Appunti relativi alla

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Spacecraft with three rotors Euler equations are written without considering external moments Ix ˙ωx +Rx ˙ρx + (Iz −Iy)ωyωz +Rzρzωy −Ryρyωz = 0 Iy ˙ωy +Ry ˙ρy + (Ix −Iz)ωxωz +Rxρxωz −Rzρzωx = 0 Iz ˙ωz +Rz ˙ρz + (Iy −Ix)ωyωx +Ryρyωx −Rxρxωy = 0 Rx ˙ρx = 0 Ry ˙ρy = 0 Rz ˙ρz = 0 We assume Iz > Iy > Ix. The R are the inertia moments of the rotors, and ρ their angular velocities. The equilibrium condition is found considering the system (Iz −Iy)ωyωz +Rzρzωy −Ryρyωz = 0 (Ix −Iz)ωxωz +Rxρxωz −Rzρzωx = 0 (Iy −Ix)ωyωx +Ryρyωx −Rxρxωy = 0 We assumed that the angular velocities of the spacecraft are assigned, and the unknowns are the angular velocities of the rotors. This system is singular, which means we can impose one of the ρ and the others are found solving the system. Once the equilibrium condition is found we linearize, neglecting the three rotor equations (no internal moments) Ix ˙ωx + (Iz −Iy)¯ωyωz + (Iz −Iy)¯ωzωy +Rz ¯ρzωy +Rz ¯ωyρz −Ry ¯ρyωz −Ry ¯ωzρy = 0 Iy ˙ωy + (Ix −Iz)¯ωxωz + (Ix −Iz)¯ωzωx +Rx¯ρxωz +Rx¯ωzρx −Rz ¯ρzωx −Rz ¯ωxρz = 0 Iz ˙ωz + (Iy −Ix)¯ωxωy + (Iy −Ix)¯ωyωx +Ry ¯ρyωx +Ry ¯ωxρy −Rx¯ρxωy −Rx¯ωyρx = 0 We de ne KY = Iz −Iy Ix KR = −Ix −Iz Iy KP = Iy −Ix Iz To study stability we use a state-space representation. ˙ωx = ( −KY ¯ωz − Rz Ix ¯ρz ) ωy + ( −KY ¯ωy + Ry Ix ¯ρy ) ωz − Rz Ix ¯ωyρz + Ry Ix ¯ωzρy = 0 ˙ωy = ( KR¯ωz + Rz Iy ¯ρz ) ωx + ( KR¯ωx − Rx Iy ¯ρx ) ωz − Rx Iy ¯ωzρx + Rz Iy ¯ωxρz = 0 ˙ωz = ( −KP ¯ωy − Ry Iz ¯ρy ) ωx + ( −KP ¯ωx + Rx Iz ¯ρx ) ωy − Ry Iz ¯ωxρy + Rx Iz ¯ωyρx = 0 where the three equation of the rotors are decoupled 1 ˙ρx = 0 ˙ρy = 0 ˙ρz = 0 Considering the state variables as x = [ωx ωy ωz ρx ρy ρz]T the system is written in the canonical form ˙x =Ax and stability can be analyzed studying the eigenvalues of A.   0 ( −KY ¯ωz…