Robotic Manipulators: Lecture 6
Department of Electrical Engineering
Dr. H. A. Talebi
July 24, 2008
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Outline
• Jacobian
–
–
–
–
–
Singularities
Static Force
Jacobian in Force Domain
Inverse Velocity
Inverse Acceleration
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Singularities
• Jacobian is a linear mapping:
ν = J(θ)θ̇
• All possible EE velocities are linear combination of the columns of Jacobian
matrix.
ν = J1θ̇1 + J2θ̇2 · · · + Jnθ̇n
• For EE to achieve any arbitrary velocity, Jacobian must have 6 independent
columns.
• The rank of a matrix is the number of linearly independent columns (rows).
• If rank(J) = 6, EE can execute any velocity.
• For J ∈ R6×n, we have rank(J) ≤ min(6, n).
• The rank of Jacobian is configuration dependent.
• The configurations for which the Jacobian looses rank is called singular
configurations
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Singularities
• At singularities:
– Certain directions of motion unattainable
– Bounded EE velocity may correspond to unbounded joint velocities
– Bounded joint torques may correspond to unbounded EE forces and
torques
– Correspond to points on the boundary of workspace
• Most manipulators have configuration for which Jacobian becomes singular.
• All manipulators have singularities in their boundaries
• Many have singularities in the workspace, too.
– Workspace-boundary singularities: Fully stretched or folded back:
end-effector near the boundary
– Workspace-interior singularities: two or more joint axes line up :
end-effector away from the boundary
• Rank deficiency for square matrix: det(J(θ)) = 0.
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Decoupling Singularities
• For decoupled manipulators, decoupled singularities is defined
• Decoupling Singularities:
– Recall that for manipulators having 3 intersecting axes, the Kinematic
problem is decoupled.
– Hence, we define arm singularities and wrist singularities
• Consider a 6 DOF manipulator (3 DOF arm and 3 DOF spherical wrist)
• Jacobian is a 6 × 6 matrix and configuration θ is singular if and only if (iff)
det(J(θ)) = 0
• If the Jacobian is partitioned into 3 × 3 blocks:
J11 J12
J = [JP |JO ] =
J21 J22
• Since the final 3 joints are always revolute:
z4 × (oN − o4) z5 × (oN − o5) z6 × (oN − o6)
JO =
z4
z5
z6
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Decoupling Singularities
• Since all axes intersect, if we select o4 = o5 = o6 = oN = o, then:
0 0 0
JO =
z4 z5 z6
J11 0
J =
J21 J22
∴ det(J) = det(J11)det(J22)
• J11 and J22 are 3 × 3 matrices
• Manipulator singularities is the union set of arm configurations for which
det(J11) = 0 and set of wrist configurations for which det(J22) = 0
• Wrist Singularities:
– Singular position if the vectors z4 , z5 , z6 are linearly dependent
– Happens if θ5 = 0 or 180 for which z4 and z6 line up
– The only singularity of the wrist and is unavoidable
• Arm Singularities:
– Need to compute det(J11 ) = 0
– Origin of EE ON , replaced by wrist center o in Jacobian formulation
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Singularities: Example
• Two-link planar manipulator:
– Obtain the Jacobian and examine the determinant:
det(J(Θ)) = det
l1s2
0
l1c2 + l2 l2
= l2l1s2
(1)
– Singular position at θ2 = 0 or 180.
– θ2 = 0 the arm is stretched out and the motion is possible in only one
direction (lost 1 DOF)
– Same is true for θ2 = 180.
– These are workspace-boundary singularities.
• Example 2:
– The EE moves in X̂ direction at 1.0 m/s, find the joint rates
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Singularities
• Example 2
– Get the Jacobian in frame {0}:
1
J −1(Θ) =
l1l2s2
l2c12
l2s12
−l1c1 − l2c12 −l1s1 − l2s12
(2)
– Then by given Cartesian velocity and using Θ̇ = J −1 Ẋ :
c12
l1s2
c1
c12
−
θ̇2 = −
l2s2 l1s2
θ̇1 =
– At singularity (θ2 = 0), both joint rates tend to infinity.
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Static Forces
• How forces and moments propagates from one link to another
• Manipulator in contact or holding a load
• What are the joint torques required to keep the manipulator in static equilibrium
• In doing so, manipulator is locked (acts like a structure)
• Write the force-moment balance relationship in terms of the link frames
• Get the joint torque must be acting about the joint axis to keep the manipulator in static equilibrium
• Torque needed to support a static load acting at the end-effector
• Gravity is not considered (will be, in Dynamics)
• We consider the static force and torque acting on the last link (for example,
when the manipulator is in contact with the environment)
• Define:
– fi = force exerted on link i by link i − 1
– ni = moment exerted on link i by link i − 1
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Static Forces
• Newtonian mechanics:
1. For every action, there is a reaction. Hence, if body 1 applies the force
f and the moment (torque) n to the body 2, then body 2 applies a force
−f and the moment −n to the body 1
2. The rate of change of the linear momentum equals the total forces applied to the body:
d(mv)
= f
dt
3. The rate of change of the angular momentum equals the total moments
applied to the body:
d(I0ω0)
= n0
dt
• The moment exerted by a force f about a point is given by:
n = f ×r
where r is the radial vector from the point where the force applied to the
point about which the moment is calculated
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Static Forces
• Balance of forces:
i
fi − ifi+1 = 0
• Balance of torques (moments):
i
ni − ini+1 − iPi+1 × ifi+1 = 0
• We can get the relationship starting from the last link to the base:
i
fi = ifi+1
i
ni = ini+1 + iPi+1 × ifi+1
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Static Forces
• Use the rotation matrix to write all forces/moments in their own frames
fi = iRi+1 i+1fi+1
i
ni = iRi+1 i+1ni+1 + iPi+1 × ifi
i
• What joint torques balance reaction forces/moments acting on the
links?
• The joint torque is applied to the Ẑ axis of the ith frame, so the
forces/moments need to be projected on Ẑ axis.
• The dot product of the joint axis vector with the moment vector gives us the
torque required about the joint axis:
τi = iniT Ẑi
• if joint i is prismatic, joint actuator force is:
τi = ifiT Ẑi
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Static Forces: Example
• A two-link planar manipulator:
• A force vector 3F acting on the end-effector. Find the required joint torques.
– Propagate the force starting from the last link


  

fx
fx
0
2
f2 =  fy  , 2n2 = l2X̂2 ×  fy  =  0 
0
0
l2fy
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Static Forces: Example
• Propagate the force starting from the last link

  

c2 −s2 0
fx
c2fx − s2fy
1
f1 =  s2 c2 0   fy  =  s2fx + c2fy 
0 0 1
0
0




0
0
1
1

0
n1 =  0  + l1X̂1 × f1 = 
l2fy
l1s2fx + l1c2fy + l2fy
• Hence,
τ1 = l1s2fx + (l1c2 + l2)fy
τ2 = l2fy
fx
l1s2 (l1c2 + l2)
τ =
0
l2
fy
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Jacobian in Force Domain
• When forces act on a mechanism, work is done if the mechanism moves
through a displacement.
• Work:
– Force acting through a distance
– A scalar with unit of energy
– the dot product of force/torque vector and displacement vector:
F.δx = τ.δΘ Or F T δx = τ T δΘ
F : a 6 × 1 Cartesian force-moment vector acting at EE
δx : a 6 × 1 infinitesimal Cartesian displacement of EE
τ : a 6 × 1 vector of torque at the joint
δΘ : a 6 × 1 vector of infinitesimal joint displacements
• The Jacobian relates δx to δΘ: δx = JδΘ
• Therefore, for all δΘ, we have: F T JδΘ = τ T δΘ
• Hence,
τ = JTF
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Jacobian in Force Domain
• When the Jacobian is in frame {0}
the force vectors written in {0} can
be transformed to joint torques:
τ = 0J T 0F
(3)
• (3) allow us to convert a Cartesian quantity into a joint-space quantity without calculating any inverse kinematic functions
• If Jacobian loses full rank (i.e., Jacobian is Singular):
– EE can not exert static forces even if desired.
– F could be increased or decreased in a certain directions (those defining
the null-space of the Jacobian) without any effect on τ .
– Near singularity:
with small joint torque, large forces could be generated at EE
– ∴ singularities manifast themselves in the force domain as well as position domain.
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Inverse Velocity
ν = J θ̇
(4)
• The inverse velocity problem is the problem of finding the joint velocities θ̇
that produce the desired end-effector velocity
• If the Jacobian is square and nonsingular:
θ̇ = J −1ν
• For manipulators that do not have six joints, Jacobian cannot be inverted
• In this case, (4) has solution iff ν lies in the range space of the Jacobian.
• This can be done by the following rank test:
rank(J(θ)) = rank ([J(θ) | ν])
• If n > 6, we can solve for θ̇ using the pseudoinverse:
• For J ∈ Rm×n, if m < n and rank(J) = m, then (JJ T )−1 exist, since
(JJ T ) ∈ Rm and has rank m.
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Inverse Velocity
T −1
T
T −1
∴ I = (JJ )(JJ ) = J J (JJ )
= JJ + J + = J T (JJ T )−1
T
• J + = J T (JJ T )−1 is called right pseudoinverse of J since JJ + = I
• Note that J +J ∈ Rn×n and in general J +J 6= I .
• Hence, to solve for θ̇, we can use:
θ̇ = J +ν + (I − J +J)b
• where b is an arbitrary vector
• In general, for m < n, (I − J +J) 6= 0 and all vectors of the form (I −
J +J)b lie in the null space of J .
• This means that if θ̇0 is a joint velocity s.t. θ̇0 = (I − J +J)b, then when the
joint move with the velocity of θ̇0 , EE remains fixed since J θ̇0 = 0
• We can apply a similar approach when the analytical Jacobian is used:
Ẋ = Ja(θ)θ̇
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Inverse Velocity
J θ̇ = ν
• There are infinitely many solutions (m < n)
• Get the solution which results in minimum of kθ̇k
minθ̇ θ̇T θ̇
subject to J θ̇ = ν
• Using the lagrange multiplier:
min θ̇T θ̇ + λT (J θ̇ − ν)
θ̇
• Differentiation wrt θ̇:
2θ̇ + J T λ = 0 =⇒ θ̇ = −J T λ/2
• Differentiation wrt λ:
J θ̇ = ν =⇒ − JJ T λ/2 = ν
λ = −2(JJ T )−1ν
+
θ̇ = J T (JJ T )−1•νFirst •Prev
= J•Next
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Inverse Acceleration
• Differentiating the above equation results in an expression for acceleration:
d
Ja(θ) θ̇
Ẍ = Ja(θ)θ̈ +
dt
• Given a vector Ẍ of EE acceleration, find the joint acceleration by solving:
d
Ja(θ)θ̈ = Ẍ −
Ja(θ) θ̇
dt
• Hence, for a 6 DOF manipulator in a nonsingular position, the inverse velocity and acceleration can be written as:
θ̇ = Ja(θ)−1Ẋ
d
θ̈ = Ja(θ)−1 Ẍ −
Ja(θ) θ̇
dt
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Acceleration of a Rigid Body
• Linear and angular accelerations are the derivatives of linear and angular
velocities:
B
A
V̇Q
Ω̇B
B
VQ(t + ∆t) − BVQ(t)
dB
=
V = lim
dt Q ∆t−→0
∆t
A
dA
ΩB (t + ∆t) − AΩB (t)
=
Ω = lim
dt B ∆t→0
∆t
• If the reference frame of differentiation is understood to be the universal
frame {U }:
v̇A =
ẇA =
U
V̇AORG
U
Ω̇A
• Linear Acceleration
– Recall the velocity of a vector BQ as seen from {A} when the origin
are coincident:
A
VQ =
A B
BR V Q
+ AΩB × BAR BQ
d A B
( R Q) =
dt B
A B
BR V Q
+ AΩB × BAR BQ
(5)
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Acceleration of a Rigid Body
• To get the acceleration, differentiate the velocity vector: (acceleration of
B
Q as viewed from {A} when the origins coincide
A
V̇Q =
d A B
d
( BR VQ) + AΩ̇B × BAR BQ + AΩB × ( BAR BQ )
dt
dt
• Use (5) twice (one for the first term and one for last):
A B
BR V̇ Q
+
+
A
ΩB × BAR BVQ + AΩ̇B × BAR BQ
A
ΩB × ( BAR BVQ + AΩB × BAR BQ )
(6)
• Combining the terms, we arrive at:
A B
BR V̇ Q
+ 2 AΩB × BAR BVQ + AΩ̇B × BAR BQ + AΩB × ( AΩB × BAR BQ )
• Generalize to the case where the two frames do not coincide:
A
B V̇ BORG
+
+
A B
A
A B
BR V̇ Q + 2 Ω B × BR V Q
A
ΩB × ( AΩB × BAR BQ )
+ AΩ̇B × BAR BQ
(7)
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Acceleration of a Rigid Body
• When the point ( BQ ) is constant:
B
B
VQ = BV̇Q = 0
V̇Q = AV̇BORG + AΩB × ( AΩB × BAR BQ ) + AΩ̇B × BAR BQ
• When the joint is prismatic, the general form is used.
• Angular Acceleration:
– Consider when {B} is rotating wrt {A} with AΩB and {C} is rotating
wrt {B} with BΩC . Find AΩC
– We express all velocities in frame {A} and add them together:
A
ΩC = AΩB + BAR BΩC
– Differentiating:
A
Ω̇C = AΩ̇B +
d A B
( R ΩC )
dt B
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Angular Acceleration
• Applying (5) to the last term:
A
Ω̇C = AΩ̇B + BAR BΩ̇C + AΩB × BAR BΩC
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