Fig. 3: A ball-beam system with the palm circle task. Fig. 5: beam. Geometric parameters of the ball and the Fig. 4: Butterfly task of Cotnact jugglin

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A Control Method of Juggling Task for Ball-beam System Akira Nakashima (Nanzan University) Abstract This paper deals with the realization of Palm circle task for a ball-beam system in a 2- dimensional space. The palm circle means the situation where the spactial position of the ball is fixed at a point while the relative position of the ball on the beam tracks a cyclic trajectory. First, the equations of motion of the ball and the beam are derived via the Lagrange equation with the constraints that stand for the rolling contact between the ball and the beam. Second, the equations of motion are transformed to the ones in terms of the control variables, i.e. the position of the ball, the relative position of the ball on the beam and the orientation angle of the beam. Third, the LQ optimal servo controller is derived based on the linear system around the origin of the equations of motion for the control. Finally, the effectiveness of the controller with a cyclic trajectory of ball on the beam is verified by numerical simulations. Key Words: Ball-beam system, contact juggling, dynamic manipulation, LQ servo control ) 2) 3) 4) 5) 6) 7) 8) 3 9) ) ) 2) Fig. : A task of Contact juggling -Palm circle. Fig. 2: Palm circle task by a ball-plate system. 3) Fig. Fig. 2

Fig. 3: A ball-beam system with the palm circle task. Fig. 5: beam. Geometric parameters of the ball and the Fig. 4: Butterfly task of Cotnact juggling 4). 2 3 2 Fig. 3 2 4) Fig. 4 3 5) Fig. 6: Configuration of the ball-beam system. LQ 2 2. Fig. 6 l F m O J F = 2 m F lf 2 kg m2 m m O kg J O = 2 m OrO 2 kg m2 Finger Object Fig. 6 Σ B 2 y Σ F x Σ O Σ B

( B p F, B R F ) SE(2) B p F R 2 Σ B Σ F B R F SO(2) Σ B Σ F Σ O ( B p O, B R O ) SE(2) B R F B R O 2 ϕ F ϕ O B R F := R(ϕ F ), B R O := R(ϕ O ) () ϕ F ϕ O R Σ B x Σ F Σ O x R(ϕ) SO(2) cos ϕ sin ϕ R(ϕ) :=, ϕ R (2) sin ϕ cos ϕ (2) A B qf q := R 6 p B, q F := F p, q O := O R 3 q O (3) Fig. 6 Σ CF Σ CO y Σ CF Σ CO Σ F Σ O ( F p CF, F R CF ) ( O p CO, O R CO ) SE(2) F αf p CF =, F R CF = I 2 (4) O ro cos α p CO = O, O R CO = R(α O π sin α 2 ) (5) O ϕ F ϕ O α F R Σ F Σ CF α O R Σ O x Σ CO y α F α O Σ F B f F R 2 τ F R B f u := F R 3 (6) 2.2 2.2. τ F B p F + B R F F p CF = B p O + B R O O p CO (7) B R F F R CF C F R CO = B R O O R CO (8) Fig. 6 F R CF = I 2, C F R CO = R(±π) (9) (7) (7) (8) α F α O q F q O α F (7) F p CF (8) (9) (5) F p CF = B R T F ( B p O B p F ) + B R T F B R O O p CO = B R T F ( B p O B p F ) O R T C O O p CO = B R T F ( B p O B p F ) e y (4) α F α F = e T x B R T F ( B p O B p F ) () e x :=, T e y :=, T e T x e y = α O (8) () A α O = ϕ F ϕ O π 2 2 2.2.2 () (7) B R F F ṗ CF = B R OOṗ CO (2) (2) (7) (8) α F = α O (3) (2) (7) (2) Bṗ F + B Ṙ F F p CF = B ṗ O + B Ṙ O O p CO A A(q) q = (4) A := A F (q) A O (q) R 2 6 A F := I 2 R( π 2 )B R F F p CF R 2 3 A O := I 2 R( π 2 )B R O O p CO R 2 3 (8) 2C F R CO = R( π)

2.3 L = 2 m F B ṗ F 2 + 2 J F ϕ 2 F + 2 m O B ṗ O 2 + 2 J O ϕ 2 O ( m F ge T y B p F + m O ge T y B p O ) (5) g = 9.8 m/s 2 (5) d dt ( ) T L = q ( ) T L A T λ + Υ (6) q λ R 2 F C := A T λ (7) λ Σ B λ = B f C = B f Cx B f Cy (8) Υ R 6 I3 Υ = Bu, B := (9) 3 3 MF M := 3 3 3 3 M O mf I M F := 2 2 2 J F FgF F g := R 6 F gf := F go mf e y M q + F g = A T λ + Bu (2), F go := 2.4 R 6 6, M O := mo e y R 3 mo I 2 2 2 J O R 3 3 (2) (4) 6) (4) A q + A q = (2) q λ (q, q) λ = ( AM A T) { AM (Bu F g ) + A q (2) (2) (2) (q, q) 7) 3 3 3. (7) 2 6 2 = 4 B p O x c := α F ϕ F } R 4 (22) q (22) B p F (7) (8) (9) (4) (5) B p F = B p O B R F F p CF + B R O O p CO = B p O B R F F p CF B R F O R T C O p CO = B p O B R F F p O (23) F p O = αf x c F p O R 2 ϕ F B p O ϕ O () ϕ O = ϕ F α O π 2 α O (3) ϕ O = ϕ F α F + α F () α O () π 2 (24) α F () α O () α F α O (23) Bṗ F = B ṗ O B Ṙ F F p O B R F F ṗ O = B ṗ O R( π 2 )B R F F p O ϕf B R F e x α F (25) 3 (7) (3) 8)

(24) ϕ O = ϕ F α F (26) (3) (3) M c ẍ c + F gc (x c ) = B c u (32) (25) (26) q ẋ c q = S(x c )ẋ c (27) M c := m c I 2 m c2 m c3 m T c 2 m c22 m c23 I 2 B R F e x R( π 2 )B R F F p O 2 S := I 2 2 2 R6 4 2 S (4) A AS = (28) (28) (27) (4) ASẋ c = x c 3.2 (2) (22) (27) q = Ṡẋ c + Sẍ c (29) (29) (2) S T S T MSẍ c + S T MṠẋ c + S T F g = S T A T λ + S T Bu (28) x c M c (x c )ẍ c +C c (x c, ẋ c )ẋ c +F gc (x c ) = B c (x c )u (3) M c := S T MS, C c := S T MṠ F gc := S T F g, B c := S T B C 3.3 (3) (x c, ẋ c ) = (, ) 2 R(ϕ) ϕ ϕ = I 2 + R( π 2 )ϕ (3) F gc := m T c 3 m c23 m c33 (m F + m O )ge y m F gϕ F m F g ϕ F I 2 2 B c := e T x e T x M c m c := m F I 2 m c2 := m F e x m c3 := m F e x m c22 := m F + J ro 2 O m c23 := (m F + J O ) m c33 := m F ro 2 + J F + J O (32) M c ẍ c + K c x c + d gc = B c u (33) 2 2 2 2 K c := 2 m F g 2 m F g (m F + m O )ge y d gc := 3.4 LQ (33) (33) d gc R 4 u = ū + u g (34) (mf + m u g := O )ge y (34) (33) B c u g = d gc M c ẍ c + K c x c = B c ū (35) xc x := R 8 (36) ẋ c

LQ Servo Controller Fig. 7: The extended system with the linear one and the integrator. (35) ẋ = Ax + Bū (37) A := 4 4 K c I 4 4, 4 4 B := M c 4 3 M c B c B p O α F 4 ϕ F B p O α F y = Cx, C := I 3 3 5 (38) y R 3 r R 3 9) w = t (r y)dt (39) Fig. 7 (37) (38) { ẋe = A e x e + B e ū y = C e x e (4) x x e := R w A 8 3 A e := R C 3 3 B B e := R 3 3 3 C e := C 3 3 R 3 (4) 4 A B (4) C 3 3 4 Fig. 8: The LQ servo controller based on the extended system. (4) J := min ū Jū (42) (x T e Qx e + ū T Rū)dt, Q, R > LQ ū = K F K I xe = K F x K I w (43) Fig. 8 (34) (39) (43) t u = K F x K F (r y)dt + u g (44) 4 4. l F =.5 m, m F =.5 kg, J F = 2 m F l 2 F kg m 2 =.5 m, m O =. kg, J O = 2 m Or 2 O kg m 2 B p F () ϕ F α F α O B p O ϕ O (23) (24) B p F () =, T m, ϕ F () = 5 deg α F =. m, α O = 9 deg Fig. 9 LQ Q R Q p Q = Q v, Q p = Q y Q v = I 4, Q y = I 3

.4.3.2. -. -.2 -.3 -.4 -.4 -.3 -.2 -...2.3.4 Fig. 9: The initial situation of the ball-beam system. 2 r(t) = A αf cos(2πf αf t) α F A αf f αf A αf =. m, f αf =.2 Hz 4.2 Figs. 3 2 sec Figs. q y Fig. B p O y x α F α F LQ Fig. 2 Fig. 3 Σ CF C F f C = F R T C F B R T F λ (45) y (43) u = K F (x x d ) x d := t (r y)dt + u g (46) r ṙ (46) Figs. 4 7 Fig. 5 α F B p O x Fig. 8 8 5 LQ 2 3

..2.2 p Fx m p Ox m p Ox m -. 5 5 2 -.2 5 5 2.5 -.2 2 4 6 8 2 4 6 8 2.5 p Fy m -.5 p Oy m p Oy m -. 5 5 2 -.5 5 5 2 4 -.5 2 4 6 8 2 4 6 8 2.2 φ F deg φ O deg 2 α F m - -2 5 5 2 5 5 2 Fig. : Simulation result: Finger & object. -.2 2 4 6 8 2 4 6 8 2 Fig. : Simulation result: Ctrl. variables..2.6 f x N f y N. -. 2 4 6 8 2 4 6 8 2 τ N m 7 6 5 4 2 4 6 8 2 4 6 8 2 4 2-2 2 4 6 8 2 4 6 8 2 Fig. 2: Simulation result: Ctrl. inputs. 2) 2, 22) 25 I-A-2 ) T. Yoshikawa. Dexterous mechanical hand. Trans. RSJ, Vol. 8, No. 6, pp. 763 766, 2. (in japanease). 2) A. Nakashima, Y. Sugiyama, and Y. Hayakawa. Paddle juggling of one ball by robot manipulator with visual servo. In 9th International Conference on Control, Automation, Robotics and Vision, 26. ICARCV 6, pp. 6, 26. 3),,,,.., Vol. 3, No. 9, pp. 924 93, 22. 4) Chunfang Liu, Yoshikazu Hayakawa, and Akira Nakashima. Racket control for a table tennis robot F Cx N.4.2 -.2 -.4 2 4 6 8 2 4 6 8 2 F Cy N 4 3 2 2 4 6 8 2 4 6 8 2 Fig. 3: Simulation result: Cotact forces. to return a ball. SICE Journal of Control, Measurement, and System Integration, Vol. 6, No. 4, pp. 259 266, 23. 5) M.T. Mason and K.M. Lynch. Dynamic manipulation. In Proc. IEEE/RSJ IROS, Vol., pp. 52 59, 993. 6).., 2. 7),. Optimal trajectory design for nonlinear mechanical systems : Acrobot example., Vol. 59, p. 4p, may 25. 8).., Vol. 46, No. 9, pp. 73 76, 27. 9) T. Ibuki, Y. Tadokoro, Y. Fujita, and M. Sampei. 3d inverted pendulum stabilization on a quadrotor via bilinear system approximations. In Control Applications (CCA), 25 IEEE Conference on, pp. 53 58, Sept 25. ),,. 3., Vol. 5, No. 6, pp. 943 952, 997. ).., Vol., No. 4, pp. 52 528, 993. 2) R. W. Brockett. Asymptotic stability and feedback stabilization. In R. W. Brocket, R. Millman, and H. Sussman, editors, Differential Geometric Control Theory: Progress in Mathematics, Vol. 27, pp. 8 28. Springer-Verlag, 983.

.2.2.2 p Fx m p Ox m p Ox m -.2 5 5 2 -.2 5 5 2.5 -.2 2 4 6 8 2 4 6 8 2.5 p Fy m -.5 p Oy m p Oy m -. 5 5 2 -.5 5 5 2 4 -.5 2 4 6 8 2 4 6 8 2.4 φ F deg - 5 5 2 φ O deg 2-2 5 5 2 Fig. 4: Simulation result(modified): Finger & object. 2 α F m.2 -.2 2 4 6 8 2 4 6 8 2 Fig. 5: Simulation result(modified): Ctrl. variables..6 f x N f y N τ N m - 2 4 6 8 2 4 6 8 2 7 6 5 4 2 4 6 8 2 4 6 8 2 5 F Cx N.4.2 -.2 -.4 2 4 6 8 2 4 6 8 2 4 3 F Cy N 2-5 2 4 6 8 2 4 6 8 2 Fig. 6: Simulation result(modified): Ctrl. inputs. 3) James Ernest. Contact Juggling. 3rd edition, 2. 4) K. M. Lynch, N. Shiroma, H. Arai, and K. Tanie. The roles of shape and motion in dynamic manipulation: the butterfly example. In Proceedings. 998 IEEE International Conference on Robotics and Automation, Vol. 3, pp. 958 963, 998. 5). Arduino MATLAB! Physical computing lab. TechShare, 22. 6) R. M. Murray, Z. Li, and S. S. Sastry. A Mathematical Introduction to ROBOTIC MANIPULATION. CRC Press, New York, U.S.A., 994. 7)..,, Japan, 7 23. 8). ().,, Japan, 4 26. 9),. MATLAB., 2. 2). :. Mobiligence.,, Japan, 2. 2) S. Katsumata, S. Ichinose, T. Shoji, S. Nakaura, and M. Sampei. Throwing motion control based on output zeroing utilizing 2-link underactuated arm. In Proc. of American Control Conference, pp. 357 364, 29. 22) Leonid Freidovich, Anders Robertsson, Anton Shiriaev, and Rolf Johansson. Periodic motions of the pendubot via virtual holonomic constraints: Theory and experiments. Automatica, Vol. 44, No. 3, pp. 785 79, 28. 2 4 6 8 2 4 6 8 2 Fig. 7: Simulation result(modified): Cotact forces. A 2 R(ϕ) SO(2) ϕ R B R (ϕ) = R T (ϕ) = R( ϕ) R(ϕ + ϕ 2 ) = R(ϕ )R(ϕ 2 ) = R(ϕ 2 )R(ϕ ) d dt R(ϕ) = R( π 2 )R(ϕ) ϕ (A.3) (28) (A.) (A.2) (27) S S F S =, S F R 3 6, S O R 3 6 (B.) S O (28) AS = A F S F A O S O (B.2) A F S F = ( I 2 B R F e x R( π 2 )B R F ) F p CF F p O A O S O = I 2 R( π 2 )B R O O p CO R( π 2 )B R O O p CO (B.2) 2 2

Fig. 8: Simulation result(modified): Snapshots of the animation. (8) (9) B R O O p CO = B R F O R T C O O p CO = B R F e y 2 3 ( ) B R F e x R( π 2 )B R O O p CO = B R F e x R( π 2 )B R F e y = (B.3) R( π 2 )B R F ( F p CF F p O ) R( π 2 )B R O O p CO =R( π 2 )B R F ( F p CF F p O + B e y ) = (B.4) (B.2) (28) C (3) (3) S (B.2) M c C c F gc B c M c = S T F M F S F + S T OM O S O C c = S T F M F Ṡ F F gc = S T F F gf + S T OF go B c = S T F m c m c2 m c3 M c = m T c 2 m c22 m c23 m T c 3 m c23 m c33 2 2 c c2 c c3 C c = 2 c c23 2 c c23 c c33 (m F + m O )ge y F gc = m F ge T x B RF Te y m F g( e x α F e y ) TB RF Te y I 2 2 B c = e T x B RF T e T x ( e x α F e y ) T

M c C c m c := (m F + m O )I 2 m c2 := m B F R F e x m c3 := m B F R F ( e x α F e y ) m c22 := m F + J ro 2 O m c23 := (m F + J O ) m c33 := m F ro 2 + J F + J O c c2 := m F ϕf B { R F e y } c c3 := m B F R F α F ϕf e x + ( ϕf α F )e y c c23 c c33 := m F α F ϕf := m F α F α F

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