Kinematic Modelling of continuum robots following constant...

Kinematic Modelling of continuum robots following

constant curvature

Centre for Robotics Research – School of Natural and Mathematical Sciences – King’s College London

Hongbin Liu

Compliant Robotics Peking University, Globex, July 20182

• Continuum robots are increasing popular in modern robotics

• Continuous design can be highly advantageous for

– Compliant adaptation to unknown environments

– Safe manipulation of fragile objects and in human-robot interaction

Introduction to Constant Curvature Model


• Unlike rigid-linked, hyper-redundant robots no discrete links

• Continuous deformation of robot’s body allows for motions such as

– Bending around unknown objects for manipulation

– Dexterous movements for locomotion in uncertain terrain



• Until now:

– Robot fully defined for a given set of joint angles and link lengths



• Until now:


• Now Continuum robots:

– Underactuated system; infinite dofs have to be addressed while only limited number of dofs can be controlled

– Forces and moments have to be considered due to inherent elastic behaviour



• Until now:


• Now Continuum robots :

– Underactuation of the system; infinite dofs have to be addressed while only limited number of dofs can be controlled

– Forces and moments have to be considered due to inherent elastic behaviour

Assumptions necessary to recreate shape of robot



Definition of Constant Curvature

• Shape of robot can be recreated assuming a Constant Curvature

• The reason will be discussed in Mechanical Modelling

• CC allows for geometrical description of robot following a curve with curvature k, bending radius r, angle of bending plane φ and arch length l


• Model can be generated based on mappings between three spaces

– Actuator (tendon driven) space

– Configuration space

– Task space

• Each space comprises set of descriptive variables and spaces changed using mapping functions

Mapping


• Robot configuration can be expressed as a set of previously defined arc parameters (k, φ and l)

• Given these parameters the kinematic description can be achieved applying universal modelling techniques, independent from robot type

Mapping


• Robot configuration can be expressed as a set of previously defined arc parameters (k, φ and l)

• Given these parameters the kinematic description can be achieved applying universal modelling techniques, independent from robot type

• Mapping from actuator to configuration space is robot-specific

Mapping


Example Tendon driven:

Continuously bending actuators

• Tendon driven leading to length change

• Placement of actuators allows bending motions of robot

• Robot tip position and orientation can be described as a

function of the actuator lengths

Mapping: Actuator to Config. space


Example: Tendon driven

CMU snake robot



OC snake robot, UK



Wire-Driven Flexible Robot Arm

The china university of Hong Kong


Example: fluidic actuation:

Continuously bending actuators

• Inflatable chambers leading to length change upon pressurization

• Placement of actuators allows elongation and bending motions of robot

• Robot tip position and orientation can be described as a

function of the actuator lengths

Mapping: Actuator to Config. space


Example: fluidic actuation

Festo arm, Germany



Stiff-Flop manipulatorEU FP7 King’s College London, UK

• Silicone-based soft body

• Multi-segment continuum robot with

3 dofs per element

• Fluidic actuation (air pressure or

hydraulic)

• Multi-purpose platform for minimally-

invasive surgery



L1L3L2


Constant curvature in 2D

𝑙1 = 𝜃𝑟𝑙2 = 𝜃 𝑟 + 2𝑑𝑙 = 𝜃(𝑟 + 𝑑)

𝑙2 = 𝜃 𝑟 + 2𝑑 + 𝜃𝑟 − 𝜃𝑟

𝑙2 = 2𝜃 𝑟 + 𝑑 − 𝜃𝑟𝑙2 = 2𝑙 − 𝑙1

𝑑𝑟

𝜃

𝑑

𝑙1

𝑙2

x

y

x1

y1

−𝜃

o1

o 𝛼𝑙 =

𝑙2 + 𝑙12



𝑙2 − 𝑙1 = 2𝜃𝑑

𝑑𝑟

𝜃

𝑑

𝑙1

𝑙2

x

y

x1

y1

−𝜃

o1

o 𝛼

𝜃 =𝑙2 − 𝑙12𝑑

𝑟 =1

𝑘=

2𝑙1𝑑

𝑙2−𝑙1

𝑙1 = 𝜃𝑟𝑙2 = 𝜃 𝑟 + 2𝑑𝑙 = 𝜃(𝑟 + 𝑑)



bending plane

base plane


Arc parameters: Length

• Given the bending geometry it can be seen that


𝑑 ∙ cosΦi

𝑟

𝜃

𝑟𝑖

bending plane base plane



• Substituting the arc lengths and this relation becomes




• The relations between and the bending planecan be denoted as

• Substituting these relations into yields the expression



Arc parameters: Bending angle

• To obtain the bending angle, the previously derived equation

is to be rearranged for actuators 1&2 and equated leading to

• This procedure can be repeated for all actuator pairs and rearranged, leading to the expression



Proof

sin ( α ± β ) = sinα cosβ ± cosα sinβ cos ( α ± β ) = cosα cosβ ∓ sinα sinβ

cos 𝜙1 = cos𝜋

2− 𝜙 = 𝑠𝑖𝑛𝜙

cos 𝜙3 = cos11𝜋

6− 𝜙 = 𝑐𝑜𝑠

11𝜋

6𝑐𝑜𝑠𝜙 + sin

11𝜋

6𝑠𝑖𝑛𝜙

3

2𝑐𝑜𝑠𝜙 −

1

2𝑠𝑖𝑛𝜙

cos 𝜙2 = cos7𝜋

6− 𝜙 = 𝑐𝑜𝑠

7𝜋

6𝑐𝑜𝑠𝜙 + sin

7𝜋

6𝑠𝑖𝑛𝜙

−3

2𝑐𝑜𝑠𝜙 −

1

2𝑠𝑖𝑛𝜙

Recall


Proof

𝜃𝑑 =𝑙2 − 𝑙1

𝐶𝑂𝑆𝜙1 − 𝐶𝑂𝑆𝜙2

𝜃𝑑 =𝑙3 − 𝑙1

𝐶𝑂𝑆𝜙1 − 𝐶𝑂𝑆𝜙3

3

2𝑠𝑖𝑛𝜙 +

3

2𝑐𝑜𝑠𝜙 = (𝑙2 − 𝑙1)/𝜃𝑑

3

2𝑠𝑖𝑛𝜙 −

3


(1)

(2)

(1)+(2)

(1)-(2)

3𝑠𝑖𝑛𝜙 = (𝑙2 + 𝑙3 − 2𝑙1)/𝜃𝑑


𝑡𝑎𝑛𝜙 =3

3

(𝑙2 + 𝑙3 − 2𝑙1)

(𝑙2 − 𝑙3)


Arc parameters: Curvature

Based on geometric observations it can be derived that

Substituting into and (r=1/K) yields

Consider actuator 1, 𝜙1 = 90 − 𝜙

𝜅 =𝑙2 + 𝑙3 − 2𝑙1

𝑙1 + 𝑙2 + 𝑙3 𝑑 c𝑜𝑠Φ1



Consider actuator 1, 𝜙1 = 90 − 𝜙

𝜅 =𝑙2 + 𝑙3 − 2𝑙1

𝑙1 + 𝑙2 + 𝑙3 𝑑 sin𝛷



Arc parameters: Curvature

Given the curvature derivation and previous expression for the bending with the identity

sin(tan−1𝑦

𝑥) = 𝑦/ 𝑥2 + 𝑦2

It can be seen that



Overview

Arc parameters of a 3dof continuum robot with actuator lengths 𝑙𝑖 can be represented as

Forward mapping


Arc geometry

• Shape of the robot can be expressed applying the transformation

Where R represent rotation matrix and p is the translation vector

Mapping: Config to Task space-2D

𝐴10 =

𝑹 𝒑0 1



𝑙 =𝑙2 + 𝑙12

𝜃 =𝑙2 − 𝑙12𝑑

𝑑𝑟

𝜃

𝑑

𝑙1

𝑙2

x

y

x1

y1

−𝜃

o1

o 𝛼

𝛼 =𝜋

2−𝜃

2𝜃

2



𝑙 =𝑙2 + 𝑙12

𝜃 =𝑙2 − 𝑙12𝑑

𝑑𝑟

𝜃

𝑑

𝑙1

𝑙2

x

y

x1

y1

−𝜃

o1

o 𝛼

𝛼 =𝜋

2−𝜃

2

𝑜𝑜1 = 2𝑙

𝜃𝑐𝑜𝑠𝛼

𝑜1 =2𝑙

𝜃𝑐𝑜𝑠𝛼 2,

2𝑙

𝜃𝑐𝑜𝑠𝛼𝑠𝑖𝑛𝛼

𝑇

𝜃

2



𝑑𝑟

𝜃

𝑑

𝑙1

𝑙2

x

y

x1

y1

−𝜃

o1

o 𝛼

𝛼 =𝜋

2−𝜃

2

𝑜1 =2𝑙


2𝑙


𝑇

𝑅 =cos(−𝜃) −sin(− 𝜃)sin(−𝜃) cos(−𝜃)

𝐴10 = 𝑅 𝑜1

0 1

𝑝01

= 𝐴10 𝑝1

1


Mapping: Config to Task space-2D- Singularity

𝑑𝑟

𝜃

𝑑

𝑙1

𝑙2

x

y

x1

y1

−𝜃

o1

o 𝛼

𝛼 =𝜋

2−𝜃

2

𝑜1 =2𝑙


2𝑙


𝑇

𝜃 =𝑙2 − 𝑙12𝑑

Singularity problem:

When θ is zero, o1 become undefined


Programming exercise in class

change name “transcc2D_q.m” to “transcc2D.m”

complete this function and type in the commend window:

[A o1]=transcc2D(pi/4)

results:A = 0.7071 0.7071 3.7292

-0.7071 0.7071 9.0032 0 0 1.0000

o1 = 3.7292 9.0032



Programming exercise in class

change name “transcc2D_q.m” to “transcc2D_L.m”

change this function to : [A o1]=transcc2D_L(L1, L2)

Input variables are the two tendon lengths L1 and L2



Mapping: Config to Task space-2D- Singularity

𝑑𝑟

𝜃

𝑑

𝑙1

𝑙2

x

y

x1

y1

−𝜃

o1

o 𝛼

𝛼 =𝜋

2−𝜃

2

𝑜1 =2𝑙


2𝑙


𝑇

𝜃 =𝑙2 − 𝑙12𝑑

(𝑟 + 𝑑) =1

𝑘=

(𝑙1+𝑙2)𝑑

𝑙2−𝑙1

Recall: 𝑟 =1

𝑘=

2𝑙1𝑑

𝑙2−𝑙1


Mapping: Config to Task space-2D- better solution

𝑑𝑟

𝜃

𝑜1 = 𝑟 − 𝑟𝑐𝑜𝑠𝜃, 𝑟𝑠𝑖𝑛𝜃 𝑇

z

x

o1

𝑟 =1

𝑘=

(𝑙1+𝑙2)𝑑

𝑙2−𝑙1


In-plane translational vector p


𝑑 ∙ cosΦ

𝑟

𝜃

𝑟𝑖

𝒑 = 𝑟 − 𝑟𝑐𝑜𝑠𝜃, 0, 𝑟𝑠𝑖𝑛𝜃 𝑇

z

x’

x’

⦿x


Arc geometry

Shape of the robot can be expressed applying the transformation

Where Ri represent rotation matrices about axis i and




𝑅𝑦 𝜃 =cos 𝜃 0 sin 𝜃0 1 0

−sin 𝜃 0 cos 𝜃𝑅𝑧 𝜙 =

cos𝜙 −sin𝜙 0sin𝜙 cos 𝜙 00 0 1



Aac= Aab Abc =Recap

𝑅𝑧 𝜙 =cos𝜙 −sin𝜙 0sin𝜙 cos𝜙 00 0 1

𝑅𝑦 𝜃 =cos 𝜃 0 sin 𝜃0 1 0

−sin 𝜃 0 cos 𝜃

Rt =cos𝜙𝑐𝑜𝑠𝜃 −sin 𝜃 𝑐𝑜𝑠𝜙𝑠𝑖𝑛𝜃sin𝜙𝑐𝑜𝑠𝜃 cos𝜙 𝑠𝑖𝑛𝜙𝑠𝑖𝑛𝜃−𝑠𝑖𝑛𝜃 0 𝑐𝑜𝑠𝜃

𝑝𝑡 =𝑐𝑜𝑠𝜙𝑟(1 − 𝑐𝑜𝑠𝜃)𝑠𝑖𝑛𝜙𝑟(1 − 𝑐𝑜𝑠𝜃)

𝑟𝑠𝑖𝑛𝜃



Rt =cos𝜙𝑐𝑜𝑠𝜃 −sin 𝜃 𝑐𝑜𝑠𝜙𝑠𝑖𝑛𝜃sin𝜙𝑐𝑜𝑠𝜃 cos𝜙 𝑠𝑖𝑛𝜙𝑠𝑖𝑛𝜃−𝑠𝑖𝑛𝜃 0 𝑐𝑜𝑠𝜃

𝑝𝑡 =𝑐𝑜𝑠𝜙𝑟(1 − 𝑐𝑜𝑠𝜃)𝑠𝑖𝑛𝜙𝑟(1 − 𝑐𝑜𝑠𝜃)

𝑟𝑠𝑖𝑛𝜃

T =

cos𝜙𝑐𝑜𝑠𝜃 −sin 𝜃 𝑐𝑜𝑠𝜙𝑠𝑖𝑛𝜃sin𝜙𝑐𝑜𝑠𝜃 cos𝜙 𝑠𝑖𝑛𝜙𝑠𝑖𝑛𝜃−𝑠𝑖𝑛𝜃0

00

𝑐𝑜𝑠𝜃0

𝑐𝑜𝑠𝜙𝑟(1 − 𝑐𝑜𝑠𝜃)𝑠𝑖𝑛𝜙𝑟(1 − 𝑐𝑜𝑠𝜃)

𝑟𝑠𝑖𝑛𝜃1



𝑘𝑙 = 𝜃

𝑑 ∙ cosΦ

𝑟

𝜃

𝑟𝑖

𝑟 =1

𝑘

S

θs

𝑘𝑠 = 𝜃𝑠

Let a point along the manipulator central

curve with arc length s

Arbitrary location on the continuum body


The manipulator-independent mapping function from configuration to task space can be summarized by

Mapping: Config to Task space


Recap: derivation from DH

• Same expression can be derived following DH convention

• Element along curve can be expressed by number of transformations

Mapping: Config to Task space


• Inverse kinematic formulation is not trivial, particularly for multi-segment manipulators with large number of segments

• For Task to Configuration space mapping solution exist based on

– A closed geometric formulation (Neppalli et al. 2008)

– Jacobian derivation ( “differential kinematics”)

Inverse mapping


• The Configuration to Actuator space mapping is similarly to FK highly individualized

• The solution exist based on

– A analytical geometric formulation

– Jacobian derivation ( “differential kinematics”)

Inverse mapping


Programming exercise

cc_1seg_execise_q.m

put cc_1seg_execise_q.m and transcc2D.m into the same folder

%tendon1

l1=12;

%tendon2

l2=8;

%pex is the coordinates of

unit vector of x axis in

the local frame of the tip

pex=[1 0]'

central bending axis of the arm


Try different combination of l1 and l2

What happens if l1=l2?

Testing the limit

Kinematic Modelling of continuum robots following constant...

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