Tangent Space and Jacobian Transformations

Một phần của tài liệu a journey from robot to digital human (Trang 92 - 97)

In robotic kinematics, one often intends to find a velocity for both position and orientation of the robotic end-effector frame referred to the base. In general robotic applications, a TCP (Tool Center Point) is usually the origin of the end-effector frame. If we have a unified vector ξ = p

ρ

R6 to represent the 6 d.o.f. end-effector rigid motion, then the velocity is its time- derivative: ˙ξ = p˙

˙ ρ

. Suppose that a robot has n joints from the base to the end-effector. This 6-dimensional Cartesian position/orientation vectorξ is obviously a function of all the joint positions that form ann-dimensional vectorq= (q1 ã ã ã qn)T Rn, and its velocity

ξ˙=∂ξ

∂qq˙=Jq,˙ where

J = ∂ξ

∂q = (g1 ã ã ã gn)

is called the Jacobian matrix of the vectorξ and each column gi =∂ξ/∂qi is a 6-dimensional vector field for i= 1,ã ã ã, n. Using a differential form, we can rewrite it as

= n i=1

gidqi= n i=1

∂ξ

∂qidqi.

Based on the exterior calculus in Chapter 2, the above differential 1-form is clearly exact because its 2-formd2ξ= 0. Namely,

∂gi

∂qj = 2ξ

∂qj∂qi = 2ξ

∂qi∂qj = ∂gj

∂qi for everyi, j= 1,ã ã ã, n. (3.19) However, since all the definitions of the 3 by 1 orientation vectorρso far have a common fundamental defect in performing rotations, such an ideal exact 1-form may not be a reality.

Of course, we cannot take a time-derivative directly on a homogeneous transformation matrix

H = R p 03 1

to directly find its velocity, where 03is the 1 by 3 zero vector. Instead, inspired by the angular velocityΩ= ˙RRT, as given in equation (3.10), let us try to post-multiply the time-derivative of the above 4 by 4 homogeneous matrix by its inverse. Note that taking the transpose of a homogeneous transformation matrix will destroy its group membership, and thus it can only take the inverse. Namely, according to (3.5),

HH˙ 1= R˙ p˙ 03 0

RT −RTp

03 1

= Ω p˙−Ωp

03 0

. (3.20)

As we can see, the fourth column of the result is ˙p−ω×p. In mechanics, the absolute linear velocity of a rigid body motion at a point on the body is generally equal to vab =vre+ω×p, where vab is the absolute velocity,vre

is the relative velocity, andω is the angular velocity of the body. Therefore, the last column of the resulting matrix represents a relative velocity of the origin of the robot end-effector frame. In other words, this velocity is just the sliding velocity along the direction of the position vectorp.

Coincidentally, the dual number representation will reach exactly the same result. Let ˆR=R+S be a 3 by 3 special orthogonal dual matrix to repre- sent both the position and orientation of the robotic end-effector frame with respect to the base, where, based on equation (2.24),SRT =P =is the skew-symmetric matrix of the position vectorp, as introduced in Chapter 2.

Likewise, we try

R˙ˆRˆT = ( ˙R+S)(R˙ T +ST) = ˙RRT +( ˙RST + ˙SRT).

The real part of the above dual matrix is obviously ˙RRT = Ω. For its dual part, let us first compare it with the time-derivative of P =SRT, i.e., P˙ =SR˙T+ ˙SRT. We find that the second term ˙SRT of the dual part is exactly

the same as the second term of ˙P so that the dual part can be ˙RST+ ˙SRT = RS˙ T −SR˙T + ˙P. Now, replacing eachS byP R and noticingΩ= ˙RRT as well as that bothP andΩare skew-symmetric, we obtain the dual part

RS˙ T + ˙SRT = ˙RRTPT −P RR˙T+ ˙P = ˙P−ΩP+P Ω= ˙P−[Ω, P], where [Ω, P] = ΩP −P Ω is the Lie commutator and [Ω, P] = (ω×p)× according to (3.16). Finally,

R˙ˆRˆT =Ω+( ˙P−[Ω, P]) =Ω+( ˙P−(ω×p)×). (3.21) Comparing it with (3.20) in the above study on homogeneous transformation, it turns out to be so consistent that both the two most promising approaches to unifying the representation for every 6 d.o.f. rigid motion, the 4 by 4 ho- mogeneous matrix and the 3 by 3 special orthogonal dual matrix, all suggest to augment a linear velocityv= ˙pand an angular velocityω as a unification of the 6-dimensional Cartesian velocity representation. Namely,V = v

ω

, even though the differential 1-formσ=ωdtis not exact in general.

Under this suggestion, a 6-dimensional Cartesian velocity is simply defined as

V = v ω

= n i=1

¯

gidqi=Jq,˙ (3.22) where the top 3 by 1 part of each 6 by 1 column ¯gi in the matrix J = (¯g1 ã ã ã g¯n) is ∂p/∂qi while the bottom 3 by 1 part may not be a partial derivative of some vector. Therefore, more precisely, the above matrix J is not a genuine mathematical Jacobian matrix in general. However, since tradi- tionally such a transformation matrix from joint velocity to Cartesian velocity is named as a Jacobian matrix in the robotics community, let us just keep this tradition in all of our later discussions. Once again, it must be emphasized that for the linear velocity v = ˙p, the position vector pmust be referred to and projected onto the fixed base before taking a time-derivative on it.

Actually, equation (3.19) can also be adopted as a criterion to test anm bynmatrix J to see if it is a genuine Jacobian matrix. Namely, by picking up any two columnsgi andgj fromJ, if

∂gi

∂qj = ∂gj

∂qi for everyi, j= 1,ã ã ã, n, (3.23) then,J is a genuine Jacobian matrix, i.e., there is anm-dimensional smooth function ξ(q) forq∈Rn such that

J =∂ξ

∂q.

This criterion implies that if one single pair of columnsgk and gl from the matrixJ violates the criterion, the genuine Jacobian conclusion is blown up.

As a typical example, let us look at a 6-joint Stanford robot arm with five revolute joints and one prismatic joint. The joint position vector is given by q= (θ1θ2d3θ4θ5θ6)T. Its detailed kinematic modeling will be discussed in the next chapter. Here, we just show its Jacobian matrixJ(3)that is projected onto frame 3, the frame of link 3, as follows:

J(3)=

⎜⎜

⎜⎜

⎜⎝

d2c2 d3 0 0 0 0

−d3s2 0 0 0 0 0 d2s2 0 1 0 0 0 s2 0 0 0 −s4 c4s5

0 1 0 0 c4 s4s5

−c2 0 0 1 0 c5

⎟⎟

⎟⎟

⎟⎠ ,

where eachci= cosθi andsi= sinθifor a short notation.

According to the criterion given in (3.23) to test if a differential 1-form is closed, this 6 by 6 matrixJ(3)is not a genuine Jacobian matrix at all, because by just picking up the second and third columnsg2 andg3, we have

∂g3

∂θ2 =

⎜⎜

⎜⎜

⎜⎝ 0 0 0 0 0 0

⎟⎟

⎟⎟

⎟⎠

, while ∂g2

∂d3 =

⎜⎜

⎜⎜

⎜⎝ 1 0 0 0 0 0

⎟⎟

⎟⎟

⎟⎠ ,

and thus,

∂g3

∂θ2

= ∂g2

∂d3

, which has already violated the criterion (3.23).

Now, let the matrixJ(3) be converted toJ(0) by pre-multiplying a 6 by 6 augmented rotation matrix:

R03 O O R03

with

R30=

c1c2 s1 c1s2 s1c2 −c1 s1s2 s2 0 −c2

for re-projecting the matrix onto the base. Then, the top half ofJ(0)becomes

J(0)(1 : 3,:) =

d2c1−d3s1s2 d3c1c2 c1s2 0 0 0 d2s1+d3c1s2 d3s1c2 s1s2 0 0 0 0 d3s2 −c2 0 0 0

.

This is, indeed, a genuine Jacobian matrix, because the criterion (3.23) holds for everyi, j= 1,ã ã ã,6. For instance,

∂g1

∂θ2

= ∂g2

∂θ1

=

−d3s1c2 d3c1c2

0

, ∂g1

∂d3

= ∂g3

∂θ1

=

−s1s2 c1s2

0

,

and

∂g2

∂d3

= ∂g3

∂θ2

=

c1c2

s1c2 s2

.

In fact, the integral of J(0)(1 : 3,:) is just the position vector of frame 6 underd6= 0, i.e.,

p60=

d3c1s2 d3s1s2 d1−d3c2

such that the top half ofJ(0) is exactly equal to J(0)(1 : 3,:) = ∂p60

∂q .

This also means that the differential 1-form σv =v0dt=dp60 for the linear velocity is exact.

In contrast, the bottom half of eitherJ(3) orJ(0)directly violate the crite- rion (3.23) and thus, it will never be a genuine Jacobian matrix. For instance, since

J(0)(4 : 6,1 : 3) =R30J(3)(4 : 6,1 : 3) =

⎝0 s1 0 0 −c1 0

1 0 0

,

for the first and second columnsg1 andg2 inJ(0)(4 : 6,:),

∂g1

∂θ2 = 0, while ∂g2

∂θ1 =

c1

s1

0

= 0.

Therefore, this Stanford robot example demonstrates that the linear ve- locity part can achieve a genuine Jacobian matrix, provided that the linear velocity and the Jacobian matrix are all projected onto the base. Whereas the angular velocity part is not, or the differential 1-form σω =ωdtfor the angular velocity is neither closed, nor exact. Reality once again exposes the complication of rotation.

In future studies on robotic system control models and applications, since each robot is a nonlinear dynamic system, we will have to exactly linearize it before developing its control strategy. Often, in a feedback linearization proce- dure, a complete velocity to cover all the d.o.f (degrees of freedom) of motion

is required to be the time-derivative of a Cartesian position/orientation, so that the Jacobian matrix has to be genuine. If each velocity could be an ex- act differential 1-form, developing an effective control law for the nonlinear robotic system would be much easier. The detailed discussion and extensive development of robotic systems control as well as the human-machine dy- namic interactive control will be addressed and covered in later chapters.

Một phần của tài liệu a journey from robot to digital human (Trang 92 - 97)

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