IEEE TRANSACTIONS ON ROBOTICS AND AUTOMATION, VOL. 17, NO. 3, JUNE 2001

361

Analytical Time Optimal Control Solution for a Two-Link
Planar Acrobot with Initial Angular Momentum
Tsutomu Mita, Sang-Ho Hyon, and Taek-Kun Nam

Abstract—In order to control gymnastic and jumping robots, we will
derive the complete analytical solution to the posture control problem of
a two-link free flying object with initial angular momentum. We will showthat the solution involves singular control and derive formulas to calculate
the optimal switching condition, optimal terminal time and optimal trajectories. As an application, a high diving motion is simulated.
Index Terms—Acrobat robot, initial angular momentum, nonholonomic
control, time optimal control.

Fig. 1. Two-link free flying robot.

II. STATEMENT OF PROBLEMS AND CANONICAL FORM
I. INTRODUCTION
The purpose of this paper is to derive the analytical time optimal posture control law for free flying planar objects having nonzero angularmomentum. We plan to use it to build a parallel bar gymnastic robot
and running robot in the next stage of our research. The solution also
can be applied to the posture control problem of a two-link planar manipulator in the horizontal plane with a passive first link and an initial
angular velocity.
There is much research treating the control of free flying objects
with zero initial angular momentum, to name but a few, [1]–[6], whereat least three degree of freedom (dof) is required to guarantee controllability. Especially, Mukherjee et al. [3] introduced the concept of geometric phase to control a three link space robot.
However, the problems treating in this paper are a little bit different
from these since the motion of the free flying objects having nonzero
angular momentum is described by an affine nonlinear system with thedrift term having no equilibrium. Kamon et al. [7] formulated the posture control problem as a path planning problem and derived a minimum energy trajectory by numerical optimization method to simulatea three–dimensional (3–D) somersault motion. Godhavn et al. [8] converted the posture control problem to a bang-bang control problem andproposed a potential numerical computation algorithm to show a somersault motion of a planar diver. However, since these are numericalsolutions, we cannot obtain the closed form control formulas which
are needed for experiments. In addition, we cannot know the nature of
the optimal solutions, e.g., we cannot tell when a particular solution
becomes singular. Berkemeier et al. [9] dealt with a two-link hopping
robot and solved the posture control problem by controlling the twist
angle as a particular periodic time function.
The obtained time optimal solutions in this paper include simple
closed form formulas of the control law. They also show that the
problem leads to a singular optimal control problem depending upon
the initial posture; the switching time is once when the singularsolution does not occur while is twice when the singular solution isused. As an application, the somersault motion of a diver approximatedby the two-link system is simulated.

Manuscript received June 19, 2000; revised November 27, 2000 and March 5,
2001. This paper was recommended for publication by Associate Editor H. Araiand Editor S. Hutchinson upon evaluation of the reviewers’ comments.
The authors are with the Department of Control and Systems Engineering,Tokyo Institute of Technology, Meguro-ku, Tokyo, 152-0033 Japan (e-mail:mita@ctrl.titech.ac.jp; sangho@ctrl.titech.ac.jp;nam@ctrl.titech.ac.jp).
Publisher Item Identifier S 1042-296X(01)06736-2.

Consider the posture control problem of a planar free flying robot asshown in Fig. 1, where the robot is composed of the body and leg;  isthe absolute angle of the body measured counterclockwise relative tothe frame of inertia; is the relative angle between the body and leg
measured counterclockwise; L and l are the distances between the joint
and the centers of mass (CM) of the body and the leg, respectively; M
and JL are the weight of the body and the moment of inertia of the
body around its CM; m and Jl are those for the leg.
Suppose that the robot has a nonzero constant angular momentum P0
which is provided from the ground before takeoff as an initial angularmomentum. Then the conservation law of angular momentum around
CM of the whole robot becomes [9]P0 = (M1 + A1 cos )_ + (M2 + A2 cos ) _

(1)

where
mM (L2 + l2 ); A1 = 2A2
m+M
mM l2
mM lL
M2 =Jl +; A2 =:
m+M
m+M
M1 =JL + Jl +(2)

In the sequel, we assume that M1 > A1 and A2 6= 0. Then, under theassumption P0 6= 0, (1) cannot be integrated and becomes a nonholonomic constraint. Note that when P0 = 0, (1) turns to be an algebraic
equation and  and cannot be controlled independently. With a little
modification of the parameters, (1) also describes a two-link manipulator in the horizontal plane with an unactuated first joint and initial
angular velocity.
Defining _ as a control, (1) can be described by
_

_

=

0

P
M +A cos+

10 MM ++AA cos
cos

u

(3)

which is a nonlinear system with a drift term:
q_ = f (q ) + b(q )u:q = ( ; )T(4)having no equilibrium because f (q ) 6= 0 (8 q ). Apparently, this system
satisfies the locally accessible condition [6].
Since the rotational motion of the robot cannot be stopped when
P0 6= 0, the control problem is to make q passing through a givenreference state qr = ( r ; r )T at a given time T . Only can be settled at r by putting u = 0 after arrival to the reference state.
Since the translated motion of the robot cannot be controlled at all,we will not mention about it. Animals and gymnasts control their posture within the falling time.
If we can transform (3) to a system where the second entry of b(q )is zero, the corresponding state can be characterized as the state inde-

1042-296X/01$10.00 © 2001 IEEE

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III. ANALYTICAL SOLUTION OF THE TIME OPTIMAL CONTROLPROBLEM
A. Optimal Control Problem and Solutions
We will solve the time optimal control problem for (5) and investigate the optimal trajectories. First of all, we defineFig. 2.

g(x ) and 0g(x ) whenp(x1 ) :=
= 0.

0

P
M +A cos(x +

=

1

+

0

)

u

(5)

J=

0 r ; x =  + w( ) 0 (r + w( r ))
2

1

1

1

2

0

(6)

M1 0 A1 tan
M1 + A1

0 2k
2

(7)

0 1)  < (2k + 1):

(9)

(10)

As the result, w( ) becomes a single-valued function just the same asg(x1 ) depicted in Fig. 2.
Let’s examine the nature of the canonical form (5) first by assuming
P0 > 0. Then, from M1 > A1 and P0 > 0, we can see that x2 (t)cannot move in the negative direction and the accessible region to theorigin must satisfies

x2 (0) < 0:

0

+

1

u

0

(14)

and the principle of optimality yields the following necessary conditions(11)

Actually, this is shown to be sufficient below, when P0 < 0, (11) is
replaced by x2 (0) > 0.

(15)
(16)

8 t):

(

(17)

Note that the bang-bang control law (15) holds only when the problem
is regular, i.e., 1 is not identically zero for some time interval [10].
When it becomes singular, only (16) and (17) hold.
Since _ 2 = 0 in (16) yields 2 = constant := 0 , the regularproblem reduces to find 1 (0) and satisfying x(T ) = 0 under theconditions:x_ 1 = 0 sgn(1 )um
x_ 2 =p(x1 )_ 1 = @p(x1 )@x1
H =1 + 1 u 0 p(x1 ) = 0(8)depending upon
(2k

(13)

u = 0 sgn(1 )um
0 @p@x(x ) 2
_ = 0 @H =0@x
H (t) =1 + 1 u + 2 p(x1 ) = 0

and M1 + A1 cos(x1 + r ) is the moment of inertia around CM of thewhole robot, x2 can be interpreted as the rotational angle around the
CM of the whole robot.
Since the domain of is defined as R1 = (01; 1) for derivationof the control law and computer simulations, the continuity of the function tan01 (3 1 tan( =2)) in (7) at = 6n becomes critical. To thisend, as introduced in [4], we propose to replace tan01 ( ) in (7) byk + tan01

dt = Tp(x1 )=1 + 1 u + 2 p(x1 )Then the control problem is transformed to drive x(0) to the origin 0
at the time T . Since (5) yields

fM + A cos(x + r )gx_ = P

0

H =1 + T

and [11]

M2 + A2 cos p dp
w( ) =
M1 + A1 cos p
0A2(M2 A1 0 M1 A2 )2=+
A1
A1 M12 0 A21
1 0 A1
2 tan01 M
tan:
M1 + A1
2

T

while bringing x(0) to x(T ) = 0 under the constraint juj  um .
The Hamiltonian of this problem iswhere x = (x1 ; x2 )T is the new coordinate defined by

x1 =

(12)

in (5). The control problem is to minimize

pendent of u. This can be possible only for the two-link robot and an
immediate calculation derives the following canonical form:

x_ 1
x_ 2

P0
M1 + A1 cos(x1 + r )

(18)
(19)
(20)
(

8 t):

(21)

However, this problem can be solved analytically without seeking

1 (0) and explicitly if we pay attention to the two integral manifolds
made by u = 6um .
Examine the case u = um (1 < 0) first. In this case, it follows fromx_ 1 = um that x1 increases monotonically. Besides, using x_ 2 =x_ 1 =dx2 =dx1 = p(x1 )=um , x2 satisfies the following manifold
x2 = g(x1 ) + C1 ;(22)
um
where g1 (x1 ) is described byg(x1 ) ==x0p(y)dy2P0
01tan
M12 0 A21

M1 0 A1 tan x1 + r
M1 + A1
2(23)

provided all constants are collected to make one integral constant C1 .
When calculating tan01 (:), we will use (9) and (10).
Similarly, when u = 0um (1 > 0), we have another manifold

x2 = 0g(x1 ) + C2 :
um

(24)

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363

Fig. 4.

Optimal trajectories when

= 0.

Fig. 5.

Optimal trajectories when

6= 0.

Fig. 3. Switching line I, II, and optimal trajectories.

The integral constants C1 and C2 must be adjusted for (22) and (24)to pass through designated states.
We will treat the case r = 0 to save the length of the paper, theng (x1 ) is given by a single-valued function passing through the origin as
depicted in Fig. 2. Furthermore, as shown in Fig. 3, (22) and (24) draw
a bunch of curves rising in the right-hand side (RHS) and left-hand side(LHS) directions, respectively, when the integral constants C1 and C2vary. We’ll refer to the two manifolds, denoted by I and II in Fig. 3,
passing through the origin as the switching line hereafter. From Fig. 3,
we can see that we can find a controlled trajectory only when x2 (0) exists in the region under the switching line I and II (the region is marked
by slashes). The control strategy is as follows. As is in Case A, whenx1 (0) > 0, we first choose u = um to ride the state on the manifold
rising in the RHS direction, then switch the control to u = 0um when
the state reaches the switching line I. Similarly, as is in Case B, whenx1 (0) < 0, we first choose u = 0um , followed by u = um when thestate reaches the switching line II. When x1 = 0, we can choose either
controls in Case A or Case B, x1 = 0 gives the boundary between Case
A and Case B.
Note that we have infinitely number of choices of the control sequences if we don’t care about the optimality. However, the final choicemust be riding the state on the switching line I or II in the fourth quadrant or third quadrant, respectively.
In Section IV, we will prove the following theorem.
Theorem 1:1) when 0 < x1 (t) <  , the mentioned control strategy is optimal;
2) when x1 (t) = 0 or x1 (t) =  happens, which correspondsto= 6 , the control problem becomes singular, and, as isshown in Fig. 4, the optimal control is given by u = 0 before
riding the state on the switching lines I or II;3) when r 6= 0, the boundary of Case A and Case B is given by
x1 = 02 r as shown in Fig. 5.
From this theorem, we can see that the control strategy explained inFig. 3 is optimal only when x(0) is located inside the area enclosed by
the bold face curves depicted in Figs. 4 and 5. We call this area a basic
region.
Note also that the robot rotates by the fastest angular velocity withholding the leg upon the body to minimize the moment of inertia of therobot in the singular period because becomes 6 .
Before concluding this section, let’s examine the role of um and accessible region. When um is chosen bigger, it follows from (22) and(24) that the lines I and II becomes flat and close to x1 line. Therefore, the region of x2 (0) which can be brought to the origin approaches

x2 < 0. Since (11) has been shown to be necessary before, (11) is thenecessary and sufficient condition of the initial state which can accessthe origin by the control.
However, in the physical world, x2 is counted by mod 2 . Therefore,2 =even when x2 (0) > 0, predetermining an integer k satisfying x
x2 0 2k < 0 and treating (x1 (0); x2 (0)) as a new initial state will
get around the limitation (11) at the expense of more time to reach theorigin. We have to introduce a similar shift to x1 to express it in theinterval 0   .
B. Optimal Switching Conditions
We will derive the switching condition as well as the optimal
switching time when x(0) lies in the basic region by making use of
Fig. 3. This will be used for practical implementation of the control
law.
When u = 6um , x_ 1 = 6um yields the solution

x1 (t) = 6um t + x1 (0)(25)

which shows that (the change of x1 =um ) gives the transition time whenu = 6um . This fact will be used frequently without mentioning.

Now let us define the switching time and the terminal time of the
trajectory in Case A by t1 and TA , thenx1 (t1 ) =um t1 + x1 (0)x1 (TA ) = 0 um (TA 0 t1 ) + x1 (t1 )

(26)

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(a)

(b)

Fig. 6. Simulation results (a) x coordinate (b) q coordinate.

hold and from the condition x1 (TA ) = 0, we havex (0) 2x1 (t1 ) 0 x1 (0)TA = 2t1 + 1
=
:
um
um

Therefore, we can calculate x1 (t2 ) from (35). Then, (34) leads to TB .
The t2 can be derived from the first equation of (33) as(27)

On the other hand, since x(0) lies on the manifold (22) and theswitching line I passes through the origin, we have

C1 = x2 (0) 0g(x1 (0))
;
um

C2 =

g(0)
;
um

(28)

respectively. Furthermore, since x(t1 ) lies on the both of the manifoldsg(x (t ))g(x1 (0))
x2 (t1 ) = 1 1 + x2 (0) 0umumg(x1 (t1 )) g(0)x2 (t1 ) = 0
+
um
um

(29)

(30)
(31)

Using the fact that g (x1 ) is a single-valued function, solving (30) gives
x1 (t1 ) and (27) provides TA .
Moreover, the first equation in (26) gives the following switching
time

t1 =

x1 (t1 ) 0 x1 (0)
:
um

(32)

Similarly, in the Case B, if the switching time and terminal time aredefined by t2 and TB , fromx1 (t2 ) = 0 um t2 + x1 (0)x1 (TB ) =um (TB 0 t2 ) + x1 (t2 ) = 0(33)

we will obtain

TB =02x1 (t2 ) + x1 (0)
um

(34)

and

u
g(x1 (0)) g(0)g(x1 (t2 )) = m x2 (0) ++222x2 (0) g(x1 (0)) g(0)x2 (t2 ) =0 2um :
+
2
2um

(35)
(36)

0x1 (t2 ) + x1 (0) :
um

(37)

In the practical implementation of the control law, making use ofx2 (t1 ) and x2 (t2 ) as the switching conditions may be more robust thanusing t1 and t2 .
As for the singular control period, it follows from x1 = 6 , u = 0
and (19) that we havex_ 2 =

hold. Therefore, solving (29) to obtainug(x1 (0)) g(0)g(x1 (t1 )) = 0 m x2 (0) ++222x2 (0) g(x1 (0)) g(0)x2 (t1 ) =0 2um + 2um :
2

t2 =

P0
:= K:
M1 0 A1(38)

Therefore, the transition time in the singular period is given by the division of the change of x2 by K . Furthermore, even Fig. 3 corresponds
to the case r = 0, the above derivation is independent of r and we
can apply all the results, from (25)–(38), to the case r 6= 0.
Example: We will simulate the motion of a planar diver approximated by two links with the parameters

M =30 (kg); L = 0:75 (m); JL = 5:0 (kg 1 m2 )m =25 (kg); l = 0:8 (m); Jt = 4:5 (kg 1 m2 )P0 =170kg 1 m2s; um = 51sTwhen q0 = (0; =2) and qr = (0; 7=2)T , which correspond tox(0) = (0; 03), the optimal control law provides the switching conditions x1 (t2 ) = 02:9694, x2 (t2 ) = 04:7124 and t2 = 0:5939 (s)(corresponding to Case B) and the terminal time T = 1:1878 (s). In thiscase, the robot performs one and half somersault. The time responses
in terms of x and q are depicted in Fig. 6(a) and (b), respectively, while
Fig. 7 shows the animation. The translated motions have been simulated under the conditions VX = 1:5 (m/s) and VY = 1:8 (m/s), where
VX and VY are initial speeds of the CM of the whole robots.
In this problem setting, the minimum time control tells how long
does it take at least to complete a specified motion within a given input
magnitude um .
Remark 1: In the case where a real robot cannot bend the leg 6=2,
the following method may be applied if we do not care about the optimality much. Draw the singular line on =  0 a (a > 0) instead of=  . When x1 comes across this new line, keep u = 0 until the state
rides on the switching line I followed by u = 0um . The maximum velocity um of the joint depends on the requested acrobatic performance,
P0 and the falling time. Some compromise will be required in the experiment.

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365

(a)

Fig. 7. The animation.

IV. PROOF OF THEOREM 1
A. Proof of the Item 1 and Item 2
We will prove the item 1 and item 2 in Theorem 1.
To this end, we will analyze the response of 1 and (21).
Let’s consider the case where 1 is not identically zero, first.
If 1 < 0, (18) leads to u = um and it follows from x_ 1 = um !
dt = dx1 =um that x1 increases with the time and we can integrate(20) as=dp(x1 ) dt =dx1
ump(x1 (t)) + K1 :

x (t)

t

1 =
um

um

(40)

from which K1 = 01=um is obtained. Therefore, 1 (t) is expressedby1 (t) = 010 p(x (t))
um

1

(41)

as a function of x1 (t).
Similarly, when 1 > 0, u = 0um leads to that x1 decreases withthe time and we can conclude

1 (t) =10 p(x (t)) :
um

1

(42)

It follows from the continuity of 1 (t) that (41) must intersect (42).
Since the two 1 ’s are the same except for their signs, they intersect only on the x1 axis as shown in Fig. 8(a). Therefore, the input
is switched at these intersection points. Let us denote one of the intersection point by x1 = x1s (x1s corresponds to x1 (t1 ) or x1 (t2 ) in theprevious chapter). Equating two 1 ’s at these points, we havep(x1s ) = 1

Fig. 8. Function  versus x (a) x 6= 6 (ordinary solution). (b) x =6 (singular solution). (c) x = 0 (trivial solution).

jx (t)j  jx j  (39)p(x1 ) + K1 0 p(x1 ) = 0(c)

If x1 would move across x1s , it should move to the next intersection
point which increases the travelling time to x1 = 0. Therefore

dp(x1 )

Substituting this into (21) yields
1 + um(b)(43)

where may be determined by the initial condition. It will be seen from
Fig. 8(a) that x1s must satisfy jx1s j   and only one switching instantis allowed before x1 arrive at 0 from x1 (0). The reason is as follows.

1

1s

(44)

must hold.
Next, suppose that 1 becomes identically zero for some finite timeinterval. In this case, since H is independent of 1 , u cannot be determined from (15) and the problem becomes singular [10]. However,from Fig. 8(a), if there exists such a 1 , it should be at the intersection
point such that 1 (t) = 0 and x1 (t) = x1s hold for some period.
Besides, in the singular period, the derivatives of 1 (t) of any ordermust be zero. Its first and second derivatives are given by2
_ 1 = dp(x1 ) ; 1 = d p(x2 1 ) x_ 1
dx1
dx1

(45)

dp(x1 ) = P0 A1 sin x1 :dx1
(M1 + A1 cos x1 )2

(46)

where

Therefore, the following conditions

x1 =x1s = 6; and x_ 1 = u = 0
x1 =x1s = 0; and x_ 1 = u = 0(47)

make them zero and the converse also holds. It is direct to check that

(1i) (t) (8i  3) = 0 hold under (47). In addition, when 1 (t) = 0,
Hamiltonian (21) becomes

H = 1 0 p(x1 )(48)

which is ensured to be zero under the conditions (43) and (47). Therefore (43) and (47) are the necessary conditions for singular control.

366IEEE TRANSACTIONS ON ROBOTICS AND AUTOMATION, VOL. 17, NO. 3, JUNE 2001

Let us consider the case x1 = x1s = 6 and u = 0. This corresponds to the 1 depicted in Fig. 8(b), where the robot rotates by thefastest speed as stated before.
Then consider the case x1 = x1s = 0 and u = 0 which corresponds to the 1 shown in Fig. 8(c). It will be a singular solution only
if x2 (0) = 0 because, when x2 (0) < 0, u = 0 leads to a trajectory
keeping x1 = 0 before the convergence which contradicts the optimal
trajectory in the basic region shown in Fig. 3. However, the solution
x1 = x2 = 0 and u = 0 is trivial and we cannot count this as a singular solution since itself is on the reference state.
Therefore, the following conclusions are obtained. When x(0) 6=
0 and jx1 j <  , the singular solution does not occur and one timeswitching is optimal; when the state comes across jx1 j =  , u = 0
gives the singular control and the optimal trajectory is produced by twotime switchings as shown in Fig. 4, since the continuity of the trajectorymust hold.
B. Proof of the Item 3
We will prove that x1 (0) = 02 r renders TA = TB when r 6= 0.
Since Fig. 3 can be applied even to the case r 6= 0 as mentioned at theend of Section III, it follows (27) and (34) that TA = TB holds only if

x1 (t1 ) + x1 (t2 ) = x1 (0)(49)

where x1 (t1 ) and x1 (t2 ) will be provided by solving (30)–(35). Thus,
the problem is to show that (49) holds when x(0) = 02 r .
An addition of (30) to (35) gives

g(x1 (t2 )) + g(x1 (t1 )) = g(x1 (0)) + g(0):(50)

To make description simpler, we denote g (x) = g0 (x+ r ), then g0 (x)becomes an oddly symmetrical function passing through the origin as
we may imagine in Fig. 2.
When x1 (0) = 02 r , it follows fromg(x1 (0)) =g(02 r ) = g0 (0 r ) = 0g0 ( r )g(0) =g0 ( r )(51)that the RHS of the first equation in (50) is zero, and (50) can be
rewritten as

g0 [x1 (t1 ) + r ] = 0g0 [x1 (t2 ) + r ](52)

using g0 . However, since g0 (x1 ) is an oddly symmetrical single-valued
function, (52) is satisfied only if

x1 (t1 ) + r = 0(x1 (t2 ) + r ):(53)

This becomes (49) when x1 (0) = 02 r .
In [12], we have even proven that TA < TB holds when 02 r <x1 (0) <  0 r .

V. CONCLUSION
We have derived an analytical solution to the minimum time optimal
control problem of a two-link flying robot having nonzero angular momentum. For the posture control of general n link robots, we can apply
obtained results after fixing n 0 2 joint angles. The optimality is notguaranteed in this methodology. We are now constructing a parallel bar
gymnastic robot which can perform a somersault.
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