Control of systems by parallel actuators

STEPHEN MONTGOMERY-SMITH

Department of Mathematics

University of Missouri, Columbia, MO 65211

U.S.A.

Abstract: We describe a particular control method for a system controlled by several actuators with the same

control constants. We show under certain assumptions that the control constants for the whole system can be

obtained immediately from the control constants for a single actuator, and that no gain scheduling is required.

Key-Words: Parallel actuator, linear control theory, Riemannian manifold, tangent space, cotangent space,

wrench set.

Received: May 18, 2021. Revised: March 7, 2022. Accepted: April 10, 2022. Published: May 4, 2022.

1 Introduction

In this paper, we describe a system whose state, which

we will call the end eector position, is given by an el-

ement of a dierential manifold η∈ M. We suppose

that the end eector position is determined by the val-

ues of nactuators:

ℓ= (ℓ1, ℓ2, . . . , ℓn)∈Rn.(1)

We write L:M → Rnfor the function which maps

the end eector position to the actuator values. Note

that we can allow the problem to be over-constrained,

that is, ncan be bigger or equal to the dimension of

M. Note that the elements of Mare generalized po-

sitions in the sense of the Euler-Lagrange formalism

[1].

Examples of these are cable-driven parallel robots,

consisting of a xed rigid frame, and a oating rigid

body called the end eector. The end eector is ma-

nipulated via eight cables attached to actuators. Each

actuator is clamped to the xed frame. The value,

ℓk, of the kth actuator is the length of cable issued

by the actuator. The manifold Mis the six dimen-

sional space of poses, that is, positions with orienta-

tions. See [2, 3, 10] for an introduction to parallel

robots. See [8, 9] for information about cable-driven

parallel robots. The algorithm described in this paper

was tested upon a cable-driven parallel robot built by

the NASA Johnson Space Center, which we describe

in another paper [7].

We assume that the only measurements we can

take are the values of the actuators, and that the only

method of control is to command a force at each actu-

ator, which converts into a force on the end eector.

We assume that once the end eector force is known,

one can determine the trajectory of the end eector

via a frictionless, quadratic Hamiltonian system. The

force is a generalized force obeying the principle of

virtual work, and we will call it the end eector force.

The method of control is to calculate the end ef-

fector position using the actuator values. Then, from

the dierence of the actual end eector position from

the requested end eector position, is calculated the

required acceleration of the end eector. From this is

calculated the required end eector force. Finally, we

nd the actuator forces to eect this.

Our main contribution is to show that the control

constants for computing the required acceleration of

the end eector are the same as the control constants

used to control a single actuator, making it easier to

determine the control method. This result requires

two things. First that the actuator response is linear.

This would be violated if, for example, if there is sig-

nicant non-linear friction. Second, it requires that

the each actuator by itself can be controlled by the

same linear controller. For example, with the cable-

driven parallel robot, if there is signicant exing or

stretching of cables, this would require a dierent con-

trol method for each actuator because the orientations

and/or lengths of the cables aren’t necessarily identi-

cal.

Since the control constants can be found by con-

sidering a controller for a single actuator, then the

same control constants can be used irrespective of the

end eector position, that is, no gain scheduling is re-

quired. Thus the formula for resonant frequencies can

also be calculated from the resonant frequencies of a

single actuator.

We believe the methods and results in Sections 4

and 5, where we show that the control constants come

from those described in Section 3, are new. But it is

possible that these methods were developed in a com-

pletely dierent context, and that the author is simply

unaware of them.

2 Mathematical description of the

system

Denote by TηMthe tangent space of Mat η, by

TM=∪η∈MTηMthe whole tangent space, and by

T∗M=∪η∈MT∗

ηMthe cotangent space. See [6]

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for information on manifolds.

Given a position that is a function of time, η(t), we

have its velocity and acceleration which are functions

of time taking their values in TMgiven by

φ= ˙η, α = ˙φ. (2)

We dene the linear operator Λη:TηM → Rnby the

directional derivative, which in local coordinates is

Ληθ=θ·dL(η)

dη .(3)

Often we simply write Λfor Ληwhen there is no con-

fusion.

Then, from the velocity φ, we can calculate the rate

of change of the actuator values:

ℓ= Ληφ. (4)

We suppose that there is a linear operator T=Tη:

Rn→T∗M, which converts actuator forces f=

(f1, f2, . . . , fn)to the end eector force τ:

τ=Tηf.(5)

By the principle of virtual work, we have

φ·Tηf=˙

ℓ·f= Ληφ·f.(6)

Hence

Tη= ΛT

η.(7)

Next we describe the equations of motion. Sup-

pose that the system is given by a Lagrangian

l(η, φ) = 1

2Mη(φ, φ)−v(η),(8)

where Mη=Mis a positive denite bilinear operator

on TηM. This includes kinetic energy of the actuators

2m0|˙

ℓ|2,(9)

where m0is the eective mass of the each actuator

(the notion of ‘eective’ is explained in Section 3 be-

low). In local coordinates we can describe Λand M

as matrices, then the kinetic energy of the actuators is

given by

2m0φTΛTΛφ, (10)

and so we must have that the matrix

M−m0ΛTΛ(11)

is positive semi-denite.

Solving the Euler-Lagrange equations [1], we ob-

tain the equations of motion

τ=Mα +µ(η),(12)

where in local coordinates

µ(η) = φ·∂Mη

∂η (φ, ·)−∂

∂η v(η).(13)

We dene the no-load forces to be the actuator

forces if the actuators are not attached to the system,

that is, the Lagrangian is given simply by equation (9):

f0=m0¨

ℓ.(14)

Thus dierentiating equation (4), we obtain (in local

coordinates)

f0=m0Ληα+m0φ·dΛη

dη φ. (15)

For the example of the cable-driven parallel robot, the

cable tensions are given by

cable tensions =f0−f.(16)

Next, we need inverse functions to Land T, which

we call Yand F. We dene the set of admissible ac-

tuator values,L ⊂ Rn, to be the range of the func-

tion L. We suppose that we have a forward kinemat-

ics function, Y:L → M, which is a left inverse to

L. Because of possible measurement errors, Yshould

produce decent answers even if the actuator values are

merely close to L. For example, this could be imple-

mented using the Newton-Raphson Method.

For the inverse function of T, we need some more

denitions. Given fb,f0∈Rn, we suppose that we

have a predened set Cfb,f0⊂Rn. Here fbis the

command force required to overcome actuator resis-

tance such as back-EMF, f0is the no-load actuator

forces, and Cfb,f0is the set of those fsuch that it is

permissible to command forces f+fbto the actua-

tors.

Typically this is a convex set dened by a nite

number of linear constraints. For the example of

cable-driven parallel robots, we might say f∈ Cfb,f0

if and only if the tensions in the cables, f0−f, is

never below a given predened value, and the com-

mand forces ±(f+fb)don’t exceed the actuator hard-

ware limits.

Then we dene the wrench set to be the set of

achievable forces:

W={(τ, fb,f0)∈T∗M × Rn×Rn:

∃f∈ Cfb,f0such that Tf=τ}.(17)

We suppose that we have a function F:W × Rn×

Rn→Rnthat provides a right inverse to the map

dened by Tin the following manner:

T(F(τ, fb,f0)) = τ, (18)

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such that

F(τ, fb,f0)∈ Cfb,f0.(19)

Because there are more actuators than the number

of degrees of freedom (that is, the dimension of M,

computing the actuator forces in the last step is an

over-constrained problem. For the example of cable-

driven cable robots, there are many approaches in the

literature [4, 5, 8].

Finally we need a way to approximate the dier-

ence between two end eector positions by an ele-

ment of the tangent space. That is, there is a function

△:M×M → TMsuch that △(η1, η2)is in Tη1M,

so that with respect to a ‘reasonable’ coordinate sys-

tem about η1that

η2≈η1+△(η1, η2).(20)

For example, if Mis a Riemannian manifold, we

could dene it as the direction of a geodesic from η1

to η2. If Mis a Lie group, we could dene it as the

direction of a one parameter subgroup from η1to η2.

3 Control of a single actuator

Let ℓdenote actuator value, let fcbe the command

force given to a single actuator, and fbe the actual

force supplied by this actuator.

Each actuator has an eective no-load mass m0,

which is the ratio f/¨

ℓwhen there is no load placed

upon the actuator.

For the remainder of this section, we suppose that

the actuator is carrying a passive load. We denote

by mto be the eective mass of the actuator with

this load. Thus no load corresponds to m=m0,

and we always have m≥m0. If the actuator is

clamped so that it cannot move, this corresponds to

m=∞, which is a mathematical idealization repre-

senting when the passive load is very large.

For the purpose of making the analogue of these

equations and the system controller equations clearer,

we shall replace the actuator actual and command

forces by actual and command accelerations

a=f

m=¨

ℓ, (21)

ac=fc

m.(22)

We look for a controller such that, given an actua-

tor value ℓr, attempts to create a command accelera-

tion acsuch that the actual actuator value, ℓ, is close

to ℓr.

First we describe an open-loop controller:

fc=m¨

ℓr+k0˙

ℓr,(23)

ac=¨

ℓr+k0

m˙

ℓr.(24)

We call k0the back-EMF constant, since for electric

motors this is a likely source of this term. This con-

troller fails badly if there is any drift or noise in the

system, because it makes no attempt to correct for er-

ror.

Next, suppose we also have a good homogeneous

closed-loop controller. Denote the vector containing

all time derivatives of order less than lof ℓby

ℓ= [ℓ, ˙

ℓ, ¨

ℓ, . . . , ℓ(l−1)]T.(25)

The controller is dened by appropriately sized con-

stant matrices A,B,C, and Das:

x=Ax+Bℓ (26)

ac=Cx+Dℓ.(27)

By a good homogeneous closed-loop controller, we

mean that if this is used to control the passively loaded

actuator, then ℓconverges to 0in a manner that is ex-

peditious enough for our application.

An example is a PID controller

ac=−(ki∫ℓ+kpℓ+kd˙

ℓ),(28)

But it could be something more complex, such as a

cascaded controller, or a linear quadratic Gaussian

controller.

Note that if cable stretching or sagging plays a sig-

nicant role in the cable-driven parallel robot, then

this should be able to control a single actuator with

a cable with similar stretching or sagging characteris-

tics attached.

The results of this paper don’t depend upon what

denition of ‘good’ we use. Our assertion is that the

parallel actuator driven robot controller behaves as

well as the single actuator controller.

We combine the open-loop and homogeneous

closed-loop controller to obtain a good closed-loop

feed-forward controller, that is, given a requested ac-

tuator value ℓr, and its vector of derivatives

ℓr= [ℓr,˙

ℓr,¨

ℓr, . . . , ℓ(l−1)

r]T,(29)

we nd the command acceleration acsuch that ℓcon-

verges to ℓrin an expeditious manner. This can be cre-

ated by applying the homogeneous closed-loop con-

troller to

ℓd=ℓ−ℓr(30)

ℓd=ℓ−ℓr(31)

to obtain

x=Ax+Bℓd(32)

ac=¨

ℓr+k0

m˙

ℓr+Cx+Dℓd.(33)

For example, with the PID controller it is

ac=¨

ℓr+k0

m˙

ℓr−(ki∫ℓd+kpℓd+kd˙

ℓd).(34)

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4 The controller for the system by

parallel actuators

For the controller, we introduce the state vector ξ,

which is a vector of elements from the tangent space,

with the same number of components as the state vec-

tor xdescribed in equation (26). We denote a matrix

multiplied by a vector of elements from the tangent

space as giving another vector of elements of the tan-

gent space as follows:

(Aξ)i:= ∑

Ai,j ξj,(35)

where ξjmeans the jth component of ξ.

The control loop is as follows.

1. Obtain the requested end eector position ηr, and

compute the requested acceleration

αr= ¨ηr.(36)

2. Measure actuator values ℓ.

3. Calculate the actual end eector position:

η=Y(ℓ).(37)

4. Find the △dierence between the actual end ef-

fector position and the requested end eector po-

sition:

θd=△(η, ηr)(38)

and compute the vector of derivatives

θd= [θd,˙

θd,¨

θd, . . . , θ(l−1)

d]T.(39)

5. Calculate the command end eector acceleration:

ξ=Aξ+Bθd(40)

αc=αr+Cξ+Dθd.(41)

For example, the PID controller would be:

αc=αr−(ki∫θd+kpθd+kd˙

θd).(42)

6. Determine the command end eector force to be

applied to the end eector:

τc=µ+Mαc.(43)

7. Calculate the requested rate of change of the

lengths of the cables

ℓr= Λ(ηr)φr,(44)

and nd the resistance overcoming part of the

command force for the actuators

fb=k0˙

ℓr.(45)

ηr

△

Controller

θd

αc

End eector

force

calculator

µ+M×

Actuator

command

calculator

Actuators

Forward

kinematics

ℓ

dt2

αr

Compute

no-load

actuator

force

Rate of change

of actuator values

from end eector

position

Figure 1: Block diagram of the controller for the robot

driven by parallel actuators.

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8. Use equation (15) to calculate the no-load actua-

tor forces.

f0=m0Ληα+m0φ·dΛη

dη φ. (46)

9. Determine whether (τc,fb,f0)∈ W. If it is, cal-

culate part of the actuator forces using

fp=F(τc,fb,f0),(47)

and command the actuators with

fc=fp+fb.(48)

Otherwise, declare that the end eector is out of

its workspace, command the actuators to brake,

and quit.

10. Go back to Step 1.

Note that in Steps 6, 8, and 9, the quantities M,µ,

Λ,W, and Fare calculated from ηand φ, but they

could just as well be calculated from ηrand φr, the

only change required being to equation (38) to

θd=− △ (ηr, η)(49)

The algorithm is shown as a block diagram in Fig-

ure 1.

5 Theoretical justication for the

controller

Assumption 1 The interaction between the command

force fc, the actual force f, and the actuator value ℓ

of a single actuator, is given by the linear system

f+c2˙

f+· · · +cnf(n−1)

+k0˙

ℓ+k1¨

ℓ+· · · +knℓ(n+1)

=fc+ ˜c2˙

fc+· · · + ˜cpf(p−1)

c.(50)

Note this linearity can be dicult to achieve if fric-

tion is signicant in the actuators. Friction is highly

non-linear, especially when ˙

ℓand fswitch between

having the same sign and having dierent signs, as

could happen with an active load.

If the actuator has a passive load mas described in

Section 3, then equation (50) becomes

ℓ+c2ℓ(3) +· · · +cnℓ(n+1)

+k0

m˙

ℓ+k1

m¨

ℓ+· · · +kn

mℓ(n+1)

=ac+ ˜c2˙ac+· · · + ˜cpa(p−1)

c.(51)

The original open-loop controller was derived from

the assumption that equation (50) is a perfect descrip-

tion of the passively loaded actuator:

fc+ ˜c2˙

fc+· · · + ˜cpf(p−1)

=m(¨

ℓr+c2ℓ(3)

r+· · · +cnℓ(n+1)

+k0˙

ℓr+k1¨

ℓr+· · · +knℓ(n+1)

r,(52)

which in Section 3 was approximated with equa-

tion (23).

Assumption 2 For every m∈[m0,∞], equa-

tions (30),(31),(32) and (33) provide a good closed-

loop feed-forward controller for system (51).

This assumption can either be tested theoretically

using eigenvalue analysis, or experimentally by load-

ing various passive loads onto a single actuator. The

latter approach doesn’t require any knowledge of the

coecients in equation (50). We simply need to be-

lieve that such an equation exists.

Assumption 3 The time scale of the corrections θd

is much smaller than the time change of ηr, and the

magnitude of θdand its derivatives are much smaller

than that of ηand its corresponding derivatives. This

means we can assume that the time derivatives of η

are negligible compared to η, and hence we can as-

sume the matrices M,F, and the covector µ, and

their derivatives, are constant in the time scales in

which the controller operates. We also assume that

equation (20) holds for time derivatives of both sides,

and that the constants are uniformly controlled in the

ranges achieved.

The main result of this paper, which we state be-

low, is described as an ‘assertion’ rather than a ‘theo-

rem,’ as the proofs are not very rigorous.

Assertion 1 Given Assumptions 1, 2, and 3, the algo-

rithm described in Section 4 is a good controller for

the parallel actuator driven robot.

Pick a time t0which is in the range of times in

which the controller performs the required correc-

tions. Let η0=η(t0). Dene

θ=△(η, η0)(53)

θr=△(ηr, η0).(54)

Thus

η≈η0+θ(55)

ηr≈η0+θr,(56)

and if we set

φr= ˙η(57)

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then

φ≈˙

θ(58)

φr≈˙

θr,(59)

and we have

θd≈θ−θr.(60)

Let nMbe the dimension of M, and dene the

(nM×nM)matrix

N=M−1ΛTΛ.(61)

Assertion 2 With the same assumptions as Asser-

tion 1, if the end eector is controlled by the algorithm

given in Section 4, then we have

θ+c2θ(3) +· · · +cnθ(n+1)

+N(k0˙

θ+k1¨

θ+· · · +knθ(n+1))

≈αc+ ˜c2˙αc+· · · + ˜cpα(p−1)

c(62)

where

αc≈ˆαc+k0N˙

θr(63)

and ˆαcis calculated thus:

θd= [θd,˙

θd,¨

θd, . . . , θ(l−1)

d]T(64)

ξ=Aξ+Bθd(65)

ˆαc=αr+Cξ+Dθd.(66)

Proof: Rewrite equation (50) for the jth actuator:

fj+c2˙

fj+· · · +cnf(n−1)

+k0˙

ℓj+k1¨

ℓj+· · · +knℓ(n+1)

=fc,j + ˜c2˙

fc,j +· · · + ˜cpf(p−1)

c,j .(67)

From equations (4), (45), (47), (48), and (59), we ob-

tain

fc,j =Fj(τc,¯

fc) + k0Λ˙

θ, (68)

and applying T= ΛTwe obtain

Tfc=τc+k0ΛTΛ˙

θr.(69)

Similarly, from equations (4) and (58), we have

Tf=τ, (70)

where τis the actual end eector force, and

T˙

ℓ= ΛTΛ˙

θ. (71)

Hence from Assumption 3, we obtain

τ+c2˙τ+· · · +cnτ(n−1)

+ ΛTΛ(k0˙

θ+k1¨

θ+· · · +knθ(n+1))

≈τc+k0ΛTΛ˙

θr+ ˜c2( ˙τc+k0ΛTΛ¨

θr)+

· · · + ˜cp(τ(p−1)

c+k0ΛTΛθ(p)

r).(72)

Next, subtracting µ, and then left multiplying by

M−1, and using Assumption 3 again, we obtain equa-

tion (62). The rest of the assertion follows by equa-

tion (60). Q.E.D.

Lemma 1 The matrix Nhas a basis of eigenvectors,

with eigenvalues in [0, m−1

0].

Proof: Let

N=M−1/2ΛTΛM−1/2.(73)

Clearly ˜

Nis symmetric and positive semi-denite,

and since M−m0ΛTΛis positive denite, we have

m−1

0I−˜

N=m−1

0M−1/2(M−m0ΛTΛ)M−1/2

(74)

is positive denite. (Here Idenotes the (nM×nM)

identity matrix.) Hence ˜

Nhas a basis of eigenvectors,

with eigenvalues in [0, m−1

0]. Also,

N=M−1/2 ˜

NM1/2,(75)

and hence Nand ˜

Nare similar matrices. Q.E.D.

Proof of Assertion 1: We use Lemma 1 to obtain σ1,

σ2, . . . , σnMa basis of eigenvectors of N, with cor-

responding eigenvalues m−1

1,m−1

2, . . . , m−1

nM, where

m1,m2, . . . , mnM∈[m0,∞]. We also form a dual

basis π1,π2, . . . , πnMthat satises

σi·πj=®1if i=j

0if i=j,(76)

so that for any ψ∈RnMwe have

ψ=

∑

i=1

(ψ·πi)σi(77)

πi·Nβ =m−1

i(πi·β).(78)

By Assumption 3, we can assume that all these vectors

and eigenvalues are constant. For each 1≤i≤nM,

dot product the equations in Assertion 2 by πi. Then

it may be seen that the resulting equations satisfy the

hypotheses of Assumption 2, with yreplaced by θ·πi,

with yrreplaced by θr·πi, and with mreplaced by mi.

Thus θ·πiis well controlled by θr·πi.

Then it follows by equation (77) that θis well con-

trolled by θr. Therefore by equations (55) and (56),

we have that ηis well controlled by ηr. Q.E.D.

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Contribution of individual authors to

the creation of a scientic article

(ghostwriting policy)

Stephen Montgomery-Smith is responsible for the en-

tire paper.

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WSEAS TRANSACTIONS on SYSTEMS and CONTROL

DOI: 10.37394/23203.2022.17.24

Stephen Montgomery-Smith

E-ISSN: 2224-2856

213

Volume 17, 2022