1.Introduction The asteroid explorer HAYABUSA was launched on May,9,2003. Along the way, HAYABUSA has lost two of three wheels, its attitude is currently stabilized by only one-axis bias-momentum wheel. Based on an ideal theory, such satellites are stable for 3 axes since one axis is stabilized through exchange of angular momentum with the wheel under a PD control, and the other two axes are stabilized by the gyroscopic stiffness derived from the momentum of the wheel. HAYABUSA had been regarded as such an ideal satellite as well. However, with a little products of inertia caused by the fuel leakage, the wheel axes are different from the principal axes, and thus the dynamics differs from ideal ones. Because of this misaligment, a nutational divergence happened to HAYABUSA in March 2007. This paper analyzes the cause of this phenomenon and discuss the influences of such misalignment upon a bias-momentum one wheel satellite like HAYABUSA. 2.Theory 2-1 Equation of motion To analyze the dynamics of HAYABUSA, consider a typical bias-momentum satellite with only one momentum wheel. We define an inertial reference frame [a0] and a body-fixed reference frame [a1] and a principal axes-fixed reference frame [a2]. A direction cosine vector of the wheel axis which expresses a difference with the principal axis is described by T = (1, 2, 3) in [a2](Fig.1). We suppose the wheel is pointed at around z-axis of [a2]. Fig.1: Tilt of the wheel from the principal axes-fixed frame Consider Aij,ij,ij as a direction cosine matrix, an angular velocity vector and an angular vector, respectively. They are described in [ai] relative to [aj]. The whole angular momentum vector L in [a1] is written as L = J110 + Hw (1) where J1 is the inertia matrix in [a1] and Hw is the angular momentum vector of the wheel. For simplicity, neglecting disturbance torque, Euler equation is written as J1? 10 + ? 10(J110 + Hw) = ! ?H w (2) where ?Hw is the control torque and ? x is a matrix describing cross product x. We use rotation vector to express the attitude. Using e as a direction cosine of Eulerfs Eigenaxis and 0 E E as rotation angle about the axis, rotation vector and its time derivative is defined as1) T = (x, y, z) = eT (3) ? = + 1 2 ? + ?2 = 2 sin 2 ! cos 2 22 cos 2 1 12 E E 1 2 (0 E E ) (4) Eqs.(2)(4) is written as J1? 10 + ? 10(J110 + Hw) = ! ?Hw (5) ? 10 = 10 + 1 2 ? 1010 + ?2 1010 (6) in [a1]. In general, we can use below equations for transforms among reference frames. A21 J1A12 = J2 A2110 + 21 = 20 A2110 + 21 = 20 A21 ? xA12 = gA21x (7) Since the definitions about [a1], [a2] and , J2 is a principal inertia matrix which consists of principal moments of inertia Jx, Jy, Jz and we can write 21 = 0 ? 21 = 0 (8) A21Hw = h A21 ?Hw = ?h (9) where h is the magnitude of Hw. By combining A21 and Eqs.(5)(6), and using above equations, we obtain J2? 20 + ? 20(J220 + h) = !?h (10) d dt (20 ! 21) =20 + 1 2 (20 ! 21) 20 +[(20 ! 21) f(20 ! 21) 20g] (11) 2-2 Control law Consider a control law of the wheel rotation. The control law of HAYABUSA is PD control using angular velocity and angle about z-axis of [a1]. That is to say, the feedback values for the PD control are the z-component of the angular velocity and the angle observed in [a1], or 10z, 10z. We can write the magnitude of control torque ?h = j ?Hwj as ?h = k110z + k210z (12) where k1, k2 are the feedback gains. We define a vector T = (1, 2, 3) such as A12 = ( ? ? ) (13) This describes products of inertia in [a1]. Using and , we can express the tilt of the wheel (or wheel array error) and the products of inertia independently(Fig.2). With above equations about transform of the references, ?h is written as ?h = k1T20 + k2T(20 ! 21) (14) Fig.2: Description of the tilt and the products of inertia in the principal axes-fixed frame 2-3 Characteristic equation To obtain the characteristic equation, we linearize Eqs.(10)(11) by restricting a range of motion properly. Defining J = J2 = 20 = 20 ! 21 (15) we obtain the below system of differential equations. J? + ? (J + h) = !k1ʃT ! k2ʃT (16) ? = + 1 2 ? + ?2 (17) Consider restricting the range of angular velocity, angle and the wheel rotation as near its zero, a certain angle 0 and bias angular momentum h0, respectively. Namely we suppose = ƒ = 0 + ƒ h = h0 + h (18) Substituting to Eqs.(16)(17) and neglecting 2 terms, we obtain the below system of linear differential equations about ,. J? + h0? = !k1ʃT ! k2ʃT (19) ? = + 1 2 ? 0 + 0 ?0 2 (20) Its matrix expression is d dt ( ) = ( !J!1(k1ʃT ! h?) !J!1k2ʃT E + 12 ?0 + ?0 2 0 ) (21) Finally, we obtain the characteristic equation det ( sE + J!1(k1ʃT ! h?) J!1k2ʃT !E ! 12 ?0 ! ?0 2 sE ) = 0 (22) where E is a unit matrix. 3.Stability analysis of the system In this section, we analyze the stability of the system (,) by eigenvalue analysis using Eq.(22). 3-1 Conditions about stability Though Eq.(22) is a sixth-order equation about s, it has double root at s = 0. Its eigenvectors are xT 1 = ( 0 0 0 3 0 !1 ) xT 2 = ( 0 0 0 2 !1 0 ) (23) which means some deviation about angle remain and the double root do not affect the stability. We can consider the stability by analysis of the other 4 eigenvalues. The characteristic equation is now fourth-order like s4 + a1s3 + a2s2 + a3s + a4 = 0 (24) Using Hurwitz criterion, we can say all real parts of the eigenvalues are negative if and only if2) ai > 0 (25) and 3 = det ? ? a1 a3 0 1 a2 a4 0 a1 a3 ? ? > 0 (26) 3-2 Approximation by supposing magnitudes It is quite hard to analyze the sign of ai, 3 because actual expanded forms are quite complex and large with some characteristic constants and the angle 0, where characteristic constants mean J, , , k1, k2 and h0. So we suppose the magnitudes of these constants and extract only dominant terms in ai and 3. When the tilt of the wheel and the products of inertia is small, and is approximated as T = ( 1 2 1 ) T = ( 1 2 1 ) (27) where 1, 2, 1 and 2 is also approximated small. Besides we suppose j0j < 1[rad]. Based on at Eq.(4) as the basis of the magnitudes, we suppose i ? i i ? i h0 Ji ? k1 ? Ji k2 ? h0 (28) and neglect terms which are less or equal to 3i . With these suppositions, ai is always positive. Meanwhile, 3 is not necessarily positive. It is written as first-order form about T 0 = (0x, 0y, 0z) like 3 = A0x + B0y + C0z + D (29) where A = h0k1k2 2Jy Jz [ xz21 + yz 1 Jx 2 ] B = ! h0k1k2 2Jx Jz [ yz22 + xz 1 Jy 1 ] C = ! h0k1k2 2Jz [ xz2 Jy 21 + yz2 Jx 22 + h0k1 Jx Jy Jz xy12 ] D = k1h01 Jz [ h30 k11 J2 x J2 y xz + ( ! h20 k2 1 J2 x Jy Jz + h20 k2 J2 x Jy ! k2 2 Jx Jz ) yz2 ] + k1h02 Jz [ h30 k12 J2 x J2 y yz + ( h20 k2 1 Jx J2 y Jz ! h20 k2 Jx J2 y + k2 2 Jy Jz ) xz1 ] ! h20 2 1 Jx Jy [ xz Jy 21 + yz Jx 22 ] + h0k1yx12 J2 z [ Jz2 1 Jx Jy ! k22 ] (30) where ij = 1 Ji ! 1 Jj 1 = h0k1 Jz 2 = k2 Jz ! h20 Jx Jy (31) We can not determine the sign of 3 unique, which means a boundary about the stability exists depending on the constants in Eq.(29). Some examples of the boundaries are shown at Fig.3. We describe them only on a plane (0x, 0y) approximately, because C is smaller than A and B. The boundary varies its shape depending on the signs of A and B. Based on above analysis, a system (or an attitude) of certain satellite with and may go into stable or unstable depending on its 0 and it may converge or diverge. We can say that the cause of these phenomena is the composition of the feedback value. It includes x and yangle (see Eq.(14)) which does not relate to control low if the satellite does not have some misalignment i, i, and this results changing the dynamics of the system. The biased feedback value about angle takes place these boundaries. So we can also say that the way of the definition of 0 ,or the definition of inertial frame varies the dynamics. Fig.3: Examples of boundary about stability 4.Influence of the stability upon the motion 4-1 Analysis of the divergent motion of HAYABUSA Based on the above stability analysis, we investigate the divergent motion of HAYABUSA shown in Section 1. When this phenomenon happened, HAYABUSAfs characteristic constants were J = ? ? 352.4 0 0 0 268.2 0 0 0 428.3 ? ?[kgm2] h0 = !2.90[kgm2/s] k1 = 114 k2 = 15.35 T = ( 0.0823 !0.0100 0.9966 ) T = ( 0.0823 !0.0100 0.9966 ) (32) Here, = means the wheel axis is equal to the z-axis of [a1]. And it happened after an attitude maneuver about the x-axis, its rotation vector was 0 = ( 0.393 0.021 0.000 ) [rad] (33) In this settings, 3 in Eq.(29) is calculated as 3 = A0x + B0y + C0z + D = !3.34 10!80x + 6.37 10!90y@ + 2.81 10!90z + 1.58 10!13 = !1.30 10!8 (34) which means the boundary shape is like upper right of Fig.3, and the angle (0x, 0y) = (0.393, 0.021) is placed in the unstable region. We can say this is why the unstable nutation divergence occurred. To compare a numerical calculation using Eqs.(16)(17)(32)(33) with HAYABUSAfs angle record, we show the result of the calculation at Fig.4. The uppers of Fig.4 is the records, the lower is the calculation. These graph consist of time biased to zero when the divergence occurred for the transverses and z-angle described by rotation vector for verticals. Compared with the record, the calculation corresponds well. -0.4 -0.35 -0.3 -0.25 -0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0 50000 100000 150000 200000 250000 angle_z [deg] time [sec] -0.45 -0.4 -0.35 -0.3 -0.25 -0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0 50000 100000 150000 200000 250000 angle_z [deg] time [sec] Fig.4: Comparison of HAYABUSA and simulation about angle z 4-2 Nutational damping using the stability We showed unstable motion was caused because its angle was placed in the unstable region. In reverse, it may be possible to damp some nutational motion when we keep its angle in the stable region. To show numerically, we place the angle as 0 = ( !0.800 0.800 0.000 ) [rad] (35) and run a numerical calculation with the constants Eq.(32) again. We saw that Eq.(32) draws its stability like upper right of Fig.3. And Eq.(35) instead of Eq.(33) means the angle is in the stable region. For example, the result is shown in Fig.5, where nutation damping can be observed. It can be said that one momentum wheel can act as a nutation damper if the attitude is kept in the stable region. 0.755 0.76 0.765 0.77 0.775 0.78 0.785 0.79 0.795 0 1 2 3 4 5 6 7 8 9 10 angle_y [rad] time [10^4 sec] Fig.5: simulation in stable region 4-3 Stabilization strategy at usual operation of HAYABUSA The unstable motion of HAYABUSA occurred after the large attitude maneuver. As is seen in section 4-2, the targeted angle 0 was in the unstable region and the attitude motion diverged. This is an unusual case, and at usual operation, a target angle is set to zero after every attitude maneuver, and the difference from zero is recognized as error angle to be feedbacked. This section shows that the unstable motion can be avoided around this angle (0 = 0). Consider the stability in this case. Eq.(29) is now written as 3 = D (36) The stability depends on only the sign of D. D = 1.58 10!13 on HAYABUSA means stable at 0 = 0. We consider the magnitude of angle d which expresses the distance between 0 = 0 and the boundary. It is written as d = D pA2 + B2 + C2 (37) Here, d = 2.65 10!4[deg] for HAYABUSA, and this means HAYABUSA is always quite close to the unstable region. This means it is possible that the angle goes into the unstable region easily because the zero 0 = 0 is always near the boundary since generally jDj ? jAj, jBj. Fig.6: the distance between 0 = 0 and the boundary However, the speed of divergence or convergence is quite slow around the boundary. For example, some numerical calculations show the amplitude of the nutational motion grows only less than 1[deg] in one week. At usual operation, HAYABUSA is not be kept left for a week without attitude correction. Based on this assumption, the amplitude of the nutational motion does little increase or decrease around 0 = 0, and thus HAYABUSA will be stabilized enough if its angle is reset to 0 = 0 after every attitude maneuver. Here, resetting to 0 = 0 means resetting the inertial frame properly, we can do the maneuver easily. It is a guarantee that no problems happened when the angle had been around 0 = 0 at usual operation. 5.Use for nutational damper In section4-2, we saw that taking an intentional angle 0 in stable region can damp nutational motion. We also saw the speed of the amplitude of the nutational motion decreasing is very slow around 0 = 0. Of course it is suitable as a nutational damper that the speed is fast, and this becomes possible by setting 0 at center of the stable region like Eq.(35). However, the result of the numerical calculation at section4-2 showed that the angle converges to around T = (!0.800, 0.776, 0.074)[rad] (only y-angle is shown at Fig.5) which has no concerns with the initial angle Eq.(35). The reason is that when the system converges, the control torque ?h also converges to zero, or satisfies T = 0 (38) (see Eq.(14)). This shapes a plane in the space (x, y, z) and it is possible to converge wherever on the plane. We can not control the convergent point, which takes place some unforeseen errors. It is not suitable for satellites when the onboard equipments require high attitude accuracy. We should consider better method of taking the intentional angle 0 to minimize the error. 6.Conclusion To analyze the nutational divergence on HAYABUSA, we analytically derived the stability condition of a satellite which is controlled by one bias-momentum wheel. We showed that when the wheel is controlled under PD control using the attitude-angle and angular velocity about the wheel axis, the feedback value is biased according to products of inertia, and this results in the change of stability. Because feedbacked attitude-angle varies the stability, its nutational motion diverge or converge depending on its attitude-angle. We showed angle-boundaries about stability, and based on this assumption, we concluded the cause of the divergence on HAYABUSA is that its attitude-angle was placed in the unstable region. To avoid this unstability, we considered two strategies. One was to set the attitude-angle zero, the other was to set the attitude-angle in the stable region. Setting the attitude-angle is done by changing the definition of the inertial frame. The former resulted eliminating the biased feedback value and the nutational motion does not diverge and converge. The latter, we could see the motion converges, and we proposed one of the way to use one momentum wheel as a nutatinal damper. References [1] B.Sacleux, : Rotation Vector-Based Attitude Control Design,AIAA Guidance, Navigation,and Control Conference and Exhibit, Portland, OR; UNITED STATES;9- 11 Aug. 1999. pp. 1941-1948. (1999) [2] gPv, 䑺:㐧_, W, pp.94 (1994)