on the rigid body in the direction of gravity (positive direction) at the fulcrum due to gravity. .
It is considered that this moment is equal to the deviation moment.
"Relationship between angular momentum and equation derived by vector representation" The distance between the center point of rotation of the rigid body and the instantaneous center point of rotation is r ₀ , the angular velocity of the rigid body is ω, the distance from the fulcrum to the center of gravity of the rigid body is L, The angle between the rotation axis and the vertical line is α, the radius of inertia is r, and the deviation angle is θ. Mr ² ω ² · sin θ = WL · sin α and the precession velocity V is V = r ₀ · ω ──────── (a) r ₀ = Ltan θ ──────── (b) From equations a and b, V = Lω · tanθ The precession angular velocity Ω is Ω = Lω · tanθ / Lsinα = ω · tanθ / sinα ───────────── formula From the above formula, ω = Ω · sinα / tanθ───────────── From the formula, Mr ² ωΩ · ^{sinαsinθ / tanθ = Mr 2 ωΩ ·} sinαcosθ = WL · sinα both sides of the divided by ^{sinα Mr 2 ωΩ · cosθ = WL} IωΩ · cosθ = WL (I = Mr 2) ─────── the equation equation

(Equation 2) In, the angular velocity of Ω is very small compared to the angular velocity of ω, and the angle θ is a very small angle, ignoring the angular velocity of Ω, and considering cos θ as 1. In the formula, the angular momentum around Ω is very small compared to the angular momentum around ω, and is consistent with the conventional formula, taking into account that it is derived by ignoring the angular momentum around Ω. .
However, since each mass point moves at right angles to the instantaneous center line of rotation, the angular momentum of Iω can be considered as the angular momentum about the instantaneous center line of rotation. Assuming that the deviation angle is θ, the angular momentum and the angular momentum about the rotation axis derived by the vector display should be multiplied by Iω by cosθ, and when multiplied, it is in agreement with the equation. It can be considered that Mr ² ω ² · sin θ = WL · sin α in the formula is a calculation formula based on centrifugal force during precession. From the "limit of deviation moment" formula

(Equation 3)

(Equation 4) "Deviation moment and vector" The y-axis in FIG.
The x-axis is a horizontal line, G is the center of gravity of the rigid pair, W is the weight, the line AG is the rotation axis, the distance between AGs is L, the intersection A of the y-axis and the x-axis is the fulcrum during precession, the y-axis and rotation When the angle with the axis AG is α, Mr ² ω ² · sin θ = WL · sin α

(Equation 5)

(Equation 6) Vector AD of resultant force of vector AB and vector AC
Is

(Equation 7) The reaction force of vector AB is vector GS, the reaction force of vector AC is vector GT, vector GS, and the resultant force of vector GT is vector GU.

(Equation 8) [Conclusion] When the direction of the rotating rigid body is changed, the instantaneous center line of rotation intersects with the rotational center line of the rigid body, and the intersection angle (deviation angle) generates a deviation moment. Since the centrifugal force acts in a direction perpendicular to the instantaneous center line of rotation, the deviation moment produces a phase of 90 degrees from the direction in which the force was run. The reason that the rotating rigid body changes its fulcrum and continues precession is that the force of statically changing the direction is exerted by the gravity of the rigid body etc., the precession occurs by 90 degrees phase, and the precession It is assumed that the deviation moment acts in the negative direction and the moment in the positive direction acting on the fulcrum by gravity is balanced, and precession continues.
Since the deviation moment works to reduce the deviation angle, it is considered that the function of the deviation moment is not to change the direction when the rotating rigid body is supported by the center of gravity. The characteristic of the rotating rigid body motion is considered to be due to the deviation moment. "After all" In the early stage, when the rotating top makes a conical turning motion at the top of the vertex, it rises in the vertical direction because of the friction with the rounded rotating shaft of the ground contact surface The deviation moment in the negative direction is larger than the moment in the negative direction due to the gravity of the top, and the vertical rises. When turning right while supporting the end of the rotating shaft of the rigid body that rotates to the right, the instantaneous center line of rotation is made inward, and the outer working moment caused by the centrifugal force due to turning and the inner knitting difference moment balance inward. Stable turning is possible, but when turning to the right while turning left, the knitting moment acts on the outside and stable turning is not possible. In figure skating on ice, jumping while turning to the left while turning to the right makes landing difficult because the moment of deviation acts outward in the moment of landing. “References” published by Kogakusha Applied Mechanics “Dynamics” S. Tymoshenko / D. H. Written by Young ■ Shigeru Watanabe and Hirofumi Miura As described above, this theory determines that the moment of deviation (hereinafter referred to as angular momentum vector) acts in a direction perpendicular to the instantaneous center line of rotation.

A z-axis is provided in the vertical direction as an operation for obtaining a control torque in the same axis direction with respect to the main body of the artificial satellite, and the rotation axes of the two wheels are symmetrical with respect to the Y axis at right angles to the Y axis and in the X axis direction. And the intersection angle between the rotation axis and the X axis is α,
(α = 0 degrees) Both rotate the wheel in the + Y + Z -Y -Z direction and make the angular momentum the same. + X side wheel is + X + Z -X -Z -X side wheel is the same around -X + Z + X -Z When the angle α is changed from 0 to 90 to 180 to 270 to 360 degrees at angular velocity and the angular momentum vector when the direction of one set of wheels changes is M, the angular momentum vector is 90 degrees from the direction in which the force is applied. 2M · sinα around the late X axis working in phase
To obtain the composite angular momentum vector of. α is 360 to 27
When the direction is changed from 0 to 180 to 90 to 0 degrees, -2M
The angular momentum vector of sinα acts around the X axis. The angular momentum vectors around the Z axis work in opposite directions to cancel each other, and the vectors around the Y axis cancel each other because they change the direction of the wheel in opposite directions. Applying this effect, a control torque in the positive and negative directions is obtained for the artificial satellite body.
Embodiments of the present invention will be described based on examples with reference to the drawings. In FIG. 6A, the gimbal rotation shafts 4a, 4b
Is parallel to the y-axis, the directions of the rotation axes 2a and 2b of the wheels are parallel to the Z-axis, the wheel 1a is in the + x-y direction, the wheel 1b is in the -x-y direction, and the same predetermined angular momentum. Spinning at. Gimbal clutch 5a, 5b
And the gimbal drive motor 6a and the gimbal shaft 4a,
The gimbal drive motor 6b and the gimbal shaft 4b are linked. When changing the direction of the wheel at the same angular velocity in the -x + z direction by the gimbal drive motor 6a and in the + x + z direction by the gimbal drive motor 6b, the torque in the + z -y direction around the x axis until the rotation axis becomes parallel to the x axis. Get. From (a) of FIG. 6 to (b) of FIG.
The operation by the gimbal drive motor is referred to as operation 1.
When the rotation axis is not parallel to the z-axis, the gimbal clutches 5a and 5b are released and the bracket 8a and the gimbal 7a, and the bracket 8b and the gimbal 7
The bracket drive motor 10a that supports b is + x
In the −y direction, the bracket driving motor 10b is −x −
By rotating the brackets 8a and 8b around the support axis at the same angular velocity in the y direction, the direction of the wheel is changed and the rotation axis is made parallel to the z axis. 6 (b) to FIG. 6 (c)
The operation from the operation 1 to the bracket drive motor up to this point is referred to as operation 2. When the direction of the wheel is changed by the operation 2, the angular momentum vector acts in the direction perpendicular to the instantaneous center line of rotation from the rigid body theory "generation of deviation moment", so the direction of the wheel is around the gimbal axis,
The rotation direction of the wheel is the rotation direction of the bracket. FIG. 7A is a front view of one set of the wheel of the present invention, and FIG. 7B is a plan view. 7, the angle of intersection of the direction of the wheel rotation axis 2a and the x axis with respect to a plane around the z axis is α _z , the angle of intersection of the direction of the gimbal rotation axis 4a and the x axis is α _y , and the bracket drive motor of operation 2 the 10a by torque T _m, with T _m · a wheel rotation shaft 2a direction when
The torque of cosα _z is applied. The direction of the wheel 1a is changed by this torque, and the angular momentum vector acts with the same force in the direction of the gimbal axis 4a in a phase of 90 degrees, and T _m ·
A set of torques of cosα _z · sin α _y works. The set of torques by the gimbal drive motor 6a in operation 1 is represented by T _m ,
And two combined torques in the + z-y direction around the x-axis
2 · T _m · sinα _y works, + z around x axis by operation 2
Two combined torques in the + y direction 2 · T _m · cosα _z · sin α _y
The bracket 8 is controlled by the bracket drive motors 10a and 10b which perform control around the x-axis by acting more.
Initially, a and 8b are set to the state shown in FIG. 6A, and the gimbal clutches 5a and 5b are closed to prepare for the next control. In the initial state, if the rotation directions of the gimbal drive motors 6a and 6b are reversed, the torque around the x axis acts in the opposite direction. Or,
If the gimbal shafts 4a and 4b are made parallel to the x-axis and the gimbal drive motors 6a and 6b are rotated in a symmetrical direction, a torque around the y-axis can be obtained. Especially enemy to satellites

In FIG. 8A, reference numeral 14 designates x
This is a wheel for unloading in the axial direction (hereinafter referred to as wheel 14). The gimbal clutches 5a and 5b are closed in a state where the rotation of the wheels 1a and 1b having the same moment of inertia is stopped, and the directions of the wheel rotation shafts 2a and 2b are parallel to the x-axis and the gimbal shaft 4
The directions of a and 4b are parallel to the y-axis. Wheel 14
The reaction torque is generated by lowering the rotation speed of the motor. The reaction torque is increased by the wheel 1a or 1b (or simultaneously) in the rotation direction of the wheel 14, and the wheel 1a or 1b is rotated.
The reaction torque is absorbed until the rotation of 4 stops. Next, the wheel 1a, 1b is lowered by the wheel drive motors 3a, 3b by a known technique so as not to cause a reaction torque, and the lower speed is increased by the wheel drive motors 3a, 3b, and the same speed is achieved. , 6b, the directions of the wheels are changed in the opposite directions at the same angular velocity to make the wheel rotation axes 2a, 2b parallel to the z-axis (at this time, the torque acting around the x-axis is controlled by the wheel 14). The gimbal clutches 5a and 5b are released to release the bracket 8a, the gimbal 7a and the bracket 8b.
And the gimbal 7b, and the bracket drive motor 10
If the directions of the wheels are changed in the opposite directions at the same angular velocity by a and 10b to make the wheel rotation shafts 2a and 2b parallel to the z-axis, the torque acting around the x-axis becomes small (from FIG. 8A). 8 (b)). Next, the rotation speeds of the wheels 1a and 1b are lowered and stopped so that the reductions in the rotation speeds become the same, and the gimbal drive motors 6a and 6b return to the initial state, and return to FIG. 8A to prepare for the next unloading. .

When the spin satellite precesses, the disturbance torque is T _D , the moment of inertia around the spin axis is I _S , the angular velocity of the spin is ω _S , and the deviation angle in the rigid body theory is θs. She is exercising in the relationship of _S · ω _S ² · sin θs = T _D.
Since the angular momentum vector acts in a direction perpendicular to the instantaneous center line of rotation, when the above angular momentum vector I _S · ω _S ² · sin θ acts more than the disturbance torque T _D , the precession stops. When the moment of inertia about the rotation axis of the wheel is I _H , the angular velocity of the wheel is ω _H , and the intersection angle between the direction of the wheel rotation axis and the assumed X axis is α, the angular momentum about the X axis is
I _H · ω _H · cos α. Assuming the X axis as the spin axis,
The attitude recovery of the spin satellite is performed by adding the angular momentum I _S · ω _S of the spin of the satellite and the combined angular momentum 2 · I _H · ω _H · cos α of the two wheels to increase the angular momentum about the spin axis.
In FIG. 9A, reference numeral 15 denotes a spin portion of a spin satellite, 1
Reference numeral 6 denotes a platform, 17 denotes a spin axis, and when the spin unit 15 spins in the + x-y direction and performs precession about the z-axis, the gimbal clutches 5a and 5b are closed, and the wheel rotation axis is set. The directions of 2a and 2b and the directions of the gimbal rotation axes 4a and 4b are parallel and perpendicular to the spin axis 17, and the wheels 1a and 1b are rotated at predetermined rotation speeds so that the increase in angular momentum is the same in the opposite rotation directions. I do. The wheels 1a, 1 are driven by the gimbal drive motors 6a, 6b.
The direction is changed at a predetermined angular velocity so that the rotation direction of b becomes the rotation direction of the spin unit 15. When the direction is changed, the angular momentum vectors of the wheels 1a and 1b act in the opposite direction of the spin of the spin unit 15, and the home plate 16 defines the angular momentum around the spin that starts to rotate in the opposite direction of the spin as a positive angular momentum. Then, the angular momentum of the home plate 16 is a negative angular momentum. However, as long as the angular velocity that changes the direction of the wheels 1a and 1b does not exceed the angular velocity at which the angular momentum vectors are balanced, from the rigid body theory “limit of deviation moment” described above. Negative angular momentum cannot be larger than positive combined angular momentum. In order not to change the direction of the home plate 16, the rotation of the spin unit 15 is slowed down by a known technique. During this operation
The relative angular momentum about the spin axis 17 is the wheel 1a, 1
b, the angular momentum vector due to precession is larger than the disturbance torque, and its direction acts in the z-axis direction with the center of gravity of the satellite body as a fulcrum. As a result, the precession subsides, and the attitude recovery of the spin satellite is achieved.
((B) of FIG. 9) It is also effective in restoring the posture of the nutation movement. Gimbal drive motors 6a, 6 when precession has subsided
b, the directions of the wheels 1a, 1b are changed at the same angular velocity in the initial rotation direction of the wheels 1a, 1b so that the wheel rotation axes 2a, 2b are parallel and perpendicular to the spin axis 17. When returning this direction, the angular momentum vector acts in the direction of spin rotation of the spin unit 15. In order not to change the direction of the home plate 16 and to return to the original number of spins, the rotation of the spin unit 15 is reduced by a known technique. Be prepared for the next attitude control. If the clutches 5a and 5b are released and the brackets are returned to the initial state by the bracket driving motors 10a and 10b, the torque around the spin decreases.

As shown in FIG. 9A, the wheel rotation axis 2 is restored as the initial state for restoring the attitude of the precession motion of the spin satellite.
When the directions of a, 2b become the z-axis direction, the wheels 1a,
When the rotational speed of 1b is increased or decreased and torque is applied in the direction of precession about the center of gravity of the satellite as a fulcrum, the angular momentum vector has a phase of 90 degrees, the spin axis is shifted in the z-axis direction, and the attitude recovery of the spin satellite is achieved. Is done.

[0017]

【The invention's effect】

(Common effects: All the above-mentioned embodiments) 1: Operation by electric power, and electric power can be supplied by a solar cell. 2: The mass can be reduced by rotating the wheel at high speed. 3: By making the operation faster, the time of the posture control can be shortened. 4: Since there is no need for gas injection, the function of the satellite can be extended. 5: Since it is not necessary to mount fuel for attitude control, the mass of the satellite is reduced. 6: Unlike the control of the gas injection, the change of the trajectory and the vibration do not occur.

Effect of attitude control device using angular momentum vector 1: When fixed in parallel with the rotation axis, angular momentum vectors cancel each other even if the direction is changed in all directions. Attitude control becomes easy.

Effect of posture restoring device at the time of precession exercise 1: The posture can be restored without changing the direction of the platform. 2: The rotation speed of the spin part can be changed by using the reaction torque due to the increase / decrease of the rotation speed of the wheel.

[Brief description of the drawings]

FIG. 1 is a diagram of a rigid body moving linearly.

Fig. 2 Precession of a rigid body

FIG. 3 is an explanatory view of a diagram during a precession motion of a rigid body.

FIG. 4 is a vector diagram of a deviation moment.

FIG. 5 is a perspective view of the present invention.

FIG. 6 is an explanatory diagram of a control torque based on an angular momentum vector.

FIG. 7 is a diagram showing the inclination of a wheel.

FIG. 8 is an explanatory view of unloading of a wheel.

FIG. 9 is an explanatory diagram of attitude recovery of a spin satellite.

[Explanation of symbols]

1a and 1b are wheels, 2a and 2b are wheel rotation axes, 3a and 3b are wheel drive motors,
4a and 4b are gimbal shafts, 5a and 5b are gimbal clutches, 6a and 6b are gimbal drive motors 7a and 7b are gimbal, 8a and 8b are brackets, 9a and 9b are bracket shafts, and 10
a and 10b are bracket drive motors 11a and 11b are gimbal bearings, 12a and 12b are wheel rotation bearings, 13 is a satellite main body,
14 is a wheel to unload 15 is a spin part of a spin satellite 16 is a platform of a spin satellite 17 is a spin axis A is a direction of a wheel rotation axis B is a direction of a gimbal rotation axis C is a direction of a spin axis D is a direction of a precession x , Y, z are the coordinate axes α _y is the intersection angle between the direction of the gimbal rotation axis and the x axis α _z is the intersection angle between the direction of the rotation axis of the wheel and the plane of the x axis around the z axis

Claims (3)

[Claims] 1. Wheels 1a and 1b having a predetermined amount of angular momentum substantially symmetrically mounted on a surface thereof are mounted on gimbals 7a and 7b, and brackets 8a and 8b are provided via clutches 5a and 5b.
Gimbal 7a, 7b, and gimbal drive motors 6a, 6b and bracket drive motors 10a, 10b.
(b) means for rotating the gimbal (7a, 7b) around a support axis in a symmetrical direction by means of a b. 2. Wheels 1a, 1b having rotating shafts 2a, 2b substantially parallel and having a predetermined moment of inertia are mounted on gimbals 7a, 7b, and the gimbals 7a, 7b are turned around symmetrical support axes. An unloading device for a wheel by a wheel, comprising a means for moving the wheel. 3. A wheel 1a, 1b having rotating shafts 2a, 2b substantially parallel and having a predetermined amount of angular momentum
An attitude restoring device for a wheeled spin satellite, comprising means mounted on a and b for rotating the gimbals 7a and 7b around support axes in symmetric directions.