Model

(a) (b)

Generalized coordinates:

q = [x0, y0, , L, R, L, R]T (1)

Generalized forces corresponding to the coordinates:

Fq = [Fx0 , Fy0 , F, FL , FR , FL , FR ]T . (2)

Positions of the thigh center of mass in cartesian coordinates: xL1, yL1 andxR1, yR1.Positions of the shin center of mass in cartesian coordinates: xL2, yL2 andxR2, yR2.Leg tip positions in cartesian coordinates: xLG, yLG and xRG, yRG.These coordinates can be stated using the generalized coordinates (correspond-ingly for the right leg):

xL1 = x0 r0 sin r1 sin( L)yL1 = y0 r0 cos r1 cos( L)xL2 = x0 r0 sin l1 sin( L) r2 sin( L + L)yL2 = y0 r0 cos l1 cos( L) r2 cos( L + L)xLG = x0 r0 sin l1 sin( L) l2 sin( L + L)yLG = y0 r0 cos l1 cos( L) l2 cos( L + L).

(3)

Translational energy of the system:

Tt =12

(m0(x

20 + y

20) + m1(x

2L1 + y

2L1 + x

2R1 + y

2R1)

+m2(x2L2 + y

2L2 + x

2R2 + y

2R2)

).

(4)

1

Rotational energy of the system (missing from the original model):

Tr =12

[J0

2 + J1(( L)2 + ( R)2

)+ J2

(( L + L)2

+( R + R)2)]

,(5)

where J0, J1 and J2 are the inertia of the torso, thigh and shin, respectively.

Total kinetic energy:T = Tt + Tr. (6)

Formulas of the generalized forces (left leg):

Fx0 = FLx + FRx

Fy0 = (m0 + 2m1 + 2m2)g + FLy + FRyF = (yL1 m1 + yL2 m2 + yR1 m1 + yR2 m2)g + yLG FLy

+yRG

FRy +xLG

FLx +xRG

FRx

FL = (yL1L m1 +yL2L

m2)g +yLGL

FLy +xLGL

FLx + ML1

FL = yL2L m2g +yLGL

FLy +xLGL

FLx + ML2.

(7)

Substitute (3) to (4) and to (7).

Lagrangian equations:d

dt

(T

qr

) T

qr= Fqr . (8)

Now for each element qr in q and Fqr in Fq (r = 1, 2, . . . , 7) calculate theLagrangian equation, and collect the coefficients of the second time derivatesof the generalized coordinates to A. This gives the seven rows in the finaldynamic equation:

A(q)q = b(q, q, M, F ), (9)

whereM = [ML1,MR1,ML2,MR2]

T (10)

andF = [FLx, FLy, FRx, FRy]

T . (11)

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