medical robotics - mnrlabmedical robotics islam s. m. khalil, omar shehata, nour ali, and ali medhat...

Medical Robotics

Islam S. M. Khalil, Omar Shehata, Nour Ali, and Ali Medhat

German University in Cairo

April 4, 2018

Islam S. M. Khalil, Omar Shehata, Nour Ali, and Ali Medhat Soft Robots

Dynamics and Control for Planar Continuum Robots

Figure: Spring-steel backbone is used to construct a soft robot and wiresare pulled to control its tip. Four stepper motors are used in the jointspace (q1, . . . , q4) to control the parameter space (κ, φ, `) through thespecific mapping fspecific. The parameter space is mapped onto the taskspace using the independent mapping findependent.



Figure: Spring-steel backbone can bend in excess of 180◦.Gravagne et al. Trans. on Mech. 2003.



Figure: Differential tension load cell system. Cables, pulleys, and winchare emphasized in white. Gravagne et al. Trans. on Mech. 2003.



The set C of allowable beam planar configuration contains{x(s),Q(s)} where

C ={{x(s),Q(s)} : [0, L]→ R2 × SO(2)

}x(s): is the position vector of the beam centerline

Q(s): is the rotation matrix

s: is an independent parameterization variable (If the beam isinextensible, represents the arc length as measured from theorigin)

L: is the beam length at rest.



For beams with high aspect ratios (length to thickness), potentialenergy can be stored via three basic types of elastic deformations:

Bending

Axial

Shear

For a beam of cross-sectional area A, cross-sectional moment ofinertia I , Youngs modulus E and shear modulus G , each elasticdeformation has an associated stiffness.

Bending stiffness: EI

Axial stiffness: EA

Shear stiffness: GA



For a beam of cross-sectional area A,cross-sectional moment of inertia I , Youngsmodulus E and shear modulus G

Bending stiffness: EI

Axial stiffness: EA

Shear stiffness: GA

The figure illustrates the difference betweenshear and bending deformations, in thecase that there is no axial compression ortension. Note that, in case a), theorientation vector q1 does not vary alongthe length of the beam. The beam hasbeen divided into (infinitesimally) smallsegments, and Q reflects the orientation ofeach segment.

Figure: Beam in (a) experiencessignificant shear deformations,but no bending. Conversely, in(b) shear effects are not present.Note that the curve tangent isthe same in both cases.Gravagne et al. Trans. on Mech.2003.



A good measure of the axial and shearstretch of each segment would be thedifference of the tangent vector and theprincipal orientation vector

v = x′ − q1 and x′ = ∂x/∂s

Applying Hooke’s Law, the resultantdeformation energies for that segmentwould be

Resul. def. energies =1

2EAv21 +

1

2GAv22

Resul. def. energies =1

2vTCv

where C = diag {EA,GA}

Figure: Beam in (a) experiencessignificant shear deformations,but no bending. Conversely, in(b) shear effects are not present.Note that the curve tangent isthe same in both cases.Gravagne et al. Trans. on Mech.2003.



To generalize this initial formulation,imagine that the columns of C representthe principal axes of a stiffness ellipsoidcentered around the segment. As theorientation of a segment changes, theellipsoid rotates also, to keep the principalaxes properly aligned. A simple similaritytransformation will accomplish thisrotation, so that shear/axial energy for thatsegment becomes

1

2vTQCQTv

Figure: Illustration of a smallbeam segment experiencingaxial, shear, and flexuraldeformation. Note how thestiffness ellipsoid changesorientation along the length ofthe beam. Gravagne et al.Trans. on Mech. 2003.



Adding up the segment energies gives thepotential energy

PE =1

2

∫ L

0

{1

2vTQCQTv + EIθ′2

}ds

The kinetic energy is more straightforward.We simply endow each quantity above witha time dependency, and sum the kineticenergies of each infinitesimal segment toget

KE =1

2

∫ L

0

{ρm‖x(s, t)‖2 + ρj θ(s, t)2

}ds

where ρm is the mass density of the beam,and ρj is the angular inertia density.

Figure: Illustration of a smallbeam segment experiencingaxial, shear, and flexuraldeformation. Note how thestiffness ellipsoid changesorientation along the length ofthe beam. Gravagne et al.Trans. on Mech. 2003.



In order to apply Hamilton’s principle tothe energy expressions above, we firstformulate the work due to the applicationof tension F (t) to the cables on the beam.As the beam bends, the cable tensioninduces both shear forces (perpendicular tothe backbone centerline) and moments atthe points where the cable passes througha guide standoff. The external workfunction can depend only upon theboundary moment

W = τ(t)θ(L) = aF (t)θ(L)

Figure: Illustration of the variousgeometric quantities in thecable/pulley system.Gravagne et al. Trans. on Mech.2003.



Applying Hamilton’s principle to theexpressions for KE and PE , and yields theresulting dynamical equations andboundary conditions

ρmx− (QCQTv)′ = 0

ρj θ − EIθ′′ − x′TSQCQTv = 0

EIθ′(L, t) = τ(t)

QCQTv |L = 0

θ(0, t) = 0

x(0, t) = 0

where S is the skew-symmetric matrix

S =

[0 1−1 0

]

The boundary conditionsQCQTv |L= 0 reveal theabsence of applied axial orshear forces on the free end ofthe beam; θ(0, t) = 0 andx(0, t) = 0 indicate that thebeam is clamped at the origin.



The control cables attach to apulley of radius b, driven by amotor with gear ratio r >> 1.Consider following motor model

J θm+B θm+τp = τm τp = (b

r)F

where θm is the angle of themotor and θp is the angle of thepulley. J and B are the rotationalinertia and viscous friction. τp isthe torque due to the cabletension F . the geometric relationbetween the motor, pulley andthe backbone boundary angle

θm = rθp =ra

bθ(L, t)

Now the motor is governed by

Jra

bθ(L, t)+B

ra

bθ(L, t)+

b

rF = τm

choose the feedback control law

τm = −kp θ(L, t)−kd θ(L, t)−kcF



τm = −kp θ(L, t)−kd θ(L, t)−kcF

θ(L, t) = θ(L, t)− θd

withkp, kd , kc > 0

and θd is the desired boundaryangle setpoint. Solving forproduces boundary torque

τ = aF = −Jeff θ(L, t)−kd θ(L, t)−kp θ(L, t)

kc is a coupling factor thatincreases the effectiveback-driveability of themotor-gear system.

kd =Br a2

b + akd

kc + br

kp =akp

kc + br

Jeff =Jr a2

b

kc + br



The stability proof for the control system employs an energy-basedLyapunov functional, consisting of both the distributed energies

V = Ke + PE +1

2kp θ(L, t)2 +

1

2Jeff θ(L, t)2

The first lumped-parameter represents a virtual torsional spring,attached to the free boundary with spring constant kp. The secondrepresents the effect of the motor and gear inertias. After somecalculation, the power flow from the system is

V = −kd θ(L, t)2

which is negative semi-definite, proving system stability.


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